   # Draw the influence lines for the vertical reactions at supports A and E and the shear and bending moment at point D of the frame shown in Fig. P8.16. FIG. P8.16

#### Solutions

Chapter
Section
Chapter 8, Problem 16P
Textbook Problem
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## Draw the influence lines for the vertical reactions at supports A and E and the shear and bending moment at point D of the frame shown in Fig. P8.16. FIG. P8.16

To determine

Draw the influence lines for the vertical reactions at supports A and E.

Draw the influence lines for the shear and bending moment at point D.

### Explanation of Solution

Calculation:

Apply a 1 kN 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.

Find the horizontal support reaction (Ax) at A using equilibrium equation:

Apply Force equilibrium along horizontal.

Consider the force acting towards right side as positive (+) and the force acting towards left side as negative ().

ΣFx=0Ax=0

The horizontal reaction at A is zero. Therefore, there is no influence line diagram for horizontal reaction (Ax) at A.

Find the vertical equation of support reaction (Ay) at A using equilibrium equation:

Consider moment equilibrium at point E.

Consider clockwise moment as positive and anticlockwise moment as negative

Sum of moment at point E is zero.

ΣME=0Ay(10)1(13x)=010Ay=13xAy=13x10 (1)

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

Substitute 0 for x in Equation (1).

Ay=13010=1.3kN/kN

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

 x (m) Points Influence line ordinate of Ay(kN/kN) 0 B 1.3 3 C 1 8 D 0.5 13 E 0 16 F ‑0.3

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

Refer Figure 1.

Find the equation of support reaction (Ey) at E using equilibrium equation:

Consider moment equilibrium at point A.

Consider clockwise moment as negative and anticlockwise moment as positive.

Sum of moment at point A is zero.

ΣMA=0Ey(10)1(x3)=010Ey=x3Ey=x310 (2)

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

Substitute 0 for x in Equation (2).

Ey=0310=0.3kN/kN

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

 x (m) Points Influence line ordinate of Ey(kN/kN) 0 B ‑0.3 3 C 0 8 D 0.5 13 E 1 16 F 1.3

Sketch the influence line diagram for the vertical reaction at supportE using Table 1 as shown in Figure 3.

Find the equation of shear force at D of portion BCD(0x<8m).

Sketch the free body diagram of the section BCD as shown in Figure 4.

Refer Figure 4.

Apply equilibrium equation of forces.

Consider upward force as positive (+) and downward force as negative ().

ΣFy=0

AySD1=0SD=Ay1

Substitute 13x10 for Ay.

SD=13x101=13x1010=3x10

Find the equation of shear force at D of portion DEF(8m<x16m).

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

Refer Figure 5.

Apply equilibrium equation of forces.

Consider upward force as positive (+) and downward force as negative ()

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