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CIVL 8/7117

Chapter 2 - The Stiffness Mehtod

The Stiffness (Displacement) Method • This section introduces some of the basic concepts on which the direct stiffness method is based. • The linear spring is simple and an instructive tool to illustrate the basic concepts. • The steps to develop a finite element model for a linear spring follow our general 8 step procedure. 1. Discretize and Select Element Types - Linear spring elements 2. Select a Displacement Function - Assume a variation of the displacements over each element. 3. Define the Strain/Displacement and Stress/Strain Relationships - use elementary concepts of equilibrium and compatibility.

The Stiffness (Displacement) Method 4. Derive the Element Stiffness Matrix and Equations Define the stiffness matrix for an element and then consider the derivation of the stiffness matrix for a linearelastic spring element. 5. Assemble the Element Equations to Obtain the Global or Total Equations and Introduce Boundary Conditions - We then show how the total stiffness matrix for the problem can be obtained by superimposing the stiffness matrices of the individual elements in a direct manner. The term direct stiffness method evolved in reference to this method.

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CIVL 8/7117

Chapter 2 - The Stiffness Mehtod

The Stiffness (Displacement) Method 6. Solve for the Unknown Degrees of Freedom (or Generalized Displacements) - Solve for the nodal displacements. 7. Solve for the Element Strains and Stresses - The reactions and internal forces association with the bar element. 8. Interpret the Results

The Stiffness (Displacement) Method 1. Select Element Type - Consider the linear spring shown below. The spring is of length L and is subjected to a nodal tensile force, T directed along the axis.

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Chapter 2 - The Stiffness Mehtod

The Stiffness (Displacement) Method 2. Select a Displacement Function - A displacement function is assumed. u  a1  a2 x

In general, the number of coefficients in the displacement function is equal to the total number of degrees of freedom associated with the element. We can write the displacement function in matrix forms as: a  u  1 x   1  a2 

The Stiffness (Displacement) Method We can express u as a function of the nodal displacements ui by evaluating u at each node and solving for a1 and a2.

u( x  0)  u1  a1 u( x  L )  u2  a2L  a1 Solving for a2: a2 

u2  u1 L

Substituting a1 and a2 into u gives:

 u  u1  u  2  x  u1  L 

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Chapter 2 - The Stiffness Mehtod

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The Stiffness (Displacement) Method In matrix form:

Or in another form:

x   u1    L  u2 

 x u   1    L  u  N1

u  N2   1  u2 

Where N1 and N2 are defined as: x N1  1  L

N2 

x L

The functions Ni are called interpolation functions because they describe how the assumed displacement function varies over the domain of the element. In this case the interpolation functions are linear.

The Stiffness (Displacement) Method u  a1  a2 x

 x u   1    L 

u  N2   1  u2 

u  N1 N1  1 

N2 

x L

x   u1    L  u2 

x L

CIVL 8/7117

Chapter 2 - The Stiffness Mehtod

The Stiffness (Displacement) Method 3. Define the Strain/Displacement and Stress/Strain Relationships - Tensile forces produce a total elongation (deformation)  of the spring. For linear springs, the force T and the displacement u are related by Hooke’s law:

T  k

where deformation of the spring  is given as:

  u2  u1

  u(L )  u(0)

The Stiffness (Displacement) Method 4. Step 4 - Derive the Element Stiffness Matrix and Equations - We can now derive the spring element stiffness matrix as follows: f1x  T

f2 x  T

Rewrite the forces in terms of the nodal displacements: T  f1x  k  u2  u1 

T  f2 x  k  u2  u1 

We can write the last two force-displacement relationships in matrix form as:  f1x   k k   u1      f2 x   k k  u2 

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Chapter 2 - The Stiffness Mehtod

The Stiffness (Displacement) Method This formulation is valid as long as the spring deforms along the x axis. The coefficient matrix of the above equation is called the local stiffness matrix k:

 k k  k   k k  5. Step 4 - Assemble the Element Equations and Introduce Boundary Conditions The global stiffness matrix and the global force vector are assembled using the nodal force equilibrium equations, and force/deformation and compatibility equations. N

K  K    k e 1

N

(e )

F  F    f ( e ) e 1

where k and f are the element stiffness and force matrices expressed in global coordinates.

