Hibbeler R.C. Structural Analysis

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170 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

In a similar manner, we can also simplify construction of the

“resultant” moment diagram for a beam by using a superposition of

“simply supported” beams. For example, the loading on the beam shown

at the top of Fig. 4–22a is equivalent to the beam loadings shown below

it. Consequently, the separate moment diagrams for each of these three

beams can be used rather than drawing the resultant moment diagram

shown in Fig. 4–22b.

5 kN/m

12 m

20 kNm

40 kNm

ⴝ

ⴙ

5 kN/m

12 m

ⴙ

12 m

20 kNm

12 m

40 kNm

superposition of simply supported beams

(a)

Fig. 4–22

M (kNm)

20

40

x (m)

resultant moment diagram

M (kNm)

x (m)

ⴝ

ⴙ

60.3

M (kNm)

x (m)

–20

ⴙ

M (kNm)

x (m)

40

superposition of associated moment diagrams

(b)

4.5 MOMENT DIAGRAMS CONSTRUCTED BY THE METHOD OF SUPERPOSITION 171

EXAMPLE 4.16

Draw the moment diagrams for the beam shown at the top of Fig. 4–23a

using the method of superposition. Consider the beam to be can-

tilevered from the support at B.

SOLUTION

If the beam were supported as a cantilever from B, it would be

subjected to the statically equivalent loadings shown in Fig. 4–23a.The

superimposed three cantilevered beams are shown below it together

with their associated moment diagrams in Fig. 4–23b. (As an aid to

their construction, refer to Fig. 4–20.) Although not needed here, the

sum of these diagrams will yield the resultant moment diagram for the

beam. For practice, try drawing this diagram and check the results.

ⴙ

superposition of associated moment diagrams

(b)

M (kft)

187.5

x (ft)

M (kft)

150

337.5

Fig. 4–23

22.5 k

15 ft

superposition of cantilevered beams

(a)

ⴙ

5 k/ft

15 ft

5 k/ft

150 kft

5 ft 15 ft

22.5 k

15 k

5 ft 15 ft

22.5 k

15 k

150 kft

ⴝ

ⴙ

172 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

Draw the moment diagrams for the beam shown at the top of Fig. 4–24a

using the method of superposition. Consider the beam to be cantilevered

from the pin at A.

SOLUTION

The superimposed cantilevered beams are shown in Fig. 4–24a

together with their associated moment diagrams, Fig. 4–24b. Notice

that the reaction at the pin (22.5 k) is not considered since it produces

no moment diagram. As an exercise verify that the resultant moment

diagram is given at the top of Fig. 4–24b.

EXAMPLE 4.17

5 k/ft

15 k

superposition of cantilevered from A beams

(a)

225 kft

37.5 k

ⴝ

ⴙ

15 ft5 ft

15 k

150 kft

22.5 k

15 ft

375 kft

5 ft

150 kft

15 ft

Fig. 4–24

24.3

x (ft)

150

375

225

M (kft)

superposition of associated moment diagrams

(

)

ⴝ

ⴙ

x (ft)

M (kft)

x (ft)

M (kft)

x (ft)

16.6 ft

4.5 MOMENT DIAGRAMS CONSTRUCTED BY THE METHOD OF SUPERPOSITION 173

4–38. Draw the shear and moment diagrams for each of

the three members of the frame. Assume the frame is pin

connected at A, C, and D and there is a fixed joint at B.

*4–40. Draw the shear and moment diagrams for each

member of the frame. Assume A is a rocker, and D is

pinned.

PROBLEMS

15 kN/m

50 kN

40 kN

1.5 m 1.5 m2m

4 m

Prob. 4–38

0.6 k/ft

0.8 k/ft

20 ft

16 ft

Prob. 4–39

2 k/ft

8ft 4ft

15 ft

4 k

3 k

Prob. 4–40

4–39. Draw the shear and moment diagrams for each

member of the frame. Assume the support at A is a pin and

D is a roller.

4–41. Draw the shear and moment diagrams for each

member of the frame.Assume the frame is pin connected at

B, C, and D and A is fixed.

3 k

6 k 6 k

3 k

15 ft

0.8 k/ft

8 ft

8 ft 8 ft

Prob. 4–41

174 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

4–42. Draw the shear and moment diagrams for each

member of the frame. Assume A is fixed, the joint at B is a

pin, and support C is a roller.

*4–44. Draw the shear and moment diagrams for each

member of the frame. Assume the frame is roller supported

at A and pin supported at C.

0.5 k/ft

20 k

8 ft

6 ft

Prob. 4–42

4 ft

15 k

4 k/ft

10 k

4 ft

10 ft

Prob. 4–43

4–43. Draw the shear and moment diagrams for each

member of the frame.Assume the frame is pin connected at

A, and C is a roller.

6 ft

10 ft

1.5 k/ft

2 k

Prob. 4–44

4–45. Draw the shear and moment diagrams for each

member of the frame.The members are pin connected at A,

B, and C.

45

15 kN

10 kN

4 kN/m

2 m 2 m 2 m

6 m

Prob. 4–45

4.5 MOMENT DIAGRAMS CONSTRUCTED BY THE METHOD OF SUPERPOSITION 175

4–46. Draw the shear and moment diagrams for each

member of the frame.

*4–48. Draw the shear and moment diagrams for each

member of the frame. The joints at A, B, and C are pin

connected.

