The beam AB is pin supported at A and supported by a cable BC . A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E .

Question

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Answer 1

Textbook Question

Chapter 1, Problem 1.1RP

The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E.

Chapter 1, Problem 1.1RP, The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold

Expert Solution & Answer

To determine

The resultant internal loadings acting on cross sections located at D and E.

Answer to Problem 1.1RP

The resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

The resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Explanation of Solution

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is $120 lb/ft$ .

The weight of the column FC is $180 lb/ft$ .

Calculation:

Find the loading at the center of the beam AB $(PAB)$ :

$PAB=Weight of beam AB×Length of beam AB$

Substitute $120 lb/ft$ for the weight of beam AB and 12 ft for the length of beam AB.

$PAB=120×12=1,440 lb$

Convert the unit from lb to kip.

$PAB=1,440 lb×1 kip1,000 lb=1.44 kip$

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 1

Refer to Figure 1.

Find the angle of cable BC to the horizontal $(θ)$ :

$sinθ=112+32sinθ=13.1623θ=sin−1(0.3162)θ=18.43°$

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

$∑MA=0(FBCsin18.43°×12)+1.44×6=03.794FBC+8.64=03.794FBC=−8.64$

$FBC=−2.277 kip$

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0Ax−2.277cos18.43°=0Ax−2.16=0Ax=2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0Ay−1.44+2.277sin18.43°=0Ay−0.72=0Ay=0.72 kip$

Find the loading at the center of the beam AD $(PAD)$ :

$PAD=Weight of beam AD×Length of beam AD$

Substitute $120 lb/ft$ for the weight of beam AD and 6 ft for the length of beam AD.

$PAD=120×6=720 lb$

Convert the unit from lb to kip.

$PAD=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 2

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0ND+2.16=0ND=−2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0VD+0.72−0.72−=0VD=0$

Take moment about D is Equal to zero.

$∑MD=0MD−0.72×3=0MD−2.16=0MD=2.16 kip⋅ft$

Hence, the resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

Find the loading at the center of the column FC $(PFC)$ :

$PFC=Weight of column FC×Length of column FC$

Substitute $180 lb/ft$ for the weight of column FC and 16 ft for the length of column FC.

$PFC=180×16=2,880 lb$

Convert the unit from lb to kip.

$PFC=2,880 lb×1 kip1,000 lb=2.88 kip$

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 3

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

$sinθ=342+32sinθ=35θ=sin−1(0.6)θ=36.87°$

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0FCGcos36.87°−2.277sin18.43°=00.8FCG−0.72=00.8FCG=0.72$

$FCG=0.9 kip$

Find the loading at the center of the column FE $(PFE)$ :

$PFE=Weight of column FE×Length of column FE$

Substitute $180 lb/ft$ for the weight of column FE and 4 ft for the length of column FC.

$PFE=180×4=720 lb$

Convert the unit from lb to kip.

$PFE=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 4

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0VE−0.54=0VE=0.54 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0NE+0.72−5.04=0NE−4.32=0NE=4.32 kip$

Take moment about E is Equal to zero.

$∑ME=0−ME+0.54×4=0−ME+2.16=0ME=2.16 kip⋅ft$

Therefore, the resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

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11:16

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Answer 2

Textbook Question

Chapter 1, Problem 1.1RP

The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E.

Chapter 1, Problem 1.1RP, The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold

Expert Solution & Answer

To determine

The resultant internal loadings acting on cross sections located at D and E.

Answer to Problem 1.1RP

The resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

The resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Explanation of Solution

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is $120 lb/ft$ .

The weight of the column FC is $180 lb/ft$ .

Calculation:

Find the loading at the center of the beam AB $(PAB)$ :

$PAB=Weight of beam AB×Length of beam AB$

Substitute $120 lb/ft$ for the weight of beam AB and 12 ft for the length of beam AB.

$PAB=120×12=1,440 lb$

Convert the unit from lb to kip.

$PAB=1,440 lb×1 kip1,000 lb=1.44 kip$

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 1

Refer to Figure 1.

Find the angle of cable BC to the horizontal $(θ)$ :

$sinθ=112+32sinθ=13.1623θ=sin−1(0.3162)θ=18.43°$

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

$∑MA=0(FBCsin18.43°×12)+1.44×6=03.794FBC+8.64=03.794FBC=−8.64$

$FBC=−2.277 kip$

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0Ax−2.277cos18.43°=0Ax−2.16=0Ax=2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0Ay−1.44+2.277sin18.43°=0Ay−0.72=0Ay=0.72 kip$

Find the loading at the center of the beam AD $(PAD)$ :

$PAD=Weight of beam AD×Length of beam AD$

Substitute $120 lb/ft$ for the weight of beam AD and 6 ft for the length of beam AD.

