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

Question

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Answer 1

Textbook Question

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

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+Cy−10=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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08:37

Students have asked these similar questions

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

Using the method of joints, calculate the force in each member of the truss shown. State whether each member is in tension or compression.

Using the method of joints, calculate the force in each memberof the trusses shown. State whether each member is in tension or compression.

Answer 2

Textbook Question

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

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+Cy−10=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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08:37

Answer 3

Textbook Question

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

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+Cy−10=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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08:37

Answer 4

Textbook Question

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

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+Cy−10=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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08:37

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

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+Cy−10=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) .

Answer 8

Expert Solution & Answer

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+Cy−10=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) .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The forces in the member from method of joints.

Answer 12

Answer to Problem 5.1P

FAC=5kips(tension)

FBC=FAB=7.07kips(compression)

Answer 13

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+Cy−10=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) .

Answer 14

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

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Answer to Problem 5.1P

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