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Computational Engineering for Reinforced Concrete Structures U. Häußler-Combe mailto:[email protected] Institute of Concrete Structures http://www.tu-dresden.de/biwitb/mbau Technische Universität Dresden http://www.tu-dresden.de

PRELIMINARY DRAFT State April 4, 2013

I

Introductory Remarks • The following notes1 serve as accompanying study material for the lecture Computational Engineering for Reinforced Concrete Structures of the Institute of Concrete Structures, Technische Universität Dresden. • The notes are not yet finished. Some existing chapters deserve a completion. More chapters – not yet included – are under work. Thus, a Preliminary Draft is given. • The actual state has been worked out according to best knowledge. Nevertheless, it may contain formal errors or wrong facts. Corresponding hints are highly appreciated by the author (→ mailto:[email protected]).

1

©2013( Ulrich Häußler-Combe. All rights preserved.

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State April 4, 2013

Contents 1

2

3

Finite Elements in a Nutshell 1.1 Modeling Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Finite Element Basics . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Material Behavior . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Weak Equilibrium and Discretization . . . . . . . . . . . . . . . . . 1.6 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Numerical Integration and Solution Methods for Algebraic Systems

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1 1 1 2 6 7 10 14

Uniaxial Structural Concrete Behavior 2.1 Short Term Stress-Strain Behavior of Concrete . . . . 2.2 Long Term Effects - Creep, Shrinkage and Temperature 2.3 Strain-Rate Effects . . . . . . . . . . . . . . . . . . . 2.4 Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Reinforcing Steel Stress-Strain Behavior . . . . . . . . 2.6 Bond between Concrete and Reinforcing Steel . . . . . 2.7 Reinforced Tension Bar . . . . . . . . . . . . . . . . . 2.8 Tension Stiffening for Reinforced Tension Bar . . . . . 2.9 Cyclic Loading of Reinforced Tension Bar . . . . . . .

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20 20 26 30 31 33 34 35 38 40

2D Structural Beams and Frames 3.1 General Cross Sectional Behavior . . . . . . . . . . . . . . . 3.1.1 Kinematics . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Linear elastic behavior . . . . . . . . . . . . . . . . . 3.1.3 Cracked reinforced concrete behavior . . . . . . . . . 3.1.3.1 Compressive zone and internal forces . . . . 3.1.3.2 Linear concrete compressive behavior . . . 3.1.3.3 Nonlinear concrete compressive behavior . . 3.2 Equilibrium of Bars . . . . . . . . . . . . . . . . . . . . . . . 3.3 Structural Beam Elements for 2D . . . . . . . . . . . . . . . . 3.4 System Building and Solution Methods . . . . . . . . . . . . 3.5 Further Aspects . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.1 Creep . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5.2 Temperature and Shrinkage . . . . . . . . . . . . . . 3.5.3 Tension Stiffening . . . . . . . . . . . . . . . . . . . 3.5.4 Prestressing . . . . . . . . . . . . . . . . . . . . . . . 3.5.5 Shear stiffness for reinforced cracked concrete sections 3.6 Application Case Studies . . . . . . . . . . . . . . . . . . . . 3.6.1 Transient Dynamics of Beams . . . . . . . . . . . . . 3.6.2 More Case Studies . . . . . . . . . . . . . . . . . . .

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42 42 42 44 45 45 47 49 52 55 60 66 66 69 73 76 81 83 83 86

II

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III

CONTENTS 4

5

6

7

8

Strut-and-Tie Models 4.1 Linear Elastic Panel Solutions . . . . . . . . . . . 4.2 Truss Modeling . . . . . . . . . . . . . . . . . . . 4.3 Computation of Plane Elasto-Plastic Truss Models 4.4 Ideal Plastic Truss Models . . . . . . . . . . . . . 4.5 Application Aspects . . . . . . . . . . . . . . . . .

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88 88 89 90 96 102

Multiaxial Concrete Material Behavior 5.1 Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Some Basics of Continuum Mechanics . . . . . . . . . . . . 5.3 Basic Linear Material Behavior Description . . . . . . . . . 5.4 Basics of Nonlinear Material Behavior . . . . . . . . . . . . 5.4.1 Tangential Stiffness . . . . . . . . . . . . . . . . . . 5.4.2 Stress Limit States . . . . . . . . . . . . . . . . . . 5.4.3 Phenomenological Approach for Biaxial Anisotropic lation . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Isotropic Damage . . . . . . . . . . . . . . . . . . . . . . . 5.6 Isotropic Plasticity . . . . . . . . . . . . . . . . . . . . . . 5.7 Microplane . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Localization and Regularization . . . . . . . . . . . . . . . 5.9 Long Term Behavior . . . . . . . . . . . . . . . . . . . . . 5.10 Short Term Behavior . . . . . . . . . . . . . . . . . . . . .

104 104 105 110 114 114 114

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119 121 126 130 130 131 131

Deep Beams 6.1 Limit Analysis . . . . . . . . . . . . . . . . 6.2 2D Crack Modeling . . . . . . . . . . . . . . 6.3 2D Modeling of Reinforcement and Bond . . 6.4 Biaxial Concrete Material Laws . . . . . . . 6.5 Further Aspects and Application Case Studies

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133 133 140 145 151 152

Slabs 7.1 Cross Sectional Behavior . . . . . . . . . . . . . . . . 7.1.1 Kinematic Basics . . . . . . . . . . . . . . . . 7.1.2 Linear elastic behavior . . . . . . . . . . . . . 7.1.3 Reinforced cracked sections . . . . . . . . . . 7.2 Equilibrium of slabs . . . . . . . . . . . . . . . . . . . 7.3 Structural Slab Elements . . . . . . . . . . . . . . . . 7.3.1 Area coordinates . . . . . . . . . . . . . . . . 7.3.2 A triangular Kirchhoff element . . . . . . . . . 7.4 System Building and Solution Methods . . . . . . . . 7.5 Reinforcement Design with linear elastic internal forces

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154 154 154 156 157 158 162 162 162 164 168

Shells 8.1 Preliminary Remarks . . . . . . . . . . . . . . 8.2 Approximation of Geometry and Displacements 8.3 Approximation of Deformations . . . . . . . . 8.4 Shell Stresses and Material Laws . . . . . . . . 8.5 System Building . . . . . . . . . . . . . . . . . 8.6 Slabs and Beams as a Special Case . . . . . . . 8.7 Locking . . . . . . . . . . . . . . . . . . . . . 8.8 Reinforced Concrete Shells . . . . . . . . . . .

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177 177 177 179 181 183 185 186 189

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IV

CONTENTS A ConFem

196

B Transformations of coordinate systems

197

C Linear regression analysis applications 199 C.0.1 Determination of moments in triangular slab elements . . . . . . . . . 199 D Numerical Integration of Elastoplastic Material Laws

201

E ACCESS lectures in summer term 2013

204

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Chapter 1

Finite Elements in a Nutshell 1.1

Modeling Basics • See Fig. 1.1

Figure 1.1: Modeling Basics according to [CF08, Fig. 7.31] • Reality of interest • Conceptual model • Computational model

1.2

Finite Element Basics • Subdivision of area of interest into smaller units – elements – nodes • Interpolation of displacement fields elementwise – → Trial functions – displacement compatibility • Integral equilibrium at nodes 1

2

1.3 Elements

Figure 1.2: FE basic idea

1.3

Elements • Basic concepts – Coordinates T * global x = ( x y z ) T * local r = ( r s t ) * element number I and local coordinate r in element – Displacement variables T * translation u = ( u v w ) T * rotation ϕ = ( ϕx ϕy ϕz )

– Deformation variables * 1D strain (small strain)

∂u ∂x * 2D strain (small strains – Voigt notation) 
T ∂u  = x y xy = ∂x =

(1.1)

∂v ∂y

∂u ∂v + ∂y ∂x

T (1.2)

* Curvature (small deflections, 2D) κ=

∂ϕy ∂2w = 2 ∂x ∂x

(1.3)

– Generalized force variables, e.g. *  → σ with Cauchy stress tensor σ where dw = σ d gives a mechanical work increment per unit volume in a body. * κ → M with bending moment M where dw = M dκ gives a mechanical work increment per unit length of a 2D bending bar. – Interpolation * Coordinates x = N(I, r) · xI

(1.4)

with vector xI collecting global coordinates of nodes of element I and row vector N collecting shape functions. State April 4, 2013

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3 * Displacements u = N(I, r) · uI

(1.5)

with vector uI collecting global displacements of nodes of element I and row vector N collecting shape functions. * Deformations, e.g. small 2D engineering strains  = B(I, r) · uI

(1.6)

with vector uI collecting global displacements of nodes of element I and matrix B collecting derivatives of shape functions with respect to global coordinates. Index I within (I, r) will be omitted in the following, as the functions N, B are the same for all elements of one (isoparametric) element type. • Bar elements – Kinematic assumption * cross section displacements constant in longitudinal direction – Interpolation functions linear 1D, 2 nodes, 2 degrees of freedom    1  xI1 1 x(r) = 2 (1 − r) 2 (1 + r) ·  xI2    1 uI1 1 u(r) = 2 (1 − r) 2 (1 + r) · uI2         2 u uI1 I1 1 1 1 1 du dr  = dx = − 2 2 dx · = − 2 2 LI · uI2 uI2

(1.7)

with a bar length LI = xI2 − xI1 and a Jacobian J=

LI ∂x = ∂r 2

(1.8)

– Interpolation functions linear 2D, 2 nodes, 4 degrees of freedom  

x(r) y(r)



u(r) v(r)



 =

1 2 (1

− r) 0

0 − r)

1 2 (1

0 1 (1 − r) 2

1 2 (1

1 2 (1

+ r) 0

1 2 (1

0 + r)



0 + r)



 ·  



 =

1 2 (1

− r) 0

+ r) 0

1 2 (1

 · 

xI1 yI1 xI2 yI2 uI1 vI1 uI2 vI2

       

(1.9) Small strain is taken in the bar direction, i.e. in a rotated coordinate system. The rotation angle (counterclockwise positive) and the transformation matrix are given by   yI2 − yI1 xI2 − xI1 cos α sin α cos α = , sin α = , T= (1.10) − sin α cos α LI LI p with a bar length LI = (yI2 − yI1 )2 + (xI2 − xI1 )2 . Reusing Eq. (1.7)3 we get   uI1    vI1  2  1 1  cos α sin α 0 0  −2 2 · = · (1.11)  uI2  0 0 cos α sin α LI vI2 mailto:[email protected]

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4

1.3 Elements • Spring elements – Kinematic assumption * Displacement difference between two nodes irrespective of their distance. The two nodes share the same location as special but common case. – "‘Interpolation"’ function 1D     uI1 ∆u = −1 1 · (1.12) uI2 – "‘Interpolation"’ function 2D 

 u I1      vI1  ∆u −1 −1 0 0  = ·  uI2  ∆v 0 0 1 1 vI2

(1.13)

• 2D continuum elements (e.g. plate with loading in plane → panel, deep beam) – Kinematic assumptions * Displacements u(x, y), v(x, y) are continuous functions. * Lateral displacement w is w = 0 (→ plane strain) or determined from a condition σz = 0 (→ plane stress) – Interpolation functions 4-node quadrilateral x(r, s) =

4 X

NJ xIJ ,

y(r, s) =

J=1

4 X

NJ yIJ ,

J=1

u(r, s) =

4 X

NJ uIJ ,

v(r, s) =

J=1

J=1

(1.14) with

1 (1.15) NJ (r, s) = (1 + rJ r)(1 + sJ s) 4 with the local node coordinates rJ , sJ . For the following we need the Jacobian  ∂x ∂y  ∂r ∂r J = ∂x , J = det J (1.16) ∂y ∂s

∂s

which is easily derived from Eq. (1.14)1,2 and relates the partial derivatives of a function • with respect to local and global coordinates1  ∂•   ∂•   ∂•   ∂•  −1 ∂x ∂x ∂r ∂r → =J · =J· (1.17) ∂• ∂• ∂• ∂• ∂s

∂y

∂y

∂s

Small engineering strains with Eqns. (1.2), (1.14), (1.17)2     ∂u ∂r ∂u ∂s + x (r, s) ∂r ∂x ∂s ∂x  ∂v ∂r ∂v ∂s  y (r, s)  =    ∂r ∂y + ∂s ∂y ∂u ∂r ∂u ∂s ∂v ∂r ∂v ∂s γxy (r, s) + + + ∂r ∂y  P  ∂s ∂y ∂r ∂x ∂r ∂s ∂x  4 1 ∂s J=1 4 hrJ (1 + sJ s) ∂x + (1 + rJ r)sJ ∂x i uIJ  P4 1  ∂s ∂r   rJ (1 + sJ s) ∂y + (1 + rJ r)sJ ∂y vIJ   PJ=1 4 nh i =  4 1 ∂r ∂s   r (1 + s s) + (1 + r r)s u J J ∂y J J ∂y IJ J=1 4   o   ∂s ∂r + rJ (1 + sJ s) ∂x + (1 + rJ r)sJ ∂x vIJ (1.18) which after evaluation leads to a form like Eq. (1.6). 1

State April 4, 2013

4 X

Closed forms of J and J−1 are available for the 4-node quadrilateral, see e.g. [???].

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NJ vIJ

5 • 2D beam elements – Displacement variables * Transverse displacement / deflection w of reference axis · → rotation angle of reference axis * Rotation angle of cross section – Kinematic assumptions * Plane cross sections remain plane * Bernoulli beam: rotation of reference axis = rotation of cross section * Timoshenko beam: Rotation of cross section decouples from rotation of reference axis and shear angle serves as independent variable – Interpolation functions Bernoulli beam, see Page 56. – Interpolation functions Timoshenko beam, see Page 58. • Plate elements (loading perpendicular to plane → slab) – Displacement variables – Kinematic assumptions – Kirchhoff plate (thin plate) – small displacements, mid-surface without strains, rotation of cross section can be derived from deflection of mid-surface – Reissner-Mindlin plate (thick plate) • 3D continuum elements • Additional aspects concerning elements – Continuity of displacements or deformations along element boundaries – Element locking – Independent interpolation of displacement, deformations, generalized forces

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6

1.4 Material Behavior

1.4

Material Behavior • In the following, a material is considered as homogeneous. • Material behavior is constituted by a force response in reaction to a deformation. • Linear elastic material law – Uniaxial σx = E x with Young’s modulus E. – plane strain    σx E(1 − ν)  σy  =  (1 + ν)(1 − 2ν) σxy

(1.19)

ν 1−ν

1 ν 1−ν

1 0

0

0 0 1−2ν 2(1−ν)

with Young’s modulus E and Poisson’s ratio ν. This is isotropic linear elastic Hooke law. – plane stress      1 ν 0 σx  σy  = E  ν 1 0 · 1 − ν2 1−ν 0 0 σxy 2

 

 x  ·  y  xy

(1.20)

a subset of the triaxial

 x y  xy

(1.21)

with Young’s modulus E and Poisson’s ratio ν ensuring σz = 0 for every combination x , y , xy – 2D bending M = −EJ κ (1.22) with Young’s modulus E and cross-sectional moment of inertia J. – Eqns. (1.19)-(1.21) are a special case of σ =C·

(1.23)

with the constant material stiffness matrix C. • Uniaxial perfect elasto-plastic (as a simple example for a nonlinear constitutive equation) → physical nonlinearity  E (x − xp ) for − xp ≤ x ≤ xp σx = (1.24) signx fy otherwise and ∆xp = ∆x for |σx | = fy

(1.25)

with a yield stress fy and an internal state parameter xp . This value is the actual remaining strain upon unloading, i.e. σx = 0 for x − xp . • In case of nonlinear material equations at least an incremental form dσ = CT · d or

σ˙ = CT · ˙

(1.26)

should exist, where the tangential material stiffness might depend on stress, strain, internal state variables. • On occasion the flexibility is needed, as counterpart of stiffness, i.e. =D·σ State April 4, 2013

(1.27)

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7

1.5

Weak Equilibrium and Discretization • Weak forms of equilibrium condition including dynamic parts – Preceding remark: given a point on a boundary either a kinematic boundary condition or a force boundary condition (zero force is also a condition!) has to be prescribed for this point. – Body (continuum) in space with a volume V , a specific mass %, prescribed dis¯ , prescribed displacements u ¯ on the surface part Au , prescribed tractributed loads p tions ¯t on surface part At (Au together with At give the whole surface A), acceleration u ¨ , virtual displacements δu and associated virtual strains δ Z Z Z Z T T T ¯ dV + δ · σ dV + δu · u ¨ %dV = δu · p δuT · ¯t dA (1.28) V

V

V

At

under the conditions ¯ on Au , u=u

δu = 0 on Au

– Unixial bar along 0 ≤ x ≤ L Z Z Z δx σx Adx + δu u ¨ %Adx = δu p¯x dx + [δu t¯]L 0 L

L

(1.29)

(1.30)

L

under the conditions u0 = u ¯0 , δu0 = 0 or uL = u ¯L δuL = 0

(1.31)

with a cross section area A and a load per length p¯x , whereby the formulation of the last term indicates the boundary term of a partial integration where traction is prescribed at either x = 0 or x = L (or none, but not both at the same time). – 2D Bernoulli beam along 0 ≤ x ≤ L Z Z Z L δw w ¨ mdx ¯ + δκ M dx = δw p¯ dx − [δϕ M ]L 0 + [δw Q]0 L

L

(1.32)

L

with a distributed mass m ¯ per length and a distributed load p¯ per length. with corresponding pairs (ϕ, M ) and (w, Q). Only one quantity out of a pair can be prescribed at a boundary. Furthermore, at least two kinematic boundary conditions should be given with at least one deflection w0 and/or wL . All these forms are variations of the principle of virtual displacements and immediately allow for physical nonlinearities.

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8

1.5 Weak Equilibrium and Discretization • Discretization – The following is performed using Eq. (1.28). Other weak formulations use the same procedure. The general approach of G ALERKIN is followed here. – Small displacements are used, if not otherwise stated. – Steps of discretization 1. Interpolation approach for displacements in the spatial domain, e.g. with Eq. (1.14)3,4 * An infinite number of degrees of freedom is reduced to a finite number of nodal values of degree of freedom (→ discretization). * This leads to discretized strains e.g. with Eq. (1.18) and further to stresses σ with a constitutive law, e.g. Eq. (1.20) or (1.21) in the linear elastic case. 2. Interpolation approach for virtual displacements * The same interpolation Eq. (1.14)3,4 is used as for the displacement (→ approach of B UBNOV-G ALERKIN). * This leads to discretized virtual strains with Eq. (1.18). 3. Evaluation of integrals * This is performed on a element by element base Z δT · σ dV = δuTI · fI , fI Z VI ¨I , δuT · u ¨ %dV = δuTI · MI · u MI VZI ¯I , ¯I ¯ dx p δu · p = δuTI · p Z VI ¯tI δuT · ¯t dA = δuTI · ¯tI ,

=

R VI

Z = ZVI =

BT · σ dV NT · N %dV ¯ dV NT · p

ZVI =

AtI

NT · ¯t dA

AtI

(1.33)

with an element index I. For integration methods see Section 1.7.

Figure 1.3: Schematic flow of nonlinear calculation 4. Assembling of element contributions to the whole system, see Figure 1.2. State April 4, 2013

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9 * Regarding, e.g. internal nodal forces f , it consists of entries for every degree of freedom of every global node. On the other hand, every discretization should have a table, which connects every element to the global nodes belonging to it. * This table relates the position of the entries of fI to a position in f . * As a node generally gets contributions from more than one element, the value of an entry in fI has to be added to the corresponding entry in f . This is symbolically described by Z X δuTI · fI (1.34) δT · σ dV = δuT · f = I

V

* The same argumentation holds for δuI (→ δu), uI (→ u), MI (→ M), ¯ I (→ p ¯ ), tI (→ t). p * Regarding arbitray values of δu a spatially discretized system ¨ + f (u) = p ¯ + ¯t M·u

(1.35)

finally results. This is a system of ordinary differential equations of second order in time t. It might be nonlinear due to nonlinear dependence of internal nodal forces f on nodel displacements u. – The linear case σ = C ·  leads to internal nodal forces Z fI = BT · C · B dV · uI = KI · uI

(1.36)

VI

see Eqns. (1.33)1 , (1.6), with a constant element stiffness matrix KI . Assembling leads to a system stiffness matrix K f (u) = K · u

(1.37)

¨ +K·u=p ¯ + ¯t M·u

(1.38)

and regarding Eq. (1.35) to

which embodies a system of linear ordinary differential equations of second order in time t. – The system Eqns. (1.35) or (1.38) should be supported with appropriate kinematic boundary conditions to prevent rigid body displacements. • Tangential stiffness – The system tangential stiffness matrix is needed for the solution of the system and furthermore reveals characteristic properties of a system, i.e. in particular its stability properties. – Element tangential stiffness dfI =

∂fI · duI = KT I · duI ∂uI

with

Z KT I = VI

∂σ ∂ · dV = B · ∂ ∂uI T

f˙I = KT I · u˙ I

or Z

BT · CT · B dV

(1.39)

(1.40)

VI

see Eqns. (1.33)1 , (1.26) (1.6), and a system tangential stiffness KT df = KT · du

or

f˙ = KT · u˙

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(1.41)

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10

1.6 Convergence

1.6

Convergence • The linear elastic, quasistatic small displacement case is regarded in the following. Given a linear material law σ =C· (1.42) weak, integral equilibrium Eq. (1.28) for a structural problem can be written as Z Z Z T T ¯ δ · C ·  dV = δu · p dV + δuT · ¯t dA V

V

(1.43)

At

with appropriate displacement boundary conditions preventing rigid body motions, given ¯ , ¯t and arbitrary virtual displacements δu. values for p – The ‘strains’ , δ are derived from the ‘displacements’ u, δu by a differential operator depending on the type of the structural problem under consideration. – The interpolation functions u, , δu, δ belong to an appropriate function space2 H defined over the problem domain V (→ space occupied by the the structure). – The boundary A of V is composed of Au and At , i.e. A = At ∪ Au and At ∩ Au = 0. u is prescribed on Au (→ Dirichlet conditions) and ¯t on At (→ Neumann conditions) whereby ¯t = n · σ with the boundary’s normal n. • The mathematical model of the structural problem Eq. (1.43) can be written as a(u, v) = (f , v)

∀v ∈ H

(1.44)

with a symmetric, bilinear a(·, ·), a linear (f , ·) and v formally replacing δu . – Symmetry a(u, v) = a(v, u)

(1.45)

a(γ1 u1 + γ2 u2 , v) = γ1 a(u1 , v) + γ2 a(u2 , v) a(u, γ1 v1 + γ2 v2 ) = γ1 a(u, v1 ) + γ2 a(u, v2 )

(1.46)

(f , γ1 v1 + γ2 v2 ) = γ1 (f , v1 ) + γ2 (f , v2 )

(1.47)

– Bilinearity

– Linearity • A norm maps a function v into a non-negative number. Sobolev norms ||v||i of order i are used in this context [Bat96, 4.3.4,(4.76)]. It is assumed that i = 1 is appropriate for the following. It can then be shown that a has the properties – Continuity3 ∃M > 0 :

|a(v1 , v2 )| ≤ M kv1 k1 kv2 k1

∀v1 , v2 ∈ H

(1.48)

– Ellipticity ∃α > 0 :

a(v, v) ≥ α kvk21

∀v ∈ H

(1.49)

where M, α depend on problem type and material values but not on v1 , v2 , v. – Due to Eq. (1.49) a(v, v) ≥ 0, i.e. a is a norm and may be physically interpreted as energy, i.e. it is twice the internal strain energy. 2 3

State April 4, 2013

Square integrable functions fulfilling the kinematic boundary conditions, see, e.g., [Bat96, 4.3.4]. ∃ means: it exists a (real number).

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11 • It can be shown that the problem Eq.(1.44) – i.e. determine a function u ∈ H such that Eq. (1.44) is fulfilled for all v ∈ H – has a unique solution u, see, e.g., [Bat96, 4.3]. This solution is exact for the mathematical model of the structural problem. • Discretization uses functions uh , vh ∈ Hh of a subset Hh ⊂ H based upon the concept of meshes with elements and nodes, see Section 1.3. – A ‘uniform’ mesh4 of elements is assumed and h is a mesh size parameter, e.g. a diameter or length of a generic element. The approximate solution uh ∈ Hh of Eq. (1.44) is determined by a(uh , vh ) = (f , vh ) ∀vh ∈ Hh

(1.50)

– The difference between approximate and exact solution gives the error eh = u − uh

(1.51)

– The approximation uh is known for Hh given, the error eh has to be estimated. • Some properties of the approximate solution – Orthogonality of error, see [Bat96, (4.86)] a(eh , vh ) = 0

∀vh ∈ Hh

(1.52)

– Energy of approximation is smaller than exact energy, see [Bat96, (4.89)] a(uh , uh ) ≤ a(u, u)

(1.53)

– Energy of error is minimized, see [Bat96, (4.91)] a(eh , eh ) ≤ a(u − vh , u − vh ) ∀vh ∈ Hh

(1.54)

– Combining Eqs. (1.49), (1.54), (1.48) leads to α keh k21 = α ku − uh k21 ≤ a(eh , eh ) = inf a(u−vh , u−vh ) ≤ M inf ku − vh k21 vh ∈Hh

vh ∈Hh

(1.55) where inf is infimum, the largest lower ku − uh k1 ≤ c d(u, Hh ),

bound5 .

This is rewritten as

d(u, Hh ) = inf ku − vh k1 , c = vh ∈Hh

p

M/α

(1.56) d is a "distance" of functions in Hh to the exact solution u, c depends on the structural problem type and the values of its parameters, but not on Hh . • Convergence means uh → u or ku − uh k1 → 0 with mesh size h → 0. – This can generally be reached with an appropriate selection of function spaces Hh whereby reducing the distance d(u, Hh ), see Eq. (1.56). – A more precise statement is possible utilizing interpolation theory. It introduces the interpolant6 ui ∈ Hh of the exact solution u. 4

The following considerations may be transferred to non-uniform meshes, see [Bat96, 4.3.5]. ku − vh k1 , vh ∈ Hh is a subset of real numbers. inf vh ∈Hh ku − vh k1 is the largest number less or equal to the numbers in this subset. 6 u and ui coincide at nodes, but generally not apart from nodes. Generally it is ui 6= uh . 5

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12

1.6 Convergence – Complete polynomials7 of degree k are used for discretization and interpolation, respectively. Interpolation theory estimates the interpolation error with ku − ui k1 ≤ cˆ hk kukk+1

(1.57)

with the mesh size h and a constant cˆ which is independent of h, see [Bat96, (4.99)]. kukk+1 is the k + 1-order Sobolev norm of the exact solution. – On the other hand it must be inf vh ∈Hh ku − vh k1 ≤ ku − ui k1 as ui ∈ Hh . Using this and Eqs. (1.56), (1.57) yields ku − uh k1 ≤ cˆ c hk kukk+1

(1.58)

The value cˆ c can be merged to c, which depends on the structural problem type and the values of its parameters, but not on h. A further merger of c and kukk+1 leads to the well known formulation ku − uh k1 ≤ c hk

(1.59)

where c depends on the structural problem type, the values of its parameters and the norm of the exact solution. • Conditions for convergence, see also [Bat96, 4.3.2] – A basic prerequisite is integrability of all quantities. This leads to requirements for the integrands of the energy a and the Sobolov norms8 , which are uh , vh , u or derivatives thereof. * It corresponds to the requirement of compatibility of finite element interpolation functions along inter element borders. – According to Eq, (1.59) a sequence of approximate solutions uh with h → 0 will converge9 with respect to ku − uh k1 if k ≥ 1. * The case k = 1 is covered by the so called patch test, i.e. the ability to model fields with constant first derivatives of finite element interpolation functions in arbitrary element configurations, see, e.g., [BLM00, 8.3.2]). – The convergence rate will be higher for larger values of k, i.e. if the finite element interpolation has a higher order of completeness. • Limitations – The coefficient c may become so large under certain conditions that accepable solutions, i.e. a sufficiently small values ||u − uh ||, cannot be reached with realizable values h small enough. * A particular occurence is given with locking of approximate solutions with incompressible or nearly incompressible materials. • Extended weak forms 7

A polynomial in x, y is complete of order 1 if it includes x, y, complete of order 2 if of order 1 and including x2 , xy, y 2 , complete of order 3 if complete of order 2 and including x3 , x2 y, xy 2 , y 3 and so on. 8 Sobolev norms are built from integration of squares of functions and squares of their derivatives. 9 Converge with respect to first order Sobolev norm ku − uh k1 may not be sufficient if generalized strains are derived from higher derivatives of displacements, e.g. with beams, slabs, shells. The theory has to be extended for this case.

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13 – Eqs. (1.43,1.44) are weak forms of displacement based methods, as a solution is given by a displacement field. Strains and stresses are derived from this solution. – Extended weak forms allow to involve fields for stresses and strains as independent solution variables. Most prominent are the principles of Hu-Washizu and HellingerReissner [Bat96, 4.4.2]. * An abstract extended problem definition analogous to Eq. (1.50) given by, see [Bat01, (16)] a(uh , vh ) + b(h , vh ) = (f , vh ) b(wh , uh ) − c(h , eh ) = 0

∀vh ∈ Hh ∀wh ∈ Wh

(1.60)

in which a, c are symmetric bilinear forms, b is a bilinear form, f is a linear form, Hh , Wh are appropriate functions spaces, uh ∈ Hh , h ∈ Wh are the approximate solutions. In most cases h stands for an independent field of strains or stresses. * The foregoing theory concerning convergence has to be extended, see [Bat01]. Such an extension includes the widely referenced inf-sup condition. – These approaches may solve, e.g., locking problems. • Nonlinear problems – The foregoing considerations relate to linear problems. They cannot be strictly applied to nonlinear probems – physically nonlinear and/or geometrically nonlinear. But the conclusions to be drawn regarding element selection and discretization should also be considered for nonlinear problems.

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14

1.7 Numerical Integration and Solution Methods for Algebraic Systems ni 0 1 2 3 .. .

ξi 0.0 ±0.57735 02691 89626 ±0.77459 66692 41483 0.0 ±0.86113 63115 94053 ±0.33998 10435 84856 .. .

ηi 2.0 1.0 0.55555 55555 55556 0.88888 88888 88889 0.34785 48451 37454 0.65214 51548 62546 .. .

Table 1.1: Sampling points and weights for Gaussian numerical integration (up to 15 digits shown)

1.7

Numerical Integration and Solution Methods for Algebraic Systems • Integration of Eqs. (1.34) for the two-dimensional case as an example – Integration of isoparametric elements Z

Z

y2

Z

x2 (y)

f (x, y) dV =

+1 Z +1

f (x, y) tdxdy = y1

VI

Z

f (r, s) J(r, s) tdrds −1

x1 (y)

−1

(1.61) with the determinant J of the Jacobian, see Eq. (1.16)2 , and a constant thickness t. – Numerical integration Z

+1 Z +1

f (r, s) J(r, s) tdrds = t −1

−1

ni X ni X

ηi ηj f (ξi , ξj ) J(ξi , ξj )

(1.62)

i=0 j=0

with integration order ni , sampling points ξ and weighting factors η. – Gaussian scheme Pni * Sampling points and weighting factors see Table 1.1. Generally we have i=1 ηi = 2. Also for other integration schemes. * Integration accuracy: an integration order ni gives exact (within the scope of numerical accuracy) results for polynoms of order 2ni + 1, e.g. a uniaxial integration of order 1 with two sampling points integrates exactly a polynom of order 3. – Other schemes: Simpson, Newton-Cotes, Lobatto – Integration of triangular elements * Remains to be added.

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15 • Solution method for quasistatic problems – Regarding Eq. (1.35) we have to solve r(u) = f (u) − p = 0,

¯ + ¯t p=p

(1.63)

with internal nodal forces f depending on displacements u and external nodal load p, which are assumed to be independent of u. – The general case is nonlinear dependence of f on u. So the solution of Eq. (1.63) has to be determined with an iteration with a sequence u(0) . . . u(ν) . – Looking at an arbitrary iteration step (ν) we have r(u(ν) ) 6= 0 and seek for a correction δu. A linear Taylor-expansion is used as a basic approach (ν)

r(u(ν) + δu) ≈ r(u(ν) ) + KT · δu, =0

(ν)

KT =

∂r ∂u u=u(ν)



=

∂f ∂u u=u(ν)



(1.64) for the Newton-Raphson method with the tangential stiffness matrix KT . This leads to (ν) δu = −[KT ]−1 · r(u(ν) ) (1.65) u(ν+1) = u(ν) + δu and an (hopefully) improved value u(ν+1) . Iteration may stop if kr(u(ν+1) )k  1 and kδuk  1 with a suitable norm k · k. The method generally has a fast convergence, but is relatively costly. In every step (ν) the tangential stiffness matrix has to be computed and a LU-decomposition has to be performed. – Other iteration methods use variants of the iteration matrix (modified Newton Raphson, secant methods like BFGS) – Iterative methods like Newton-Raphson generally are embedded in an incrementally iterative scheme, so loading is given as a history: p = p(t). Often we choose 0 ≤ t ≤ 1 for the load history time10 , without restrictions to generality. * We look at discrete time values ti , t0 = 0 and have pi = p(ti ). This is known before a solution. * Accordingly, we have ui = u(ti ), fi = f (ui ). These are not known before a solution, except u0 , f0 . We assume r0 = f0 − p0 = 0. t1 , where u1 has to be determined. This is done with * The solution starts with(0) (ν) an iteration sequence u1 . . . u1 , e.g. with the Newton-Raphson method and (0) u1 = u0 . * A converged u1 is used as a base for t2 and so on.

10

Load history time is different to real time.

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16

1.7 Numerical Integration and Solution Methods for Algebraic Systems • Solution method for transient problems, in particular creep problems – Extended constitutive law σ˙ = CT · ˙ + Σ

(1.66)

with the tangential material stiffness CT , see Eq. (1.26), and an additional term depending Σ on stress and strain, see e.g. Section 2.2. We have to solve for σ(t), whereupon (t), (t) ˙ are given as functions of time. – A numerical method is used with the trapezoidal rule for derivatives of stress and strain σ i+1 = σ i + ∆t [ασ˙ i+1 + (1 − α) σ˙ i ] (1.67) i+1 = i + ∆t [α˙ i+1 + (1 − α) ˙ i ] with time discretization parameters ∆t, α. Eqs. (1.66), Eq. (1.67)1 lead to

σ i+1 = σ i + (1 − α) ∆t CT,i · ˙ i + Σi + α ∆t CT,i+1 · ˙ i+1 + Σi+1 (1.68) The parameters α, ∆t rule stability and accuracy of the temporal discretization. Regarding numerical methods for systems of first order differential equations see [BSMM00, 19.4], [Hug00, 8]. – In the following, two approaches will be considered: (1) the implicit Euler approach with α = 1, (2) the explicit Euler approach with α = 0. Regarding Eq. (1.67)2 , the expicit approach leads to σ i+1 = σ i + CT,i · (i+1 − i ) + ∆tΣi

(1.69)

σ i+1 = σ i + CT,i+1 · (i+1 − i ) + ∆tΣi+1

(1.70)

and the implicit to

The latter may be a nonlinear equation in case Σ depends on  and/or σ. – A particular case is given with Σ=V·−W·σ

(1.71)

with a constant Kelvin viscosity V and a constant Maxwell viscosity W, see e.g. Section 2.2. Thus, Σi+1 = V · i+1 − W · σ i+1 and the implicit scheme is σ i+1 = σ i + CT,i+1 · (i+1 − i ) + ∆t V · i+1 − ∆t W · σ i+1  = [I + ∆t W]−1 · σ i + CT,i+1 · (i+1 − i ) + ∆t V · i+1  = [I + ∆t W]−1 · σ i + [CT,i+1 + ∆t V] · (i+1 − i ) + ∆t V · i (1.72) – Internal nodal forces according to Eq. (1.33)1 Z fi+1 = BT · σ i+1 dV = fi + KT,i+1 · ∆u + ¯fi+1 (1.73) V

with

R T fi = RV B T · σ i dV ¯fi+1 = R∆t V B · Σi+1 dV ¯ T,i+1 · B dV KT,i+1 = V BT · C

(1.74)

– This is embedded in a time stepping scheme, compare Eq. (1.63) ri+1 = fi+1 − pi+1 = 0

→

KT,i+1 · ∆u = pi+1 − fi − ¯fi+1

(1.75)

Starting with initial conditions f0 , ¯f0 and a given load p(t) displacements may be determined time step by time step. But Eq. (1.75) might be nonlinear in a time step and need an iteration, see Eq. (1.65). State April 4, 2013

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17 • Solution methods for dynamic problems – Based on Eq. (1.35) we have in analogy to Eq. (1.63) ¯ (t) + ¯t(t) p(t) = p

¨ + f − p(t) = 0, r=M·u

(1.76)

where the loading p is a prescribed function of the time t. The displacements u(t) ˙ ¨ (t), internal nodal forces f ) and all derived values (velocities u(t), accelerations u are unknown before solution. Eq. (1.76) is discretized in the spatial domain, but not yet in the time domain, i.e. it is system of ordinary differential equations of 2nd order in time. – This problem needs appropriate boundary conditions and initial conditions for the ˙ displacements u0 = u(0) and velocities u˙ 0 = u(0). – A solution may be computed with a discretization in time via a difference scheme. A widespread approach is given with the Newmark method h i ¨ i+1 + (1 − γ)¨ u˙ i+1 = u˙ i + ∆t γ u ui h i (1.77) ¨ i+1 + ( 21 − β)¨ ui+1 = ui + ∆t u˙ i + ∆t2 β u ui ˙ i+1 ), u ¨ i+1 = u ¨ (ti+1 ) a time step length ∆t = with ui+1 = u(ti+1 ), u˙ i+1 = u(t ti+1 −ti and integration parameters γ, β. Eqns. (1.77) are solved for the acceleration and velocity in time step i + 1. We get ¨ i+1 = u

1 ˜ i+1 ] [ui+1 − u β∆t2

(1.78)

with an auxiliary quantity ˜ i+1 = ui + ∆t u˙ i + u

∆t2 ¨i (1 − 2β) u 2

(1.79)

and the velocity u˙ i+1

    γ γ γ ¨i = [ui+1 − ui ] + 1 − u˙ i + ∆t 1 − u β∆t β 2β

(1.80)

Finally, dynamic equilibrium Eq. (1.76) is applied for time step i + 1 with the acceleration according to Eq. (1.78): r=

1 ˜ i+1 ] + fi+1 − pi+1 = 0 M · [ui+1 − u β∆t2

(1.81)

¨ i , given mass matrix With given parameters γ, β, ∆t, a given previous state ui , u˙ i , u M and load Pi+1 , Eq. (1.81) has to be solved for ui+1 , where the dependence of fi+1 on ui+1 is crucial and might be nonlinear. – We apply the Newton-Raphson method Eq. (1.65). The Jacobian matrix is given with ∂f 1 1 (ν) (ν) M+ = M + KT (1.82) AT = 2 β∆t ∂u u=u(ν) β∆t2 i+1

leading to an iteration scheme (ν+1) ui+1

=

(ν) ui+1

 h i (ν) −1 (ν) + AT · pi+1 − fi+1 −

h i 1 (ν) ˜ M · ui+1 − ui+1 β∆t2

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 (1.83)

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18

1.7 Numerical Integration and Solution Methods for Algebraic Systems – In the linear case with (ν)

(ν)

(ν)

fi+1 = K · ui+1 ,

AT = A =

1 M+K β∆t2

(1.84)

the Eq. (1.83) simplifies to ui+1 = A

−1



1 ˜ i+1 · pi+1 + M·u β∆t2

 (1.85)

with no iteration necessary, compare [Bat96, 9.2.4]. – Numerical integration parameters γ, β rule consistency and numerical stability of the method. * Stability means that an amount of error introduced in a certain step due to a finite time step length ∆t is not is not increased in the subsequent steps. Consistency means that the iteration scheme converges to the differential equa* tion for ∆t → 0. Stability and consistency are necessary to ensure that the error of the numerical method remains within some bounds for a finite time step length ∆t Generally a choice β = 41 , γ = 12 is reasonable for the Newmark method to reach consistency and stability.

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Bibliography [Bat96]

BATHE, K.J.: Finite Element Procedures. Englewood Cliffs, New Jersey : Prentice Hall, 1996

[Bat01]

BATHE, K.J.: The inf-sup condition and its evaluation for mixed finite element methods. In: Computers and Structures 79 (2001), S. 243–252

[BLM00]

B ELYTSCHKO, T. ; L IU, W.K. ; M ORAN, B.: Nonlinear Finite Elements for Continua and Structures. Chichester : John Wiley & Sons, 2000

[BSMM00] B RONSTEIN ; S EMENDJAJEW ; M USIOL ; M UEHLIG: Taschenbuch der Mathematik. 5. Auflage. Frankfurt/Main : Verlag Harri Deutsch, 2000 [CF08]

CEB-FIP: Practitioners’ guide to finite element modelling of reinforced concrete structures. Bd. Bulletin Nr. 45. Lausanne : International Federation for Structural Concrete fip, 2008

[Hug00]

H UGHES, T.R.: The Finite Element Method - Linear Static and Dynamic Finite Element Analysis. Mineola, New York : Dover Publications, Inc., 2000

19

Chapter 2

Uniaxial Structural Concrete Behavior 2.1

Short Term Stress-Strain Behavior of Concrete • Short term behavior means, that loading is applied within a few minutes, hours, or days, and that one regards the immediate reaction of the structural material. – Time-dependent behavior, e.g. creep and shrinkage, is not considered. • Unconstrained compression – Unconstrained → spatially homogeneous loading – Stress-strain behavior under monotonic loading, see Fig. 2.1a * Linear part * Nonlinear hardening part * Nonlinear softening part

Figure 2.1: a) Uniaxial compressive stress strain behavior b) mesoscale view * Analytical description · Approach by Saenz, see [CS94, 8.8.1] σ= 1+



Ec0 Ec1

Ec0    2   − 2 c1 + c1

(2.1)

with initial Young’s modulus Ec0 of concrete, its secant modulus Ec1 = −fc /c1 at compressive strength fc (unsigned), strain c1 (signed) at strength. With  = c1 Eq. (2.1) leads to σ = −fc . 20

21 · The derivative with respect to  (→ tangential material stiffness) is given with   2 E 1 − c0 2 dσ c1 Et = (2.2) =  2  d  2 c0 − 2 + 1+ E Ec1 c1 2 c1

which leads to Et = Ec0 for  = 0, furthermore Et = 0 for  = c1 and Et < 0 for  < c1 , || > |c1 |. · Alternatives by Modelcode 90 [Com93, 2.1.4.4.1] and DIN 1045-1 [din08, 9.1.5, 8.5.1]. These approaches may all be classified as ‘phenomenological’, · i.e. free parameters of a polynomial form are chosen such, that characteristic points of measured stress-strain behavior are reproduced. – Mesoscale view with a distinction of * * * *

aggregates, mortar, interface layer between aggregates and mortar, → highly inhomogeneous stress distribution within a specimen with internal lateral tensile stresses locally.

– Failure modes * Lateral splitting / cracking with respect to compressed longitudinal direction * Volume expansion * Diffuse failure • Unconstrained tension

Figure 2.2: a) Uniaxial tensile stress strain behavior b) mesoscale cracking – Homogeneous end loading of a uniform bar – Stress-strain behavior, see Fig. 2.2a * Linear part * Nonlinear hardening part * Nonlinear softening part

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22

2.1 Short Term Stress-Strain Behavior of Concrete – Mesoscale view, see Fig. 2.2b, with successive states * micro cracking, * crack bridges, * macro cracking.

Figure 2.3: Scheme of localization – Localization and crack band * Weak cross section within specimen where tensile strength is reached * Narrow band of high strains → crack band * Strain distribution c (x) across crack band with x1 ≤ x ≤ x2 and a crack band width bw = x2 − x1 * Softening of crack band → increasing strains with decreasing stresses * Unloading of specimen areas beyond crack zone · → Snap back behavior of whole bar depending on ratio of crack band width to total length · → Localized failure in contrast to diffuse failure * Process band ending up in macro crack with zero stress and zero strain beyond crack band. * The whole process may be unstable under quasistatic conditions. – Fictitious crack width

Z

x2

w=

c (x) dx = bw ¯c

(2.3)

x1

with mean strain ¯c in the crack band. * In the following, c ≈ const. is assumed within the crack band for simplification. * Accordingly, the fictitious crack width, see Eq. (2.3), is given with w = bw c

(2.4)

* Furtmermore, this assures a σ −c -relation in the softening part of Fig. 2.2a and a critical strain cr with σ(cr ) = 0. * Regarding Eq. (2.4), a σ − c -relation may be transformed into a σ − w-relation with a fictitious critical crack width wcr with σ(wcr ) = 0.

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23 – Crack energy * Energy in a material element within crack band depending on c or w Z c Z w 1 0 0 σ( ) d = g= σ(w0 ) dw0 bw 0 1

(2.5)

with a lower bound e.g. 1 = ct and σ(0 ) according to Fig. 2.2a. * With c = cr or w = wcr Eq. (2.5) leads to the specific crack energy gf (related to volume), which corresponds to the shaded area in Fig. 2.2a. * Sum along crack band leads to crack energy (related to surface) Z wcr Z cr 0 0 σ(w0 ) dw0 (2.6) σ( ) d = Gf = bw gf = bw 1

0

* Gf measures energy dissipation due to creation of new surfaces. * Gf is assumed as (constant) material parameter, so that Eq. (2.6) leads to a constraint for a σ − w-relation or σ − c -relation, respectively. • Unloading to stress free state with a reduction of stresses and strains – Reduction of stiffness → damage – Remaining strains in stress free state → plasticity Unloading is hard to realize for softening structural elements, as it requires total displacement control of softening areas.

Example 2.1 Simple concrete tensile bar with localization

Figure 2.4: Example 2.1 a) geometry scheme b) force-displacement curve • Geometry and Discretization, see Fig. 2.4a – Bar Length L = 0.5 m, cross sectional area Ac = 0.1 × 0.1 m2 – Discretization with Ne = 500 1D-bar elements with two nodes, see Eq. (1.7). • Material – Concrete grade C40 according to [Com93, 2.1] mailto:[email protected]

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24

2.1 Short Term Stress-Strain Behavior of Concrete – Initial Young’s modulus Ec = 36 000 MN/m2 , tensile strength fct = 3.5 MN/m2 • Material model – Damage model with gradient damage, see [HC07]. • Boundary conditions – Left end with zero displacements,right end with prescribed displacement. – Nonlocal damage assumed as zero at both ends (→ failure in mid point). • Solution method – Incrementally iterative with arc length control of load step size and Newton-Raphson iteration within each increment. • Results

Figure 2.5: Example 2.1 strain distribution a) load step A b) load step B – Load displacement curve see Fig. 2.4b. * Initial elastic part followed by increasing damage * Limit load where the stresses reach the tensile strength of the material * Softening part with increasing displacements and decreasing load · Bar has two types of material behavior in its longitudinal direction. Hardening and softening tangential material stiffness. In the softening part the bar elongates with decreasing stresses (→ localization) in the hardening part it becomes shorter due to reduced stresses. * Snap back part with decreasing displacements and decreasing load · Totally, it shortens in the example case, but this depends on the ratio of length of softening part to hardening part. · Longer bars show a more pronounced snap back behavior. – Strain distribution along bar * Before limit load see Fig. 2.5a. Due to prescribed zero nonlocal damage on both ends strain moderately increases in the mid range of the bar. Stress is σ = 2.90 MN/m2 and mean strain ¯ = 0.94 · 10−4 . * After limit load in the snap back range see Fig. 2.5b. Strains strongly increase in a short mid range and otherwise decrease due to the load decrease. Stress is σ = 1.64 MN/m2 and mean strain ¯ = 1.08 · 10−4 . State April 4, 2013

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25 end example 2.1

• Remarks – In a real specimen the localization will not center exactly in the mid point of a bar, but in the weakest cross section. This cross section will arise due to the stochastic variation of material strength. Its cross-sectional strength and location cannot be exactly determined, but only with statistical parameters.