$PAD=120×6=720 lb$

Convert the unit from lb to kip.

$PAD=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 2

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0ND+2.16=0ND=−2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0VD+0.72−0.72−=0VD=0$

Take moment about D is Equal to zero.

$∑MD=0MD−0.72×3=0MD−2.16=0MD=2.16 kip⋅ft$

Hence, the resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

Find the loading at the center of the column FC $(PFC)$ :

$PFC=Weight of column FC×Length of column FC$

Substitute $180 lb/ft$ for the weight of column FC and 16 ft for the length of column FC.

$PFC=180×16=2,880 lb$

Convert the unit from lb to kip.

$PFC=2,880 lb×1 kip1,000 lb=2.88 kip$

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 3

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

$sinθ=342+32sinθ=35θ=sin−1(0.6)θ=36.87°$

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0FCGcos36.87°−2.277sin18.43°=00.8FCG−0.72=00.8FCG=0.72$

$FCG=0.9 kip$

Find the loading at the center of the column FE $(PFE)$ :

$PFE=Weight of column FE×Length of column FE$

Substitute $180 lb/ft$ for the weight of column FE and 4 ft for the length of column FC.

$PFE=180×4=720 lb$

Convert the unit from lb to kip.

$PFE=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 4

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0VE−0.54=0VE=0.54 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0NE+0.72−5.04=0NE−4.32=0NE=4.32 kip$

Take moment about E is Equal to zero.

$∑ME=0−ME+0.54×4=0−ME+2.16=0ME=2.16 kip⋅ft$

Therefore, the resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

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11:16

Answer 3

Textbook Question

Chapter 1, Problem 1.1RP

The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E.

Chapter 1, Problem 1.1RP, The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold

Expert Solution & Answer

To determine

The resultant internal loadings acting on cross sections located at D and E.

Answer to Problem 1.1RP

The resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

The resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Explanation of Solution

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is $120 lb/ft$ .

The weight of the column FC is $180 lb/ft$ .

Calculation:

Find the loading at the center of the beam AB $(PAB)$ :

$PAB=Weight of beam AB×Length of beam AB$

Substitute $120 lb/ft$ for the weight of beam AB and 12 ft for the length of beam AB.

$PAB=120×12=1,440 lb$

Convert the unit from lb to kip.

$PAB=1,440 lb×1 kip1,000 lb=1.44 kip$

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 1

Refer to Figure 1.

Find the angle of cable BC to the horizontal $(θ)$ :

$sinθ=112+32sinθ=13.1623θ=sin−1(0.3162)θ=18.43°$

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

$∑MA=0(FBCsin18.43°×12)+1.44×6=03.794FBC+8.64=03.794FBC=−8.64$

$FBC=−2.277 kip$

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0Ax−2.277cos18.43°=0Ax−2.16=0Ax=2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0Ay−1.44+2.277sin18.43°=0Ay−0.72=0Ay=0.72 kip$

Find the loading at the center of the beam AD $(PAD)$ :

$PAD=Weight of beam AD×Length of beam AD$

Substitute $120 lb/ft$ for the weight of beam AD and 6 ft for the length of beam AD.

$PAD=120×6=720 lb$

Convert the unit from lb to kip.

$PAD=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 2

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0ND+2.16=0ND=−2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0VD+0.72−0.72−=0VD=0$

Take moment about D is Equal to zero.

$∑MD=0MD−0.72×3=0MD−2.16=0MD=2.16 kip⋅ft$

Hence, the resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

Find the loading at the center of the column FC $(PFC)$ :

$PFC=Weight of column FC×Length of column FC$

Substitute $180 lb/ft$ for the weight of column FC and 16 ft for the length of column FC.

$PFC=180×16=2,880 lb$

Convert the unit from lb to kip.

$PFC=2,880 lb×1 kip1,000 lb=2.88 kip$

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 3

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

$sinθ=342+32sinθ=35θ=sin−1(0.6)θ=36.87°$

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0FCGcos36.87°−2.277sin18.43°=00.8FCG−0.72=00.8FCG=0.72$

$FCG=0.9 kip$

Find the loading at the center of the column FE $(PFE)$ :

$PFE=Weight of column FE×Length of column FE$

Substitute $180 lb/ft$ for the weight of column FE and 4 ft for the length of column FC.

$PFE=180×4=720 lb$

Convert the unit from lb to kip.

$PFE=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 4

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0VE−0.54=0VE=0.54 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0NE+0.72−5.04=0NE−4.32=0NE=4.32 kip$

Take moment about E is Equal to zero.