300 lb/ft

500 lb/ft

30

7 ft

3.5 ft

7 ft

Prob. 4–47

6 ft 6 ft

8 ft

120 lb/ft

250 lb/ft

Prob. 4–48

5 kN 5 kN

10 kN

2 kN/m

3 m 3 m

4 m

2 m 2 m

Prob. 4–46

4–47. Draw the shear and moment diagrams for each

member of the frame. Assume the joint at A is a pin and

support C is a roller.The joint at B is fixed.The wind load is

transferred to the members at the girts and purlins from the

simply supported wall and roof segments.

6 k

0.8 k/ft

3 k

8 ft 8 ft 8 ft

15 ft

Prob. 4–49

4–49. Draw the shear and moment diagrams for each of

the three members of the frame. Assume the frame is pin

connected at B, C, and D and A is fixed.

176 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

4–50. Draw the moment diagrams for the beam using the

method of superposition. The beam is cantilevered from A.

4–54. Draw the moment diagrams for the beam using

the method of superposition. Consider the beam to be

cantilevered from the pin support at A.

4–55. Draw the moment diagrams for the beam using

the method of superposition. Consider the beam to be

cantilevered from the rocker at B.

3 ft 3 ft3 ft

600 lb 600 lb 600 lb

1200 lbft

Prob. 4–50

80 lb/ft

12 ft12 ft

600 lb

Prob. 4–51

20 ft

150 lbft150 lbft

A B

250 lb/ft

Probs. 4–52/4–53

4–51. Draw the moment diagrams for the beam using the

method of superposition.

4–52. Draw the moment diagrams for the beam using

the method of superposition. Consider the beam to be

cantilevered from end A.

4–53. Draw the moment diagrams for the beam using the

method of superposition. Consider the beam to be simply

supported at A and B as shown.

Probs. 4–54/4–55

30 kN

80 kNm

4 kN/m

8 m 4 m

Prob. 4–56

30 kN

80 kNm

4 kN/m

8 m 4 m

4–57. Draw the moment diagrams for the beam using the

method of superposition. Consider the beam to be simply

supported at A and B as shown.

200 lb/ft

100 lbft 100 lbft

20 ft

Prob. 4–57

*4–56. Draw the moment diagrams for the beam using

the method of superposition. Consider the beam to be

cantilevered from end C.

PROJECT PROBLEMS 177

4–1P. The balcony located on the third floor of a motel is

shown in the photo. It is constructed using a 4-in.-thick

concrete (plain stone) slab which rests on the four simply

supported floor beams, two cantilevered side girders AB

and HG, and the front and rear girders. The idealized

framing plan with average dimensions is shown in the

adjacent figure.According to local code, the balcony live load

is 45 psf. Draw the shear and moment diagrams for the front

girder BG and a side girder AB. Assume the front girder is a

channel that has a weight of 25 lb兾ft and the side girders are

wide flange sections that have a weight of 45 lb兾ft. Neglect

the weight of the floor beams and front railing. For this

solution treat each of the five slabs as two-way slabs.

PROJECT PROBLEMS

Prob. 4–1P

BCDEF

6 ft

4 ft

4 ft 4 ft 4 ft 4 ft

4–2P. The canopy shown in the photo provides shelter

for the entrance of a building. Consider all members to be

simply supported. The bar joists at C, D, E, F each have a

weight of 135 lb and are 20 ft long. The roof is 4 in. thick

and is to be plain lightweight concrete having a density of

Live load caused by drifting snow is assumed to

be trapezoidal, with 60 psf at the right (against the wall)

and 20 psf at the left (overhang). Assume the concrete slab

is simply supported between the joists. Draw the shear and

moment diagrams for the side girder AB. Neglect its

weight.

102 lb>ft

1.5 ft 1.5 ft 1.5 ft 1.5 ft 1.5 ft

CDE F

Prob. 4–2P

178 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

4–3P. The idealized framing plan for a floor system

located in the lobby of an office building is shown in the

figure. The floor is made using 4-in.-thick reinforced stone

concrete. If the walls of the elevator shaft are made from

4-in.-thick lightweight solid concrete masonry, having a

height of 10 ft, determine the maximum moment in beam

AB. Neglect the weight of the members.

8 ft 6 ft 6 ft

8 ft

Elevator

shaft

Prob. 4–3P

Structural members subjected to planar loads support

an internal normal force N, shear force V, and bend-

ing moment M. To find these values at a specific point

in a member, the method of sections must be used.

This requires drawing a free-body diagram of a seg-

ment of the member, and then applying the three

equations of equilibrium.Always show the three inter-

nal loadings on the section in their positive directions.

The internal shear and moment can be expressed as a

function of x along the member by establishing the

origin at a fixed point (normally at the left end of the

member, and then using the method of sections, where

the section is made a distance x from the origin). For

members subjected to several loads, different x coor-

dinates must extend between the loads

CHAPTER REVIEW

positive sign convention

CHAPTER REVIEW 179

Shear and moment diagrams for structural members can be drawn by plotting the shear and moment functions. They also

can be plotted using the two graphical relationships.

Slope of

Moment Diagram

f = 5Shear

= V

Slope of

Shear Diagram

f = e

Intensity of

Distributed Load

= w(x)

Change in

Moment

f = e

Area under

Shear Diagram

¢M =

V(x)

Change in

Shear

f =

Area under

Distributed Loading

Diagram

¢V =

w(x) dx

Note that a point of zero shear locates the point of maximum moment since .V = dM>dx = 0

A force acting downward on the beam will cause the shear diagram to jump downwards, and a counterclockwise couple

moment will cause the moment diagram to jump downwards.

Using the method of superposition, the moment diagrams for a member can be represented by a series of simpler shapes.

The shapes represent the moment diagram for each of the separate loadings. The resultant moment diagram is then the

algebraic addition of the separate diagrams.

M¿

M¿P