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26

2.2 Long Term Effects - Creep, Shrinkage and Temperature

2.2

Long Term Effects - Creep, Shrinkage and Temperature • Besides the immediate reaction of a structural material, its deferred reaction upon an applied loading is regarded. This is typically creep or relaxation, respectively. The following approach is chosen for the uniaxial strain depending on time t Z t J(t, τ )σ(τ ˙ ) dτ, 0 ≤ τ ≤ t (2.7) (t) = 0

for a creeping material exposed to a uniaxial load history σ(τ ). The compliance function J(t, τ ) is specific for every material type. Eq. (2.7) describes linear creep. • The following form has proven to be appropriate for the compliance function J(t, τ ) =

N X

Jµ (t, τ ),

Jµ (t, τ ) =

µ=1

 1  1 − e[yµ (τ )−yµ (t)] , Eµ (τ )

yµ (τ ) = (τ /τµ )qµ

(2.8) with material parameters τµ , qµ and material functions Eµ (τ ). The parameter τµ has a dimension of time and the function Eµ a dimension of stress. – The case N = 1, E1 = E = const., τ1 → 0 gives J(t, τ ) = 1/E with Eq. (2.8), and with Eq. (2.7) a linear elastic law σ(t) = C (t). Combining Eqns. (2.7), (2.8) leads to (t) =

t

Z

XN µ=1

µ (t),

µ (t) =

Jµ (t, τ ) σ(τ ˙ ) dτ

(2.9)

0

• Determination of Jµ → literature

Figure 2.6: a) Kelvin-Voigt element, Maxwell element b) Simple chains • Simplifications are used in the following, i.e. Eµ = const., qµ = 1. Using Eqns. (2.8), (2.7) leads to − t Z τ σ(t) e τµ t (2.10) µ (t) = − σ(τ ˙ ) e τµ dτ Eµ Eµ 0 with −

˙µ (t) =

t

1 e τµ σ(t) ˙ + Eµ τµ Eµ

Z

t

σ(τ ˙ ) e 0

τ τµ

−

t

−

t

τ e τµ 1 e τµ dτ − σ(τ ˙ ) e τµ = Eµ τµ Eµ

Z

t

τ

σ(τ ˙ ) e τµ dτ 0

(2.11) State April 4, 2013

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27 Thus, the strain Eq. (2.10) fulfills the differential equation ηµ ˙µ (t) + Eµ µ (t) = σ(t),

ηµ = Eµ τµ

(2.12)

with an initial condition µ (0) = 0. Eq. (2.12) describes a Kelvin-Voigt element with a spring and a damper in parallel, see Fig. 2.6a. – An alternative is given with the Maxwell element with a spring and a damper in series, see Fig. 2.6a. Eq. (2.9)1 yields a Kelvin-Voigt chain, i.e. a number of Kelvin-Voigt elements in series. These systems of springs and dampers add up to rheological models. • A simple Kelvin-Voigt chain N = 2, µ = 0 . . . 1, η0 = 0 is considered E0 0 (t) = σ(t),

η1 ˙1 (t) + E1 1 (t) = σ(t)

(2.13)

see Fig. 2.6b, with (t) = 0 (t) + 1 (t),

(t) ˙ = ˙0 (t) + ˙1 (t)

(2.14)

σ(t) ˙ E0

(2.15)

First of all, we have 1 (t) = (t) −

σ(t) , E0

˙1 (t) = (t) ˙ −

Thus, Eq. (2.13)2 after a few calculations leads to σ(t) ˙ +

E0 + E1 E0 E1 σ(t) = E0 (t) ˙ + (t) η1 η1

(2.16)

We introduce a final stiffness 1/E = 1/E0 +1/E1 , consider E0 as an initial stiffness, and introduce a creep coefficient ϕ defined by E = E0 /(1 + ϕ). This leads to E1 = E0 /ϕ and Eq. (2.16) may be reformulated as σ(t) ˙ +

(1 + ϕ)E0 E2 σ(t) = E0 (t) ˙ + 0 (t) ϕ η1 ϕ η1

(2.17)

Thus, prescribing a constant stress σ(t) = σ0 , σ˙ = 0 with an initial strain (0) = σ0 /E0 leads to a strain varying with time (t) =

 i σ0 h 1 + ϕ 1 − e−ζ t , E0

ζ=

E0 ϕ η1

(2.18)

The asymptotical final value is asym = (1 + ϕ)σ0 /E0 , the creep portion is ϕσ0 /E0 . This provides a method to describe η1 , as the creep portion has a characteristic time t? upon a part α with 0 ≤ α < 1 of the asymptotic creep part. This leads to ?

1 − e−ζ t = α

→

ζ=−

ln(1 − α) t?

(2.19)

The values E0 , t? , ϕ, α may be determined based on experiments. This results in η1 and a differential visco-elastic material law Eq. (2.17) to describe creep.

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28

2.2 Long Term Effects - Creep, Shrinkage and Temperature • Incremental material law – See Section 1.4, Eq. (1.26). This has to be extended with respect to viscous parts, like e.g. Eq. (2.17). For following uniaxial applications the form     σ(t) ˙ = E0 (t) ˙ + ζ (t) − 1 + ϕ ζ σ(t),

ζ=

ln α E0 =− ? ϕ η1 t

(2.20)

is chosen. • Shrinkage and temperature – Temperature and shrinkage impose strains independent from loads. Uniaxial temperature strains, e.g., are given by T = αT ∆T

(2.21)

with thermal expansion coefficient αT and a temperature change ∆T . Concrete shrinkage strains cs mainly depend on time, humidity conditions and ratio of surface to volume, see [din08, 9.1.4] or [Com93, 2.1.6.4.4]. – Stress is induced by total, measured strain less imposed strains. Thus, in the uniaxial linear elastic case stress is given by σ = E ( − I ) ,

I = T + cs

– This is transferred to the case including linear creep with Eq. (2.20) h  i   σ(t) ˙ = E0 (t) ˙ − ˙I (t) + ζ (t) − I (t) − 1 + ϕ ζ σ(t)       = E0 (t) ˙ − ˙I (t) + ζ E0 (t) − I (t) − ζ 1 + ϕ σ(t)

(2.22)

(2.23)

or     σ(t) ˙ = CT (t) ˙ − ˙I (t) + V (t) − I (t) − W σ(t)

(2.24)

with CT = E0 ,

V = ζ E0 ,

  W =ζ 1+ϕ

(2.25)

where I (t), ˙I (t) should be a known function of time. – A transient problem arises – first of all without inertial terms –, which needs some care regarding integration of systems equations in time, see Section (1.7).

Example 2.2 Simple concrete tensile bar with creep and imposed strains • Geometry and Discretization – Bar Length L = 1.0 m, cross sectional area Ac = 1.0 m2 . Discretization with Ne = 5 1D-bar elements with two nodes, see Eq. (1.7). – As a homogeneous state – along longitudinal direction – is considered in this example, the length of the bar is irrelevant and one element would be sufficient. • Material – Concrete with an initial Young’s modulus E0 = 30 000 MN/m2 and a tensile strength fct = 3.5 MN/m2 . State April 4, 2013

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29 – Material creep is assumed with ϕ = 2.0 and t? = 100 [d] for α = 0.5, i.e. half of total creep occurs after 100 days for a constant stress load. With Eqns. (2.20), (2.25) this leads to ζ = 0.006931 1/d,

CT = 30 000 MN/m2 ,

V = 207.94 MN/m2 d,

W = 0.020794 1/d (2.26)

• Boundary conditions – Displacement of left node is prescribed with zero in all following cases. • Solution method – Incrementally iterative with Newton-Raphson iteration within each increment, if neccessary. The method is described on Page 16. – Following time discretization values are chosen: implicit and ∆t = 10 days. A period of 500 days is regarded.

Figure 2.7: a) Strain depending on time b) Stress depending on time • Case 1: validation with stress loading σ0 = 3.0 MN/m2 constant in time – Computed strain depending on time see Fig. 2.7a. An exact solution is given with Eq. (2.18) for this case. Differences between exact solution and numerically computed solution are small and are not visible in the Figure. • Case 2: prescribed immediate strain 0 = 3.0/30 000 = 0.0001 constant in time – Computed stress depending on time see Fig. 2.7b. – In exact technical terms, this is not creep anymore but relaxation. • Case 3: prescribed imposed contraction – A slow contraction of 0.15 ‰ is prescribed over a period of t = 100 d linear in time and then hold constant. As total strain  is prescribed with zero, a tensile stress is induced, which would lead to tensile failure without relaxation. Stress with relaxation is shown in Fig. 2.7b. end example 2.2

• While analytical, exact solutions are availabe for cases 1, 2, the numerical approach is necessary for arbitrarily prescribed loads or displacements. Furthermore, more complex creep models, see e.g. [Mal69, 6.4], [JB01, 28, 29], can only be solved with numerical models. mailto:[email protected]

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30

2.3 Strain-Rate Effects

2.3

Strain-Rate Effects • Remains to be added.

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31

2.4

Cracks • This is strongly connected to the notions of localization, crack band and fictitious crack width, see Page 21. • We consider an instant with a fictitious crack width w given according to Eq. (2.3). Due to equilibrium reasons a constant cross sectional stress σ occurs along the crack band, which gives a pair σ, w. Considering all instants of tension softening a relation σ(w) can be derived, i.e. a stress across crack band depending on fictitious crack width.

Figure 2.8: Constitutive law for cohesive crack model • Cohesive crack model (general spatial view) – Fictitious crack * Implies a geometric plane corresponding to one of the two crack surfaces. * This plane supports a reference system, where the distance between the crack surfaces is measured by a normal crack width component wx and two sliding crack width components wy , wz . · A high idealization in the area of e.g. crack bridges. – Crack-tractions tx , ty , tz → Cauchy stress tensor σ projected onto crack plane * Crack-tractions tx , ty , tz are transferred across crack plane. – Material law connecting crack-traction and crack-width * A common format is tx = fn (wx ),

ty = fs (wy ),

tz = fs (wz )

(2.27)

with different laws fn for the normal component and fs for the sliding component. * This material law uses crack tractions as force variable and crack widths as deformation variable, compare page 2 * All material frameworks like plasticity, damage etc. may be used.

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32

2.4 Cracks – A common simplified approach: Set fs = 0. fn is given by a cut off and a scaling of Fig. 2.2a, which results in Fig. 2.8. This reflects the uniaxial case. – The cohesive crack model is a macroscopic model for the complex crack development process. • Smeared crack model for the simplified approach – Approach for strain of a uniaxial black box element1 with a length Lc with one crack =

1 [(Lc − bw ) m + bw c ] = (1 − ξ) m + ξ c , Lc

ξ=

bw Lc

(2.28)

with the strain m of the uncracked material, see Fig. (2.9). For c , bw see Eq. (2.4).

Figure 2.9: Smeared crack concept – Connection to stresses / forces * The strain part u is connected to a homogeneous material, see Section 1.4, leading to material stresses. * The strain part c or w is connected to a crack or discontinuity, leading to crack stresses, see Eq. (2.27)1 . * Both stress parts are connected through equilibrium, e.g. have the same value in an uniaxial element. – If element length is adjusted to crack band width bw , i.e. Lc = bw from Eqns. (2.4), (2.28) we get a crack width w = Lc  (2.29) In case of σ = 0, u = 0 adjusting of element length is not necessary, i.e. w = Lc  for any selection of Lc . – The smeared crack model is a model for the fictitious crack preserving continuity of displacements, which is important for ordinary finite elements. There may be alternative models for the fictitious crack, i.e. for the implemention of the cohesive crack model.

1

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We cannot see inside.

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33

2.5

Reinforcing Steel Stress-Strain Behavior • Uniaxial Compression / Tension, see 2.10a, b – Linear range – Transition range – Yielding range – Unloading – Reloading – Cyclic loading – Energy dissipation

Figure 2.10: a) Uniaxial stress-strain behavior for reinforcing steel b) hardening behavior • Simple uniaxial constitutive law with isotropic hardening – Stress-strain law ( σs =

Es (s − p ) sign fy

if p −

fy Es

≤  ≤ p + else

fy Es

(2.30)

with Young’s modulus Es of reinforcing steel, its yield stress fy , the sign function sign and the actual plastic strain p as internal state parameter. – Evolution law for internal state parameter  ˙p = ˙s if s ˙s > 0 and |σs | = fy f˙y = ET |˙s |

(2.31)

with a hardening or tangential modulus ET . Eqns. (2.30), (2.31) cover loading, unloading and reloading for tension and compression for uniaxial ideal elastoplastic behavior. The yield stress fy may increase due to hardening. • Hardening, see Fig. 2.10b – Isotropic – Kinematic • Codes – DIN 1045-1 [din08, 9.2, 9.3], Modelcode 90 [Com93, 2.2, 2.3]. mailto:[email protected]

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34

2.6 Bond between Concrete and Reinforcing Steel

2.6

Bond between Concrete and Reinforcing Steel • Bond mechanisms, see Fig. 2.11a, with – reinforcement ribs, – conical struts, – circumferential ties, – → complex triaxial problem.

Figure 2.11: a) Triaxial bond mechanisms b) Bond law • Modeling and model variables, compare page 2 – Uniaxial orientation along rebar direction. – Kinematic variable → slip s with a dimension of length. – Force variable → bond stress τ with a dimension of stress. – Bond law τ = fτ (s). * All constitutive frames like plasticity, damage etc. may basically be used. – Bond flow t = U τ with a rebar cirumference U and dimension force per length. • Bond laws – Monotonous loading. * See e.g. [Com93, 3.1.1]. * A smoothed version τ = fτ (s), see Fig. 2.11b. · Characterized by maximum bond stress τmax , a corresponding slip s1 , a residual bond stress τf and a corresponding slip s2 . · This is composed of a quadratic, cubic and linear polynom with continuous derivatives at the connection points. * As long as monotonous loading is considered, approaches with plasticity and damage approaches are dispensable. – Unloading and Reloading * Remains to be added. State April 4, 2013

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35

2.7

Reinforced Tension Bar • A finite element model – Concrete * Element type: 1D linear bar element, see Eq. (1.7). * Constitutive law: uniaxial linear elastic, see Eq. (1.19) with a limited tensile strength fct  Ec ,  ≤ fEctc (2.32) σ= 0, else – Concrete with Cracks * Smeared crack approach, see Page 32. – Reinforcing steel * Element type: 1D linear bar element, see Eq. (1.7). * Constitutive law: uniaxial elastoplastic, see Eqns. (2.30), (2.31). – Bond * Element type: 1D spring element, see Eq. (1.12), with slip s instead of ∆u. * Constitutive law: bond law, see Fig. 2.11b. – Geometry scheme see Fig. 2.12a. – Boundary conditions / loading * Base with fixed zero displacement on one end. * Prescribed displacement on the other end.

Figure 2.12: a) Geometry scheme of reinforced tension bar b) force-displacement curve of example 2.3 • Solution methods – Incrementally iterative. Equilibrium iteration within each loading increment with Newton-Raphson method, see Page 15. • Computation of crack width – See Eq. (2.29) and the following remark. As cracks are assumed as stress free immediately after cracking we have w = LI  with the length LI of the cracked element and its computed strain . mailto:[email protected]

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36

2.7 Reinforced Tension Bar

Example 2.3 Simple reinforced concrete tension bar • Model input values – Bar length L = 1.0 m, cross sectional area of concrete Ac = 0.1 × 0.1 m2 , reinforcement 1  16, As = 2.01 cm2 , circumference Us = 5.02 cm. – Discretization with two-node bar elements, see Eq. (1.7), length LI = 0.01 m. * 100 Elements for concrete with 101 nodes, 100 Elements for reinforcement with 101 nodes, 101 bond elements connecting concrete nodes and reinforcement nodes. * Totally 202 nodes. – Material parameters, see Table 2.1 concrete Young’s modulus Ec MN/m2 tensile strength fct MN/m2 reinforcing steel Young’s modulus Es MN/m2 yield strength fsy MN/m2 bond strength τmax MN/m2 slip at strength mm residual strength τres MN/m2 slip at residuum mm

35 000 3.5 200 000 500 6.0 0.1 3.0 1.0

Table 2.1: Material parameters of RC tensile bar example 2.3 * Concrete material model see Eq. (2.32). * Reinforcement material model see Eqns. (2.30), (2.31). * Bond model see see Fig. 2.11b. This corresponds to a material behavior per unit surface of bond. To gain the corresponding values for a whole spring, a multiplication by Us LI is neccessary. – Boundary conditions * Zero displacement reinforcement node on left boundary, prescribed displacement uR = 2.4 mm for reinforcement node on right side incrementally applied in 100 steps corresponding to a mean strain mean = 2.4 ‰. • Results: monotonic loading – Load-displacement curve see Fig. 2.12b * state I: Uncracked * state IIa: Ongoing cracking. Each sudden load decrease corresponds to a crack. Stiffness is reduced after each crack. * state IIb: Final cracking state before rebar yielding. * state III: Limit state with rebar yielding with a limit load Pu = As fsy = 0.1005 MN at a displacement uR = 2.14 mm State April 4, 2013

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37

Figure 2.13: Example 2.3 a) concrete stresses b) steel stresses * Stiffness of bare reinforcement EA = 40 MN. Limit load of Pu = 0.1005 MN applied on steel alone leads to a displacement uR = 2.5 mm compared to the actual displacement of uR = 2.14 mm (→ tension stiffening effect). – Stress distributions along bar at beginning limit state (uR = 2.14 mm). * Concrete stresses see Fig. 2.13a with zero stresses at a crack. Three cracks according to three peaks in the load-displacement curve, see Fig. 2.12b. * Reinforcement stresses see Fig. 2.13b with peak stresses at a crack. Yield strength is reached at each crack. * In between cracks forces are transferred among concrete and reinforcement by bond stresses, see Fig. 2.14a. Bond stresses have maximum absolute values in cracks and change sign across a crack.

Figure 2.14: Example 2.3 a) bond stresses b) displacements – Displacements at beginning limit state * See Fig. 2.14b. Displacements of concrete and reinforcement are different at a bar’s cross section due to flexible bond. The difference results in slip, which connected to bond stresses, see Section 2.6. – Crack widths by Eq. (2.29), where element length LI and crack band width are chosen to be equal. Thus, in a cracked concrete element crack width is given by the displacement difference of the end nodes. According to Fig. 2.14b here we get typically w ≈ 0.4 mm in the beginning limit state. end example 2.3

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38

2.8 Tension Stiffening for Reinforced Tension Bar

2.8

Tension Stiffening for Reinforced Tension Bar • Mean strain of tension bar – There is some contribution of concrete between cracks with reduced reinforcement stresses between cracks, see Fig. 2.13b. – It is assumed that reinforcement has not yet reached its yield limit. Therefore, reinforcement strain has the same shape as the reinforcement stress. – We consider the mean strain of the reinforcement which corresponds to the mean strain of the tension bar. * Mean strain refers to some reference length in the center area of the tension bar spanning several cracks. It results from the integration of strains in the reference length (→ displacement of reference length) divided by reference length itself. • Stress depending on mean strain – Stress of tension bar is given by reinforcement stress in cracks. This value also corresponds to the force of the tension bar. – Stress depending on mean strain corresponds to the course of Fig. 2.12b. It shows a higher stiffness compared to the stiffness of the reinforcement alone. The difference is called tension stiffening effect. • Stabilized cracking is distinguished from ongoing cracking with single cracks in the following. For a discussion of crack types see e.g. [HCH09].

Figure 2.15: Models for a) cracking b) tension stiffening

• Estimation of tension stiffening effect for stabilized cracking – The mean value of reinforcement strain between cracks may be estimated with sm =

1 (σsr − βt ∆σs ) Es

(2.33)

with reinforcement stress σsr in a crack, reinforcement stress difference ∆σs from crack to minimum value between cracks, and a parameter βt for the shape of reinforcement stress distribution, see Fig. 2.15a. For a discussion of βt see e.g. [HCH09]. Common values are assumed in the range 0.4 ≤ βt ≤ 0.6. State April 4, 2013

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39 – In case of stabilized crack pattern the stress difference is given by ∆σs =

fct %ef f

(2.34)

with the concrete tensile strength fct and the effective reinforcement ratio %ef f . Eqns. (2.33), (2.34) lead to σsr = Es sm + βt

fct %ef f

(2.35)

– Eqn. (2.35) leads to a shift of the pure reinforcement stiffness to the left, see Fig. 2.15a, i.e. a given mean strain has a higher stress with tension stiffening. This is limited by the yielding plateau of the reinforcement. • Ongoing cracking with single cracks – In case of single cracks a suitable reference length is hard to determine, as crack spacing is irregular and areas affected by cracks alternate with such not affected by cracks which have a full utilization of concrete. – The envelope2 of the a stress-strain curve is relatively flat. Starting with the initial reinforcement stress σsr,i = fct /%ef f in a crack immediately after cracking and a corresponding mean strain fct /Ec immediately before cracking an assumption   fct fct σsr = k sm − (2.36) + Ec %ef f is made with a free parameter k. This line should meet the line Eq. (2.35) at a stress value ασsr,i leading to k = Ec

α−1 , α − βt − n %ef f

n=

Es Ec

(2.37)

whereby experience shows that α ≈ 1.3 can be assumed. – The strain 0sm belonging to the meeting point can be determined with Eq. (2.35), which leads to 1 fct 0sm = (α − βt ) (2.38) Es %ef f • A simplified version might straighten the initial kink, see Fig. 2.15b. – This leads to σsr =

ασsr,i sm 0sm

(2.39)

in the initial range. • Further references: [Com93, 3.2], [DAf03, zu 8.5.1].

2

Envelope to saw tooth behavior during ongoing cracking.

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40

2.9 Cyclic Loading of Reinforced Tension Bar

2.9

Cyclic Loading of Reinforced Tension Bar • Reinforcement should not yield in case of cyclic loading. Thus, linear elastic behavior of reinforcement will be assumed. Furthermore, a stabilized cracking state is assumed. • First unloading – Initially along loading path of stabilized crack state – Upon further unloading crack closure resistance follows as rough crack surfaces do not fit exactly together. – Residual displacements / strains remain compared to uncracked state in stress free state.

Figure 2.16: Model for cyclic loading • Reloading – Degradation of bond → initially lower reloading stiffness compared to final unloading stiffness. – After full utilization of concrete same tangential stiffness as reinforcement. But the connection point / area to this tangential stiffness is shifted compared to that of the previous unloading path. • Further unloading-reloading cycles follow the same pattern with lower initial reloading stiffness compared to previous final unloading stiffness. But this effect becomes less pronounced. • After a large number of loading cycles the stress-strain behavior of tensions bar should converge against stress-strain behavior of reinforcement alone, with some horizontal shift due to crack closure effect. – A corresponding stress-strain relation is given by σsr = Es sm − ∆σs,c

(2.40)

see Fig. 2.15b, but the value ∆σs,c is hard to estimate. – Total crack closure w = 0 is connected with ein eigenstress state, i.e. concrete compressive stresses and reinforcement tensile stresses which upon integration equilibrate and have no resulting tensile bar force and stress, respectively. • See also Fig. (2.16) and [Com93, ?].

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Bibliography [Com93] C OMITE E URO -I NTERNATIONAL DE B ETON: CEB-FIP Model Code 1990. London : Thomas Telford, 1993 [CS94]

C HEN, W.F. ; S ALEEB, A.F.: Constitutive Equations for Engineering Materials, Volume 1: Elasticity and Modeling. 2. Auflage. Amsterdam : Elsevier Science B.V., 1994

[DAf03] DA F S TB: Erläuterungen zu DIN 1045-1 / Deutscher Ausschuss für Stahlbeton. Berlin, 2003 (Heft 525). – Forschungsbericht [din08]

DIN 1045-1: Tragwerke aus Beton, Stahlbeton und Spannbeton. Teil 1: Bemessung und Konstruktion. : DIN 1045-1: Tragwerke aus Beton, Stahlbeton und Spannbeton. Teil 1: Bemessung und Konstruktion, August 2008

[HC07]

H ÄUSSLER -C OMBE, U.: Zur Verwendung von Stoffgesetzen mit Entfestigung in numerischen Rechenverfahren. In: Bauingenieur 82 (2007), S. 286–298

[HCH09] H ÄUSSLER -C OMBE, U. ; H ARTIG, J.: Rissbildung von Stahlbeton bei Zwangbeanspruchungen. In: Bauingenieur 84 (2009), S. 546–556 [JB01]

J IRAZEK, M. ; BAZANT, Z.: Inelastic Analysis of Structures. 1. New York : John Wiley & Sons, 2001

[Mal69] M ALVERN, L. E.: Introduction to the Mechanics of a Continuous Medium. 1. Auflage. Englewood Cliffs, New Jersey : Prentice-Hall, 1969

41

Chapter 3

2D Structural Beams and Frames 3.1

General Cross Sectional Behavior

3.1.1

Kinematics

• Straight reference axis along beam direction – → Coordinate system x, z with x in the longitudinal direction and z in the transverse direction, see Fig. 3.1. • Bernoulli-hypothesis (→ cross sections remain plane during deformation) – Displacements w(x, z) = = u(x, z) =

w(x, 0) w(x) ¯ u ¯(x)   − z ϕ(x) ∂ w(x) ¯ − γ(x) = u ¯(x) − z ∂x

(3.1)

with a cross section rotation angle ϕ(x), a shear angle γ(x) and a displacement u ¯(x), w(x) ¯ of the reference axis. z,w

z2

w´ x,u

γ φ w´ u(x,z)

z1

w(x,z)

Figure 3.1: Beam kinematics – Strains, compare Eqns. (1.1)-(1.3) x (x, z)

=

xz (x, z) =

∂u ∂x ∂u ∂w + ∂z ∂x 42

=

∂u ¯ ∂x

−z

h

∂2w ¯ ∂x2

w ¯ = − ∂∂x +γ+

−

∂w ¯ ∂x

∂γ ∂x

i

=γ

(3.2)

43

3.1.1 Kinematics • In the following a notation ∂ • /∂x = •0 , ∂ 2 • /∂x2 = •00 is used for abbreviation. • The overbars on u ¯, w ¯ will be omitted in the following. To simplify the notation u, w will be written instead. • Considering shear deformations, deformations are advantageously defined as (x) =

∂u , ∂x

κ(x) = ϕ0 = w00 − γ 0

(3.3)

With Eq. (3.2) this leads to strains x (x, z) = (x) − z κ(x)

(3.4)

linearily varying along the beam height with extremal values at the top and bottom of the cross section. • In the following , κ, γ are chosen as deformation variables. –  → strain of the reference axis – κ → curvature of deformed cross sections1 – γ → shearing angle of deformed cross sections relative to reference axis • To describe material behavior, deformation variables have to be connected to force variables, which are moment M , normal force N and shear force Q in case of 2D beams. Generally, the following dependencies are assumed M = M (, κ),

N = N (, κ),

Q = Q(γ)

(3.5)

• Basics of beam theory look quite simple, but there are some little inconsistencies. – A shear strain xz , which is constant over the cross section leads to non-vanishing shear stress at the lower and upper side of a beam. But this contradicts the local equilibrium conditions. – On the other hand, a parabolic or other nonlinear course of shear stresses according to equilibrium conditions with linear normal stresses leads to a curved course of shear strains with vanishing values on top and bottom sides. – These contradictions can be resolved with plate theory. Plane beam theory is its limiting case or a very useful approximation, respectively.

1

This is different compared to the curvature of the deflection w of the reference axis. Both are related by Eq. (3.3).

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3.1 General Cross Sectional Behavior

3.1.2

Linear elastic behavior

• We consider Eq. (1.21) with σy = 0, i.e. y = −ν x , and gain as normal stresses σx (x, z) = E x = E [(x) − z κ(x)]

(3.6)

and as shear stresses with the shear modulus G σxz (x) = G xz , G =

E 2(1 + ν)

(3.7)

• Internal forces Z N

z2

Q

Z

z2

bdz (x) − E

z bdz κ(x) z1 Z z2 = − σx z bdz = −E z bdz (x) + E z 2 bdz κ(x) Z z1z2 Zz1z2 Z zz12 bdz γ(x) xz bdz = α G σxz bdz = G = σx bdz

=

=

E

Z zz12

Zz1z2

M

z2

Z

z1

z1

z1

(3.8)

– The shear correction factor α is used to compensate the difference between mean shearing strain/stress over the cross section and the shearing strain/stress γ, Gγ in the reference point with z = 0. – In case of a rectangular cross section shape with a reference axis through the mid points it is α = 5/6. This is only valid in the linear elastic case. Other cross section shapes have other values α. • Section properties with cross-sectional area A, sectional modulus S and second moment of area J Z z2 Z z2 Z z2 A= bdz, S = z bdz, J = z 2 bdz (3.9) z1

z1

z1

Generally it is assumed, that the reference axis x coincides with the centre of area, which has a condition Z z2 z bdz = 0 (3.10) S= z1

which formally simplifies a lot in the linear elastic case, but is not mandatory. • Finally, in the linear elastic case under the condition of Eq. (3.10) we get the well known relations2 N = EA , M = EJ κ, Q = α GA γ (3.11) or material stiffness which equals the tangential material stiffness   EA 0 0 0  C = CT =  0 EJ 0 0 α GA or in case of a theory neglecting shear deformations   EA 0 C = CT = 0 EJ

(3.12)

(3.13)

which altogether is the simple linear elastic form of Eqns. (3.5). 2

Sign for moment is different compared to classical structural beam. The difference results from a different orientation of the z-axis, see Fig. 3.1.

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3.1.3 Cracked reinforced concrete behavior

3.1.3

Cracked reinforced concrete behavior

3.1.3.1

Compressive zone and internal forces

• For cross sections of reinforced concrete structures the linear elastic relations3 hold only until the tensile strength of the concrete is reached. In the following cracked sections are treated, i.e. beam strains x exceed the tensile limit strain ct of concrete. • More kinematic items – The reference axis is placed in the centre of the cross section, thus the bottom side has a coordinate z = z1 = −h/2 and the top side z = z2 = h/2. – We look at the strain 1 at the bottom side z1 = −h/2 and 2 at the top side z2 = h/2 with the cross section height h. Eq. (3.4) leads to 1 =  − z1 κ =  +

h κ, 2

2 =  − z2 κ =  −

h κ 2

(3.14)

and further to

1 − 2 , h = z2 − z1 (3.15) h Correct signs for strains have to be considered, e.g. 1 > 0, 2 < 0 in bending with compression on the top side. κ=

– The coordinate of a line (in width direction) with a given strain x = 0 is determined with Eq. (3.4)  − 0 z0 = (3.16) κ e.g. the zero line 0 = 0 is determined with z0 =

 κ

(3.17)

The vertical z-coordinates of the concrete’s tensile or compressive limit strain may be determined in the same way. – Strains in cracked concrete sections are fictitious values regarding concrete. – The lower reinforcement has a coordinate zs1 = −h/2+d1 und the upper reinforcement zs2 = h/2 − d2 , where d1 , d2 give the edge distances of both reinforcements. Eq. (3.4) leads to reinforcement strains     h h s1 =  − − d1 κ, s2 =  + − d2 κ (3.18) 2 2 → Perfect bond is implicitely assumed.

3

The reinforcement part has then to be considered with a weighting factor n = ES /Ec .

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3.1 General Cross Sectional Behavior • Concrete compressive zone – Edge strains according to Eq. (3.14) and zero line Eq. (3.17) determine the size of the concrete compressive zone: z0 < −h/2 and 1 < 0 z0 < −h/2 and 1 ≥ 0 −h/2 ≤ z0 ≤ h/2 and 2 < 0 −h/2 ≤ z0 ≤ h/2 and 1 < 0 z0 > h/2 and 2 < 0 z0 > h/2 and 2 ≥ 0

cross section totally under compression totally under tension upper bending compressive zone lower bending compressive zone totally under compression totally under tension

Cases with compression or tension of total cross section may also include bending moments and normal forces. Concrete contribution is assumed within a cross section range zc1 , zc2 : cross section totally under compression totally under tension upper bending compressive zone lower bending compressive zone

zc1 = z1 , zc2 = z2 no concrete contribution zc1 = z0 , zc2 = z2 zc1 = z1 , zc2 = z0

To consider a concrete’s restricted compressive limit strain or an tensile limit strain larger zero due to a tensile strength the value z0 has to be replaced according to Eq. ( 3.16) with appropriate values chosen for 0 . • Resulting internal forces – Normal force

Z

zc2

N = As1 σs1 + As2 σs2 +

σc bdz

(3.19)

zc1

where uniaxial reinforcement stresses σs1 , σs2 and conrete stresses σc are functions of x , see Eq. (3.4). Correct stress signs have to be considered. – Bending moment Z

zc2

M = −As1 σs1 zs1 − As2 σs2 zs2 −

σc z bdz

(3.20)

zc1

for sign conventions see Fig. 3.3. • A variable cross section width b(z) may be regarded within a numerical integration of Eqns. (3.19), (3.20), (3.43), (3.44).

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3.1.3 Cracked reinforced concrete behavior 3.1.3.2

Linear concrete compressive behavior

• A linear stress-strain behavior is assumed for concrete in its compressive zone  Ec ≤0 0 0 σ = Ec , Ec = 0 else

(3.21)

excluding a tensile strength. A limited tensile strength may easily be incorporated. • A cross section with height h is considered. A beam strain , κ given, a compressive zone is determined according to Page 46. The compressive zone has an extension zc1 ≤ z ≤ zc2 with a height hc = zc2 − zc1 and edge strains ce = B ·  with

 ce =

c1 c2



 ,

B =

(3.22)

1 −zc1 1 −zc2



 ,

 κ

=

 (3.23)

This leads to edge stresses σ ce = Ec0 ce = Ec0 B · ,

 σ ce =

• Following cross sectional values will be used Z zc2 Z zc2 Ac = bdz, Sc = z bdz, zc1



σc1 σc2

(3.24)

Z

zc2

Jc =

zc1

z 2 bdz

(3.25)

zc1

• Extremal stresses at zc1 , zc2 are given by σc1 , σc2 with a linear course in between according to Eq. (3.21). They are interpolated with σc1 zc2 − σc2 zc1 σc2 − σc1 + z (3.26) σc (z) = zc2 − zc1 zc2 − zc1 Thus, concrete contributions to internal forces are given by Z zc2 σc1 zc2 − σc2 zc1 σc1 − σc2 σc (z) bdz = Ac − Sc Nc = hc hc zc1 Z zc2 (3.27) σc1 zc2 − σc2 zc1 σc1 − σc2 Mc = − σc (z)z bdz = − Sc + Jc hc hc zc1 see also Eq. (3.8), or σ c = Aσ · σ ce = A · Bσ · σ ce

(3.28)

with  σc =

Nc Mc



 ,

A=

Ac −Sc −Sc Jc

 ,

1 Bσ = hc



zc2 −zc1 1 −1

 (3.29)

whereby σ c and σ ce have to be clearly distinguished in the following! • Eq. (3.21) is applied to stresses σ ce , ce at zc1 , zc2 , i.e. leading to a material stiffness σc = C ·  with C = Ec0 A · Bσ · B = Ec0 A = Ec0

(3.30) 

Ac −Sc −Sc Jc

 (3.31)

In case of rectangular cross sections with width b and a height h this evaluates to Ac = 2 − z 2 ), J = b (z 3 − z 3 ), in case of z = −h/2, z = h/2 to b(zc2 − zc1 ), Sc = 2b (zc2 c c1 c2 c1 c1 3 c2 Ac = bh, Sc = 0, Jc = bh3 /12. Contrarily to Eq. (3.11) the Eq. (3.30) couples a normal force Nc to  and a bending moment Mc to κ. mailto:[email protected]

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3.1 General Cross Sectional Behavior • As the compressive zone limits zc1 , zc2 may change due to change of in , κ the relation Eq. (3.30) is basically nonlinear. To consider this effect Eq. (3.28) is written as   " ∂Nc ∂Nc #   " ∂Nc ∂Nc #   N˙ c σ ˙ z˙c1 c1 ∂σ ∂σ ∂z ∂z c1 c2 c1 c2 = ∂Mc ∂Mc · + ∂Mc ∂Mc · (3.32) σ˙ c2 z˙c2 M˙ c ∂σc1 ∂σc2 ∂zc1 ∂zc2 or σ˙ c = Aσ · σ˙ ce + Az · z˙ c

(3.33)

with Aσ according to Eqs. (3.28,3.29). To simplify Az a linear variation of width b is assumed with b1 = b(zc1 ), b2 = b(zc2 ). This yields   −σc1 b2 − 2σc1 b1 − 2σc2 b2 − σc2 b1 σc1 b2 + 2σc1 b1 + 2σc2 b2 + σc2 b1 1  (b2 zc1 + zc2 b2 + zc1 b1 )σc2 + (b2 zc1 − 3zc2 b2 − zc2 b1 )σc2 − Az = 6 (b2 zc1 − zc2 b1 + 3zc1 b1 )σc1 (zc1 b1 + zc2 b2 + zc2 b1 )σc1 (3.34) • The values zc1 , zc2 stand for zero lines given by  ˙  (3.35) , z˙0 = − 2 κ˙ κ κ κ see Eq. (3.17), or upper or lower edges of the concrete compression zone. The following cases have to be considered: z0 =

1. Dominating bending with lower compression zone zc1 = −h/2, zc2 = z0 < h/2 and z˙c1 = 0, σc2 = 0, σ˙ c2 = 0 and   h         zc1 0 0 z˙c1 ˙ −2 = = 1 , (3.36) ·   zc2 z ˙ κ ˙ − c2 κ κ κ2 2. Dominating normal forces with fully compressed cross section zc1 = −h/2, z˙c1 = 0, zc2 = h/2, z˙c2 = 0, hc = h and   h         −2 zc1 ˙ 0 0 z˙c1 = (3.37) · = , h zc2 κ ˙ 0 0 z ˙ c2 2 3. Dominating bending with upper compression zone zc1 = z0 > −h/2, zc2 = h/2 and z˙c2 = 0, σc1 = 0, σ˙ c1 = 0 and         1    zc1 z˙c1 ˙ − κ2 κ κ = , = · (3.38) h zc2 z˙c2 κ˙ 0 0 2 Anyway, we set z˙ c = Bz · ˙

(3.39)

• We use Eqs.(3.22,3.39) and (3.24) to formulate the edge strains and stresses ˙ ce = B · ˙ − κ z˙ c = B · ˙ − κBz · ˙ = (B − κBz ) · ˙ σ˙ ce = Ec0 ˙ ce = Ec0 (B − κBz ) · ˙

(3.40)

This yields a tangential material stiffness together with Eq. (3.33) 0 ˙ + Az · Bz · ˙ σ˙ c = E σ · (B − κBz ) ·  c A = Ec0 Aσ · (B − κBz ) + Az · Bz · ˙ = CT · ˙

(3.41)

also considering the change of the extension of the concrete compressive zone. • A reinforcement can be superposed.

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3.1.3 Cracked reinforced concrete behavior 3.1.3.3

Nonlinear concrete compressive behavior

• Material behavior – Concrete uniaxial stress σc see Eq. (2.1) or alternatives Modelcode 90 [Com93, 2.1.4.4.1] and DIN 1045-1 [din08, 9.1.5, 8.5.1]. – Reinforcement uniaxial stress σs see Eq. (2.30) or alternatives Modelcode 90 [Com93, 2.2.4] and DIN 1045-1 [din08, 9.2.3, 8.5.1] • Tangential material stiffness for bending moment and normal force – Strain dependence on deformation variables according to Eq. (3.4) leads to ∂x = 1, ∂

∂x = −z ∂κ

(3.42)

– With Eq. (3.19) ∂N ∂

= =

∂N ∂κ

= =

Z zc2 ∂σs1 ∂x ∂σs2 ∂x ∂σc ∂x As1 + As2 + bdz ∂x ∂ ∂x ∂ Z zc1 ∂x ∂ zc2 ∂σc ∂σs2 ∂σs1 + As2 + bdz As1 ∂x ∂x ∂x zc1 Z zc2 ∂σc ∂x ∂σs1 ∂x ∂σs2 ∂x As1 + As2 + bdz ∂x ∂κ ∂x ∂κ zc1 ∂x ∂κ Z zc2 ∂σs1 ∂σs2 ∂σc −As1 zs1 − As2 zs2 − z bdz ∂x ∂x zc1 ∂x

(3.43)

– With Eq. (3.20) Z zc2 ∂σs1 ∂x ∂σs2 ∂x ∂σc ∂x = −As1 zs1 − As2 zs2 − z bdz ∂x ∂ ∂x ∂ Z zc2zc1 ∂x ∂ ∂σs2 ∂σc ∂σs1 zs1 − As2 zs2 − z bdz = −As1 ∂x ∂x zc1 ∂x ∂N = ∂κ Z zc2 ∂M ∂σs1 ∂x ∂σs2 ∂x ∂σc ∂x = −As1 zs1 − As2 zs2 − z bdz ∂κ ∂x ∂κ ∂x ∂κ Z zc1 ∂x ∂κ zc2 ∂σc 2 ∂σs2 2 ∂σs1 2 = As1 zs1 + As2 zs2 + z bdz ∂x ∂x zc1 ∂x (3.44) Basically the derivatives of the longitudinal stresses ∂σs1 /∂x , ∂σs2 /∂x , ∂σc /∂x are needed, the latter varying with z in the limits zc1 , zc2 , see Fig. (3.1), whereby dependency of integration limits zc1 , zc2 on , κ has been neglected. – The tangential material stiffness, see Eq. (1.26), is given with  ∂N ∂N  ∂ ∂κ CT = ∂M (3.45) ∂M ∂M ∂

∂

∂κ

which again couples normal force to κ and moment to . – Linear elastic, symmetric ∂σs1 /∂x = ∂σs2 /∂x = R system as Rspecial case Rwith 2 Es , ∂σc /∂x = Ec and bdz = A, z bdz = 0, z bdz = J. • To include shear forces Eq. (3.45) has to be extended with  ∂N ∂N ∂N   CT = 

∂ ∂M ∂ ∂Q ∂

∂κ ∂M ∂κ ∂Q ∂κ

∂γ ∂M ∂γ ∂Q ∂γ

 

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(3.46)

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3.1 General Cross Sectional Behavior • Calculation types – Specification of strain , curvature κ → calculation of M, N * Normal force with Eq. (3.19), moment with Eq. (3.20). Integration of concrete stresses analytically or numerically. – Specification of curvature κ, normal force N → calculation of M,  * With a given N, κ the nonlinear equation f (, κ) − N = 0 has to be solved for κ. This may efficiently be done with a Newton-Raphson iteration, i.e. (ν+1) = (ν) −

1 ∂f ∂



f ((ν) , κ) − N



(3.47)

=(ν)

see Eq. (1.65). A starting value is (0) = 0. ∂f /∂ is given with Eq. (3.43)1 . * With κ given and  calculated, M = f (, κ) can be determined. – Specification of moment M , normal force N → calculation of κ,  * From Eqns. (3.8,3.10) a nonlinear algebraic system arises for unknowns κ, ε0 after evaluation of integrals of σs1 , σs2 , σc . This may principally be solved with the Newton-Raphson method, see Page 15.

Example 3.1 Prescribed curvature κ, normal force N → calculation M,  • Parameters – Cross section h = 0.4 m, b = 0.2 m. Concrete C30/37, fcR = 21.675 MN/m2 , Ec0 = 31 900 MN/m2 . Uniaxial concrete behavior with DIN 1045-1 [din08, 9.1.5, 8.5.1]. – Upper and lower reinforcement each 4  20, As2 = As1 = 12.57 cm2 , d2 = d1 = 5 cm. Reinforcement behavior with DIN 1045-1 [din08, 9.2.3, 8.5.1].

Figure 3.2: Moment-curvature curve Example 3.1 • Initial bending stiffness concrete alone EJc = 34.0 MN/m2 . Initial bending stiffness reinforcement alone EJs = 11.3 MN/m2 . Total initial stiffness EJ = 45.3 MN/m2 . • Given values of normal forces NEd = 0, −1, −2 MN. State April 4, 2013

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3.1.3 Cracked reinforced concrete behavior • Curvature is varied within certain bounds. Therefore, N, κ serve as input values. The computed moments are of primary interest and lead to moment-curvature curves, see Fig. 3.2. – The initial stiffness is ruled by the following factors: (1) initial height of the compression zone, (2) initial values of concrete stress, higher compression leads to lower tangential stiffness due to the nonlinearity of the stress-strain curve, see Fig. 2.1. – The upper kinks result from the yielding of reinforcement. – The signed end points roughly mark the state, when ultimate concrete compressive stress of c1u = −0.0035 is reached. • The behavior of the uncracked, linear elastic cross section is shown as reference, furthermore the case with reinforcement alone. End Example 3.1

• Generalization – Fiber models – Every line along the beam axis cut by the cross section may be regarded as a fiber. Each fiber is strained according to beam kinematics a leads to a stress. – With the numerical integration of stresses to result in internal forces any type of a uniaxial material law may be used for fibers. – Efficiency of fiber models is influenced by numerical integration methods. * Commonly used methods like trapezoidal rule or Simpson rule may be used. * Gauss integration is not optimal, as the important upper and lower edges are not captured. Lobatto integration is an alternative.

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3.2 Equilibrium of Bars

3.2

Equilibrium of Bars • Preliminaries – We regard a loading p¯x (x, t), p¯z (x, t) varying with place x and time t. Inertial effects will be taken into consideration. Therefore, the bar’s (inertial) mass per length m and the 2nd moment of inertia Θ have to be regarded. ˙ ∂ 2 • /∂t2 = ¨• is used for abbreviation. – In the following a notation ∂ • /∂t = •, – To simplify the formulation another kinematic variable ϕ = w0 − γ,

ϕ0 = κ

(3.48)

is used in the following, see Eq. (3.3),which describes the cross section rotation due to bending.

Figure 3.3: Beam equilibrium • Strong differential equilibrium of an infinitesimal bar element is described by p¯x + N 0 = m · u ¨ p¯z + Q0 = m · w ¨ 0 Q + M = Θ · ϕ¨

(3.49)

according to Newton’s law – force = mass × acceleration – with the accelerations u ¨, w, ¨ γ¨ , ϕ¨ = w ¨ 0 − γ¨ . • The linear elastic case – From Eq. (3.11) we have N = EA  = EA u0 ,

M = EJ κ = EJ ϕ0 ,

Q = GA? γ

(3.50)

with A? = αA. – Combining Eqns. (3.49), (3.50) leads to mu ¨ − EA u00 = p¯x ? 0 mw ¨ − GA γ = p¯z Θ (w ¨ 0 − γ¨ ) − EJ (w000 − γ 00 ) = GA? γ

(3.51)

Eq. (3.51)1 represents the one dimensional wave equation4 . 4

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We can set m = % A with the specific mass % and px = 0 to gain the familiar form.

