The forces in each member of the truss.

Question

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Answer 1

Question

Chapter 12, Problem 12.1P

To determine

The forces in each member of the truss.

Expert Solution & Answer

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Explanation of Solution

Check the determinacy of the truss as shown below.

$m+r−2j=0$

Here, number of members are $m$ , number of reactions are $r$ and number of joints are $j$ .

Substitute $11$ for $m$ , $3$ for $r$ and $6$ for $j$ .

$11+3−2×6>02>0$

Hence, the given truss is statically indeterminate by $2$ degree.

The following diagram shows the free body diagram of the truss.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 1

Figure-(1)

Write the equation for moment about A.

$ΣMA=0(Cy×16 ft)−(4 k×16 ft)−(5 k×8 ft)=0Cy×16 ft=104 k⋅ftCy=6.5 k$

Here, vertical reaction at C is $Cy$ .

Write the equation for sum of vertical forces.

$ΣFy=0Ay+Cy−3 k−4 k−5 k=0Ay+Cy=12 k$ ...... (I)

Here, vertical reaction at A is $Ay$ .

Substitute $6.5 k$ for $Cy$ in Equation (I).

$6.5 k+Ay=12 kAy=(12−6.5)kAy=5.5 k$

Consider the force in diagonal members FB is in tension and AE is in compression.

$FFB=FAE$

Here, force in member AE and FB are $FAE$ and $FFB$ .

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 2

Figure-(2)

Calculate the angles as shown below.

$cosθ=35sinθ=45$

Here, the angle between the members FA and FB is $θ$ .

Consider the joint F.

Write the Equation for sum of vertical forces.

$ΣFy=05.5 k−3 k−FFBcosθ−FAEcosθ=0$

Substitute $35$ for $cosθ$ .

$5.5 k−3 k−2FFB×35=02FFB×35=2.5 kFFB=2.08 k(Tension)FAE=2.08 k(Compression)$

Write the equation for the sum of horizontal forces.

$ΣFx=0FFE+FFBsinθ=0$

Here, force in member FE is $FFE$ .

Substitute $2.08 k$ for $FFB$ and $45$ for $sinθ$ .

$FFE+(2.08 k×45)=0FFE=1.67 k(Compression)$

Consider joint A.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 3

Figure-(3)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the equation for sum of vertical forces.

$ΣFy=0FFA+FAEsinθ−5.5 k=0$

Substitute $2.08 k$ for $FAE$ and $35$ for $sinθ$ .

$FFA+(2.08 k×35)−5.5 k=0FFA=4.25 k(Compression)$

Here, force in member FA is $FFA$ .

Write the Equation for sum of horizontal forces.

$ΣFx=0FAB−FAEcosθ=0$

Substitute $2.08 k$ for $FAE$ and $45$ for $cosθ$ .

$FAB−2.08 k×45=0FAB=1.67 k(Tension)$

Here, force in member AB is $FAB$ .

Consider the force in diagonal members BD is in tension and EC is in compression.

$FBD=FEC$

Here, force in member BD and EC are $FBD$ and $FEC$ .

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 4

Figure-(4)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the Equation for sum of vertical forces.

$ΣFy=05.5−3−5+FBDsinθ+FECsinθ=02FBDsinθ=2.5$

Substitute $35$ for $sinθ$ .

$2FBD×35=2.5FBD=2.08 k(Tension)FCE=2.08(Compression)$

Here, force in the member BD and CE are $FBD$ and $FCE$ .

Write the equation for moment about B.

$(FED×6+FECcosθ×6+4×8−6.5×8)k⋅ft=0$

Here, force in the member ED is $FED$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FEC$ .

$(FED×6+2.08×45×6+4×8−6.5×8) k⋅ft=0FED×6 ft=10.01 k⋅ftFED=1.67 k(Compression)$

Write the Equation for moment about E.

$(FBC×6+FBDcosθ×6−6.5×8+4×8)=0$

Here, force in the member BC is $FBC$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FBD$ .

$(FBC×6+2.08 k×45×6−6.5×8+4×8) k⋅ft=0FBC×6 ft=10.01 k⋅ftFBC=1.67 k(Tension)$

Write the Equation for sum of vertical forces.

