UCD FUND OF STRUCTURAL ANALYSIS 5E
UCD FUND OF STRUCTURAL ANALYSIS 5E
5th Edition
ISBN: 9781264843923
Author: Leet
Publisher: MCG
Question
Book Icon
Chapter 12, Problem 40P

(a)

To determine

Draw the influence lines for the horizontal and vertical reactions at support A and the member forces in members BC, CM, and ML.

(a)

Expert Solution
Check Mark

Explanation of Solution

Influence line for the vertical reaction at A.

The influence line ordinate of (Ay) are determined by applying a 1-kip load at successive panel points.

Apply 1 kip at D.

Find the vertical reaction at support A.

The reaction is determined by applying a 1-kip load at successive panel points.

Apply 1 kip load at C.

Draw the free body diagram as shown in Figure 1.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  1

Refer Figure 1.

Consider moment equilibrium at I.

ΣMI=0Ay(144)=1(96)Ay=0.667k

Thus, the influence line ordinate of vertical reaction at A is 0.667.

Similarly find the influence line ordinate of Ay by applying 1 kip at A, B, D, E, F, G, and H and tabulate the values in Table 1.

Pointsx (ft)Influence line ordinate of Ay
B01
C240.833
D480.667
E720.5
F960.333
G1200.167
H1440

Draw the influence line ordinate of Ay using Table 1 as in Figure 2.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  2

Influence line for the horizontal reaction  at A.

The influence line ordinate of (Ax) are determined by applying a 1-kip load at successive panel points.

Apply 1 kip at D.

Find the horizontal reaction at support A.

The reaction is determined by applying a 1-kip load at successive panel points.

Apply 1 kip load at C.

Draw the free body diagram as shown in Figure 3.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  3

Refer Figure 4.

Consider moment equilibrium at E.

ΣMI=0Ay(72)Ax(36)1(24)=00.667(72)36Ax=2436Ax=24Ax=0.667k

Thus, the influence line ordinate of horizontal reaction at A is 0.667.

Similarly find the influence line ordinate of Ax by applying 1 kip at A, B, D, E, F, G, and H and tabulate the values in Table 2.

Pointsx (ft)Influence line ordinate of Ax
B00
C240.333
D480.667
E721
F960.667
G1200.333
H14400

Draw the influence line ordinate of Ax using Table 2 as in Figure 4.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  4

Influence line for the force in member BC.

Find the force (FBC) in member BC.

The influence line ordinate of FBC are determined by applying a 1-kip load at successive panel points.

Apply 1 kip at C and consider a section passes through members BC, CM, and ML.

Sketch the free body diagram of the section as in Figure 5.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  5

Refer Figure 5.

Find the member force BC.

Consider moment equilibrium at M.

ΣMM=00.667(20)0.667(24)FBC(16)=0FBC=0.167k

Thus, the influence line ordinate of member force BC at D is ‑0.167.

Similarly find the influence line ordinate of FBC by applying 1 kip at B, C, D, E, F, G, and H and tabulate the values in Table 4.

Pointsx (ft)Influence line ordinate of FBC
B00
C24‑0.833
D48‑0.167
E720.5
F960.333
G1200.167
H1440

Draw the influence line ordinate of FBC using Table 4 as in Figure 6.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  6

Influence line for the force in member ML.

Find the force (FML) in member ML.

The influence line ordinate of FML are determined by applying a 1-kip load at successive panel points.

Apply 1 kip at C and consider a section passes through members BC, CM, and ML.

Sketch the free body diagram of the section as in Figure 7.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  7

Refer Figure 7.

The slope of member ML is 24 ft horizontal and 12 ft vertical.

Find the member force ML.

Consider horizontal equilibrium equation.

ΣFx=00.667+(0.167)+FML(24242+122)=00.894FML=0.5FML=0.56kips

Thus, the influence line ordinate of member force ML at D is ‑0.56.

Similarly find the influence line ordinate of FML by applying 1 kip at B, C, D, E, F, G, and H and tabulate the values in Table 5.

Pointsx (ft)Influence line ordinate of FML
B00
C240.56
D48‑0.56
E72‑1.68
F96‑1.12
G120‑0.56
H1440

Draw the influence line ordinate of FML using Table 5 as in Figure 8.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  8

Influence line for the force in member CM.

Find the force (FCM) in member CM.

The influence line ordinate of FCM are determined by applying a 1-kip load at successive panel points.

