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The frame of this building is statically indeterminate. The force analysis can be done using the stiffness method.

Plane Frame Analysis Using the Stiffness Method

16

The concepts presented in the previous chapters on trusses and beams will be extended in this chapter and applied to the analysis of frames. It will be shown that the procedure for the solution is similar to that for beams, but will require the use of transformation matrices since frame members are oriented in different directions.

16.1 Frame-Member Stiffness Matrix In this section we will develop the stiffness matrix for a prismatic frame member referenced from the local x¿, y¿, z¿ coordinate system, Fig. 16–1. Here the member is subjected to axial loads qNx¿, qFx¿, shear loads qNy¿, qFy¿, and bending moments qNz¿, qFz¿ at its near and far ends, respectively. These loadings all act in the positive coordinate directions along with their associated displacements. As in the case of beams, the moments qNz¿ and qFz¿ are positive counterclockwise, since by the right-hand rule the moment vectors are then directed along the positive z¿ axis, which is out of the page. We have considered each of the load-displacement relationships caused by these loadings in the previous chapters. The axial load was discussed in reference to Fig. 14–2, the shear load in reference to Fig. 15–5, and the bending moment in reference to Fig. 15–6. By superposition, if these

595

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THE

STIFFNESS METHOD

qFy¿ dFy¿

x¿

qFz¿ dFz¿

qFx¿ dFx¿ F y¿ qNy¿ dNy¿

qNx¿ dNx¿

This pedestrian bridge takes the form of a “Vendreel truss.” Strictly not a truss since the diagonals are absent, it forms a statically indeterminate box framework, which can be analyzed using the stiffness method.

N qNz¿ dNz¿ positive sign convention

Fig. 16–1

results are added, the resulting six load-displacement relations for the member can be expressed in matrix form as 16

Nx¿

Ny¿

AE qNx¿ 0

qNz¿

0 

qFx¿

0

L

qNy¿



AE

0

qFz¿

0



Fx¿



0

12EI

6EI

L3

L2

6EI

4EI

L2

L

0

L

qFy¿

Nz¿

L3



AE

L2

6EI

2EI

L2

L

0

dNx¿

0



12EI

6EI

L3

L2

0



6EI

2EI

L2

L

0

L

6EI

Fz¿

0

L

AE

0

12EI

Fy¿

12EI

0

0

0

L3 



dNy¿ dNz¿ dFx¿

6EI L2

6EI

4EI

L2

L

dFy¿ dFz¿

(16–1) or in abbreviated form as q = k¿d

(16–2)

The member stiffness matrix k¿ consists of thirty-six influence coefficients that physically represent the load on the member when the member undergoes a specified unit displacement. Specifically, each column in the matrix represents the member loadings for unit displacements identified by the degree-of-freedom coding listed above the columns. From the assembly, both equilibrium and compatibility of displacements have been satisfied.

16.2

DISPLACEMENT AND FORCE TRANSFORMATION MATRICES

597

16.2 Displacement and Force Transformation Matrices As in the case for trusses, we must be able to transform the internal member loads q and deformations d from local x¿, y¿, z¿ coordinates to global x, y, z coordinates. For this reason transformation matrices are needed.

Displacement Transformation Matrix. Consider the frame

y¿

member shown in Fig. 16–2a. Here it is seen that a global coordinate displacement DNx creates local coordinate displacements dNx¿ = DNx cos ux

uy ux

dNy¿ = - DNx cos uy

DNx

Likewise, a global coordinate displacement DNy, Fig. 16–2b, creates local coordinate displacements of dNx¿ = DNy cos uy

x¿

y

dNy¿  DNx cos uy

x

dNx¿  DNx cos ux

dNy¿ = DNy cos ux

(a)

16

Finally, since the z¿ and z axes are coincident, that is, directed out of the page, a rotation DNz about z causes a corresponding rotation dNz¿ about z¿. Thus, dNz¿ = DNz

y

In a similar manner, if global displacements DFx in the x direction, DFy in the y direction, and a rotation DFz are imposed on the far end of the member, the resulting transformation equations are, respectively, dFx¿ = DFx cos ux

dFy¿ = - DFx cos uy

dFx¿ = DFy cos uy

dFy¿ = DFy cos ux

DNy

uy ux x

dNx¿  DNy cos uy (b)

