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

through 5.7 Calculate the forces in all members of the trusses shown, using the method of joints.

Chapter 5, Problem 5.1P, through 5.7 Calculate the forces in all members of the trusses shown, using the method of joints.

Expert Solution & Answer
Check Mark
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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Chapter 5 Solutions

Applied Statics and Strength of Materials (6th Edition)

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