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53 – In case of a normal, slender beam the shear rotation γ is small compared to the total rotation w0 . The same holds for the derivatives. Thus, we have ϕ00 ≈ w000 and ϕ¨ ≈ w ¨0. – Furthermore, the term Θ (w ¨ 0 − γ¨ ) is neglected as Θ  1. – Combining the derivative of Eq. (3.51)3 and Eq. (3.51)2 finally leads to mw ¨ + EJ w0000 = p¯z

(3.52)

representing the equation of dynamic beam bending. • Weak integral equilibrium – Weak equilibrium formulations utilize test functions or so called virtual displacements δu, δw, δγ, δϕ = δw0 − δγ. These test functions have to be kinematically compatible, i.e. the first derivative of δu, δw, δγ, δϕ should exist. – An equivalent to the strong differential equilibrium formulation Eq. (3.49) is given by a weak integral equilibrium formulation regarding a bar of length L ZL

ZL δu m¨ u dx −

0

ZL

0

L

=

0

Z

δw p¯z dx + 0

ZL

0

ZL δϕ Θϕ¨ dx −

δw Q dx + 0

ZL

L

δu p¯x dx + 0

δw mw ¨ dx −

δu N dx + 0

Z

ZL

0

δϕ M 0 dx

0

(δw0 − δγ) Q dx

0

(3.53) Generally, the solutions of Eq. (3.53) solve also Eq. (3.49) and vice versa. – Those terms with derivatives of internal forces are partially integrated: Z L Z L   0 δu N dx = δu(L) N (L) − δu(0) N (0) − δ N dx Z0 L Z 0L   δϕ M 0 dx = δϕ(L) M (L) − δϕ(0) M (0) − δκ M dx Z0 L Z 0L   δw Q0 dx = δw(L) Q(L) − δw(0) Q(0) − δw0 Q dx 0

(3.54)

0

with virtual deformations δ = δu0 , δκ = δϕ0 . Combining Eq. (3.53) leads to Z L Z L Z L δu m¨ u dx + δϕ Θϕ¨ dx + δw mw ¨ dx+ 0 Z0 L Z L0 Z L + δ N dx + δκ M dx + δγ Q dx 0 0 0 Z L Z L  L  L  L = δu p¯x dx + δw p¯z dx + δu N 0 + δϕ M 0 + δw Q 0 0

0

(3.55) where boundary terms have been abbreviated. The last equation may be interpreted as virtual work principle and has the following parts: * * * *

Inertial forces with the first three terms of the left hand side. Internal forces with the last three terms of the left hand side. Distributed loading with the first two terms of the right hand side. Finally, the boundary terms with the forces at the beam’s ends with the last three terms of the right hand side.

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3.2 Equilibrium of Bars – A generalizing matrix notation of Eq. (3.55) is given with Z L Z L Z L T T ¨ dx + ¯ dx + δUT · ¯t δu · m · u δ · σ dx = δuT · p 0

0

(3.56)

0

with T m 0 0
T
T ¯ = p¯x p¯z 0 u= u w ϕ , p , m= 0 m 0  0 0 Θ
T
T , σ= N Q M =  γ κ
T U = u0 w0 ϕ0 uL wL ϕL
¯t = N0 Q0 M0 NL QL ML T (3.57) 0 0 00 0 with •L = •(L), •0 = •(0) and ϕ = w − γ, κ = ϕ = w − γ . * Actually Eqn. (3.54) requires a formulation
¯t = −N0 −Q0 −M0 NL QL ML T , but regarding finite element discretization a global sign convention is more appropriate: internal forces on the left cross section are assumed as positive with the same directions as on the right cross section. – As to the beam’s ends, i.e. its boundary conditions, they have to be prescribed with one element out of each the pairs (u, N ), (w, Q), (ϕ, M ) for every end. In case of ¯ ... prescribed forces, they are marked with an overbar, i.e. N → N – Disregarding shear deformations (→Bernoulli-Beam) * In case of a normal, slender beam the shear rotation γ is small compared to the total rotation w0 . The same holds for the derivatives and it is set 

ϕ = w0 ,

κ = ϕ0 = w00

* With γ = 0 a contribution δγ Q vanishes in Eq. (3.55). Thus, we set
T
T =  κ , σ= N M

(3.58)

(3.59)

Shear force Q is determined via Eq. (3.49)3 , which also holds for γ = 0. – Disregarding of rotational inertia * The inertial mass moment Θ generally is relatively small. Thus, we may neglect the corresponding contributions and set  T
T
T m 0 ¯ = p¯x p¯z (3.60) u= u w , p , m= 0 m – The quasistatic case: m = 0 in Eq. (3.56). • Boundary conditions – Kinematic boundary conditions: It is easy to choose displacement trial functions such that kinematic boundary conditions are fulfilled by them. Thus, test functions or virtual displacement may be set to zero along boundaries with kinematic boundary conditions. This will simplify Eqns. (3.55) or (3.56). – Force boundary conditions: Force boundary conditions may arise with the term ¯t in Eq. (3.57). It may be shown that the solutions of the integral equation (3.56) fulfill the differential equation (3.49) and prescribed force boundary conditions, under the assumption that kinematic boundary conditions are fulfilled by the trial function choice.

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3.3

Structural Beam Elements for 2D • In the following, specific forms will be chosen for the interpolation of displacement variables u, w, ϕ with Finite Elements, see Sections 1.3, 1.5. – The beam with a coordinate range 0 ≤ x ≤ L is subdivided into Ne elements. In a first approach each element has two nodes. Totally, there are N = Ne + 1 nodes. – An element I, I = 1 . . . Ne , has global nodal coordinates xI1 , xI2 and a length LI = xI2 − xI1 . – Moreover, an element I has a local coordinate r. The relation between local and global coordinates is given with    1  xI1 1 x = 2 (1 − r) 2 (1 + r) · (3.61) xI2 with x(−1) = xI1 , x(1) = xI2 . The inverse relation is r = (2x − xI2 − xI1 )/LI leading to a Jacobian J ∂r 2 J= (3.62) = ∂x LI which is used for numerical integration, see Eq. (1.61). – To interpolate displacement variables within an element polynomial forms Xn y(r, t) = ai (t) · ri i=0

(3.63)

are chosen, where the coefficients ai are functions of time t, while y may stand for the displacement variables u, w and ϕ. A particular formulation yields a trial function. – Test functions are selected in the same way as the trial functions, i.e. Xn δy(r, t) = δai (t) · ri i=0

(3.64)

(→ Bubnov-Galerkin-approach). • Deformation variables – Deformation variables , γ, κ are defined from displacement variables according to Section 3.1.1. With a given trial function they are derived on base of the derivatives of Eqns. (3.63), (3.62). • Due to the required integrability of Eq. (3.56), see also Section 1.6 and the compatibility requirement, some constraints for trial and test functions arise, as their derivatives should be finite across elements. – In case of Bernoulli kinematics ϕ = w0 and κ = w00 , i.e. deformation results from 2nd derivative of displacements. Thus, first derivations of test functions have to be continuous across elements (C 1 -continuity). – In case of Timoshenko kinematics κ = ϕ0 and ϕ and w are decoupled by the shear deformation γ. Deformations result from 1st derivatives of displacements and continuity of test functions across elements is sufficient (C 0 -continuity).

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56

3.3 Structural Beam Elements for 2D • 2D Bernoulli beam element – For Bernoulli beam see Page 54. u, w, ϕ = w0 remain as variables. The latter enforces a coupling of w and ϕ. A reasonable approach is given with     uI1 (t) u(r, t) = 12 (1 − r) 12 (1 + r) · uI2 (t) h 3 2 3 3 L r L r LI r3 LI r2 r 3r 1 1 I w(r, t) = 4 − 4 + 2 − I8 − L8I r + L8I − r4 + 3r 4 + 2 8 + 8 −  8  wI1 (t)  ϕI1 (t)   ·  wI2 (t)  h ϕI2 (t) w0 (r, t) =

∂w ∂r ∂r ∂x

=

3r2 4

2 LI

− 34 

3LI r2 − 2L8I r 8 

−

2

LI 8

− 3r4 +

3 4

3LI r2 8

+

2LI r 8

wI1 (t)  ϕI1 (t)   ·  wI2 (t)  ϕI2 (t)

(3.65) yielding u(−1, t) = uI1 (t), u(1, t) = uI2 (t) and w(−1, t) = wI1 (t), w(1, t) = wI2 (t) and w0 (−1, t) = ϕI1 (t), w0 (1, t) = ϕI2 (t). – Eq. (3.65) is abbreviated with u(r, t) = N(r) · uI (t)

(3.66)

with  N(r)

=

uI (t)

=


T

u(r, t) w(r, t)

u(r, t) = 1 2 (1

− r) 0 0 ··· LI r3 LI r2 LI r LI r3 3r 1 0 ··· 4 − 4 + 2 8 − 8 − 8 + 8
T uI1 (t) wI1 (t) ϕI1 (t) uI2 (t) wI2 (t) ϕI2 (t)



(3.67) – Deformation variables are derived with Eqns. (3.3), (3.58), (3.62), (3.65)     uI1 (t) ∂u ∂r 1 (r, t) = ∂r ∂x = LI −1 1 · uI2 (t)  κ(r, t) =

∂w0 ∂r ∂r ∂x

=

4 L2I



6r 4

6LI r 8

−

2LI 8

− 6r 4

6LI r 8

+

2LI 8

 wI1 (t)   ϕI1 (t)   ·  wI2 (t)  ϕI2 (t) (3.68)

– This is abbreviated with (r, t) = B(r) · uI (t)

(3.69)

with (r, t) = B(r)

=

1 LI



−1 0


T (r, t) κ(r, t) 0 0 1 0 6r 6r 3r − 1 0 − LI LI

0 3r + 1



(3.70)

– Potential locking risks * Longitudinal strains are constant within this element, while curvature is linear. Thus, artificial constraints arise to the coupling of normal forces to curvature and moments to longitudinal strains. State April 4, 2013

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−

LI r 8

−

LI 8

i

LI 8

i

57 • Enhanced 2D two-node Bernoulli beam element – As mentioned before the 2D two-node Bernoulli beam element holds some locking risks. This should be resolved by an enhanced two-node Bernoulli beam element to a large degree. – The enhancement is performed with the parabolic extension of longitudinal displacement with a third interior node at a local position r = 0. This node is not used for interpolation of geometry and lateral deflections and carries only one degree of freedom. – The additional degree of freedom leads to a linear longitudinal strain within an element, corresponding to the linear curvature. – Extension of Eq. (3.67) u(r, t) =  N(r)

=

uI (t)

=


T u(r, t) w(r, t)  1 0 0 1 − r2 · · · 2 r(r − 1) LI r3 LI r2 LI r LI 3r 1 r3 0 ··· 0 4 − 4 + 2 8 − 8 − 8 + 8
T uI1 (t) wI1 (t) ϕI1 (t) uI2 (t) uI3 (t) wI3 (t) ϕI3 (t) (3.71)

and Eq. (3.70) (r, t) = B(r)

=

1 LI



2r − 1 0

0 6r LI


T (r, t) κ(r, t) 0 −4r 2r + 1 0 3r − 1 0 0 − L6rI

0 3r + 1

 (3.72)

– This type of element may not be found in every finite element package.

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58

3.3 Structural Beam Elements for 2D • 2D Timoshenko beam element – Trial function choice for displacement variables =



1 2 (1

− r)

w(r, t) =



1 2 (1

− r)

ϕ(r, t)



1 2 (1

− r)

u(r, t)

=

 uI1 (t)  uI2 (t)   wI1 (t) 1 2 (1 + r) ·  wI2 (t)   ϕI1 (t) 1 2 (1 + r) · ϕI2 (t) 1 2 (1

+ r)





·

(3.73)

This may again be abbreviated with u(r, t) = N(r) · uI (t)

(3.74)

with
T u(r, t) w(r, t) ϕ(r, t)  1 1 0 0 0 0 2 (1 − r) 2 (1 + r) 1 1  0 0 0 0 =  2 (1 − r) 2 (1 + r) 1 1 0 0 + r) 0 0 2 (1 − r) 2 (1
T uI1 (t) wI1 (t) ϕI1 (t) uI2 (t) wI2 (t) ϕI2 (t) = (3.75)

u(r, t) =  N(r) uI (t)

– Deformation variables with Eq. (3.3), (3.48), (3.62)     uI1 (t) 1 ∂u ∂r = LI −1 1 · (r, t) = ∂r ∂x  uI2 (t)    ϕI1 (t) ∂r κ(r, t) = ∂ϕ = L1I −1 1 · ∂r ∂x ϕI2 (t)  wI1 (t)    ϕI1 (t)   −1 − L2I (1 − r) 1 − L2I (1 + r) ·   wI2 (t)  ϕI2 (t) (3.76) 

γ(r, t) =

∂w ∂r ∂r ∂x

−ϕ =

1 LI

This may be abbreviated with (r, t) = B(r) · uI (t)

(3.77)

with (r, t) =

(r, t) κ(r, t) −1 0 0  0 0 −1 0 −1 − L2I (1 − r) 

B(r)

=

1 LI


T γ(r, t)  1 0 0  0 0 1 LI 0 1 − 2 (1 + r)

(3.78)

– Potential locking risks * In case of pure bending or slender beams with a low bending stiffness compared to shear stiffness it is γ ≈ 0. Thus, Eq. (3.76)3 imposes a constraint on the nodal variables, which does not result from physics but from discretization. For an analysis of Timoshenko beam locking see ???. * A remedy is given with reduced integration, see Pages 60, 63.

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59 • Enhanced 2D Timoshenko beam element – Trial function choice – Extension of Eq. (3.75)
T u(r, t) w(r, t) ϕ(r, t)  r(r−1) r(1+r) 2 0 0 1 − r 0 0 0 2 2   r(1+r) r(r−1) 0 0 1 − r2 0 0   0 2 2 (1+r) (1−r) 0 0 0 0 0 0 2 2
T uI1 (t) wI1 (t) ϕI1 (t) uI2 (t) wI2 (t) uI3 (t) wI3 (t) ϕI3 (t) (3.79)

u(r, t) =

 N(r)

=

uI (t)

=

– and Eq. (3.78) (r, t) = B(r)

=

1 LI


T (r, t) κ(r, t) γ(r, t)   2r − 1 0 0 −4r 0 2r + 1 0 0   0 0 −1 0 0 0 0 1 LI LI 0 2r − 1 − 2 (1 − r) 0 −4r 0 2r + 1 − 2 (1 + r) (3.80)

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60

3.4 System Building and Solution Methods

3.4

System Building and Solution Methods • As a key element the weak equilibrium condition, see Eqn. (3.56), has to be evaluated. This performed element by element, see Page 8 and Eqn. (1.33). • For elementwise integral evaluation see Eqn. (1.61). In case of structural beam elements as described before the Jacobian is given by Eq. (3.62). – Internal nodal forces, e.g., are determined by Z

BT (x) · σ(x) dx =

fI = LI

LI 2

Z

1

BT (r) · σ(r) dr

(3.81)

−1

where B is given according to the element type chosen and σ according to Eq. (3.57) or (3.59)5 . The integration is performed numerically, see Eq. (1.62) fI =

LI Xni ηi BT (ξi ) · σ(ξi ) i=0 2

(3.82)

– The same procedure is applied to gain the tangential element stiffness matrix KT I =

LI Xni ηi BT (ξi ) · CT (ξi ) · B(ξi ) i=0 2

(3.83)

see Eq. (1.40), with the tangential material stiffness according to Eqs. (3.45) or (3.13) in case of Bernoulli beams, or (3.46), (3.12) in case of Timoshenko beams. Furthermore the element mass matrix MI =

LI Xni ηi NT (ξi ) · m · N(ξi ) i=0 2

(3.84)

see Eqn. (1.33)2 , with m according to Eq. (3.57) and the element distributed loading ¯I = p

LI Xni ¯ (ξi ) ηi NT (ξi ) · p i=0 2

(3.85)

¯ according to Eq. (3.57). The elements boundary6 loading see Eqn. (1.33)3 , with p can be directly taken from Eq. (3.57)7 ¯tI =

¯I1 Q ¯ I1 M ¯ I1 N ¯I2 Q ¯ I2 M ¯ I2 N


T

(3.86)

in case that any nodal loads are prescribed for a particular element. – Required integration order ni , see Eq. (1.62), to have an exact numerical integration depends on polynomial degree of trial functions for deformation variables. * The Bernoulli beam element with strains according to Eq. (3.70) yields a stiffness matrix KT I with highest polynomial degree 2 in case of constant material stiffness within an element. This requires ni = 1 and two sampling points with the Gaussian integration scheme, see Page 14 * The same argument bascially holds for the Timoshenko beam element, see Eq. (3.78), due to the shear deformation parts. Reduced integration ni = 1 neglects the linear contribution of shear deformations but might solve the locking problem in case of slender beams, see Page 63. 5 6

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Internal forces dimension has to match to row dimension of B. Boundaries are embodied by nodes for one-dimensional elements.

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61 – A few words about components. According to components of nodal displacements, see Eq. (3.67)3 , the vector of internal nodal forces has components fI =

NI1 QI1 MI1 NI2 QI2 MI2


T

(3.87)

in case of the two-node elements, which corresponds to number of rows of the matrix BT . Moreover, the nodal normal force NIi , shear force QIi and bending moment MIi correspond to the nodal longitudinal displacement uIi , the nodal lateral displacement wIi and the nodal rotation ϕIi . In a similar way Eq. (3.85) leads to ¯I = p

¯p ¯p M ¯p Q ¯p N ¯p M ¯p Q N I2 I2 I2 I1 I1 I1


T

(3.88)

¯ p . . . should not be confused with N ¯I1 . . . from ¯tI , see Eq. (3.86). where N I1 According to the rules of matrix multiplication the element stiffness matrix KT I and the element mass matrix MT I have a dimension 6 × 6 for 2D two node elements, with appropriate dimensions for CT and m. In case of the enhanced two-node Bernoulli element an additional component arises according to Eq. (3.71)3 with a corresponding nodal force NI for the central node. • Transformation to global system

Figure 3.4: Beam orientation in 2D space

– Up to now we assumed that the longitudinal axis or local x–axis of a beam element and the global x-axis have the same direction. But a 2D structural beam may have an orientation in 2D space. Thus, we have to consider a transformation of vectors in 2D cartesian coordinate systems. – The direction of a straight element is assumed with the first element node as start and the last element node as end. A rotation angle α is assumed starting from global x–direction to element direction, see Fig. 3.4. – Regarding Eq. (B.2) and considering two nodes and a perpendicular rotation direction of the displacement, at first the following relations hold uI = T · ugI ,

ugI = TT · uI ,

fI = T · fIg ,

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fIg = TT · fI

(3.89) State April 4, 2013

62

3.4 System Building and Solution Methods with     T=   

cos α sin α 0 0 0 0 − sin α cos α 0 0 0 0 0 0 1 0 0 0 0 0 0 cos α sin α 0 0 0 0 − sin α cos α 0 0 0 0 0 0 1

       

(3.90)

for two-node elements, where the upper index g indicates that the nodal displacements and internal nodal forces are measured with respect to the global coordinate system and •T denotes the transpose of •. The Eqns. (3.89) are also valid for incre¯I , p ¯ gI , ¯tI , ¯tgI . ments duI , dugI , dfI , dfIg and for loadings p – The tangential element stiffness matrix is at first defined in the local system of a beam by dfI = KT I · duI (3.91) see Eq. (1.41). Using the transformation rules of Eq. (3.89) and considering T−1 = TT this leads to T · dfIg = KT I · T · dugI

→

dfIg = TT · KT I · T · dugI

(3.92)

and finally results in a transformation rule for the tangential element stiffness matrix KgT I = TT · KT I · T

(3.93)

It has to be pointed out, that this rule is valid only in connection with the sign conventions of Fig. 3.4. Similar arguments lead to the transformation rule for the element mass matrix MgT I = TT · MT I · T (3.94) – In case of the enhanced two-node Bernoulli element the transformation matrix T is given with   cos α sin α 0 0 0 0 0  − sin α cos α 0 0 0 0 0     0 0 1 0 0 0 0    0 0 0 1 0 0 0  T= (3.95)     0 0 0 0 cos α sin α 0    0 0 0 0 − sin α cos α 0  0 0 0 0 0 0 1 The central 1 belongs to the longitudinal degree of freedom of the interior node. This value should not be rotated as it does not have a lateral component by definition. Beyond the change in T, all transformations remain unchanged. • Assembling of all element contributions – See Page 8. The procedure is basically the same for all element types. Regarding data organization it has to be considered that different nodes may have a different number of degrees of freedom with the beam elements, e.g. in case of enhanced elements. – Regarding a particular node, which is shared by several elements, the elements contributions to nodal forces and nodal loads basically have to sum up to zero with respect to equilibrium. State April 4, 2013

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63 • Consideration of kinematic boundary conditions – Kinematic boundary conditions are applied by prescribing nodal values of uIi , wIi , ϕIi . To simplify the procedure the quasistatic linear case K·u=p

(3.96)

is considered, see Eqns. (1.37), (1.63). – We consider, e.g., the transverse displacement wIi of a node i with a prescribed value w. ¯ If necessary, the value has to be transformed to the global system according to Eq. (3.89). The global index of this degree of freedom be k, see Page 8. Let N be the total number of degrees of freedom. To apply the particular boundary condition one may set pi pk Kkk Kkj Kik

:= = = = =

pi − Kik w ¯ i = 1 . . . k − 1, k + 1 . . . N w ¯ 1 0 j = 1 . . . k − 1, k + 1 . . . N 0 i = 1 . . . k − 1, k + 1 . . . N

(3.97)

– There must be enough boundary conditions to prevent rigid body motions. – The particular degree of freedom is released from balancing equilibrium by summing up nodal force contributions from elements and loadings. The corresponding internal nodal forces give a support reaction. – The same procedure may principally be used in nonlinear and dynamic cases. • Integration aspects for Timoshenko beam elements – Regarding Eq. (3.83) with ni = 1 this leads to a stiffness matrix KT I of dimension 6 × 6 with rank three, as B has a dimension 3 × 6 acc. to Eq. (3.78) and CT has a dimension 3 × 3 acc. to, e.g. Eq. (3.12). Thus, an analysis of eigenforms leads to 3 rigid body modes and 3 deformation modes, which makes sense. – The portion of the shear deformation mode is ruled by the parameter k = αGA L2 /EI, where L is a measure of the total beam span. – Adding another integration sampling point, i.e. ni = 2 increases the rank of KT I . This introduces non-physical constraints with restriction of rigid body displacements. This restriction is proportional to k. • Solution method – Cracked reinforced concrete leads to nonlinear models due to the nonlinear relations between moment, normal force, curvature and longitudinal strain. – Generally, an incremental iterative method is used to solve nonlinear systems, i.e. a loading history is regarded where a real time (→ dynamic or transient case) or a pseudo time (→ quasistatic case) is used to control the load. Thus, increments are defined by time steps. – Regarding quasistatic problems, the Newton-Raphson method may be an appropriate method for equilibrium iteration within every time step, see Eq. (1.65).

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64

3.4 System Building and Solution Methods Example 3.2 Simple reinforced concrete beam • Geometry, discretization an boundary conditions – Span L = 5.0 m, cross section width b = 0.2 m, height h = 0.4 m. Discretization with Ne = 10 enhanced two-node Bernoulli beam elements, see Page 57. – Hinge bearing of left and right node, i.e. lateral displacements are zero but rotations are not restricted. Longitudinal displacement of left node is zero, right node not restricted in longitudinal direction. • Material and loading – Concrete without tensile strength with an initial Young’s modulus Ec0 = 31 900 MN/m2 in the compressive range. Concrete compressive strength is assumed with fcd = 21.675 MN/m2 with c1 = −0.0023, c1u = −0.0035 according to DIN 1045-1 [din08, 9.1.5, 8.5.1]. – Reinforcing steel 4  20, As1 = 12.57 cm2 , d1 = 5 cm. Uniaxial reinforcing steel behavior acc. to DIN 1045-1 [din08, 9.2.3, 8.5.1]. No compression reinforcement. – Altogether an ultimate moment Mu ≈ 0.18 MNm is given with N = 0 leading to uniform ultimate load of qu = 8Mu /L2 = 58 kN/m. A load qu = 40 kN/m is chosen for the computation. • Solution method – Incrementally iterative with Newton-Raphson iteration within each loading increment as described on Page 16. A loading step ∆t = 0.1 is chosen for time discretization with the final target 1.

Figure 3.5: Example 3.2 a) System b) Deflection curve (scaled by 100) • Results – As a statically determined system is given the course of the bending moment can 2 be computed with M = 2q (l − x)x and a midspan maximum value Mmax = ql8 = 0.125 MNm in this particular case. Linear statics can only be approximately applied in case of statically indeterminate systems as the bending stiffness is not constant anymore but depending on internal forces. – Deflections of cracked concrete beams – even for statically determinate systems – cannot be determined with linear elastic statics and require the methods as described State April 4, 2013

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65 before. Fig. 3.5b shows the computed deflection curve. The uncracked linear elastic case with a bending stiffness EJ = 31 900 · 0.2 · 0.43 /12 = 34.03 MNm is shown as comparison. The difference roughly amounts to a factor of 2.4. – A special property of cracked reinforced cross sections is given with the fact, that the reference axis changes its length even if there is no resulting normal force. In case of this example the reference axis and with it the beam becomes slightly longer after loading, see Fig. 3.5b. This is connected with longitudinal strains of the reference axis, see Fig. 3.6a. It is caused by a coupling of Normal forces to curvature, i.e.

Figure 3.6: Example 3.2 a) Strain of reference axis b) Upper concrete strain and lower reinforcement strain N = N (κ, ), see Section 3.1.3, which leads to  6= 0 for κ 6= 0 even in case N = 0. – Finally, the computed strains of the upper compressed concrete edge and the strains of the reinforcement, as they can be determined on base of Eq. (3.4), are shown in Fig. 3.6b. They have to be compared to the concrete strain corresponding to strength c1 = −0.0023 and to steel strain corresponding to yield y = fy /Es = 0.0025. end example 3.2

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66

3.5 Further Aspects

3.5

Further Aspects

3.5.1

Creep

• Uniaxial material laws regarding creep, see Section 2.2, can be directly incorporated in the fiber model. We start with collecting previous items. Eq. (2.20) is applied to concrete   σ˙ c = Ec0 ˙x + ζc Ec0 x − ψc σc , ψc = 1 + ϕc ζc (3.98) with Ec0 according to Eq. (3.21). Reinforcing steel is treated as usual σ˙ s = Es ˙x

(3.99)

with longitudinal strains x . Kinematics is ruled by Eq. (3.4) x =  − z κ,

˙x = ˙ − z κ˙

(3.100)

Strain of reinforcement is considered with s1 =  − zs1 κ,

˙s1 = ˙ − zs1 κ, ˙

s2 =  − zs2 κ,

˙s2 = ˙ − zs2 κ˙

(3.101)

according to Eq. (3.18). Finally, regarding Eqs. (3.19,3.20) internal forces are Z zc2 Nc = σc bdz zc1

Ns

=

Mc =

As1Zσs1 + As2 σs2 zc2

−

(3.102)

σc z bdz zc1

Ms = −As1 σs1 zs1 − As2 σs2 zs2 and N = Nc + Ns , M = Mc + Ms . The coordinates zc1 , zc2 indicate the range of the concrete’s compression zone. • Internal forces of concrete σ c = ( Nc Mc )T will be connected to the concrete edge stresses of the compression zone σ cedge = ( σc1 σc2 )T . Eqs. (3.98,3.22) yield rates of edge stresses for fixed coordinates zc1 , zc2             σc1  σ˙ c1 1 −zc1 ˙ 1 −zc1 0 0 = Ec · + ζc Ec · − ψc σc2 κ 1 −zc2 σ˙ c2 1 −zc2 κ˙ (3.103) or σ˙ cedge = Ec0 B · ˙ + ζc Ec0 B ·  − ψc σ cedge (3.104) They are connected to rates of internal forces by Eq. (3.33) σ˙ c = Aσ · σ˙ cedge + Az · z˙ c

(3.105)

with zc = ( zc1 zc2 )T . • Combining edge stresses Eq. (3.104), internal forces Eq. (3.105) and regarding the relation Eq. (3.39) between rates of zc and generalized beam strains  leads to h i σ˙ c = Aσ · Ec0 B · ˙ + ζc Ec0 B ·  − ψc σ ce + Az · Bz · ˙ h i (3.106) = Ec0 Aσ · B + Az · Bz · ˙ + ζc Ec0 Aσ · B ·  − ψc Aσ · σ ce State April 4, 2013

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67

3.5.1 Creep whereby Aσ , B , Bz are functions of zc1 , zc2 or  and Az is a function of zc1 , zc2 , σc1 , σc2 . To consider the reinforcement, Eqs. (3.99,3.101,3.102) give σ˙ s = CT,s · ˙ with

 CT,s = Es

As1 + As2 −As1 zs1 − As2 zs2 2 + A z2 −As1 zs1 − As2 zs2 As1 zs1 s2 s2

(3.107) 

Combining all together yields   σ˙ = CT,s + Ec0 Aσ · B + Az · Bz · ˙ + ζc Ec0 Aσ · B ·  − ψc Aσ · σ ce

(3.108)

(3.109)

Eqns. (3.104), (3.109) form a coupled system of ordinary differential equations of first ˙ are given as a function of time. order for σ, σ ce depending on t, while (t), (t) • Eq. (3.104), e.g., is a specification of Eq. (1.66)

with σ  CT Σ

σ˙ = CT · ˙ + Σ

(3.110)


T σc1 σc2 =
T  κ = = Ec0 B 0 = ζc Ec B ·  − ψc σ

(3.111)

Applying Eq. (1.70) yields

σ i+1 = σ i + Ec0 B,i+1 · i+1 − i + ∆t ζc Ec0 B,i+1 · i+1 − ψc σ i+1

(3.112)

h i
1 σ i + Ec0 B,i+1 · i+1 − i + ∆tζc Ec0 B,i+1 · i+1 1 + ∆t ψc

(3.113)

or σ i+1 =

This is embedded into system equations Eqs. (1.73)-(1.75).

Example 3.3 Creep deformations of reinforced concrete beam • Geometry, boundary conditions and discretization is adopted from Example 3.2, see Figure 3.5. • Material – Concrete without tensile strength with a Young’s modulus Ec0 = 31 900 MN/m2 in the compressive range. Creep properties with ϕ = 2.0 and t? = 100 [d] for α = 0.5, i.e. half of total creep occurs after 100 days for a constant stress load. With Eqns. (2.20), (3.98) ζc = 0.006931 1/d, ψc = 0.020793 1/d. – Reinforcing steel 4  20, As1 = 12.57 cm2 , d1 = 5 cm. Uniaxial reinforcing steel behavior acc. to DIN 1045-1 [din08, 9.2.3, 8.5.1].. – Concrete compressive strength is assumed with fcd = 21.675 MN/m2 with c1 = −0.0023, c1u = −0.0035 according to DIN 1045-1 [din08, 9.1.5, 8.5.1]. Regarding the reinforcement an ultimate bending moment Mu ≈ 0.18 MNm with N = 0 is given. mailto:[email protected]

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68

3.5 Further Aspects • Loading and characteristics of material behavior – A total uniform load of qu = 8Mu /L2 ≈ 58 kN/m can be sustained in the ultimate limit state. – Roughly a third of this load is assumed to occur in the service state, qs = 20 kN/m. This loading leads to a moment Ms = 0.0625 MNm with a maximum concrete strain c = −0.39 ‰ in the cracked cross section. – Thus, a linear visco-elastic behavior may be assumed for the concrete’s compressive zone. • Solution method – Incrementally iterative with Newton-Raphson iteration within each loading or time increment as described on Page 16. A time step ∆t = 10 days is chosen for time discretization. A period of 500 days is regarded.

Figure 3.7: Example 3.3 a) System b) Mid span deflection during time • Results – Fig. 3.7 shows the computed midspan deflection in the course of time. The short term uncracked linear elastic deflection value (EJ = 31 900 · 0.2 · 0.43 /12 = 34.03 MNm) is given for comparison to cracked short term and long term deflection values. – Influence of concrete creeping can be characterized as follows: concrete contraction becomes larger with a (roughly) constant concrete stress, while reinforcement extension (roughly) remains the same. This leads to an increasing curvature, see Eq. (3.15), and an increasing deflection. But increase factor is less than 1 + ϕ due to bending and reinforcement. – As a variation, the influence of a compressive reinforcement 2  20, As2 = 6.28 cm2 , d2 = 5 cm is regarded. It can be seen, that a compressive reinforcement further reduces long term creep deflections, as concrete deformations are constrained to some extent. end example 3.3

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69

3.5.2 Temperature and Shrinkage

3.5.2

Temperature and Shrinkage

• Uniaxial considerations for temperature and shrinkage, see Section 2.2 Page 28, can be directly incorporated in the fiber model. Increments of measurable strain are given by 1 σ˙ x + ˙tem x CT

(3.114)

σ˙ x = CT ˙x − ˙tem x

(3.115)

˙x = leading to




• According to Eq. (3.4) a linear variation of imposed longitudinal strains is assumed over cross section height tem tem − z κtem (3.116) x (z) =  Imposed strain of the reference axis I and imposed curvature κI can be determined with tem of lower and upper edges prescribed imposed strains tem 1 , 2 tem =

tem + tem 1 2 , 2

κtem =

tem − tem 1 2 , h

h = z2 − z1

(3.117)

tem tem tem Thus, tem x (−h/2) = 1 , x (h/2) = 2 .

• Eq. (3.115) can be directly integrated in the linear case CT = E0 = const., i.e.
σx = E0 x − tem x

(3.118)

Using kinematics according to Eq. (3.4), internal forces relations Eq. (3.8) and cross section properties Eq. (3.9) yield
σ = C ·  − tem (3.119) with  σ=

N M



 ,

C = E0

Ac −Sc −Sc Jc



 ,

=

 κ

 ,

t

 =



tem κtem

 (3.120)

• Eq. (3.119) is sufficient for implementation in programs. The effect is as follows: if Eq. (3.119) is regarded in Eq. (3.81) a split of internal nodel forces like Z Z Z fI = BT · σ dx = BT · C ·  dx − BT · C · tem dx = fI − fItem (3.121) LI

LI

LI

occurs. As imposed strains tem are prescribed as, e.g., function of time, the part fItem of nodal forces is principally known and automatically included as part of the load vector p, see Eq. (1.63).

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3.5 Further Aspects

Example 3.4 Effect of temperature actions on a reinforced concrete beam • We refer to Example 3.2 with essentially the same system with the exception of boundary conditions7 and increased upper reinforcement. – Left and right node are totally constrained, i.e. lateral and longitudinal displacements and rotations are set to zero. – Upper and lower reinforcement with As1 = As2 = 12.57 cm2 , d1 = d2 = 5 cm. – Thermal expansion coefficient is chosen with αT = 1 · 10−5 K−1 both for concrete and reinforcement. A corresponding linear elastic case with E = 31 900 MN/m2 and the same geometry is regarded as a reference case.

Figure 3.8: Computed bending moments for Example 3.4 a) Linear elastic b) RC • Two load cases are investigated: 1. Service loading with q = 20 kN/m as in Example 3.3. 2. Service loading + Temperature loading8 on lower / upper edge with T1 = −10 K, T2 = 10 K. As serviceability states occur with these loadings a linear concrete behavior can be assumed in the compressive range. Concrete tensile strength is neglected. • Solution method – Incrementally iterative with BFGS-method for equilibrium iteration in each load increment. • Results – Fig. 3.8a shows the bending moments for the linear elastic reference case, where the rReinforcement is neglected. 7 8

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Statically determinate systems will have no imposed forces in case of temperature actions. In case of physical nonlinearities all loadings have to considered together.

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3.5.2 Temperature and Shrinkage * Load 1 case has a fixed end moment Ms = −0.0416 MNm and a field moment Mf = 0.0208, which accords to the analytic solution −qL2 /12 and qL2 /24, respectively. tem = −EJ κtem * The temperature load case can be superposed with a constant M tem −3 tem with κ = αT (T1 − T2 )/h = −0.5 · 10 leading to M = 0.0170. – Fig. 3.8b shows the bending moments for the case of reinforced concrete (RC) without tensile strength for concrete. * First of all, distribution of moments required for equilibrium in a statically indetermintate system depends on stiffness relations. Principally, those depend on loading in case of RC. * For RC bending stiffness is generally lower compared to the linear elastic case, but the stiffness relations do not differ very much. Thus, we get Ms = −0.0406, Mf = 0.0219 in load case 1 for RC, i.e. the field area is somehow stiffer and attracts bending moments. · Computed mid span deflection is wmax = 1.26 cm in load case 1 for RC and is considerably higher compared to the linear elastic case with wmax = qL4 /384EJ = 0.96 cm due to overall reduced stiffness. * Superposing is not allowed for load case 2 due to the physical nonlinearities of RC. Computed total moments are Ms = −0.0278, Mf = 0.0333 leading to imposed moments Mstem = 0.0406 − 0.0278 = 0.0128, Mftem = −0.0219 + 0.0333 = 0.0114. The additional temperature moment is lower compared to the linear elastic case and not constant along the beam. · A computation with temperature loading alone without other loading would lead to a constant M tem = 0.0227, i.e. a superposition would be definitely wrong.

Figure 3.9: Example 3.4 RC a) computed curvature b) computed strain of reference axis – Fig. 3.9 shows the deformation state for RC for load case 2. * Curvature along the beam see Fig. 3.9a. The course is not anymore analog to the bending moment course, as it would be in the linear elastic case. * Fig. 3.9b shows the strain of the reference or central axis, respectively. Such strains arise in contrast to the linear elastic case, as cracked RC beams tend to elongate without longitudinal displacement restrictions. * An overall elongation is not allowed in the example due to boundary conditions, i.e. a normal compressive force is induced on one hand, on the other hand strain values occur depending on M/N -ratio. But integral of strains must sum up to zero. mailto:[email protected]

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3.5 Further Aspects end example 3.4

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3.5.3 Tension Stiffening

3.5.3

Tension Stiffening

• For basics of tension stiffening see Section 2.8. – Tension stiffening is a matter of tensile stress transfer between reinforcement and concrete via bond. – Experimental observations and detailed calculations show, that the affected concrete area is restricted to a neighbourhood of the the main tensile reinforcement. – → Concept of area of action of reinforcement, see DIN 1045-1 [din08, 11.2.3], or effective reinforcement ratio %ef f respectively. These considerations have to be modified, as there is not a homogeneous behavior in a cross section in case of bending. Thus, the tension zone of bending is regarded separately from the compression zone and treated as an isolated tension bar, whereby the concept of effective concrete cross section is applied. • Tension stiffening may be quantitatively described as a reduction of reinforcement strains between cracks, i.e. reinforcement strains in cracks are replaced by reinforcement mean strains while reinforcement stresses in cracks are kept. – This leads to the following modified stress-strain relations for the reinforcement, see Eqns. (2.35), (2.40)  αfct for s ≤ 0s   %ef f 0s s σs = (3.122) Es s + βt %fefctf 0s < s ≤ y   fy + Et (s − y ) y <  with the reinforcement yield stress fy and 0s

1 fct = (α − βt ), Es %ef f

1 y = Es

  fct fy + βt %ef f

(3.123)

and with k according to Eq. (2.37). For discussion of the parameters βt , α see Section 2.8. A reasonable choice is βt = 0.4, α = 1.3. – These relations are controlled by βt , e.g. βt = 0 leads to 0s = αfct /%ef f , y = fy /Es and retains the relation σs = Es s . • Eq. (3.122) may replace σs = Es s in Eqs. (3.19), (3.20) and other of the similar type. The tangential material stiffness has to be adjusted in a corresponding way to reach convergence in equilibrium iterations.

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3.5 Further Aspects

Example 3.5 Effect of tension stiffening on a reinforced concrete beam exposed to external loading and temperature loading • We refer to Example 3.3 with the same system. The influence of tension stiffening on displacements will be examined. • Additionally to the material values of Example 3.3 the following values are assumed to model tension stiffening fct = 3.0MN/m2 ,

Ac,ef f = 0.02 m2 ,

βt = 0.4,

α = 1.3

(3.124)

0sm = 0.2147971360 · 10−3

(3.125)

leading to %ef f =

0.1257 · 10−2 = 0.063, 0.02

with 0sm according to Eq. (2.38). The modified stress-curve for the tensile reinforcement

Figure 3.10: Example 3.5 a) Modfied stress-strain law for tensile reinforcement b) bending moments in case of gravity + temperature loading with and without tension stiffening is shown in Fig. 3.10. The difference to the original curve is relatively low due to the high effective reinforcement ratio. • Results 1 – The computed immediate deformation at midspan with tension stiffening is w0 = 0.87 cm compared to 0.97 cm without tension stiffening, compare Fig. 3.7b. This difference of 0.1 cm is maintained during the 500 day creep phase, i.e. w500 = 1.33 cm versus 1.43 cm. • As beam stiffness is increased due to tension stiffening an influence on imposed forces may be supposed. To examine this item we refer to Example 3.4. – All system and loading parameters are kept but with tension stiffening included. Load case 2 – combined gravity and temperature – is regarded. • Results 2 State April 4, 2013

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3.5.3 Tension Stiffening – Fig. 3.10b shows the computed bending moments for the cases with and without tension stiffening. The difference is quite low, but this may not be generalized to other systems. end example 3.5

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3.5 Further Aspects

3.5.4

Prestressing

• Originally, generalized stress σ is formulated as a function of generalized deformations , e.g. σ =C· (3.126) in the most simple linear elastic case. For C see, e.g., Eq. (3.120). • This concept is extended with respect to prestressing, i.e. an additional part is assigned to generalized stresses resulting from prestressing tendons σ = C ·  + σp This additional part depends on the tendon profile, i.e.  p    N cos αp p p σ = =F Mp −zp cos αp

(3.127)

(3.128)

compare Eq. (3.59), with the prestressing force F p , the height coordinate or lever arm zp of the tendon and the inclination αp = dzp /dx of the tendon against the beam reference axis. Internal shear forces are neglected in a first approach. – The prestressing force may vary due to loss of prestress from friction of the tendon in a conduit. – Tendon profile parameters zp , αp may vary with the beam coordinate x according to prestressing design. – Furthermore, a beam deformation may lead to a change in the tendon profile after application and fixing of prestressing. Two cases have to be considered: 1. Prestressing without bond: total length of the tendon changes. This leads to a global change of the prestressing force. 2. Prestressing with bond: length of the tendon changes locally to keep the geometric compatibility with the concrete. This leads to locally varying changes in the prestressing force. – Eq. (3.127) may be generalized to nonlinear material laws. An additional remark: – The very basic approach to prestressing of beams is based on Eq. (3.49). We consider the quasistatic case, split internal forces into a part • from beam deformation, a part •p from prestressing and eliminate shear forces −N 0 − N p0 = p¯x −M 00 − M p00 = p¯z

(3.129)

Furthermore, Eq. (3.128) is used leading to −N 0 = p¯x + (P cos αp )0 00 −M = p¯z + (−zp P cos αp )00

(3.130)

Eqs. (3.127), (3.128) can be used in the weak forms of Page 53 to gain weak forms corresponding to Eq. (3.130). – A common approximation is F p ≈ const., cos αp ≈ 1 resulting in −M 00 = p¯z − zp00 F p , wherein zp00 F p is a redirection force from curvature zp00 . State April 4, 2013

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3.5.4 Prestressing • Regarding Eq. (3.127) in Eq. (3.81), a split of internal nodel forces like Z Z Z fI = BT · σ dx = BT · C ·  dx + BT · σ p dx = fI + fIp LI

LI

(3.131)

LI

occurs. The part −fIp may be regarded as a further contribution to the load vector, see Eq. (1.63). Two stages have to be considered: – Application of prestressing with prescribed prestressing force including friction losses. – The period after fixing the tendon with changes of the prestressing force according to the enumeration above. Different approaches are necessary to describe the global or local change of the tendon geometry for this stage. • Tendon geometry may be described for each finite beam element as follows: – An approach analog to the Bernoulli beam shape function Eq. (3.65) is chosen h 2 3 2 3 3 LI r3 1 zp = r4 − 3r − LI8r − L8I r + L8I − r4 + 3r + 12 LI8r + LI8r − 4 + 2 8 4   zp,I1  αp,I1   ·  zp,I2  αp,I2 dzp dr dzp 0 αP = dx = dr dx = zp (3.132) with the element length LI , zp,I1 , αp,I1 at the start node and zp,I2 , αp,I2 at the end node, and the local element coordinate in the range −1 ≤ r ≤ 1. This approach reproduces zp (−1) = zp,I1 , zp0 (−1) = αp,I1 and zp (1) = zp,I2 , zp0 (1) = αp,I2 . – Geometric length of tendon in an element I, see [BSMM00, 8.2.2.2] Z LI 1 q 0 2 LPI = (xp ) + (zp0 )2 dr 2 −1

LI r 8

−

LI 8

i

(3.133)

where also the derivative of the tendon position xp in the longitudinal direction has to be regarded. Eq. (3.133) has to be integrated numerically for each element, e.g. with a Gaussian quadrature, see Page 14. – Nominal tendon geometry according to design x0p =

dxp dx

=1
T zp,I1 αp,I1 zp,I2 αp,I2 = zp0,I1 αp0,I1 zp0,I2 αp0,I2 (3.134) with prescribed values zp0,I1 , αp0,I1 , zp0,I2 , αp0,I2 .
T

– Tendon geometry in deformed beam x0p
T zp,I2 αp,I2

=1+
T zp,I1 αp,I1 = zp0,I1 + wI1 αp0,I1 + ϕI1 zp0,I2 + wI2 αp0,I2 + ϕI2 (3.135) with the longitudinal strain  of the beam’s reference axis and the beams nodal displacements wI1 , ϕI1 , wI2 , ϕI2 , see Eq. (3.65). – Summing up all element contributions gives the total length LP of a tendon. • A prestressing force F0p is given by definition with the application of prestressing. mailto:[email protected]

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3.5 Further Aspects • Regarding prestressing without bond, tendon length can be determined separately for ap0 plication of prestressing, lets say LP , and for the states following the fixing of tendons, 00 lets say LP , using Eqs. (3.133), (3.135) with the appropriate deformations or displacements, respectively. Regarding Eq. (3.128) this leads to a prestressing force 00

LP p 0 F0 P L in the states following the fixing of tendons. Fp =

(3.136)

• Regarding prestressing with bond a tendon gets a local elongation after fixing of prestressing due to bond. This local elongation is ruled by the beam’s deformation kinematics Eq. (3.4), i.e. the additional strain of the tendon is given by ∆p (x) = ∆(x) − zp ∆κ(x)

(3.137)

with the changing ∆, ∆κ of the beam deformations after fixing of prestressing. This leads to to a prestressing force F p (x) = F0p + Ep Ap ∆p (x)

(3.138)

with the Young’s modulus Ep of the prestressing steel, the cross section Ap of prestressing and F p (x) used in Eq. (3.128). • The procedure can be summarized as follows: – – – – –

Define tendon geometry and prestressing force. Compute internal forces from prestressing. Compute nodal loads from prestressing. Compute system reaction including prestressing loads. Iterate if neccessary to capture changing in prestressing force.

