   # 13.37 through 13.45 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.37–P13.45 using the method of consistent deformations. FIG.P13.38

#### Solutions

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Chapter 13, Problem 38P
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## 13.37 through 13.45 Determine the reactions and draw the shear and bending moment diagrams for the structures shown in Figs. P13.37–P13.45 using the method of consistent deformations. FIG.P13.38

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 and the number of equilibrium equations is 2.

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

Select the vertical reaction By and Dy at the supports B and D as redundant.

Release the support B and D and consider the notation of moments due to external load as MO.

Sketch the free body diagram of primary beam subjected to external loading without support B and D as shown in Figure 2.

Refer Figure 2.

Find the reactions at the supports without considering support B and D using equilibrium equations:

Summation of moments of all forces about E is equal to 0.

ME=0MEO1(10)(102+40)35(30)2(20)(202)=0MEO=1,900k-ft()

Summation of forces along y-direction is equal to 0.

+Fy=0EyO1(10)352(20)=0EyO=85k()

For unit value of the unknown redundant Dy:

Consider the notation of moments due to external load as mD.

Sketch the free body diagram of primary beam Subjected to unit value of redundant Dy as shown in Figure 3.

Refer Figure 3.

Find the support reaction and moment at E when 1 k vertical load applied at B.

Summation of moments of all forces about A is equal to 0.

ME=0MEB=1(40)MEB=40k-ft()

Summation of forces along y-direction is equal to 0.

+Fy=0EyB+1=0EyB=1k()

For unit value of the unknown redundant By:

Consider the notation of moments due to external load as mB.

Sketch the free body diagram of primary beam Subjected to unit value of redundant By as shown in Figure 4.

Refer Figure 4.

Find the support reaction and moment at E when 1 k vertical load applied at D.

Summation of moments of all forces about E is equal to 0.

MD=0MED=1(20)MED=20k-ft()

Summation of forces along y-direction is equal to 0.

+Fy=0EyD+1=0EyD=1k()

Find the moment equation of the beam for different sections on the beam.

Consider a section XX in the portion AB of the primary structure at a distance of x from A.

Refer Figure 1.

Draw the primary structure with section XX as shown in Figure 5.

Refer Figure 5.

Portion AB.

Consider origin as A. (0x10ft).

Find the moment at section XX in the portion AB as shown in Figure 5.

MO=1(x)(x2)=x22

Tabulate the moment equation of different segment of beam as in Table 1.

 Segment x-coordinate MO (k-ft) mBY (k-ft/k) mDY (k-ft/k) Origin Limits (ft) AB A 0−10 x22 0 0 BC E 10−20 1(10)(x−102) −1(x−10) 0 CD G 20−30 1(10)(x−5)+35(x−20) −1(x−10) 0 DE D 30−50 [1(10)(x−5)+35(x−20)+(x−30)2] −1(x−10) −1(x−30)

The vertical deflection at point B due to external loading is ΔBO, vertical deflection at point D due to external loading is ΔDO,  and the flexibility coefficient representing the deflection at B due to unit value of redundant By is fBB, the redundant Dy is fDD, and the both redundant By and Dy is fBD=fDB.

Calculate the value of ΔBO using the equation as follows:

ΔBO=ΣMOmBYEIdx=1EI{010(x22)(0)dx+1020[1(10)(x102)][1(x10)]dx+2030[1(10)(x5)+35(x20)][1(x10)]dx+3050[1(10)(x5)+35(x20)+(x30)2][1(x10)]dx}=1EI[05,833.3360,000753,333.33]=819,166.67EIk-ft3

Calculate the value of ΔDO using the equation as follows:

ΔDO=ΣMOmDYEIdx=1EI{010(0)dx+1020(0)dx+2030(0)dx+3050[1(10)(x5)+35(x20)+(x30)2][1(x30)]dx}=1EI[0+0+0280,000]=280,000EIk-ft3

Calculate the value of fBB using the equation as follows:

fBB=ΣmBY2EIdx=1EI{010(0)2dx+1020[1(x10)]2dx+2030[1(x10)]2dx+3050[1(x10)]2dx}=1EI[0+333

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