Letting lx = cos ux, ly = cos uy represent the direction cosines of the member, we can write the superposition of displacements in matrix form as ly lx 0 0 0 0

y¿

dNy¿  DNy cos ux

dFz¿ = DFz

lx dNx¿ dNy¿ -ly dNz¿ 0 V = F F dFx¿ 0 dFy¿ 0 dFz¿ 0

x¿

0 0 1 0 0 0

0 0 0 lx -ly 0

0 0 0 ly lx 0

0 DNx 0 DNy 0 DNz VF V 0 DFx 0 DFy 1 DFz

(16–3)

or d = TD

(16–4)

By inspection, T transforms the six global x, y, z displacements D into the six local x¿, y¿, z¿ displacements d. Hence T is referred to as the displacement transformation matrix.

Fig. 16–2

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STIFFNESS METHOD

Force Transformation Matrix. If we now apply each component of load to the near end of the member, we can determine how to transform the load components from local to global coordinates. Applying qNx¿, Fig. 16–3a, it can be seen that QNx = qNx¿ cos ux

x¿

y

QNy = qNx¿ cos uy

y¿ uy

If qNy¿ is applied, Fig. 16–3b, then its components are ux

QNx  qNx¿ cos ux

QNx = - qNy¿ cos uy

x

QNy = qNy¿ cos ux

QNy  qNx¿ cos uy qNx¿

Finally, since qNz¿ is collinear with QNz, we have

(a)

16

y¿

In a similar manner, end loads of qFx¿, qFy¿, qFz¿ will yield the following respective components:

uy qNy¿ QNx  qNy¿ cos uy

QNz = qNz¿

x¿

y

ux

x QNy  qNy¿ cos ux

QFx = qFx¿ cos ux

QFy = qFx¿ cos uy

QFx = - qFy¿ cos uy

QFy = qFy¿ cos ux

QFz = qFz¿

(b)

Fig. 16–3

These equations, assembled in matrix form with lx = cos ux, ly = cos uy, yield

QNx lx QNy ly Q 0 F Nz V = F QFx 0 QFy 0 QFz 0

-ly lx 0 0 0 0

0 0 1 0 0 0

0 0 0 lx ly 0

0 0 0 -ly lx 0

0 qNx¿ 0 qNy¿ 0 q V F Nz¿ V 0 qFx¿ 0 qFy¿ 1 qFz¿

(16–5)

or Q = TTq

(16–6)

Here, as stated, TT transforms the six member loads expressed in local coordinates into the six loadings expressed in global coordinates.

16.3

599

FRAME-MEMBER GLOBAL STIFFNESS MATRIX

16.3 Frame-Member Global Stiffness Matrix The results of the previous section will now be combined in order to determine the stiffness matrix for a member that relates the global loadings Q to the global displacements D. To do this, substitute Eq. 16–4 1d = TD2 into Eq. 16–2 1q = k¿d2. We have q = k¿TD

(16–7)

Here the member forces q are related to the global displacements D. Substituting this result into Eq. 16–6 1Q = TTq2 yields the final result, Q = TTk¿TD

(16–8)

or Q = kD where k = TTk¿T

16

(16–9)

Here k represents the global stiffness matrix for the member. We can obtain its value in generalized form using Eqs. 16–5, 16–1, and 16–3 and performing the matrix operations. This yields the final result, Nx

