A Pratt roof truss is loaded as shown. Determine the force members FH , FI , and GI . Fig. P6.55 and P6.56

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Answer 1

Textbook Question

Chapter 6.2, Problem 6.56P

A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI.

Chapter 6.2, Problem 6.56P, A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI. Fig. P6.55 and

Fig. P6.55 and P6.56

Expert Solution & Answer

To determine

The force in the members FH,FI and GI.

Answer to Problem 6.56P

The force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Explanation of Solution

The Pratt roof truss is loaded as shown in the figure. There is load at all the point on the top. The free body diagram of the given arrangement is given by Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 1

Figure 1

The total load on the truss is given from the figure as,

Total load=5(3 kN)+2(1.5 kN)=18 kN

By symmetry the y component of the reaction at A will be half the total load.

Ay=12(18 kN)=9 kN

The section’s free body diagram is given as,

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 2

The tangent of the angle α can be written from the diagram as,

tanα=6.753

The angle is derived from the tangent value as,

α=tan−16.753=66.04°

From the diagram the sum of the moments in the counter clockwise direction about I is given as,

45FFH(23×6.75 m)+Ay(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)=0 (I)

From the diagram the sum of the moments in the counter clockwise direction about L is given as,

FFI(sinα)(6 m)−(3 kN)(6 m)−(3 kN)(3 m)=0 (II)

From the diagram the sum of the moments in the counter clockwise direction about H is given as,

FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−Ay(9 m)=0 (III)

Conclusion:

Solve for FFH from (I) by substituting 9 kN for Ay.

{45FFH(23×6.75 m)+(9 kN)(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)}=0FFH={−(9 kN)(6 m)+(1.5 kN)(6 m)+(3 kN)(3 m)}45(23×6.75 m)=−10.0 kN

Solve for FDE from (II) by substituting 66.04° for α.

{FFI(sin(66.04°))(6 m)−(3 kN)(6 m)−(3 kN)(3 m)}=0FFI=(3 kN)(6 m)+(3 kN)(3 m)(sin(66.04°))(6 m)=+4.92 kN

Solve for FDF from (III) by substituting 9 kN for Ay.

{FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−(9 kN)(9 m)}=0FGI={−(1.5 kN)(9 m)−(3 kN)(3 m)−(3 kN)(6 m)+(9 kN)(9 m)}(6.75 m)=+6.00 kN

Therefore, the force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

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Answer 2

Textbook Question

Chapter 6.2, Problem 6.56P

A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI.

Chapter 6.2, Problem 6.56P, A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI. Fig. P6.55 and

Fig. P6.55 and P6.56

Expert Solution & Answer

To determine

The force in the members FH,FI and GI.

Answer to Problem 6.56P

The force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Explanation of Solution

The Pratt roof truss is loaded as shown in the figure. There is load at all the point on the top. The free body diagram of the given arrangement is given by Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 1

Figure 1

The total load on the truss is given from the figure as,

Total load=5(3 kN)+2(1.5 kN)=18 kN

By symmetry the y component of the reaction at A will be half the total load.

Ay=12(18 kN)=9 kN

The section’s free body diagram is given as,

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 2

The tangent of the angle α can be written from the diagram as,

tanα=6.753

The angle is derived from the tangent value as,

α=tan−16.753=66.04°

From the diagram the sum of the moments in the counter clockwise direction about I is given as,

45FFH(23×6.75 m)+Ay(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)=0 (I)

From the diagram the sum of the moments in the counter clockwise direction about L is given as,

FFI(sinα)(6 m)−(3 kN)(6 m)−(3 kN)(3 m)=0 (II)

From the diagram the sum of the moments in the counter clockwise direction about H is given as,

FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−Ay(9 m)=0 (III)

Conclusion:

Solve for FFH from (I) by substituting 9 kN for Ay.

{45FFH(23×6.75 m)+(9 kN)(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)}=0FFH={−(9 kN)(6 m)+(1.5 kN)(6 m)+(3 kN)(3 m)}45(23×6.75 m)=−10.0 kN

Solve for FDE from (II) by substituting 66.04° for α.