Example 3.6 Prestressed reinforced concrete beam • We refer to Example 3.2 with principally the same system, but the span is doubled to L = 10 m. Thus, the load bearing capacity is strongly reduced with the given dimensioning. Prestressing is used to increase the bearing capacity. • System parameter items – Concrete cross section b = 0.2, h = 0.4, compressive strength fcd = 21.675 MN/m2 , and lower and upper reinforcement As1 = As2 = 12.57 cm2 , d1 = d2 = 5 cm result in an ultimate bending moment Mu ≈ 0.19 MNm with N = 0, see Example 3.3. This corresponds to a uniform loading qu = 8Mu /L2 = 15.2 kN/m. The question is, whether this loading can be increased by prestressing. – A nominal uniform concrete prestressing stress of σ0c = −10 MN/m2 is chosen in a first approach leading to F0p = 0.8 MN. Nominal tendon geometry of the whole beam is given with a parabola starting and ending in the centerline with a downward camber hp . This is described by  2  x x zp = 4hp − (3.139) L2 L A value hp = 0.15 m is chosen in this example. State April 4, 2013

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3.5.4 Prestressing

Figure 3.11: Example 3.6 a) system b) midspan load-deflection curve – Tendon properties with cross section Ap = 6 cm2 , strength fp0,1 = 1600 MN/m2 , fp = 1800 MN/m2 and Young’s modulus Ep = 200 000 MN/m2 . Nominal initial steel stress is σ0p = 1 333 MN/m2 with a strain p0 = 6.67 ‰. – Self weight loading is assumed with qg = 5 kN/m. – Nonlinear uniaxial concrete behavior acc. to DIN 1045-1 [din08, 9.1.5, 8.5.1], see also Fig. 2.1a. • Loading is applied in two steps: (1) Application of prestressing and self weight, (2) fixing of prestressing and additional application of a service load qp = 25 kN/m. Frictional losses are neglected to simplify this example. Two cases are distinguished – Prestressing without bond – Prestressing with bond • Solution method is incrementally iterative with BFGS iteration within increments. • Results for prestressing without bond – The computed increase in prestressing force after load step 2 according to Eq. (3.136) is minimal with F p /F0p = 1.002. This results from the low ratio hp /L = 1/67. – Displacements, see Fig. 3.11. Deflection starts with an uplift during application of prestressing. Final mid span deflection is quite large with 0.145 m (≈ 1/70 of the span), but the load carrying capacity is not yet exhausted with an upper midspan concrete compressive strain of -3.17 ‰(limit strain is -3.5 ‰). Serviceability is obviously not given due to high slenderness (1/25). – Bending moment and normal force in reinforced concrete (RC) cross section, see Fig. 3.12. Total moment moment from self weight and service load is Mq = 0.03 · 102 /8 = 0.375 MNm. The computed RC midspan contribution is Mc = 0.255 and the contribution from prestressing Mp = 0.12. The increased RC moment compared to the initial estimation results from the compressive normal force. • Results for prestressing with bond – In case of bond the tendon gets a local additional strain due to the locally varying deformation of the beam, see Eq. (3.137). This lead to an additional prestressing force, see Eq. (3.138) and Fig. 3.12b, and in the end to higher contribution of prestressing to load bearing capacity and a higher total load bearing capacity. mailto:[email protected]

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3.5 Further Aspects

Figure 3.12: Example 3.6 a) RC bending moment b) RC normal force under total loading – Detail results are given with final midspan deflection 0.12 m, see Fig. 3.11b, RC moment contribution Mc = 0.220, see Fig. 3.12a, contribution from prestressing Mp = 0.155. • The comparison between prestressing with and without bond in this example is somehow academic, as in practice prestressing with bond is exposed to more non-mechanical effects, which might lead to some restrictions to fully utilize the load carrying capacities of the prestressing steel. Details are ruled in codes. end example 3.6

• Relaxation of prestressing steel follows basically the approach given in Section 2.2, which may be transferred to other materials as concrete. – Details remain to be added. • Prestressing may be superposed with – creep (Section 3.5.1), – temperature and shrinkage (Section 3.5.2), – tension stiffening (3.5.3). All desribed approaches are principally compatible and may be used in any combination. • Stresses σs of reinforcing steel in the tensile regime are generally computed under the assumption of a state II, i.e. a cracked cross section occurs. These values can directly be used for a crack width calculation, see e.g. [din08, 11.2.4].

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3.5.5 Shear stiffness for reinforced cracked concrete sections

3.5.5

Shear stiffness for reinforced cracked concrete sections

Figure 3.13: Shear stiffness • The following considerations base on the truss model for shear. • Simple truss with shear deformations only – Undeformed, deformed length of concrete strut 1 s 2  2 p h h h l1u = , l1d = + h+ γ = l1u 1 + 2γ sin θ cos θ sin θ tan θ tan θ (3.140) 2 where γ has been neglected. Bar strain is given with 1 =

l1d − l1u p = 1 + 2γ sin θ cos θ − 1 l1u

(3.141)

The root term is expanded with a Taylor series. As γ  1, this results in 1 = sin θ cos θ γ

(3.142)

– Force in bar 1 with linear elastic behavior with Young’s modulus E and a bar cross section height h1 and a beam width b F1 = bh1 E 1 = bh1 E sin θ cos θ γ

(3.143)

Summing up all struts intersected by a cross section leads to F = n F1 =

h F1 = bh E sin θ cos2 θ γ h1 / cos θ

(3.144)

– Due to equilibrium reasons the shear force Q and the total concrete strut forces are related by Q = F sin θ. This leads to

and a shear stiffness

Q = bh E sin2 θ cos2 θ γ

(3.145)

∂Q = EA sin2 θ cos2 θ ∂γ

(3.146)

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3.5 Further Aspects – Reinforcement ties basically follow the same argumentation. A strut 2 is considered. It is s 2  2 p h h h l2u = , l2d = + h− γ = l2u 1 − 2γ sin α cos α sin α tan α tan α (3.147) and 2 =

l2d − l2u p = 1 − 2γ sin α cos α − 1 ≈ − sin α cos α γ l2u

(3.148)

and F2 = As2 Es 2 = −As2 Es sin α cos α γ

(3.149)

Cross section spacing sc and longitudinal spacing of rebars s are related by sc /s = tan α. This leads to n = h/sc = h/(s tan α) rebars in a cross section. Summing up all ties intersected by a cross section leads to h As2 F2 = −h Es sin2 α cos α γ = −h as2 Es sin2 α cos α γ sc / tan α s (3.150) Q = −F sin α finally leads to F =

Q = h as2 Es sin3 α cos α γ

(3.151)

– We consider θ = π/4 as special case. This leads to a shear stiffness EA/4 = GA/2 for ν = 0, see Eq. (3.7) for the shear modulus G. Now, the linear elastic case is compared with Eq. (3.11)3 . In the current argumentation concrete tensile struts are disregarded due to the limited tensile strength of concrete and shear reinforcement was implicitely assumed as vertical stirrups, which do not directly contribute to Q. Thus, the current argumentation and the linear elastic case confirm for θ = π/4. • Concrete shear strut angle – There is some margin to choose the concrete shear strut angle, see also [Com93, 6.3.3.1], [din08, 10.3.4], roughly in the range 20° ≤ θ ≤ 45°. As a first estimation, the same strut angle should be used as for the design of the stirrups.

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3.6

Application Case Studies

3.6.1

Transient Dynamics of Beams

• To begin with, the idea of inertial mass has to be discussed with respect to beams. Basics are given with Eqs. (3.49), (3.56), which introduce the inertial mass m per unit length of a beam and the inertial mass moment Θ. These are given with9 m = ρ A,

Θ = ρJ

(3.152)

with the material’s specific mass ρ, the cross section area A and the second moment of area J. Some care should be given to the units as specific mass has to be distinguished from specific weight, i.e. a given specific weight has to be divided by earth acceleration. – Contributions connected with the inertial mass moment Θ are often neglected, as they are relatively small. • Regarding dynamics, appropriate methods have already been laid down. A basic approach is given with Eq. (1.76) ¨ + f − p(t) = 0 r=M·u (3.153) with element contributions from – element mass matrices MI according to Eq. (3.84), – element nodal displacements according to Eq. (3.67)3 or Eq. (3.71)3 , – element internal nodal forces according to Eq. (3.82), – element nodal loads according to Eq. (3.85). Nodal loads p(t) are generally prescribed as a function of time t. Eq. (3.153) forms a system of ordinary differential equations of 2nd order in time t. Furthermore, initial conditions have have to be given for the nodal displacements at a time t = 0. ¨ may have a strong influence on the structural behavior • The additional inertial term M · u in case of loads with a rapid change of loading in time10 . • Two fundamental solution approaches are given: – Modal decomposition – Integration in time • A well known method for numerical integration in time is given with the Newmarkmethod, see Section 1.7 Eq. (1.83). This may immediately be applied to Eq. (3.153).

9 10

Has to be modified, if the centre line of area does not coincide with the beam’s reference line. A reference value of rapidness may be given by the largest natural period of a system.

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3.6 Application Case Studies

Example 3.7 Beam under impact load • We refer to Example 3.3 with the same geometry and boundary conditions. A sudden concentrated single point load (→ impact) is applied in midspan. A linear elastic behavior is assumed in a first approach to demonstrate basic characteristics of dynamic behavior under impact. – Young’s modulus is assumed with E = 31 900 MN/m2 . – Furthermore, the specific weight is taken as 25 kN/m3 . With an earth acceleration g ≈ 10 m/s2 this leads to a specific mass % = 0.025/10 = 2.5 · 10−3 MNs2 /m4 and with the cross section A = 0.2 · 0.4 m2 of this example to a beam mass m = 0.2 · 10−3 MNs2 /m2 . 2p m – With the given parameters the longest natural period is given with T = 2L π EJ = 0.0386 s and a frequency ν = 25.9 Hz. • The point load is characterized by magnitude and a time variation function P (t) = P0 f (t) The time function is chosen as a step function with limited duration  1 for t ≤ td f (t) = 0 t > td

(3.154)

(3.155)

Thus, loading is characterized by the parameters P0 , td . • Discretization with Ne = 20 enhanced two-node Bernoulli beam elements, see Page 57, time step ∆t = 0.001s • Results

Figure 3.14: Example 3.7 a) Deflection time curve b) course of moments in 10 time steps until 1st maximum – Values P0 = −0.07 MN and td = 0.1 s are used for a linear elastic reference case. Fig. (3.14)a shows the mid span deflection in the course of time. Following items are characteristic: (1) a cosine shaped oszillation, (2) a doubled maximum deflection 0.0106 m compared to the quasistatic value with 0.0053 m. State April 4, 2013

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3.6.1 Transient Dynamics of Beams – In same way maximum internal forces are doubled compared to the quasistatic case. Fig. (3.14)b shows the bending moments along the beam in certain time steps up to the time 0.02 s when first midspan maximum is reached. * Moment does not immediately follow the load due to inertial effects. * There is no triangular course anymore due to a moment wave effect. In the beginning moments are initiated in the impact point, while the support areas are unaffected. In the following period moment waves travel along the beam and bring the whole beam into action.

Figure 3.15: Example 3.7 a) related maximum deflection depending on related load duration time b) RC deflection time curve – Impact loads often have a short duration. Thus, a small parameter study is performed with varying load duration td and constant load amplitude P0 = 0.07 MN. – Fig. 3.15a shows the computed maximum midspan deflections related to the quasistatic deflection 0.0053 m depending on td related to the longest natural period 0.0386 s. * Very short loadings are compensated by inertia and only partially result in internal forces. If we consider a very short load duration time, e.g. td = 0.001 s, the beam gets only roughly 20 % of the quasistatic moment, or in other words, it may sustain five times the original load to have the same internal forces. – Finally, we consider the original reference case P0 = −0.07 MN and td = 0.1 s but with a nonlinear reinforced concrete section (RC) instead of linear elastic behavior. Material properties and reinforcement are chosen as in Example 3.6. * Fig. 3.15b shows the mid span deflection in the course of time. Due to the reduced stiffness the period of the oscillation grows, compare Fig. 3.14a. Maximum midspan deflection grows to 0.019 m, roughly a doubling occurs compared to the linear elastic case. * Fig. 3.16a shows the bending moments along the beam in 10 time steps up to the time 0.03 s when first midspan maximum is reached. Roughly the same moments occur with some time shift compared to the linear elastic case, see Fig. 3.14a. * In case of RC some normal forces arise without normal force loading, see Fig. 3.16b. This is caused by the beam’s movement in the longitudinal direction due to cracking, which is constrained by the beam’s inertia. end example 3.7

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3.6 Application Case Studies

Figure 3.16: Example 3.7 RC a) course of bending moments in 10 time steps until 1st maximum b) course of normal forces in 10 time steps until 1st maximum

3.6.2

More Case Studies

• Heavy multi-span RC-beam with prestressing (post-tension) under truck load • Light single span RC-beam with prestressing (post-tension) under pedestrian load • Prefabricated long span binder with prestressing (pre-tension) • Damaged RC girder with harmonic loading to identify progress of damage remain to be appended.

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Bibliography [BSMM00] Bronstein, Semendjajew, Musiol, and Muehlig. Taschenbuch der Mathematik. Verlag Harri Deutsch, Frankfurt/Main, 5. auflage edition, 2000. [Com93]

Comite Euro-International de Beton. CEB-FIP Model Code 1990. Thomas Telford, London, 1993.

[din08]

DIN 1045-1: Tragwerke aus Beton, Stahlbeton und Spannbeton. Teil 1: Bemessung und Konstruktion, August 2008.

87

Chapter 4

Strut-and-Tie Models 4.1

Linear Elastic Panel Solutions • Typical example – Single-span panel / deep beam with large opening, see e.g. Fig. 4.1a. • FE modeling – 2D continuum elements, e.g. 4-node quadrilateral, see page 4. – Linear elastic constitutive law, see Page 6.

Figure 4.1: a) Deep beam with opening b) principal stresses in deep beam with opening • Prinicipal stress / force flow, see Fig. 4.1b – Plane stress state σx , σy , σxy → principal stress state σ1 , σ2 , ϕ. – ϕ → stress trajectories → compression force flow & tensile force flow. * How to determine a stress trajectory: let y(x) be the function describing the course of a stress trajectory in the x, y-plane and x1 , y1 with y1 = f (x1 ) a starting point. Then, dy/dx = ϕ = f (σx (x, y), σy (x, y), σxy (x, y)) provides an ordinary differential equation, which at least can be solved numerically for y(x). * Practically, an estimation of trajectories using a principal stress visualization is sufficient.

88

89 – Compression force flow → ≈ concrete struts. – Tensile force flow → ≈ reinforcement ties. • Lattice truss model for a panel with – concrete struts, – reinforcement ties, – nodes connecting struts and ties forms a so called strut-and-tie model.

4.2

Truss Modeling • System determination – Short force flow path from load to support. – Minimization of total potential energy with elastic internal energy (→ minimization of displacement of loading point). – Approximation of stress trajectories by polygons. Basic rule: any change of a force direction needs needs an extra diversion force. • Property determination – Ties: reinforcement cross section, Young’s modulus of reinforcement steel (tension stiffening!) – Struts: a constructed cross section along system line according to a fan out and fan in of concrete stresses. There is some margin of discretion in the determination of strut cross sections. – Nodes: area of nodes is determined by adjoining struts and supports. • Statically determinate systems vs. indeterminate systems – Internal forces of determinate systems can be calculated by equilibrium conditions alone. – Indeterminate system need bar stiffness values for internal forces calculation.

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90

4.3 Computation of Plane Elasto-Plastic Truss Models

4.3

Computation of Plane Elasto-Plastic Truss Models • We consider a plane truss with m bars and n nodal degrees of freedom. – Initially, there are two degrees of freedom per node, but some nodes should be supported and degrees of freedom constrained to prevent a rigid body movement of the system. – It is assumed, that the system is stable, i.e. an arbitrary nodal load can be transferred to the supports. • Kinematic compatibility – Bar k with end nodes i, j has an orientation angle α from i to j and node displacements u ˜jk , u ˜ik in the longitudinal bar direction. We define a strain of bar k as1 ek = u ˜jk − u ˜ik

(4.1)

– Displacements in the longitudinal bar direction have to be transformed into the global system. This leads to global displacements of nodes i, j !     i   j u cos αk u cos α x k i j i x (4.2) u ˜jk = u ˜k , u = u = = uiy sin αk sin αk ujy resulting in u ˜jk =

· uj

(4.3)

– Combining Eqns. (4.1), (4.3) gives the strain of bar k

ek = − cos αk − sin αk · ui + cos αk sin αk · uj .

(4.4)

u ˜ik =

cos αk sin αk




· ui ,

cos αk sin αk




– Collection of all bars in a vector leads to a formulation

.. .. .. .. .. . . . .  .    ek  =  · · · − cos αk − sin αk · · · cos αk sin αk    .. .. .. .. .. . . . . . 





 . ..  i    uxi  uy    .. ···  ·  .j  ux   uj  y .. .

            

(4.5) or in a matrix notation e = B · u,

B ∈ Rm×n , u ∈ Rn , e ∈ Rm

(4.6)

where the vector e collects all m bar strains and u all n nodal displacements. There is a similarity compared to Eq. (1.6).

1

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This differs from the conventional definition of strain, but is more convenient for the following argumentation.

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91 • Equilibrium – A bar k has end nodes i, j and a bar force sk (tension positive, compression negative). Bar k contributes forces both to the nodes i, j, see Fig. 4.2a !  i      j fk,x fk,x − cos αk cos αk j i fk = = sk , fk = = sk (4.7) j i fk,y − sin αk sin αk fk,y

Figure 4.2: a) Truss equilibrium b) Example truss system – Local equilibrium condition for nodes i, j with external nodal loads pi , pj X X j fki = pi , fk = pj k

k

– Global equilibrium condition through assembling of all nodes    . .. .. .    i  · · · − cos αk · · ·        pxi ..  · · · − sin αk · · ·      .   py     .. ..   ·  sk  . =    .j ..  · · · cos αk · · ·   px .     · · · sin αk   pj ···    y .. .. . .

(4.8)

            

(4.9)

or in a matrix notation BT · s = p,

BT ∈ Rn×m , s ∈ Rm , p ∈ Rn

(4.10)

where the vector s collects all m bar forces and the vector f all n nodal loads. Here we see again matrix B of Eq. (4.6), but now it has been transposed. There is a similarity compared to Eq. (1.33)1 . – A case m = n indicates a statically determinate and m > n a statically indeterminate system. Bar forces may be determined directly from Eq. (4.10) for a statically determinate system. This is not possible for an indeterminate systems as the number of unknowns m is larger as the number of equations n.

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92

4.3 Computation of Plane Elasto-Plastic Truss Models • Linear constitutive law and linear equations of systems – To begin with, a uniaxial linear elastic material behavior is assumed, see Eq. (1.19). AI is used for the cross sectional area of a bar I, LI as its length and EI as its Young’s modulus. With sI = AI σxI , where σxI is stress of bar I, and xI = eI /LI , where xI is the conventional strain of bar I, bar forces are given by sI = CI eI ,

CI =

EI AI , LI

I = 1...m

(4.11)

– In a matrix notation this can be written as s = C · e,

C ∈ Rm×m

(4.12)

where the material matrix C is diagonal with coefficients CI . – Combination of Eqs. (4.10), (4.12) and (4.6) leads to K · u = p,

K = BT · C · B,

K ∈ Rn×n

(4.13)

with a constant symmetric stiffness matrix K. There is a similarity of Eq. (4.13) compared to Eq. (1.36).

Example 4.1 Deep beam with strut and tie model • We refer to the problem given with Fig. 4.1a and linear elastic principal stresses shown in Fig. 4.1b. This problem is also discussed in [CF08, 8.8].

Figure 4.3: Example 4.1 a) member stresses b) proposed reinforcement scheme – The chosen strut and tie model is shown in Fig. 4.2b. * The system has m = 36 bars and 21 nodes. Seven nodal degrees of freedom are restrained by boundary conditions leading to n = 2 · 21 − 7 = 35. Thus, we have ’nearly’ a statically determinate system. Internal forces are influenced by the stiffness of the members to a small extent. – Obviously there is some effort to circumvent the central rectangular hole. – The resulting total load has been distributed to upper nodes according to the loading scheme of Fig. 4.1a. State April 4, 2013

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93 – A preliminary computation reveals a distinction between struts (→ compressive, concrete) and ties (→ tensile, reinforcing steel). The following reference values for Young’s modulus and cross section area are chosen for the ties: Es = 200 000 MN/m2 , As = 10·10−3 m2 , and for the struts: Ec = 30 000 MN/m2 , Ac = 0.12 m2 . The last value corresponds to a deep beam thickness t = 0.6 m and an assumed strut width of 0.2 m. • Results – Computed stresses of struts and ties are shown in Fig. 4.3a. Negative values with small amount correspond to struts, positive values with relatively large amount correspond to ties. According to each individual stress value the reference value of cross section area may be increased or reduced for the corresponding member to gain some target stress. – Target stress values may be derived from strength values in codes, see e.g. [din08] • Design of reinforcement – A reinforcement scheme as proposed in [CF08, Fig. 8.49] is shown in Fig. 4.3b. end example 4.1

• Nonlinear constitutive law and nonlinear equations of systems – Actually struts and ties have limited bearing capacities. – This may in a first approach described with an uniaxial elastoplastic constitutive law, see Section 2.5 applied to both the reinforcement and the concrete in the compressive range. * With member strains eI → xI given this leads to member stresses σxI and member forces sI . * Stress determination is more elaborated compared to the linear elastic case and requires housekeeping with an internal state variable, i.e. the actual plastic strain Ip . – Nodal forces from members or internal nodal forces from Eq. (4.10) f (u) = BT · s(u)

(4.14)

– Equilibrium condition in analogy to Eq. (1.63) r(u) = f (u) − p

(4.15)

– Tangential material stiffness  CT I =

EsI ET I

|σxI | < fyI |σxI | = fyI

(4.16)

All these coefficients are collected in a diagonal matrix CT . – Tangential system stiffness2 in analogy to Eq. (4.13) KT = BT · CT · B,

KT ∈ Rn×n

(4.17)

– Solution with Newton-Raphson method according to Eqns. (1.63) – (1.65). 2

Distinguish upper T for transposed from lower T for tangential.

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94

4.3 Computation of Plane Elasto-Plastic Truss Models

Example 4.2 Corbel with elasto-plastic strut and tie model • The corbel problem is discussed in [FS07, Fig. 1b], see Fig. 4.4a. • In the following a slightly modified problem is regarded, which is statically indeterminate. Corbel geometry, loading and strut-and-tie model are shown in Fig. 4.4. We have m = 14 bar members and n = 12 degrees of freedom, i.e. the system is twofold statically indeterminate.

Figure 4.4: a) Corbel acc. to [FS07] b) model of Example 4.2 • A provisional first calculation has to be performed to distinguish between tensile and compressive members. Basing upon this calculation member properties are chosen as follows: – Tensile members 1 – 7 are chosen as ties with Es = 200 000 MN/m2 , fy = 500 MN/m2 and a reference cross section area As0 = 10 cm2 = 10 · 10−4 m2 . Cross section factor of each tie is shown in Fig. 4.4b. Compressive members 8 – 14 are chosen as struts with Ec = 30 000 MN/m2 and a yield limit3 fyc = 40 MN/m2 . Cross section of all struts is assumed with Ac = 4 · 50 cm2 = 0.02 m2 . Finally, a slight hardening is assumed for reinforcement steel and concrete, see Section 2.5. • System buildung and solution method – The system equations may be built according to Eqns. (4.14) - (4.17). – Alternatively, the formal finite element approach with 2D-bar elements according to Page 3, material law according to Section 2.5 and system building according to Section 3.4 may be applied. This leads to the same result. – Loading is applied with prescription of vertical displacement of the loaded node. – The solution is gained incrementally iterative with the Newton-Raphson approach, see Page 15. • Results – The computed load displacement curve is shown in Fig. 4.5a. 3

Concrete is assumed as elasto-plastic in this context. The assumed value is quite high, codes ordinarily restrict allowed concrete stresses to a high extent.

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95 * Initially linear elastic up to reaching first yielding of a tie. * Decrease of stiffness with first yielding of tie, but some more loading can be applied. * Ultimate limit state with yielding of other ties in the statically indeterminate system such that the system becomes "‘nearly kinematic"’. * The system does not become "‘really kinematic"’ with the hardening assumption, but otherwise the determinant of the tangential system stiffness would become zero without the hardening assumption and the Newton-Raphson method would not work anymore.

Figure 4.5: a) member stresses b) model of Example 4.2 • Member stresses in last computed state see Fig. 4.5b. – Yield limit is reached in most ties and slightly exceeded due to hardening. – Struts are not critical in this particular case, but not far away from reaching nominal yield limit. end example 4.2

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96

4.4 Ideal Plastic Truss Models

4.4

Ideal Plastic Truss Models • Definition of the ideal plastic problem – Kinematic compatibility see Eq. (4.6), m equations. – Equilibrium see Eq. (4.10), n equations. – Material behavior * Limit state condition – flow condition |sI | ≤ suI ,

suI = fyI AI , I = 1 . . . m

(4.18)

with unsigned bearing capacities suI and signed forces sI of members. * Assumption about strains – flow rule  =0 for |sI | < suI |eI | , I = 1 . . . m. (4.19) >0 |sI | = suI * Dissipation condition – Kuhn-Tucker conditions sI eI ≥ 0,

s · e ≥ 0,

s, e ∈ Rm

(4.20)

– There is no way for elastic deformations in the ideal plastic approach. • Balance of equations vs. unknowns and solution approach – m bar forces s, n displacements u, m bar strains e vs. 2 × m + n equations (4.6), (4.10), (4.19) plus 2 × m constraint equations (4.18), (4.20). – The solution of this problem is not straightforward. It is provided by methods of optimization and linear programming, see e.g. [Lue84]. – For sake of simplicity we restrict to a loading type f = λ f0 ,

λ>0

(4.21)

with a unit load f0 which is fixed and a loading factor λ. – The basic solution idea is regarding a maximization problem λ → max

(4.22)

while fulfilling equilibrium and the limit state function, i.e. constraints BT · s = λ f0

and |sI | ≤ suI , I = 1 . . . n

(4.23)

The quantities B, su , f0 are known and s, λ are unknown in this formulation.

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97 – Furthermore, linear programming theory states a so called dual problem associated to the maximization problem. This is a minimization problem ¯su · e ¯ → min

(4.24)

¯ collects4 all |eI | and ¯su all suI . The minimization has constraints where e B·e=u

and f0 · u = 1

(4.25)

i.e. essentially kinematic compatibility. The quantities B, su , f0 are known and e, u are unknown in this formulation. Eq. (4.24) can be interpreted as minimization of energy. – Linear programming theory also states that solutions s? , λ? exist for the maximization and e? , u? for the minimization. Both have the same optimal value ¯? λ? = ¯s?u · e

(4.26)

with λ? > 0. It may finally be shown that the solutions fulfill the conditions e?I = 0 if |s?I | < suI

and s? · e? > 0

(4.27)

Thus, the solutions s? , λ? , e? , u? of the associated optimization problems are also solutions of the ideal plastic problem as has been formulated before. – The ideal plastic problem is also fulfilled by s? , λ? and e.g. βe? , βu? with an arbitrary scalar5 β > 0. * → The absolute values of the deformations are indeterminate. The relation between the deformation components is determined to some extent. • The solution s? , λ? f0 is also a solution for the corresponding elasto-plastic problem if case no hardening occurs in the state of yielding. – Young’s moduli of bars do not influence this limit state for forces and loads. – But the deformations of the ideal elastoplastic problem are influenced by Young’s moduli, due to elastic deformations prior to plastic deformations. • Solution methods – Basically, we are interested in loads λ f0 and member bar forces s. Thus, we have to solve the maximization problem Eqns. (4.22), (4.23). – The solution is simple for statically determinate systems with n = m and a square B. For small statically indeterminate systems it may be found by inspection or trial and error. – Larger statically problems require systematic methods of linear programming, e.g. the simplex method, or general methods of optimization.

4

¯· has to be interpreted as operator on ·. But this may violate the constraint f0 · u = 1 and the maximization problem is not solved anymore.

5

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98

4.4 Ideal Plastic Truss Models – To determine deformations, the following approach is appropriate instead of solving the minimization problem. * First of all, from analysis of forces and Eq. (4.19) we can determine a set of bars with strains eR 6= 0, R = 1 . . . me and all other bars with eS = 0, S = me+1 . . . m. * We consider kinematic compatibility Eq. (4.6) and reformulate it with a unit matrix Im ∈ Rm×m B · u = Im · e,

e ∈ Rm , u ∈ Rn , B ∈ Rm×n , m ≥ n

(4.28)

* As B has n linear independent rows, Eq. (4.28) can be transformed into     In P ·u = ·e, In ∈ Rn×n , 0 ∈ R(m−n)×n , P ∈ Rn×m , Q ∈ R(m−n)×m 0 Q (4.29) by Gauss elimination, see [ZF97], with a unit matrix In , a zero matrix 0 and generally fully occupied matrices P, Q. Eq. (4.29) leads to u = P · e,

Q · e = 0.

(4.30)

The last equation describes kinematic compatibility in a tighter sense. In case m = n the matrix P is the inverse of the matrix B. * As eS = 0, S = me+1 . . . m a submatrix Qe of Q is given with Qe · ee = 0,

ee ∈ Rme , Qe ∈ R(m−n)×me

(4.31)

where ee collects all yielding eS . Two cases have to be distinguished. · me ≤ m − n: Eq. (4.31) can generally only be fulfilled with ee = 0, i.e. no deformation occurs at all. · me > m − n: a number of me − (m − n) components of ee may be chosen arbitrarily under the condition Eq. (4.20), while the rest is determined with Eq. (4.31). But this leaves the absolute deformation undetermined. * Only the relations of the deformation components are determined to some extent. This is a characteristic property of ideal plastic systems. – In case of a statically determinate system m = n things simplify. Kinematic compatibility in a tighter sense Eq. (4.30)2 vanishes and strains of yielding bars may be given arbitrary values, while all other strains are zero. – With given strains e displacements u can be determined directly by Eq. (4.30)1 . • Limit theorems of plasticity 1. Any equilibrium state which fulfills the limit state condition, see Eqns. (4.22), (4.23), gives a lower bound for the loading factor. 2. The work of load bearing capacities on kinematically admissable deformations with a constraint f0 · u = 1, see Eqns. (4.24), (4.25), gives an upper bound for the loading factor. These theorems are generally stated as a kind of postulates. Actually, their exact formulation and proof correspond to linear programming theory.

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99

Example 4.3 Corbel for ideal-plastic strut and tie model • We refer to Example 4.2. The same system is computed as ideal plastic on basis of the 1st limit theorem of plasticity. This requires the knowledge of the effective cross section areas of members and their yield stress, which are taken from Example 4.2. – Young’s modulus and hardening behavior after reaching yield limit are not relevant in this approach. – Basic parameters are number of bars m = 14, 9 nodes, 3 boundary nodes, number of degrees of freedom n = 12. – All the following bases upon a linear elastic precalculation distinguishing between compressive and tensile members. • To start with, states fulfilling 1st limit theorem of plasticity are determined. – Equilibrium is described by Eq. (4.10) BT · s = λ f0

(4.32)

with the vector of member forces s, a loading factor λ, a unit load f0 =

0 0 0 0 0 0 0 0 0 −1 0 0


T

(4.33)

applied vertically downward on node 7, see Fig. 4.5b, and   0 0 0 0 0 0 0 0 1 0 0 −.707 0 0  −1 0 0 0 0 0 0 0 0 0 0 −.707 0 0     0 0 −1 0 0 0 1 .243 0 0 0 0 −.707 0     0 0 0 0 0 0 0 .971 0 0 0 0 −.707 0     0  0 0 1 0 0 0 0 0 0 0 .707 0 −.243     0 0 0 0 0 0 0 0 0 0 .707 0 −.971   0 T B =  0 0 0 1 −1 0 0 0 0 0 0 .707 0   0    0 1 0 0 0 0 0 0 0 0 0 0 .707 0     0 0 0 0 0 1 0 0 0 0 .781 0 0 .243     0 0 0 0 0 0 0 0 0 0 .625 0 0 .971     0 0 1 −1 0 0 0 0 0 .781 −.781 0 0 0  1 −1 0 0 0 0 0 0 0 .625 −.625 0 0 0 (4.34) where the boundary nodes 1, 5 and 9 and their degrees of freedom are disregarded. – With prescription of m ¯ = m − n + 1 = 3 more or less arbitrary member forces, Eq. (4.32) constitutes a system of n = 12 equations for the remaining n − 1 = 11 member forces plus the loading factor λ, i.e. n equations for n unknowns. This is principally solvable for a well behaved structural system.

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100

4.4 Ideal Plastic Truss Models

Figure 4.6: Example 4.3 a) stresses in ultimate limit state b) a displacement mode – The members 1, 7, 11, see Fig. 4.5b, are chosen by chance. * They are prescribed with their bearing capacities, i.e. s1 = 0.05 MN, s7 = 0.1 MN, s11 = −0.8 MN. * Solution of the modified system Eq. (4.32) as described before leads to s2 = 0, s3 = 0.1, s4 = −0.375, s5 = 0.6372, s6 = 0.6372, s8 = 0, s9 = −0.05, s10 = −0.88, s12 = −0.0707, s13 = 0, s14 = 0.0515 and a loading factor λ = 0.5498. 2 * The computed member forces lead to stresses σ5 = 637 MN/m , σ6 = 637, σ10 = −44 in the members 5, 6 and 10. This is not admissable, as the stresses exceed the material strength and the limit state condition Eq. (4.18) is violated for these three members. * Thus, all member forces are reduced by the less of the factors 500/637 = 0.785 and −40/ − 44 = 0.909 which is α = 0.785. This leaves the members 5 and 6 in a yielding state. * To maintain equilibrium according to Eq. (4.32) the loading factor also has to be reduced by the same factor leading to a final loading factor of λ = α 0.5498 = 0.43. This is a lower bound for the real loading factor, i.e. the real loading is not less but may be larger. * Deformations can be determined as described on Page 98. With two yielding members me = 2 and me = m − n. Thus, the matrix Q has 2 × 2 dimension and Eq. (4.31) can only be fulfilled with zero deformations. Furthermore, this solution does not fulfill Eq. (4.19) of the ideal plastic problem. – Another choice by chance with the members 9, 11, 13 leads to lower bound of λ = 0.081 and leaves only member 1 in a yielding state. No deformation can be found fulfilling the kinematic compatibility conditions. – Using the binomial coefficient there are   m! m = = 364 m ¯ m!(m ¯ − m)! ¯

(4.35)

possibilities for a choice of m ¯ = 3 out of m = 12. This trial and error method to determine the largest value λ and therewith a solution of the ideal plastic problem is obviously cumbersome.

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101 • Application of simplex method. – The theory of the simplex method has to be omitted here for sake of brevity. It is a standard method to solve linear constrained optimization problems, where the problem formulation Eqns. (4.22) - (4.23) comes from, and is used in all disciplines of science, technology and economics. – The optimization methods yields the member stresses shown in Fig. 4.6a. The corresponding loading factor is λ = 0.44. * The horizontal tie in half height is fully utilized which leads to some increase of the limit load. * Compare Fig. 4.5 for the results of the elasto plastic calculation, which leads to the same ultimate limit load and a similar stress distribution. – Displacements can again be determined as described on Page 98. . * With the five yielding members 1, 3, 5, 6 and 7 it is me = 5 > m − n = 2, i.e. three member strains may principally chosen arbitrarily while the remaining two are determined on base of Eq. 4.31. A choice e1 = 0.1, e3 = 0.03, e5 = 0.1 leads to
T e = 0.1 0 .03 0 0.1 0.1225 0.03125 0 0 0 0 0 0 0 (4.36) ¯ = 0.122375 Energy according Eq. (4.24) amounts to ¯su · e * Displacements can be determined with Eq. (4.30)1 . Results are shown in Fig. 4.6b. The vertical displacement of the loaded node 7 is computed with −0.278125. A displacement scaling factor of 1/0.278125 = 3.5955 has to be chosen to fulfill ¯ = 0.44 Eq. (4.25) with f0 according to Eq. (4.33). This leads to an energy ¯su · e and corresponds to the maximum loading factor as stated by the limit theorem of plasticity. In contrast to the elasto-plastic model of Example 4.2 determination of absolute deformations is not possible with the ideal plastic approach due to arbitrary choice of some member strains. A deformation mode can be determined to some extent. end example 4.3

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102

4.5 Application Aspects

4.5

Application Aspects • Strut limit forces – A nominal value is determined by cross section of struts and uniaxial compression strength of concrete. – There is some uncertainty in the effective value of cross section of struts. – Effective compression strength may be reduced to lateral tensile effects. – → Some caution has to be considered for the determination of limit forces su of struts. • Ductility demands – Before reaching its limit state a system has to undergo more or less more deformations. * This starts with pure elastic deformations. * Additional plastic deformations will follow in one or more yielding bars. This is accompanied by a force redistribution in statically indeterminate system with further increasing load before reaching the system’s limit load. – Thus, all bars have to deform to some extent without rupture or brittle failure before reaching the system’s limit load. – This imposes ductility requirements on bars (→ minimum strains required) and nodes (→ minimum relative rotations of adjoining bars required). • Nodes – Nodes connect struts and ties and have to handle an exchange of forces. – Nodes occupy some extension determined by the adjoining bars.

Figure 4.7: a) compressione node b) compression-tension node c) reinforcement redirection node – This area of extension may considered as a plate with a two dimensional stress state. – Nodes may fail as bearing elements of a strut-and-tie system. – Basically, their bearing capacity has to be checked.

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Bibliography [CF08] CEB-FIP. Practitioners’ guide to finite element modelling of reinforced concrete structures, volume Bulletin Nr. 45. International Federation for Structural Concrete fip, Lausanne, 2008. [din08] DIN 1045-1: Tragwerke aus Beton, Stahlbeton und Spannbeton. Teil 1: Bemessung und Konstruktion, August 2008. [FS07] F. Fingerloos and G. Stenzel. Konstruktion und bemessung von details nach din 1045. In Betonkalender, pages 325–374, Berlin, 2007. Ernst & Sohn. [Lue84] David G. Luenberger. Linear and Nonlinear Programming. Addison-Wesley, Reading, Massachusetts, 2. Auflage edition, 1984. [ZF97] R. Zurmühl and S. Falk. Matrizen und ihre Anwendungen. Springer-Verlag, Berlin, 7. Auflage edition, 1997.

103

Chapter 5

Multiaxial Concrete Material Behavior 5.1

Scales

Continuum mechanics provides a framework to describe the behaviour of solids. Within this framework two scales are considered to describe structural concrete, i.e. mesoscale and macroscale. The mesoscale distinguishes the cement matrix, aggregates and the interface transition zone. Each of these material phases is regarded as a continuous isotropic solid with its own constitutive law and its own material parameters. A continuous displacement is assumed along contact surfaces of different phases in case of deformations. Regarding a concrete specimen composed of phases its internal geometric characteristics are random as a matter of principle, e.g. size, shape, position and orientation of aggregates are random. Thus, two samples chosen out of a collection of specimens of same geometry and with the same material parameters of phases generally will show different reactions under the same imposed action. The variation of reactions depends on the size of the specimen relative to the size of the largest aggregates and the type of the action. The variations of reactions tend to become smaller for larger specimen sizes. If they are considered to be negligible for the relevant types of action the specimen size constitutes a representative volume element (RVE). A mixture of phases may be homogenized for a RVE. Regarding concrete, cement matrix, aggregates and interface transition zone are homogenized within the macroscale with a common type of constitutive law and a corresponding single set of material parameters. Assuming material types the material parameters of the homogenized continuum may basically be determined analytically from the material parameters of the phases by mixture theories, see e.g. (Mori & Tanaka, 1973), (Wriggers & Moftah, 2006). But such approaches are limited regarding the mesoscale randomness of concrete. As an alternative, numerical multiscale methods may provide macroscale parameters derived from numerical mesoscale calculations considering particular actions and phase properties. Finally, parameters of macroscale constitutive laws may directly be chosen according to the experimental behaviour of specimens with at least RVE-size using parameter identification and calibration methods. Considering numerical simulation methods with a discretization of space like Finite-ElementMethods a single element generally should not be exposed to larger gradients of reactions within itself. This leads to upper bounds for elements sizes or fineness of discretization, respectively. Thus, due to local inhomogeneities discretizations of a given structure require an order of ten times more elements for each spatial dimension within the mesoscale compared to discretizations within the macroscale. As a further consequence numerical calculations of larger structures are generally performed in a macroscale using homogenized constitutive laws. 104

105

5.2

Some Basics of Continuum Mechanics • In the following space is measured in a 3D cartesian coordinate system (orthogonal, right-handed) if not otherwise stated. A space point x has a vector of coordinates1 ( x1 x2 x3 )T . A body of material occupies an area of space in a configuration, see Fig. 5.1a. Due to a loading history this configuration changes with time t. A material point is identified by the space point X it occupies in a reference configuration at a time t0 , i.e. X = x for t = t0 . Displacements2 are defined with u = x − X and have a vector of components u = ( u1 u3 u3 )T . • Small strain measurement – A notion of small: displacements have a magnitude of millimetres while the body has dimensions in the magnitude of metres. – Strain components 1 ij = 2



∂uj ∂ui + ∂xj ∂xi

 = ji ,

i, j = 1 . . . 3

(5.1)

are defined involving two directions, i.e. a displacement direction and a reference direction. – Strain components form a symmetric tensor of second order   11 12 13  =  12 22 23  13 23 33

(5.2)

A second order tensor obeys particular transformation laws based upon transformation laws for vectors (→ first order tensors) in case of a transformation of the underlying coordinate system. – Strain tensor in so-called Voigt notation    11  22       33     =  γ23  =      γ13   γ12

11 22 33 223 213 212

       

(5.3)

utilizing symmetry of small strain. – Incremental changes of strain during loading history with progressing time t ˙ = lim

∆t→0

∆ ∆t

(5.4)

• Stress – Definition of Cauchy stress σ with reference to an infinitesimal tetrahedron at a position x with a tetrahedron base area dA, base normal n = ( nx ny nz )T exposing a force df and a stress t = df /dA. 1 2

Change in notation compared to Section 1.3. Per definition displacements become zero in the reference configuration.

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106

5.2 Some Basics of Continuum Mechanics

Figure 5.1: a) Continuous body in reference and deformed configuration b) Infinitesimal stress tetrahedron * Force df is equilibrated by three exposed forces on each coordinate plane, see Fig. 5.1b, each force with an exposed stress, each stress with three components. * Thus, nine stress components σij related to the coordinate planes with i, j = 1, 2, 3 are given. The first index denotes the plane normal, the second index the global direction component. * Considering an inifinitesimal cube it can be shown that σij = σji due to equilibrium considerations. * Finally, Cauchy stress is given as a symmetric tensor of second order. Its components may be written as a matrix   σ11 σ12 σ13 σ =  σ12 σ22 σ23  (5.5) σ13 σ23 σ33 – Relation between exposed stress t and tensorial stress σ t=σ·n

(5.6)

using matrix algebra with vectors t, n and the matrix arrangement of σ. – Stress written as vector in so called Voigt notation   σ11  σ22     σ33   σ=  σ23     σ13 

(5.7)

σ12 utilizing symmetry of Cauchy stress. – Incremental changes of stress during loading history with progressing time t σ˙ = lim

∆t→0

∆σ ∆t

(5.8)

• Voigt notation is used in the following if not otherwise stated. • General description of material behavior State April 4, 2013

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107 – The following general form is assumed3 σ˙ = CT · ˙

(5.9)

where the tangential material stiffness CT may depend on stress σ, strain  and internal state variables ξ. – Internal state variables ξ comprise a loading history. They are necessary as an actual state σ,  for a given ˙ may lead to different reponses σ˙ for different loading histories. * Internal state variables require laws describing their evolution depending on the evolution of stress and strain. – If the components of CT are constant, i.e. CT = C , Eq. (5.9) may be integrated in time t to give σ =C· (5.10) i.e. a linear material law. • Change of coordinate system – Another cartesian coordinate system is regarded, with the same origin but different orientation or different directions of axes, respectively. The relation between these systems is ruled by three rotation angles, see Fig. 5.2a, and a transformation matrix Q depending on the rotation angles. The matrix Q is orthogonal, i.e. Q−1 = QT . – Values of the components of stress and strain differ in the two systems. Given a strain  in the original system there is a strain 0 = Q · 

(5.11)

in the transformed system4 . A stress σ in the initial system is related by σ = QT · σ 0

(5.12)

to the stress σ 0 in the transformed system. With σ 0 = C0 · 0 the material stiffness transforms according to C0 = Q · C · QT ,

C = QT · C0 · Q

(5.13)

– Regarding a point of a material body the states of stress and strain essentially remain the same, but are measured in different coordinate systems5 .

3

Voigt notation allows to write CT as a matrix, otherwise tensor notation is required. This and the following is slightly more complicated as actually written as the transformations for stress and strain differ slightly when using the Voigt notation. But the essentials remain unchanged. 5 Look a vehicle moving in space: it has a velocity vector, which may be measured in different inertial systems and has different components. But the velocity essentially remains the same. 4

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108

5.2 Some Basics of Continuum Mechanics

Figure 5.2: a) Rotation of coordinate system b) Principal stress space • Principal values, principal orientations – A strain or stress tensor has principal orientations depending on its tensorial components, i.e. a coordinate transformation with vanishing shear components, i.e. σ12 = σ13 = σ23 = 0 or 12 = 13 = 23 = 0. – It remain principal values σ1 , σ2 , σ3 or 1 , 2 , 3 as components in three orthogonal directions which are the principal directions. * Sign convention for the following: compression negative, tension positive. – Principal values and directions are by definition independent from coordinate system orientation. * Tensors measured in different coordinate systems having the same principal values and having rotation angles leading to the same principal directions describe the same state. Thus, regarding a material point its state of stress and strain is preferably characterized by the corresponding principal values and directions. – Invariants are often used as alternative formulation for principal values, i.e. in case of stress6 I1 = J2 = = J3 = with σm

σ1 + σ2 + σ3   2 2 + (σ − σ )2 (σ − σ ) + (σ − σ ) 1 2 2 3 3 1   1 2 2 2 2 (σ11 − σm ) + (σ22 − σm ) + (σ33 − σm ) + σ23 σ32 + σ13 σ31 + σ12 σ21 (σ1 − σm )(σ2 − σm )(σ3 − σm ) (5.14) = (σ1 + σ2 + σ3 )/3. 1 6

– Principal directions of stress and strain coincide only under restrictive assumptions, see the following Section 5.3. • Deviatoric and pressure parts – Regarding stress pressure is defined as 1 1 p = − (σ11 + σ22 + σ33 ) = − I1 3 3 6

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(5.15)

Exactly spoken: I1 1st invariant of stress, J2,3 2nd/3rd invariant of stress deviator.

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109 – Furthermore, deviatoric part σ 0 of stress σ is derived as   2  0   1 1 σ11 σ11 + p 3 σ11 − 3 σ22 − 3 σ33 2 1 1 0     σ22   0   σ22 + p   32 σ22 − 13 σ11 − 31 σ33  σ33   σ33 + p   σ33 − σ11 − σ22 3 3     3 σ0 =   σ 0  =  σ23  =  σ23    23    σ 0   σ13   σ13 13 0 σ12 σ12 σ12

       

(5.16)

This may also be written as  σ 0 = Idev · σ,

Idev

   =   

2 3

− 13 − 13 0 0 0

− 13 2 3

− 31 − 13

0 0 0

0 0 0

− 31

2 3

0 0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1

       

(5.17)

with the deviatoric unit matrix Idev . – Similar relations are valid for strain.

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110

5.3 Basic Linear Material Behavior Description

5.3

Basic Linear Material Behavior Description • Isotropy of materials – Notion of action directions, e.g. principal values of imposed strain and their orientation (→ principal strain directions). – Notion of reaction directions, e.g. principal values of resulting stresses and their orientation (→ principal stress directions). – Notion of material directions, i.e. a cartesian coordinate system being spanned by four material points in the reference configuration. – An isotropic material behaves in the same way in every loading direction: * Principal stress directions change in the same way as principal strain directions, i.e. principal stress directions coincide with principal strain directions7 . * Principal stress values are independent from principal strain directions, i.e. principal stress values do not change with a change of principal strain directions relative to material directions but unchanging principal strain values. – A mathematical description * We consider a loading  with an associated response σ = C0 · . 0 0 * The loading direction is rotated into another arbitrary direction  with  = Q· according to Eq. (5.11). 0 * Isotropy requires, that the rotated associated response σ has the the same ma0 0 0 terial law, i.e. σ = C0 ·  , while rotating σ according to Eq. (5.12). With σ = QT · σ 0 = QT · C0 · 0 = QT · C0 · Q ·  a requirement C0 = QT · C0 · Q

(5.18)

follows for arbitrary rotations Q. • Triaxial isotropy – We assume an abstract8 material matrix C0  C11 C12 C13  C21 C22 C23   C31 C32 C33 C0 =   C41 C42 C43   C51 C52 C53 C61 C62 C63

C14 C24 C34 C44 C54 C64

C15 C25 C35 C45 C55 C65

C16 C26 C36 C46 C56 C66

       

(5.19)

It can be shown that for arbitrary rotations Q the isotropy requirement Eq. (5.18) can only be fulfilled with a form   C1 C2 C2 0 0 0  C2 C1 C2  0 0 0    C2 C2 C1  0 0 0  C0 =  (5.20) 1  0  0 0 2 (C1 − C2 ) 0 0   1  0  0 0 0 0 2 (C1 − C2 ) 1 0 0 0 0 0 2 (C1 − C2 ) 7 8

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More precisely they should keep their same relative orientation, but this extension is theoretical. Without a specific material in background.