Ny

冢Ll  L l冣 AE 12EI 冢 L  L 冣l l AE

12EI

2 x

2 y

3

x y

3



6EI L2

Nz

冢 L  L 冣l l AE 12EI 冢Ll  L l冣 AE

12EI

x y

3

2 y

6EI

ly

2 x

3

L2

lx

6EI



L2

Fx ly

6EI

lx L2 4EI

Fy

冢Ll  L l冣 AE 12EI 冢  ll L L 冣 

AE

2 x

12EI

x y

3

6EI L2

L

2 y

3

Fz

冢 L  L 冣l l AE 12EI 冢 l  l冣 L L



AE

12EI

x y

3

2 y



ly

2 x

3

6EI L2

lx

6EI



L2

ly

6EI

lx L2 2EI L

Nx Ny Nz

k

冢Ll  L l冣 AE 12EI 冢  ll L L 冣 

AE

12EI

2 x

3

3



6EI L2

ly

2 y

x y

冢 L  L 冣l l AE 12EI 冢 l  l冣 L L 

AE

12EI

x y

3

2 y

6EI L2

3

lx

2 x

6EI L2 

ly

6EI L2 2EI L

lx

冢Ll  L l冣 AE 12EI 冢 L  L 冣l l AE

12EI

2 x

3

3

6EI L2

ly

2 y

x y

冢 L  L 冣l l 12EI AE 冢Ll  L l冣 AE

12EI

x y

3

2 y



3

6EI L2

lx

2 x

6EI L2 

ly

6EI L2 4EI L

lx

Fx Fy Fz

(16–10) Note that this 6 * 6 matrix is symmetric. Furthermore, the location of each element is associated with the coding at the near end, Nx, Ny, Nz, followed by that of the far end, Fx, Fy, Fz, which is listed at the top of the columns and along the rows. Like the k¿ matrix, each column of the k matrix represents the coordinate loads on the member at the nodes that are necessary to resist a unit displacement in the direction defined by the coding of the column. For example, the first column of k represents the global coordinate loadings at the near and far ends caused by a unit displacement at the near end in the x direction, that is, DNx.

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THE

STIFFNESS METHOD

16.4 Application of the Stiffness Method for Frame Analysis Once the member stiffness matrices are established, they may be assembled into the structure stiffness matrix in the usual manner. By writing the structure matrix equation, the displacements at the unconstrained nodes can be determined, followed by the reactions and internal loadings at the nodes. Lateral loads acting on a member, fabrication errors, temperature changes, inclined supports, and internal supports are handled in the same manner as was outlined for trusses and beams.

Procedure for Analysis The following method provides a means of finding the displacements, support reactions, and internal loadings for members of statically determinate and indeterminate frames. 16

Notation

• Divide the structure into finite elements and arbitrarily identify each

• •

•

element and its nodes. Elements usually extend between points of support, points of concentrated loads, corners or joints, or to points where internal loadings or displacements are to be determined. Establish the x, y, z, global coordinate system, usually for convenience with the origin located at a nodal point on one of the elements and the axes located such that all the nodes have positive coordinates. At each nodal point of the frame, specify numerically the three x, y, z coding components. In all cases use the lowest code numbers to identify all the unconstrained degrees of freedom, followed by the remaining or highest code numbers to identify the constrained degrees of freedom. From the problem, establish the known displacements Dk and known external loads Qk. When establishing Qk be sure to include any reversed fixed-end loadings if an element supports an intermediate load.

Structure Stiffness Matrix

• Apply Eq. 16–10 to determine the stiffness matrix for each element expressed in global coordinates. In particular, the direction cosines lx and ly are determined from the x, y coordinates of the ends of the element, Eqs. 14–5 and 14–6.

• After each member stiffness matrix is written, and the six rows and columns are identified with the near and far code numbers, merge the matrices to form the structure stiffness matrix K. As a partial check, the element and structure stiffness matrices should all be symmetric.

16.4

601

APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS

Displacements and Loads

• Partition the stiffness matrix as indicated by Eq. 14–18. Expansion then leads to Qk = K11Du + K12Dk Qu = K21Du + K22Dk The unknown displacements Du are determined from the first of these equations. Using these values, the support reactions Qu are computed from the second equation. Finally, the internal loadings q at the ends of the members can be computed from Eq. 16–7, namely q = k¿TD If the results of any of the unknowns are calculated as negative quantities, it indicates they act in the negative coordinate directions. 16

EXAMPLE 16.1 Determine the loadings at the joints of the two-member frame shown in Fig. 16–4a. Take I = 500 in4, A = 10 in2, and E = 2911032 ksi for both members.

