Calculate the support reactions for the given structures.

Question

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Answer 1

Question

Chapter 3, Problem 29P

To determine

Calculate the support reactions for the given structures.

Expert Solution & Answer

Answer to Problem 29P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=7.5 k↑_$ .

The vertical reaction at B is $By=90 k↑_$ .

The vertical reaction at C is $Cy=90 k↑_$ .

The vertical reaction at D is $Dy=7.5 k↑_$ .

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 $(∑Mat a point=0)$ , 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 $By$ , $Cy$ , and $Dy$ be the vertical reaction at the roller support B, C, and D.

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

Structural Analysis, 5th Edition, Chapter 3, Problem 29P

Use equilibrium equations:

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

$+→∑Fx=0Ax=0$

Therefore, the horizontal reaction at A is $Ax=0_$ .

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

$∑MEAE=0−Ay(15)+12(3)(15)(13×15)=0−Ay(15)+112.5=0Ay=7.5 k↑$

Therefore, the vertical reaction at A is $Ay=7.5 k↑_$ .

For the member DF, the summation of moments about F is equal to 0.

$∑MFDF=0Dy(15)−12(3)(15)(13×15)=0Dy(15)−112.5=0Dy=7.5 k↑$

Therefore, the vertical reaction at D is $Dy=7.5 k↑_$ .

Summation of moments about C is equal to 0.

$∑MC=0−Ay(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(20)+Dy(30)=0$

Substitute 7.5 k for $Ay$ and 7.5 k for $Dy$ .

$−7.5(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(153+15)+7.5(30)=0−20By+1,800=0By=90 k↑$

Therefore, the vertical reaction at B is $By=90 k↑_$ .

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

$+↑∑Fy=0Ay+By+Cy+Dy−12(3)(15)−3(50)−12(3)(15)=0$

Substitute 7.5 k for $Ay$ , 90 k for $By$ , and 7.5 k for $Dy$ .

$7.5+90+Cy+7.5−12(3)(15)−3(50)−12(3)(15)=0Cy−90=0Cy=90 k↑$

Therefore, the vertical reaction at C is $Cy=90 k↑_$ .

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Answer 2

Question

Chapter 3, Problem 29P

To determine

Calculate the support reactions for the given structures.

Expert Solution & Answer

Answer to Problem 29P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=7.5 k↑_$ .

The vertical reaction at B is $By=90 k↑_$ .

The vertical reaction at C is $Cy=90 k↑_$ .

The vertical reaction at D is $Dy=7.5 k↑_$ .

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 $(∑Mat a point=0)$ , 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 $By$ , $Cy$ , and $Dy$ be the vertical reaction at the roller support B, C, and D.

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

Structural Analysis, 5th Edition, Chapter 3, Problem 29P

Use equilibrium equations:

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

$+→∑Fx=0Ax=0$

Therefore, the horizontal reaction at A is $Ax=0_$ .

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

$∑MEAE=0−Ay(15)+12(3)(15)(13×15)=0−Ay(15)+112.5=0Ay=7.5 k↑$

Therefore, the vertical reaction at A is $Ay=7.5 k↑_$ .

For the member DF, the summation of moments about F is equal to 0.

$∑MFDF=0Dy(15)−12(3)(15)(13×15)=0Dy(15)−112.5=0Dy=7.5 k↑$

Therefore, the vertical reaction at D is $Dy=7.5 k↑_$ .

Summation of moments about C is equal to 0.

$∑MC=0−Ay(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(20)+Dy(30)=0$

Substitute 7.5 k for $Ay$ and 7.5 k for $Dy$ .

$−7.5(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(153+15)+7.5(30)=0−20By+1,800=0By=90 k↑$

Therefore, the vertical reaction at B is $By=90 k↑_$ .

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

$+↑∑Fy=0Ay+By+Cy+Dy−12(3)(15)−3(50)−12(3)(15)=0$

Substitute 7.5 k for $Ay$ , 90 k for $By$ , and 7.5 k for $Dy$ .

$7.5+90+Cy+7.5−12(3)(15)−3(50)−12(3)(15)=0Cy−90=0Cy=90 k↑$

Therefore, the vertical reaction at C is $Cy=90 k↑_$ .

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Answer 3

Question

Chapter 3, Problem 29P

To determine

Calculate the support reactions for the given structures.

Expert Solution & Answer

Answer to Problem 29P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=7.5 k↑_$ .

The vertical reaction at B is $By=90 k↑_$ .

The vertical reaction at C is $Cy=90 k↑_$ .

The vertical reaction at D is $Dy=7.5 k↑_$ .

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 $(∑Mat a point=0)$ , 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 $By$ , $Cy$ , and $Dy$ be the vertical reaction at the roller support B, C, and D.

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

Structural Analysis, 5th Edition, Chapter 3, Problem 29P

Use equilibrium equations:

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

$+→∑Fx=0Ax=0$

Therefore, the horizontal reaction at A is $Ax=0_$ .

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

$∑MEAE=0−Ay(15)+12(3)(15)(13×15)=0−Ay(15)+112.5=0Ay=7.5 k↑$

Therefore, the vertical reaction at A is $Ay=7.5 k↑_$ .