{FFI(sin(66.04°))(6 m)−(3 kN)(6 m)−(3 kN)(3 m)}=0FFI=(3 kN)(6 m)+(3 kN)(3 m)(sin(66.04°))(6 m)=+4.92 kN

Solve for FDF from (III) by substituting 9 kN for Ay.

{FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−(9 kN)(9 m)}=0FGI={−(1.5 kN)(9 m)−(3 kN)(3 m)−(3 kN)(6 m)+(9 kN)(9 m)}(6.75 m)=+6.00 kN

Therefore, the force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

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Answer 3

Textbook Question

Chapter 6.2, Problem 6.56P

A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI.

Chapter 6.2, Problem 6.56P, A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI. Fig. P6.55 and

Fig. P6.55 and P6.56

Expert Solution & Answer

To determine

The force in the members FH,FI and GI.

Answer to Problem 6.56P

The force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Explanation of Solution

The Pratt roof truss is loaded as shown in the figure. There is load at all the point on the top. The free body diagram of the given arrangement is given by Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 1

Figure 1

The total load on the truss is given from the figure as,

Total load=5(3 kN)+2(1.5 kN)=18 kN

By symmetry the y component of the reaction at A will be half the total load.

Ay=12(18 kN)=9 kN

The section’s free body diagram is given as,

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 2

The tangent of the angle α can be written from the diagram as,

tanα=6.753

The angle is derived from the tangent value as,

α=tan−16.753=66.04°

From the diagram the sum of the moments in the counter clockwise direction about I is given as,

45FFH(23×6.75 m)+Ay(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)=0 (I)

From the diagram the sum of the moments in the counter clockwise direction about L is given as,

FFI(sinα)(6 m)−(3 kN)(6 m)−(3 kN)(3 m)=0 (II)

From the diagram the sum of the moments in the counter clockwise direction about H is given as,

FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−Ay(9 m)=0 (III)

Conclusion:

Solve for FFH from (I) by substituting 9 kN for Ay.

{45FFH(23×6.75 m)+(9 kN)(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)}=0FFH={−(9 kN)(6 m)+(1.5 kN)(6 m)+(3 kN)(3 m)}45(23×6.75 m)=−10.0 kN

Solve for FDE from (II) by substituting 66.04° for α.

{FFI(sin(66.04°))(6 m)−(3 kN)(6 m)−(3 kN)(3 m)}=0FFI=(3 kN)(6 m)+(3 kN)(3 m)(sin(66.04°))(6 m)=+4.92 kN

Solve for FDF from (III) by substituting 9 kN for Ay.

{FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−(9 kN)(9 m)}=0FGI={−(1.5 kN)(9 m)−(3 kN)(3 m)−(3 kN)(6 m)+(9 kN)(9 m)}(6.75 m)=+6.00 kN

Therefore, the force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

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Answer 4

Textbook Question

Chapter 6.2, Problem 6.56P

A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI.

Chapter 6.2, Problem 6.56P, A Pratt roof truss is loaded as shown. Determine the force members FH, FI, and GI. Fig. P6.55 and

Fig. P6.55 and P6.56

Expert Solution & Answer

To determine

The force in the members FH,FI and GI.

Answer to Problem 6.56P

The force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Explanation of Solution

The Pratt roof truss is loaded as shown in the figure. There is load at all the point on the top. The free body diagram of the given arrangement is given by Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 1

Figure 1

The total load on the truss is given from the figure as,

Total load=5(3 kN)+2(1.5 kN)=18 kN

By symmetry the y component of the reaction at A will be half the total load.

