Hibbeler R.C. Structural Analysis

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160 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

FUNDAMENTAL PROBLEMS

F4–13. Draw the shear and moment diagrams for the

beam. Indicate values at the supports and at the points

where a change in load occurs.

F4–17. Draw the shear and moment diagrams for the

beam. Indicate values at the supports and at the points

where a change in load occurs.

F4–19. Draw the shear and moment diagrams for the

beam. Indicate values at the supports and at the points

where a change in load occurs.

2 m

3 kN

8 kN

F4–13

4 m 2 m

2 m

6 kN

6 kN · m

8 kN

F4–14

2 k/ft

10 ft

30 k·ft

F4–15

F4–14. Draw the shear and moment diagrams for the

beam. Indicate values at the supports and at the points

where a change in load occurs.

F4–15. Draw the shear and moment diagrams for the

beam. Indicate values at the supports and at the points

where a change in load occurs.

F4–16. Draw the shear and moment diagrams for the

beam. Indicate values at the supports and at the points

where a change in load occurs.

12 ft 12 ft

6 k/ft

18 k

F4–16

2 kN/m

4.5 m

F4–17

4 kN

1.5 m

2 m

1.5 m

F4–18

6 kN/m

2 m 2 m 2 m

6 kN/m

F4–19

F4–18. Draw the shear and moment diagrams for the

beam. Indicate values at the supports and at the points

where a change in load occurs.

F4–20. Draw the shear and moment diagrams for the

beam. Indicate values at the supports and at the points

where a change in load occurs.

2 k/ft

6 ft 6 ft

F4–20

4–23. Draw the shear and moment diagrams for the beam.

4–26. Draw the shear and moment diagrams for the beam.

PROBLEMS

6 ft

2 ft

4 ft 4 ft

500 lb

200 lb

300 lb

Prob. 4–23

4 ft4 ft4 ft4 ft4 ft

2 k 2 k 2 k

4 ft4 ft4 ft4 ft4 ft

2 k

Prob. 4–24

*4–24. Draw the shear and moment diagrams for the beam.

4–25. Draw the shear and moment diagrams for the beam.

2 m

1 m

2 m

1 m

0.4 m

6 kN

0.6 m

20 kN · m

Prob. 4–25

10 k 8 k

ACB

0.8 k/ft

6 ft 6 ft12 ft 12 ft

Prob. 4–26

15 ft

600 lb⭈ft

400 lb/ft

Prob. 4–27

L/3 L/3 L/3

Prob. 4–28

2 m

3 m

1.5 kN/m

Prob. 4–29

4–27. Draw the shear and moment diagrams for the beam.

*4–28. Draw the shear and moment diagrams for the

beam (a) in terms of the parameters shown; (b) set

L = 8 m.M

= 500 N

4.3 S

HEAR AND MOMENT DIAGRAMS FOR A BEAM 161

4–29. Draw the shear and moment diagrams for the beam.

162 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

4–30. Draw the shear and bending-moment diagrams for

the beam.

4–34. Draw the shear and moment diagrams for the beam.

20 ft 10 ft

50 lb/ft

200 lb⭈ft

Prob. 4–30

––

Prob. 4–31

250 lb/ft

150 lb

⭈ ft150 lb ⭈ ft

20 ft

Prob. 4–32

40 kN/m

20 kN

150 k

N⭈m

8 m 3 m

Prob. 4–33

4–31. Draw the shear and moment diagrams for the beam.

4–36. Draw the shear and moment diagrams of the beam.

Assume the support at B is a pin and A is a roller.

4–33. Draw the shear and moment diagrams for the beam.

*4–32. Draw the shear and moment diagrams for the beam.

4 ft 4 ft 4 ft 4 ft

200 lb/ft

CD E F G

A B

Prob. 4–34

200 lb/ft

30 ft

800 lb

1200 lb⭈ ft

Prob. 4–35

800 lb · ft

100 lb/ft

16 ft

4 ft

Prob. 4–36

8 kN/m

1.5 m 6 m

Prob. 4–37

4–35. Draw the shear and moment diagrams for the beam.

4–37. Draw the shear and moment diagrams for the beam.

Assume the support at B is a pin.

