   # Determine the reactions for the beam shown in Fig. P13.48 due to a small settlement Δ at the roller support C . FIG. P13.48

#### Solutions

Chapter
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Chapter 13, Problem 48P
Textbook Problem
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## Determine the reactions for the beam shown in Fig. P13.48 due to a small settlement Δ at the roller support C.FIG. P13.48 To determine

Calculate the reactions at the supports due to the small settlement at C for the given beam.

### Explanation of Solution

Given information:

The settlement at support C is Δ.

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 given 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 C is denoted by Cy.

The reactions acting in the beam is 4.

The number of Equilibrium reaction is 3. Then,

The degree of indeterminacy of the beam is 1.

Take the vertical reaction at C as the redundant.

The deflection at C due to the external moment is zero.

Modify the Figure 1 as shown in Figure 2.

Refer Figure 2.

The deflection coefficient representing the deflection at C due to unit value of redundant Cy is fCC.

The deflection at C due to unknown redundant Cy is fCCCy.

Determine the bending moment at x for segment CB.

Mx=1×x=x

Determine the bending moment at x for segment BA.

Mx=1×x=x

Calculate the value of fCC using the virtual work method as shown below.

fCC=(Mx)2EIdx

Substitute x for Mx for the span CB and x for Mx and 2I for I for the span BA.

fCC=0L2(x)2EIdx+L2L(x)22EIdx=1EI(0L2x2dx+12L2Lx2dx)=1EI[(x33)0L2+12(x33)L2L]=1EI[((L2)330)+12(L33(L2)33)

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