   # Solve Problem 13.16 by the method of least work. See Fig. P13.16. 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. P13.16, P13.59

#### Solutions

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Chapter 13, Problem 59P
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## Solve Problem 13.16 by the method of least work. See Fig. P13.16.13.13 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. P13.16, P13.59 To determine

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

### 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 horizontal and vertical reaction at B is denoted by By.

Consider the moment at A is denoted by MA.

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 reaction at B as the redundant.

Consider a section at a distance x from support C as shown in Figure 2.

Refer to Figure 2.

Determine the bending moment at x for segment CB.

Mx=100x

Differentiate with respect to By.

MxBy=0

Determine the bending moment at x for segment BA.

Mx=100×x+By(x6)10×(x6)×(x6)2=100x+By(x6)5(x6)2

Differentiate with respect to By.

MxBy=x6

Calculate the reaction at B (By) using the method of least work as shown below.

UBy=(MxBy)MxEIdx=0        (1)

Substitute 100x for Mx and 0 for MxBy for the span CB and 100x+By(x6)5(x6)2 for Mx and x6 for MxBy for the span BA in Equation (1).

06(0)100xE(2I)dx+618(x6)(100x+By(x6)5(x6)2)EIdx=01EI618(100x(x6)+By(x6)25(x6)3)dx=0618(100x2+600x+By(x2+3612x)5(x321618x2+108x))dx=0618(100x2+600x+By(x2+3612x)5x3+1,080+90x2540x)dx=0

618(10x2+60x5x3+1,080+By(x2+3612x))dx=06</

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