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111 This is equivalent to the triaxal isotropic linear elastic law  E(1−ν) Eν Eν 0 (1+ν)(1−2ν) (1+ν)(1−2ν)  (1+ν)(1−2ν) E(1−ν) Eν Eν  0  (1+ν)(1−2ν) (1+ν)(1−2ν) (1+ν)(1−2ν)  E(1−ν) Eν Eν 0  C0 =  (1+ν)(1−2ν) (1+ν)(1−2ν) (1+ν)(1−2ν) E  0 0 0 2(1+ν)   0 0 0 0  0 0 0 0

0

0



0

0

0 0 E 2(1+ν)

0 0 0

0

E 2(1+ν)

         

(5.21) with Young’s modulus E and Poisson’s ratio ν. • Triaxial orthotropy – If Eq. (5.18) is fulfilled for three rotations Q, with an angle 180° around each coordinate axis, then an orthotropic material is given. Eq. (5.18) leads to   C11 C12 C13 0 0 0  C21 C22 C23 0 0 0     C31 C32 C33 0 0 0    C0 =  (5.22)  0 0 0 C 0 0 44    0 0 0 0 C55 0  0 0 0 0 0 C66 with 12 nonzero components, see [Mal69, (6.2.24)]. This form requires that the material’s symmetry directions coincide with the coordinate directions910 . – Symmetry is assumed due to thermodynamic reasoning, i.e.  C11 C12 C13 0 0 0  C12 C22 C23 0 0 0   C13 C23 C33 0 0 0 C0 =   0 0 0 C 0 0 44   0 0 0 0 C55 0 0 0 0 0 0 C66

       

(5.23)

with 9 nonzero components. A flexibility form with some physical evidence – material coefficents are directly taken from experimental results – is given by  1  − νE121 − νE131 0 0 0 E1 1  − ν21 − νE232 0 0 0   E2  E  − ν31 − ν232  1 0 0 0   E3 E3 D0 =  E3 (5.24)  1 0 0 0 0  0  G 4    0 0 0 0 G15 0  0 0 0 0 0 G16 where νij is a measure for the deformation in i-direction caused by a stress in jdirection. Relations ν12 /E1 = ν21 /E2 , . . . and so on must hold to have symmetry.

9

Otherwise it will be fully occupied, but 12 independent material parameters remain. Stresses and strains will have the same principal directions if they coincide with the the material principal directions, otherwise not. 10

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112

5.3 Basic Linear Material Behavior Description • Plane strain, plane stress for the isotropic linear elastic law derived from Eq. (5.21) – In both cases γ23 = γ13 = 0 holds. – Plane strain is given with 33 = 0, i.e.    1−ν σ11 1−2ν  σ22  = E  ν 1−2ν 1+ν σ12 0

ν 1−2ν 1−ν 1−2ν

0

   0 11 0  ·  22  1 γ12 2

(5.25)

– Plane stress is given with σ33 = 0. Using this condition with Eqs. (5.10), (5.21) leads to       σ11 1 ν 0 11  σ22  = E  ν 1 0  ·  22  (5.26) 1 − ν2 1−ν σ12 0 0 γ 12 2 • Plane stress for the symmetric orthotropic case – It is assumed that the material’s symmetry directions coincide with the coordinate directions with σ33 = 0 and γ23 = γ13 = 0. Thus, the flexibility form Eq. (5.24) leads to    1    − νE121 0 11 σ11 E1 1  22  =  − ν21 0  ·  σ22  (5.27) E2 E2 1 γ12 σ12 0 0 G6 – The inversion yields the stiffness form       11 E1 ν12 E2 0 σ11 1  ·  22  (5.28)  ν21 E1  σ22  = E2 0 1 − ν12 ν21 0 0 (1 − ν12 ν21 )G γ12 σ12 with G = G6 , see also [CS94, (6.110)]. A notation ν12 = ν1 (→ deformation in 1-direction caused by a lateral stress in 2-direction) and ν21 = ν2 (→ deformation in 2-direction caused by a lateral stress in 1-direction) is used in the following. – Requiring symmetry ν1 E2 = ν2 E1 and using a modified Poisson’s ratio √ √ ν2 E1 ν1 E2 ν¯ E1 E2 ν¯ E1 E2 ν¯ = √ , ν1 = , ν1 ν2 = ν¯2 (5.29) =√ → ν2 = E1 E2 E1 E2 E1 E2 leads to

  σ11  σ22  =   σ12 

E1 1−¯ ν2 √ ν¯ E1 E2 1−¯ ν2

√ ν¯ E1 E2 1−¯ ν2 E2 1−¯ ν2

0

0

   11  0  ·  22  γ12 G 0

(5.30)

whereby E1 , E2 ≥ 0 has been assumed.

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113 • Invariance of shear flexibility for the symmetric orthotropic plane stress case – It is assumed that the shear coefficient G does not change for all plane transformations QT · C0 · Q, see also Eq. (5.18), with   cos2 ϕ sin2 ϕ − cos ϕ sin ϕ sin2 ϕ cos2 ϕ cos ϕ sin ϕ  Q= (5.31) 2 cos ϕ sin ϕ −2 cos ϕ sin ϕ cos2 ϕ − sin2 ϕ i.e. a rotation of stresses and strains around the plane’s normal with an arbitrary angle ϕ. This requirement can be fulfilled with a form √    E ν¯ E1 E2 E1 ν2 E1 1 0 0 2 ν 1−¯ ν2 1−ν ν 1−ν1 ν2  ν √1−¯   ν1 E12 2 E2 E E E 1 2 2  0 C0 =  1−ν1 ν2 1−ν1 ν2 0  =  1−¯ν 2 1−¯ ν2 √ E1 +E2 −2ν1 E2 E1 +E2 −2¯ ν E1 E2 0 0 0 0 4(1−ν1 ν2 ) 2

   

4(1−¯ ν )

(5.32) where the coefficient C33 is invariant11 . Notice, ν1 E2 = ν2 E1 and C33 = E/(2(1 + ν)) for E1 = E2 , ν1 = ν2 , compare Eq. (5.26). Still the assumption holds, that symmetry and coordinate directions coincide. – Three material parameters remain with the form Eq. (5.32). This is also a major motivation for the assumption of invariance of shear flexibility, which by the way absolves from the burden of experimental determination of G. – The unixial behavior in 1-direction with σ22 = 0 is given by σ11 = E1 11 ,

˜22 = − √

ν¯ σ11 E1 E2

(5.33)

˜11 = − √

ν¯ σ22 E1 E2

(5.34)

and in 2-direction with σ11 = 0 by σ22 = E2 22 ,

This basically allows to derive an orthotropic material parameters from uniaxial experimental data, which are given by σ11 , 11 , ˜22 and σ22 , 22 , ˜11 . Notice, ˜22 in Eq. (5.33) is different to 22 in Eq. (5.34). The same holds for ˜11 , 11 . * The set of four Eqs. (5.33,5.34) is overdetermined – with σ11 , 11 , ˜22 and σ22 , 22 , ˜11 given – to determine the values of E1 , E2 , ν¯. Maybe a fest fit has to be found, e.g. with methods of regression.

11

Other coefficients are not necessarily invariant!

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114

5.4 Basics of Nonlinear Material Behavior

5.4

Basics of Nonlinear Material Behavior

5.4.1

Tangential Stiffness

• Regarding the uniaxial case nonlinear material behavior is characterized by a decreasing tangential material stiffness. This basic idea is generalized to the multiaxial case. – A formulation of nonlinear material behavior has been given with Eq. (5.9): σ˙ = CT · ˙

(5.35)

The tangential stiffness CT is subject to change depending on state of stress, strain and on loading history. We will consider such materials which in some way become softer during a load history. • Isotropic nonlinear behavior is characterized in the same way as on Page 110: – An isotropic material behaves in the same way in every action direction. Values of principal stress increments are independent from principal strain directions. For a given material state directions of principal stress increments coincide with directions of principal strain increments. – An isotropic tangential material stiffness Eq. (5.9) obeys Eq. (5.18), i.e. CT = QT · CT · Q

(5.36)

Hence, the tangential material stiffness CT has to follow a form like Eq. (5.20) which allows only for two independent coefficients. • Starting with the initial unloaded state, all isotropic solid materials – including such with essentially nonlinear behavior – may be described initially by Eq. (5.9), whereby the tangential stiffness is given by Eq. (5.21) with an initial Young’s modulus and an initial Poisson’s ratio. – But initially isotropic materials may become anisotropic in higher loading regimes due to the type of loading. – Thus, forms like Eq. (5.23) or Eq. (5.32) may become appropriate for the tangential stiffness in case of load induced anisotropy, with the coefficients depending on state of stress, strain and on loading history.

5.4.2

Stress Limit States

• Stress limit states mark the other end compared to initial states. They describe strength of materials. For initially isotropic materials such strength limit states may be described in a form f (σ1 , σ2 , σ3 ) = 0 (5.37) using principal stress values σ1 , σ2 , σ3 . Stress states with f (σ1 , σ2 , σ3 ) ≤ 0 are admissable, states f (σ1 , σ2 , σ3 ) > 0 cannot be sustained. Orientation of principal stress directions has no influence in case of isotropic materials. – In a general case the strength limit may be a function of principal stress values and principal stress directions relative to the material directions. • Stress-strain behavior Eq. (5.35) is often separated from strength limit states Eq. (5.37) and formally both are treated independently. State April 4, 2013

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115

5.4.2 Stress Limit States – To have a consistent material description the integration of Eq. (5.35) should not lead to stress states violating the strength limit state Eq. (5.37). – Regarding isotropic strength limit states, they principally allow for anisotropic stress strain behavior. – Regarding concrete a load induced anisotropy – e.g. tensile strength is reached in one direction while a compressive strength is utilized in orthogonal directions – may be combined with an isotropic strength limit state formulation. • Eq. (5.37) forms a surface in triaxial principal stress space. A stress point on this surface can be described by pressure Eq. (5.39), deviatoric length Eq. (5.40) and deviatoric angle12 Eq. (5.42). – Principal stress values span a cartesian coordinate system (→ principal stress space) and the corresponding state may be described by a vector. The following elements are significant the in principal stress space: √ 13 T * Hydrostatic axis: a space diagonal with a direction nξ = ( 1 1 1 ) / 3. T * Projection of a stress√vector σ = ( σ1 σ2 σ3 ) on the√hydrostatic axis: ξ = ξ ( 1 1 1 )T / 3 with length14 ξ = (σ1 + σ2 + σ3 )/ 3. * With ξ given, its corresponding deviatoric √ plane: this plane has ξ as normal, i.e. (σ − ξ) · ξ = 0 or σ1 + σ2 + σ3 = 3ξ. T * Projection of a stress vector σ = ( σ1 σ2 σ3 ) on its deviatoric plane   2σ1 − σ2 − σ3 1 ρ = σ − ξ =  −σ1 + 2σ2 − σ3  (5.38) 3 −σ1 − σ2 + 2σ3 T * Projection of the particular stress vector σ = ( 1 0 0 )p on the deviatoric ¯ 1 = 2/3( 1 − 12 − 21 )T plane: ρ1 = 2/3( 1 − 12 − 12 )T . It has a direction ρ called Rendulic direction in the following. – Description of the principal stress state by Haigh-Westergaard coordinates: * Length of the hydrostatic projection ξ √ 1 I1 ξ = √ (σ1 + σ2 + σ3 ) = √ → I1 = 3 ξ (5.39) 3 3 * Length of the deviatoric projection ρ (→ Eq. (5.38)) p ρ2 ρ = |ρ| = 2J2 → J2 = (5.40) 2 * Deviatoric angle from Rendulic direction to deviatoric direction θ

cos θ =

1 ¯1 ρ·ρ ρ

A commonly used variation of this formulation is given with √ 3 3 J3 3 p cos 3θ = 4 cos θ − 3 cos θ = 2 J23

(5.41)

(5.42)

with the second and third invariant J2 , J3 of the stress deviator, see Eq. (5.14). Eq. (5.42) yields one solution in the range 0° ≤ θ ≤ 60°. But any interchanging of σ1 , σ2 , σ3 in invariants leads to the same solution θ. 12

Also called Lode angle. A direction is a vector of length 1 by definition. 14 This is different to mean stress (σ1 + σ2 + σ3 )/3 or hydrostatic pressure −(σ1 + σ2 + σ3 )/3. 13

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5.4 Basics of Nonlinear Material Behavior

Figure 5.3: Deviatoric length and angle in deviatoric plane * ξ and ρ span a Rendulic plane. • Regarding isotropic materials principal stress values may interchange their position in the strength limit condition as the orientation of principal stress directions with respect to material directions is not relevant: f (σ1 , σ2 , σ3 ) = f (σ1 , σ3 , σ2 ) = f (σ2 , σ3 , σ1 ) = f (σ2 , σ1 , σ3 ) = f (σ3 , σ1 , σ2 ) = f (σ3 , σ2 , σ1 ). Thus, σ1 ≥ σ2 ≥ σ3

(5.43)

(signed!) may be set without loss of generality. From this point of view the deviatoric angle Eq. (5.42) uniquely characterizes a state of stress. • Strength limit conditions have their own special elements in principal stress space: – Looking at a cylindrical specimen – with confining stresses in radial direction and a distinguished stress in longitudinal direction – is useful for the following. – Compressive meridian σ1 = σ2 > σ3 : cylindrical specimen with compression σ3 < 0 in the longitudinal direction and circumferential confining pressure σ1 = σ2 < 0, |σ1 | < |σ3 |. From Eq. (5.14) we get J2 = (σ1 − σ3 )2 /3 and J3 = −2(σ1 − σ3 )3 /27 and from Eq. (5.42) cos 3θ = −1 or θ = 60°. – Tensile meridian σ1 > σ2 = σ3 : cylindrical specimen with circumferential confining pressure σ2 = σ3 < 0 and a longitudinal compression σ1 < 0, |σ1 | < |σ3 |. From Eq. (5.14) we get J2 = (σ1 − σ3 )2 /3 and J3 = 2(σ1 − σ3 )3 /27 and from Eq. (5.42) cos 3θ = 1 or θ = 0°. – Compressive and tensile meridian are determined as the intersection of stress limit surface with the Rendulic planes with θ = 60° and θ = 0°. • Strength limit surfaces of concrete form a smoothed, curved tetrahedron, see Fig. 5.4. – Its tip is located in the positive octant (σ1 > 0, σ2 > 0, σ3 > 0) near to the origin (→ triaxial tensile strength). – It opens in the negative octant(σ1 < 0, σ2 < 0, σ3 < 0), i.e. generally strength ‘grows’ under pressure. * More precisely, the admissable deviatoric length increases with the amount of pressure for a certain range of pressures. This also depends on the deviatoric angle. State April 4, 2013

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5.4.2 Stress Limit States * Deviatoric concrete strength under high pressures is not yet really known. 15 * From a theoretical point of view there is no strength limit for a pure pressure . But a material is highly compacted under high pressures with large deformations. – Its compressive and tensile meridian are slightly curved. The tensile meridian falls below the the compressive meridian. – Stress-strain behavior within this surface is initially linear elastic and becomes increasingly nonlinear when approaching the limit surface.

Figure 5.4: Stress limit surfaces a) general view direction b) pressure axis view direction • Selection of some popular isotropic strength limit functions for concrete: – Ottosen [Ott77]

√ I1 J2 J2 +b −1=0 f =a 2 +λ fc fc fc

(5.44)

where fc is compression strength (unsigned), a, b are constants, and λ is a function of 3θ   λ = k1 cos  31 arccos(k2 cos 3θ) , for cos 3θ ≥ 0  (5.45) λ = k1 cos π3 − 13 arccos(−k2 cos 3θ) , for cos 3θ ≤ 0 in which k1 , k2 are constants. The four parameters a, b, k1 , k2 are determined from tensile strength fct , biaxial strength, and points on the compressive meridian. – Hsieh/Ting/Chen [HTC82] √ J2 J2 σ1 I1 f =a 2 +b +c +d −1=0 fc fc fc fc

(5.46)

with constants a, b, c, d and largest principal stress σ1 . This may be written as  f =a ¯

ρ fc

2


ρ ξ + ¯b cos θ + c¯ + d¯ − 1 = 0 fc fc

(5.47)

with hydrostatic length ξ Eq. (5.39), deviatoric length ρ Eq. (5.40) and deviatoric angle θ Eq. (5.42). 15

Practically, pure pressure is not reachable in experimental setups. Small deviatoric parts cannot be avoided.

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5.4 Basics of Nonlinear Material Behavior – Willam / Warnke [WW75], [CS94, Section 5.5]. p 2ρc (ρ2c − ρ2t ) cos θ + ρc (2ρt − ρc ) 4(ρ2c − ρ2t ) cos2 θ + 5ρ2t − 4ρt ρc ρ¯ = 4(ρ2c − ρ2t ) cos2 θ + (ρc − 2ρt )2 (5.48) with ξ¯ = a0 + a1 ρt + a2 ρ2t , ξ¯ = b0 + b1 ρc + b2 ρ2c (5.49) and ξ¯ = ξ/fc , ρ¯ = ρ/fc . The parameters ρt describe the normalized tensile meridian, i.e. θ = 0° and ρc the normalized compressive meridian, i.e. θ = 60°. The parameters a0 , a1 , a2 , b0 , b1 , b2 are material constants. As compressive and tensile meridian should meet at the same point on the ξ-axis a0 = b0 . With ξ given the values of ρt , ρc are determined from Eq (5.49). This may be used to determine ρ with Eq. (5.48) depending on θ. These approaches in a first view provide the same shapes of stress limit state surfaces, see Fig. 5.4. Differences are given with details, e.g. – Number of material constants – Exact course of compressive and tensile meridian – Simplicity of formulation – Occurence of sharp edges, i.e. lines with non-unique normals * Hsieh has a sharp compressive meridian, Willam/Warnke has no sharp edges. • Biaxial strength as a special case – Biaxial strength is combined with a plane stress state, i.e. one principal stress component is zero.

Figure 5.5: a) Biaxial stress limit state b) stress paths – Stress limit states under this condition are given as a special case of the triaxial limit surface, i.e. the intersection of the triaxial limit surface with any of the planes σ1 = 0 or σ2 = 0 or σ3 = 0. An example is given with Fig. 5.5. We have a closed line in stress plane instead of a surface in stress space. – For biaxial experimental results see e.g. [KH69]. • Uniaxial strength as a further special case State April 4, 2013

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5.4.3 Phenomenological Approach for Biaxial Anisotropic Stress-Strain Relation – Uniaxial strength is combined with a uniaxial stress state, i.e. two principal stress components are zero. – Stress limit states under this condition are given as a special case of the biaxial limit curve, i.e. the intersection of the biaxial limit curve with the stress coordinate axis cutting the uniaxial compressive and tensile strength. • Remarks involving stress strain behavior of concrete – Concrete may be regarded as isotropic in its unloaded initial state. – As far as it is known stress limit states are independent from the stress path, i.e. different loading histories with different ways aiming at the same point of the stress limit surface actually reach it as final point. – But the stress strain behavior along these ways may be different. * Regarding the stress paths D1 and D2 of Fig. 5.5b path D1 will lead to a load induced anisotropy due to approaching the tensile strength and cracking first, while some amount of compressive strength and stiffness remains in the orthogonal direction. * On the other hand path D2 will have a more or less isotropic stress strain behavior. * A simple formulation for load induced anisotropy will be given in the following Section 6.2. – Approaching the stress limit states stress strain behavior becomes increasingly nonlinear. This may be be described by the formats of damage and plasticity.

5.4.3

Phenomenological Approach for Biaxial Anisotropic Stress-Strain Relation

• Load induced anisotropy, which is characteristic for concrete, shall be covered for the plane stress case. • Basic approach – Consider a material point during its loading history with its principal strains. – These principal strain directions span a coordinate system which will be used in the following. Each of the two coordinate direction aligns to a distinguished material direction called principal material direction in the following. • Each principal material direction is regarded as uniaxial. The uniaxial behavior is described by a generalized form of Eq. (2.1), see [CS94, 6.8.2] σ= 1+



a p σp

a  −2

 p

+

 2

(5.50)

 p

where σ,  are stress and strain in a principal direction, σp , p experimentally determined values of maximum stress and corresponding strain in a principal direction under biaxial conditions and an experimentally determined value a representing the initial tangent modulus. – The effective initial modulus is assumed with a=

E0 1 − να

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(5.51) State April 4, 2013

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5.4 Basics of Nonlinear Material Behavior with the initial16 Young’s modulus E0 under uniaxial loading, the corresponding initial Poisson’s ratio ν and the ratio α of the principal stress in the orthogonal direction to principal stress in the direction considered. This approach makes a material stiffer under biaxial compression. – The parameter a is not known a priori. It influences the values of σp , p . Its value has to be estimated and to be corrected if necessary. – The tangential stiffness corresponding to Eq. (5.50) is given by   2  a 1 − p ET =  2 2 a  1 − p + σp

(5.52)

whereby the parameters a, p , σp are assumed as constant. The tangential material stiffness in a principal directions depends on the strain in that direction. A value  = 0 yields ET = a and  = p yields ET = 0. • A combination of Eqs. (5.35), (5.32), and (5.52) is proposed for anisotropic stress-strain behavior of concrete under plane stress conditions, see [CS94, 6.8.2], [MK96, 3.4.2.3], and the coefficients of Eq. (5.32) are replaced by values acc. to Eq. (5.52). – This includes a stress limit condition. – Regarding ν1 , ν2 some margin of discretion remains. An approach is given in the following Example 5.1

Example 5.1 Modeling of biaxial stress strain behavior with orthotropic hypoelasticity • A stress ratio value α=

σ2 = 0.5 σ1

(5.53)

is assumed, which basically rules the nonlinear material behavior. • Corresponding experimental results of [KH69] provide σp1 = −36 MN/m2 , p1 = −0.003 and σp2 = −18 MN/m2 , p2 = −0.001. Furthermore, E0 = 30 000 MN/m2 , ν = 0.2 are assumed. Firstly, this leads to a = 30 000/(1 − 0.2 · 0.5) = 33 333 MN/m2 with Eq. (5.51). • Using these values Eq. (5.52) yields for the compressive range    2   2  1 2 33 333 1 − −0.003 33 333 1 − −0.001 ET 1 =  , E = 2  2 (5.54) T2 2 2 1 1 33 333 33 333 1 − −0.003 − 36 1 1 − −0.001 − 18 2 where strains 1 , 2 have to be considered with their signs. The computed values depending on the absolute value of 1 , 2 are shown in Fig. 5.6a. 16

State April 4, 2013

The initial material is assumed as isotropic.

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Figure 5.6: Example 5.1 a) b) • Regarding the values of the Poisson’s ratio, see Eq. (5.29) with E1 , E2 replacing ET 1 , ET 2 , ν1 = ν is assumed leading to ET 2 (5.55) ν2 = ν ET 1 This may yield special effects in case of e.g. ET 1 ≤ 0, but shall not be examined with all details. • With values determined for ET 1 , ET 2 , ν1 , ν2 depending on 1 , 2 , a tangential material stiffness for plane stress conditions may be determined according to Eq. (5.32). end example 5.1

• Limitations – The stress ratio parameter α, which has been assumed as constant in deriving the coefficients of the tangential material stiffness, generally is not constant in applications. Thus, the values a, p , σp will change during a load history. – Principal directions of stress and strain generally will not coincide due to variations in α. Thus, shear stresses will arise in the principal directions of strains, which are assmued as natural coordinate system for load induced anisotropy. • It has to be concluded that the extension of phenomenological approaches, which work well for uniaxial behavior, may become awkward for biaxial and especially triaxial behavior. This motivates for the application of frameworks like damage and plasticity, which generalize important aspects of mechanical behavior of solids.

5.5

Isotropic Damage • Basic approach for isotropic damage σ = (1 − D) C0 · 

(5.56)

This can be applied for the triaxal case with C0 according to Eq. (5.21), or for plane strain according to Eq. (5.25) or plane stress according to Eq. (5.26). • Eq. (5.56) introduces a state parameter D. This scalar damage variable by definition has a range 0≤D≤1 (5.57) mailto:[email protected]

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5.5 Isotropic Damage D = 0 denotes a fully undamaged material, D = 1 a fully damaged material leading to σ = 0 for every . The value of D is not allowed to decrease. It may retain its value during a loading process or increase, i.e. D˙ ≥ 0. • The damage variable D needs an evolution law describing its development from 0 to 1 during a loading history. – Stress based damage approach → D depends on stress history – Strain based damage approach → D depends on strain history In the following some simple forms of evolution laws are given for D. A lot of alternative forms exist, se e.g. [???]. The conventions σ1 ≥ σ2 ≥ σ3 or 1 ≥ 2 ≥ 3 with signed values have to be followed. • Strain based damage

Figure 5.7: Graph of Eq. (5.58) – Damage is connected to an equivalent strain κ, e.g. ( 0 κ ≤ e0   κ−e0 gd D(κ) = , − e d 1−e κ > e0

(5.58)

see Fig. 5.7, with constant material parameters e0 , ed , gd . This form guarantees the condition 0 ≤ D ≤ 1 for arbitrary values κ ≥ 0. – Equivalent strain is connected to strain with a so called damage function, e.g. * Approach 1

 F =

α1 − κ 0

1 > 0 else

with the larged principal strain 1 and a material constant α. * Approach 2  p  F = c1 J2, + κ c2 J2, + c3 1 + c4 I1, − κ2

(5.59)

(5.60)

with the largest principal strain 1 , the 1st strain invariant I1, of  and the 2nd invariant J2, of the deviator of , see Eq. (5.14) where strain principal values are used instead of stress principal values. The coefficients c1 . . . c4 are further constant material parameters. State April 4, 2013

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123 · In case of uniaxial tension is 2 = 3 = −ν 1 leading to

F =

2 2 J2, = 13 (1 + ν)  1 ,

2 c1 (1+ν) 21 3

+κ

√ c2 1+ν 3

I1, = (1 − 2ν) 1

+ c3 + c4 (1 − 2ν) 1 − κ2

(5.61)

· In case of uniaxial compression is 1 = 2 = −ν 3 leading to I1, = (1 − 2ν) 3 J2, = 31 (1 + ν)2 23 , (1+ν)2 23 1+ν √ F = c1 + κ −c2 3 − c3 ν + c4 (1 − 2ν) 3 − κ2 3

(5.62)

* A general form is given with F = F (, κ) whereby the dependence on  has to be isotropic, i.e. only strain invariants are involved. The equivalent strain κ is connected to damage D by, e.g., Eq. (5.58). Such approaches may be used for triaxial behavior and include biaxial and uniaxial behavior as a special cases. – So called Kuhn-Tucker constraints relate a change of the damage variable to the damage function and distinguish loading from unloading: F ≤ 0,

D˙ ≥ 0,

F D˙ = 0

(5.63)

* In case F < 0 is D˙ = 0, i.e. unloading occurs and damage will not change. * In case F = 0 is D˙ ≥ 0, i.e. loading occurs and damage may increase. This implies a consistency condition ∂F ∂F · ˙ + κ˙ = 0 F˙ = ∂ ∂κ

→

1 ∂F κ˙ = − ∂F · ˙ ∂ ∂κ

(5.64)

• Stress based damage – Definition of an equivalent stress η. Possible approaches: * Determination from known yielding functions, e.g. Mises, Drucker-Prager * Definitions derived from thermodynamic considerations, e.g. as the complementary energy norm of the stresses p η = 2Λ0 (σ) (5.65) with the elastic complementary energy function Λ0 of the virgin material. – Equivalent stress η is connected to stress with a damage function, e.g. F =η−r

(5.66)

with the equivalent damage threshold r. The threshold r is given by e.g. r = max [r0 ; max ηs ]

(5.67)

with an initial threshold r0 and the maximum value max ηs of the equivalent damage threshold reached during loading history so far. – The connection of r to the damage variable D is given by relatively complex relations, in most cases not explicitely. – Stress based damage has also to be completed by Kuhn-Tucker constraints, see Eq. (5.63). mailto:[email protected]

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5.5 Isotropic Damage E0 [MN/m2 ] ν [−]

30 0000 0.2

e0 [−] ed [−] gd [−]

2.836 · 10−4 1.910 · 10−3 2

c1 [−] c2 [−] c3 [−] c4 [−]

1.738 0.608 7.700 2.993

Table 5.1: Material parameters of Example 5.2

Example 5.2 Modeling of uniaxial stress strain behavior with strain based damage approach 2 • The material parameters given in Table 5.1 including the initial value E0 of Young’s modulus. • Considering continuously increasing compression with ˙3 < 0, D˙ > 0 according to Kuhn-Tucker constraints Eq. (5.62) yields F = −3 − κ = 0 and κ = −3 . Eqns. (5.56), (5.58) in case 3 ≤ −e0 lead to −



σ3 = (1 − D) E0 3 = e

−3 −e0 ed

g

d

E0 3

(5.68)

• Considering continuously increasing tension with ˙1 > 0, D˙ > 0 Eq. (5.61) yields F = α 1 − κ = 0 and κ = α 1 with α depending on ν, c1 . . . c4 . The values given in Table 5.1 result in α = 10. Eqns. (5.56), (5.58) in case 1 ≥ e0 /α lead to −

σ1 = (1 − D) E0 1 = e



α1 −e0 ed

g

d

E0 1

(5.69)

Figure 5.8: Example 5.2 a) Uniaxial stress strain curve loading only b) loading, unloading, reloading

• The stress-strain curve resulting from Eqns. (5.68), (5.69) is shown in Fig. (5.8). – The computed initial value of Young’s modulus – ratio of stress and strain in the uniaxial case – exactly reproduces the prescribed value of Table 5.1. State April 4, 2013

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125 – Uniaxial compressive strength is computed with fc = 30 MN/m2 at a strain c1 = −0.0015 = −1.5‰, uniaxial tensile strength with fct = 3 MN/m2 . – With given values of E0 , ν the values of fc , c1 , fct are determined by choosing e0 , ed , α. There is systematic method to determine the values of e0 , ed , α with chosen values of fc , c1 , fct [???]. • Loading, unloading and reloading – Loading in a range 0 ≥ 3 ≥ −1.4 · 10−3 with ˙3 < 0, D˙ > 0, F = 0. – Unloading in a range −1.4 · 10−3 ≤ 3 ≤ 0 with ˙3 > 0, D˙ = 0, F = −3 − κ0 < 0, i.e. 3 > −κ0 with κ0 = 1.4 · 10−3 . – Change of index in stress and strain from 3 to 1 due to change from compressive into tensile regime. Actually the physical direction is not changed. – Elastic reloading ranging 0 ≤ 1 ≤ κ0 /α with ˙1 > 0, D˙ = 0, F = 1 α − κ0 < 0. – Resumed loading ranging κ0 /α ≤ 1 ≤ 0.5 · 10−3 with ˙1 > 0, D˙ > 0, F = 0. end example 5.2

• Tangential material stiffness for isotropic strain based approach – From Eq. (5.56) σ˙ = (1 − D) C0 · ˙ − C0 ·  D˙ = (1 − D) C0 · ˙ − σ 0 D˙

(5.70)

– On the other hand damage D is a function of the equivalent damage strain κ, see e.g. Eq. (5.58), leading to dD dD dκ ∂F D˙ = κ˙ = − ∂F · ˙ dκ ∂ ∂κ

(5.71)

σ˙ = CT · ˙

(5.72)

with Eq. (5.64). Finally is with17 ( CT = The quantities ∂F ∂ ,

(1 − D) C0 +

∂F dD ∂κ , dκ

dD dκ ∂F ∂κ

σ 0 ∂F ∂

(1 − D) C0

for loading

(5.73)

unloading

have to be computed from the forms for F, D.

17

cij

The form σ 0 ∂F is an outer product of two vectors. An outer product a b yields a matrix c with components ∂ = ai bj . This is not necessarily symmetric.

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126

5.6 Isotropic Plasticity

5.6

Isotropic Plasticity • General format for stress based plasticity – General triaxial approach for stresses σ = C0 · ( − p )

(5.74)

with the isotropic linear elastic material matrix C0 according to Eq.(), total strains18  and permanent strains p . This leads to zero stresses σ = 0 in case of total strains equalizing permanent plastic strains  = p . The rate form σ˙ = C0 · (˙ − ˙ p )

(5.75)

has to be used for general purposes. – Plastic strains are variable. They are derived with a flow rule19 ∂G ˙ p = λ˙ ∂σ

(5.76)

using a flow potential G(σ, κp ), a plastic multiplier λ and a state variable κp comprising the load history. – Plastic flow occurs in the case yielding or loading, respectively. Yielding is ruled by a yield function F (σ, κp ) = 0 (5.77) Loading is distinguished from unloading by Kuhn-Tucker conditions similar to Eq. (5.63) F ≤ 0,

λ˙ ≥ 0,

F λ˙ = 0

(5.78)

* In case F < 0 is λ˙ = 0, i.e. elastic loading / reloading or unloading occurs and permanent strains will not change. * In case F = 0 is λ˙ ≥ 0, i.e. plastic loading may occur and permanent strains may chance. This implies a consistency condition similar to Eq. (5.64) ∂F ∂F F˙ = · σ˙ + κ˙ p = 0 ∂σ ∂κp

(5.79)

The state variable κp has an initial value κp0 for the unloaded virgin material. – Finally, the formalism has to be completed with an evolution law for the internal state variable. This is assumed with κ˙ p = λ˙ H(σ, κp )

(5.80)

– The functions F (σ, κp ), G(σ, κp ), H(σ, κp ) are material functions specific for a particular material. They have to be known and to be defined in advance. Furthermore, with an actual state of the material described by a given stress σ and a given state variable κd the loading condition may be controlled. * In case of unloading F < 0 the stress increment is given by σ˙ = C0 · ˙

(5.81)

according to Eq. (5.75). 18 19

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Actually measured with a strain gauge or so. This has shear components 23 , 13 , 12 instead of γ23 , γ13 , γ12 , see Eq. (5.3).

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127 * In case of loading, i.e. F = 0, Eq. (5.79) is combined with Eqs. (5.75), (5.80) to yield ∂F ∂F ˙ ˙ − ˙ p )] + ∂κ λH = 0 (5.82) ∂ σ · [C0 · ( p Using Eq. (5.76) this can solved for λ˙ leading to 1 ∂F ˙ · C0 · , λ˙ = A ∂σ

A=−

∂F ∂F ∂G H+ · C0 · ∂κp ∂σ ∂σ

(5.83)

Combining Eq. (5.83), (5.76) and (5.75) leads to an incremental material according to Eq. (5.35) σ˙ = CT · ˙ (5.84) with a tangential material stiffness20 CT = C0 −

∂G ∂F 1 C0 · · C0 A ∂σ ∂σ

(5.85)

with C0 according to Eq. (5.21). * A correct evaluation of the tangential material stiffness is essential for nonlinear equation solution, see Section 1.7 and Eq. (1.40). Regarding incremental approaches or discretization in time algorithms like radial return and algorithmic material modulus are appropriate, [BLM00, 5.9]. Regarding Mises plasticity, see Example 5.3 these aspects are discussed in Appendix D. – Associated plasticity with identity of flow potential and yield condition G = F is often assumed. This simplifies the whole formalism.

Example 5.3 Modeling of uniaxial stress strain behavior with Mises plasticity • Mises plasticity is a simple approach to describe plastic behavior of metals. It is stress based and has an associated flow rule with a yield function limiting the deviatoric length, see Eq. (5.40)21 r r p ∂G 1 3 0 ∂F 3 ∂F F =G= ρσ − κp = 3J2 − κp , = = σ = −1 (5.86) 2 ∂σ ∂σ 2 J2 ∂κp with the 2nd invariant J2 of the stress deviator, see Eq. (5.14)2 , and the stress deviator σ 0 , see Eq. (5.16). • Regarding the tangential stiffness according to Eq. (5.85) it may be shown that r ∂F 3 0 E · C0 = G σ, G= ∂σ J2 2(1 + ν)

(5.87)

with the shear modulus G, and22 A = H + 3G,

CT = C0 −

G 1 0 0 σ σ H J 1 + 3G 2

(5.88)

20

∂F The form ∂∂G σ ∂ σ is again an outer product of two vectors. An outer product a b yields a matrix c with components cij = ai bj . This is not necessarily symmetric. 21 The tensor components σ23 , σ32 and so on have to be distinguished while building ∂F/∂σ. 22 Full tensor notation for stress and elasticity tensor is required to derive Eq. (5.88)1 .

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5.6 Isotropic Plasticity • In case of a uniaxial stress state with σ22 = σ33 = σ23 = σ13 = σ12 = 0 is J2 =

1 2 σ , 3 11

F = σ11 − κp

(5.89)

with the initial Young’s modulus E and Poisson’s ratio ν, see Eq. (5.21). For this case the tangential stiffness, see Eq. (5.85), after some rearrangement is given with   (3α+1−3να+ν) 1 1 (3να+ν+1) 1 (3να+ν+1) 0 0 0 3 (1−2ν)(1+α) 3 (1−2ν)(1+α)   3 1 (1−2ν)(1+α) (3να+ν+1) 1 (6α+5−6να−4ν) 1 (6να+8ν−1)  0 0 0  6 (1−2ν)(1+α) 6 (1−2ν)(1+α)   3 (1−2ν)(1+α)  E  1 (3να+ν+1) 1 6να+8ν−1) 1 (6α+5−6να−4ν) 0 0 0   3 (1−2ν)(1+α) CT = 6 ( (1−2ν)(1+α) 6 (1−2ν)(1+α)   1+ν  1  0 0 0 0 0 2   1  0 0 0 0 2 0  0 0 0 0 0 12 (5.90) 2 H with α = (1 + ν) 3 E . • In case of uniaxial stress also is σ˙ 22 = σ˙ 33 = σ˙ 23 = σ˙ 13 = σ˙ 12 = 0. Thus, regarding Eqs. (5.84), (5.90) with the Voigt-notation according to Eqs. (5.3), (5.7) leads to γ23 = γ13 = γ12 = 0 and 3να + 1 + ν ˙11 (5.91) ˙22 = ˙33 = − 3α + 2 + 2ν This in turn may be used to determine σ˙ 11 = CT,11 ˙11 + CT,12 ˙22 + CT,13 ˙33 =

αE H ˙11 ˙11 = 2 α + 3 (1 + ν) 1+ H E

(5.92)

in case of loading. The elastic part of longitudinal strain may be determined with ˙el,11 = σ˙ 11 /E yielding a plastic strain ˙p,11 = ˙11 − ˙el,11 =

1 σ˙ 11 H

→

σ˙ 11 = H ˙p,11

(5.93)

The material function H is assumed as constant for Mises plasticity and plays a role as hardening modulus. Regarding Eq. (5.89)2 with F = 0 in case of loading leads to σ11 = κp . But the longitudinal stress σ11 also corresponds to a current uniaxial yield stress fy . Thus κp = fy and fy replaces κp for Mises plasticity. • Pure shear with σ11 = σ22 = σ33 = σ23 = σ13 = 0 and σ12 6= 0 may be treated in an analogous way leading to σ˙ 12 =

H γ12 ˙ , 3+ H G

G=

E , 2(1 + ν)

γp,12 = γ12 − γel,12 =

3 σ˙ 12 H

(5.94)

• Mises plasticity is characterized by four material constants: initial Young’s modulus E, Poission’s ratio ν, hardening modulus H and initial uniaxial yield stress fy0 . The value of fy0 may be directly taken from a given unixial bilinear stress-strain relation, see e.g. Fig. 2.10a. The value of H may be indirectly determined from such a relation by transforming Eq. (5.92) into H=

1

∆σ11 ∆11 11 − E1 ∆σ ∆11

(5.95)

The current yield stress fy starts with the initial value fy0 and changes in case of plastic loading as is ruled by the hardening modulus. This applies in the same way to tension and compression. State April 4, 2013

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129 • The values for H, fy0 derived from a uniaxial bilinear stress strain relation may also be used for the general triaxial √ case whereby the current yield stress is determined by means of Eq. (5.86)1 , i.e. fy = 3J2 . • Ideal Mises plasticity is given with H = 0 and fy = fy0 = const. as a special case. end example 5.3

• For isotropic materials the material functions can be described as functions of stress invariants or principal stress values F = F (σ1 , σ2 , σ3 , κp ),

G = G(σ1 , σ2 , σ3 , κp ),

H = H(σ1 , σ2 , σ3 , κp )

(5.96)

Orientation of principal stress directions has no influence for isotropic materials. This allows the representation of yield criteria as surfaces in principal stress space. • Yield criteria are strongly related to stress limit states, see Section 5.4.2, as the latter are fixed and not depending upon a state variable and furthermore form a boundary for yield criteria, i.e. F ≤ f . • Yield criteria for concrete

Figure 5.9: Surfaces of Mohr-Coulomb and Drucker-Prager yield criteria in principal stress space

– Mises plasticity is not appropriate for concrete as it is independent from pressure and treats compression and tension in the same way. – Extending the mises plasticity the the Drucker-Prager criterion assumes p F = κp a I1 + 3J2 − κp

(5.97)

For stress invariants I1 , J2 see Eq. (5.14). The parameter a is assumed as material constant and κp as state variable. Using Haigh-Westergaard coordinates this may be reformulated as p √ F = κp a 3 ξ + 3/2 ρ − κp (5.98) With p F = 0 this forms a circular cone in principal stress space with a radius ρ = 2/3 √ κp in the deviatoric plane ξ = 0 and an apex (→ ρ = 0) located at ξ = 1/( 3 a). With a > 0 the yield cone opens in the compressive octant. The parameter a is measure for the allowed tensile stress. mailto:[email protected]

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5.8 Localization and Regularization – The Mohr-Coulomb criterion based on Coulomb friction makes the bearable shear stress τ in a plane dependent on the plane’s normal stress σ τ = κp (c − σ tan φ) ,

0 ≤ κp tan φ ≤ 1

(5.99)

with a nominal cohesion c and a nominal angle of internal friction23 φ. Regarding triaxial stress states with Eq. (5.43) the maximum shear stress and its attached normal stress are given by τm = (σ1 − σ3 )/2, σm (σ1 + σ3 )/2, see [Mal69, 3.4]. They are related to a pair τ, σ with the largest ratio |τ /σ|, see Mohr’s circle, by σ = σm + τm sin φ, τ = τm cos φ. This leads Eq. (5.99) to κp sin φ 1 F = (σ1 − σ3 ) + (σ1 + σ3 ) − κp c cos φ 2 2

(5.100)

F = 0 spans a plane between the compressive meridian F = 0, σ2 = σ1 > σ3 (signed!) and the tensile meridian F = 0, σ2 = σ3 < σ1 (signed!). Regarding compressive and tensile meridians see Page 116. Cyclic interchanging of principal stresses leads to totally six planes forming a hexangular cone with an apex at σ1 = σ2 = σ3 = c cot φ, see Fig. 5.9. * The Mohr-Coulomb criterion has the Rankine criterion as a special case with κp tan φ = 1. This restricts tensile sustainable tensile stresses but allows unbounded compressive stresses. – Both former approaches have characteristic drawbacks regarding concrete strength, see Page 116: Drucker-Prager has identical compressive and tensile meridians, MohrColoumb has sharp edges along the meridians with undefined yield surface gradients. A yield criterion to cover these drawbacks may base on the the stress limit function of Willam / Warnke, see Page 118. • Flow rules for concrete –

5.7

Microplane

5.8

Localization and Regularization • Softening • Mesh dependence of energy • Overview regularization methods • Crack band approach • Nonlocal damage • Gradient damage

23

Notice: friction angle φ here is positive in clockwise direction. A nominal value is considered as material constant. To cover a hardening the variable multiplier κp is introduced which serves as an internal state variable.

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131

5.9

Long Term Behavior • Creep • Shrinkage

5.10

Short Term Behavior

• Strain rate effect

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Bibliography [BLM00] T. Belytschko, W.K. Liu, and B. Moran. Nonlinear Finite Elements for Continua and Structures. John Wiley & Sons, Chichester, 2000. [CS94]

W.F. Chen and A.F. Saleeb. Constitutive Equations for Engineering Materials, Volume 1: Elasticity and Modeling. Elsevier Science B.V., Amsterdam, 2. Auflage edition, 1994.

[HTC82] S.S. Hsieh, E. Ting, and W.F. Chen. A plasticity fracture-model for concrete. Int. J. Solids Structures, 18:181–197, 1982. [KH69]

H. Kupfer and H.K. Hilsdorf. Behavior of concrete under biaxial stresses. ACI Journal, 66:656–666, 1969.

[Mal69]

L. E. Malvern. Introduction to the Mechanics of a Continuous Medium. PrenticeHall, Englewood Cliffs, New Jersey, 1. auflage edition, 1969.

[MK96]

G. Mehlhorn and J. Kollegger. Anwendung der finite elemente methode im stahlbetonbau. In G. Mehlhorn, editor, Der Ingenieurbau - Rechnerorientierte Baumechanik, pages 293–425. Ernst & Sohn, 1996.

[Ott77]

N.S. Ottosen. A failure criterion for concrete. Journal of Engineering Mechanics, 103:527–535, 1977.

[WW75] K.J. Willam and E. Warnke. Constitutive model for the triaxial behavior of concrete. In IABSE Proceedings Vol. 19. International Association for Bridge and Structural Engineering, 1975.

132

Chapter 6

Deep Beams 6.1

Limit Analysis • Basic ideas – A deep beam with plane stress conditions is given with geometry, boundary conditions, material properties, a unit loading1 . * Material properties in a first approach are described by Young’s modulus E and Poisson’s ratio ν. * Material properties are complemented with values for the uniaxial strength of concrete and reinforcement. * A multiplier ruling load intensity (→ loading factor) is used as variable. – A state of equilibrium can be determined with a linear elastic analysis. The corresponding state of stress can be linearily scaled with the loading factor. – In a further step, the loading factor is determined such that the internal stress state does not exceed the strength of the materials (→ limit state condition, compare Eq. (4.18)) all around the deep beam. * Thus, the actual stress state in any plate point has to be compared to the material strength. This includes strength of concrete as well as strength of reinforcement. * The crucial point is to combine resistance of concrete and reinforcement while regarding the restricted tensile strength of concrete. The determined loading factor corresponds to a load lower or equal to the limit load of the system regarding the assumed unit loading type. * This procedure is justified by the first limit theorem of plasticity. It has been demonstrated for truss models on page 98, but it is principally valid for every type of structural element. The procedure conforms to a proof of the limit load for a given system. • A variant is given with the design procedure for a given load, i.e. compressive strength of concrete or amount of reinforcement are adjusted such, that the limit state condition is not violated. – The shape and dimension of the concrete body are generally assumed as given in this context.

1

Load or largest load in a load combination has a value 1.