The Stiffness (Displacement) Method 6. Step 6 - Solve for the Nodal Displacements Solve the displacements by imposing the boundary conditions and solving the following set of equations:

F  Kd 7. Step 7 - Solve for the Element Forces Once the displacements are found, the forces in each element may be calculated from:

T  k

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 1 Consider the following two-spring system shown below:

where the element axis x coincides with the global axis x. For element 1:

 f1x   k1 k1   u1      f3 x   k1 k1  u3 

For element 2:

f3 x   k 2 k2  u3      f2 x   k 2 k 2  u2 

The Stiffness Method – Spring Example 1 Both continuity and compatibility require that both elements remain connected at node 3.

u3(1)  u3(2) We can write the nodal equilibrium equation at each node as:

F1x  f1x (1)

F3 x  f3 x (1)  f3 x (2)

F2 x  f2 x (2)

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Chapter 2 - The Stiffness Mehtod

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The Stiffness Method – Spring Example 1 Therefore the force-displacement equations for this spring system are:

F2 x  k 2u3  k2u2

F1x  k1u1  k1u3

F3 x  k1u1  k1u3  k2u3  k2u2 In matrix form the above equations are: 0  F1x   k1    k2 F2 x    0 F   k k 2  3x   1

k1   u1    k 2  u2   k1  k2  u3 

F  Kd

where F is the global nodal force vector, d is called the global nodal displacement vector, and K is called the global stiffness matrix.

The Stiffness Method – Spring Example 1 Assembling the Total Stiffness Matrix by Superposition Consider the spring system defined in the last example:

The elemental stiffness matrices may be written for each element. For element 1: For element 2: u1

u3

 k k1  k (1)   1   k1 k1 

u3

u2

 k k 2  k (2)   2   k 2 k 2 

CIVL 8/7117

Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 1 Write the stiffness matrix in global format for element 1 as follows: (1)  1 0 1  u1  f1x      k1  0 0 0  u2   f2(1)x       (1)   1 0 1  u3  f3 x  For element 2: (1) 0 0 0   u1  f1x      k 2 0 1 1 u2   f2(1)x    0 1 1  u3  f3(1)x 

The Stiffness Method – Spring Example 1 Apply the force equilibrium equations at each node. f1(1)   0   F1x  x    (2)     0   f2 x   F2 x  f (1)  f (2)  F   3x   3x   3x 

The above equations give: 0  k1  0 k2   k1 k 2

k1   u1   F1x      k 2  u2   F2 x   k1  k 2  u3  F3 x 

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 1 To avoid the expansion of the each elemental stiffness matrix, we can use a more direct, shortcut form of the stiffness matrix. u1 u3 u3 u 2  k k1  k (1)   1   k1 k1 

 k k 2  k (2)   2   k 2 k 2 

The global stiffness matrix may be constructed by directly adding terms associated with the degrees of freedom in k(1) and k(2) into their corresponding locations in the K as follows: u u u 1

2

 k1 0 K   0 k2   k1 k2

3

k1  k 2   k1  k2 

The Stiffness Method – Spring Example 1 Boundary Conditions In order to solve the equations defined by the global stiffness matrix, we must apply some form of constraints or supports or the structure will be free to move as a rigid body. Boundary conditions are of two general types: 1. homogeneous boundary conditions (the most common) occur at locations that are completely prevented from movement; 2. nonhomogeneous boundary conditions occur where finite non-zero values of displacement are specified, such as the settlement of a support.

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 1 Consider the equations we developed for the two-spring system. We will consider node 1 to be fixed u1 = 0. The equations describing the elongation of the spring system become: 0 k1   0   F1x   k1      0 k2 k 2  u2   F2 x     k1 k 2 k1  k 2  u3  F3 x  Expanding the matrix equations gives:

F1x  k1u3 F2 x  k2u3  k2u2 F3 x  k 2u2   k1  k 2  u3

The Stiffness Method – Spring Example 1 The second and third equation may be written in matrix form as: k 2  u2  F2 x   k2  k k  k  u   F  2 3  2 1  3x  Once we have solved the above equations for the unknown nodal displacements, we can use the first equation in the original matrix to find the support reaction.

F1x  k1u3 For homogeneous boundary conditions, we can delete the row and column corresponding to the zero-displacement degrees-of-freedom.