Apply 1 kip at C and consider a section passes through members BC, CM, and ML.

Sketch the free body diagram of the section as in Figure 9.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  9

Refer Figure 9.

The slope of member ML is 24 ft horizontal and 12 ft vertical.

Find the member force CM.

Consider vertical equilibrium equation.

ΣFy=00.667+FMC+FML(12242+122)=00.667+FMC+(0.56)(12242+122)=00.667+FCM+(0.56)(0.447)=0FCM=0.416k

Thus, the influence line ordinate of member force CM at D is ‑0.416.

Similarly find the influence line ordinate of FCM by applying 1 kip at B, C, D, E, F, G, and H and tabulate the values in Table 5.

Pointsx (ft)Influence line ordinate of FCM
B00
C24‑1.083
D48‑0.416
E720.25
F960.167
G1200.083
H1440

Draw the influence line ordinate of FCM using Table 5 as in Figure 10.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  10

(b)

To determine

Find the forces (compression and tension) in bars CM and ML produced by the dead load.

(b)

Expert Solution
Check Mark

Answer to Problem 40P

The dead load force in member CM is 115.1k(C)_

The dead load force in member ML is 387.07k(C)_

Explanation of Solution

Given Information:

The uniform dead load, wD is 4.8 kip/ft.

Calculation:

Refer Figure 11.

Sketch the influence line diagram of member CM as in Figure 11.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  11

Refer Figure 11.

Find the length x1 using similar triangle.

x10.416=x1+LCD1.0831.083x1=0.416x1+0.416LCD1.083x10.416x1=0.416(24)x1=15ft

Find the length x2 using similar triangle.

x20.25=x10.416x2=150.416(0.25)x2=9ft

Refer Figure 11.

Find the dead load force in member CM using the equation.

FCM=wD(Areaofinfluence line diagram)=wD[12(1.083)(LBD+x1)+12(0.25)(x2+LEH)]=4.8[0.5(1.083)(48+15)+0.5(0.25)(9+72)]=4.8(34.112+10.125)=115.1k=115.1k(Compresion)

Therefore, the dead load force in member CM is 115.1k(C)_

Refer Figure 9.

Sketch the influence line diagram of member ML as in Figure 12.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  12

Refer Figure 12.

Find the length x3 and x4 using similar triangle.

x40.56=x4+LDE1.68x40.56=x4+241.681.68x4=0.56x4+13.44x4=12ft

Find the length x3 and x4 using similar triangle.

x30.56=x4+LDE1.681.68x3=0.56(x4)+0.56LDE1.68x3=0.56(12)+0.56(24)x3=12ft

Refer Figure 11.

Find the dead load force in member CM using the equation.

FML=wD(Areaofinfluence line diagram)=wD[12(0.56)(LBC+x3)+12(1.68)(x4+LDH)]=4.8[0.5(0.56)(24+12)+0.5(1.68)(12+96)]=4.8(10.0890.72)=387.07k=387.07k(Compresion)

Therefore, the dead load force in member ML is 387.07k(C)_

(c)

To determine

Find the forces (compression and tension) in bars CM due to live load.

(c)

Expert Solution
Check Mark

Answer to Problem 40P

The maximum compression force in member CM due to live load is 48.95k_.

The maximum tension force in member CM due to live load is 13.1k_

Explanation of Solution

Given Information:

The uniform live load, wL is 0.8 kip/ft.

The concentrated live load, P is 20 k.

Calculation:

Refer Figure 11.

Sketch the influence line diagram of member CM as in Figure 13.

UCD FUND OF STRUCTURAL ANALYSIS 5E, Chapter 12, Problem 40P , additional homework tip  13

Refer Figure 13.

Find the maximum compression force in member CM using the equation.

FCM=wL(negative areaofinfluence line diagram)+P(maximumnegativeordinate)=0.8[12(1.083)(LBD+x1)]+20(1.083)=0.8[0.5(1.083)(48+15)]+21.66=48.95k

Therefore, the maximum compression force in member CM due to live load is 48.95k_.

Find the maximum tension force in member CM using the equation.

FCM=wL(positive areaofinfluence line diagram)+P(maximumpositiveordinate)=0.8[12(0.25)(LEH+x2)]+20(0.25)=0.8[0.5(0.25)(72+9)]+5=13.1k

Therefore, the maximum tension force in member CM due to live load is 13.1k_.

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