Ay=12(18 kN)=9 kN

The section’s free body diagram is given as,

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 2

The tangent of the angle α can be written from the diagram as,

tanα=6.753

The angle is derived from the tangent value as,

α=tan−16.753=66.04°

From the diagram the sum of the moments in the counter clockwise direction about I is given as,

45FFH(23×6.75 m)+Ay(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)=0 (I)

From the diagram the sum of the moments in the counter clockwise direction about L is given as,

FFI(sinα)(6 m)−(3 kN)(6 m)−(3 kN)(3 m)=0 (II)

From the diagram the sum of the moments in the counter clockwise direction about H is given as,

FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−Ay(9 m)=0 (III)

Conclusion:

Solve for FFH from (I) by substituting 9 kN for Ay.

{45FFH(23×6.75 m)+(9 kN)(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)}=0FFH={−(9 kN)(6 m)+(1.5 kN)(6 m)+(3 kN)(3 m)}45(23×6.75 m)=−10.0 kN

Solve for FDE from (II) by substituting 66.04° for α.

{FFI(sin(66.04°))(6 m)−(3 kN)(6 m)−(3 kN)(3 m)}=0FFI=(3 kN)(6 m)+(3 kN)(3 m)(sin(66.04°))(6 m)=+4.92 kN

Solve for FDF from (III) by substituting 9 kN for Ay.

{FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−(9 kN)(9 m)}=0FGI={−(1.5 kN)(9 m)−(3 kN)(3 m)−(3 kN)(6 m)+(9 kN)(9 m)}(6.75 m)=+6.00 kN

Therefore, the force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The force in the members FH,FI and GI.

Answer to Problem 6.56P

The force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Explanation of Solution

The Pratt roof truss is loaded as shown in the figure. There is load at all the point on the top. The free body diagram of the given arrangement is given by Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 1

Figure 1

The total load on the truss is given from the figure as,

Total load=5(3 kN)+2(1.5 kN)=18 kN

By symmetry the y component of the reaction at A will be half the total load.

Ay=12(18 kN)=9 kN

The section’s free body diagram is given as,

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 2

The tangent of the angle α can be written from the diagram as,

tanα=6.753

The angle is derived from the tangent value as,

α=tan−16.753=66.04°

From the diagram the sum of the moments in the counter clockwise direction about I is given as,

45FFH(23×6.75 m)+Ay(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)=0 (I)

From the diagram the sum of the moments in the counter clockwise direction about L is given as,

FFI(sinα)(6 m)−(3 kN)(6 m)−(3 kN)(3 m)=0 (II)

From the diagram the sum of the moments in the counter clockwise direction about H is given as,

FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−Ay(9 m)=0 (III)

Conclusion:

Solve for FFH from (I) by substituting 9 kN for Ay.

{45FFH(23×6.75 m)+(9 kN)(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)}=0FFH={−(9 kN)(6 m)+(1.5 kN)(6 m)+(3 kN)(3 m)}45(23×6.75 m)=−10.0 kN

Solve for FDE from (II) by substituting 66.04° for α.

{FFI(sin(66.04°))(6 m)−(3 kN)(6 m)−(3 kN)(3 m)}=0FFI=(3 kN)(6 m)+(3 kN)(3 m)(sin(66.04°))(6 m)=+4.92 kN

Solve for FDF from (III) by substituting 9 kN for Ay.

{FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−(9 kN)(9 m)}=0FGI={−(1.5 kN)(9 m)−(3 kN)(3 m)−(3 kN)(6 m)+(9 kN)(9 m)}(6.75 m)=+6.00 kN

Therefore, the force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Answer 8

Expert Solution & Answer

To determine

The force in the members FH,FI and GI.

Answer to Problem 6.56P

The force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Explanation of Solution

The Pratt roof truss is loaded as shown in the figure. There is load at all the point on the top. The free body diagram of the given arrangement is given by Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 1

Figure 1

The total load on the truss is given from the figure as,

Total load=5(3 kN)+2(1.5 kN)=18 kN

By symmetry the y component of the reaction at A will be half the total load.