$ΣFy=0FBE+FFBsinθ+FBDsinθ=0$

Substitute $35$ for $sinθ$ , $2.08 k$ for $FFB$ and $2.08 k$ for $FBD$ .

$FBE+2.08 k×35+2.08 k×35=0FBE+1.25 k+1.25 k=0FBE=2.5 k(Compression)$

Here, force in the member BE is $FBE$ .

Consider the joint C.

Angles will be same as calculated at joint B.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 5

Figure-(5)

Write the Equation for sum of vertical forces.

$ΣFy=0FCD+FECsinθ−6.5 k=0$

Here, force in the member CD is $FCD$ .

Substitute $35$ for $sinθ$ , and $2.08 k$ for $FEC$ .

$FCD+(2.08 k×35)−6.5 k=0FCD=5.25 k(Compression)$

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

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A k 000 6 ft A kips Bl D ft C C kips 10 ft 12 ft E B k/ft D E ft t

Answer 2

Question

Chapter 12, Problem 12.1P

To determine

The forces in each member of the truss.

Expert Solution & Answer

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Explanation of Solution

Check the determinacy of the truss as shown below.

$m+r−2j=0$

Here, number of members are $m$ , number of reactions are $r$ and number of joints are $j$ .

Substitute $11$ for $m$ , $3$ for $r$ and $6$ for $j$ .

$11+3−2×6>02>0$

Hence, the given truss is statically indeterminate by $2$ degree.

The following diagram shows the free body diagram of the truss.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 1

Figure-(1)

Write the equation for moment about A.

$ΣMA=0(Cy×16 ft)−(4 k×16 ft)−(5 k×8 ft)=0Cy×16 ft=104 k⋅ftCy=6.5 k$

Here, vertical reaction at C is $Cy$ .

Write the equation for sum of vertical forces.

$ΣFy=0Ay+Cy−3 k−4 k−5 k=0Ay+Cy=12 k$ ...... (I)

Here, vertical reaction at A is $Ay$ .

Substitute $6.5 k$ for $Cy$ in Equation (I).

$6.5 k+Ay=12 kAy=(12−6.5)kAy=5.5 k$

Consider the force in diagonal members FB is in tension and AE is in compression.

$FFB=FAE$

Here, force in member AE and FB are $FAE$ and $FFB$ .

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 2

Figure-(2)

Calculate the angles as shown below.

$cosθ=35sinθ=45$

Here, the angle between the members FA and FB is $θ$ .

Consider the joint F.

Write the Equation for sum of vertical forces.

$ΣFy=05.5 k−3 k−FFBcosθ−FAEcosθ=0$

Substitute $35$ for $cosθ$ .

$5.5 k−3 k−2FFB×35=02FFB×35=2.5 kFFB=2.08 k(Tension)FAE=2.08 k(Compression)$

Write the equation for the sum of horizontal forces.

$ΣFx=0FFE+FFBsinθ=0$

Here, force in member FE is $FFE$ .

Substitute $2.08 k$ for $FFB$ and $45$ for $sinθ$ .

$FFE+(2.08 k×45)=0FFE=1.67 k(Compression)$

Consider joint A.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 3

Figure-(3)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the equation for sum of vertical forces.

$ΣFy=0FFA+FAEsinθ−5.5 k=0$

Substitute $2.08 k$ for $FAE$ and $35$ for $sinθ$ .

$FFA+(2.08 k×35)−5.5 k=0FFA=4.25 k(Compression)$

Here, force in member FA is $FFA$ .

Write the Equation for sum of horizontal forces.

$ΣFx=0FAB−FAEcosθ=0$

Substitute $2.08 k$ for $FAE$ and $45$ for $cosθ$ .

$FAB−2.08 k×45=0FAB=1.67 k(Tension)$

Here, force in member AB is $FAB$ .

Consider the force in diagonal members BD is in tension and EC is in compression.

$FBD=FEC$

Here, force in member BD and EC are $FBD$ and $FEC$ .

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 4

Figure-(4)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the Equation for sum of vertical forces.

$ΣFy=05.5−3−5+FBDsinθ+FECsinθ=02FBDsinθ=2.5$

Substitute $35$ for $sinθ$ .