133

134

6.1 Limit Analysis • Linear elastic analysis – Closed form analytical solutions are available for simple cases, see e.g. [Gir74]. – Finite element solutions are appropriate for more complex situations, an example has been demonstrated on Page 88. A suitable element type is given with the 4-node quadrilateral, see page 4. Young’s modulus and Poisson’s ratio, see Eq. (1.21), can be chosen according to the initial values of concrete. – The calculation yields a stress state σx , σy , σxy for every point of a plate or for every integration point of every finite element, respectively. • Plane principal stresses – A given plane stress state σx , σy , σxy has principal stresses σ1 , σ2 with an angle ϕ measured positive counterclockwise from the x-axis σx −σy 2 cos 2ϕ = q σx −σy 2 2 ( 2 ) + σxy

(6.1)

This has one solution for ϕ in the range 0 . . . π/2, whereupon ϕ multiplied by sign σxy indicates the direction of σ1 . The direction of σ2 is perpendicular. The principal stress values are given by r r   σx −σy 2 σx −σy 2 σx +σy σ +σ x y 2 , 2 (6.2) σ1 = 2 + − + σ σ = + σxy 2 xy 2 2 2 This has a well known representation by Mohr cycles, see Fig. 6.1a. – An alternative and more general representation of stresses is given with σ1 = σx cos2 ϕ + σy sin2 ϕ + σxy 2 cos ϕ sin ϕ σ2 = σx sin2 ϕ + σy cos2 ϕ − σxy 2 cos ϕ sin ϕ

(6.3)

whereby ϕ indicates the direction of σ1 and ϕ + π/2 the direction of σ2 . This form allows to control the stress values depending on the direction ϕ. – Solving Eq. (6.3)2 for σx , σy , σxy results in σx = σ1 cos2 ϕ + σ2 sin2 ϕ σy = sin2 ϕ σ1 + cos2 ϕ σ2 σxy = sin ϕ cos ϕ (σ1 − σ2 )

(6.4)

– Please note: equations in the sets Eq. (6.1), (6.2) cannot be mixed with or complemented to equations in the set Eq. (6.4).

Requiring additionally (σy − σx ) cos ϕ sin ϕ + σxy (cos2 ϕ − sin2 ϕ) = 0 as condition for vanishing of shear stresses. 2

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135

Figure 6.1: Mohr circles a) common view b) collecting applied, concrete and reinforcement stress • Concrete stress and reinforcement stresses at a point have to be derived from a given stress state σx , σy , σxy . – It is assumed that concrete can sustain a compression in a direction ϕc , but no stresses are transferred in the orthogonal direction. This implies cracking in the orthogonal direction and a principal stress state with σc,1 < 0, σc,2 00. This can be assumed without loss of generality with an appropriate choice of ϕc . Therefore, using Eq. (6.4) concrete contribution in the global directions is given by σc,x = σc,1 cos2 ϕc ,

σc,y = σc,1 sin2 ϕc ,

σc,xy = σc,1 sin ϕc cos ϕc

(6.5)

A notation σc,1 = σc is used in the following. – Furthermore, two sets of (distributed, smeared) reinforcement are considered with prescribed directions ϕs1 , ϕs2 and geometric reinforcement ratios ρs1 , ρs2 . Each reinforcement set can sustain stresses in its own direction. Eq. (6.4) yields also reinforcement contributions in the global directions σs1,x = cos2 ϕs1 σs1,1 , σs2,x = cos2 ϕs2 σs2,1 ,

σs1,y = sin2 ϕs1 σs1,1 , σs2,y = sin2 ϕs2 σs2,1 ,

σs1,xy = sin ϕs1 cos ϕs1 σs1,1 σs2,xy = sin ϕs2 cos ϕs2 σs2,1 (6.6)

with σs1,2 = σs2,2 = 0. – Summed contributions are in equilibrium with the given stress state σx , σy , σxy ρs1 σs1,x + ρs2 σs2,x + σc,x = σx ρs1 σs1,y + ρs2 σs2,y + σc,y = σy ρs1 σs1,xy + ρs2 σs2,xy + σc,xy = σxy

(6.7)

Inserting Eqns. (6.5), (6.6) into Eq. (6.7) yields three equations for the reinforcement stresses σs1,1 , σs2,1 , the concrete stress direction ϕc and the concrete stress σc . Thus, one parameter out of these may be prescribed, and the others result from Eqns. (6.7). Eq. (6.7) is illustrated by Mohr cycles in Fig. 6.1b. • A remark concerning a stress state parameter open for prescription: reality has a unique solution. We might miss this solution in the current approach as deformation behavior is not taken into account. By choosing an appropriate value for, e.g., the concrete compression direction ϕc we hope to yield a good approximation for the real solution. • A common special case mailto:[email protected]

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136

6.1 Limit Analysis – It is assumed that reinforcement directions are aligned to global coordinate axes, i.e. ϕs1 = 0, ϕs2 = π/2. Regarding Eq. (6.6) this leads to σs1,x = σs1,1 ,

σs2,y = σs2,1 ,

σs1,y = σs1,xy = σs2,x = σs2,xy = 0

(6.8)

A notation σs1,x = σsx , σs2,y = σsy , ρs1 = ρx , ρs2 = ρy is used in the following. – From Eqns. (6.5)3 , (6.7)3 sin ϕc cos ϕc =

σxy σc

(6.9)

With given values3 of σc and σxy this provides up to two solutions for ϕc in the range −π/2 ≤ ϕc ≤ π/2 under the condition |σxy /σc | ≤ 0.5, see Fig. 6.2.

Figure 6.2: Course of sin ϕc cos ϕc indicating solution range of Eq. (6.9) – From Eqns. (6.7)1,2 , (6.8) ρx σsx = σx − σc cos2 ϕc ,

ρy σsy = σy − σc sin2 ϕc

(6.10)

Eqns. (6.9), (6.10) are the basis for the following. • A further special case – A cross section, e.g. of a beam, of height z is considered with the conditions ρs2 = 0, σy = 0. – Eqs. (6.7), (6.5), (6.6) yield with omitted index for reinforcement ρs cos2 ϕs σs + σc cos2 ϕc = σx 2 2 ρs sin ϕs σs + σc sin ϕc = 0 ρs sin ϕs cos ϕs σs + σc sin ϕc cos ϕc = σxy

(6.11)

– Let us prescribe σxy , ϕs , ϕc and σs . The remaining values ρs , σc , σx may be determined from the equations. This will be performed for ϕs = π/2, σxy = V /z, σs = fy and leads to σc =

V , z sin ϕc cos ϕc

ρs = −

V , fy z cot ϕc

∆N = V cot ϕc

(6.12)

with the assumption 0 < ϕc < π/2, V < 0 and σx = ∆N/z. Eq. (6.12)2 corresponds to the design rules for stirrups. From Eq. (6.12)3 the shifting distance a for the longitudinal reinforcement derives with V = (z∆N/2)/a leading to a = z cot ϕc /2. 3

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Please note: σc < 0 by definition and σxy has to be taken with the correct sign.

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137 • Design strategy for the common special case – With the concrete stress σc limited by concrete compressive strength fc (unsigned) Eq. (6.9) leads to a constraint |σxy | ≤ 0.5fc . The case |σxy | = 0.5fc is connected with |ϕc | = π/4. Larger deviations from this direction make the constraint more restrictive. – To begin with, ϕc may freely be chosen within the admissable range not violating the constraint. This leads to an admissable concrete stress and with Eq. (6.10) to values ρx σsx and ρy σsy . * We assume ρx σsx ≥ 0, ρy σsy ≥ 0 and desire to choose ϕc such that the total amount of reinforcement determined by adding Eqns. (6.10)
ρtot = ρx + ρy = f1y σx − σc cos2 ϕc + σy − σc sin2 ϕc (6.13) = f1y (σx + σy − σc ) is minimized with a choice σsx = σsy = fy . The contribution −σc is positive by definition, i.e. σc = σxy /sin ϕc cos ϕc has to be minimized. * This is reached with ϕc = ±π/4, see Fig. 6.2, depending on the sign of σxy and leading to 1 (σx + σy + 2|σxy |) (6.14) ρtot,min = fy Please note: this result is valid only for orthogonal reinforcement meshes with coordinate directions aligned to reinforcement directions. – Further constraints arise with ρx σsx ≥ 0 and ρy σsy ≥ 0. We consider a case that the foregoing procedure leads to a result of, e.g. ρx σsx < 0. * Setting ρx σsx = 0 leaves ρy σsy and σc and ϕc as unknown values to be determined from three Eqs. (6.9), (6.10). * This remaining set of equations is nonlinear and cannot be solved directly. A numerical method like the Newton-Raphson method may be used instead, see Eq. (1.65). Collecting Eqns. (6.10), (6.9) by     σx − σc cos2 ϕc σc u =  ρy σsy  , f (u) =  σy − σc sin2 ϕc − ρy σsy  = 0 (6.15) σxy ϕc σc − sin ϕc cos ϕc the application of Newton-Raphson, see Eq. (1.65), leads to  ∂f −1 ∂f ∂f  u(ν+1) = u(ν) − 

1

1

1

∂u1 ∂f2 ∂u1 ∂f3 ∂u1

∂u2 ∂f2 ∂u2 ∂f3 ∂u2

∂u3 ∂f2 ∂u3 ∂f3 ∂u3

 

· f (u(ν) )

(6.16)

u=u(ν)

With an appropriate start value u(0) this generally converges leading to a solution σc , ρy σsy , ϕc . * An analogous method can be used in case with prescribed ρy σsy = 0 and unknown ρx σsx , σc , ϕc . The design strategy has to be performed for all relevant points of a deep beam with a loading σx , σy , σxy given for each point.

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138

6.1 Limit Analysis Example 6.1 Limit design of reinforcement for deep beam • We refer to the example of Fig. 4.1 with the same system and loading. A state of stress has been calculated for each integration point of each element with a linear elastic calculation. – Uniaxial concrete compressive strength (unsigned) is assumed with fc = 15 MN/m2 , reinforcement yield strength with fy = 500 MN/m2 . The width of the deep beam is t = 0.6 m. Safety factors are not regarded.

Figure 6.3: Example 6.1 a) characteristic stress points b) upper right reinforcement • Four points with different characteristic types of stress state are considered. • Biaxial principal compression – Element 61 at lower right integration point with σx = −0.86, σy = −3.66, σxy = −0.87. This leads to prinicpal compressive stresses with values −3.91, −0.61. A reinforcement is theoretically not necessary. • Biaxial principal tension – Element 118 at upper right integration point with σx = 3.90, σy = 2.27, σxy = 2.76. This leads to principal tensile stresses with values 5.96, 0.21. – The direction of concrete compressive stress is chosen with ϕc = −π/4. Using Eq. (6.9) this leads to σc = −2σxy = −5.52 MN/m2 . Setting σsx = σsy = fy yields from Eq. (6.10)
ρx = f1y σx − σc cos2 ϕc = 0.0133
(6.17) = 0.0101 ρy = f1y σy − σc sin2 ϕc This corresponds to reinforcement cross sections with t in cm units asx = t ρx = 79.8 cm2 /m and asy = t ρy = 60.4 cm2 /m. These related values are required locally only, not necessarily over a cross sectional length of 1 m. – To actually reach this state with both reinforcement directions under tensile yielding and the diagonal concrete direction under compression may require a considerable cracking. Crack directions should be aligned to concrete compression direction.

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139 • Mixed principal stresses – Element 91 at upper left integration point with σx = −0.61, σy = −0.30, σxy = 2.24. This leads to principal stresses −2.70, 1.78. – Concrete compressive direction is again chosen with ϕc = −π/4 leading to σc = −4.48 MN/m2 . This yields ρx = 0.0033, ρy = 0.0039 in the same way like Eq. (6.17). • Mixed principal stresses with constraint violation – Element 48 at upper left integration point with σx = 4.01, σy = −0.11, σxy = 0.02. This leads to principal stress values 4.01, −0.12. – Concrete compressive direction ϕc = −π/4 leads to ρy σsy < 0. Thus, ρy σsy = 0 is prescribed and an iteration is performed according to Eq. (6.16). This yields σc = −0.10 MN/m2 , ϕc = −0.43π and ρx = 0.0080 with σsx = fy . • Reinforcement design – A minimum constructive reinforcement ratio is assumed with ρx,min = ρy,min = ρmin = 0.0015 = 0.15 %, see [din08, 13.6]. This value is not explicitely taken into account for the computation, but the shown results are modified insofar, as only those reinforcement ratios are shown in Fig. 6.3b, which exceed ρmin . – Additional mesh / bar reinforcement: has to be completed end example 6.1

• Analogy – The basic approach – perform a linear elastic calculation for internal forces and perform a reinforcement design and concrete proof with methods of limit analysis – has strong analogies with the common practice for design and proof of reinforced concrete beams with respect to bending, shear and torsion. • Concrete strength – The concrete model used within this scope is uniaxial. Actually, there may be tension in the direction orthogonal to the principal compression direction. Lateral tension may lead to a decrease in concrete strength. This is regarded by reduction factors applied on fc , see e.g. [CF08, Section 2.2.7] to provide a "‘safe"’ value, which is used as concrete strength in the limit analysis. – There may be more reasons to reduce concrete strength for such calculations. • Ductility requirements – In order to reach stress limit states a larger redistribution of internal forces may be necessary. This may require larger deformations or a sufficient ductility, respectively. – Construction rules to ensure ductility demands remain to be added. • Limitations – Limit analyis does not regard deformations, i.e. serviceability is not proven. • Examples – See [CF08, Section 2.2.8]

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140

6.2 2D Crack Modeling

6.2

2D Crack Modeling • Cracking has already been discussed in Section 2.1, Page 21 and in Section 2.4. These basic concepts are adapted to the 2D-case in the following. • Rankine cracking criterion – A crack initiates when the largest principal stress reaches the tensile strength. – The crack direction is given with the direction perpendicular to the direction of the largest principal stress. – Determination of crack length depends on a scan method. • Scan method – A scan method defines how points for testing the cracking criterion are selected. – A mesh based method scans a whole domain in discrete points constituting a regular or irregular mesh. * Regarding 2D finite elements, such a testing mesh is naturally given by the element integration points. * If the testing mesh is sufficiently dense, crack propagation is described by an increasing number of cracked points, e.g. points with a stress state fulfilling the Rankine criterion. * As a consequence, the crack length question is not explicitely addressed in the mesh based scan method. – A boundary based method distinguishes crack initiation and crack propagation. * A crack in most cases initiates from a boundary. Thus, boundary points are scanned for testing the cracking criterion. If a scanned point fulfills the criterion for crack initiation, it becomes a crack tip. * A crack may propagate with the movement of its crack tip. Thus, a criterion for the event of crack tip movement has to be defined4 . * Upon fulfilling such a criterion, a rule to determine the direction of crack tip movement has to be defined. An additional criterion has to be defined to determine the length of the crack tip movement. * Altogether, this requires quite sophisticated models to describe crack propagation. The mesh based method will be used in the following. This method has proven its practicability and avoids dealing with the crack length problem. A drawback may be considered insofar, as a crack geometry is not precisely captured, but cracking of concrete is a diffuse matter anyway.

4

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The Rankine criterion is not suitable in all cases for this particular purpose.

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141 • Rotating crack vs. fixed crack – Testing of a particular point with a known stress state is performed with, e.g., the Rankine criterion. This decides whether this point has to be regarded as cracked, and if so, to determine its orientation. Cracking requires a reformulation of the material stiffness, which in the 2D case depends on the crack orientation. – Regarding a load history, principal stress orientations may change. This leads to two alternative concepts regarding stress orientations, fixed crack and rotating crack. * Fixed crack: crack orientation is fixed with that value occurring in crack initiation. This corresponds to clearly separated opposite crack surfaces or macro cracks, see Fig. 2.2b. * Rotating crack: crack orientation follows the direction of principal stresses, i.e. it may change during a load history. This corresponds to crack bands and micro cracking, where the orientation of a bunch of micro cracks may change upon changing of principal tensile stress direction, see Fig. 2.2b. – Although the fixed crack concept seems to be more realistic from a phenomenological point of view it shows some disadvantages. * In practical applications it yields a too stiff behavior of the cracked material in many cases. * Furthermore, it introduces a relative sliding of crack surfaces resulting in friction forces. This requires the formulation of a friction-slip law which is difficult to formulate. Thus, in the following we will start with the rotating crack concept which is easier to handle. • Cohesive crack – An uncracked point with a computed stress state is considered. A local coordinate system is chosen aligned to principal stress directions with the x-axis in principal tensile direction.

Figure 6.4: a) Bilinear cohesive crack model b) smeared crack concept – A crack is assumed to initiate if the computed principal tensile stress exceeds the tensile strength in a certain point. Principal tensile stress is orthogonal to crack surfaces by definition in the rotating crack concept. Therefore, the immediate cracking situation can be regarded as uniaxial and direction indices are dropped temporarily. mailto:[email protected]

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142

6.2 2D Crack Modeling – A process zone arises on crack initiation. It is characterized by high strains within a band width bw , see Page 22. The distance bw accounts roughly to a small multiple of the largest aggregate diameter and is kept constant. – A crack width w is defined as the distance change between left and right border of the process zone. This distance starts with a value bw . Assuming a uniform strain cb within the process zone the crack width is given by w = bw cb

(6.18)

see also Eq. (2.4). With a developing macro crack the assumption of a cb is not strictly realistic anymore, but the definition of a crack width still holds. – Uniaxial cracking behavior of concrete is characterized by the cohesive model with crack traction depending on crack width, see Section 2.4 and Fig. 2.8. – The crack traction-crack width relation will be simplified as bilinear in the following, see Fig. 6.4a. This leads to a relation   ( w ct fct 1 − w−b bw ct < w ≤ wcr wcr σ= (6.19) 0 w > wcr with the concrete strain ct = fct /E upon cracking, the crack traction σ and the crack width w. The gray shaded area corresponds to the crack energy Gf , see Eq. (2.6), which is assumed as material constant. The smallest crack width with vanishing crack tractions, the so called critical crack width wcr is determined by wcr = 2Gf /fct + bw ct as a material constant. • Single Smeared crack – We introduce the notion of a black box embracing of piece of material with an orientation in the crack normal direction. The box is chosen such, that one crack arises in it, but the exact point of cracking is not localized. – Furthermore, the concept of a crack band is incorporated, see Page 22, with a crack band width bw within the box. – Assigning the box with a length Lc ≥ bw leads to a box strain in crack normal direction i 1 h bw = (Lc − bw ) m + bw cb = (1 − ξ) m + ξ cb , ξ = (6.20) Lc Lc see Eq. (2.28) and Fig. 6.4b, with the strain m of the uncracked bulk material and the strain cb within the crack band. – Essentially, the smeared crack approach allows to use conventional continuum 2D finite elements with a continuous displacement approach as described on Page 4 to model cracks, which are are more or less discontinuous in nature.

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143 • Stress strain relation for single smeared crack with monotonuous loading – Direction indices are reused again. – Because of equilibrium considerations the stress component in the bulk material in crack normal direction has to be the same as the crack traction σ1 from Eq. (6.19). m Bulk material stress in crack tangential direction is σ2 . Thus, m 1 = 1 (σ1 , σ2 ) in a 2D case. For the sake of brevity a linear elastic plane stress behavior according to Eq. (1.21) or (5.26) is assumed in the following5 , i.e. m 1 =

1 (σ1 − ν σ2 ) , E

2 =

1 (σ2 − ν σ1 ) . E

(6.21)

with the bulk material strain 2 in the crack tangential direction and its Young’s modulus E and Poisson’s ratio ν. – Furthermore, from Eq. (6.20) 1 =

i 1 h cb m cb (Lc − bw ) m 1 + bw 1 = (1 − ξ) 1 + ξ 1 , Lc

ξ=

bw Lc

(6.22)

– Strains 1 , 2 are generally given from a superordinated calculation. With given concb stant values E, ν, bw , Lc , fct , wcr there remain five unknowns σ1 , σ2 , w1 , m 1 , 1 and five equations (6.18), (6.19), (6.21), (6.22). – Furthermore, bw  Lc or ξ  1, i.e. element size is large compared to crack band width, and ct bw  wcr , i.e. elastic energy until cracking is small to crack energy, will be assumed to simplify the relations. – Regarding a range 1 > ct , w1 < wcr with ct = fct /E, α = Lc /wcr a solution for stresses is given by σ1 = fct

1 − α1 − αν2 , 1 − αct (1 − ν 2 )

σ2 = Ec

(1 − αct )2 − αct ν 1 + ct ν 1 − αct (1 − ν 2 )

(6.23)

and for the crack width w=

1 + ν 2 − ct (1 − ν 2 ) Lc 1 − αct (1 − ν 2 )

Eq. 6.23 leads to a tangential material stiffness   −αct −αct ν E CT = 1−αct (1−ν 2 ) −αct ν 1 − αct

(6.24)

(6.25)

The parameter αct somehow plays a keyrole. It should be αct < 1, which leads to restrictions to Lc for a given material. – The relations Eq. (6.21), (6.25) are derived in a local coordinate system aligned to the principal axes of stress and strain6 . This is based on a strain state x , y , xy leading to principal strains 1 , 2 and a principal strain orientation7 ϕ. Stresses and tangential material stiffness have to be transformed to the global system for usage with, e.g., Eqns. (1.33)1 , (1.40) according to Eqns. (5.12), (5.13).

5

Plane strain, nonlinear or anisotropic approaches essentially follow the same procedure. Both coincide within this frame due to the rotating crack approach. 7 This is calculated by Eq. (6.1) with σx , σy , σxy replaced by x , y , xy /2 as xy defined by Eq. (1.2). 6

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144

6.2 2D Crack Modeling • Extension to multiple cracks – Remains to be completed. • Un- and reloading – Unloading of a crack band, i.e. a reduction of crack width, leads to decreasing strains and decreasing stresses within the crack band. A simple model is shown in Fig. 6.4a. – Reloading of a crack band approximately goes along the unloading path until the latest maximum crack width is reached. Then it follows the loading path, see Fig. 6.4a. – Eqns. (6.23) - (6.25) have to be modified for these cases. • Softening, regularization and characteristic length of elements – The cohesive crack approach, e.g. Eq. (6.19), implies softening, i.e. increascing deformations with decreasing stresses. This leads to a localization in nature, see Fig. (2.3), and also in numerical models with finite elements. – Localization is generally given with relatively narrow bands of high strains compared to the adjacent areas. – If no special measures are taken a localization band in a finite element mesh spreads along one element in its normal direction. This holds in the same way for small element sizes and for large element sizes. – A key point for numerical models arises with the proper description of the crack energy, see Page 22. It must be the same for small finite elements and large finite elements. * Regarding the smeard crack concept this is reached with Eqns. (6.19) - (6.18) and the incorporation of Lc , wcr , fct . The latter both yield a prescribed crack energy Gf . • Discrete Crack modeling – The smeared crack crack approach allows to model discontinuous cracks with continuous displacement fields. Cracks are in a black box, which can be seen as a whole only. – For an explicit modeling of cracks or displacement discontinuities finite element extensions like enhancements with internal degrees of freedom or extended finite elements (XFEM) or other interpolation techniques like elementfree Galerkin methods (EFG) are necessary.

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6.3

2D Modeling of Reinforcement and Bond • To start with a single reinforcement bar may be modeled with 2D bar elements, see Page 3. – Material behavior of reinforcement is uniaxial and may be described by the uniaxial elasto-plastic law, see Section 2.5. – Thus, modeling of a reinforcement bar may be considered as a special case of a elasto-plastic truss, which has already been discussed in Section 4.3 – But this truss is embedded in concrete and interacts with it via bond. • Bond – For bond mechanisms see Section 2.6.

Figure 6.5: Bond a) rigid b) flexible – Rigid bond model * It is assumed that slip between concrete and reinforcement can be disregarded. * As a consequence, finite elements for reinforcement on one hand and concrete on the other hand share the same nodes, see Fig. 6.5a. * This enforces the same displacements of concrete and reinforcement in nodes. There might be minor displacement differences along a rebar axis between nodes, but this is not significant if the mesh is not too coarse. – Flexible bond model * Slip between concrete and reinforcement in the longitudinal direction is regarded while both have the same displacement in the lateral direction basically. * As a consequence finite elements for reinforcement on one hand and concrete on the other hand have to have their own nodes, see Fig. 6.5b. * Concrete nodes and reinforcement nodes have to be connected by a special type of spring elements, see Page 4. * This type of element is a so called bond element, which is constrained or very stiff in the lateral rebar direction and has a bond law, see Fig. 2.11b, in the longitudinal direction. – For sake of brevity and simplicity a rigid bond will be assumed in the following, if not otherwise stated. • Reinforcement mesh mailto:[email protected]

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146

6.3 2D Modeling of Reinforcement and Bond – To model each bar of a reinforcement mesh with a finite element bar is very elaborate. A so called smeared model is used instead, i. e. reinforcement bars of one direction are modelled as a sheet. The cross section of a reinforcement mesh is charcterized by single bar cross section As and bar distance s. This leads to a sheet thickness As t= (6.26) s – A more common notation is abbreviated with as which denotes the total reinforcement cross section in cm2 related to a width of 1 m, i.e. as = 100 t if t is measured in m. – A reinforcement sheet is regarded as a plate with an anisotropic behavior. * It is assumed that a reinforcement sheet has a direction ϕs given by the direction8 of its bars. A rotated cartesian coordinate system is assigned with the x0 -axis in the bar direction.
T can be transformed to the * A given global strain state  = x y xy rotated system using Eq. (5.11) 0 = Q · 

(6.27)

with  cos2 ϕs sin2 ϕs cos ϕs sin ϕs sin2 ϕs cos2 ϕs − cos ϕs sin ϕs  Q= −2 cos ϕs sin ϕs 2 cos ϕs sin ϕs cos2 ϕs − sin2 ϕs 

(6.28)

This leads to a strain 0x which can be used to determine a stress σx0 according to the uniaxial material law Eqs. (2.30), (2.31) appropriate for the reinforcement.
T The rotated stress state is given with σ 0 = σx0 0 0 . This is transformed back to the global system with Eq. (5.12) σ = QT · σ 0 * The tangential material stiffness in the rotated system is given by   CT 0 0 C0T =  0 0 0  0 0 0

(6.29)

(6.30)

with CT = Es in case of loading below yield limit and unloading and CT = ET at the yield limit, see Fig. 2.10. * This is transformed to the global system with Eq. (5.13)   cos4 ϕs cos2 ϕs sin2 ϕs cos3 ϕs sin ϕs sin4 ϕs sin3 ϕs cos ϕs  CT = QT ·C0T ·Q = CT  cos2 ϕs sin2 ϕs 3 3 cos ϕs sin ϕs sin ϕs cos ϕs cos2 ϕs sin2 ϕs (6.31) leading to a symmetric but fully occupied matrix CT . – With Eqs. (6.27)-(6.31) the necessary components are given for modeling a reinforcement mesh with 2D continuum elements, e.g. 4-node quadrilateral elements as described on Page 4. 8

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Measured positive counterclockwise starting from the global x-direction.

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147 • Overlay of elements

Figure 6.6: Overlay of elements – Each reinforcement direction may be modeled separately with its own finite element mesh as has been described before. – In the same way the concrete part may be modeled separately. – An appropriate discretization approach is given by an element overlay, i.e. finite elements for several reinforcement directions and finite elements for concrete share the same geometry and – assuming a rigid bond – the same nodes, see Fig. 6.6. – Finally bar elements for single reinforcement bars may be superposed.

Example 6.2 Nonlinear simulation for deep beam • We refer to the deep beam example given with Fig. 4.1. In contrast to Example 6.1 the nonlinear stress-strain behavior of concrete and reinforcement will be considered to some extent. – Thickness of deep beam is t = 0.6 m. – Concrete material model according to Page 143ff. The following material properties are assumed: Young’s modulus Ec = 31 900 MN/m2 , Poisson’s ratio ν = 0.2, tensile strength fct = 1.0 MN/m2 , crack energy Gf = 50 Nm/m2 . – Reinforcement material model according to Section 2.5. The following properties are assumed: Young’s modulus Es = 200 000 MN/m2 , yield limit fy = 500 MN/m2 , ultimate strength ft = 550 MN/m2 , strain at ultimate strength u = 0.05. The latter values lead to a hardening modulus ET = 1053 MN/m2 . – Smeared reinforcement mesh model according to Page 145. An orthogonal mesh is chosen with ϕs1 = 0 °, ts1 = 0.006 m and ϕs2 = 90 °, ts2 = 0.006 m which corresponds to a reinforcement ratio of ρs1 = ρs2 = 1 %. – Plane stress conditions are assumed throughout the example. • Loading – Loading is applied by displacement control of a node, where a concentrated load acts upon, see Fig. 4.1a. Distributed loads are neglected. Actual loading is given by the reaction force of the displaced node. mailto:[email protected]

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6.3 2D Modeling of Reinforcement and Bond * In contrast to load control the control of displacements allows to model limit states of a structure, i.e. limited loading with increasing displacements. The term loading is maintained also for control of displacements. – The target displacement of the loaded node is prescribed with 0.036 m. • Discretization and computation method – Discretization is shown in Fig. 6.7a. It consists of 209 nodes and 3 × 176 + 1 fournode quadrilateral elements. Thereof, 176 elements are used for the concrete layer and 176 elements each for the two reinforcement directions.

Figure 6.7: Example 6.2: a) discretization b) load – deflection curve – Integration order for 4-node quadrilateral elements, see Eqs. (1.28), (1.33) and (1.62), is generally chosen with 2 × 2. – Special care has to be given for the lower left hand support. A nodal support of the RC deep beam itself is not appropriate as this leads to concentrated high tensile stresses. A more realistic approach is given with a support by an extra element, whereby this element is assumed as linear elastic with Ec = 31 900 M/m2 , ν = 0.2 and with a vertical nodal support of its lower two nodes. – As the problem is physically nonlinear with a non-smooth, rough characteristic9 special care has to be given to the iteration matrix in an incrementally-iterative solution approach, see Page 15. The BFGS method in combination with line search is used instead of the Newton-Raphson method, as the latter will not lead to convergence during the equilibrium iterations. – In case of a new crack detection an equilibrium iteration sequence is performed without applying a load increase. This procedure is separately performed for each new crack. New loading is only applied on a cracked system in equilibrium. – Proper selection of load or pseudo time step increments is a matter of experience and sometimes intuition is required in such problems. A value ∆t = 0.01 is chosen here with a target value t = 1.0. Thus, 100 iteration sequences arise caused by load increments. But there are roughly 500 more iteration sequences caused by cracking. Each iteration sequence may have a large number of single iterations before it reaches convergence or equilibrium, respectively. Taking all together the computation time is quite high. • Results of numerical computation 9

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This is mainly caused by the multi-linear post cracking behavior, see Fig. 6.4a, leading to derivatives with jumps.

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149 – The computed curve for the load depending on the deflection of the loaded node is shown in Fig. 6.7a. Four stages can be recognized as is typical for reinforced concrete: * * * * *

Uncracked state I. Ongoing cracking state IIa with elastic reinforcement behavior. Finally cracked state IIb with elastic reinforcement behavior. Finally cracked state III with yielding reinforcement. The transitions are gradual in the current example, as a number of elements are included in a state change.

The computed loading for a displacement of 4 cm is computed with P = 14.3 MN. This should be near to the ultimate limit load due to the horizontal characteristic of the load-deflection curve. – Concrete stresses in the finally computed step are shown in Fig. 6.8a.

Figure 6.8: Example 6.2: a) concrete principal stresses b) reinforcement principal stresses * Low tensile stresses remain according to the prescribed tensile strength. 2 * Minimum concrete stress (→ maximum compression) is computed with −96 MN/m near the load application point. Remember, a linear elastic behavior is assumed for compressive concrete in the current model. · This computed value is beyond the compressive strength of concrete. · On the other hand, a point load as assumed is not realistic. · Furthermore, a biaxial compressive state is given, which leads to a strength increase of concrete. – Reinforcement stresses in the finally computed step are shown in Fig. 6.8b. * Horizontal lines belong to the stresses of the horizontal reinforcement direction, vertical lines to the stresses of the vertical stress direction. * Horizontal stresses reach the yield strength in the lower midspan area and the upper right side support area. * Small reinforcement areas have compression due to bond with concrete.

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150

6.3 2D Modeling of Reinforcement and Bond • Control and comparative calculations – External loading is computed with P = 14.3 MN in the final state and vertical support reaction on the left hand lower support with A = 7.46 MN. – A comparative calculation with a simple beam model – span L = 10.5 m, loaded at Lp = 3.5 m, hinge support on left side, fully restrained on right side – leads to a left hand support reaction of A = 7.41 MN and a span moment Mf = 22.2 MNm.

Figure 6.9: Simple beam model of Example 6.2 – An internal lever arm of d = 3 m between resulting compression and tension forces is assumed regarding the cross section below the loaded node. This leads to a reinforcement force of Fs = 22.3/3 = 7.4 MN leading to a reinforcement of As ≈ 150 cm2 . With a reinforcement ratio ρ = 1 % and a deep beam thickness t = 0.6 m this requires a reinforcement area of 150/0.6 ≈ 2.5 m in vertical direction beginning from the bottom. This roughly matches to the area of computed reinforcement yielding, see Fig. 6.8b. • Summary – The model of this example combines equilibrium (in an integral form), nonlinear material behavior with limited material strength and, in contrast to the companion Example 6.1, kinematic compatibility. – An issue remains with the actually limited compressive strength of concrete, which has not been considered. end example 6.2

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6.4

Biaxial Concrete Material Laws • Concrete under biaxial compression – plane strain – Biaxial orthotropic extension of Saenz proposal, see Eq. (2.1), according to [LNS72], se also [CS94, 6.8]. * * * *

Classification: Fit of Measurements (FoM) approach. Monotonic loading without un- and reloading. The coordinate system is coinciding with the principal strain directions. We assume stresses and a stress ratio σ11 ≤1 (6.32) σ22 ≤ σ11 < 0, α= σ22

* The stress ratio α can be related to a pair of strength values fc2 , fc1 (unsigned), compare Fig. 5.5, and corresponding strain values c2 , c1 (signed). – The transfer of Eq. (2.1) for the x-direction of the biaxial case is given with σ22 = 1+



˜ E Ec2

˜ 22 E   2 , 22 − 2 22 + c2 c2

˜= E

Ec0 fc2 , Ec2 = − 1 − να c2

(6.33)

with the initial Young’s modulus Ec0 and Poisson’s ratio ν, which is assumed as constant. This leads to a corresponding tangential stiffness, see Eq. (2.2)   2 ˜ 1 − 22 E 2 dσ22 c2 Et2 = (6.34) =    ˜ 222 2 d22 1 + EEc2 − 2 22 + 2  c2 c2

where the stress ratio α is assumed as constant with respect to differentiation10 . – In the same way we get a relation between σ11 , 11 , where index 2 is replaced by index 1 in the Eqs. (6.33), (6.34). This leads to different behavior in the 1- and 2coordinate directions depending on the loading ratio α (→ load induced anisotropy). – Special cases ˜ = Ec0 and Eqs. (2.1), (6.33) are the * Uniaxial loading α = 0: this leads to E same. * Biaxial isotropic loading α = 1 and 11 = 22 and 11 /c1  1: σi = Ec0 /(1− ν) i , i = 1, 2. This corresponds to Eq. (5.26). – Both tangential stiffnesses Et2 , Et1 are used to construct in incremental material law like Eq. (5.9), with an orthotropic tangential material stiffness formally according to Eq. (5.32)  Et1  β Et2 βν 0  , β = Et1 ˙ β 0 σ˙ = CT · , CT =  βν Et1 2 Et1 Et2 Et2 − ν 0 0 Et1 +Et2 +2νEt2

(6.35)

which regards also a shear strain increment γ˙ 12 . – The tangential material stiffness with respect to directions of principal strain / orthortopy has to be transformed into the global coordinate system using QT · CT · Q (???) with Q according to Eq. (5.31). 10

This should at least be approximately true for proportional loading, i.e. 22 = λ 20 , 11 = λ 10 with fixed values for 20 , 10 .

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6.5 Further Aspects and Application Case Studies

6.5

Further Aspects and Application Case Studies • Loading by external load control vs. displacement control • Arc length method • Benchmarks • Comparison of different approaches for concrete material laws • Complex loading paths • Extension of 2D panels to 3D folded plate – thin walled girders

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Bibliography [CF08]

CEB-FIP. Practitioners’ guide to finite element modelling of reinforced concrete structures, volume Bulletin Nr. 45. International Federation for Structural Concrete fip, Lausanne, 2008.

[CS94]

W.F. Chen and A.F. Saleeb. Constitutive Equations for Engineering Materials, Volume 1: Elasticity and Modeling. Elsevier Science B.V., Amsterdam, 2. Auflage edition, 1994.

[din08]

DIN 1045-1: Tragwerke aus Beton, Stahlbeton und Spannbeton. Teil 1: Bemessung und Konstruktion, August 2008.

[Gir74]

K. Girkmann. Flächentragwerke. Springer-Verlag, Wien, 6. Auflage edition, 1974.

[LNS72] T.C.Y. Liu, A.H. Nilson, and F.O. Slate. Biaxial stress-strain relations for concrete. Journal of the Structural Division, 98:1025–, 1034 1972.

153

Chapter 7

Slabs 7.1

Cross Sectional Behavior

7.1.1

Kinematic Basics

• Reference plane – Coordinates x, y in the reference plane directions and z in the transverse direction. – The reference plane is placed in the midst of a slab. Bottom and top coordinates are given by z1 = −h/2, z2 = h/2 with the slab thickness h. • Bernoulli-hypothesis (→ cross sections remain plane during deformation) • Displacements w(x, y, z) =

w(x, y,h 0) = w(x, y)

u(x, y, z)

= u ¯(x, y) − z

v(x, y, z)

= v¯(x, y) − z

∂w(x,y) h ∂x ∂w(x,y) ∂y

i − γx (x, y) i − γy (x, y)

(7.1)

with the displacements u ¯(x, y), v¯(x, y) of the reference plane. This inlcudes plates as a special case with the inclusion of displacement components u, v. • Strains, compare Eqns. (1.1)-(1.3) x

=

y

=

xy = xz yz

= =

∂u ∂x ∂v ∂y ∂u ∂y ∂u ∂z ∂v ∂z

+ + +

= = ∂v ∂x ∂w ∂x ∂w ∂y

= = =

h

i

∂ 2 w(x,y) (x,y) ∂u ¯ − ∂γx∂x ∂x − z h ∂x2 i ∂γy (x,y) ∂ 2 w(x,y) ∂¯ v − z − 2 ∂y ∂y ∂yh i ∂γy ∂γx ∂u ¯ ∂¯ v ∂2w + − z 2 − − ∂y ∂x ∂x∂y ∂y ∂x ∂w − ∂w + γ + = γ x x ∂x ∂x ∂w − ∂w + γ + = γ y y ∂y ∂y

(7.2)

• The strains of the reference plane are given by ¯x =

∂u ¯ , ∂x

¯y =

∂¯ v , ∂y

154

¯xy =

∂u ¯ ∂¯ v + ∂y ∂x

(7.3)

155

7.1.1 Kinematic Basics • With the inclusion of shear deformations curvature is advantageously defined as1 κx κy

∂γx ∂2w 2 − ∂x ∂x 2 ∂γ ∂ w − ∂yy ∂y 2 ∂γy ∂2w x − ∂γ 2 ∂x∂y ∂y − ∂x

= =

κxy =

(7.4)

• With Eq. (3.2) this leads to strains x = ¯x − z κx ,

y = ¯y − z κy ,

xy = ¯xy − z κxy

(7.5)

with x , y , xy linearily varying along the beam height with extremal values at the top and bottom of the cross section and constant xz = γy , yz = γy along the beam height. Nevertheless, all these strains are varying with the reference plane coordinates x, y. – At a given location x, y these strains may be transformed to other directions leading to values 0x , 0y , 0xy . Direction angle is measured by ϕ counterclockwise against the x-axis. The transformation rule is given by      0x cos2 ϕ sin2 ϕ cos ϕ sin ϕ x  0y  =  sin2 ϕ cos2 ϕ − cos ϕ sin ϕ  ·  y  0 xy −2 cos ϕ sin ϕ 2 cos ϕ sin ϕ cos2 ϕ − sin2 ϕ xy (7.6) 

– This transformation is connected with a principal direction ϕ with 0xy = 0 and principal strains 1 , 2 . For a given x, y the principal values 1 , 2 , ϕ may vary along the height z. – For a given x, y the reference axis strain ¯x , ¯y , ¯xy and curvature κx , κy , κxy have the same transformation rule Eq. (7.6). Thus, they also have principal values and directions ¯1 , ¯2 , ϕ¯ and κ1 , κ2 , ϕκ , respectively. – The orientation of the principal systems generally must not be the same, i.e. ϕ , ϕ¯ and ϕκ generally may have different values. But ϕ = ϕ¯ in case of κx = κy = κxy = 0 and ϕ = ϕκ in case of ¯x = ¯y = ¯xy = 0 due to Eq. (7.5). • Regarding kinematics there are displacement variables u ¯, v¯, w and deformation variables ¯x , ¯y , ¯xy , κx , κy , κxy , γx , γy . Corresponding are internal forces nx , ny , nxy , mx , my , mxy , qx , qy , see Fig. (7.1). At a slab location x, y these are given by Z

h/2

Z

h/2

Z

h/2

nx = σx dz, ny = σy dz, nxy = σxy dz −h/2 −h/2 −h/2 Z h/2 Z h/2 Z h/2 mx = − σx z dz, my = − σy z dz, mxy = − σxy z dz −h/2 −h/2 −h/2 Z h/2 Z h/2 qx = σxz dz, qy = σyz dz −h/2

(7.7)

−h/2

where σx , σy , σxy , σxz , σyz depend on x , y , xy , xz , yz . For comparisons with beams see Eq. (3.8). 1

x , x , xy and κx , κy , κxy each form the component of a second order tensor which are each defined by a force direction and the direction of a reference plane.

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156

7.1 Cross Sectional Behavior

7.1.2

Linear elastic behavior

• We consider plane stress Eq. (1.21) and with Eq. (7.5) gain as normal stresses σx = σy =

E 1−ν 2 E 1−ν 2

E (x + ν y ) = 1−ν x + ν¯ y ) − z (κx + ν κy )] 2 [(¯ [(¯ y + ν¯ x ) − z (κy + ν κx )]

(7.8)

and as a shear stress in planes parallel to the reference plane with G = E/2(1 + ν). σxy = G xy = G (¯ xy − z κxy )

(7.9)

The both remaining shear stress components are given similarily σxz = G xz = G γx ,

σyz = G yz = G γy

(7.10)

R h/2 • Internal forces ( −h/2 zdz = 0!) nx ny nxy mx my

=

Z

h/2

dz

= Kn (¯ x + ν ¯y )

−h/2

=

Kn (¯ y + ν ¯x ) E ¯xy 2(1 + ν)

= =

Z

h/2

= (1 − ν) Kn

dz −h/2 Z h/2

E (κx + ν κy ) 1 − ν2

z 2 dz

¯xy 2

= K (κx + ν κy )

−h/2

=

mxy = qx

E (¯ x + ν ¯y ) 1 − ν2

(7.11)

K (κy + ν κx ) E κxy 2(1 + ν)

=

Z

h/2

κxy z 2 dz = (1 − ν) K 2 −h/2 Z h/2 αG γx dz

= (1 − ν) K

∂2w ∂x∂y

= αGh γx

−h/2

qy

=

αGh γy

Eh with the slab stiffnesses Kn = 1−ν 2, K = the shear reduction factor α, see Page 44.

E h3 , 12(1−ν 2 )

see also [Gir74, Section 189.], and

• Balance of equations2 – Variables: 3 displacement variables u ¯, v¯, w, 8 deformation variables ¯x , ¯y , ¯xy , κx , κy , κxy , γx , γy , 8 force variables nx , ny , nxy , mx , my , mxy , qx , qy , which makes together 19 variables. – Equations: 8 force-displacement relations Eq. (7.11), 6 kinematic relations Eqs. (7.3), (7.4), which makes together 14 equations. This is complemented with 5 equilibrium conditions, see Eqs. (7.12)-(7.14).

2 The slopes ∂w/∂x, ∂w/∂y need not to be adressed in this particular balance. In case the referred variables are directly determined from the referred equations the slopes result from the derivatives of the deflection surface w(x, y).

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157

7.1.3 Reinforced cracked sections

7.1.3

Reinforced cracked sections

• The following considerations regard a slab point x, y, i.g. an integration point. • Layer model – Subdivision of slab thickness into layers. – Strain of each layer is determined by the kinematic assumptions Eq. (7.2) – Stress of each layer is computed from strains with an appropriate material law. This has some aspects. * Separation of in-plane reaction – stresses from x , y , xy – from vertical shear reaction – stresses from xz , yz . * In-plane reaction is biaxial. Thus, a biaxial material law can be used. * Basically, each layer may have its own material law which allows to distinguish between concrete layers and reinforcement layers3 . * Regarding concrete, a first approach to model cracks in a biaxial setting has been described in Section 6.2. · This includes limited tensile strength, the Rankine crack criterion, smeared cracks, crack orientation, crack width, crack tractions and softening. · Generally, cracking states may vary along the slab thickness. This concerns the cracking event itself, but furthermore actual values for crack orientation and crack width in case a crack occurs. * Regarding reinforcement, smeared layers are appropriate as have been described in Section 6.3, Page 145. – Numerical integration of stresses along thickness to determine resulting internal forces according to Eq. (7.7) – The layer model is computationally expensive. – More issues * Treatment of shear forces * Tangential stiffness • Decoupled model – This is based on principal deformations for reference plan strains ¯x , ¯y , ¯xy and curvature κx , κy , κxy . It is assumed that the corresponding principal directions are at least approximately the same. Thus, a transformation to a common principle system is possible. – Each principal direction is treated like a beam cross section with unit width, see Section 3.1.3. * Lateral expansion effects seem to be disregarded. – Integration of stresses over slab thickness in each principal direction leads to principal moments m1 , m2 , which are transformed back to the global system to yield moments mx , my mxy . – The same procedure may be basically applied to vertical shear deformations and shear forces.

3

This implies rigid bond due to the kinematic approach.

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158

7.2 Equilibrium of slabs

7.2

Equilibrium of slabs • u, v are used as symbols for reference plane displacements u ¯, v¯ in the following.

Figure 7.1: Slab equilibrium • Strong differential equilibrium – We regard a place x, y of the reference plane with a distributed loading px , py , pz . – Equilibrium of normal forces ∂nx ∂nxy + + px = 0, ∂x ∂y

∂ny ∂nxy + + py = 0 ∂y ∂x

(7.12)

– Equilibrium of shear forces ∂qx ∂qy + + pz = 0 ∂x ∂y

(7.13)

– Equilibrium of moments ∂mx ∂mxy + + qx = 0, ∂x ∂y

∂my ∂mxy + + qy = 0 ∂y ∂x

(7.14)

• The linear elastic Kirchhoff case4 – Shear deformations are neglected. Then Eq. (7.4) leads to κx =

∂2w , ∂x2

κy =

∂2w , ∂y 2

κxy = 2

∂2w ∂x∂y

(7.15)

– From Eqns. (7.14), (7.13) ∂ 2 mxy ∂ 2 my ∂ 2 mx + + 2 = pz ∂x2 ∂x∂y ∂y 2

(7.16)

– From Eq. (7.11)4−6 ∂ 2 mx ∂x2 ∂ 2 my ∂y 2 ∂ 2 mxy ∂x∂y

=K =K = (1





∂ 2 κy ∂ 2 κx 2 + ν ∂x2 ∂x  2  ∂ κy ∂ 2 κx + ν 2 2 ∂y ∂y 2 K ∂ κxy − ν) 2 ∂x∂y

=K =K



∂4w 4  ∂x 4 ∂ w ∂y 4

= (1 −



∂4w ∂x2 ∂y 2  4w + ν ∂y∂2 ∂x 2 4 w ν) K ∂x∂2 ∂y 2

+ν

(7.17)

4

Other derivations use an opposite sign convention for z, w compared to Fig. 7.1. This reverses the sign in the moment-curvature relations.