$∑ME=0−ME+0.54×4=0−ME+2.16=0ME=2.16 kip⋅ft$

Therefore, the resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

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11:16

Answer 4

Textbook Question

Chapter 1, Problem 1.1RP

The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E.

Chapter 1, Problem 1.1RP, The beam AB is pin supported at A and supported by a cable BC. A separate cable CG is used to hold

Expert Solution & Answer

To determine

The resultant internal loadings acting on cross sections located at D and E.

Answer to Problem 1.1RP

The resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

The resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Explanation of Solution

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is $120 lb/ft$ .

The weight of the column FC is $180 lb/ft$ .

Calculation:

Find the loading at the center of the beam AB $(PAB)$ :

$PAB=Weight of beam AB×Length of beam AB$

Substitute $120 lb/ft$ for the weight of beam AB and 12 ft for the length of beam AB.

$PAB=120×12=1,440 lb$

Convert the unit from lb to kip.

$PAB=1,440 lb×1 kip1,000 lb=1.44 kip$

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 1

Refer to Figure 1.

Find the angle of cable BC to the horizontal $(θ)$ :

$sinθ=112+32sinθ=13.1623θ=sin−1(0.3162)θ=18.43°$

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

$∑MA=0(FBCsin18.43°×12)+1.44×6=03.794FBC+8.64=03.794FBC=−8.64$

$FBC=−2.277 kip$

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0Ax−2.277cos18.43°=0Ax−2.16=0Ax=2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0Ay−1.44+2.277sin18.43°=0Ay−0.72=0Ay=0.72 kip$

Find the loading at the center of the beam AD $(PAD)$ :

$PAD=Weight of beam AD×Length of beam AD$

Substitute $120 lb/ft$ for the weight of beam AD and 6 ft for the length of beam AD.

$PAD=120×6=720 lb$

Convert the unit from lb to kip.

$PAD=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 2

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0ND+2.16=0ND=−2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0VD+0.72−0.72−=0VD=0$

Take moment about D is Equal to zero.

$∑MD=0MD−0.72×3=0MD−2.16=0MD=2.16 kip⋅ft$

Hence, the resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

Find the loading at the center of the column FC $(PFC)$ :

$PFC=Weight of column FC×Length of column FC$

Substitute $180 lb/ft$ for the weight of column FC and 16 ft for the length of column FC.

$PFC=180×16=2,880 lb$

Convert the unit from lb to kip.

$PFC=2,880 lb×1 kip1,000 lb=2.88 kip$

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 3

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

$sinθ=342+32sinθ=35θ=sin−1(0.6)θ=36.87°$

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0FCGcos36.87°−2.277sin18.43°=00.8FCG−0.72=00.8FCG=0.72$

$FCG=0.9 kip$

Find the loading at the center of the column FE $(PFE)$ :

$PFE=Weight of column FE×Length of column FE$

Substitute $180 lb/ft$ for the weight of column FE and 4 ft for the length of column FC.

$PFE=180×4=720 lb$

Convert the unit from lb to kip.

$PFE=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 4

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0VE−0.54=0VE=0.54 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0NE+0.72−5.04=0NE−4.32=0NE=4.32 kip$

Take moment about E is Equal to zero.

$∑ME=0−ME+0.54×4=0−ME+2.16=0ME=2.16 kip⋅ft$

Therefore, the resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

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11:16

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The resultant internal loadings acting on cross sections located at D and E.

Answer to Problem 1.1RP

The resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

The resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Explanation of Solution

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is $120 lb/ft$ .

The weight of the column FC is $180 lb/ft$ .

Calculation:

Find the loading at the center of the beam AB $(PAB)$ :

$PAB=Weight of beam AB×Length of beam AB$

Substitute $120 lb/ft$ for the weight of beam AB and 12 ft for the length of beam AB.

$PAB=120×12=1,440 lb$

Convert the unit from lb to kip.

$PAB=1,440 lb×1 kip1,000 lb=1.44 kip$

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 1

Refer to Figure 1.

Find the angle of cable BC to the horizontal $(θ)$ :

$sinθ=112+32sinθ=13.1623θ=sin−1(0.3162)θ=18.43°$

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

$∑MA=0(FBCsin18.43°×12)+1.44×6=03.794FBC+8.64=03.794FBC=−8.64$

$FBC=−2.277 kip$

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0Ax−2.277cos18.43°=0Ax−2.16=0Ax=2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0Ay−1.44+2.277sin18.43°=0Ay−0.72=0Ay=0.72 kip$

Find the loading at the center of the beam AD $(PAD)$ :

$PAD=Weight of beam AD×Length of beam AD$

Substitute $120 lb/ft$ for the weight of beam AD and 6 ft for the length of beam AD.