$2FBD×35=2.5FBD=2.08 k(Tension)FCE=2.08(Compression)$

Here, force in the member BD and CE are $FBD$ and $FCE$ .

Write the equation for moment about B.

$(FED×6+FECcosθ×6+4×8−6.5×8)k⋅ft=0$

Here, force in the member ED is $FED$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FEC$ .

$(FED×6+2.08×45×6+4×8−6.5×8) k⋅ft=0FED×6 ft=10.01 k⋅ftFED=1.67 k(Compression)$

Write the Equation for moment about E.

$(FBC×6+FBDcosθ×6−6.5×8+4×8)=0$

Here, force in the member BC is $FBC$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FBD$ .

$(FBC×6+2.08 k×45×6−6.5×8+4×8) k⋅ft=0FBC×6 ft=10.01 k⋅ftFBC=1.67 k(Tension)$

Write the Equation for sum of vertical forces.

$ΣFy=0FBE+FFBsinθ+FBDsinθ=0$

Substitute $35$ for $sinθ$ , $2.08 k$ for $FFB$ and $2.08 k$ for $FBD$ .

$FBE+2.08 k×35+2.08 k×35=0FBE+1.25 k+1.25 k=0FBE=2.5 k(Compression)$

Here, force in the member BE is $FBE$ .

Consider the joint C.

Angles will be same as calculated at joint B.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 5

Figure-(5)

Write the Equation for sum of vertical forces.

$ΣFy=0FCD+FECsinθ−6.5 k=0$

Here, force in the member CD is $FCD$ .

Substitute $35$ for $sinθ$ , and $2.08 k$ for $FEC$ .

$FCD+(2.08 k×35)−6.5 k=0FCD=5.25 k(Compression)$

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

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Answer 3

Question

Chapter 12, Problem 12.1P

To determine

The forces in each member of the truss.

Expert Solution & Answer

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Explanation of Solution

Check the determinacy of the truss as shown below.

$m+r−2j=0$

Here, number of members are $m$ , number of reactions are $r$ and number of joints are $j$ .

Substitute $11$ for $m$ , $3$ for $r$ and $6$ for $j$ .

$11+3−2×6>02>0$

Hence, the given truss is statically indeterminate by $2$ degree.

The following diagram shows the free body diagram of the truss.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 1

Figure-(1)

Write the equation for moment about A.

$ΣMA=0(Cy×16 ft)−(4 k×16 ft)−(5 k×8 ft)=0Cy×16 ft=104 k⋅ftCy=6.5 k$

Here, vertical reaction at C is $Cy$ .

Write the equation for sum of vertical forces.

$ΣFy=0Ay+Cy−3 k−4 k−5 k=0Ay+Cy=12 k$ ...... (I)

Here, vertical reaction at A is $Ay$ .

Substitute $6.5 k$ for $Cy$ in Equation (I).

$6.5 k+Ay=12 kAy=(12−6.5)kAy=5.5 k$

Consider the force in diagonal members FB is in tension and AE is in compression.

$FFB=FAE$

Here, force in member AE and FB are $FAE$ and $FFB$ .

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 2

Figure-(2)

Calculate the angles as shown below.

$cosθ=35sinθ=45$

Here, the angle between the members FA and FB is $θ$ .

Consider the joint F.

Write the Equation for sum of vertical forces.

$ΣFy=05.5 k−3 k−FFBcosθ−FAEcosθ=0$

Substitute $35$ for $cosθ$ .

$5.5 k−3 k−2FFB×35=02FFB×35=2.5 kFFB=2.08 k(Tension)FAE=2.08 k(Compression)$

Write the equation for the sum of horizontal forces.

$ΣFx=0FFE+FFBsinθ=0$

Here, force in member FE is $FFE$ .

Substitute $2.08 k$ for $FFB$ and $45$ for $sinθ$ .

$FFE+(2.08 k×45)=0FFE=1.67 k(Compression)$

Consider joint A.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 3

Figure-(3)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the equation for sum of vertical forces.

$ΣFy=0FFA+FAEsinθ−5.5 k=0$

Substitute $2.08 k$ for $FAE$ and $35$ for $sinθ$ .

$FFA+(2.08 k×35)−5.5 k=0FFA=4.25 k(Compression)$

Here, force in member FA is $FFA$ .