20 ft

5k

SOLUTION

20 ft

Notation. By inspection, the frame has two elements and three nodes, which are identified as shown in Fig. 16–4b. The origin of the global coordinate system is located at ①. The code numbers at the nodes are specified with the unconstrained degrees of freedom numbered first. From the constraints at ① and ➂, and the applied loading, we have 0 0 Dk = D T 0 0 Structure Stiffness Matrix. element stiffness matrices:

6 7 8 9

5 0 Qk = E 0 U 0 0

1 2 3 4 5

The following terms are common to both

(a)

y 2

6

3

5 1

4

1

8 7

9 3 (b)

Fig. 16–4

x 5k

2 2

10[2911032] AE = = 1208.3 k>in. L 201122 12[291103215002] 12EI = = 12.6 k>in. L3 [201122]3

1

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THE

STIFFNESS METHOD

EXAMPLE 16.1 (Continued) 6[291103215002] 6EI = = 1510.4 k L2 [201122]2

4[291103215002] 4EI = = 241.711032 k # in. L 201122 2[291103215002] 2EI = = 120.8311032 k # in. L 201122 Member 1: 20 - 0 0 - 0 = 1 ly = = 0 20 20 Substituting the data into Eq. 16–10, we have lx =

16

4 1208.3 0 0 k1 = F - 1208.3 0 0

6 0 12.6 1510.4 0 -12.6 1510.4

5 0 1510.4 241.711032 0 - 1510.4 120.8311032

1 -1208.3 0 0 1208.3 0 0

2 0 - 12.6 - 1510.4 0 12.6 - 1510.4

3 0 1510.4 120.8311032 V 0 - 1510.4 241.711032

4 6 5 1 2 3

The rows and columns of this 6 * 6 matrix are identified by the three x, y, z code numbers, first at the near end and followed by the far end, that is, 4, 6, 5, 1, 2, 3, respectively, Fig. 16–4b. This is done for later assembly of the elements. Member 2: 20 - 20 - 20 - 0 = 0 ly = = -1 20 20 Substituting the data into Eq. 16–10 yields lx =

1 12.6 0 k2 = F 1510.4 - 12.6 0 1510.4

2 0 1208.3 0 0 -1208.3 0

3 1510.4 0 241.711032 - 1510.4 0 120.8311032

7 -12.6 0 -1510.4 12.6 0 -1510.4

8 0 -1208.3 0 0 1208.3 0

9 1510.4 0 120.8311032 V -1510.4 0 241.711032

1 2 3 7 8 9

As usual, column and row identification is referenced by the three code numbers in x, y, z sequence for the near and far ends, respectively, that is, 1, 2, 3, then 7, 8, 9, Fig. 16–4b.

16.4

603

APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS

The structure stiffness matrix is determined by assembling k1 and k2. The result, shown partitioned, as Q = KD, is 1 1220.9

5

0

0 0

1510.4

0

1208.3

0 Q6



3 1510.4

1220.9

0

12.6

12.6

Q8

0

0

0

0

1208.3

0 241.7(103)

0

12.6 1510.4

8

7 12.6 1510.4

0

D2 D3 D4

1510.4

0

0

0

D5

12.6

0

0

0

0

0

0

12.6

0

0

0

0

0

0

0

120.83(103)

120.83(103)

D1

0

0

0

0

0

1510.4 0

9 1510.4

0

0

0

1510.4

0 1208.3

0

1510.4

1208.3

1510.4

6

0

120.83(103)

0

120.83(103)

5 1510.4

0

483.3(103)

0 1510.4

4 1208.3

1510.4

1510.4

0

Q7

Q9

2 0

0 1208.3

1510.4

0

1510.4 0 241.7(103)

(1)

0 0 0

Displacements and Loads. Expanding to determine the displacements yields 5 1220.9 0 0 E 0 U = E 1510.4 0 -1208.3 0 0

0 1220.9 -1510.4 0 -1510.4

1510.4 - 1510.4 483.311032 0 3 120.83110 2

-1208.3 0 0 1208.3 0

16

0 D1 0 -1510.4 D2 0 3 120.83110 2 U E D3 U + E 0 U 0 D4 0 3 241.7110 2 D5 0

Solving, we obtain D1 0.696 in. D2 -1.55110-32 in. E D3 U = E -2.488110-32 rad U D4 0.696 in. D5 1.234110-32 rad Using these results, the support reactions are determined from Eq. (1) as follows: 1 Q6 0 Q -12.6 D 7T = E Q8 0 Q9 1510.4

2 - 12.6 0 - 1208.3 0

3 1510.4 - 1510.4 0 120.8311032

4 0 0 0 0

5 1510.4 0.696 0 -1.87 k -3 0 - 1.55110 2 0 -5.00 k + D T = D T 0 U E -2.488110-32 U 0 1.87 k 0 0.696 0 750 k # in. -3 1.234110 2

Ans.

