Hibbeler R.C. Structural Analysis

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Determine the deflection at ① and the reactions on the beam shown

in Fig. 15–12a. EI is constant.

590

CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD

EXAMPLE 15.5

SOLUTION

Notation. The beam is divided into two elements and the nodes and

members are identified along with the directions from the near to far

ends, Fig. 15–12b. The unknown deflections are shown in Fig. 15–12c.

In particular, notice that a rotational displacement does not occur

because of the roller constraint.

Member Stiffness Matrices. Since EI is constant and the members

are of equal length, the member stiffness matrices are identical. Using

the code numbers to identify each column and row in accordance with

Eq. 15–1 and Fig. 15–12b, we have

= EI

1256

1.5 1.5 -1.5 1.5

1.5 2 -1.5 1

-1.5 -1.5 1.5 -1.5

1.5 1 -1.5 2

= EI

34 12

1.5 1.5 -1.5 1.5

1.5 2 -1.5 1

-1.5 -1.5 1.5 -1.5

1.5 1 -1.5 2

2 m 2 m

(a)

(b)

1 2

Fig. 15–12

15.4 A

PPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 591

Displacements and Loads. Assembling the member stiffness

matrices into the structure stiffness matrix, and applying the structure

stiffness matrix equation, we have

Solving for the displacements yields

Ans.

Note that the signs of the results match the directions of the

deflections shown in Fig. 15–12c. Using these results, the reactions

therefore are

2.667P

1.667P

0 =-1.5D

+ 1.5D

0 = 0D

+ 4D

+ 1.5D

= 3D

+ 0D

- 1.5D

-P

V= EI

1 2 3456

30-1.5 -1.5 -1.5 1.5

0 4 1.5 1 -1.5 1

-1.5 1.5 1.5 1.5 0 0

-1.5 1 1.5 2 0 0

-1.5 -1.5 0 0 1.5 -1.5

1.5 1 0 0 -1.5 2

Q = KD

(c)

Ans.

Ans.=-1.5P

= 1.5EIa-

1.667P

b+ 1EIa

b+ 0a-

2.667P

= P

=-1.5EIa-

1.667P

b- 1.5EIa

b+ 0a-

2.667P

=-0.5P

=-1.5EIa-

1.667P

b+ 1EIa

b+ 1.5EIa-

2.667P

592 CHAPTER 15 BEAM ANALYSIS U SING THE STIFFNESS METHOD

Probs. 15–1/15–2

Prob. 15–3

Prob. 15–4

Prob. 15–5

Prob. 15–6

Prob. 15–7

15–1. Determine the moments at ① and ③. Assume ② is a

roller and ① and ③ are fixed. EI is constant.

15–2. Determine the moments at ① and ③ if the support ②

moves upward 5 mm. Assume ② is a roller and ① and ③ are

fixed. .EI = 60(10

) N

15–5. Determine the support reactions. Assume ② and ③

are rollers and ① is a pin. EI is constant.

15–3. Determine the reactions at the supports. Assume

the rollers can either push or pull on the beam. EI is

constant.

15–6. Determine the reactions at the supports. Assume ①

is fixed ➁ and ③ are rollers. EI is constant.

*15–4. Determine the reactions at the supports. Assume ①

is a pin and ② and ③ are rollers that can either push or pull

on the beam. EI is constant.

15–7. Determine the reactions at the supports. Assume

① and ③ are fixed and ② is a roller. EI is constant.

PROBLEMS

4 m

6 m

25 kN/m

1 3

4 6

6 m 8 m

15 kN/m

6 kN/m

9 kN/m

6 m

4 m

6 m 8 m

10 kN/m

10 ft 10 ft10 ft

1 2

3 k

12 m

8 m

6 kN/m

463

20 kNm

15.4 A

PPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 593

Prob. 15–8

Prob. 15–9

Prob. 15–10

Prob. 15–11

Prob. 15–13

Prob. 15–12

*15–8. Determine the reactions at the supports. EI is

constant.

15–11. Determine the reactions at the supports. There is a

smooth slider at ①. EI is constant.

15–9. Determine the moments at ② and ③. EI is constant.

Assume ①, ②, and ③ are rollers and ④ is pinned.

*15–12. Use a computer program to determine the reactions

on the beam. Assume A is fixed. EI is constant.

15–10. Determine the reactions at the supports. Assume ②

is pinned and ① and ③ are rollers. EI is constant.

15–13. Use a computer program to determine the reactions

on the beam. Assume A and D are pins and B and C are

rollers. EI is constant.

4 m

15 kN/m

3 m

12 m 12 m 12 m

4 kN/m

2 31

234

3 4

AB CD

8 ft

8 ft 20 ft

3 k/ft

8 ft 8 ft4 ft 4 ft

3 k/ft

456

1 2

8 ft8 ft15 ft20 ft

12 k

4 k/ft

30 kN/m

4 m

The frame of this building is statically indeterminate. The force analysis can

be done using the stiffness method.

