Applied Statics and Strength of Materials (6th Edition)
Applied Statics and Strength of Materials (6th Edition)
6th Edition
ISBN: 9780133840544
Author: George F. Limbrunner, Craig D'Allaird, Leonard Spiegel
Publisher: PEARSON
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Chapter 5, Problem 5.1P
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To determine

The forces in the member from method of joints.

Answer to Problem 5.1P

  FAC=5kips(tension)

  FBC=FAB=7.07kips(compression)

Explanation of Solution

Given:

Two reactions at the pin support is Ax and Ay.

At roller support one reaction is Cy.

Force on the member =10kips

Free body diagram of the truss.

Applied Statics and Strength of Materials (6th Edition), Chapter 5, Problem 5.1P , additional homework tip  1

Taking moment about A

  MA=0CY(L)10(L2)=0CY(L)5L=0CY=5kips

From equilibrium equations

  Fx=0Ax=0

  Fy=0Ay+Cy10=0Ay=5kips

Considering joint A

Applied Statics and Strength of Materials (6th Edition), Chapter 5, Problem 5.1P , additional homework tip  2

From the equilibrium equation

  Fy=0Ay+FABsin45=05+FABsin45=0FAB=7.07kips(compression)Fx=0Ax+FAC+FABcos45=00+FAC+(7.07)cos45=0FAC=5kips(tension)

Considering joint C

Applied Statics and Strength of Materials (6th Edition), Chapter 5, Problem 5.1P , additional homework tip  3

  Fy=0Cy+FBCsin45=05+FBCsin45=0FBC=7.07kipsFBC=7.07kips(compression)

Conclusion:

Forces in the member FBC=FAB=7.07kips(compression) and FAC=5kips(tension) .

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