For the member DF, the summation of moments about F is equal to 0.

$∑MFDF=0Dy(15)−12(3)(15)(13×15)=0Dy(15)−112.5=0Dy=7.5 k↑$

Therefore, the vertical reaction at D is $Dy=7.5 k↑_$ .

Summation of moments about C is equal to 0.

$∑MC=0−Ay(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(20)+Dy(30)=0$

Substitute 7.5 k for $Ay$ and 7.5 k for $Dy$ .

$−7.5(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(153+15)+7.5(30)=0−20By+1,800=0By=90 k↑$

Therefore, the vertical reaction at B is $By=90 k↑_$ .

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

$+↑∑Fy=0Ay+By+Cy+Dy−12(3)(15)−3(50)−12(3)(15)=0$

Substitute 7.5 k for $Ay$ , 90 k for $By$ , and 7.5 k for $Dy$ .

$7.5+90+Cy+7.5−12(3)(15)−3(50)−12(3)(15)=0Cy−90=0Cy=90 k↑$

Therefore, the vertical reaction at C is $Cy=90 k↑_$ .

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Answer 4

Question

Chapter 3, Problem 29P

To determine

Calculate the support reactions for the given structures.

Expert Solution & Answer

Answer to Problem 29P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=7.5 k↑_$ .

The vertical reaction at B is $By=90 k↑_$ .

The vertical reaction at C is $Cy=90 k↑_$ .

The vertical reaction at D is $Dy=7.5 k↑_$ .

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 $(∑Mat a point=0)$ , 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 $By$ , $Cy$ , and $Dy$ be the vertical reaction at the roller support B, C, and D.

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

Structural Analysis, 5th Edition, Chapter 3, Problem 29P

Use equilibrium equations:

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

$+→∑Fx=0Ax=0$

Therefore, the horizontal reaction at A is $Ax=0_$ .

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

$∑MEAE=0−Ay(15)+12(3)(15)(13×15)=0−Ay(15)+112.5=0Ay=7.5 k↑$

Therefore, the vertical reaction at A is $Ay=7.5 k↑_$ .

For the member DF, the summation of moments about F is equal to 0.

$∑MFDF=0Dy(15)−12(3)(15)(13×15)=0Dy(15)−112.5=0Dy=7.5 k↑$

Therefore, the vertical reaction at D is $Dy=7.5 k↑_$ .

Summation of moments about C is equal to 0.

$∑MC=0−Ay(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(20)+Dy(30)=0$

Substitute 7.5 k for $Ay$ and 7.5 k for $Dy$ .

$−7.5(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(153+15)+7.5(30)=0−20By+1,800=0By=90 k↑$

Therefore, the vertical reaction at B is $By=90 k↑_$ .

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

$+↑∑Fy=0Ay+By+Cy+Dy−12(3)(15)−3(50)−12(3)(15)=0$

Substitute 7.5 k for $Ay$ , 90 k for $By$ , and 7.5 k for $Dy$ .

$7.5+90+Cy+7.5−12(3)(15)−3(50)−12(3)(15)=0Cy−90=0Cy=90 k↑$

Therefore, the vertical reaction at C is $Cy=90 k↑_$ .

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Answer 5

Chapter 3, Problem 29P

To determine

Calculate the support reactions for the given structures.

Expert Solution & Answer

Answer to Problem 29P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=7.5 k↑_$ .

The vertical reaction at B is $By=90 k↑_$ .

The vertical reaction at C is $Cy=90 k↑_$ .

The vertical reaction at D is $Dy=7.5 k↑_$ .

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 $(∑Mat a point=0)$ , 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 $By$ , $Cy$ , and $Dy$ be the vertical reaction at the roller support B, C, and D.

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

Structural Analysis, 5th Edition, Chapter 3, Problem 29P

Use equilibrium equations:

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

$+→∑Fx=0Ax=0$

Therefore, the horizontal reaction at A is $Ax=0_$ .

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

$∑MEAE=0−Ay(15)+12(3)(15)(13×15)=0−Ay(15)+112.5=0Ay=7.5 k↑$

Therefore, the vertical reaction at A is $Ay=7.5 k↑_$ .

For the member DF, the summation of moments about F is equal to 0.

$∑MFDF=0Dy(15)−12(3)(15)(13×15)=0Dy(15)−112.5=0Dy=7.5 k↑$

Therefore, the vertical reaction at D is $Dy=7.5 k↑_$ .

Summation of moments about C is equal to 0.

$∑MC=0−Ay(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(20)+Dy(30)=0$

Substitute 7.5 k for $Ay$ and 7.5 k for $Dy$ .

$−7.5(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(153+15)+7.5(30)=0−20By+1,800=0By=90 k↑$

Therefore, the vertical reaction at B is $By=90 k↑_$ .

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

$+↑∑Fy=0Ay+By+Cy+Dy−12(3)(15)−3(50)−12(3)(15)=0$

Substitute 7.5 k for $Ay$ , 90 k for $By$ , and 7.5 k for $Dy$ .