595

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 coordinate system,

Fig. 16–1. Here the member is subjected to axial loads shear

loads and bending moments 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 and are positive counterclockwise, since by the

right-hand rule the moment vectors are then directed along the positive

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

z¿

Fz¿

Nz¿

Fz¿

Nz¿

Fy¿

Ny¿

Fx¿

Nx¿

z¿y¿,x¿,

Plane Frame Analysis

Using the Stiffness

Method

results are added, the resulting six load-displacement relations for the

member can be expressed in matrix form as

596

CHAPTER 16 PLANE FRAME ANALYSIS U SING THE STIFFNESS METHOD

x¿

y¿

Ny¿

Nx¿

Nz¿

Fy¿

Fz¿

Fx¿

positive sign convention

12EI

6EI



0000

Nx¿

x¿

y¿

z¿

x¿

y¿

z¿

Nx¿

Ny¿

6EI

4EI

2EI

6EI



Nz¿



12EI

6EI



6EI



00 00

Fx¿

Fy¿

6EI

2EI

4EI



Fz¿



Fig. 16–1

(16–1)

or in abbreviated form as

(16–2)

The member stiffness matrix 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.

k¿

q = k¿d

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.

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 coordinates to

global x, y, z coordinates. For this reason transformation matrices are

needed.

Displacement Transformation Matrix. Consider the frame

member shown in Fig. 16–2a. Here it is seen that a global coordinate

displacement creates local coordinate displacements

Likewise, a global coordinate displacement Fig. 16–2b, creates local

coordinate displacements of

Finally, since the and z axes are coincident, that is, directed out of the

page, a rotation about z causes a corresponding rotation about

Thus,

In a similar manner, if global displacements in the x direction,

in the y direction, and a rotation are imposed on the far end of the

member, the resulting transformation equations are, respectively,

Letting represent the direction cosines of the

member, we can write the superposition of displacements in matrix

form as

(16–3)

(16–4)

By inspection, T transforms the six global x, y, z displacements D into

the six local displacements d. Hence T is referred to as the

displacement transformation matrix.

z¿y¿,x¿,

d = TD

Nx¿

Ny¿

Nz¿

Fx¿

Fy¿

Fz¿

V= F

0000

-l

0000

001000

000l

000-l

000001

= cos u

Fz¿

= D

Fx¿

= D

cos u

Fy¿

= D

cos u

Fx¿

= D

cos u

Fy¿

=-D

cos u

Nz¿

= D

z¿.

Nz¿

z¿

Nx¿

= D

cos u

Ny¿

= D

cos u

Nx¿

= D

cos u

Ny¿

=-D

cos u

z¿y¿,x¿,

(a)

Ny¿

 D

cos u

Nx¿

 D

cos u

y¿

x¿

(b)

Ny¿

 D

cos u

Nx¿

 D

cos u

y¿

x¿

Fig. 16–2

598 CHAPTER 16 PLANE FRAME ANALYSIS U SING THE 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 Fig. 16–3a, it can be

seen that

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

Finally, since is collinear with we have

In a similar manner, end loads of will yield the following

respective components:

These equations, assembled in matrix form with

yield

(16–5)

(16–6)

Here, as stated, transforms the six member loads expressed in local

coordinates into the six loadings expressed in global coordinates.

Q = T

V= F

-l

00 0 0

001000

000l

-l

000l

000001

V F

Nx¿

Ny¿

Nz¿

Fx¿

Fy¿

Fz¿

= cos u

= q

Fz¿

=-q

Fy¿

cos u

= q

Fy¿

cos u

= q

Fx¿

cos u

= q

Fx¿

cos u

Fz¿

Fy¿

Fx¿

= q

Nz¿

=-q

Ny¿

cos u

= q

Ny¿

cos u

Ny¿

= q

Nx¿

cos u

= q

Nx¿

cos u

Nx¿

(a)

Nx¿

 q

cos u

 q

cos u

x¿

y¿

(

)

Ny¿

 q

Ny¿

cos u

 q

Ny¿

cos u

x¿

y¿

Fig. 16–3

16.3 FRAME-MEMBER GLOBAL STIFFNESS MATRIX 599

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

into Eq. 16–2 We have

(16–7)

Here the member forces q are related to the global displacements D.

Substituting this result into Eq. 16–6 yields the final result,

(16–8)

where

(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,

k = T

k¿T

Q = kD

Q = T

k¿TD

1Q = T

q = k¿TD

1q = k¿d2.1d = TD2

12EI

冢冣



12EI

冢冣



6EI



12EI

冢冣



12EI

冢冣



6EI

12EI

冢冣



12EI

冢冣



6EI

12EI

冢冣



12EI

冢冣



6EI

12EI

冢冣



12EI

冢冣



6EI



6EI



6EI

4EI

6EI



6EI

2EI

6EI



6EI

4EI

6EI



6EI

2EI

12EI

冢冣



12EI

冢冣



6EI



12EI

冢冣



12EI

冢冣



6EI



12EI

冢冣



12EI

冢冣



6EI



k 

(16–10)

Note that this matrix is symmetric. Furthermore, the location of

each element is associated with the coding at the near end,

followed by that of the far end, which is listed at the top of the

columns and along the rows. Like the 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, .D

k¿

6 * 6