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Structural Analysis

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

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Chapter
Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 13, Problem 42P
Textbook Problem
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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. P13.42

Chapter 13, Problem 42P, 13.37 through 13.45 Determine the reactions and draw the shear and bending moment diagrams for the

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.
  • Consider the positive sign indicates the clockwise moment the negative sign indicates the counterclockwise moment.

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 Dx and Dy at the supports D as redundant.

Release the support 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 D using equilibrium equations:

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

mAO=0MO=25(9)(92)75(3)=1,237.5kN-mMO=1,237.5kN-m()

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

+Fy=0AyO=25(9)AyO=225kN()

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

+Fx=0AxO+75=0AxO=75kN()

For unit value of the unknown redundant Dy:

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

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

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

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

MA=0MADY=1(9)MADY=9kN-m()

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

+Fy=0AyDY+1=0AyDY=1kN()

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

+Fx=0AxDY=0

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 3.

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

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

MA=0MADX=1(6)MADX=6kN-m()

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

+Fy=0AyDX=0AyDX=0

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

+Fx=0AxDX+1=0AxDX=1kN()

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. (0x3m).

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

MO=1,237.5+75x

Similarly calculate the moment of the remaining section in the external loading and redundant loading structures.

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

Segmentx-coordinateMO (kN-m)mDX (kN-m/kN)mDX (kN-m/kN)
OriginLimits (m)
ABA031,237.5+75x6x9
BCA361,012.56x9
DCD0912.5x20x

Let the horizontal deflection at point D due to external loading is ΔDXO, vertical deflection at point D due to external loading is ΔDYO,  and the flexibility coefficient representing the deflection at D due to unit value of redundant Dx is fDX,DX, the redundant Dy is fDY,DY, and the both redundant DxandDy is FDY,DX.

Calculate the value of ΔDXO using the equation as follows:

ΔDXO=ΣMOmDXEIdx=1EI[03(1,237.5+75x)(6x)dx+36(1,012.5)(6x)dx+09(12.5x2)(0)dx]=1EI[(15,356.3)+(4,556.25)+0]=19,912.5EIkN-m3

Calculate the value of ΔDYO using the equation as follows:

ΔDYO=ΣMOmDYEIdx=1EI[03(1,237.5+75x)(9)dx+36(1,012.5)(9)dx+09(12.5x2)(x)dx]=1EI[(30,375)+(27,337.5)+(20,503.1)]=78,215

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Chapter 13 Solutions

Structural Analysis
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