$7.5+90+Cy+7.5−12(3)(15)−3(50)−12(3)(15)=0Cy−90=0Cy=90 k↑$

Therefore, the vertical reaction at C is $Cy=90 k↑_$ .

Answer 6

Chapter 3, Problem 29P

To determine

Calculate the support reactions for the given structures.

Expert Solution & Answer

Answer to Problem 29P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=7.5 k↑_$ .

The vertical reaction at B is $By=90 k↑_$ .

The vertical reaction at C is $Cy=90 k↑_$ .

The vertical reaction at D is $Dy=7.5 k↑_$ .

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 $(∑Mat a point=0)$ , 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 $By$ , $Cy$ , and $Dy$ be the vertical reaction at the roller support B, C, and D.

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

Structural Analysis, 5th Edition, Chapter 3, Problem 29P

Use equilibrium equations:

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

$+→∑Fx=0Ax=0$

Therefore, the horizontal reaction at A is $Ax=0_$ .

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

$∑MEAE=0−Ay(15)+12(3)(15)(13×15)=0−Ay(15)+112.5=0Ay=7.5 k↑$

Therefore, the vertical reaction at A is $Ay=7.5 k↑_$ .

For the member DF, the summation of moments about F is equal to 0.

$∑MFDF=0Dy(15)−12(3)(15)(13×15)=0Dy(15)−112.5=0Dy=7.5 k↑$

Therefore, the vertical reaction at D is $Dy=7.5 k↑_$ .

Summation of moments about C is equal to 0.

$∑MC=0−Ay(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(20)+Dy(30)=0$

Substitute 7.5 k for $Ay$ and 7.5 k for $Dy$ .

$−7.5(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(153+15)+7.5(30)=0−20By+1,800=0By=90 k↑$

Therefore, the vertical reaction at B is $By=90 k↑_$ .

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

$+↑∑Fy=0Ay+By+Cy+Dy−12(3)(15)−3(50)−12(3)(15)=0$

Substitute 7.5 k for $Ay$ , 90 k for $By$ , and 7.5 k for $Dy$ .

$7.5+90+Cy+7.5−12(3)(15)−3(50)−12(3)(15)=0Cy−90=0Cy=90 k↑$

Therefore, the vertical reaction at C is $Cy=90 k↑_$ .

Answer 7

Chapter 3, Problem 29P

To determine

Calculate the support reactions for the given structures.

Answer 8

Expert Solution & Answer

Answer to Problem 29P

The horizontal reaction at A is $Ax=0_$ .

The vertical reaction at A is $Ay=7.5 k↑_$ .

The vertical reaction at B is $By=90 k↑_$ .

The vertical reaction at C is $Cy=90 k↑_$ .

The vertical reaction at D is $Dy=7.5 k↑_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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 $(∑Mat a point=0)$ , 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 $By$ , $Cy$ , and $Dy$ be the vertical reaction at the roller support B, C, and D.

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

Structural Analysis, 5th Edition, Chapter 3, Problem 29P

Use equilibrium equations:

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

$+→∑Fx=0Ax=0$

Therefore, the horizontal reaction at A is $Ax=0_$ .

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

$∑MEAE=0−Ay(15)+12(3)(15)(13×15)=0−Ay(15)+112.5=0Ay=7.5 k↑$

Therefore, the vertical reaction at A is $Ay=7.5 k↑_$ .

For the member DF, the summation of moments about F is equal to 0.

$∑MFDF=0Dy(15)−12(3)(15)(13×15)=0Dy(15)−112.5=0Dy=7.5 k↑$

Therefore, the vertical reaction at D is $Dy=7.5 k↑_$ .

Summation of moments about C is equal to 0.

$∑MC=0−Ay(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(20)+Dy(30)=0$

Substitute 7.5 k for $Ay$ and 7.5 k for $Dy$ .

$−7.5(50)−By(20)+12(3)(15)(153+35)+3(50)(10)−12(3)(15)(153+15)+7.5(30)=0−20By+1,800=0By=90 k↑$

Therefore, the vertical reaction at B is $By=90 k↑_$ .

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

$+↑∑Fy=0Ay+By+Cy+Dy−12(3)(15)−3(50)−12(3)(15)=0$

Substitute 7.5 k for $Ay$ , 90 k for $By$ , and 7.5 k for $Dy$ .

$7.5+90+Cy+7.5−12(3)(15)−3(50)−12(3)(15)=0Cy−90=0Cy=90 k↑$

Therefore, the vertical reaction at C is $Cy=90 k↑_$ .

Answer 12

Ch. 3 - Prob. 1P Ch. 3 - Prob. 2P Ch. 3 - Prob. 3P Ch. 3 - Prob. 4P Ch. 3 - Prob. 5P Ch. 3 - Prob. 6P Ch. 3 - Prob. 7P Ch. 3 - Prob. 8P Ch. 3 - Prob. 9P Ch. 3 - Prob. 10P

Calculate the support reactions for the given structures.

Concept explainers

Calculate the support reactions for the given structures.

Answer to Problem 29P

Explanation of Solution

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Chapter 3 Solutions