$PAD=120×6=720 lb$

Convert the unit from lb to kip.

$PAD=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 2

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0ND+2.16=0ND=−2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0VD+0.72−0.72−=0VD=0$

Take moment about D is Equal to zero.

$∑MD=0MD−0.72×3=0MD−2.16=0MD=2.16 kip⋅ft$

Hence, the resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

Find the loading at the center of the column FC $(PFC)$ :

$PFC=Weight of column FC×Length of column FC$

Substitute $180 lb/ft$ for the weight of column FC and 16 ft for the length of column FC.

$PFC=180×16=2,880 lb$

Convert the unit from lb to kip.

$PFC=2,880 lb×1 kip1,000 lb=2.88 kip$

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 3

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

$sinθ=342+32sinθ=35θ=sin−1(0.6)θ=36.87°$

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0FCGcos36.87°−2.277sin18.43°=00.8FCG−0.72=00.8FCG=0.72$

$FCG=0.9 kip$

Find the loading at the center of the column FE $(PFE)$ :

$PFE=Weight of column FE×Length of column FE$

Substitute $180 lb/ft$ for the weight of column FE and 4 ft for the length of column FC.

$PFE=180×4=720 lb$

Convert the unit from lb to kip.

$PFE=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 4

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0VE−0.54=0VE=0.54 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0NE+0.72−5.04=0NE−4.32=0NE=4.32 kip$

Take moment about E is Equal to zero.

$∑ME=0−ME+0.54×4=0−ME+2.16=0ME=2.16 kip⋅ft$

Therefore, the resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Answer 8

Expert Solution & Answer

To determine

The resultant internal loadings acting on cross sections located at D and E.

Answer to Problem 1.1RP

The resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

The resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Explanation of Solution

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is $120 lb/ft$ .

The weight of the column FC is $180 lb/ft$ .

Calculation:

Find the loading at the center of the beam AB $(PAB)$ :

$PAB=Weight of beam AB×Length of beam AB$

Substitute $120 lb/ft$ for the weight of beam AB and 12 ft for the length of beam AB.

$PAB=120×12=1,440 lb$

Convert the unit from lb to kip.

$PAB=1,440 lb×1 kip1,000 lb=1.44 kip$

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 1

Refer to Figure 1.

Find the angle of cable BC to the horizontal $(θ)$ :

$sinθ=112+32sinθ=13.1623θ=sin−1(0.3162)θ=18.43°$

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

$∑MA=0(FBCsin18.43°×12)+1.44×6=03.794FBC+8.64=03.794FBC=−8.64$

$FBC=−2.277 kip$

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0Ax−2.277cos18.43°=0Ax−2.16=0Ax=2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0Ay−1.44+2.277sin18.43°=0Ay−0.72=0Ay=0.72 kip$

Find the loading at the center of the beam AD $(PAD)$ :

$PAD=Weight of beam AD×Length of beam AD$

Substitute $120 lb/ft$ for the weight of beam AD and 6 ft for the length of beam AD.

$PAD=120×6=720 lb$

Convert the unit from lb to kip.

$PAD=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 2

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0ND+2.16=0ND=−2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0VD+0.72−0.72−=0VD=0$

Take moment about D is Equal to zero.

$∑MD=0MD−0.72×3=0MD−2.16=0MD=2.16 kip⋅ft$

Hence, the resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

Find the loading at the center of the column FC $(PFC)$ :

$PFC=Weight of column FC×Length of column FC$

Substitute $180 lb/ft$ for the weight of column FC and 16 ft for the length of column FC.

$PFC=180×16=2,880 lb$

Convert the unit from lb to kip.

$PFC=2,880 lb×1 kip1,000 lb=2.88 kip$

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 3

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

$sinθ=342+32sinθ=35θ=sin−1(0.6)θ=36.87°$

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0FCGcos36.87°−2.277sin18.43°=00.8FCG−0.72=00.8FCG=0.72$

$FCG=0.9 kip$

Find the loading at the center of the column FE $(PFE)$ :

$PFE=Weight of column FE×Length of column FE$

Substitute $180 lb/ft$ for the weight of column FE and 4 ft for the length of column FC.

$PFE=180×4=720 lb$

Convert the unit from lb to kip.

$PFE=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 4

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0VE−0.54=0VE=0.54 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0NE+0.72−5.04=0NE−4.32=0NE=4.32 kip$

Take moment about E is Equal to zero.

