   # Draw the influence lines for the reaction moment at support A , the vertical reactions at supports A and B , and the shear at the internal hinge C of the frame shown in Fig. P8.38. FIG. PB.38

#### Solutions

Chapter
Section
Chapter 8, Problem 38P
Textbook Problem
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## Draw the influence lines for the reaction moment at support A, the vertical reactions at supports A and B, and the shear at the internal hinge C of the frame shown in Fig. P8.38. FIG. PB.38

To determine

Draw the influence lines for the reaction moment at support A, the vertical reactions at supports A and B and the shear at the internal hinge C.

### Explanation of Solution

Calculation:

Influence line for vertical reaction at support B.

Apply a 1 k unit moving load at a distance of x from left end B.

Sketch the free body diagram of frame as shown in Figure 1.

Refer Figure 1.

Apply 1 k load just left of C (0x10ft).

Consider section BC.

Consider moment equilibrium at C.

Take moment at C from B.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMC=0By(10)1(10x)=010By=10xBy=1x10

Apply 1 k load just right of C (10ftx40ft).

Consider section BC.

Consider moment equilibrium at C.

Take moment at C from B.

Consider clockwise moment as positive and anticlockwise moment as negative.

ΣMC=0By(10)=0By=0

Thus, the equation of vertical support reaction at B as follows,

By=1x10, (0x10ft)        (1)

By=0, (10ftx40ft)        (2)

Find the influence line ordinate of By at B using Equation (1).

Substitute 0 for x in Equation (1).

By=1010=1k

Thus, the influence line ordinate of By at B is 1k/k.

Similarly calculate the influence line ordinate of By at various points on the frame and summarize the values in Table 1.

 x (ft) Points Influence line ordinate of By(k/k) 0 B 1 10 C 0 20 D 0 30 E 0 40 F 0

Sketch the influence line diagram for vertical support reaction at B using Table 1 as shown in Figure 2.

Influence line for vertical reaction at support A.

Apply a 1 k unit moving load at a distance of x from left end C.

Refer Figure 1.

Find the vertical support reaction (Fy) at F using equilibrium equation:

Apply 1 k load just left of E (0x30ft).

Consider section EF.

Consider moment equilibrium at point E.

Consider clockwise moment as positive and anticlockwise moment as negative

ΣME=0Fy(10)=0Fy=0

Apply 1 k load just right of E (30ftx40ft).

Consider section EF.

Consider moment equilibrium at point E.

Consider clockwise moment as positive and anticlockwise moment as negative

ΣME=0Fy(10)+1(x30)=010Fy=x+30Fy=3x10

Thus, the equation of vertical support reaction at F as follows,

Fy=0, (30ftx40ft)        (3)

Fy=3x10, (30ftx40ft)        (4)

Apply a 1 k unit moving load at a distance of x from left end B.

Refer Figure 1.

Apply vertical equilibrium in the system.

Consider upward force as positive and downward force as negative.

Ay+ByFy1=0Ay=1By+Fy        (5)

Find the equation of vertical support reaction (Ay) from B to C (0x10ft) using Equation (5).

Substitute 1x10 for By and 0 for Fy in Equation (5).

Ay=1(1x10)+0=11+x10=x10

Find the equation of vertical support reaction (Ay) from C to E (10ftx30ft) using Equation (5).

Substitute 0 for By and 0 for Fy in Equation (5).

Ay=100=1k

Find the equation of vertical support reaction (Ay) from E to F (30ftx40ft) using Equation (5).

Substitute 0 for By and 3x10 for Fy in Equation (5).

Ay=10+(3x10)=1+3x10=4x10

Thus, the equation of vertical support reaction at A as follows,

Ay=x10, (0x10ft)        (6)

Ay=1k, (10ftx30ft)        (7)

Ay=4x10, (30ftx40ft)        (8)

Find the influence line ordinate of Ay at F using Equation (8).

Substitute 40 ft for x in Equation (8).

Ay=44010=44=0

Thus, the influence line ordinate of Ay at F is 0k/k.

Similarly calculate the influence line ordinate of Ay at various points on the beam and summarize the values in Table 2.

 x (ft) Points Influence line ordinate of Ay(k/k) 0 B 0 10 C 1 20 D 1 30 E 1 40 F 0

Sketch the influence line diagram for the vertical reaction at support A using Table 3 as shown in Figure 3

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