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STIFFNESS METHOD

EXAMPLE 16.1 (Continued) The internal loadings at node ➁ can be determined by applying Eq. 16–7 to member 1. Here k1œ is defined by Eq. 16–1 and T1 by Eq. 16–3. Thus, 4 1208.3 0 0 q1  k1T1D  1208.3 0 0

6 0 12.6 1510.4 0 12.6 1510.4

1

5 0 1510.4 241.7(103) 0 1510.4 120.83(103)

1208.3 0 0 1208.3 0 0

2

3

0 12.6 1510.4 0 12.6 1510.4

0 1510.4 120.83(103) 0 1510.4 241.7(103)

1 0 0 0 0 0

0 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 0 0

0 0 0 0 1 0

0.696 0 1.234(103) 0.696 1.55(103) 2.488(103)

0 0 0 0 0 1

4 6 5 1 2 3

Note the appropriate arrangement of the elements in the matrices as indicated by the code numbers alongside the columns and rows. Solving yields q4 0 q6 - 1.87 k q 0 F 5V = F V q1 0 q2 1.87 k q3 - 450 k # in.

16

Ans.

The above results are shown in Fig. 16–4c. The directions of these vectors are in accordance with the positive directions defined in Fig. 16–1. Furthermore, the origin of the local x¿, y¿, z¿ axes is at the near end of the member. In a similar manner, the free-body diagram of member 2 is shown in Fig. 16–4d. 1.87 k 450 k  in.

5k

y¿

y¿ 1.87 k x¿ 450 k  in. 1.87 k

5k 750 k  in.

1.87 k x¿

(c)

(d)

Fig. 16–4

16.4

605

APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS

EXAMPLE 16.2 Determine the loadings at the ends of each member of the frame shown in Fig. 16–5a. Take I = 600 in4, A = 12 in2, and E = 2911032 ksi for each member. SOLUTION

3 k/ ft

15 ft

Notation. To perform a matrix analysis, the distributed loading acting on the horizontal member will be replaced by equivalent end moments and shears computed from statics and the table listed on the inside back cover. (Note that no external force of 30 k or moment of 1200 k # in. is placed at ➂ since the reactions at code numbers 8 and 9 are to be unknowns in the load matrix.) Then using superposition, the results obtained for the frame in Fig. 16–5b will be modified for this member by the loads shown in Fig. 16–5c. As shown in Fig. 16–5b, the nodes and members are numbered and the origin of the global coordinate system is placed at node ①. As usual, the code numbers are specified with numbers assigned first to the unconstrained degrees of freedom. Thus, 0 0 0 Dk = F V 0 0 0

20 ft

20 ft (a)

Fig. 16–5

30 k 2

y

2 5

4 5 6 7 8 9

6

0 1 Qk = C -30 S 2 -1200 3

1

8

3

9 7

1 2 1200 k  in.

4

3

x

1 (b)

ⴙ

Structure Stiffness Matrix

3 k/ ft

Member 1: AE = L

12[2911032] 251122

= 1160 k>in.

12[29110 2]600 12EI = = 7.73 k>in. 3 L [251122]3 3

6[291103)]600 6EI = = 1160 k L2 [25112)]2

4[2911032]600 4EI = = 23211032 k # in. L 251122 2[2911032]600 2EI = = 11611032 k # in. L 251122 lx =

20 - 0 = 0.8 25

ly =

15 - 0 = 0.6 25

30 k

30 k

1 (3)(20)2  100 k  ft 20 ft __ 12 (1200 k  in.)

100 k ft (1200 k  in.)

