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Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

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BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 3, Problem 42P
Textbook Problem
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3.16 through 3.42 Determine the reactions at the supports for the structures shown.

Chapter 3, Problem 42P, 3.16 through 3.42 Determine the reactions at the supports for the structures shown. FIG. P3.42

FIG. P3.42

To determine

Calculate the support reactions for the given structure.

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, Ay, and MA be the horizontal reaction, vertical reaction, and moment at the fixed support A.

Let Bx, By, and MB be the horizontal reaction, vertical reaction, and moment at the fixed support B.

Let Cx and Cy be the horizontal and vertical reaction at the internal hinge C.

Let Dx and Dy be the horizontal and vertical reaction at the internal hinge D.

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

Find the angle of the inclined member with respect to horizontal axis:

Refer Figure 1.

tanθ=310θ=tan1(310)=16.7°

Convert the 15kN/m uniformly distributed load into point load:

15kN/m =15×3sin16.7°=156.6kN

Use Figure 1 to find the length of the inclined member:

L=32+102=10.44m

Use equilibrium equations:

Find the forces at the internal hinges C and D:

Consider the free body diagram of the portion CED.

Summation of moments about D is equal to 0.

MD=0Cy(20)(25+15)(4.5)(4.52)156.6(cos16.7°)(15+5)=020Cy=3,404.9Cy=170.25kN

For the member CE, the summation of moments about E is equal to 0.

MECE=0Cy(10)Cx(7.5)+25(4.5)(4.52+3)156.6(10.442)=0

Substitute 170.25kN for Cy.

170.25(10)Cx(7.5)+25(4.5)(4.52+3)156.6(10.442)=07.5Cx+1,475.673=0Cx=196.75kN

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

+Fx=0Cx+(25+15)(4.5)+Dx=0

Substitute 196.75kN for Cx.

196.75+(25+15)(4.5)+Dx=0Dx=16.75kN

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

+Fy=0Cy+2(156.6)(cos16.7°)Dy=0

Substitute 170.25kN for Cy.

170

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Structural Analysis
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