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 1 Let’s again look at the equations we developed for the twospring system. However, this time we will consider a nonhomogeneous boundary condition at node 1: u1 = . The equations describing the elongation of the spring system become: 0 k1      F1x   k1      0 k2 k 2  u2   F2 x         k1 k 2 k1  k 2  u3  F3 x  Expanding the matrix equations gives:

F1x  k1  k1u3

F2 x  k2u3  k2u2

F3 x  k1  k1u3  k2u3  k2u2

The Stiffness Method – Spring Example 1 By considering the second and third equations because they have known nodal forces we get:

F2 x  k2u3  k2u2

F3 x  k1  k1u3  k2u3  k2u2

In matrix form the above equations are: k 2  u2   F2 x   k2  k k  k  u   F  k   2 3 1   2 1  3x

For nonhomogeneous boundary conditions, we must transfer the terms from the stiffness matrix to the right-hand-side force vector before solving for the unknown displacements.

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The Stiffness Method – Spring Example 1 Once we have solved the above equations for the unknown nodal displacements, we can use the first equation in the original matrix to find the support reaction.

F1x  k1  k1u3

The Stiffness Method – Spring Example 2 Consider the following three-spring system:

The elemental stiffness matrices for each element are: 1

3

3

 1 1 k (1)  1000    1 1 4

1 3

2

 1 1 k (3)  3000    1 1

4 2

4

 1 1 k (2)  2000    1 1

3 4

CIVL 8/7117

Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 2 Using the concept of superposition (the direct stiffness method), the global stiffness matrix is: 0 1000 0   1000  0 3000  3000 0  K  1000 0 3000 2000    3000 2000 5000   0

The global force-displacement equations are: 0 1000 0   u1   F1x   1000  0 3000  u2  F2 x  3000 0        1000 0 3000 2000  u3  F3 x    3000 2000 5000  u4  F4 x   0

The Stiffness Method – Spring Example 2 We have homogeneous boundary conditions at nodes 1 and 2 (u1 = 0 and u2 = 0). The global force-displacement equations reduce to: 0 1000 0   u1   F1x   1000  0 3000 0 3000  u2  F2 x         1000 0 3000 2000  u3  F3 x    3000 2000 5000  u4  F4 x   0

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 2 Substituting for the known force at node 4 (F4x = 5,000 lb) gives:    3000 2000  u3    0   2000 5000  u  5,000     4  Solving for u3 and u4 gives:

u3 

10 in 11

u4 

15 in 11

The Stiffness Method – Spring Example 2 To obtain the global forces, substitute the displacement in the force-displacement equations. 0 1000 0  0   F1x   1000 F   0 3000 0 3000   0   2x        10 F     1000 0 3000 2000 3 x    11   3000 2000 5000  15 11 F4 x   0

Solving for the forces gives: 10,000 45,000 F1x   lb F2 x   lb 11 11

F3 x  0

F4 x 

55,000 lb 11

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The Stiffness Method – Spring Example 2 Next, use the local element equations to obtain the force in each spring. For element 1:  f   1000 1000   0  1x    10  f3 x   1000 1000   11 The local forces are:

f1x  

10,000 lb 11

f3 x 

10,000 lb 11

A free-body diagram of the spring element 1 is shown below.

The Stiffness Method – Spring Example 2 Next, use the local element equations to obtain the force in each spring. For element 2: f3 x   2000 2000  10 11     f4 x   2000 2000  15 11 The local forces are:

f3 x  

10,000 lb 11

f4 x 

10,000 lb 11

A free-body diagram of the spring element 2 is shown below.

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The Stiffness Method – Spring Example 2 Next, use the local element equations to obtain the force in each spring. For element 3: f   3000 3000  15  11 4x      f2 x   3000 3000   0  The local forces are:

f4 x 

45,000 lb 11

f2 x  

45,000 lb 11

A free-body diagram of the spring element 3 is shown below.

The Stiffness Method – Spring Example 3 Consider the following four-spring system:

The spring constant k = 200 kN/m and the displacement  = 20 mm. Therefore, the elemental stiffness matrices are:  1 1 k (1)  k (2)  k (3)  k ( 4)  200   kN / m  1 1

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 3 Using superposition (the direct stiffness method), the global stiffness matrix is: Element 1

Element 2

Element 3

Element 4

0 0 0  200 200  200 400 200 0 0   K 0 200 400 200 0   0 0 200 400 200    0 0 0 200 200 

The Stiffness Method – Spring Example 3 The global force-displacement equations are: 0 0 0   u1   F1x   200 200  200 400 200 0 0  u2  F2 x        0 200 400 200 0  u3   F3 x    0 0 200 400 200  u4  F4 x        0 0 0 200 200  u5  F5 x 

Applying the boundary conditions (u1 = 0 and u5 = 20 mm) and the known forces (F2x, F3x, and F4x equal to zero) gives: 0 0   u2   0   400 200  200 400 200 0   u3   0       0 200 400 200   u4   0    0 200 200  0.02  F5 x   0