(c)

16

606

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THE

STIFFNESS METHOD

EXAMPLE 16.2 (Continued) Applying Eq. 16–10, we have 4 745.18 553.09 -696 k1 = F -745.18 - 553.09 - 696

5 553.09 422.55 928 -553.09 - 422.55 928

6 - 696 928 23211032 696 - 928 11611032

1 - 745.18 -553.09 696 745.18 553.09 696

2 - 553.09 -422.55 -928 553.09 422.55 -928

3 -696 928 11611032 V 696 -928 23211032

Member 2: 16

12[2911032] AE = = 1450 k>in. L 201122

12[2911032]600 12EI = = 15.10 k>in. L3 [201122]3 6[2911032]600 6EI = = 1812.50 k L2 [201122]2

4[2911032]600 4EI = = 2.9011052 k # in. L 201122

23291103)4600 2EI = = 1.4511052 k # in. L 320112)4 lx =

40 - 20 = 1 20

ly =

15 - 15 = 0 20

Thus, Eq. 16–10 becomes

1 1450 0 k2 = F 0 - 1450 0 0

2 0 15.10 1812.50 0 -15.10 1812.50

3 0 1812.50 29011032 0 - 1812.50 14511032

7 - 1450 0 0 1450 0 0

8 0 - 15.10 -1812.50 0 15.10 -1812.50

9 0 1812.50 14511032 V 0 -1812.50 29011032

1 2 3 7 8 9

4 5 6 1 2 3

16.4

APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS

607

The structure stiffness matrix, included in Q = KD, becomes

1

2

3

4

5

6

7

8

9

0

2195.18

553.09

696

745.18

553.09

696

1450

0

0

D1

30

553.09

437.65

884.5

553.09

422.55

928

0

15.10

1812.50

D2

696

884.5

522(103)

696

0

1812.50

1200 Q4 

Q5

928

116(103)

145(103)

D3

745.18

553.09

696

745.18

553.09

696

0

0

0

0

553.09

422.55

928

553.09

422.55

928

0

0

0

0

928

116(103)

696

928

232(103)

0

0

0

0

0

0

0

1450

0

0

0

0

0

0

15.10

0

0

0

0

1812.50

Q6 Q7

696 1450

Q8

0

15.10

Q9

0

1812.50

0

0 1812.50 145(103)

0

1812.50 290(103)

(1)

0 0

Displacements and Loads. Expanding to determine the displacements, and solving, we have 16

C

0 2195.18 -30 S = C 553.09 -1200 696

553.09 437.65 884.5

696 D1 0 884.5 S C D2 S + C 0 S 52211032 D3 0

D1 0.0247 in. C D2 S = C -0.0954 in. S D3 -0.00217 rad

Using these results, the support reactions are determined from Eq. (1) as follows:

Q4 -745.18 Q5 -553.09 Q 696 F 6V = F Q7 -1450 Q8 0 Q9 0

-553.09 -422.55 -928 0 -15.10 1812.50

-696 0 35.85 k 928 0 24.63 k 0.0247 3 116110 2 0 -145.99 k # in. V C - 0.0954 S + F V = F V 0 - 35.85 k 0 - 0.00217 -1812.50 5.37 k 0 14511032 0 - 487.60 k # in.

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THE

STIFFNESS METHOD

EXAMPLE 16.2 (Continued) The internal loadings can be determined from Eq. 16–7 applied to members 1 and 2. In the case of member 1, q = k1œ T1D, where k1œ is determined from Eq. 16–1, and T1 from Eq. 16–3. Thus, 4

5

6

1

2

q4 1160 q5 0 q6 0 F V = F q1 - 1160 q2 0 q3 0

0 7.73 1160 0 - 7.73 1160

0 1160 23211032 0 - 1160 11611032

- 1160 0 0 1160 0 0

0 - 7.73 - 1160 0 7.73 - 1160

16

x¿ 1.81 k

3 0 0.8 1160 -0.6 11611032 0 VF 0 0 - 1160 0 23211032 0

0 0 1 0 0 0

0 0 0 0.8 - 0.6 0

0 0 0 0.6 0.8 0

0 0 0 0 0 0 VF V 0 0.0247 0 -0.0954 1 -0.00217

4 5 6 1 2 3

Here the code numbers indicate the rows and columns for the near and far ends of the member, respectively, that is, 4, 5, 6, then 1, 2, 3, Fig. 16–5b. Thus,

43.5 k 398 k in.

q4 43.5 k q5 - 1.81 k q6 - 146 k # in. F V = F V q1 - 43.5 k q2 1.81 k q3 - 398 k # in.

y¿ 146 kin. 43.5 k

0.6 0.8 0 0 0 0

1.81 k (d)

Ans.

These results are shown in Fig. 16–5d. A similar analysis is performed for member 2. The results are shown at the left in Fig. 16–5e. For this member we must superimpose the loadings of Fig. 16–5c, so that the final results for member 2 are shown to the right.

