   # 13.5 through 13.8 Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P13.1–P13.4 by using the method of consistent deformations. Select the reaction moment at the fixed support to be the redundant. FIG. P13.2, P13.6

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Chapter 13, Problem 6P
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## 13.5 through 13.8 Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P13.1–P13.4 by using the method of consistent deformations. Select the reaction moment at the fixed support to be the redundant.FIG. P13.2, P13.6 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 free body diagram of the beam as shown in Figure 1.

Refer Figure 1.

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

Consider the moment at A is denoted by MA.

Consider the vertical reaction at B is denoted by By.

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 moment at A as the redundant.

Modify the Figure 1 as shown in Figure 2.

Refer Figure 2.

The slope at A due to the external loading is denoted by θAO.

The flexibility coefficient representing the slope at A due to unit value of redundant MA is fAA.

Slope of the Primary Beam:

Show the expression for the slope at the hinge support of a simply supported beam due to the uniformly distributed load w as follows:

θ=wL324EI (1)

Here, L is the length of the beam, E is the modulus of Elasticity, and I is the moment of inertia.

Show the expression for the slope at the hinge support of a simply supported beam due to the unit moment M as follows:

fAA=Ml3EI (2)

Here, L is the length of the beam, M is the moment acting at the hinge support, E is the modulus of Elasticity, and I is the moment of inertia.

Substitute 1kNm for M in Equation (2).

fAA=L3EI (3)

Show the compatibility Equation as follows:

θAO+fAAMA=0 (4)

Modify Equation (4) using Equation (2) and (3).

wL324EI+L3EI×MA=0MA=wL324EI×3EILMA=wL28

Refer Figure 1.

Take the sum of the moment at A as follows:

MAwL×L2+By×L=0wL28wL22+ByL=0ByL=3wL28By=3wL8()

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

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