7-48 Draw the Shear and Moment Diagrams for the

170 C HAPTER 4I NTERNAL Fifty OADINGS D EVELOPED IN S TRUCTURAL M EMBERS

four

In a similar manner, nosotros can also simplify structure 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 pinnacle of Fig. 4–22a is equivalent to the beam loadings shown beneath

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

beams can be used rather than cartoon the resultant moment diagram

shown in Fig. 4–22b.

5 kN/ grand

12 grand

twenty kN 1000

40 kN chiliad

v kN/ yard

12 grand

12 thou

twenty kN 1000

12 g

40 kN chiliad

superposition of simply supported beams

(a)

Fig. 4–22

M (kN m)

twenty

xl

x (m)

resultant moment diagram

M (kN m)

x (m)

60.3

90

Thousand (kN chiliad)

x (grand)

–20

M (kN g)

ten (m)

40

superposition of associated moment diagrams

(b)

4.v GrandOMENT DIAGRAMS CONSTRUCTED BY THE METHOD OF SUPERPOSITION 171

four

Example four.16

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

using the method of superposition. Consider the axle to be tin can-

tilevered from the support at B.

SOLUTION

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

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

superimposed 3 cantilevered beams are shown below it together

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

their construction, refer to Fig. iv–20.) Although not needed hither, the

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

beam. For practice, endeavor drawing this diagram and cheque the results.

superposition of associated moment diagrams

(b)

M (k ft)

187.5

x (ft)

10 (ft)

x (ft)

Grand (thousand ft)

150

337.v

Fig. 4–23

22.v yard

15 ft

superposition of cantilevered beams

(a)

5 k/ ft

15 ft

5 k/ ft

5 thou/ ft

A

B

150 grand ft

150 k ft

v ft 15 ft

22.five k

15 m

5 ft fifteen ft

22.five thou

15 k

150 k ft

A

B

172 C HAPTER 4I NTERNAL 50 OADINGS D EVELOPED IN S TRUCTURAL Yard EMBERS

4

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

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

from the pivot 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 practice verify that the resultant moment

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

Instance 4.17

5 g/ ft

A

B

five 1000/ ft

15 k

15 thousand

superposition of cantilevered from A beams

(a)

B

225 k ft

37.5 k

15 ftfive ft

xv m

150 m ft

22.5 k

15 ft

375 g ft

five ft

150 k ft

150 grand ft

15 ft

Fig. 4–24

24.3

x (ft)

150

150

375

225

Chiliad (1000 ft)

M (k ft)

superposition of associated moment diagrams

b

x (ft)

M (k ft)

x (ft)

1000 (g ft)

x (ft)

16.6 ft

4.v MOMENT DIAGRAMS CONSTRUCTED By THE METHOD OF SouthUPERPOSITION 173

4

4–38. Depict 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 articulation at B.

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

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

pinned.

Issues

15 kN/ m

50 kN

forty kN

A

D

BC

1.5 thousand 1.5 1000 2m

four m

6m

A

B

C

D

0.6 k/ft

0.viii k/ft

20 ft

16 ft

2 k/ ft

8ft 4ft

15 ft

D

A

BC

4 chiliad

iii one thousand

Prob. 4–40

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

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

D is a roller.

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

member of the frame.Presume the frame is pin continued at

B, C, and D and A is fixed.

B

C

D

A

3 k

6 g vi k

3 k

15 ft

0.8 k/ft

8 ft

8 ft 8 ft

174 C HAPTER 4I NTERNAL L OADINGS D EVELOPED IN S TRUCTURAL Yard EMBERS

iv

four–42. Describe the shear and moment diagrams for each

member of the frame. Assume A is stock-still, the joint at B is a

pin, and support C is a roller.

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

fellow member of the frame. Assume the frame is roller supported

at A and pin supported at C.

B

A

C

0.five grand/ft

20 thou

8 ft

6 ft

6 ft

3

4

5

Prob. 4–42

4 ft

fifteen 1000

4 thousand/ft

10 m

iv ft

10 ft

A

B

C

Prob. four–43

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

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

A, and C is a roller.

A

B

C

six ft

6 ft

10 ft

1.5 k/ ft

2 thousand

Prob. iv–44

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

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

B, and C.

45

A

B

C

fifteen kN

10 kN

4 kN/ m

2 grand 2 one thousand ii m

six one thousand

Prob. 4–45

4.five MOMENT DIAGRAMS CONSTRUCTED BY THE KETHOD OF SUPERPOSITION 175

4

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

member of the frame.