Write the Equation for sum of horizontal forces.

$ΣFx=0FAB−FAEcosθ=0$

Substitute $2.08 k$ for $FAE$ and $45$ for $cosθ$ .

$FAB−2.08 k×45=0FAB=1.67 k(Tension)$

Here, force in member AB is $FAB$ .

Consider the force in diagonal members BD is in tension and EC is in compression.

$FBD=FEC$

Here, force in member BD and EC are $FBD$ and $FEC$ .

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 4

Figure-(4)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the Equation for sum of vertical forces.

$ΣFy=05.5−3−5+FBDsinθ+FECsinθ=02FBDsinθ=2.5$

Substitute $35$ for $sinθ$ .

$2FBD×35=2.5FBD=2.08 k(Tension)FCE=2.08(Compression)$

Here, force in the member BD and CE are $FBD$ and $FCE$ .

Write the equation for moment about B.

$(FED×6+FECcosθ×6+4×8−6.5×8)k⋅ft=0$

Here, force in the member ED is $FED$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FEC$ .

$(FED×6+2.08×45×6+4×8−6.5×8) k⋅ft=0FED×6 ft=10.01 k⋅ftFED=1.67 k(Compression)$

Write the Equation for moment about E.

$(FBC×6+FBDcosθ×6−6.5×8+4×8)=0$

Here, force in the member BC is $FBC$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FBD$ .

$(FBC×6+2.08 k×45×6−6.5×8+4×8) k⋅ft=0FBC×6 ft=10.01 k⋅ftFBC=1.67 k(Tension)$

Write the Equation for sum of vertical forces.

$ΣFy=0FBE+FFBsinθ+FBDsinθ=0$

Substitute $35$ for $sinθ$ , $2.08 k$ for $FFB$ and $2.08 k$ for $FBD$ .

$FBE+2.08 k×35+2.08 k×35=0FBE+1.25 k+1.25 k=0FBE=2.5 k(Compression)$

Here, force in the member BE is $FBE$ .

Consider the joint C.

Angles will be same as calculated at joint B.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 5

Figure-(5)

Write the Equation for sum of vertical forces.

$ΣFy=0FCD+FECsinθ−6.5 k=0$

Here, force in the member CD is $FCD$ .

Substitute $35$ for $sinθ$ , and $2.08 k$ for $FEC$ .

$FCD+(2.08 k×35)−6.5 k=0FCD=5.25 k(Compression)$

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

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Answer 4

Question

Chapter 12, Problem 12.1P

To determine

The forces in each member of the truss.

Expert Solution & Answer

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Explanation of Solution

Check the determinacy of the truss as shown below.

$m+r−2j=0$

Here, number of members are $m$ , number of reactions are $r$ and number of joints are $j$ .

Substitute $11$ for $m$ , $3$ for $r$ and $6$ for $j$ .

$11+3−2×6>02>0$

Hence, the given truss is statically indeterminate by $2$ degree.

The following diagram shows the free body diagram of the truss.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 1

Figure-(1)

Write the equation for moment about A.

$ΣMA=0(Cy×16 ft)−(4 k×16 ft)−(5 k×8 ft)=0Cy×16 ft=104 k⋅ftCy=6.5 k$

Here, vertical reaction at C is $Cy$ .

Write the equation for sum of vertical forces.

$ΣFy=0Ay+Cy−3 k−4 k−5 k=0Ay+Cy=12 k$ ...... (I)

Here, vertical reaction at A is $Ay$ .

Substitute $6.5 k$ for $Cy$ in Equation (I).

$6.5 k+Ay=12 kAy=(12−6.5)kAy=5.5 k$

Consider the force in diagonal members FB is in tension and AE is in compression.

$FFB=FAE$

Here, force in member AE and FB are $FAE$ and $FFB$ .

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 2

Figure-(2)

Calculate the angles as shown below.

$cosθ=35sinθ=45$

Here, the angle between the members FA and FB is $θ$ .

Consider the joint F.

Write the Equation for sum of vertical forces.

$ΣFy=05.5 k−3 k−FFBcosθ−FAEcosθ=0$

Substitute $35$ for $cosθ$ .

$5.5 k−3 k−2FFB×35=02FFB×35=2.5 kFFB=2.08 k(Tension)FAE=2.08 k(Compression)$

Write the equation for the sum of horizontal forces.