4.4 SHEAR AND MOMENT DIAGRAMS FOR A FRAME 163

4.4 Shear and Moment Diagrams

for a Frame

Recall that a frame is composed of several connected members that are

either fixed or pin connected at their ends. The design of these

structures often requires drawing the shear and moment diagrams for

each of the members. To analyze any problem, we can use the

procedure for analysis outlined in Sec. 4–3. This requires first

determining the reactions at the frame supports. Then, using the

method of sections, we find the axial force, shear force, and moment

acting at the ends of each member. Provided all loadings are resolved

into components acting parallel and perpendicular to the member’s

axis, the shear and moment diagrams for each member can then be

drawn as described previously.

When drawing the moment diagram, one of two sign conventions is

used in practice. In particular, if the frame is made of reinforced concrete,

designers often draw the moment diagram positive on the tension side of

the frame. In other words, if the moment produces tension on the outer

surface of the frame, the moment diagram is drawn positive on this side.

Since concrete has a low tensile strength, it will then be possible to tell at

a glance on which side of the frame the reinforcement steel must be

placed. In this text, however, we will use the opposite sign convention

and always draw the moment diagram positive on the compression side of

the member. This convention follows that used for beams discussed in

Sec. 4–1.

The following examples illustrate this procedure numerically.

The simply supported girder of this concrete building frame was

164 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

Draw the moment diagram for the tapered frame shown in Fig. 4–17a.

Assume the support at A is a roller and B is a pin.

EXAMPLE 4.13

15 ft

5 k

5 ft

6 ft

3 k

(a)

Fig. 4–17

15 ft

5 k

5 ft

6 ft

3 k

1 k

(b)

6 k

5 ft

3 k

6 k

15 kft

3 k

6 k

5 k

1 k

3 k

15 kft

3 k

15 kft

6 k

15 kft

3 k

1 k

15 ft

1 k

3 k

(c)

15

(d)

M (kft)

member CB

member AC

M (kft)

x (ft)

SOLUTION

Support Reactions. The support reactions are shown on the

free-body diagram of the entire frame, Fig. 4–17b. Using these results,

the frame is then sectioned into two members, and the internal reac-

tions at the joint ends of the members are determined, Fig. 4–17c.

Note that the external 5-k load is shown only on the free-body diagram

of the joint at C.

Moment Diagram. In accordance with our positive sign convention,

and using the techniques discussed in Sec. 4–3, the moment diagrams

for the frame members are shown in Fig. 4–17d.

4.4 SHEAR AND MOMENT DIAGRAMS FOR A FRAME 165

EXAMPLE 4.14

Draw the shear and moment diagrams for the frame shown in

Fig. 4–18a. Assume A is a pin, C is a roller, and B is a fixed joint.

Neglect the thickness of the members.

SOLUTION

Notice that the distributed load acts over a length of

The reactions on the entire frame are calculated

and shown on its free-body diagram, Fig. 4–18b. From this diagram the

free-body diagrams of each member are drawn, Fig. 4–18c.The

distributed loading on BC has components along BC and perpendicular

to its axis of

as shown. Using these results, the shear and moment diagrams are

also shown in Fig. 4–18c.

10.1414 k>ft2 cos 45° = 10.1414 k>ft2 sin 45° = 0.1 k>ft

10 ft 22 = 14.14 ft.

10 ft

14.14 ft

1.06

5

0.354 k

10.6

0.625

0.5 k

2 k

(c)

0.5 k

1.06 k

1.77 k

2 k

0.5 k

1.06 k

0.1 k/ft

5 kft

M (kft)

V (k)

x (ft)

0.5

V (k)

x (ft)

–5

x (ft)

Fig. 4–18

10 ft

0.1414 k/ft

10 ft

(a)

20 ft

5 ft

(0.1414 k/ft)(14.14 ft)  2 k

0.5 k

2 k

(b)

166 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

EXAMPLE 4.15

Draw the shear and moment diagrams for the frame shown in Fig. 4–19a.

Assume A is a pin, C is a roller, and B is a fixed joint.

