Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Chapter 3, Problem 33P
To determine

Calculate the support reactions for the given structure.

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Answer to Problem 33P

The horizontal reaction at A is Ax=55kN_.

The vertical reaction at A is Ay=216.11kN_.

The horizontal reaction at B is Bx=45kN_.

The vertical reaction at B is By=183.89kN_.

Explanation of Solution

Given information:

The structure is given in the Figure.

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

Let Bx and By be the horizontal reaction and vertical reaction at the hinged support B.

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

Structural Analysis, Chapter 3, Problem 33P

Use equilibrium equations:

Summation of moments about A is equal to 0.

MA=0((100)(5)+12(20)(4)(23×4)20(12)(122)+20(4)(42)12(20)(4)(12+43)+By(12))=0500+106.6671440+160533.333+12By=0By=183.89kN

Therefore, the vertical reaction at B is By=183.89kN_.

For the member BC, the summation of moments about C is equal to 0.

MCBC=0(20)(6)(62)12(20)(4)(43+6)+By(6)Bx(10)=0

Substitute 183.89kN for By.

(20)(6)(62)12(20)(4)(43+6)+(183.89)(6)Bx(10)=010Bx+450.007=0Bx=45kNBx=45kN

Therefore, the horizontal reaction at B is Bx=45kN_.

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

+Fx=0Ax+Bx+100=0

Substitute 45kN for Bx.

Ax45+100=055+Ax=0Ax=55kNAx=55kN

Therefore, the horizontal reaction at A is Ax=55kN_.

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

+Fy=0Ay+By12(20)(4)20(16)12(20)(4)=0

Substitute 183.89kN for By.

Ay+183.8912(20)(4)20(16)12(20)(4)=0Ay216.11=0Ay=216.11kN

Therefore, the vertical reaction at A is Ay=216.11kN_.

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