# 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. P13.45

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

Chapter
Section

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 13, Problem 45P
Textbook Problem
390 views

## 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. P13.45

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

Therefore, the degree of indeterminacy of the frame is i=3.

Select the reaction Bx, By and MB at the support B as redundant.

Release the support B 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 B 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 A is equal to 0.

MA=0MAO=20(30)1.5(40)(402)MAO=1,800k-ftMAO=1,800k-ft()

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

+Fx=0AxO+20=0AxO=20k()

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

+Fy=0AyO1.5(40)=0AyO=60k()

For unit value of the unknown redundant Bx:

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

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

Find the support reaction at A when 1 k horizontal load applied at B.

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

+Fx=0AxBX1=0AxBX=1k()

For unit value of the unknown redundant By:

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

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

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

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

MA=0mABY=1(40)mABY=40k-ft()

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

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

For unit value of the unknown redundant MB:

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

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

Refer Figure 4.

Find the support reaction and moment at A when 1 k-ft unit moment applied at B.

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

MA=0mAMB=1mAMB=1k-ft()

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 5.

Refer Figure 5.

Portion AC.

Consider origin as A. (0x30ft).

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

MO=1,800+20(x)=1,800+20x

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

 Segment x-coordinate Moment of Inertia MO (k-ft) mBX (k-ft/k) mBY (k-ft/k) mMB (k-ft/k) Origin Limits (ft) AC A 0−30 I −1,800+20x −x 40 1 DC D 0−40 2I 1.5x(x2)=0.75x2 −30 x 1 BD B 0−30 I 0 x 0 −1

The horizontal deflection at point B due to external loading is ΔBXO, vertical deflection at point B due to external loading is ΔBYO, deflection at B due to moment is ΘBO and the flexibility coefficient representing the deflection at B due to unit value of redundant Bx,By,andMB are fBX,BX, fBY,BY, fMB,MB, fBX,BY, fBX,MB, and fBY,MB.

Calculate the value of ΔBXO using the equation as follows:

ΔBXO=ΣMOmBXEIdx=[0301EI(1,800+20x)(x)dx+04012EI(0.75x2)(30)dx+1EI030(0)(x)dx]=1EI[630,000+480,0002+0]=870,000EIk-ft3

Calculate the value of ΔBYO using the equation as follows:

ΔBYO=ΣMOmBYEIdx=[0301EI(1,800+20x)(40)dx+04012EI(0.75x2)(x)dx+1EI030(0)(0)dx]=1EI[1,800,000480,0002+0]=2,040,000EIk-ft3

Calculate the value of ΘBO using the equation as follows:

ΘBO=ΣMOmMBEIdx=[0301EI(1,800+20x)(1)dx+04012EI(0.75x2)(1)dx+1EI030(0)(1)dx]=1EI[45,00016,0002+0]=53,000EIk-ft3

Calculate the value of fBX,BX using the equation as follows:

fBX,BX=ΣmBX2EIdx=[1EI030(x)2dx+12EI040(30)2dx+1EI030(x)2dx]=1EI[9,000+36,0002+9,000]=36,000EIk-ft3/k

Calculate the value of fBY,BY using the equation as follows:

fBY,BY=ΣmBY2EIdx=[1EI030(40)2dx+12EI040(x)2dx+1EI030(0)2dx]=1EI[48,000+21,333.332+0]=58,666

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