   # 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. P13.13

#### Solutions

Chapter
Section
Chapter 13, Problem 13P
Textbook Problem
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## 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. P13.13

To determine

Find the reactions and sketch the shear and bending moment diagrams for the given beam using method of consistent deformation.

### Explanation of Solution

Given information:

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 negative and the counter clockwise moment as positive.

Calculation:

Show the beam as shown in the Figure 1.

Refer Figure 1.

Consider the horizontal and vertical reaction at A are denoted by Ax and Ay.

Consider the vertical reaction at B and D is denoted by By and Dy.

The reactions acting in the beam is 4.

The number of Equilibrium reaction is 3.

The degree of indeterminacy of the beam is 1.

Take the vertical reaction at B as the redundant.

Modify the Figure 1 as shown in Figure 2.

Refer Figure 2.

Consider a section at a distance x from both the supports A and C.

The deflection at B due to the external loading is denoted by ΔBO.

The flexibility coefficient representing the deflection at B due to unit value of redundant By is fBB.

Primary beam subjected to unit value of redundantBy.

Calculate the value of reaction at A and D as follows:

Ay=Dy=12

Calculate the value of deflection fBB as follows:

fBB=1EI[012(x2)2dx+23012(x2)2dx]=1EI[14012x2dx+16012x2dx]=1EI[14(x33)012+16(x33)012]=1EI[14(12330)+16(12330)]

fBB=1EI[144+96]=240kNm3/kNEI        (1)

Take the sum of the moment about A as zero.

MA=025×12×6250×18+Dy×24=0Dy=630024Dy=262.5kN

Take the sum of the forces in the vertical direction as zero.

Fx=0Ay+Dy=25×12+250Ay+Dy=550

Substitute 262.5kN for Dy.

Ay+262.5=550Ay=550262.5Ay=287.5kN

Calculate the value of deflection ΔBO as follows:

ΔBO=1EI[012(287.5x25x22)(x2)dx+2306(262.5x)(x2)dx+23612(262.5x250(x6))(x2)dx]=1EI[12012(287.5x2+25x32)dx1306(262.5x2)dx13612(12.5x2+1,500x)dx]=1EI[12(287.5x33+25x42×4)01213(262.5x33)0613(12.5x33+1,500x22)612]=1EI[12(287

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