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 3 Rearranging the first three equations gives: 0  u2  0   400 200  200 400 200  u   0    3   0 200 400  u4  4  

Solving for u2, u3, and u4 gives:

u2  0.005 m

u3  0.01m

u4  0.015 m

Solving for the forces F1x and F5x gives:

F1x  200(0.005)  1.0kN F5 x  200(0.015)  200(0.02)  1.0kN

The Stiffness Method – Spring Example 3 Next, use the local element equations to obtain the force in each spring. For element 1:  f1x   200 200   0      f2 x   200 200  0.005 

f1x  1.0kN

f2 x  1.0kN

For element 2: f2 x   200 200  0.005      f3 x   200 200   0.01 

f2 x  1.0kN

f3 x  1.0kN

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 3 Next, use the local element equations to obtain the force in each spring. For element 3:

f3 x   200 200   0.01      f4 x   200 200  0.015 

f3 x  1.0kN For element 4:

f4 x  1.0kN

f4 x   200 200  0.015      f5 x   200 200   0.02  f4 x  1.0kN

f5 x  1.0kN

The Stiffness Method – Spring Example 4 Consider the following spring system:

The boundary conditions are: u1  u3  u4  0 The compatibility condition at node 2 is: u2(1)  u2(2)  u2(3)  u2

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The Stiffness Method – Spring Example 4 Using the direct stiffness method: the elemental stiffness matrices for each element are: 1

2

2

 k k1  k (1)   1   k1 k1 

1 2

2

3

 k k 2  k (2)   2   k 2 k 2 

 k k3  k (3)   3   k 3 k3 

2 3

Using the concept of superposition (the direct stiffness method), the global stiffness matrix is: 1

2

3

4

  k1 0  0  k1  k k  k 2  k 3 k k 3  2 K 1 1  0  k 2 k2  0  0 k3 0 k3   

1 2 3 4

The Stiffness Method – Spring Example 4 Applying the boundary conditions (u1 = u3 = u4 = 0) the stiffness matrix becomes: 1

2

3

4

4

  k1 0  0  k1  k k  k 2  k 3 k k 3  2 K 1 1  0  k 2 k2  0  0 k3 0 k3   

1 2 3 4

2 4

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Chapter 2 - The Stiffness Mehtod

The Stiffness Method – Spring Example 4 Applying the known forces (F2x = P) gives: P   k1  k 2  k3  u2

Solving the equation gives:

u2 

P k1  k 2  k3

Solving for the forces gives: F1x  k1u2

F3 x  k2u2

F4 x  k3u2

Potential Energy Approach to Derive Spring Element Equations One of the alternative methods often used to derive the element equations and the stiffness matrix for an element is based on the principle of minimum potential energy. This method has the advantage of being more general than the methods involving nodal and element equilibrium equations, along with the stress/strain law for the element. The principle of minimum potential energy is more adaptable for the determination of element equations for complicated elements (those with large numbers of degrees of freedom) such as the plane stress/strain element, the axisymmetric stress element, the plate bending element, and the threedimensional solid stress element.

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Chapter 2 - The Stiffness Mehtod

Total Potential Energy The total potential energy is defined as the sum of the internal strain energy U and the potential energy of the external forces :

p  U   Strain energy is the capacity of the internal forces (or stresses) to do work through deformations (strains) in the structure. The potential energy of the external forces  is the capacity of forces such as body forces, surface traction forces, and applied nodal forces to do work through the deformation of the structure.

Total Potential Energy Recall the force-displacement relationship for a linear spring:

F  kx

The differential internal work (or strain energy) dU in the spring is the internal force multiplied by the change in displacement which the force moves through: dU  Fdx   kx  dx

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Total Potential Energy x

The total strain energy is:

U   dU    kx  dx  L

0

1 2 kx 2

The strain energy is the area under the force-displacement curve. The potential energy of the external forces is the work done by the external forces:

  Fx

Total Potential Energy Therefore, the total potential energy is:  p 

1 2 kx  Fx 2

The concept of a stationary value of a function G is shown below: dG 0 dx

The function G is expressed in terms of x. To find a value of x yielding a stationary value of G(x), we use differential calculus to differentiate G with respect to x and set the expression equal to zero.

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Chapter 2 - The Stiffness Mehtod

Total Potential Energy We can replace G with the total potential energy p and the coordinate x with a discrete value di. To minimize p we first take the variation of p (we will not cover the details of variational calculus):

 p 

 p d1

 d1 

 p d 2

 d 2  ... 

 p d n

 dn

The principle states that equilibrium exist when the di define a structure state such that p = 0 for arbitrary admissible variations di from the equilibrium state.