802.3 k in.

3 k/ ft

5.37 k

30 k

3 k/ ft

ⴙ

35.85 k 5.37 k

35.85 k 487.6 k in.

24.6 k

30 k

ⴝ

1200 k  in.

1200 k  in. (e)

Fig. 16–5

35.85 k 398 k  in.

35.4 k 35.85 k 1688 k  in.

16.4

609

APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS

PROBLEMS 16–1. Determine the structure stiffness matrix K for the frame. Assume ① and ➂ are fixed. Take E = 200 GPa, I = 30011062 mm4, A = 1011032 mm2 for each member. 16–2. Determine the support reactions at the fixed supports ① and ➂. Take E = 200 GPa, I = 30011062 mm4, A = 1011032 mm2 for each member.

*16–4. Determine the support reactions at ① and ➂. Take E = 200 MPa, I = 30011062 mm4, A = 2111032 mm2 for each member. 5m 8

2

8

1 2

1

3 1

7 1

1

3

1 7

12 kN/m 9

2

300 kN  m

9

2

4m

2m

4m 2

2

10 kN

6 4

5 3

3

2m

6

16 5

4

Prob. 16–4 Probs. 16–1/16–2 16–3. Determine the structure stiffness matrix K for the frame. Assume ➂ is pinned and ① is fixed. Take E = 200 MPa, I = 30011062 mm4, A = 2111032 mm2 for each member.

5m 8 9

9

2

300 kN  m

3

1

16–5. Determine the structure stiffness matrix K for the frame. Take E = 200 GPa, I = 35011062 mm4, A = 1511032 mm2 for each member. Joints at ① and ➂ are pins.

7

3

5 1

2

1

3

Prob. 16–3

1

8 1

1 2m

4m

2

60 kN

2 2m 2

4m

2 7

6 4

3

5

Prob. 16–5

4 6

610

CHAPTER 16

P L A N E F R A M E A N A LY S I S U S I N G

16–6. Determine the support reactions at pins ① and ➂. Take E = 200 GPa, I = 35011062 mm4, A = 1511032 mm2 for each member.

9

THE

STIFFNESS METHOD

*16–8. Determine the components of displacement at ①. Take E = 2911032 ksi, I = 650 in4, A = 20 in2 for each member. 2

2

60 kN

1

8 1

1 2m

5

6k

3

5

3

1

4k

6 4

1 1

2

2

2m 2

4m 2

12 ft 7 4

3

8

6 3

16

Prob. 16–6

9 7

10 ft

Prob. 16–8 16–7. Determine the structure stiffness matrix K for the frame. Take E = 2911032 ksi, I = 650 in4, A = 20 in2 for each member.

16–9. Determine the stiffness matrix K for the frame. Take E = 2911032 ksi, I = 300 in4, A = 10 in2 for each member.

2 6k 4k

1

16–10. Determine the support reactions at ① and ➂. Take E = 2911032 ksi, I = 300 in4, A = 10 in2 for each member.

5 3

6 4

1 1

2 2 k/ft

2

7 4

2 1

2

12 ft

6 2

3 20 ft

3

1

10 ft

8

9

9 7

5 1

8

10 ft

Prob. 16–7

Probs. 16–9/16–10

3

16.4

16–11. Determine the structure stiffness matrix K for the frame. Take E = 2911032 ksi, I = 700 in4, A = 20 in2 for each member.

16–13. Use a computer program to determine the reactions on the frame. AE and EI are constant.

1.5 k/ft

7 4

15 k 6

3

2 20 k 12 ft

5

12 ft

B

C

A

D

16 ft 20 ft

2

9

611

APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS

3

8

1 1

2

1

Prob. 16–11

16 24 ft

Prob. 16–13

*16–12. Determine the support reactions at the pins ① and ➂. Take E = 2911032 ksi, I = 700 in4, A = 20 in2 for each member.

16–14. Use a computer program to determine the reactions on the frame. Assume A, B, D, and F are pins. AE and EI are constant.

8 kN 7

B

C

E

A

D

F

4 6

3

8m 2 20 k 12 ft

5

16 ft

2

9 12 ft

3

8

1 1

1

2 6m

Prob. 16–12

4m

Prob. 16–14