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159 – Combining Eqns. (7.16), (7.17) leads to ∂4w ∂4w ∂4w ∂4w pz ∂4w + ν + 2(1 − ν) + + ν = ∂x4 ∂x2 ∂y 2 ∂x2 ∂y 2 ∂y 4 ∂y 2 ∂x2 K

(7.18)

or ∂4w ∂4w ∂4w pz + 2 + = ∂x4 ∂x2 ∂y 2 ∂y 4 K

(7.19)

– Boundary conditions are given with appropriate combinations of, e.g. w, ∂w/∂x, ∂ 2 w/∂x2 , ∂ 3 w/∂x3 . The physical meaning of the latter two is given by e.g. mx = −K(∂ 2 w/∂x2 + ν∂ 2 w/∂y 2 ) and ∂mx /∂x = qx − ∂mxy /∂y, respectively. • Weak integral equilibrium – Extending Eq. (7.4) the following kinematic variables are defined ϕx ϕy κx κy

= = = =

∂w ∂x ∂w ∂y

− γx − γy

κxy =

∂2w ∂x2 ∂2w ∂y 2

=

→ γx = ∂w ∂x − ϕx ∂w → γy = ∂y − ϕy ∂ϕx x − ∂γ ∂x = ∂x ∂ϕ ∂γy − ∂y = ∂yy ∂ϕx ∂y

+

(7.20)

∂ϕy ∂x

introducing two further displacement variables ϕx , ϕy and two further kinematic relations. – An equivalent to strong differential equilibrium formulation Eqs. (7.12)–(7.14) is given by a weak integral equilibrium formulation for a slab of Area A with potentially coupling normal forces and and moments R

A δu



     R R ∂n ∂qy ∂n x + px da + A δv ∂yy + ∂xxy + py da + A δw ∂q + + p da z ∂x ∂y     R ∂m ∂m ∂m xy y xy x + A δϕx ∂m ∂x + ∂y + qx da + A δϕy ∂y + ∂x + qy da = 0 (7.21)

∂nx ∂x

+ R

∂nxy ∂y

– Those terms with derivatives of internal forces are partially integrated: 

 ∂nx ∂nxy + da ∂y  ZA  ∂x ∂ny ∂nxy + da δv ∂x   ∂y ZA ∂qx ∂qy δw + da ∂y  ∂x ZA ∂mx ∂mxy + + qx da δϕx ∂x ∂y A Z

δu

Z 

 I ∂δu ∂δu nx + nxy da + δu(nx ex + nxy ey )ds ∂y  ZA  ∂x IA ∂δv ∂δv =− ny + nxy da + δv(nxy ex + ny ey )ds ∂x  ZA  ∂y IA ∂δw ∂δw =− qx + qy da + δw(qx ex + qy ey )ds ∂x ∂y A  ZA  ∂δϕx ∂δϕx =− mx + mxy − δϕx qx da ∂x A I ∂y =−

+



Z δϕy A



∂mxy ∂my + + qy da = − ∂y ∂x

Z  A

δϕx (mx ex + mxy ey )ds  ∂δϕy ∂δϕy my + mxy − δϕy qy da ∂y I ∂x A

+

δϕy (mxy ex + my ey )ds A

(7.22)

with a unit normal vector e with components ex , ey along the boundary. mailto:[email protected]

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7.2 Equilibrium of slabs – Combining Eqs. (7.22), (7.21) yields Z (δx nx + δy ny + δxy nxy + δκx mx + δκy my + δκxy mxy + δγx qx + δγy qy ) da Z A (δu px + δv py + δw pz ) da = I h A + δu(nx ex + nxy ey ) + δv(nxy ex + ny ey ) + δw(qx ex + qy ey ) A i +δϕx (mx ex + mxy ey ) + δϕy (mxy ex + my ey ) ds (7.23) with virtual deformations δx = ∂δu/∂x, δy = ∂δv/∂y, δxy = ∂δu/∂y+∂δv/∂x and δκx = ∂δϕx /∂x, δκy = ∂δϕy /∂y, δκxy = ∂δϕx /∂y + ∂δϕy /∂x and δγx = ∂δw/∂x − δϕx , δγy = ∂δw/∂y − δϕy . – A generalizing matrix notation of Eq. (7.23) is given with Z Z I T T δ · σ da = δu · p da + δUT · t ds (7.24) A

with  σ u p U t

= = = = = =

A

A


T x y xy κx κy κxy γx γy
T nx ny nxy mx my mxy qx qy
T u v w ϕx ϕy
T px py pz 0 0
T us vs ws ϕsx ϕsy
T nsx nsy qs msx msy

(7.25)

with the displacement boundary values us , vs , ws , ϕsx , ϕsy and nsx nsy qs msx msy

= nx ex + nxy ey = nxy ex + ny ey = qx ex + qy ey = mx ex + mxy ey = mxy ex + my ey

(7.26)

– A coupling of normal forces and moments – as may arise with reinforced cracked concrete – is realized with a dependence of normal forces on strains and additionally curvatures and also of moments on curvature and strains. • Decoupling of normal forces and moments – In case normal forces do not depend on curvature and moments do not depend on strains – e.g. for a linear elastic material behavior – weak equilibrium can be formulated independently for Eq. (7.12) and for Eqs. (7.13), (7.14). – In case of normal forces this leads to
T x y xy  =
T nx ny nxy σ =
T u v u = (7.27)
T px py p =
T us vs U =
T nsx nsy t = while Eq. (7.24) is still valid. Basically this corresponds to the biaxial plane strain or plane stress case. State April 4, 2013

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161 – In case of bending moments the relations  σ u p U t


T κx κy κxy γx γy
T mx my mxy qx
qy w ϕx ϕy
pz 0 0
T ws ϕsx ϕsy
T qs msx msy

= = = = = =

(7.28)

hold with Eq. (7.24) unchanged. These relations correspond to the Reissner-Mindlin slab. – w is decoupled from ϕx , ϕy by the shear deformation γx , γy . Thus, w, ϕx , ϕy may be interpolated as independent displacement variables within the Finite Element approach. As a further consequence, C 0 -continuity is sufficient for the interpolation of w, ϕx , ϕy , see also Section 1.6. • Kirchhoff-slab – Bending moments are considered only. –

∂w ∂x

= ϕx → γx = 0,

∂w ∂y

= ϕy → γy = 0.

– Eq. (7.28) is modified with  σ u p U t

= = = = = =

κx κy mx my w ϕx pz 0 ws ϕsx qs msx


T κxy
T mxy
ϕy
0
T ϕsy
T msy

(7.29)

with the same Eq. (7.24). – According to Eq. (7.4) the relations between strains and deformations are given by κx =

∂2w , ∂x2

κy =

∂2w , ∂y 2

κxy = 2

∂2w ∂x∂y

(7.30)

As a consequence, C 1 -continuity is required for test and trial functions for w in order to ensure compatibility and integrability of Eq. (7.24), see also Section 1.6. This requirement is not trivial to fulfill within the two dimensional setting. A number of ways to relax it have been investigated. – With a linear elastic material behavior according to Eqs. (7.11)4 − 6 the relation between generalized strains and stresses can be formulated as   1 ν 0 E h3 0 , K = σ = C · , C = K  ν 1 (7.31) 12(1 − ν 2 ) 1−ν 0 0 2 with Young’s modulus E, Poisson’s ratio ν and the slab height h.

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7.3 Structural Slab Elements

7.3

Structural Slab Elements

7.3.1

Area coordinates

• Basic approach x = L1 x 1 + L2 x 2 + L3 x 3 y = L1 y1 + L2 y2 + L3 y3 L1 + L2 + L3 = 1

(7.32)

L2 = a2 + b2 x + c2 y,

(7.33)

This leads to L1 = a1 + b1 x + c1 y, with a1 = a2 = a3 =

y3 x2 −y2 x3 , 2∆ y1 x3 −y3 x1 , 2∆ y2 x1 −y1 x2 , 2∆

b1 = b2 = b3 =

y2 −y3 2∆ , y3 −y1 2∆ , y1 −y2 2∆ ,

L3 = a3 + b3 x + c3 y c1 = c2 = c3 =

x3 −x2 2∆ x1 −x3 2∆ x2 −x1 2∆

(7.34)

and 1 x1 y1 1 1 ∆ = det 1 x2 y3 = area123 = (y1 x3 + y3 x2 + y2 x1 − y2 x3 − y3 x1 − y1 x2 ) 2 2 1 x2 y3 (7.35) • Nodal values of the area coordinates x = x 1 , y = y1 x = x 2 , y = y2 x = x 3 , y = y3

→ → →

L1 = 1, L2 = 0, L3 = 0 L1 = 0, L2 = 1, L3 = 0 L1 = 0, L2 = 0, L3 = 1

(7.36)

• Derivatives of area coordinates with bi , ci according to Eq. (7.34) ∂Li = bi , ∂x

7.3.2

∂Li = ci ∂y

(7.37)

A triangular Kirchhoff element

• Triangular element, three nodes, 9 DOF, see [ZT91, 1.5] with a basic approach w ∂w ∂x

∂w ∂y

= α1 L1 + α2 L2 + α3 L3 + α4 L1 L2 + α5 L2 L3 + α6 L3 L1 +α7 L21 L2 + α8 L22 L3 + α9 L23 L1 ∂w ∂L1 ∂w ∂L2 ∂w ∂L3 = ∂L1 ∂x + ∂L + ∂L 2 ∂x 3 ∂x
= α1 + α4 L2 + α6 L3 + 2α7 L1 L2 + α9 L23 b1
+ α2 + α4 L1 + α5 L3 + α7 L21 + 2α8 L2 L3 b
2 + α3 + α5 L2 + α6 L1 + α8 L22 + 2α9 L1 L3 b3 ∂w ∂L1 ∂w ∂L2 ∂w ∂L3 = ∂L1 ∂y + ∂L + ∂L 2 ∂y 3 ∂y
= α1 + α4 L2 + α6 L3 + 2α7 L1 L2 + α9 L23 c1
+ α2 + α4 L1 + α5 L3 + α7 L21 + 2α8 L2 L3 c
2 + α3 + α5 L2 + α6 L1 + α8 L22 + 2α9 L1 L3 c3

(7.38)

• This yields for node 1 with L1 = 1, L2 = L3 = 0 w1 = α 1 ϕ1x = α1 b1 + (α2 + α4 + α7 ) b2 + (α3 + α6 ) b3 ϕ1y = α1 c1 + (α2 + α4 + α7 ) c2 + (α3 + α6 ) c3 State April 4, 2013

(7.39)

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163

7.3.2 A triangular Kirchhoff element and for node 2 with L2 = 1, L2 = L3 = 0 w2 = α2 ϕ2x = (α1 + α4 ) b1 + α2 b2 + (α3 + α5 + α8 ) b3 ϕ2y = (α1 + α4 ) c1 + α2 c2 + (α3 + α5 + α8 ) c3

(7.40)

and for node 3 with L3 = 1, L1 = L2 = 0 w3 = α3 ϕ3x = (α1 + α6 + α9 ) b1 + (α2 + α5 ) b2 + α3 b3 ϕ3y = (α1 + α6 + α9 ) c1 + (α2 + α5 ) c2 + α3 c3

(7.41)

• The solution is α1 α2 α3 α4 α5 α6 α7 α8 α9

= w1 = w2 = w3 = −c3 ϕ2x + w2 − w1 + b3 ϕ2y = −ϕ3x c1 − w2 + w3 + b1 ϕ3y = −c2 ϕ1x + w1 − w3 + b2 ϕ1y = −b3 ϕ2y + c3 ϕ2x − b3 ϕ1y + c3 ϕ1x − 2w2 + 2w1 = −b1 ϕ2y + c1 ϕ2x − b1 ϕ3y + ϕ3x c1 + 2w2 − 2w3 = −b2 ϕ1y + c2 ϕ1x − b2 ϕ3y + c2 ϕ3x − 2w1 + 2w3

(7.42)

leading to w = −L1 (−1 + L2 − 2L1 L2 − L3 + 2L23 )w1 + L1 (c3 L1 L2 + L23 c2 − L3 c2 )ϕ1x + −L1 (b2 L23 + b3 L1 L2 − b2 L3 )ϕ1y + L2 (−2L21 − L3 + L1 + 1 + 2L2 L3 )w2 + L2 (−c3 L1 + c3 L21 + c1 L2 L3 )ϕ2x + −L2 (−b3 L1 + b3 L21 + b1 L2 L3 )ϕ2y + L3 (L2 + 2L3 L1 − 2L22 + 1 − L1 )w3 + L3 (c1 L22 + c2 L3 L1 − c1 L2 )ϕ3x + −L3 (b1 L22 + b2 L3 L1 − b1 L2 )ϕ3y (7.43) • Higher derivatives, see also (7.4) with γx = γy = 0 ∂2w ∂x2 ∂2w ∂y 2 ∂2w ∂x∂y ∂2w ∂y∂x

= = = =

∂2w ∂x∂L1 ∂2w ∂y∂L1 ∂2w ∂x∂L1 ∂2w ∂x∂y

∂L1 ∂x ∂L1 ∂y ∂L1 ∂y

+ + +

∂ 2 w ∂L2 ∂x∂L2 ∂x ∂ 2 w ∂L2 ∂y∂L2 ∂y ∂ 2 w ∂L2 ∂x∂L2 ∂y

2

∂w ∂L3 + ∂x∂L 3 ∂x ∂w2 ∂L3 + ∂y∂L 3 ∂y

= κx = κy

+

=

∂w2 ∂L3 ∂x∂L3 ∂y

κxy 2

(7.44)

or κx κy

κxy 2

= B11 w1 + B12 ϕ1x + B13 ϕ1y + B14 w2 + B15 ϕ2x + B16 ϕ2y + B17 w3 + B18 ϕ3x + B19 ϕ3y = B21 w1 + B22 ϕ1x + B23 ϕ1y + B24 w2 + B25 ϕ2x + B26 ϕ2y + B27 w3 + B28 ϕ3x + B29 ϕ3y = B31 w1 + B32 ϕ1x + B33 ϕ1y + B34 w2 + B35 ϕ2x + B36 ϕ2y + B37 w3 + B38 ϕ3x + B39 ϕ3y (7.45)

with B11 = −8b3 b1 L3 + 4b21 L2 + 2(4b2 b1 − 2b23 )L1 + 2(−b2 b1 + b3 b1 ) B12 = 4b1 c2 b3 L3 + 2c3 b21 L2 + 2(2c3 b1 b2 + c2 b23 )L1 − 2b1 c2 b3 ...

(7.46)

• These derivations demonstrate the basic procedure. But this particular approach has drawbacks, e.g. the patch test is not fulfilled for arbitrary element shapes. Improved forms are proposed by [Spe88] using forth order terms instead of cubic terms in Eq. (7.38)1 . – Details are omitted here and have to be taken from [Spe88], [ZT91, 1.5]. mailto:[email protected]

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7.4 System Building and Solution Methods

7.4

System Building and Solution Methods • System building and solution methods basically follow the outlines as have already been described with their general aspects in Section 1.7 and with applications for beams in Section 3.4. – Strains are approximated with Finite Element approximations  = B · uI

(7.47)

see Eq. (1.6), with uI according to Eq. (7.25)3 , (7.28)3 or (7.29)3 applied to all nodes of an element and  according to Eq. (7.25)1 , (7.28)1 or (7.29)1 for the coupled problem, the Reissner-Mindlin slab or the Kirchhoff slab, respectively. * In case of, e.g., a Kirchhoff slab the matrix B is determined by Eq. (7.45). – The weak equilibrium condition Eq. (7.24) has to be integrated element by element. As has been demonstrated before, see Section 1.5, integration results in r(u) = f (u) − p = 0

(7.48)

with nodal internal forces f , nodal displacements u of all elements and external nodal loads p, see also Eq. (1.63). – In case of linear material behavior internal nodal forces may be written as f (u) = K · u

→

K·u=p

(7.49)

with a constant stiffness matrix K assembled from element stiffness matrices Z BT · C · B dV (7.50) KI = VI

see also Eqs. (1.36), (1.37), with C according to, e.g., Eq. (7.31) in case of a Kirchhoff slab. This yields an immediate solution for the displacement with a given load p u = K−1 · p (7.51) • Numerical integration schemes for triangular 3-node elements – Triangular elements often use area coordinates to mark a spatial point, see Section 7.3.1. – The approach to integrate for internal nodal forces, external loads and stiffness matrices evaluates an integrand at a number of sampling points within an element, multiplies these sampled values with weighting factors and sums up the weighted values. * This resembles the procedure of Page 14. – Sampling points and weighting factors are given in Table 7.1 up to integration order 2. • Boundary conditions – First of all, kinematic boundary conditions have to provide a stable support and to prevent rigid body motions. – Boundary forces are basically given by Eq. (7.26) with a unit normal vector e with components ex , ey along the boundary. State April 4, 2013

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165 ni 0 1

2

L1 1/3 1/ 2 0 1/ 2 1/3 0.6 0.2 0.2

L2 1/3 1/2 1/2 0 1/3 0.2 0.6 0.2

L3 1/3 0 1/2 1/2 1/3 0.2 0.2 0.6

ηi 1 1/3 1/3 1/3 - 27/48 25/48 25/48 25/48

Table 7.1: Sampling points and weights for triangular numerical integration – In order to simplify the formulations a boundary edge parallel to the global y-axis with x = const., ex = 1, ey = 0 is considered examplarily, i.e. qs = qx ,

msx = mx ,

msy = mxy

(7.52)

along the boundary edge. * Free edge · Values qs , msx , msy are prescribed – in most cases with 0 – and go directly into the boundary force vector t, see Eqs. (7.28)6 , (7.29)6 . * Simply supported edge · Regarding Eqs. (7.28)5 , (7.29)5 kinematic boundary conditions w = ws = 0 and ϕy = ϕsy = 0 are given. As ϕx is not prescribed the corresponding force msx = mx has to be prescribed, generally with a value 0. On the other hand, as ϕy is prescribed the corresponding boundary force msy = mxy has to result from the computation and is transmitted to the slab’s support5 . * Clamped edge · Regarding Eqs. (7.28)5 , (7.29)5 kinematic boundary conditions w = ws = 0 and ϕx = ϕsx = 0, ϕy = ϕsy = 0 are given. All corresponding boundary forces are determined from the computation, whereby msy = mxy = 0 as ∂ϕx /∂y = 0. – A boundary edge parallel to the global x-axis with y = const. is treated in the same way with indices exchanged. A straight skew or curved boundary edge with, e.g., simply support leads to a prescribed coupling of ϕx , ϕy and mx , my , mxy along the boundary. These may be regarded as additional constraint conditions. – Kinematic boundary conditions are applied to nodes and directly implemented upon assembling the system, e.g. by the modification of the system’s stiffness matrix and load vector, see Page 63 for the basic approach. – Boundary force reactions are automatically computed as internal nodal forces for those boundary degrees of freedom which have kinematic boundary conditions prescribed. These particular internal nodal forces are not equilibrated by external nodal loads.

5

It is connected to a compensatory shear force which is combined with the ordinary shear force qx .

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7.4 System Building and Solution Methods

Example 7.1 A linear elastic rectangular plate with large opening and free edges • Geometry and boundary conditions are shown in Fig. 7.2a. A single support is given in the lower left corner. The left and the lower edge are not supported. The upper and right edge are sinply supported (hinged). The slab has an opening with dimensions given in Fig. 7.2a. • The material properties are given with E = 31 900 MN/m2 , ν = 0.2. The slab thickness is h = 0.3 m A uniform loading is assumed with p = 10 kN/m2 . It does not act in the opening area. • Kirchhoff theory is assumed with a neglection of shear deformations. The element type is chosen according to Section 7.3.2. The discretization is performed with 132 elements, which are indicated in Fig. 7.2b.

Figure 7.2: Example 7.1 a) system b) discretization and principal moments • Results – Principal moments are shown in Figure 7.2b. * Principal moments are aligned to free edges * Uniaxial behavior along free edges * Twisting moments in the upper right corner * Reversed twisting near lower left single support – The computed deflections are shown in Fig. 7.3. They obviously conform to prescribed boundary conditions. The maximum deflection is in the range of 3 cm, i.e. wmax /L ≈ 1/230. Thus, thickness and stiffness are too low for serviceability. – Reaction forces * The computed reaction forces in the boundary nodes are given in Fig. 7.3b. Number belongs to the lower left single support, the numbers 2-6 to the right edge, the number 7 to the right upper corner and the numbers 8-14 to the upper edge. * The sum of all reaction forces equals 330 kN and thus equals the total loading. * The by far largest reaction force is given the simple support. * The upper right reaction force corresponds to an uplift. This conforms to the theory of slabs with drilling resistance. State April 4, 2013

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Figure 7.3: Example 7.1 a) deflections b) boundary suppoer reactions end example 7.1

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7.5 Reinforcement Design with linear elastic internal forces

7.5

Reinforcement Design with linear elastic internal forces • Moment transformation and principal moments – Moments mx , my , mxy as they are defined by Eq. (7.7) – similar to stresses – behave like second order tensors. – Transformation of moments into a coordinate system rotated by an angle ϕ (positive counterclockwise, see Eq. (B.3)) m0 = T · m

(7.53)

with  m0x m0 =  m0y  , m0xy

   cos2 ϕ sin2 ϕ 2 cos ϕ sin ϕ mx sin2 ϕ cos2 ϕ −2 cos ϕ sin ϕ  , m =  my  T= − cos ϕ sin ϕ cos ϕ sin ϕ cos2 ϕ − sin2 ϕ mxy (7.54)





– The principal direction is defined with the condition m0xy = 0

→

p my − mx + cos 2ϕ · mxy = 0 1 − cos2 2ϕ · 2

(7.55)

Within a range 0 ≤ ϕ ≤ π/2 a unique solution ϕ is determined by mx −my 2 cos 2ϕ = q mx −my 2 ( 2 ) + m2xy

(7.56)

for mxy 6= 0. Principal moments are already given by mx , my for mxy = 0. – A solution ϕ multiplied by sign mxy – the sign of mxy – indicates the direction of a principal moment m1 . A second principal moment m2 is perpendicular. They have the values s s 2  mx − my mx − my 2 mx + my mx + my 2 m1 = + + mxy , m2 = − + m2xy 2 2 2 2 (7.57) • Corresponding forces – According to the definition of Eq. (7.7) and Fig. 7.1 a moment mx leads to stresses σx , my to σy and mxy to σxy . Assigning an internal lever arm z yields corresponding couple force resultants tx = ±

mx , z

ty = ±

my , z

txy = ±

mxy z

(7.58)

– Each of the two components may be attached to a lower and upper layer of the slab. • All considerations regarding reinforcement design of deep beams, see Section 6.1 Pages 134137, may be applied to the layer forces or upper or lower reinforcement, respectively. This is summarized in the following, which has to be applied to the upper and lower layer in the same way.

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169 • Reinforcement design basics – According to Eq. (6.4) forces are connected to principal forces by6 tx = t1 cos2 ϕ + t2 sin2 ϕ ty = t1 sin2 ϕ + t2 cos2 ϕ txy = (t1 − t2 ) sin ϕ cos ϕ

(7.59)

This form is alternative to the form of Eqs. (7.56,7.57) with the angle ϕ measured from the x-axis to the 1-axis (counterclockwise positive, see Eq. (B.4)). – Two reinforcement directions ϕs1 , ϕs2 and a concrete direction ϕc are considered. These directions describe the principal 1-directions for each part. The corresponding principal stress values are transformed to the global directions applying Eq. (7.59) to each part, whereby the stresses in the principal 2-directions vanish due to the uniaxial behavior of each part. This leads to tc,x = tc,1 cos2 ϕc , ts1,x = ts1,1 cos2 ϕs1 , ts2,x = ts2,1 cos2 ϕs2 ,

tc,y = tc,1 sin2 ϕc , ts1,y = ts1,1 sin2 ϕs1 , ts2,y = ts2,1 sin2 ϕs2 ,

tc,xy = tc,1 sin ϕc cos ϕc ts1,xy = ts1,1 sin ϕs1 cos ϕs1 ts2,xy = ts2,1 sin ϕs2 cos ϕs2 (7.60) see also Eqs. (6.5,6.6). A notation tc = tc,1 , ts1 = ts1,1 , ts2 = ts2,1 will be used in the following. Reinforcement forces are connected to reinforcement stresses via reinforcement cross sections as1 , as2 , i.e. ts1 = as1 σs1 ,

ts2 = as2 σs2

(7.61)

– The parts contribute to total forces according to Eq.(6.7) tc,x + ts1,x + ts2,x = tx tc,y + ts1,y + ts2,y = ty tc,xy + ts1,xy + ts2,xy = txy

(7.62)

– Usage of of Eqs. (7.60), (7.61) in Eq. (7.62) first of all yields three equations for the 8 design parameters tc , ϕc , σs1 , as1 , ϕs1 , σs2 , as2 , ϕs2 . – Three further Eqs. (7.58) connect tx , ty , txy to moments mx , my , mxy by the internal lever arm z. Thereby mx , my , txy are given, e.g., from a linear elastic FEcalculation which has been performed in advance. – It is reasonable to prescribe the four values σs1 , ϕs1 , σs2 , ϕs2 . Thus, the parameters tc , ϕc , as1 , as2 and z remain open for design.

6

May also be derived according to Eq. (7.54) with a sign reversal of ϕ and a zero mixed component.

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170

7.5 Reinforcement Design with linear elastic internal forces • A common special case and its design strategy basics – It is assumed that reinforcement directions are aligned to global coordinate axes, i.e. ϕs1 = 0, ϕs2 = π/2 and thus sin ϕs1 = 0, cos ϕs1 = 1, sin ϕs2 = 1, cos ϕs2 = 0. This leads to ts1,x = tsx = asx σs1 , ts1,y = 0, ts2,x = 0, ts2,y = tsy = asy σs2 ,

ts1,xy = 0 ts2,xy = 0

(7.63)

Insertion into Eq. (7.62) together with Eq. (7.60)1 yields tc cos2 ϕc + asx σs1 = tx tc sin2 ϕc + asy σs2 = ty tc sin ϕc cos ϕc = txy

(7.64)

To exploit reinforcement load carrying capacity stresses σs1 = σs2 = fy are used in the following with the reinforcement yield limit fy . – With Eq. (7.64)3 the concrete force – negative by definition – is given by tc =

txy sin ϕc cos ϕc

(7.65)

The bearing capacity of the compression zone can be estimated with appropriate assumptions about the values and the distribution of concrete stresses. * A constant concrete stress distribution is assumed with a value σc = −χ fc , a zero line of bending x measured from the compressed side and the compression stress zone height k x. Values χ = 0.95, k = 0.8 can be used, see DIN 1045-1 [din08, 9.1.6]. This leads to tc = −χfc k x. Furthermore, z = d − k x/2 with the structural height d. Assuming tc = mc /z yields s s ! ! d 2kmc 2kmc d x= 1− 1+ ¯ 2 , z = 1+ 1+ ¯ 2 (7.66) k 2 fc d fc d with f¯c = χfc k and mc < 0 by definition. – A reinforcement can be determined with Eq. (7.64) asx = asy =

tx −tc cos2 ϕc fy ty −tc sin2 ϕc fy

= =

±mx −mc cos2 ϕc z fy ±my −mc sin2 ϕc z fy

(7.67)

whereby the +sign is used for the lower slab side and the −sign for the upper side, see Fig. 7.1, and – derived from Eqs. (7.65), (7.58), tc = mc /z – with mc sin ϕc cos ϕc = ±mxy

(7.68)

Summarizing, four equations7 (7.68), (7.67), (7.66)2 hold for the five design parameters asx , asy , mc , z, ϕc . – Regarding ϕc , a first choice  π −4 for mxy ≥ 0 ϕc = (7.69) π mxy < 0 4 is appropriate as has been shown on Page 137. This leads to tc = −2|txy |, mc = −2|mxy | and asx = ±mx + |mxy |/(z fy ), asy = ±my + |mxy |/(z fy ). The internal lever arm z can directly be determined from Eq. (7.66)2 and therefore also asx , asy . 7

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Setting z = 1, m = σ and omitting Eq. (7.66)2 recovers the deep beam case, see Page 135.

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171

Figure 7.4: Slab reinforcement – A value asi < 0 indicates reorientation of the concrete stress direction, i.e. the angle ϕc has to be changed to gain asi = 0. * This case is covered by the nonlinear system of four equations (7.66)2 , (7.67) with asi = 0 and (7.68) for four unknowns ϕc , mc , z and asj with i 6= j. This system has to be solved iteratively in a similar way as has been demonstrated for deep beams, see Page 137. – The zero line x of bending measured should not exceed ≈ d/2. A condition x ≤ d/2 leads to (k = 0.8).
2 1 − 1 − k2 |mc | ≤ f¯c d2 = 0.32 f¯c d2 (7.70) 2

Example 7.2 Reinforcement design for slab with linear elastic internal forces • We refer to Example 7.1 with the same system and loading. Moments have been calculated for each element integration point with a linear elastic calculation. • Uniaxial concrete compressive strength (unsigned) is assumed with fc = 17.0 MN/m2 (concrete grade C30/37, f¯cd = 12.92), reinforcement strength with fy = 435 MN/m2 . The structural height of the slab is d = 0.25 m. Safety factors are not regarded for the loading.

Figure 7.5: Example 7.2 a) reinforcement of section b) selected points • Selected points with different characteristics of bending are considered. mailto:[email protected]

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172

7.5 Reinforcement Design with linear elastic internal forces • Point x = 0.17, y = 2.5 near midspan left free edge. – Computed moments are mx = 0.001 MNm/m, my = 0.051, mxy = 0.020 leading to principal moments m1 = 0.058, m2 = −0.006 with an orientation of ϕ1 = 71°. – Lower reinforcement: Assuming ϕ = −45° yields asx = 2.02 cm2 /m, asy = 6.87 with z/d = 0.95 and x/d = 0.10. Upper reinforcement: ϕc = 45° leads to asy < 0. Thus, an iteration has to be performed leading to asx = 0.67 cm2 /m, asy = 0 with z/d = 0.92, x/d = 0.16 and ϕc = 69° computed. • Point x = 6.5, y = 4.17 near upper right corner of simple support. – Computed moments are mx = 0.004 MNm/m, my = 0.004, mxy = −0.028 leading to m1 = 0.032, m2 = −0.024 with ϕ1 = −45°. – Curvature is negative in the corner’s diagonal direction and positive lateral to it. This indicates a load transfer ‘over edge’ supporting a load transfer in the corner’s diagonal and corresponds to the well known drilling effect. – Lower reinforcement: ϕ = 45° yields asx = 3.21 cm2 /m, asy = 3.14 with z/d = 0.0.93, x/d = 0.14. Upper reinforcement: ϕ = −45° yields asx = 2.32 cm2 /m, asy = 2.39 with z/d = 0.93, x/d = 0.14 (must here be the same as the upper values). • Point x = 3.5, y = 0.17 near midspan lower free edge. – Computed moments are mx = 0.060 MNm/m, my = 0.007, mxy = 0.008 leading to m1 = 0.061, m2 = 0.006 with ϕ1 = 8°. – Lower reinforcement: ϕ = 45° yields asx = 6.33 cm2 /m, asy = 1.39 with z/d = 0.98, x/d = 0.04. An upper reinforcement is not necessary as both principal stresses are positive and lead to a upper biaxial compressive stress state. • Point x = 0.5, y = 0.17 near lower left single support. – Computed moments are mx = 0.02 MNm/m, my = 0.011, mxy = 0.046 leading to m1 = 0.061, m2 = −0.030 with ϕ1 = 42°. – Curvature is positive in the corner’s diagonal direction and negative lateral to it. This corresponds to the major load transfer along the free edges compared to the diagonal. – Lower reinforcement: ϕc = −45° yields asx = 6.94 cm2 /m, asy = 5.89 with z/d = 0.88, x/d = 0.24. Upper reinforcement: ϕc = 45° yields asx = 2.63 cm2 /m, asy = 3.68 with z/d = 0.88, x/d = 0.24 (must here be the same as the upper values). • As compressive heights are computed with x/d < 0.5 for all points load bearing capacity of concrete is provided. end example 7.2

• Basically the same comments about high beams regarding concrete strength, ductility requirements and serviceability, see Page 139, are also valid for slabs. • Shear forces State April 4, 2013

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173 – Shear forces have to be calculated as derivatives of bending moments in case of Kirchhoff slabs qx = −

∂mx ∂mxy − , ∂x ∂y

qy = −

∂my ∂mxy − ∂y ∂x

(7.71)

see Eq. (7.14). * Bending moments are determined from curvature with Eq. (7.31) in case of linear elastic material behavior. Furthermore, curvature is determined from Finite Element interpolation with Eq. (7.47). To do this analytically is very elaborate, e.g. to compute third derivatives of displacements in order to determine first derivatives of moments. Furthermore, the accuracy decreases with computation of higher displacement derivatives. * A separate interpolation or approximation of moments within elements should be less expansive and yield reliable shear force values. This may base on a linear approach m = a x + b y + c,

∂m = a, ∂x

∂m =b ∂y

(7.72)

where m stands for mx , my , mxy . The coefficients a, b, c need at least three computed values, i.e. three integration points, within an element and may be determined using a linear regression analysis, see Appendix C.0.1. – Transformation rule for shear forces T * A cross section with normal vector e = ( cos ϕ sin ϕ ) is regarded, whereby ϕ denotes the angle with the global x-axis. The shear force in this cross section is calculated from qx , qy , see Fig. 7.1, by8

qϕ = qx cos ϕ + qy sin ϕ

(7.73)

– Shear forces generally become relevant near supported edges. A shear design calculation can be performed as follows: * Shear basically arises from change of longitudinal forces resulting from change of moments. This change of forces in adjacent cross sections leads to forces in horizontal sections and adjoined shear forces in cross sections. In case of reinforced concrete these forces are realized by concrete struts and reinforcement ties. The mechanisms are basically the same for beams and slabs. * Principal longitudinal forces t1 , t2 computed from tx , ty , txy are used in case of slabs. The force t1 acts at a cross section orientated with a normal direction angle ϕ, the force t2 at a cross section perpendicular to it. * Each of these forces is completed by the opposite longitudinal force – both forming a couple for moments m1 , m2 – and a shear force computed with Eq. (7.73). * Each cross section has a corresponding longitudinal section which can be considered as a section of a beam in a first approach. The strut and tie approach for beams shear should be applicable along this section. * A corresponding reinforcement force is given by all transformed reinforcement contributions. With tsx = asx fy , tsx = asx fy , see Eq. (7.63), the transforma8

This corresponds to Eq. (7.26)3 .

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174

7.5 Reinforcement Design with linear elastic internal forces tion of reinforcement forces to the ϕ-direction is basically determined according to Eq. (7.54) and leads to9 tsϕ = tsx cos2 ϕ + tsy sin2 ϕ

→

asϕ = asx cos2 ϕ + asy sin2 ϕ (7.74)

with the effective reinforcement asϕ in ϕ- direction. – Shear design itself can be performed as for beams within this setting. It has to be performed for both principal directions, if necessary.

Example 7.3 Computation of shear forces and shear design • We refer to Examples 7.1, 7.2 with the same system and loading. With four integration points given per element, see Table 7.1 and Fig. 7.2b, four values have been computed for each of mx , my , mxy . • A linear approximation of moments is performed within each element according to Eq. (7.72) with details given in Appendix C.0.1. Derivatives of moments are determined according to Eqs. C.6, C.7, which are constant within an element. Thus, shear forces qx , qy computed with Eq. (7.71) are also constant within an element but may differ from element to element. The results are shown in Fig. 7.6a.

Figure 7.6: Example 7.3 a) computed shear forces b) critical shear points – Two computed shear force values within elements adjacent to the right and upper boundary edge are specified. Regarding an element base length of 1 m these values basically match to the corresponding support reactions, see Fig. 7.3b. Their positive sign, see Fig. 7.1 for sign conventions, corresponds to the negative (downward) external loading. – Larger negative shear forces predominate around the left lower single support. This also corresponds to the negative external loading. – Shear forces distribution looks somehow confuse around the central opening. A finer discretization may presumably help to have more evidence regarding this area. • A shear design calculation can be performed according to Page 173. 9 This transformation differs from Eq. (7.73), as tsx , tsy are in plane and subject to a transformation of both direction and reference length, while qx , qy are orthogonal to plane and subject to transformation of reference length only.

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175 – A basic design value is given by the shear force admittable in case without shear reinforcement, see [din08, 10.3.3] vRd,ct = 0.10 · κ · (100ρ · fck )1/3 · d

(7.75)

with κ = 1.89, fck = 30 MN/m2 , d = 0.25 m in the current example. The parameter ρ denotes the reinforcement ratio, i.e. ρ = asϕ /d in the current context. A typical value is given by as = 5 cm2 /m and ρ = 5 · 10−4 /0.25 = 0.002. This leads to vRd,ct = 0.086 MN/m2 . – A value qϕ according to Eq. (7.73) is computed throughout the slab and compared to the admittable shear force according to Eq. (7.75) with the effective reinforcement Eq. (7.74). * The points computed with |qϕ | > vRd,ct are shown in Fig. 7.6b. Most of them are insofar not critical as the computed necessary bending reinforcement is low, i.e. the necessary shear bearing capacity may be reached with larger bending reinforcement, which is built in anyway to have minimum uniform bending reinforcement. end example 7.3

• Punching – Punching is a particular occurence of shear forces, i.e. shear forces are regarded with respect to single supports. – A design value for punching results from a computed support reaction force. – Design itself obeys the corresponding methods of ordinary reinforced concrete.

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Bibliography [din08] DIN 1045-1: Tragwerke aus Beton, Stahlbeton und Spannbeton. Teil 1: Bemessung und Konstruktion, August 2008. [Gir74] K. Girkmann. Flächentragwerke. Springer-Verlag, Wien, 6. Auflage edition, 1974. [Spe88] B. Specht. Modified shape functions for the three-node plate bending element passing the patch test. Int. J. Num. Meth. Eng., 26:705–715, 1988. [ZT91] O. C. Zienkiewicz and R. L. Taylor. The Finite Element Method, Volume 2. McGraw Hill, London, 4.Auflage edition, 1991.

176

Chapter 8

Shells 8.1

Preliminary Remarks

Shell kinematics is quite complex, see [???]. Thus, deviating from the standard way for structural elements up to now – kinematics, generalized material behavior, equilibrium formulated in generalized forces, appropriate element types, system building – a relatively short track coupled to a simple standard finite shell element is discussed in the following. The chosen element is the continuum based 4-node shell element as has been discussed, e.g., by [Bat96, 5.4.2]. Shells include plates and slabs as special cases. In particular, they can model slabs exposed to the combined action of lateral and in-plane actions. This effect has already been discussed for cracked reinforced concrete beams in a simplier setup, see Examples 3.2, 3.4.

8.2

Approximation of Geometry and Displacements • Shells are an extension of slabs where the reference plane becomes a simply or doubly curved reference surface. The geometry of a surface in space is described in a global cartesian system with base vectors e1 , e2 , e3 by its coordinates x1 = x1 (r, s),

x2 = x2 (r, s),

x3 = x3 (r, s)

(8.1)

whereby isoparametric coordinates r, s serve as independent variables. Thus, a pair r, s identifies a point of the reference surface or a shell point. Every shell point has thickness a. Reference surface and thickness occupy a shell body. • Furthermore, every shell point has a so called director. This is a unit vector Vn describing a direction of a cross section on which the Bernoulli-hypothesis is applied. Basically the shell director is chosen independently from the geometriy definition of Eq. (8.1), but more or less it coincides with the normal of the reference surface in case of smooth forms. • A local cartesian coordinate system is created for a shell point using, e.g. the unit vector ey of the global coordinate system with the vector cross product × leading to a unit vector V1   Vnz ey × Vn V1 = , ey × Vn =  0  (8.2) |ey × Vn | −Vnx and another unit vector V2 

 V1z Vny V2 = Vn × V1 =  V1x Vnz − V1z Vnx  −V1x Vny 177

(8.3)

178

8.2 Approximation of Geometry and Displacements V1 , V2 , Vn in this sequence form an orthogonal, normalized, right-handed coordinate system which more or less snuggles against the reference surface. • Shell geometry is approximated by an isoparametric Finite-Element interpolation. Nodes are placed in the reference surface spanning a mesh of quadrilaterial elements whereby each element has four nodes. The body occupied by the undeformed shell is interpolated by xi (r, s, t) =

X4 K=1

NK (r, s) xiK +

t X4 aK NK (r, s) VniK , K=1 2

i = 1 . . . 3 (8.4)

with xi NK (r, s) rK , sK xiK aK VniK r, s t

i-th coordinate of shell body = 14 (1 + rK r)(1 + sK s) acc. to Eqs. (1.4,1.15) local isoparametric coordinates of node K i-th coordinate of node K shell thickness at node K i-th component of director at node K local isoparametric coordinates within the reference surface local isoparametric coordinate lateral to the reference surface

xi , xiK , aK have dimensions of [length] while −1 ≤ r, s, t ≤ 1 and rK , sK = ±1 are dimensionless. Eq. (8.4) leads to the Jacobian similar to Eq. (1.16)    ∂x1 ∂x1 ∂x1  J11 J12 J13 ∂r ∂s ∂t ∂x2 ∂x2  2 J =  J21 J22 J23  =  ∂x (8.5) ∂r ∂s ∂t ∂x3 ∂x3 ∂x3 J31 J32 J33 ∂r ∂s ∂t connecting global coordinates with local isoparametric coordinates. Its components are given by P P4 ∂xi = 4K=1 brK XiK + 2t K=1 brK aK VniK ∂r P4 ∂xi t P4 i = 1...3 (8.6) = K=1 bsK XiK + 2 K=1 bsK aK VniK ∂s ∂xi 1 P4 = N a V k=1 K K niK ∂t 2 with brK =

∂NK 1 = rK (1 + sK s), ∂r 4

bsK =

∂NK 1 = sK (1 + rK r) ∂s 4

(8.7)

• Shell displacements have to be approximated in the next step. For this purpose we introduc a small rotation α around the vector V1 , see Eq. (8.2), and a small rotation β around the vector V2 , see Eq. (8.6). This leads to a vector v v = −α V2 + β V1

(8.8)

lying in the plane spanned by V1 , V2 . This vector is in particular given at nodes with vK = −α V2K + β V1K and V2K , V1K determined from Eqs. (8.2,8.6) using the particular director Vnk . The vector vK are used to deform the directors VnK leading to an interpolation of displacements ui (r, s, t) =

X4 K=1

NK (r, s) uiK +

t X4 aK NK (r, s) viK , K=1 2

i = 1 . . . 3 (8.9)

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179 ui uiK viK

i-th component of displacement i-th component of displacement of node K = −αK V2iK + βK V1iK i-te component of director change at node K

This approach realizes Bernoulli’s hypothesis of plane cross sections whereby the orientation of a cross section is defined by the direction of the corresponding director.

8.3

Approximation of Deformations • Shell deformations are derived from shell displacements by their derivatives with respect to spatial coordinates. We start with the local isoparametric coordinates  

∂ui ∂r ∂ui ∂s ∂ui ∂t





brK =  bsK K=1 0 4 X

t g1iK brK t g1iK bsK g1iK NK

   uiK t g2iK brK t g2iK bsK  ·  αK  , βK g2iK NK

i = 1 . . . 3 (8.10)

with scaled rotation axes 

   g1xK V2xK  g1yK  = − 1 aK  V2yK  2 g 1zK   V2zK  g2xK V1xK  g2yK  = 1 aK  V1yK  2 g2zK V1zK

g1K = − 21 aK V2K , g2K = 12 aK V1K ,

(8.11)

This is transformed into derivatives with respect to global coordinates with the inverse of the Jacobian    ∂ui  ∂ui ∂x1 ∂r  ∂ui  −1 i  i = 1...3 (8.12)  ∂x2  = J ·  ∂u ∂s ∂ui ∂t

∂ui ∂x3

with  −1 −1  −1 J13 J12 J11  −1  −1 −1 =  J21 = J22 J23 −1 −1 −1 J31 J32 J33 

J−1

∂r ∂x1 ∂s ∂x1 ∂t ∂x1

∂r ∂x2 ∂s ∂x2 ∂t ∂x2

∂r ∂x3 ∂s ∂x3 ∂t ∂x3

  

(8.13)

Eqs. (8.10,8.12,8.13) may be used to obtain the interpolation of the small strain tensor components   ∂uj 1 ∂ui + , i, j = 1 . . . 3 (8.14) ij = 2 ∂xj ∂xi This is identical to the strains of a three dimensional body. The difference arises from the constrained freedom to deform according to Eq. (8.9) including Bernoulli’s hypothesis. The second order strain tensor as a whole is given by E=

3 3 X X

ij ei ej

(8.15)

i=1 j=1

whereby ei ej is the tensor product of the global system unit vectors ei and ej . mailto:[email protected]

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180

8.3 Approximation of Deformations • Deformations according to Eq. (8.14) are measured in the global cartesian system and are inconvenient for thin curved shell bodies. A covariant or so called natural coordinate system is more suitable1 . Its base vectors are formed by the Jacobian J as 

           G11 J11 G21 J12 G31 J13 G1 =  G12  =  J21  , G2 =  G22  =  J22  , G3 =  G32  =  J23  G13 J31 G23 J32 G33 J33 (8.16) with j-th component Gij of base vector i measured in the absolute cartesian system. G1 is the tangential vector along the space curve given with varying r while s, t are hold constant, G2 the tangential vector along the curve with varying s and r, t constant and G3 with varying t and r, s constant. The covariant system generally is skew and not normalized, i.e. Gi · Gj 6= 0 for i 6= j and Gi · Gj 6= 1 for i = j. Thus, a contravariant coordinate system is introduced with base vectors   −1   21   −1   31   −1  J11 J21 J31 G 11 G G 1 2 3 −1  −1  −1  12 22 32          G G G G = = ,G = = ,G = = J12 J22 J32 −1 −1 −1 13 23 33 G G G J13 J23 J33 (8.17) −1 j utilizing the inverse J of the Jacobian. Due to the definitions of Gi , G the properties Gi ·Gj = 0 hold for i 6= j and Gi ·Gj = 1 for i = j. Contravariant and covariant systems may also formally be derived for cartesian coordinate systems, but then they coincide due to normalization and orthogonality. 

Following the approach in [Bat96, 2.4, 6.5.2], [DB84] the strain as a whole is described as 3 X 3 X E= ˜ij Gi Gj (8.18) i=1 j=1

with so called covariant strain components2 ˜ij or natural strains. Natural strain components have a dimension of [length]2 as Gi , Gj each have a dimension of [length]−1 . Identity of Eqs. (8.15,8.18) together with Eq. (8.14) leads to3

˜ij =

3 X 3 X

Gir Gjs rs =

r=1 s=1

 3  ∂uk ∂uk 1 X Gik + Gjk , 2 ∂ξj ∂ξi

i, j = 1 . . . 3

(8.19)

k=1

with ξ1 = r, ξ2 = s, ξ3 = t. A final remark: The components Gij of the covariant base form a second order tensor. But it is not symmetric, i.e. Gij 6= Gji . The same holds for the contravariant base: G ij 6= G ji . • Taking Eqs. (8.19,8.10) together the interpolation of contravariant strain components is given by 4 X ˜ = BK · uK (8.20) K=1 1

This should improve the elements behavior in case of mesh distortions [Bat96, p. 425]. Writing indices of ˜ij as subscripts and thus making these quantities ‘contravariant’ is pure convention but quite convenient in the context of tensor calculus. 3 May be shown using (1) Gi · Gj = 0 for i 6= j, (2) ej · Gi = Gi · ej = Gij and (3) Gij = Jji = ∂xj /∂ξi . 2

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181 with ˜

= 

BK uK

G11 brK  G21 bsK =  .. . =


T ˜11 ˜22 ˜33 2˜ 23 2˜ 13 2˜ 12  G12 brK G13 brK t brK H11K t brK H12K G22 bsK G23 bsK t bsK H21K t bsK H22K   .. .. .. .. . . . .
T u1K u2K u3K αK βK

(8.21)

and H11K H21K .. .

