   # 13.13 through 13.25 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.13–P13.25 using the method of consistent deformations. FIG. P13.18

#### Solutions

Chapter
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Chapter 13, Problem 18P
Textbook Problem
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## 13.13 through 13.25 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.13–P13.25 using the method of consistent deformations. FIG. P13.18

To determine

Calculate the support reactions for the given structure using method of consistent deformation.

Sketch the shear and bending moment diagrams for the given structure.

### Explanation of Solution

Given information:

The structure is given in the Figure.

Apply the sign conventions for calculating reactions, forces and moments using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
• For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
• For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as positive when moment taken from left and the counter clockwise moment as positive when moment taken from right.

Calculation:

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

Refer Figure 1.

Find the degree of indeterminacy of the structure:

Degree of determinacy of the beam is equal to the number of unknown reactions minus the number of equilibrium equations.

The beam is supported by 4 support reactions (r) and the number of equilibrium equations (e) is 3.

i=re=43=1

Therefore, the degree of indeterminacy of the beam is i=2.

Select the vertical reaction By at the supports B as redundant.

Release the support B and consider the deflection due to external load at B as ΔBO

Sketch the loading diagram with deflected shape as shown in Figure 2.

Refer Figure 2.

The formula for deflection at free end of cantilever beam with uniformly varying load as follows.

Δmax=wL430EI        (1)

Here, w is the uniformly varying load through the beam, L is the total length of the beam, and EI is the flexural rigidity.

Find the deflection at B.

Substitute h for L in Equation (1).

ΔBO=wh430EI

Find the deflection at B due to unit load at B.

Consider the upward deflection due to unit load at B as fBB

Sketch the beam with unit load at B with deflected shape due to unit load as shown in Figure 3.

Refer Figure 3.

The formula for deflection at free end of cantilever beam with point load at free end as follows.

Δmax=WL33EI        (2)

Here, W is the end point load, L is the total length of the beam, EI is the flexural rigidity.

Find the deflection at B.

Substitute 1 unit for W and h for L in Equation (2).

fBB=(1)h33EI=h33EI

Find the vertical reaction at B.

Write the compatibility equation:

ΔBO+fBBBx=0

Substitute wh430EI for ΔBO and h33EI for fBB.

wh430EI+(h33EI)Bx=0h33EIBx=wh430EIBx=wh10

Therefore, the horizontal reaction at B is Bx=wh10_.

Refer Figure 1.

Summation of forces along x-axis is zero.

ΣFx=0Ax+Bx=12(h)(w)

Substitute wh10 for Bx.

Ax+wh10=12(h)(w)Ax=wh2wh10Ax=2wh5

Therefore, the horizontal reaction at A is Ax=2wh5_

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