Structural Analysis, 5th Edition
Structural Analysis, 5th Edition
5th Edition
ISBN: 9788131520444
Author: Aslam Kassimali
Publisher: Cengage Learning
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Chapter 3, Problem 15P
To determine

Find the expression for vertical reaction at the supports in terms of distance x.

Sketch the graph for reactions as a function of x.

Expert Solution & Answer
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Answer to Problem 15P

The vertical reaction at A for the interval 0x20m is Ay=452xkN_.

The vertical reaction at B for the interval 0x20m is By=5+2xkN_.

The horizontal reaction at A for the interval 0x20m is Ax=0_.

The vertical reaction at A for the interval 20mx25m is Ay=(25x)25kN_.

The vertical reaction at B for the interval 20mx25m is By=(625x2)5kN_.

Explanation of Solution

Given information:

The structure is given in the Figure.

Length of the trolley is 5 m.

Apply the sign conventions for calculating reaction 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:

Let the roller at B exerts the vertical reaction By.

Let Ax and Ay be the horizontal and vertical reactions at the hinged support A.

Sketch the free body diagram of the beam bridge as shown in Figure 1.

Structural Analysis, 5th Edition, Chapter 3, Problem 15P , additional homework tip  1

Use equilibrium equations for the interval 0x20m:

Summation of moments about B is equal to 0.

MB=0Ay(25)+10(5)(25(x+2.5))=025Ay=50(25x2.5)25Ay=50(22.5x)

Ay=1,12550x25=452xkN

Therefore, the vertical reaction at A for the interval 0x20m is Ay=(452x)kN_.

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

+Fy=0Ay10(5)+By=0By=Ay+50

Substitute (452x)kN for Ay.

By=(452x)+50=45+2x+50=(5+2x)kN

Therefore, the vertical reaction at B for the interval 0x20m is By=(5+2x)kN_.

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

+Fx=0Ax=0

Therefore, the horizontal reaction at A for the interval 0x20m is Ax=0_.

Sketch the free body diagram of the beam bridge for the interval 20mx25m as shown in Figure 2.

Structural Analysis, 5th Edition, Chapter 3, Problem 15P , additional homework tip  2

Use equilibrium equations for the interval 20mx25m:

Summation of moments about B is equal to 0.

MB=0Ay(25)+10(25x)(25x2)=025Ay=5(25x)2Ay=5(25x)225

Ay=(25x)25kN

Therefore, the vertical reaction at A for the interval 20mx25m is Ay=(25x)25kN_.

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

+Fy=0Ay10(25x)+By=0By=Ay+10(25x)

Substitute (25x)25kN for Ay.

By=(25x)25+10(25x)=(25x)2+50(25x)5=(25250x+x2)+1,25050x5=625+50xx2+1,25050x5

By=(625x2)5kN

Therefore, the vertical reaction at B for the interval 20mx25m is By=(625x2)5kN_.

Sketch the graph of the reaction Ay vs. x as shown in Figure 3.

Structural Analysis, 5th Edition, Chapter 3, Problem 15P , additional homework tip  3

Sketch the graph of the reaction By vs. x as shown in Figure 3.

Structural Analysis, 5th Edition, Chapter 3, Problem 15P , additional homework tip  4

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