*4–48. Depict 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

A

B

C

thirty

7 ft

3.5 ft

3.5 ft

7 ft

Prob. iv–47

6 ft half-dozen ft

8 ft

120 lb/ft

250 lb/ft

A

B

C

sixty

Prob. 4–48

5 kN 5 kN

10 kN

ii kN/m

3 m 3 m

4 one thousand

2 m 2 m

A

B

C

D

Prob. 4–46

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

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

support C is a roller.The joint at B is stock-still.The air current load is

transferred to the members at the girts and purlins from the

simply supported wall and roof segments.

A

B

D

C

6 k

0.8 k/ ft

3 one thousand

8 ft 8 ft 8 ft

fifteen 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 C HAPTER 4I NTERNAL L OADINGS D EVELOPED IN S TRUCTURAL Chiliad EMBERS

4

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

method of superposition. The beam is cantilevered from A.

4–54. Describe the moment diagrams for the axle using

the method of superposition. Consider the beam to be

cantilevered from the pin support at A.

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

the method of superposition. Consider the axle to be

cantilevered from the rocker at B.

3 ft 3 ft3 ft

A

600 lb 600 lb 600 lb

1200 lbft

Prob. 4–50

80 lb/ft

12 ft12 ft

600 lb

Prob. 4–51

xx 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. Describe the moment diagrams for the beam using

the method of superposition. Consider the beam to exist

cantilevered from end A.

iv–53. Describe the moment diagrams for the beam using the

method of superposition. Consider the beam to be merely

supported at A and B as shown.

Probs. iv–54/4–55

30 kN

80 kN k

4 kN/ m

A

B

C

8 m 4 m

Prob. iv–56

30 kN

80 kN m

4 kN/ m

A

B

C

8 m 4 chiliad

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

method of superposition. Consider the axle to be just

supported at A and B equally shown.

200 lb/ft

100 lbft 100 lbft

A

B

20 ft

Prob. iv–57

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

the method of superposition. Consider the axle to exist

cantilevered from cease C.

PROJECT PROBLEMS 177

4

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

shown in the photo. It is synthetic using a four-in.-thick

concrete (plainly 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 programme with average dimensions is shown in the

adjacent effigy.According to local code, the balustrade live load

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

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

aqueduct that has a weight of 25 lbft and the side girders are

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

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

solution treat each of the v slabs as 2-way slabs.

PROJECT Bug

Prob. 4–1P

A

BCDEastwardF

H

G

six ft

iv ft

4 ft 4 ft iv 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 xx ft long. The roof is four in. thick

and is to exist patently lightweight concrete having a density of

Live load acquired by drifting snow is assumed to

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

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

is merely supported between the joists. Draw the shear and

moment diagrams for the side girder AB . Neglect its

weight.

102 lb> ft

three

.

1.5 ft ane.5 ft one.5 ft 1.5 ft 1.five ft

A

CDE F

B

Prob. 4–2P

178 C HAPTER 4I NTERNAL L OADINGS D EVELOPED IN S TRUCTURAL M EMBERS

4

iv–3P. The idealized framing program for a floor system

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

figure. The floor is fabricated using 4-in.-thick reinforced rock

physical. If the walls of the lift shaft are fabricated from

iv-in.-thick lightweight solid physical masonry, having a

height of 10 ft, make up one's mind the maximum moment in beam

AB. Fail the weight of the members.

8 ft half-dozen ft 6 ft

viii ft

8 ft

8 ft

Elevator

shaft

AB

I

C

FH

GEast

D

Prob. four–3P

Structural members subjected to planar loads support

an internal normal forcefulness N, shear strength V, and bend-

ing moment M . To discover these values at a specific signal

in a fellow member, the method of sections must exist 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 department in their positive directions.

The internal shear and moment tin be expressed as a

function of ten along the fellow member by establishing the

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

member, and and then using the method of sections, where

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

members subjected to several loads, dissimilar 10 coor-

dinates must extend between the loads

Chapter REVIEW

K

Northward

V

North

V

M

positive sign convention

P

westward

ten

1

10

2

ten

3

CHAPTER REVIEW 179

4

Shear and moment diagrams for structural members can be fatigued by plotting the shear and moment functions. They as well

can be plotted using the two graphical relationships.

Gradient of

Moment Diagram

f = 5Shear

dM

dx

= V

Slope of

Shear Diagram

f = e

Intensity of

Distributed Load

dV

dx

= w(ten)

Change in

Moment

f = eastward

Surface area under

Shear Diagram

¢ M =

L

V(x)

dx

Change in

Shear

f =

50

Area under

Distributed Loading

Diagram

¢ V =

50

w(x) dx

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

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

moment volition cause the moment diagram to jump downwards.

Using the method of superposition, the moment diagrams for a fellow member tin can be represented past 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 dissever diagrams.

One thousand

R

P

V

R

Yard

50

M

50

Five

L

5

L

VM

Five

R

xx

M

R

M¿

M

L

Thousand

R

Thousand¿ P

7-48 Draw the Shear and Moment Diagrams for the

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