Concept explainers
Find the expression for vertical reaction at the supports in terms of distance x.
Sketch the graph for reactions as a function of x.
Answer to Problem 15P
The vertical reaction at A for the interval 0≤x≤20 m is Ay=45−2x kN↑_.
The vertical reaction at B for the interval 0≤x≤20 m is By=5+2x kN↑_.
The horizontal reaction at A for the interval 0≤x≤20 m is Ax=0_.
The vertical reaction at A for the interval 20 m≤x≤25 m is Ay=(25−x)25 kN↑_.
The vertical reaction at B for the interval 20 m≤x≤25 m is By=(625−x2)5 kN↑_.
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 (∑Mat a point=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.
Use equilibrium equations for the interval 0≤x≤20 m:
Summation of moments about B is equal to 0.
∑MB=0−Ay(25)+10(5)(25−(x+2.5))=0−25Ay=−50(25−x−2.5)−25Ay=−50(22.5−x)
Ay=1,125−50x25=45−2x kN↑
Therefore, the vertical reaction at A for the interval 0≤x≤20 m is Ay=(45−2x) kN↑_.
Summation of forces along y-direction is equal to 0.
+↑∑Fy=0Ay−10(5)+By=0By=−Ay+50
Substitute (45−2x) kN for Ay.
By=−(45−2x)+50=−45+2x+50=(5+2x) kN↑
Therefore, the vertical reaction at B for the interval 0≤x≤20 m 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 0≤x≤20 m is Ax=0_.
Sketch the free body diagram of the beam bridge for the interval 20 m≤x≤25 m as shown in Figure 2.
Use equilibrium equations for the interval 20 m≤x≤25 m:
Summation of moments about B is equal to 0.
∑MB=0−Ay(25)+10(25−x)(25−x2)=0−25Ay=−5(25−x)2Ay=5(25−x)225
Ay=(25−x)25 kN↑
Therefore, the vertical reaction at A for the interval 20 m≤x≤25 m is Ay=(25−x)25 kN↑_.
Summation of forces along y-direction is equal to 0.
+↑∑Fy=0Ay−10(25−x)+By=0By=−Ay+10(25−x)
Substitute (25−x)25 kN for Ay.
By=−(25−x)25+10(25−x)=−(25−x)2+50(25−x)5=−(252−50x+x2)+1,250−50x5=−625+50x−x2+1,250−50x5
By=(625−x2)5 kN↑
Therefore, the vertical reaction at B for the interval 20 m≤x≤25 m is By=(625−x2)5 kN↑_.
Sketch the graph of the reaction Ay vs. x as shown in Figure 3.
Sketch the graph of the reaction By vs. x as shown in Figure 3.
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