$ΣFx=0FFE+FFBsinθ=0$

Here, force in member FE is $FFE$ .

Substitute $2.08 k$ for $FFB$ and $45$ for $sinθ$ .

$FFE+(2.08 k×45)=0FFE=1.67 k(Compression)$

Consider joint A.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 3

Figure-(3)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the equation for sum of vertical forces.

$ΣFy=0FFA+FAEsinθ−5.5 k=0$

Substitute $2.08 k$ for $FAE$ and $35$ for $sinθ$ .

$FFA+(2.08 k×35)−5.5 k=0FFA=4.25 k(Compression)$

Here, force in member FA is $FFA$ .

Write the Equation for sum of horizontal forces.

$ΣFx=0FAB−FAEcosθ=0$

Substitute $2.08 k$ for $FAE$ and $45$ for $cosθ$ .

$FAB−2.08 k×45=0FAB=1.67 k(Tension)$

Here, force in member AB is $FAB$ .

Consider the force in diagonal members BD is in tension and EC is in compression.

$FBD=FEC$

Here, force in member BD and EC are $FBD$ and $FEC$ .

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 4

Figure-(4)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the Equation for sum of vertical forces.

$ΣFy=05.5−3−5+FBDsinθ+FECsinθ=02FBDsinθ=2.5$

Substitute $35$ for $sinθ$ .

$2FBD×35=2.5FBD=2.08 k(Tension)FCE=2.08(Compression)$

Here, force in the member BD and CE are $FBD$ and $FCE$ .

Write the equation for moment about B.

$(FED×6+FECcosθ×6+4×8−6.5×8)k⋅ft=0$

Here, force in the member ED is $FED$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FEC$ .

$(FED×6+2.08×45×6+4×8−6.5×8) k⋅ft=0FED×6 ft=10.01 k⋅ftFED=1.67 k(Compression)$

Write the Equation for moment about E.

$(FBC×6+FBDcosθ×6−6.5×8+4×8)=0$

Here, force in the member BC is $FBC$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FBD$ .

$(FBC×6+2.08 k×45×6−6.5×8+4×8) k⋅ft=0FBC×6 ft=10.01 k⋅ftFBC=1.67 k(Tension)$

Write the Equation for sum of vertical forces.

$ΣFy=0FBE+FFBsinθ+FBDsinθ=0$

Substitute $35$ for $sinθ$ , $2.08 k$ for $FFB$ and $2.08 k$ for $FBD$ .

$FBE+2.08 k×35+2.08 k×35=0FBE+1.25 k+1.25 k=0FBE=2.5 k(Compression)$

Here, force in the member BE is $FBE$ .

Consider the joint C.

Angles will be same as calculated at joint B.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 5

Figure-(5)

Write the Equation for sum of vertical forces.

$ΣFy=0FCD+FECsinθ−6.5 k=0$

Here, force in the member CD is $FCD$ .

Substitute $35$ for $sinθ$ , and $2.08 k$ for $FEC$ .

$FCD+(2.08 k×35)−6.5 k=0FCD=5.25 k(Compression)$

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

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Answer 5

Chapter 12, Problem 12.1P

To determine

The forces in each member of the truss.

Expert Solution & Answer

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Explanation of Solution

Check the determinacy of the truss as shown below.

$m+r−2j=0$

Here, number of members are $m$ , number of reactions are $r$ and number of joints are $j$ .

Substitute $11$ for $m$ , $3$ for $r$ and $6$ for $j$ .

$11+3−2×6>02>0$

Hence, the given truss is statically indeterminate by $2$ degree.

The following diagram shows the free body diagram of the truss.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 1

Figure-(1)

Write the equation for moment about A.

$ΣMA=0(Cy×16 ft)−(4 k×16 ft)−(5 k×8 ft)=0Cy×16 ft=104 k⋅ftCy=6.5 k$

Here, vertical reaction at C is $Cy$ .

Write the equation for sum of vertical forces.

$ΣFy=0Ay+Cy−3 k−4 k−5 k=0Ay+Cy=12 k$ ...... (I)

Here, vertical reaction at A is $Ay$ .

Substitute $6.5 k$ for $Cy$ in Equation (I).