Ay=12(18 kN)=9 kN

The section’s free body diagram is given as,

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 2

The tangent of the angle α can be written from the diagram as,

tanα=6.753

The angle is derived from the tangent value as,

α=tan−16.753=66.04°

From the diagram the sum of the moments in the counter clockwise direction about I is given as,

45FFH(23×6.75 m)+Ay(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)=0 (I)

From the diagram the sum of the moments in the counter clockwise direction about L is given as,

FFI(sinα)(6 m)−(3 kN)(6 m)−(3 kN)(3 m)=0 (II)

From the diagram the sum of the moments in the counter clockwise direction about H is given as,

FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−Ay(9 m)=0 (III)

Conclusion:

Solve for FFH from (I) by substituting 9 kN for Ay.

{45FFH(23×6.75 m)+(9 kN)(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)}=0FFH={−(9 kN)(6 m)+(1.5 kN)(6 m)+(3 kN)(3 m)}45(23×6.75 m)=−10.0 kN

Solve for FDE from (II) by substituting 66.04° for α.

{FFI(sin(66.04°))(6 m)−(3 kN)(6 m)−(3 kN)(3 m)}=0FFI=(3 kN)(6 m)+(3 kN)(3 m)(sin(66.04°))(6 m)=+4.92 kN

Solve for FDF from (III) by substituting 9 kN for Ay.

{FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−(9 kN)(9 m)}=0FGI={−(1.5 kN)(9 m)−(3 kN)(3 m)−(3 kN)(6 m)+(9 kN)(9 m)}(6.75 m)=+6.00 kN

Therefore, the force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The force in the members FH,FI and GI.

Answer 12

Answer to Problem 6.56P

The force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Answer 13

Explanation of Solution

The Pratt roof truss is loaded as shown in the figure. There is load at all the point on the top. The free body diagram of the given arrangement is given by Figure 1.

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 1

Figure 1

The total load on the truss is given from the figure as,

Total load=5(3 kN)+2(1.5 kN)=18 kN

By symmetry the y component of the reaction at A will be half the total load.

Ay=12(18 kN)=9 kN

The section’s free body diagram is given as,

Vector Mechanics for Engineers: Statics, Chapter 6.2, Problem 6.56P , additional homework tip 2

The tangent of the angle α can be written from the diagram as,

tanα=6.753

The angle is derived from the tangent value as,

α=tan−16.753=66.04°

From the diagram the sum of the moments in the counter clockwise direction about I is given as,

45FFH(23×6.75 m)+Ay(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)=0 (I)

From the diagram the sum of the moments in the counter clockwise direction about L is given as,

FFI(sinα)(6 m)−(3 kN)(6 m)−(3 kN)(3 m)=0 (II)

From the diagram the sum of the moments in the counter clockwise direction about H is given as,

FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−Ay(9 m)=0 (III)

Conclusion:

Solve for FFH from (I) by substituting 9 kN for Ay.

{45FFH(23×6.75 m)+(9 kN)(6 m)−(1.5 kN)(6 m)−(3 kN)(3 m)}=0FFH={−(9 kN)(6 m)+(1.5 kN)(6 m)+(3 kN)(3 m)}45(23×6.75 m)=−10.0 kN

Solve for FDE from (II) by substituting 66.04° for α.

{FFI(sin(66.04°))(6 m)−(3 kN)(6 m)−(3 kN)(3 m)}=0FFI=(3 kN)(6 m)+(3 kN)(3 m)(sin(66.04°))(6 m)=+4.92 kN

Solve for FDF from (III) by substituting 9 kN for Ay.

{FGI(6.75 m)+(1.5 kN)(9 m)+(3 kN)(3 m)+(3 kN)(6 m)−(9 kN)(9 m)}=0FGI={−(1.5 kN)(9 m)−(3 kN)(3 m)−(3 kN)(6 m)+(9 kN)(9 m)}(6.75 m)=+6.00 kN

Therefore, the force in the members FH,FI and GI are, 10.0 kN which is a compressive force, 4.92 kN and 6.00 kN both of which are tension forces respectively.

Answer 14

A Pratt roof truss is loaded as shown. Determine the force members FH , FI , and GI . Fig. P6.55 and P6.56

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