Hibbeler R.C. Structural Analysis

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EXAMPLE 10.10

440 CHAPTER 10 ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES BY THE F ORCE METHOD

Fig. 10–24

Draw the influence line for the vertical reaction at A for the beam in

Fig. 10–24a. EI is constant. Plot numerical values every 6 ft.

SOLUTION

The capacity of the beam to resist the reaction is removed. This is done

using a vertical roller device shown in Fig. 10–24b.Applying a vertical unit

load at A yields the shape of the influence line shown in Fig. 10–24c.

In order to determine ordinates of the influence line we will use the

conjugate-beam method. The reactions at A and B on the “real beam,”

when subjected to the unit load at A, are shown in Fig. 10–24b. The cor-

responding conjugate beam is shown in Fig. 10–24d. Notice that the sup-

port at remains the same as that for A in Fig. 10–24b.This is because a

vertical roller device on the conjugate beam supports a moment but no

shear, corresponding to a displacement but no slope at A on the real

beam, Fig. 10–24c. The reactions at the supports of the conjugate beam

have been computed and are shown in Fig. 10–24d.The displacements of

points on the real beam, Fig. 10–24b, will now be computed.

A¿

18 ft

(a)

qualitative influence line

for reaction at A

(c)

1 k

6 ft

B¿

conjugate beam

(d)

6 ft6 ft

C¿ D¿

1944

____

162

___

A¿

6 ft6 ft 6 ft

real beam

(b)

1 k

18 kft

1 k

10.11 QUALITATIVE INFLUENCE LINES FOR FRAMES 441

A 1

C 0.852

D 0.481

B 0

For since no moment exists on the conjugate beam at

Fig. 10–24d, then

For Fig. 10–24e:

For Fig. 10–24f:

For Fig. 10–24d:

Since a vertical 1-k load acting at A on the beam in Fig. 10–24a will

cause a vertical reaction at A of 1 k, the displacement at A,

should correspond to a numerical value of 1 for the

influence-line ordinate at A. Thus, dividing the other computed

displacements by this factor, we obtain

= 1944>EI,

= M

A¿

1944

A¿,

= M

C¿

162

1122-

b1122142=

1656

©M

C¿

= 0;

C¿,

= M

D¿

162

162-

b162122=

936

©M

D¿

= 0;

D¿,

= M

B¿

= 0

B¿,B¿,

A plot of these values yields the influence line shown in Fig. 10–24g.

61218

0.852

0.481

quantitative influence line

for reaction at A

(g)

6 ft

D¿

162

___

(e)

12 ft12 ft

___

(f)(f)

162

____

162

____

442 CHAPTER 10 ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES BY THE F ORCE METHOD

EXAMPLE 10.11

Draw the influence line for the shear at D for the beam in Fig. 10–25a.

EI is constant. Plot numerical values every 9 ft.

SOLUTION

The capacity of the beam to resist shear at D is removed. This is done

using the roller device shown in Fig. 10–25b. Applying a positive unit

shear at D yields the shape of the influence line shown in Fig. 10–25c.

The support reactions at A, B, and C on the “real beam” when

subjected to the unit shear at D are shown in Fig. 10–25b. The corre-

sponding conjugate beam is shown in Fig. 10–25d. Here an external

couple moment must be applied at in order to cause a different

internal moment just to the left and just to the right of These

internal moments correspond to the displacements just to the left and

just to the right of D on the real beam, Fig. 10–25c.The reactions at the

supports and the external moment on the conjugate

beam have been computed and are shown in Fig. 10–25e. As an

exercise verify the calculations.

D¿

C¿B¿,A¿,

D¿.

