Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition
Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition
5th Edition
ISBN: 9781259820960
Author: Leet, Kenneth
Publisher: McGraw-Hill Education
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Question
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Chapter 12, Problem 53P

(a)

To determine

Draw the influence line for the positive moment at section B using moment distribution method.

(a)

Expert Solution
Check Mark

Explanation of Solution

Influence line for the moment at support B.

Influence lines will be constructed by placing the unit load at every 20 ft interval from A along the span of the beam. The points are indicated by the circled numbers.

To establish the influence line ordinate at the left end (point 1), the unit load is placed on the beam directly over support A. Since the entire load passes directly into the support, the beam is unstressed. Therefore,

RA=1, RC=0, and RD=0,  and MB=0.

Similarly, if the unit load is moved at C and D, RA=0, RC=1, RD=0, and MB=0 and RA=0, RC=0, RD=1, and MB=0

Case 1:

Apply unit load at 20 ft from A. (point 2).

Draw the beam with points of applying unit load at B (point 2) as in Figure 1.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  1

Refer Figure 1.

Find the fixed end moment at each end of the member CD as follows;

FEMCD=Pab2L2=(1)(20)(40)2602=8.89k-ft

Find the fixed end moment at each end of the member CA as follows;

FEMDC=Pa2bL2=(1)(20)2(40)602=4.44k-ft

Find the distribution factor.

Show the computation of distribution factor as in Table 1.

JointsMemberStiffness, KΣKDF=KΣK
CCA34(ILCA)=34(I60)=0.0125IΣK=0.0125I+0.0125I=0.025IDFCA=0.025I0.0125I=0.5
CD34(ILCD)=34(I60)=0.0125IDFCD=0.025I0.0125I=0.5

Show the moment distribution computations as in Table 2.

JointACD
MemberACCACDDC
DF 0.50.5
FEM‑8.894.4400
Released moment+8.89000
COM04.440 
Total08.8800
DEM‑4.44‑4.44
COM   
Final04.44‑4.440

Find the reaction at A using section AC.

Draw the free body diagram of section AC as in Figure 2.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  2

Refer Figure 1.

Find the reaction at A.

Consider moment at C

ΣMC=0RA(60)1(40)+4.44=060RA=35.56RA=0.593k

Find the moment at B.

Consider moment at B

ΣMB=0MB=RA(20)=0.593(20)MB=11.86k-ft

Thus, the influence line ordinate of moment at B when 1 kip applied at B is 11.86.

Case 2:

Apply unit load at 20 ft from C. (point 5).

Draw the beam with points of applying at point 5 as in Figure 3.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  3

Refer Figure 3.

Find the fixed end moment at each end of the member CD as follows;

FEMCD=Pab2L2=(1)(20)(40)2602=8.89k-ft

Find the fixed end moment at each end of the member DC as follows;

FEMDC=Pa2bL2=(1)(20)2(40)602=4.44k-ft

Show the moment distribution computations as in Table 3.

JointACD
MemberACCACDDC
DF 0.50.5
FEM00‑8.894.440
Released moment000‑4.44
COM00‑2.22 
Total00‑11.110
DEM+5.56+5.56
COM   
Final05.56‑5.560

Find the reaction at A using section AC.

Draw the free body diagram of section AC as in Figure 4.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  4

Refer Figure 4.

Find the reaction at A.

Consider moment at C

ΣMC=0RA(32)+3=032RA=3RA=0.938k

Find the moment at B.

Consider moment at B

ΣMB=0MB=RA(20)=0.0938(20)MB=1.876k-ft

Thus, the influence line ordinate of moment at B when 1 kip applied at point 5 is ‑1.876.

Similarly calculate the influence line ordinate by applying the points 3 and 5 and tabulate the values as in Table 4.

Apply 1 k at pointInfluence line ordinate of moment at B.
10
211.86
40
5‑1.876
6‑1.876
70

Draw the influence line diagram of influence line ordinate of moment at B using Table 4 as in Figure 5.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  5

(b)

To determine

Find the maximum moment at B for the given loading condition.

(b)

Expert Solution
Check Mark

Answer to Problem 53P

The maximum moment at B for the given loading condition is 887k-ft_.

Explanation of Solution

Consider the uniformly distributed load of 2 kips/ft act on the span AC only and the concentrated live load of 20 kips act at B.

Draw the free body diagram of the beam for the given loading as in Figure 6.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  6

Refer Figure 6.

Find the fixed end moment at each end of the member AC as follows;

FEMAC=wL212Pab2L2=2(60)21220(20)(40)2602=600177.78=777.78k-ft

Find the fixed end moment at each end of the member CA as follows;

FEMCA=wL212+Pa2bL2=2(60)21220(20)2(40)602=600+88.89=688.89k-ft

Show the moment distribution computations as in Table 5.

JointACD
MemberACCACDDC
DF 0.50.5
FEM‑777.78688.8900
Released moment+777.780000
COM 388.89  
Total01,077.7800
DEM‑538.89‑538.89
COM   
Final0538.89‑538.890

Find the reaction at A using section AC.

Draw the free body diagram of section AC as in Figure 7.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  7

Refer Figure 7.

Find the reaction at A.

Consider moment at C

ΣMC=0RA(60)20(40)2(60)(602)+538.89=060RA=800+3,600538.89RA=64.35k

Find the moment at B.

Consider moment at B

ΣMB=0MB=RA(20)2(20)(202)=64.35(20)400MB=887k-ft

Therefore, the maximum moment at B for the given loading condition is 887k-ft_.

(c)

To determine

Find the maximum moment at B due to wheel loads using moment distribution method.

(c)

Expert Solution
Check Mark

Answer to Problem 53P

The maximum moment at B due to wheel loads using moment distribution method is 179.8k-ft_.

Explanation of Solution

The maximum moment at B is produced when wheel load is placed at B.

Draw the free body diagram of the beam for the wheel loading as in Figure 8.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  8

Refer Figure 8.

Find the fixed end moment at each end of the member AC as follows;

FEMAC=Pab2L2PL8=9(20)(40)26029(60)8=8067.5=147.5k-ft

Find the fixed end moment at each end of the member CA as follows;

FEMCA=Pa2bL2PL8=9(20)2(40)602+9(60)8=40+67.5=107.5k-ft

Show the moment distribution computations as in Table 6.

JointACD
MemberACCACDDC
DF 0.50.5
FEM‑147.5107.500
Released moment+147.5000
COM 73.75  
Total0181.2500
DEM‑90.6390.63
COM   
Final0‑90.6390.630

Find the reaction at A using section AC.

Draw the free body diagram of section AC as in Figure 9.

Connect Access Card For Fundamentals Of Structural Analysis (one Semester Access) 5th Edition, Chapter 12, Problem 53P , additional homework tip  9

Refer Figure 7.

Find the reaction at A.

Consider moment at C

ΣMC=0RA(60)9(40)9(30)+90.63=060RA=360+27090.63RA=8.99k

Find the moment at B.

Consider moment at B

ΣMB=0MB=RA(20)=8.99(20)MB=179.8k-ft

Therefore, the maximum moment at B for the given loading condition is 179.8k-ft_.

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