   # 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. P13.43

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Chapter 13, Problem 43P
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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. P13.43 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:

Find the degree of indeterminacy of the structure:

Degree of determinacy of the frame is equal to the number of unknown reactions minus the number of equilibrium equations.

The frame is supported by 5 support reactions and the number of equilibrium equations is 3.

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

Select the vertical reaction Ax and Ay at the supports A as redundant.

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

Sketch the free body diagram of primary frame subjected to external loading without support A as shown in Figure 1.

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

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

MEO=0MO=25(16)2.5(26)(262)=445k-ftMO=445k-ft()

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

+Fy=0EyO=2.5(26)AyO=65k()

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

+Fx=0ExO25=0ExO=25k()

For unit value of the unknown redundant Ay:

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

Sketch the free body diagram of primary frame Subjected to unit value of redundant Ay as shown in Figure 2.

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

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

ME=0MEAY=1(20)MEAY=20k-ft(counterclockwise)

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

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

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

+Fx=0ExAY=0

For unit value of the unknown redundant Ax:

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

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

Find the support reaction and moment at A when 1 k horizontal load applied at D.

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

MA=0MEAX=1(24)MEAX=24k-ft()

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

+Fy=0EyAX=0

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

+Fx=0ExAX+1=0ExAX=1k()

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

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 4.

Refer Figure 4.

Consider origin as A. (0x8ft).

Find the moment at section XX in the portion AB as shown in Figure 4.

MO=0

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

 Segment x-coordinate MO (k-ft) mAX (k-ft/k) mDY (k-ft/k) Origin Limits (ft) AB A 0−8 0 −1x 0 BD A 8−24 25(x−8) −1x 0 ED E 0−20 −445+65x−1.25x2 −24 20−x

Let the horizontal deflection at point A due to external loading is ΔAXO, vertical deflection at point A due to external loading is ΔAYO,  and the flexibility coefficient representing the deflection at A due to unit value of redundant Ax is fAX,AX, the redundant Ay is fAY,AY, and the both redundant AxandAy is FAY,AX.

Calculate the value of ΔAXO using the equation as follows:

ΔAXO=ΣMOmAXEIdx=1EI[08(0)(1x)dx+82425(x8)(1x)dx+09(445+65x1.25x2)(24)dx]=1EI[0+(59,733)+(18,400)]=78,133EIk-ft3

Calculate the value of ΔAYO using the equation as follows:

ΔAYO=ΣMOmAYEIdx=1EI[08(0)(0)dx+82425(x8)(0)dx+09(445+65x1.25x2)(20x)dx]=1EI[(0)+(0)+(19,000)]=19,000EIk-ft3

Calculate the value of fAX,AX using the equation as follows:

fAX,AX=ΣmAX2EIdx=1EI[08(1x)2dx+824(1x)2dx+020(24)2dx]=1EI[(170.67)+(4,437.3)+(11,520)]=16,128EIk-ft3/k

Calculate the value of fAY,AY using the equation as follows:

fAY,AY=ΣmAY2EIdx=1EI[08(0)2dx+824(0)2dx+020(20x)2dx]=1EI[(0)+(0)+(2,666

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