$6.5 k+Ay=12 kAy=(12−6.5)kAy=5.5 k$

Consider the force in diagonal members FB is in tension and AE is in compression.

$FFB=FAE$

Here, force in member AE and FB are $FAE$ and $FFB$ .

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 2

Figure-(2)

Calculate the angles as shown below.

$cosθ=35sinθ=45$

Here, the angle between the members FA and FB is $θ$ .

Consider the joint F.

Write the Equation for sum of vertical forces.

$ΣFy=05.5 k−3 k−FFBcosθ−FAEcosθ=0$

Substitute $35$ for $cosθ$ .

$5.5 k−3 k−2FFB×35=02FFB×35=2.5 kFFB=2.08 k(Tension)FAE=2.08 k(Compression)$

Write the equation for the sum of horizontal forces.

$ΣFx=0FFE+FFBsinθ=0$

Here, force in member FE is $FFE$ .

Substitute $2.08 k$ for $FFB$ and $45$ for $sinθ$ .

$FFE+(2.08 k×45)=0FFE=1.67 k(Compression)$

Consider joint A.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 3

Figure-(3)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the equation for sum of vertical forces.

$ΣFy=0FFA+FAEsinθ−5.5 k=0$

Substitute $2.08 k$ for $FAE$ and $35$ for $sinθ$ .

$FFA+(2.08 k×35)−5.5 k=0FFA=4.25 k(Compression)$

Here, force in member FA is $FFA$ .

Write the Equation for sum of horizontal forces.

$ΣFx=0FAB−FAEcosθ=0$

Substitute $2.08 k$ for $FAE$ and $45$ for $cosθ$ .

$FAB−2.08 k×45=0FAB=1.67 k(Tension)$

Here, force in member AB is $FAB$ .

Consider the force in diagonal members BD is in tension and EC is in compression.

$FBD=FEC$

Here, force in member BD and EC are $FBD$ and $FEC$ .

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 4

Figure-(4)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the Equation for sum of vertical forces.

$ΣFy=05.5−3−5+FBDsinθ+FECsinθ=02FBDsinθ=2.5$

Substitute $35$ for $sinθ$ .

$2FBD×35=2.5FBD=2.08 k(Tension)FCE=2.08(Compression)$

Here, force in the member BD and CE are $FBD$ and $FCE$ .

Write the equation for moment about B.

$(FED×6+FECcosθ×6+4×8−6.5×8)k⋅ft=0$

Here, force in the member ED is $FED$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FEC$ .

$(FED×6+2.08×45×6+4×8−6.5×8) k⋅ft=0FED×6 ft=10.01 k⋅ftFED=1.67 k(Compression)$

Write the Equation for moment about E.

$(FBC×6+FBDcosθ×6−6.5×8+4×8)=0$

Here, force in the member BC is $FBC$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FBD$ .

$(FBC×6+2.08 k×45×6−6.5×8+4×8) k⋅ft=0FBC×6 ft=10.01 k⋅ftFBC=1.67 k(Tension)$

Write the Equation for sum of vertical forces.

$ΣFy=0FBE+FFBsinθ+FBDsinθ=0$

Substitute $35$ for $sinθ$ , $2.08 k$ for $FFB$ and $2.08 k$ for $FBD$ .

$FBE+2.08 k×35+2.08 k×35=0FBE+1.25 k+1.25 k=0FBE=2.5 k(Compression)$

Here, force in the member BE is $FBE$ .

Consider the joint C.

Angles will be same as calculated at joint B.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 5

Figure-(5)

Write the Equation for sum of vertical forces.

$ΣFy=0FCD+FECsinθ−6.5 k=0$

Here, force in the member CD is $FCD$ .

Substitute $35$ for $sinθ$ , and $2.08 k$ for $FEC$ .

$FCD+(2.08 k×35)−6.5 k=0FCD=5.25 k(Compression)$

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Answer 6

Chapter 12, Problem 12.1P

To determine

The forces in each member of the truss.

Expert Solution & Answer

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Explanation of Solution

Check the determinacy of the truss as shown below.

$m+r−2j=0$

Here, number of members are $m$ , number of reactions are $r$ and number of joints are $j$ .