D¿M

D¿

real beam

(b)

9 kft

9 ft 9 ft 18 ft

1 k 1 k

1 k

2 k 1 k

qualitative influence line

for shear at D

(c)

1 k

conjugate beam

(d)

A¿

B¿

C¿

9 ft 9 ft 18 ft

D¿

___

(e)

A¿

B¿

C¿

12 ft

6 ft

D¿

6 ft

12 ft

B¿

162

___

270

___

3888

____

162

___

108

___

108

___

(a)

9 ft

18 ft

Fig. 10–25

10.11 QUALITATIVE INFLUENCE LINES FOR FRAMES 443

A 0

0.406

B 0

C 0

-0.0938

-0.594D

A plot of these values yields the influence line shown in Fig. 10–25i.

Since there is a discontinuity of moment at the internal moment

just to the left and right of will be computed. Just to the left of

Fig. 10–25f, we have

Just to the right of Fig. 10–25g, we have

From Fig. 10–25e,

For point E,Fig.10–25b, using the method of sections at the corresponding

point on the conjugate beam, Fig. 10–25h, we have

The ordinates of the influence line are obtained by dividing each of

the above values by the scale factor We haveM

D¿

= 3888>EI.

= M

E¿

40.5

132-

192=-

364.5

©M

E¿

= 0;

E¿

= M

A¿

= 0

= M

B¿

= 0

= M

C¿

= 0

= M

40.5

132-

270

192+

3888

1579.5

©M

= 0;

D¿,

= M

40.5

132-

270

192=-

2308.5

©M

= 0;

D¿,D¿

D¿,

quantitative influence line

for shear at D

(i)

27 36

0.406

0.0938

0.594

(f)(f)

6 ft6 ft

40.5

___

40.5

___

270

___

270

___

D¿

3 ft3 ft

D¿

(g)(g)

6 ft6 ft

D¿

3 ft3 ft

D¿

40.5

___

40.5

___

270

___

270

___

3888

____

3888

____

(h)(h)

6 ft6 ft

E¿

3 ft3 ft

E¿

40.5

___

40.5

___

444 CHAPTER 10 ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES BY THE F ORCE METHOD

EXAMPLE 10.12

Draw the influence line for the moment at D for the beam in

Fig. 10–26a. EI is constant. Plot numerical values every 9 ft.

SOLUTION

A hinge is inserted at D in order to remove the capacity of the beam

to resist moment at this point, Fig. 10–26b.Applying positive unit couple

moments at D yields the influence line shown in Fig. 10–26c.

The reactions at A, B, and C on the “real beam” when subjected to

the unit couple moments at D are shown in Fig. 10–26b. The corre-

sponding conjugate beam and its reactions are shown in Fig. 10–26d.

It is suggested that the reactions be verified in both cases. From

Fig. 10–26d, note that

= M

A¿

= 0

= M

B¿

= 0

= M

C¿

= 0

Fig. 10–26

qualitative influence line for moment at D

(c)

1 kft

(d)

D¿

A¿

B¿

C¿

E¿

9 ft 9 ft 9 ft 9 ft

(a)

9 ft

18 ft

real beam

(b)

1 kft

9 ft 9 ft 9 ft 9 ft

0.111 k 0.222 k 0.111 k

10.11 QUALITATIVE INFLUENCE LINES FOR FRAMES 445

A 0

D 3.656

B 0

C 0

-0.844

(e)

6 ft

___

D¿

3 ft

D¿

4.5

___

(f)

6 ft

E¿

3 ft

E¿

4.5

___

For point Fig. 10–26e:

For point Fig. 10–26f:

The angular displacement at D of the “real beam” in Fig. 10–26c

is defined by the reaction at on the conjugate beam. This factor,

is divided into the above values to give the ordinates of

the influence line, that is,

= 48>EI,

D¿

= M

E¿

4.5

132-

192=-

40.5

©M

E¿

= 0;

E¿,

= M

D¿

4.5

132+

192=

175.5

©M

D¿

= 0;

D¿,

A plot of these values yields the influence line shown in Fig. 10–26g.

quantitative influence line

for moment at D

(g)

27 36

3.656

0.844

446 CHAPTER 10 ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES BY THE F ORCE METHOD

Prob. 10–46

Prob. 10–47

Prob. 10–41

Probs. 10–42/10–43

Prob. 10–44

Prob. 10–45

6 m 6 m

15 ft15 ft

ABC

3 m 3 m

3 m

3 m 6 m 6 m

2 m

2 m 4 m 2 m 2 m

A B

3 m 3 m

10–47. Sketch the influence line for (a) the vertical

reaction at C, (b) the moment at B, and (c) the shear at E.In

each case, indicate on a sketch of the beam where a uniform

distributed live load should be placed so as to cause a

maximum positive value of these functions. Assume the

beam is fixed at F.

