Structural Analysis, 5th Edition
Structural Analysis, 5th Edition
5th Edition
ISBN: 9788131520444
Author: Aslam Kassimali
Publisher: Cengage Learning
Question
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Chapter 10, Problem 19P
To determine

Find the force in each member of the truss using structural symmetry.

Expert Solution & Answer
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Answer to Problem 19P

The force in the member AB and BC is 95k(C)_.

The force in the member AH is 134.35k(T)_.

The force in the member HI and IJ is 160k(T)_.

The force in the member BH is 30k(C)_.

The force in the member CH is 91.92k(C)_.

The force in the member CI is 0_.

The force in the member CD and DE is 195k(C)_.

The force in the member CJ is 49.5k(T)_.

The force in the member JK and KL is 180k(T)_.

The force in the member DJ is 50k(C)_.

The force in the member EJ is 21.21k(T)_.

The force in the member EK is 0_.

The force in the member EF and FG is 115k(C)_.

The force in the member GL is 162.63k(T)_.

The force in the member FL is 50k(C)_.

The force in the member EL is 91.92k(C)_.

Explanation of Solution

Given information:

The structure is given in the Figure.

The young’s modulus E and area A are constant.

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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Calculation:

Refer the given structure.

The structure is symmetric with respect to the s axis.

Divide the magnitudes of forces and moments of the given loading by 2 to obtain the half loading.

Sketch the half loading for the given structure as shown in Figure 1.

Structural Analysis, 5th Edition, Chapter 10, Problem 19P , additional homework tip  1

Draw the reflection of half loading about the specified axis s.

Sketch the reflection of half loading as shown in Figure 2.

Structural Analysis, 5th Edition, Chapter 10, Problem 19P , additional homework tip  2

Add the half loading and reflection of half loading to find the symmetric component.

Sketch the symmetric loading component as shown in Figure 3.

Structural Analysis, 5th Edition, Chapter 10, Problem 19P , additional homework tip  3

Subtract the symmetric loading component from the given loading to obtain the antisymmetric loading component.

Sketch the antisymmetric loading component as shown in Figure 4.

Structural Analysis, 5th Edition, Chapter 10, Problem 19P , additional homework tip  4

Find the member forces due to symmetric loading component:

Sketch the substructure with symmetric boundary conditions as shown in Figure 5.

Structural Analysis, 5th Edition, Chapter 10, Problem 19P , additional homework tip  5

Find the reactions and member end forces of substructure using equilibrium equations and to the right of s axis.

The member end forces to the left of s axis are obtained by the reflection.

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

+Fy=0254040+Gy=0Gy=105k

Summation of moments about D is equal to 0.

MD=0105×6040×4040×20Jx×20=020Jx=3,900Jx=195k

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

+Fx=0DxJx=0Dx195=0Dx=195k

Find the angle θ made by the member JE with respect to the horizontal axis using the given Figure.

tanθ=2020θ=45°

Consider joint D, find the force in the member DE and DJ:

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

+Fy=025FDJ=0FDJ=25k

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

+Fx=0195+FDE=0FDE=195k

Consider joint J, find the force in the member JK and JE:

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

+Fy=0FDJ+FJEsin(45°)=025+FJEsin(45°)=0FJE=35.36k

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

+Fx=0195+FJK+FJEcos(45°)=0195+FJK+35.36cos(45°)=0FJK=170k

Consider joint K, find the force in the member KL and EK:

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

+Fy=0FEK=0

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

+Fx=0FJK+FKL=0170+FKL=0FKL=170k

Consider joint E, find the force in the member EF and EL:

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

+Fy=040FEJsin45°FELsin45°=040(35.36)sin45°FELsin45°=0FEL=91.92k

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

+Fx=0FDE+FEFFEJcos(45°)+FELcos(45°)=0(195)+FEF35.36cos(45°)+(91.92)cos(45°)=0FEF=105k

Consider joint F, find the force in the member FL and FG:

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

+Fy=040FFL=0FFL=40k

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

+Fx=0FEF+FFG=0(105)+FFG=0FFG=105k

Consider joint G, find the force in the member GL:

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

+Fy=0105FGLsin45°=0FGL=148.49k

Sketch the substructure with antisymmetric boundary conditions as shown in Figure 6.

Structural Analysis, 5th Edition, Chapter 10, Problem 19P , additional homework tip  6

Find the reactions and member end forces of substructure using equilibrium equations and to the right of s axis.

The member end forces to the left of s axis are obtained by reflecting the negatives of computed forces and moments about the axis of symmetry.

Summation of moments about D is equal to 0.

MD=0Gy×6010×4010×20=060Gy=600Gy=10k

The vertical reaction at joint D is 0.

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

+Fy=0Jy+Dy1010+Gy=0Jy+01010+10=0Jy=10k

Consider joint J, find the force in the member JK and JE:

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

+Fy=010+FJEsin(45°)=0FJE=14.14k

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

+Fx=0FJK+FEJcos45°=0FJK+14.14cos45°=0FJK=10k

Consider joint D, find the force in the member DE:

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

+Fy=0FDE=0

Consider joint K, find the force in the member KL and EK:

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

+Fy=0FEK=0

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

+Fx=0FJK+FKL=0(10)+FKL=0FKL=10k

Consider joint E, find the force in the member EF and EL:

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

+Fy=010FEKFELsin45°FJEcos45°=0100FELsin45°(14.14)cos45°=0FEL=0

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

+Fx=0FEFFJEcos(45°)=0FEF(14.14)cos(45°)=0FEF=10k

Consider joint F, find the force in the member FL and FG:

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

+Fy=010FFL=0FFL=10k

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

+Fx=0FEF+FFG=0(10)+FFG=0FFG=10k

Consider joint G, find the force in the member GL:

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

+Fy=010FGLsin45°=0FGL=14.14k

The total member end forces are obtained by superposing the member forces due to symmetric and antisymmetric components of loading.

Sketch the member end forces due to total loading as shown in Figure 7.

Structural Analysis, 5th Edition, Chapter 10, Problem 19P , additional homework tip  7

Therefore,

The force in the member AB and BC is 95k(C)_.

The force in the member AH is 134.35k(T)_.

The force in the member HI and IJ is 160k(T)_.

The force in the member BH is 30k(C)_.

The force in the member CH is 91.92k(C)_.

The force in the member CI is 0_.

The force in the member CD and DE is 195k(C)_.

The force in the member CJ is 49.5k(T)_.

The force in the member JK and KL is 180k(T)_.

The force in the member DJ is 50k(C)_.

The force in the member EJ is 21.21k(T)_.

The force in the member EK is 0_.

The force in the member EF and FG is 115k(C)_.

The force in the member GL is 162.63k(T)_.

The force in the member FL is 50k(C)_.

The force in the member EL is 91.92k(C)_.

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