$∑ME=0−ME+0.54×4=0−ME+2.16=0ME=2.16 kip⋅ft$

Therefore, the resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The resultant internal loadings acting on cross sections located at D and E.

Answer 12

Answer to Problem 1.1RP

The resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

The resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Answer 13

Explanation of Solution

Given information:

The beam AB is pin supported at A and supported by a cable BC.

The weight of the beam AB is $120 lb/ft$ .

The weight of the column FC is $180 lb/ft$ .

Calculation:

Find the loading at the center of the beam AB $(PAB)$ :

$PAB=Weight of beam AB×Length of beam AB$

Substitute $120 lb/ft$ for the weight of beam AB and 12 ft for the length of beam AB.

$PAB=120×12=1,440 lb$

Convert the unit from lb to kip.

$PAB=1,440 lb×1 kip1,000 lb=1.44 kip$

Sketch the Free Body Diagram of the beam AB shown in Figure 1.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 1

Refer to Figure 1.

Find the angle of cable BC to the horizontal $(θ)$ :

$sinθ=112+32sinθ=13.1623θ=sin−1(0.3162)θ=18.43°$

Find the tension in cable BC as shown below.

Take moment about A is Equal to zero.

$∑MA=0(FBCsin18.43°×12)+1.44×6=03.794FBC+8.64=03.794FBC=−8.64$

$FBC=−2.277 kip$

Find the support reaction at A as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0Ax−2.277cos18.43°=0Ax−2.16=0Ax=2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0Ay−1.44+2.277sin18.43°=0Ay−0.72=0Ay=0.72 kip$

Find the loading at the center of the beam AD $(PAD)$ :

$PAD=Weight of beam AD×Length of beam AD$

Substitute $120 lb/ft$ for the weight of beam AD and 6 ft for the length of beam AD.

$PAD=120×6=720 lb$

Convert the unit from lb to kip.

$PAD=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point D as shown in Figure 2.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 2

Refer to Figure 2.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0ND+2.16=0ND=−2.16 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0VD+0.72−0.72−=0VD=0$

Take moment about D is Equal to zero.

$∑MD=0MD−0.72×3=0MD−2.16=0MD=2.16 kip⋅ft$

Hence, the resultant internal loadings at cross section at D are $ND=−2.16 kip, VD=0_$ , and $MD=2.16 kip⋅ft_$ .

Find the loading at the center of the column FC $(PFC)$ :

$PFC=Weight of column FC×Length of column FC$

Substitute $180 lb/ft$ for the weight of column FC and 16 ft for the length of column FC.

$PFC=180×16=2,880 lb$

Convert the unit from lb to kip.

$PFC=2,880 lb×1 kip1,000 lb=2.88 kip$

Sketch the Free Body Diagram of the beam FC shown in Figure 3.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 3

Refer to Figure 3.

Find the angle of cable CG to the horizontal.

$sinθ=342+32sinθ=35θ=sin−1(0.6)θ=36.87°$

Find the tension in cable CG as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0FCGcos36.87°−2.277sin18.43°=00.8FCG−0.72=00.8FCG=0.72$

$FCG=0.9 kip$

Find the loading at the center of the column FE $(PFE)$ :

$PFE=Weight of column FE×Length of column FE$

Substitute $180 lb/ft$ for the weight of column FE and 4 ft for the length of column FC.

$PFE=180×4=720 lb$

Convert the unit from lb to kip.

$PFE=720 lb×1 kip1,000 lb=0.72 kip$

Sketch the Free Body Diagram of the section for point E as shown in Figure 4.

MECHANICS OF MATERIAL IN SI UNITS, Chapter 1, Problem 1.1RP , additional homework tip 4

Refer to Figure 4.

Find the internal loadings as shown below.

Apply the Equations of Equilibrium as shown below.

Summation of forces along horizontal direction is Equal to zero.

$∑Fx=0VE−0.54=0VE=0.54 kip$

Summation of forces along vertical direction is Equal to zero.

$∑Fy=0NE+0.72−5.04=0NE−4.32=0NE=4.32 kip$

Take moment about E is Equal to zero.

$∑ME=0−ME+0.54×4=0−ME+2.16=0ME=2.16 kip⋅ft$

Therefore, the resultant internal loadings at cross section at E are $NE=4.32 kip, VE=0.54 kip_$ , and $ME=2.16 kip⋅ft_$ .

Answer 14

The beam AB is pin supported at A and supported by a cable BC . A separate cable CG is used to hold up the frame. If AB weighs 120 lb/ft and the column FC has a weight of 180 lb/ft, determine the resultant internal loadings acting on cross sections located at points D and E .

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