= G11 g1xK + G12 g1yK + G13 g1zK , = G21 g1xK + G22 g1yK + G23 g1zK , .. .

H12K H22K .. .

= G11 g2xK + G12 g2yK + G13 g2zK = G21 g2xK + G22 g2yK + G23 g2zK .. .

(8.22) with Gij according to Eqs. (8.16,8.5), brK , bsK according to Eq. (8.7), both depending on r, s, and gixK , giyK according to Eq. (8.11) whereby K indicates the element’s nodes. The approach Eq. (8.21) still includes a non-zero strain component ˜33 normal to the shell’s reference surface. This results from the continuum based approach. Its absolute value should be considerably smaller compared to the other components in practical applications. • Regarding Eq. (8.21), the discretized strain state of every point in the shell body is ruled by five degrees of freedom per node. Thus, a so called five-parameter shell model is given.

8.4

Shell Stresses and Material Laws • The concept of Cauchy stresses has already been discussed in Section 5.2, Page 105. Stress components have been introduced with respect to the global cartesian system. The stress tensor as whole in analogy to Eq. (8.15) is given by S=

3 X 3 X

σij ei ej

(8.23)

i=1 j=1

Within the context of shells it is appropriate to use the covariant system Eq. (8.16) as a base for stress components 3 X 3 X S= σ ˜ ij Gi Gj (8.24) i=1 j=1

with so called contravariant stress components σ ˜ ij . Identity of Eqs. (8.23,8.24) leads to4 ij

σ ˜ =

3 X 3 X

G ir G js σrs

(8.25)

r=1 s=1

• Motivation of introducing contravariant stress components is given by formulating the rate of internal specific strain energy which is defined in the global cartesian system as e˙ =

3 X 3 X

σij ˙ij

(8.26)

i=1 j=1 4

May be shown using (1) Gi · Gj = 0 for i 6= j, (2) ej · Gi = Gi · ej = G ij .

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182

8.4 Shell Stresses and Material Laws This particular formulation of the strain energy establishes, e.g., the formulation of the principle of virtual displacements, see Eqs. (1.28,1.29), which on the other hand is a basis for the Finite-Element-Method. Using the transformation rules Eq. (8.19,8.25) and regarding Gi · Gj = 0 for i 6= j and Gi · Gj = 1 for i = j it can be shown that 3 X 3 X

σij ˙ij =

i=1 j=1

3 X 3 X

σ ˜ ij ˜˙ ij

(8.27)

i=1 j=1

i.e. contravariant stress components are complementary to covariant strain components from an energetically point of view. Thus, while using covariant or natural strain components to describe shell deformation it is mandatory to use contravariant stress components while utilizing weak equilibrium conditions like the principle of virtual displacements. • To give a comprehensible description of the material behavior of shells it is appropriate to use the local system V1 , V2 , Vn as has been introduced with the shell director Vn and Eqs. (8.2,8.3). The local system is an orthogonal, normalized and right-handed or cartesian coordinate system, respectively. On one hand it leans against the shell’s reference surface, thus it generally changes with every point of the reference surface5 . On the other hand it is appropriate for the description of material behavior due to its normalization and orthogonality. To facilitate the notation V3 = Vn is used in the following. In analogy to Eqs. (8.15,8.23) local strain and stress components ¯ij , σ ¯ij are given by6 E=

3 X 3 X

¯ij Vi Vj ,

S=

i=1 j=1

3 X 3 X

σ ¯ij Vi Vj

(8.28)

i=1 j=1

To determine local material behavior natural strains ˜ as they are derived using Eq. (8.20) have to be transformed into local strains ¯. The identity of Eqs. (8.281 ,8.18) leads to ¯ij =

3 P 3 P

Tir Tsj ˜rs

(8.29)

r=1 s=1

with Tij = Vi · Gj ,

Tij 6= Tji

(8.30)

This may be written as a matrix operation ¯ = T · ˜

(8.31)

with ¯, ˜ ordered according to Eq. (5.3) and the components of T derived by Eq. (8.30). • To simplify a linear elastic material behavior is considered. The shell body differs from the three-dimensional continuum ruled by the linear elastic law Eq. (5.21) insofar as its normal stress in a plane normal to the reference surface should be small compared to all other stress components. Thus, we use ¯ · ¯ ¯ =C σ 5 6

State April 4, 2013

(8.32)

Also called co-rotational. As they are referenced in a cartesian system no distinction between contravariant and covariant is necessary,

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183 ¯ ordered as in Eq. (5.7) and with σ  E(1−ν) Eν     ¯ C=    

(1+ν)(1−2ν) Eν (1+ν)(1−2ν)

(1+ν)(1−2ν) E(1−ν) (1+ν)(1−2ν)

0 0 0 0

0 0 0 0

0

0

0

0



0 0 0 0 0

0 0

0 0 0 E 2(1+ν)

0 0 0 0

0

E 2(1+ν)

        

E 2(1+ν)

0 0

(8.33)

according to Eq. (5.21) whereby the influence of the colateral components ¯33 , σ ¯33 has been neglected due to the assumption of thin shells. The relation Eq. (8.32) may be ¯ T · ¯˙ , see also Eq. (5.9), with the tangential ¯˙ = C generalized as incremental form σ ¯ material stiffness CT covering nonlinear behavior as desired. Local stress components σ ¯ij have to be transformed into contravariant components σ ˜ij , see following Section 8.5. It can be shown that ˜ = TT · σ ¯ σ

(8.34)

˜ ordered as in Eq. (5.7) and the transposed of the transformation matrix T from with σ Eq. (8.31). This bases on the identity of Eqs. (8.28)2 ,(8.25). • Finally, combination of Eqs. (8.34,8.32,8.31) leads to ¯ · T · ˜ = C e · ˜ ˜ = TT · C σ

(8.35)

which yields a transformation law e = TT · C ¯ ·T C

(8.36)

for the material matrix. This transformation law is also valid for a tangential and nonlinear material matrix. Material matrices are required to determine the tangential stiffness of discretized systems, see Page 9.

8.5

System Building • The theory under consideration treats the shell body as a continuum with constraints regarding deformations. Thus, in a first approach the general form Eq. (1.28) is used to ˜ with ˜ accorddescribe weak equilibrium but by the product δT · σ replaced by δ˜T · σ ˜ according to Eq. (8.34) ing to Eq. (8.21) and σ Z Z Z Z T T T ˜ dV + ¯ dV + δ˜ · σ δu · u ¨ %dV = δu · p δuT · ¯t dA (8.37) V

V

V

At

For the evaluation of integrals see Eqs. (1.33)-(1.35). Integration is performed by numerical methods, the basic approach has been described in Section 1.7 on Page 14. It will be extended to the case of continuum based shells. Integration of internal nodal forces is performed with Z fI = VI

Z+1Z+1Z+1 ˜ s, t) dV = ˜ s, t) J(r, s, t) dtdrds B (r, s, t) · σ(r, BT (r, s, t) · σ(r, T

−1 −1 −1

(8.38) mailto:[email protected]

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184

8.5 System Building ˜ from Eq. (8.34), losee Eq. (1.33), with B assembled by the BK ’s from Eq.(8.21), σ cal isoparametric coordinates r, s, t and the determinant J = detJ of the Jacobian, see Eq. (8.5). It is evaluated numerically, i.e. fI =

nu X nu X nv X

˜ i , sj , tk ) J(ri , sj , tk ) ηi ηj ηk BT (ri , sj , tk ) · σ(r

(8.39)

i=0 j=0 k=0

with integration orders nu , nv , sampling points ri , sj , tk and weighting factors ηi,j,k . Let us assmune that a Gaussian integration is used. Then it may be appropriate to use different integration orders nu , nv along the reference surface with local coordinates r, s and along the collateral direction with local coordinate t. • The collateral direction may need a different treatment compared to the in-surface directions due to shell bending with transverse linear strains. In case of linear elastic material behavior a choice u = v = 1, nu = nv = 2, see Table 1.1, is appropriate as stresses also arise linearily. In case of nonlinear material behavior stresses may vary nonlinearily with kinks and jumps. This generally requires a higher integration order for the collateral directions, e.g. u = 1, nu = 2 and v = 4, nv = 5. It may also be appropriate to choose a different integration scheme for the collateral directions, e.g. a Lobatto-scheme which yields a higher accuracy in cases with extremal integrand values on the boundary. • As every node has five kinematic degrees of freedom Eq. (8.38) leads to five components for internal forces fI at every node I which are three force components with respect to the global coordinate system and two bending moment components with respect to the local directions V1 , V2 , see Eqs. (8.2,8.3). ¯ , see Eq. (1.28), are given as forces per volume and prePrescribed distributed loads p scribed surface tractions ¯t as forces per area, each with directions related to the global coordinate system. Corresponding nodal forces, see Eq. (1.33), again have five components for each node. Element stiffness and mass matrices have 20 × 20 entries with the four node element. Assembling of element contributions is performed in the standard way, see Page 8. • Due to the continuum based approach structural response is described by strains and stresses varying with the position in the reference surface and the collateral direction distance. Regarding shells and slabs a more familiar approach is given with, e.g., moments and shear forces. It is appropriate to refer them to the local system, see Page 182. In an analogous way as for slabs, see Eq. (7.7), resulting local internal forces are derived from ¯ see Eq. (8.32), by local stresses σ, Z Z Z a 1 a 1 a 1 n ¯1 = σ ¯11 dt, n ¯2 = σ ¯22 dt, n ¯ 12 = σ ¯12 dt 2 2 2 −1 −1 −1 Z Z Z a2 1 a2 1 a2 1 m ¯1 = − σ ¯11 tdt, m ¯2 = − σ ¯22 tdt, m ¯ 12 = − σ ¯12 tdt 4 −1 4 −1 4 −1 Z 1 Z 1 a a q¯1 = σ ¯13 dt, q¯2 = σ ¯23 dt 2 −1 2 −1 (8.40) with the local shell thickness a and the isoparametric local coordinate −1 ≤ t ≤ 1. In practice the integration is again performed numerically. Thus, nonlinear materials are automatically covered. For integration method and order see the remarks above concerning integration of system integrals. State April 4, 2013

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185

8.6

Slabs and Beams as a Special Case • A rectangular slab element of constant thickness a is considered as a special case of
T the general shell element, see Fig. 8.1. Directors are given by V1 = 1 0 0 ,

Figure 8.1: Slab element as a special case of a shell
T
T . Thus, after some calculations the matrix and Vn = 0 0 1 V2 = 0 1 0 B of interpolation functions Eq. (8.20) is written as 

BK

   =   

J11 brK 0 0 0 0 J11 bsK

0 J22 bsK 0 0 0 J22 brK

0 0 0 J33 bsK J33 brK 0

0 a −t 2 J22 bsK 0 a − 2 J22 NK 0 −t a2 J22 brK

t a2 J11 brK 0 0 0 a J NK 11 2 t a2 J11 bsK

       

(8.41)

with the components Jij of the Jacobian according to Eq. (8.5), bsK , brK according to Eq. (8.7) and for NK see Eq. (8.4). • A displacement in the x1 −x3 -plane is applied with u1 = u4 =
T u2 = u3 = u2 0 w2 0 β2 leading to        

˜11 ˜22 ˜33 2˜ 23 2˜ 13 2˜ 12





      =      

u1 0 w1 0 β1

− 12 J11 u1 − t a4 J11 β1 + 12 J11 u2 + t a4 J11 β2 0 0 0 − 12 J33 w1 + a4 J11 (1 − r)β1 + 12 J33 w2 + a4 J11 (1 + r)β2 0


T

,

       

(8.42)

see Eqs. (8.20,8.4-8.7). Furthermore, we have J11 = L1 /2, J33 = a/2, see Eq. (8.5). For rectangular slabs it is convienient to transform natural strain components ˜ij back into the global cartesian coordinate system. Transformation rules may again be derived by the identity of Eqs. (8.15,8.18). For the special case under consideration the transformation is given by 11 = G 11


2

˜11 =

4 ˜11 , L21

13 = G 11 G 33 ˜13 =

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4 ˜13 L1 a

(8.43) State April 4, 2013

186

8.7 Locking This finally leads to  

11 213



1 = L1



−1 0 0 −1

−t a2

L1 2 (1

1 0 − r) 0 1

    · L1   2 (1 + r)  t a2

u1 w1 β1 u2 w2 β2

       

(8.44)

• The interpolation of strains according to Eq. (8.44) corresponds to the interpolation of strains of the 2D Timoshenko beam element, see Eq. (3.78). This becomes obvious with (1) setting 213 = γ, (2) regarding the reversed orientation of rotations, see Eq. (8.8), and (3) adopting Eq. (3.4) ruling beam kinematics to the current case with 11 =  + t a2 κ, −1 ≤ t ≤ 1.

8.7

Locking • As a special case a state of uniform bending in the longitudinal x1 -direction is applied to Eq. (8.44) with β2 = −β1 = β/2 and u1 = w1 = u2 = w2 = 0. Eq. (8.44) yields 11 = t

a β , 2 L1

γ13 =

1 β r, 2

−1 ≤ r, t ≤ 1

(8.45)

The term z = t a2 with the thickness a describes the distance from the reference plane and Lβ1 corresponds to a curvature. To simplify a linear elastic behavior is assumed with ν = 0, see Eqs. (5.3,5.7,5.10,5.21), σ11 = E 11 = E t

a β β =Ez , 2 L1 L1

σ13 = G γ13 =

1 G βr 2

(8.46)

with a normal stress σ11 in the longitudinal x1 -directions, a shear stress σ13 in the vertical x3 -direction, Young’s modulus E and G = E2 . This corresponds to stresses in beams with κ = Lβ1 , see Eqs. (3.6,3.7) and leads to a resulting moment and shear force per unit width m=E

a3 β , 12 L1

q=

1 G a βr 2

(8.47)

Thus, the applied deformation results in a constant bending moment and a linearily varying shear force along the element. This obviously violates equilibrium conditions locally, as a zero shear force is required throughout the element in case of constant bending moment. A spurious transverse shear force arises with this type of element. The local error shall be measured by L1 q = 3 2 r, −1 ≤ r ≤ 1 (8.48) m a It becomes larger with more slender elements, i.e. with decreasing thickness or increasing element length. • Local or strong equilibrium is not enforced within the Finite-Element-Method, but weak or integral equilibrium, see Section 1.5. As a consequence, nodal forces resulting from State April 4, 2013

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187 integration of internal forces have to be in equilibrium. For the special case under consideration nodal forces according to Eq. (8.38) are given by    0 −1 0  0   0 −1    Z1 Z1  a a3 β β  L a a 1   1  −t 2 2 (1 − r)  E t 2 L a L1  − 12 L1 − 24 L1 β 1 r= · dt dr = E  1  0 0  L1  2 2  1  4Eβr  −1 −1  0  1 0  L1 a a3 β t a2 (1 + r) + 2 12 L1 24 L1 β (8.49) The non-zero nodal forces are conjugate to the applied rotation angle −β/2 on the left hand side r = −1 and β/2 on the right hand side r = 1. Insofar moments are given from a mechanical point of view forming an equilibrated system. As has already been mentioned β/L1 is a curvature. Furthermore, Ea3 /12 is the bending stiffness per unit width. Thus, the first term in each entry corresponds to a reasonable mechanical behavior.

        

The second part ±EaL1 β/24 leads to an additional spurious moment resulting from the spurious shear force. This spurious moment corresponds to an additional spurious stiffness of this element. The influence of this spurious effects increase with decreasing thickness a with a constant element length L1 , i.e. with increasing element slenderness. The spurious effects reduce with finer discretizations, i.e. decreasing L1 for a constant a. • Convergence may principally be reached, but it is reached slowly with a large parameter c, see Section 1.6 Eq. (1.59). This type of element under consideration yields much too stiff models in practical applications due to spurious transverse shear forces. So called transverse shear locking occurs. A number of locking phenomena are known, e.g. – transverse shear locking, i.e. spurious transverse shear forces in case of transverse bending – in-plane shear locking, i.e. spurious in-plane shear forces in case of in-plane bending – membrane locking, i.e. spurious membrane forces in case of transverse bending and others [Bis99, 6.4]. • Transverse shear locking is a major cause for deficiences of slab and shell elements. It has the property that spurious shear stresses disappear in distinguished points of an element, e.g in the point r = 0 for the case under consideration in the previous Section. On the other hand, such shear forces that are reasonable from a mechanical point of view yield a value in these distinguished points. This motivates popular approaches to avoid locking: – Reduced integration of system integrals, see Eqs. (1.33,1.36,1.40,1.61,1.62). This corresponds to an integration at r = 0, s = 0 for the element under consideration. Reduced integration does not affect integration order along the local t-axis. Albeit, a numerical instability of results, so called hour-glassing, may occur with reduced integration. The occurence of hour-glassing depends on the discretized geometry and applied boundary and loading conditions – Transverse shear strains are approximated with their own fields applying a mixed interpolation These fields are connected to the fields given by Eq. (8.20) through the values of ˜13 , ˜23 in those distinguished points with vanishing spurious transverse shear forces. mailto:[email protected]

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188

8.7 Locking These points are given by the coordinates A : r = 0, s = 1, B : r = −1, s = 0, C : r = 0, s = −1 and D : r = 1, s = 0 for the element under consideration. The particular strains determined by Eq. (8.20) are given by ˜A ˜B ˜C ˜D ˜213 13 ,  13 ,  13 ,  13 and  correspondingly. Anchored by these values the fields for transverse shear strains are assumed with 1 ˜13 (r, s) = 12 (1 + s) ˜A ˜C 13 + 2 (1 − s)  13 (8.50) 1 1 D ˜23 (r, s) = 2 (1 + r) ˜23 + 2 (1 − r) ˜B 23 The approach is called Assumed-Natural-Strain-Method (ANS) [DB84] and leads to a modification of the rows 4 and 5 of the matrix BK , see Eq. (8.21). This modification is straightforward with evaluating Eq. (8.20) in points A, B, C, D and combining it with Eq. (8.50). – Assuming strains partially independent from displacements or applying mixed interpolations in a first instance is not covered by the principle of virtual displacements Eqs. (1.28,1.29). Thus, an extended weak form like the principle of Hu-Washizu [Bat96, 4.4.2] is required. This involves fields for stresses and strains as independent solution variables. Its applicability requires an additive split of the matrix B, see Eq. (8.20). The current approach of Eq. (8.50) leading to an extension of Eq. (8.21) as indicated above allows such a split. Independent parts of the stress field may be eliminated in advance applying reasonable assumptions. Thus, the mixed interpolation may finally be applied in the framework of the principle of virtual displacements.

Example 8.1 Convergence study for linear simple slab • We consider a quadratic linear elastic slab with a span L = L1 = L2 = 8.0 m, a thickness a = 0.25 m and material parameters E = 33 000 MN/m2 , ν = 0.2 (0.25) in accordance with common concrete grades. The slab is simply supported along its edges, i.e. hinged without vertical displacements. A constant vertical loading is assumed with q = 16 kN/m2 downward. • Assuming the Kirchhoff theory, see Page 7.2, neglection shear an exact solution for this problem is described in [Gir74, 7.8 b)]. The maximum deflection in the center point is given by an infinite double sum e wmax =

mπ 16q L4 X X sin nπ 2 sin 2 , , K π 6 m n mn(m2 + n2 )2

K=

1 E a3 1 − ν 2 12

(8.51)

e With the parameters given this yields a converged value wmax = 5.95 · 10−3 m.

• A small convergence study is performed with meshes of 1 element up to 16 elements whereby quarter symmetry is used. Fig. 8.2 shows the meshes. Boundary conditions of nodes along symmetry axes are given by prescribing appropriate zero rotations. The n maximum deflection wmax arises at right upper corner node. Computed values are Discretization wmax [m] State April 4, 2013

1×1 5.39 · 10−3

2×2 5.87 · 10−3

3×3 5.95 · 10−3

4×4 5.98 · 10−3

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189

Figure 8.2: Example 8.1 a) b) whereby the Assumed-Natural-Strain-Method, see Page 188 and Eq. (8.50), has been applied to avoid transverse shear locking7 . Inclusion of shear deformations leads to slightly larger converged displacement compared to Kirchhoff theory. end example 8.1

8.8

Reinforced Concrete Shells • Basically the approach Eq. (8.32) ¯ · ¯(r, s, t) ¯ s, t) = C σ(r,

(8.52)

with local isoparametric coordinates8 r, s, t or its incremental form ¯ T · ¯˙ (r, s, t) ¯˙ s, t) = C σ(r,

(8.53)

see also Section 5.2 Page 106, allow for an arbitrary material behavior with variable ma¯ or tangential material stiffness C ¯ T . Regarding a shell point r, s in the terial stiffness C reference surface it is appropriate to introduce the notion of layers. – A layer is a plane through the point r, s, t, i.e. thickness t is regarded in contrast to the related shell point. This plane has the shell point’s director Vn (r, s) as normal. The behavior with respect to a layer is characterized by the components σ ¯11 , σ ¯22 , σ ¯12 and ¯11 , ¯22 , ¯12 , respectively. Furthermore, for reinforced cracked concrete it is appropriate to decouple transverse shear characterized by σ ¯13 , σ ¯23 , ¯13 , ¯23 from layer behavior. These assumptions motivate a generalization   C11 C12 0 0 0 C16  C21 C22 0 0 0 C26      0 0 0 0 0 0 ¯ =  (8.54) C  0 0 0 c44 0 0     0 0 0 0 c55 0  C61 C62 0

0

0

C66

of Eq. (8.33) with component ordering according to Eqs. (5.3,5.7). Uppercase coefficients mark layer behavior while lower case coefficients mark transverse shear behavior. 7

If it is not applied, i.e. Eq. (8.21) is used as is without modifications regarding the entries for ˜23 , ˜13 , the computed maximum deflection would be wmax = 1.44 · 10−3 m for the 4 × 4-discretization. That is an error of roughly 80%. 8 Local isoparametric coordinates tranlate into physical coordinates with Eq. (8.4).

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190

8.8 Reinforced Concrete Shells • Transverse shear stiffness has already been discussed Section 3.5.5 in the context of structural beams. Structural beams and shells both rely on the assumption of plane deformed cross sections. Regarding transverse shear stiffness of shells a substancially better approach than that used for beams is not yet available. This leads to a proposal c44 = c55 = α G,

G=

E 2(1 + ν)

(8.55)

with a reduction factor α and the initial values of Young’s modulus E, Poisson’s ratio ν, shear modulus G. According to Section 3.5.5 the reduction factor may be chosen with α = 0.5. • Layer behavior is described by a general form 

     σ ¯11 C11 C12 C16 ¯11  σ ¯22  =  C21 C22 C26  ·  ¯22  σ ¯12 C61 C62 C66 ¯12

(8.56)

or the corresponding incremental form. A layer has a biaxial plane stress state. – Thus, local layer behavior corresponds to local behavior of deep beams. The thickness of a counterpart deep beam corresponds to the ‘height’ of a layer. The height of layer is a matter of t-integration of stresses into nodal forces, see Eq. (8.38), or resulting internal forces, see Eq. (8.40). Such a height is implicitely included in a numerical integration process. The modeling of reinforced concrete deep beams as already been in dicussed in Chapter 6. The approach may be directly transferred to layers of reinforced concrete shells. – Modeling of cracks due to limited tensile strength of concrete is treated in Section 6.2. This leads to a stress-strain relation in a principal local coordinate system9 , see Eq. (6.25). Its transformation to the local system leads to a form covered by Eq. (8.56). This may be transferred to the form Eq. (8.54) and finally be used in Eq. (8.32). This material relation is based on principal strains 1 , 2 derived from ¯11 , ¯22 , ¯12 . Prinicpal strain directions also rule crack directions. Such calculations are performed for each concrete layer t. Basically, principal strain direction may change with t for a given shell point r, s, i.e. crack faces may become curved surfaces in space. – Reinforcement and bond are treated in Section (6.3). Rigid bond is assumed in the following in order to simplify. Regarding thin reinforcement meshes basically the same procedure as described for deep beams, see Page 145, may be applied for reinforced shell layers. A difference is given as shell reinforcement layers are not implemented as separate elements but subject to integration along the collateral t-direction within the frame given by Eqs. (8.38,8.40). A reinforcement layer of sheet thickness aR , see Eq. (6.26), is regarded with a collateral coordinate tR . The contribution of this layer 9 This involves the coordinate system transformations ‘natural system’ → ‘local system’ → ‘principal system’ and backward. This looks elaborate but is justifiable to cover geometrical and physical complexity. This note basically also concerns the treatment of reinforcement meshes.

State April 4, 2013

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191 to, e.g. internal forces, see Eq. (8.40), is given by n ¯ R1 = aR σ ¯R11 , n ¯ R2 = aR σ ¯R22 , n ¯ R12 = aR σ ¯R12 a a a m ¯ R1 = −aR σ ¯R11 tR , m ¯ R2 = −aR σ ¯R22 tR , m ¯ R12 = −aR σ ¯R12 tR 2 2 2 (8.57) with local reinforcement stresses σ ¯R11 , σ ¯R22 , σ ¯R12 determined in analogy to Eq. (6.29) and its related equations. Thin reinforcement layers generally do not contribute to transverse shear forces. The described approach may be applied to multiple reinforcement layers while adding up their contributions. • Alternative methods to deal with reinforced concrete shells. • Larger reinforced concrete shells were built in a large number for a wide span of applications during the twenties up to the sixties of the 20th century. An excellent documentation is given by [Joe62]. Due to expensive formwork and other upcoming restrictions their application mainly reduced to cooling towers of large power plants nowadays. A future perspective may arise with, e.g., upwind solar chimneys. The relevance of the shell approach for reinforced concrete arises with the combination of bending with membrane forces or slabs with deep beams, respectively. As has already been demonstrated for structural beams bending and and membrane forces may interact in case of cracked reinforced concrete sections. This will be demonstrated with the following Example 8.2.

Example 8.2 Nonlinear calculation for simple slab • We consider the system of Example 8.1 with the same dimensions. A concrete grade C30/37 according to EC2 [EN 04, 3.1] is chosen. This gives a Young’s modulus Ecm = 33 000 MN/m2 and a characteristic concrete strength of fck = 30 MN/m2 . • Reinforcing steel properties adhere to EC2 [EN 04, 3.2] with Young’s modulus Es = 200 000 MN/m2 , yield strength fyk = 500 MN/m2 and tensile strength ft = 525 MN/m2 at a strain of uk = 25 · 10−3 . The cover to reinforcement is chosen with c = 0.02 m and the effective depth of cross section with d ≈ 0.22 m. • Self weight is given by g = 0.25 · 25 = 6.25 kN/m2 and a variable service load of q = 5.0 kN/m2 is assumed. These are characteristic values and have to be multiplied by safety factors to have a design load of p = 1.35 · 6.25 + 1.50 · 5.0 = 16 kN/m2 . • To determine a reinforcement a design is performed according to [EN 04, 6.1]. Internal forces are determined based on Eqs. (7.11)4−6 whereby deflections and curvatures are determined with an extension of Eq. (8.51) taking variable coordinates into account. Values of bending and twisting moment are given by mx = my = 45.3 kNm/m,

mxy = 38.0 kNm/m

(8.58)

Reinforcement forces are determined by fs = m/z with the internal lever arm z. A good estimation is given with z = 0.8 d. The design value for the reinforcement strength is mailto:[email protected]

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192

8.8 Reinforced Concrete Shells chosen with fyd = 500/1.15 = 435 MN/m2 . Finally, the required reinforcement cross section area as obeys fyd as = m/(0.8 d). This yields10 −3

1 45.3·10 2 asx = asy = 435 0.8·0.22 → 5.9 cm /m −3 1 38.0·10 2 asxy = 435 0.8·0.22 → 5.0 cm /m

(8.59)

The bending moments mx , my occur in the slab midpoint while the twisting moments occur in the corner points. The bending moments require a bottem reinforcement while the twisting moment also requires an upper reinforcement. A bottom reinforcement of asx,lower = asy,lower = 6.0 cm2 /m and an upper reinforcement asx,upper = asy,upper = 5.0 cm2 /m are chosen throughout the slab in the following. • A more detailed design approach shows that the failure of this slab occurs due reinforcement failure while concrete compression remains approximately in the elastic regime. But the concrete will crack due to tensile stresses. This activates the reinforcement and leads to a different behavior compared to the elastic case. The approach of Section 8.8 with cracked concrete behavior as described in Section 6.2 will be used in the following. The tension stress of concrete is restricted to a value of fct = 1 MN/m2 . A reinforcement is regarded as has been determined before. The following model parameters are chosen:

lower x-direction lower y-direction upper x-direction upper y-direction

sheet thickness aR [m] 0.6 · 10−3 0.5 · 10−3 0.5 · 10−3 0.5 · 10−3

sheet height coordinate tR [m] -0.10 -0.10 0.10 0.10

Sheet height coordinates tR are assumed to be the same in x- and y-directions to preserve symmetry. The tension stiffening effect, see Section 2.8, is regarded. • Discretization is chosen with 4 × 4 elements for the quarter slab. Due to the physical nonlinearities the loading is applied in 10 steps. An equilibrium iteration has to be performed within each loading step leading to an incrementally iterative scheme, see Section 1.7, Page 15. A Gaussian quadrature is used for integration of system matrices and vectors. Integration orders, see Section 8.5, Eq. (8.38), are chosen with nu = 2 in the reference surface directions and with nv = 4 in the collateral direction. Collateral Integration of reinforcement contributions is performed separately while considering their discrete positions tR . • Results of numerical computation – The relation between the load factor and the mid point deflection is shown in Fig. 8.3a. Again thress stages can be seen: (I) stage I with basically linear behavior and only sporadic cracking. (IIa) Stage IIa with progressing cracking leading to a nearly final state of cracking. The load level is hold constant during ths stage. Deflections increase to a multiple of the value at the end of stage I, which may happen rapidly due to constant loading. (IIb) Stage IIb of basically final cracking state with a much lower stiffness compared to stage I. The final deflection of wmax = 5.15 · 10−2 m is 10 Twisting moments may be seen as skew principal moments which are directed along the diagonals of the slab’s corners. This leads to, e.g., upper diagonal tension which is hold by reinforcement in x- and y-direction. Transformation of stresses and of their reference length leads to the relation for asxy .

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193

Figure 8.3: Example 8.2 a) load deflection curve b) principal strains in upper and lower concrete layer roughly 8 times higher compared to the linear elastic case of Example 8.1 and leads to deflection-span ratio of ≈ 1/15011 . – A smeared crack concept is basically used as described in Section 6.2, Page 143. It is extended to two directions. Smeared strains in cracked integration points are shown in Fig. 8.3b for the case under consideration. The top and bottom layer are chosen out of the four concrete layers. The lateral thickness direction is not to scale with the ground directions. Length and orientation of red bars indicate size and orientation of principal values of smeared strains. Cracks are not shown explicitely but arise orthogonal to the strain directions. Top and and bottom show a different behavior due to bending. * The bottom layer has two orthogonal cracks in the central region due to bending moments and single cracks in diagonal direction in corner regions due to twisting moments. Corresponding principal strains are across the diagonal direction for the latter. * The top layer has no cracks in the central region as it is under compression. But it has single cracks across the diagonal direction due to twisting moments. Corresponding principal strains are in diagonal direction. – Computed prinicipal moments are shown in Fig. 8.4a. Bar directions indicate the directions of the corresponding stresses. A green color indicates a positive moment, a red color a negative one. A positive moment has tension on the lower surface and compression on the upper surfacce. Skew principal moment directions with opposite signs in the corner region correspond to twisting moments. The negative moment in the diagonal direction has diagonal tension on the upper side which is compensated by upper x- and y-reinforcement. The positive moment across the diagonal direction has corresponding tension on the lower side. It is compensated by lower x- and y-reinforcement. Central areas also have skew principal moments but with nearly the same sign and size. Thus, there will be only minor twisting moments. Lower surface tension is compensated by lower x- and y-reinforcement. – A new phenomenon compared to Example 8.1 is given with membrane forces, which are determined in analogy to the internal forces n ¯1, n ¯2, n ¯ 12 of Eq. (8.40)1−3 . This 11

This ratio should be smaller than 1/250 in ordinary cases according to codes like Eurocode 2. But such a value has to be maintained with load safety factors of 1.

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194

8.8 Reinforced Concrete Shells

Figure 8.4: Example 8.2 a) principal moments b) principal membrane forces is caused by the elongation effect of cracked reinforced cross sections which has already been discussed for beams in Example 3.2, Page 65. In contrast to beams an enlongation may have a different direction for every point of the slab’s reference surface as the crack directions may be different. This leads to eigenstresses, i.e. self-equilibrating internal forces without reaction forces on supports. Due to this eigenstresses kinematic compatibility is preserved even in case local elongations or also contractions. The corresponding principal membrane forces are shown in Fig. 8.4b. In case that horizontal movements of the slab are prevented on the supporting edges horizontal reaction force would arise and the whole slab will come under compression. end example 8.2

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Bibliography [Bat96] K.J. Bathe. Finite Element Procedures. Prentice Hall, Englewood Cliffs, New Jersey, 1996. [Bis99] M. Bischoff. Theorie und numerik einer dreidimensionalen schalenformulierung. Technical report, Bericht Nr. 30, Institut für Baustatik, Universität Stuttgart, 1999. [Cze99] F. Czerny. Tafeln für rechteckplatten. In Betonkalender 1999, Band I, chapter 277-339, pages 1–145. Verlag Ernst u. Sohn, Berlin, 1999. [DB84] E.N. Dvorkin and K.J. Bathe. A continuum mechanics based four-node shell element for general nonlinear analysis. Eng. Comput., 1:77–88, 1984. [EN 04] EN 1992-1-1. Eurocode 2: Design of concrete structures - Part 1-1: General rules and rules for buildings, December 2004. [Gir74] K. Girkmann. Flächentragwerke. Springer-Verlag, Wien, 6. Auflage edition, 1974. [Joe62] J. Joedicke. Schalenbau. Dokumente der Modernen Architektur 2. Karl Krämer Verlag, Stuttgart, 1962.

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Appendix A

ConFem • Coordinate system

Figure A.1: Coordinate System

196

Appendix B

Transformations of coordinate systems • In the following rectangular cartesian coordinates are considered.

Figure B.1: Coordinate transformations • Transformation of 2D vectors – A vector  a=

ax ay

 (B.1)

is initially given within a coordinate system x, y. Another coordinate system x ˜, y˜ is rotated with an angle α (positive in the counterclockwise direction). The compo
T ˜x a ˜y nents of a in the ˜·–system are a . Then the following relations hold ˜ = T · a, a

T

˜, a=T ·a

 T=

cos α sin α − sin α cos α



• Transformation of 2D stresses       ˜11  cos2 ϕ sin2 ϕ 2 cos ϕ sin ϕ  σ  σ11  σ ˜22 sin2 ϕ cos2 ϕ −2 cos ϕ sin ϕ  · σ22 =     2 2 σ ˜12 − cos ϕ sin ϕ cos ϕ sin ϕ cos ϕ − sin ϕ σ12 Inversion       cos2 ϕ sin2 ϕ −2 cos ϕ sin ϕ ˜11   σ11   σ σ22 cos2 ϕ 2 cos ϕ sin ϕ  · σ ˜22 =  sin2 ϕ     σ12 cos ϕ sin ϕ − cos ϕ sin ϕ cos2 ϕ − sin2 ϕ σ ˜12 197

(B.2)

(B.3)

(B.4)

198

APPENDIX B. TRANSFORMATIONS OF COORDINATE SYSTEMS • Transformation of 2D strains      ˜ 11   cos2 ϕ sin2 ϕ cos ϕ sin ϕ  D  D11  ˜ sin2 ϕ cos2 ϕ − cos ϕ sin ϕ  · D22 = D  ˜22    2 2 −2 cos ϕ sin ϕ 2 cos ϕ sin ϕ cos ϕ − sin ϕ 2D12 2D12 (B.5) Inversion       ˜ cos2 ϕ sin2 ϕ − cos ϕ sin ϕ  D11   D11  ˜ D22 sin2 ϕ cos2 ϕ cos ϕ sin ϕ  · = D    ˜22  2 2 2D12 2 cos ϕ sin ϕ −2 cos ϕ sin ϕ cos ϕ − sin ϕ 2D12 (B.6)

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Appendix C

Linear regression analysis applications For basics of linear regression analysis see [BSMM00, 4.4.3.1].

C.0.1

Determination of moments in triangular slab elements

• Area coordinates, see Section 7.3.1, are used in the following. It is assumed, that four sampling points according to Table 7.1 are used for numerical integration. • An linear approximation approach m = a L1 + b L2 + c L3

(C.1)

with L1 , L2 , L3 according to Eq. (7.33) is chosen for the field m. Regarding the sampling points of numerical integration the approximation yields  m(1)  m(2)   m=  m(3)  , m(4) 

m = X · a,



 1/3 1/3 1/3  0.6 0.2 0.2   X=  0.2 0.6 0.2  , 0.2 0.2 0.6

 a a =  b  (C.2) c 

The target values of the m (→ direct results of FEM-calculation) are collected in  m ¯ (1)  m ¯ (2)   ¯ = m  m ¯ (3)  m ¯ (4) 

(C.3)

The linear regression equation to determine the coefficients a is given by    53  124 88 88 −11 −11 2 1 1  88 124 88  , A−1 =  −11 53 −11  ¯ A·a = XT ·m, A = XT ·X = 2 225 6 88 88 124 −11 −11 53 2 (C.4) and ¯ a = A−1 · XT · m (C.5) • Nodal values of m may be determined by applying Eq. (C.1) with nodal area coordinates. – Node 1: L1 = 1, L2 = 0, L3 = 0 → m = a – Node 2: L1 = 0, L2 = 1, L3 = 0 → m = b 199

200

APPENDIX C. LINEAR REGRESSION ANALYSIS APPLICATIONS – Node 3: L1 = 0, L2 = 0, L3 = 1 → m = c • With known coefficients a the derivatives with respect to global coordinate axes are determined by ∂m ∂m ∂L1 ∂m ∂L2 ∂m ∂L3 = + + = a b1 + b b2 + c b3 ∂x ∂L1 ∂x ∂L2 ∂x ∂L3 ∂x

(C.6)

and

∂m = a c1 + b c2 + c c3 ∂x with bi , ci according to Eq. (7.37).

State April 4, 2013

(C.7)

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Appendix D

Numerical Integration of Elastoplastic Material Laws • We refer to Section 5.6. The material behavior is described by a set of first order differential equations Eqs. (5.75), (5.76), (5.79), (5.80) σ˙ − C0 · (˙ − ˙ p ) ˙ p − λ˙ ∂∂G σ κ˙ p − λ˙ H(σ, κp ) F˙ (σ, κp )

=0 =0 =0 =0

(D.1)

in case of loading. This is a set of 14 equations driven by total strain (t) to determine 14 unknowns σ(t), p (t), λ(t), κp (t). This is discretized in time for t = tn+1 with the implicit Euler-scheme, i.e. 1 σ σ˙ n+1 = ∆t (σ n+1 − σ n ) = ∆∆t ∆λ 1 = ∆t λ˙ n+1 = ∆t (λn+1 − λn ) (D.2) ∆κ 1 κ˙ p,n+1 = ∆t (κp,n+1 − κp,n ) = ∆tp 1  ˙ n+1 = ∆t (n+1 − n ) =∆ ∆t and Fn+1 = Fn +

∂F ∂F · ∆σ + ∆κp = 0 ∂σ ∂κp

(D.3)

leading to a set of algebraic equations ∂G a = C−1 0 · ∆σ + ∆λ ∂ σ − ∆ = 0 b= H ∆λ − ∆κp =0 ∂F f = ∂∂F · ∆σ + ∆κ + F = 0 p n σ ∂κp

(D.4)

Herein, ∆ is given from a superordinated calculation, the values σ n , κp,n and Fn are assumed to be known1 and ∆σ, ∆λ, ∆κp remain to be determined. Depending upon ∂F ∂F the characteristics of ∂∂G σ , ∂ σ , ∂κp , H this set of equations may be nonlinear. Associated plasticity F = G with ∂F/∂σ = ∂G/∂σ = r is assumed in the following to simplify the notation. The Newton-Raphson method, see Eq. (1.65), may be used to determine solutions. It is applied with an iteration rule    ∂r ∂r    C−1 r 0 + ∂ σ ∆λ ∂κp ∆λ δσ a ∂H ∂H  H    ·  δκp  = −  b  (D.5) ∂ σ ∆λ ∂κp ∆λ − 1 2 ∂r ∂r ∂ F ∂F δλ f i r + ∂ σ · ∆σ + ∂κ ∆κp ∂κ2 ∆κp + ∂κ 0 p

1

p

p

A value Fn < 0 may occur due to a finite discrecization in time.

201

i

202

APPENDIX D. NUMERICAL INTEGRATION OF ELASTOPLASTIC MATERIAL LAWS and

     δσ ∆σ ∆σ  ∆κp  =  ∆κp  +  δκp  δλ ∆λ ∆λ i i+1 

(D.6)

• Mises plasticity, see Example 5.3, is considered as an exemplary case. The material functions are given by  0  σ11 0   σ22 r  0  2  3   σ33  , ∂r = 0, ∂F = −1, ∂ F = 0, H = const., ∂H = ∂H = 0 r=   4J2  σ23  ∂κp ∂κp ∂κ2p ∂σ ∂κp  σ13  σ12 (D.7) see Eqs. (5.86), (5.95). Thus, the residuals and the iteration matrix of Eq. (D.5) take a form    −1    −1 a r C0 + ∂∂r C0 · ∆σ + r ∆λ − ∆ σ ∆λ 0  , Ai =  ˜i = −  b  = −  0 −1 H  a H ∆λ − ∆κp ∂r f i r · ∆σ − ∆κp + Fn r + ∂ σ · ∆σ −1 0 i i (D.8) −1 Finally we have C0 derived from Eq. (5.21), and   3 2 3 ∂r bI = √ Idev − r r = √ (D.9) ∂σ 3 2 3J2 2 3J2 with the deviatoric unit matrix Idev according to Eq. (5.17). The matrix bI has the following properties bI2 = bI,

bI·r = 0,

bI·I = 0,

bI·Idev = bI and bI·∆σ = 0 → ∂r ·∆σ = 0 (D.10) ∂σ

see [BLM00, (5.9.21) and following clause], with the unit matrix I. • The foregoing relations for the Mises plasticity may be simplified to a large extent, see [BLM00, 5.9.3]. Eqs. (D.4), (D.8) are reformulated as       U11 U12 v w (D.11) = · δλ −f i U21 0 i with  U11 =

∂r C−1 0 0 + ∂ σ ∆λ 0 −1



 , U12 =

r H

 , U21 =



r −1



using Eq. (D.11)6 and      −1  δσ a C0 · ∆σ + r ∆λ − ∆ v= , w=− =− δκp b H ∆λ − ∆κp Eq. (D.11) yields a solution for the plastic multiplier increment U21 · U−1 11 · w + f δλ = −1 U21 · U11 · U12 i State April 4, 2013

(D.12)

(D.13)

(D.14)

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203 A first key point is that U−1 11 in case of Mises plasticity is given by U−1 11

 =

C0 − 2µc bI 0 0 −1

 , c=

E 2µa 3 ∆λ, µ = , a= 1 + 2µa 2κp 2(1 + ν)

(D.15)

leading to U21 · U−1 11 =





r · C0 1

,

U21 · U−1 11 · U12 = r · C0 · r + H = 3µ + H (D.16)

using Eqs. (D.10). A second key point is that the stress increment ∆σ may be generally determined with ∆σ = C0 · (∆ − ∆p ) = C0 · (∆ − ∆λ r)

→

∆ = C−1 · ∆σ + ∆λ r (D.17)

based on Eqs. (D.1)1 , (5.76) instead of using the linearized system Eqs. (D.5), (D.6). This
T and in case of Mises plasticity yields w = − 0 H ∆λ − ∆κp U21 · U−1 11 · w + f = f + ∆κp − H ∆λ = Fn + r · ∆σ − H ∆λ

(D.18)

using Eqs. (D.4)3 , (D.7)3 and completes the iteration rule Eq. (D.14). • A corresponding iteration algorithm is given as follows. It is performed on an integration point level, see Page 14. 1. i=0: Given are ∆ from a superordinated calculation and σ n , Fn , κp,n from the previous time step. The values of E, ν, H and shear modulus µ are material constants. Initial iteration values are assumed with ∆λ0 = 0, ∆p,0 = 0. 2. Stress is predicted with ∆σ i = C0 · (∆ − ∆ p,i ) ,

σ n+1,i = σ n + ∆σ i

(D.19)

This also yields a value for ri . 3. If i = 0 and Fn+1,i < 0 from Eq. (5.86) an elastic state is given. Upate Fn with Fn+1,i . Finish the iteration. 4. A Newton increment for the plastic multiplier is computed according to Eqs. (D.14), (D.18), (D.16) Fn + ri · ∆σ i − H ∆λi δλ = (D.20) 3µ + H 5. If |δλ| < T ol a convergent state has been reached within a prescribed tolerance T ol. Finish the iteration and update internal state variable by going to 8. 6. Update the plastic multiplier increment and plastic strains ∆λi+1 = ∆λi +δλ,

∆p,i+1 = ∆λi+1


T ri,1 ri,2 ri,3 2ri,4 2ri,5 2ri,6 (D.21)

7. i := i + 1. Continue with step 2. p 8. Update internal state variable κp,n+1 = 3J2,n+1 according to Eq. (5.86) using a converged stress σ n+1 . Update Fn with 0. Compute tangential material stiffness with Eq. (5.88) for a consistent system’s tangential stiffness. 9. Proceed with next equilibrium iteration step of superordinated calculation.

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Appendix E

ACCESS lectures in summer term 2013 Preliminary overview 1. Tue 09.04.2013 Lecture: Reinforced Concrete Tension Bars 2. Mon 15.04.2013 Exercise: 3. Tue 16.04.2013 Lecture: Finite Element ingredients for RC tension bars 4. Tue 23.04.2013 Lecture: 5. Mon 29.04.2013 Exercise: 6. Tue 30.04.2013 Lecture: 7. Tue 07.05.2013 Lecture: 8. Mon 13.05.2013 Exercise: 9. Tue 14.05.2013 Lecture: 10. Mon 27.05.2013 Exercise: 11. Tue 28.05.2013 Lecture: 12. Tue 04.06.2013 Lecture: 13. Mon 10.06.2013 Exercise: 14. Tue 11.06.2013 Lecture: 15. Tue 18.06.2013 Lecture: 16. Mon 24.06.2013 Exercise: 17. Tue 25.06.2013 Lecture: 18. Tue 02.07.2013 Lecture: 19. Mon 08.07.2013 Exercise: 20. Tue 09.07.2013 Lecture: 21. Tue 16.07.2013 Lecture:

204

Bibliography [BLM00]

T. Belytschko, W.K. Liu, and B. Moran. Nonlinear Finite Elements for Continua and Structures. John Wiley & Sons, Chichester, 2000.

[BSMM00] Bronstein, Semendjajew, Musiol, and Muehlig. Taschenbuch der Mathematik. Verlag Harri Deutsch, Frankfurt/Main, 5. auflage edition, 2000.

205