   # 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. P13.20

#### Solutions

Chapter
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Chapter 13, Problem 20P
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## 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. P13.20

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 4 support reactions and the number of equilibrium equations is 3.

Therefore, the degree of indeterminacy of the frame is i=1.

Select the horizontal reaction Dx at the support D as redundant.

Release the horizontal support at D 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 D as shown in Figure 1.

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

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

MAO=0DyO(15)2(15)(152)40(10)=015DyO=625DyO=41.67k()

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

+Fy=0AyO+41.672(15)=0AyO=11.67kAyO=11.67kN()

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

+Fx=0AxO+40=0AxO=40k()

For unit value of the unknown redundant Dx.

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

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

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=0DyDX(15)+1(20)=015DyDX=20DyDX=1.33kDyDX=1.33k()

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

+Fy=0AyDX1.33=0AyDX=1.33k()

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

+Fx=0AxDX=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 3.

Refer Figure 3.

Consider origin as A. (0x10ft).

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

MO=40(x)

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

 Segment x-coordinate MO(k-ft) mDX(k-ft/k) Origin Limits (ft) AB A 0−10 40x −x BC A 10−20 40x−40(x−10) −x DC D 0−15 41.67x−x2 −1.33x

Let the vertical deflection at point D due to external loading is ΔDO, and the flexibility coefficient representing the deflection at D due to unit value of redundant Dx is fDD.

Calculate the value of ΔDO using the equation as follows:

ΔDO=ΣMOmDYEIdx=1EI[010(40x)(x)dx+1020[40x40(x10)](x)dx+015(41.67xx2)(1

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