40 kN/m

4 m

3 m

2 m

80 kN

(a)

Fig. 4–19

120 kN

2 m6 m

80 kN

(b)

1.5 m

 82.5 kN

36.87

 2.5 kN

 120 kN

SOLUTION

Support Reactions. The free-body diagram of the entire frame is

shown in Fig. 4–19b. Here the distributed load, which represents wind

loading, has been replaced by its resultant, and the reactions have been

computed. The frame is then sectioned at joint B and the internal

loadings at B are determined, Fig. 4–19c. As a check, equilibrium is

satisfied at joint B, which is also shown in the figure.

Shear and Moment Diagrams. The components of the distributed

load, and

are shown on member AB, Fig. 4–19d. The associated shear and

moment diagrams are drawn for each member as shown in Figs. 4–19d

and 4–19e.

196 kN2>15 m2= 19.2 kN>m,172 kN2>15 m2= 14.4 kN>m

4.4 SHEAR AND MOMENT DIAGRAMS FOR A FRAME 167

96 kN

2 kN

36.87

1.5 kN

72 kN

96 kN

72 kN

36.87

170 kNm

2 kN

1.5 kN

36.87

170 kNm

2.5 kN

170 kNm

1.5 kN

2 kN

2.5 kN

170 kNm

80 kN

82.5 kN

(

)

14.4 kN/m

70 kN

97.5 kN

19.2 kN/m

5 m

170 kNm

1.5 kN

2 kN

x (m)

4.86

170.1

170

4.86

x (m)

V (kN)

M (kNm)

(d)

2.5 kN

170 kNm

80 kN

82.5 kN

(e)

x (m)

2.5

V (kN)

82.5

M (kNm)

170 165

x (m)

168 CHAPTER 4INTERNAL LOADINGS DEVELOPED IN S TRUCTURAL MEMBERS

4.5 Moment Diagrams Constructed by

the Method of Superposition

Since beams are used primarily to resist bending stress, it is important

that the moment diagram accompany the solution for their design. In

Sec. 4–3 the moment diagram was constructed by first drawing the shear

diagram. If we use the principle of superposition, however, each of the

loads on the beam can be treated separately and the moment diagram

can then be constructed in a series of parts rather than a single and

sometimes complicated shape. It will be shown later in the text that this

can be particularly advantageous when applying geometric deflection

methods to determine both the deflection of a beam and the reactions

on statically indeterminate beams.

Most loadings on beams in structural analysis will be a combination of

the loadings shown in Fig. 4–20. Construction of the associated moment

diagrams has been discussed in Example 4–8. To understand how to use

PL

(a)

Fig. 4–20

(

)

(c)

parabolic curve

w

______

(d)

w

______

cubic curve

4.5 MOMENT DIAGRAMS CONSTRUCTED BY THE METHOD OF SUPERPOSITION 169

the method of superposition to construct the moment diagram consider

the simply supported beam at the top of Fig. 4–21a. Here the reactions

have been calculated and so the force system on the beam produces a zero

force and moment resultant.The moment diagram for this case is shown at

the top of Fig. 4–21b. Note that this same moment diagram is produced for

the cantilevered beam when it is subjected to the same statically equivalent

system of loads as the simply supported beam. Rather than considering all

the loads on this beam simultaneously when drawing the moment diagram,

we can instead superimpose the results of the loads acting separately on

the three cantilevered beams shown in Fig. 4–21a. Thus, if the moment

diagram for each cantilevered beam is drawn, Fig. 4–21b, the superposition

of these diagrams yields the resultant moment diagram for the simply

supported beam. For example, from each of the separate moment diagrams,

the moment at end A is as verified by the

top moment diagram in Fig. 4–21b. In some cases it is often easier to

construct and use a separate series of statically equivalent moment

diagrams for a beam, rather than construct the beam’s more complicated

“resultant” moment diagram.

=-200 - 300 + 500 = 0,

4 k/ft

300 kft

10 ft 10 ft

15 k 25 k

4 k/ft

10 ft

40 k

200 kft

ⴝ

ⴙ

10 ft

300 kft

ⴙ

20 ft

500 kft

25 k

superposition of cantilevered beams

(

)

4 k/ft

300 kft

10 ft 10 ft

15 k

25 k

M (kft)

250

50

x(ft)

resultant moment diagram

M (kft)

200

ⴝ

ⴙ

x(ft)

M (kft)

x(ft)

ⴙ

300

M (kft)

x(ft)

500

superposition of associated moment diagrams

(b)

Fig. 4–21