Structural Analysis, 5th Edition
Structural Analysis, 5th Edition
5th Edition
ISBN: 9788131520444
Author: Aslam Kassimali
Publisher: Cengage Learning
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Chapter 7, Problem 50P
To determine

Find the vertical deflection at joint E of the frame using virtual work method.

Expert Solution & Answer
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Answer to Problem 50P

The vertical deflection at joint E of the frame is 23mm_.

Explanation of Solution

Given information:

The frame is given in the Figure.

The value of E is 200 GPa and I is 350×106mm4.

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 negative and the counter clockwise moment as positive.

Calculation:

Consider the real system.

Draw a diagram showing all the given real loads acting on it.

Let the bending moment due to real load be M.

Sketch the real system of the frame as shown in Figure 1.

Structural Analysis, 5th Edition, Chapter 7, Problem 50P , additional homework tip  1

Find the reactions at the supports A and B:

Summation of moments about A is equal to 0.

MA=0By(10)15(15)(4.5)65(3)=0By=120.75kN

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

+Fy=0Ay+By15(15)=0Ay+120.7515(15)=0Ay=104.25kN

Consider ACDE, summation of moments about E is equal to 0.

ME=015(8)(4)+65(3)Ax(6)Ay(5)=0480+195Ax(6)104.25(5)=06Ax+153.75=0Ax=25.625kN

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

+Fx=0AxBx+65=025.625Bx+65=0Bx=39.375kN

Sketch the member end forces of the real system as shown in Figure 2.

Structural Analysis, 5th Edition, Chapter 7, Problem 50P , additional homework tip  2

Consider the virtual system.

Draw a diagram of frame without the given real loads. For vertical deflection apply unit load at the joint in the vertical direction.

Let the bending moment due to virtual load be Mv.

Sketch the virtual system of the frame with unit load at joint E as shown in Figure 3.

Structural Analysis, 5th Edition, Chapter 7, Problem 50P , additional homework tip  3

Find the reactions at the supports A and B:

Summation of moments about A is equal to 0.

MA=0By(10)1(5)=0By=12kN

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

+Fy=0Ay+121=0Ay=12kN

Consider ACDE, summation of moments about E is equal to 0.

ME=0Ax(6)Ay(5)=0Ax(6)12(5)=0Ax=512kN

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

+Fx=0AxBx=0512Bx=0Bx=512kN

Sketch the member end forces of the virtual system as shown in Figure 4.

Structural Analysis, 5th Edition, Chapter 7, Problem 50P , additional homework tip  4

Find the equations for M and Mv for the 5 segments of the frame as shown in Table 1.

Segmentx-coordinate

M

(kNm)

Mv

(kNm)

OriginLimits (m)
AHA0325.625x5x12
HDA3625.625x65(x3)5x12
BFB0639.375x5x12
DED0559.25x108.757.5x2x252
FEF0590.75x266.257.5x2x252

Find the vertical deflection at E using the virtual work expression:

ΔE=0LMvMEIdx (1)

Moment of inertia of span AH is I, moment of inertia of span HD is I, moment of inertia of span BF is I, moment of inertia of span DE is 2I, and moment of inertia of span FE is 2I.

Rearrange Equation (1) for the limits 03, 36, 06, 05, and 05 as follows.

ΔE=1EI[03(MvM)dx+36(MvM)dx+06(MvM)dx+1205(MvM)dx+1205(MvM)dx]

Substitute 5x12 for Mv, 25.625x for M for the limits 03, 5x12 for Mv, 25.625x65(x3) for M for the limits 36, 5x12 for Mv, 39.375x for M for the limits 06, x252 for Mv, 59.25x108.757.5x2 for M for the limits 05, x252 for Mv, 90.75x266.257.5x2 for M for the limits 05.

ΔE=1EI[03(5x12)(25.625x)dx+36(5x12)(25.625x65(x3))dx+06(5x12)(39.375x)dx+1205(x252)(59.25x108.757.5x2)dx+1205(x252)(90.75x266.257.5x2)dx]=1EI[03(10.677x2)dx+36(5x12)(39.375x+195)dx+06(16.406x2)dx+1205(29.625x254.375x3.75x2148.125x+271.875+18.75x2)dx+1205(45.375x2133.125x3.75x2226.875x+665.625+18.75x2)dx]=1EI[03(10.677x2)dx+36(16.406x281.25x)dx+06(16.406x2)dx+1205(44.625x2202.5x+271.875)dx+1205(60.375x2360x+665.625)dx]=1EI[(10.677x33)03+(16.406x3381.25x22)36+(16.406x33)06+12(44.625x33202.5x22+271.875x)05+12(60.375x33360x22+665.625x)05]

ΔE=1EI[10.677(3)33+16.406(6)3381.25(6)2216.406(3)33+81.25(3)22+16.406(6)33+12(44.625(5)33202.5(5)22+271.875(5))+12(60.375(5)33360(5)22+665.625(5))05]=1,607.8125kN2m3EI

Substitute 200 GPa for E and 350×106mm4 for I.

ΔE=1,607.8125kN2m3(200GPa)(350×106mm4)=1,607.8125kN2m3×109mm31m3(200GPa×1kN/mm21GPa)(350×106mm4)=22.97mm23mm

Therefore, the vertical deflection at E is 23mm_.

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