10–45. Draw the influence line for the reaction at C. Plot

the numerical values every 5 ft. EI is constant.

10–42. Draw the influence line for the moment at A.

Plot numerical values at the peaks. Assume A is fixed and

the support at B is a roller. EI is constant.

10–43. Draw the influence line for the vertical reaction at

B. Plot numerical values at the peaks. Assume A is fixed and

the support at B is a roller. EI is constant.

*10–44. Draw the influence line for the shear at C.

Plot numerical values every 1.5 m. Assume A is fixed and

the support at B is a roller. EI is constant.

10–46. Sketch the influence line for (a) the moment at E,

(b) the reaction at C, and (c) the shear at E. In each case,

indicate on a sketch of the beam where a uniform distributed

live load should be placed so as to cause a maximum

positive value of these functions. Assume the beam is fixed

at D.

10–41. Draw the influence line for the reaction at C. Plot

numerical values at the peaks. Assume A is a pin and B and

C are rollers. EI is constant.

PROBLEMS

10.11 QUALITATIVE INFLUENCE LINES FOR FRAMES 447

Prob. 10–50

Prob. 10–51

Prob. 10–48

Prob. 10–49

10–50. Use the Müller-Breslau principle to sketch the

general shape of the influence line for (a) the moment at A

and (b) the shear at B.

10–49. Use the Müller-Breslau principle to sketch the

general shape of the influence line for (a) the moment at A

and (b) the shear at B.

*10–48. Use the Müller-Breslau principle to sketch the

general shape of the influence line for (a) the moment at A

and (b) the shear at B.

10–51. Use the Müller-Breslau principle to sketch the

general shape of the influence line for (a) the moment at A

and (b) the shear at B.

448 CHAPTER 10 ANALYSIS OF S TATICALLY INDETERMINATE STRUCTURES BY THE F ORCE METHOD

CHAPTER REVIEW

The analysis of a statically indeterminate structure requires satisfying equilibrium, compatibility, and the force-displacement

relationships for the structure.A force method of analysis consists of writing equations that satisfy compatibility and the

force-displacement requirements, which then gives a direct solution for the redundant reactions. Once obtained, the

remaining reactions are found from the equilibrium equations.

Simplification of the force method is possible, using

Maxwell’s theorem of reciprocal displacements, which

states that the displacement of a point B on a structure

due to a unit load acting at point A, , is equal to the

displacement of point A when the load acts at B,.f

actual beam

redundant B

applied

¿

 B

ⴝⴙ

primary structure



+T 0 =¢

- B

= f

CHAPTER REVIEW 449

The analysis of a statically indeterminate structure can be

simplified if the structure has symmetry of material,

geometry, and loading about its central axis. In particular,

structures having an asymmetric loading can be replaced

with a superposition of a symmetric and antisymmetric

load.

Influence lines for statically indeterminate structures will

consist of curved lines. They can be sketched using the

Müller-Breslau principle, which states that the influence

line shape for either the reaction, shear, or moment is to

the same scale as the deflected shape of the structure

when it is acted upon by the reaction, shear, or moment,

respectively. By using Maxwell’s theorem of reciprocal

deflections, it is possible to obtain specific values of the

ordinates of an influence line.

8 kN

2 kN/m

ⴝ

4 kN 4 kN

1 kN/m

ⴙ

symmetric loading

4 kN

1 kN/m

antisymmetric loading

influence line shape for moment at A