Total Potential Energy

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Total Potential Energy

To satisfy p = 0, all coefficients associated with di must be zero independently, therefore:  p d i

0

i  1, 2,, n

or

 p

 d 

0

Total Potential Energy – Example 5 Consider the following linear-elastic spring system subjected to a force of 1,000 lb. Evaluate the potential energy for various displacement values and show that the minimum potential energy also corresponds to the equilibrium position of the spring.

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Total Potential Energy – Example 5 The total potential energy is defined as the sum of the internal strain energy U and the potential energy of the external forces :

p  U  

U

1 2 kx 2

The variation of p with respect to x is:

  Fx

 p 

 p x

Since x is arbitrary and might not be zero, then:

Total Potential Energy – Example 5 Using our express for p, we get:

p   p x

1 2 1 kx  Fx  500( lb in ) x 2  1,000lb  x 2 2  0  500 x  1,000

If we had plotted the total potential energy function p for various values of deformation we would get:

x  2.0 in

x  0  p x

0

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Chapter 2 - The Stiffness Mehtod

Total Potential Energy Let’s derive the spring element equations and stiffness matrix using the principal of minimum potential energy. Consider the linear spring subjected to nodal forces shown below:

The total potential energy p

p 

1 2 k  u2  u1   f1x u1  f2 x u2 2

Expanding the above express gives:

p 





1 k u22  2u1u2  u12  f1x u1  f2 x u2 2

Total Potential Energy Let’s derive the spring element equations and stiffness matrix using the principal of minimum potential energy. Consider the linear spring subjected to nodal forces shown below:

Recall:

 p d i

0

Therefore:  p u1  p u2

i  1, 2,, n

or



k  2u2  2u1   f1x  0 2



k  2u2  2u1   f2 x  0 2

 p

 d 

0

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Chapter 2 - The Stiffness Mehtod

Total Potential Energy Let’s derive the spring element equations and stiffness matrix using the principal of minimum potential energy. Consider the linear spring subjected to nodal forces shown below:

Therefore:

k  u1  u2   f1x k  u1  u2   f2x

In matrix form the equations are:  f1x   k k   u1      f2 x   k k  u2 

Total Potential Energy – Example 6 Obtain the total potential energy of the spring system shown below and find its minimum value.

The potential energy p for element 1 is:

 p(1) 

1 2 k1  u3  u1   f1x u1  f3 x u3 2

The potential energy p for element 2 is:

 p(2) 

1 2 k2  u4  u3   f3 x u3  f4 x u4 2

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Chapter 2 - The Stiffness Mehtod

Total Potential Energy – Example 6 Obtain the total potential energy of the spring system shown below and find its minimum value.

The potential energy p for element 3 is:

 p(3) 

1 2 k3  u2  u4   f2 x u2  f4 x u4 2

The total potential energy p for the spring system is: 3

 p    p( e ) e 1

Total Potential Energy – Example 6 Minimizing the total potential energy p:  p u1  p u2  p u3

 p u4

 0  k1u3  k1u1  f1x (1)  0  k3u2  k3u4  f2 x (3)  0  k1u3  k1u1  k 2u4  k 2u3  f3 x (1)  f3 x (2)

 0  k 2u4  k 2u3  k3u2  k3u4  f4 x (2)  f4 x (3)

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Total Potential Energy – Example 6 In matrix form:  k1 0 0   u1   k1 f1x   0 k      k3 u2   f2 x 0  3      (1) (2)   k1 0 k1  k 2 k 2  u3   f3 x  f3 x    (2) (3)  0 k3 k 2 k 2  k3  u4  f4 x  f4 x  Using the following force equilibrium equations: f1x (1)  F1x f2 x (3)  F2 x

f3 x (1)  f3 x (2)  F3 x f4 x (2)  f4 x (3)  F4 x

Total Potential Energy – Example 6 The global force-displacement equations are: 0 1000 0   u1   F1x   1000  0 3000  u2  F2 x  3000 0        1000 0 3000 2000  u3  F3 x    3000 2000 5000  u4  F4 x   0

The above equations are identical to those we obtained through the direct stiffness method.

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Homework Problems 5. Do problems 2.4, 2.10, and 2.22 on pages 66 - 71 in your textbook “A First Course in the Finite Element Method” by D. Logan.

End of Chapter 2

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