Substitute $11$ for $m$ , $3$ for $r$ and $6$ for $j$ .

$11+3−2×6>02>0$

Hence, the given truss is statically indeterminate by $2$ degree.

The following diagram shows the free body diagram of the truss.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 1

Figure-(1)

Write the equation for moment about A.

$ΣMA=0(Cy×16 ft)−(4 k×16 ft)−(5 k×8 ft)=0Cy×16 ft=104 k⋅ftCy=6.5 k$

Here, vertical reaction at C is $Cy$ .

Write the equation for sum of vertical forces.

$ΣFy=0Ay+Cy−3 k−4 k−5 k=0Ay+Cy=12 k$ ...... (I)

Here, vertical reaction at A is $Ay$ .

Substitute $6.5 k$ for $Cy$ in Equation (I).

$6.5 k+Ay=12 kAy=(12−6.5)kAy=5.5 k$

Consider the force in diagonal members FB is in tension and AE is in compression.

$FFB=FAE$

Here, force in member AE and FB are $FAE$ and $FFB$ .

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 2

Figure-(2)

Calculate the angles as shown below.

$cosθ=35sinθ=45$

Here, the angle between the members FA and FB is $θ$ .

Consider the joint F.

Write the Equation for sum of vertical forces.

$ΣFy=05.5 k−3 k−FFBcosθ−FAEcosθ=0$

Substitute $35$ for $cosθ$ .

$5.5 k−3 k−2FFB×35=02FFB×35=2.5 kFFB=2.08 k(Tension)FAE=2.08 k(Compression)$

Write the equation for the sum of horizontal forces.

$ΣFx=0FFE+FFBsinθ=0$

Here, force in member FE is $FFE$ .

Substitute $2.08 k$ for $FFB$ and $45$ for $sinθ$ .

$FFE+(2.08 k×45)=0FFE=1.67 k(Compression)$

Consider joint A.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 3

Figure-(3)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the equation for sum of vertical forces.

$ΣFy=0FFA+FAEsinθ−5.5 k=0$

Substitute $2.08 k$ for $FAE$ and $35$ for $sinθ$ .

$FFA+(2.08 k×35)−5.5 k=0FFA=4.25 k(Compression)$

Here, force in member FA is $FFA$ .

Write the Equation for sum of horizontal forces.

$ΣFx=0FAB−FAEcosθ=0$

Substitute $2.08 k$ for $FAE$ and $45$ for $cosθ$ .

$FAB−2.08 k×45=0FAB=1.67 k(Tension)$

Here, force in member AB is $FAB$ .

Consider the force in diagonal members BD is in tension and EC is in compression.

$FBD=FEC$

Here, force in member BD and EC are $FBD$ and $FEC$ .

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 4

Figure-(4)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the Equation for sum of vertical forces.

$ΣFy=05.5−3−5+FBDsinθ+FECsinθ=02FBDsinθ=2.5$

Substitute $35$ for $sinθ$ .

$2FBD×35=2.5FBD=2.08 k(Tension)FCE=2.08(Compression)$

Here, force in the member BD and CE are $FBD$ and $FCE$ .

Write the equation for moment about B.

$(FED×6+FECcosθ×6+4×8−6.5×8)k⋅ft=0$

Here, force in the member ED is $FED$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FEC$ .

$(FED×6+2.08×45×6+4×8−6.5×8) k⋅ft=0FED×6 ft=10.01 k⋅ftFED=1.67 k(Compression)$

Write the Equation for moment about E.

$(FBC×6+FBDcosθ×6−6.5×8+4×8)=0$

Here, force in the member BC is $FBC$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FBD$ .

$(FBC×6+2.08 k×45×6−6.5×8+4×8) k⋅ft=0FBC×6 ft=10.01 k⋅ftFBC=1.67 k(Tension)$

Write the Equation for sum of vertical forces.

$ΣFy=0FBE+FFBsinθ+FBDsinθ=0$

Substitute $35$ for $sinθ$ , $2.08 k$ for $FFB$ and $2.08 k$ for $FBD$ .

$FBE+2.08 k×35+2.08 k×35=0FBE+1.25 k+1.25 k=0FBE=2.5 k(Compression)$

Here, force in the member BE is $FBE$ .

Consider the joint C.

Angles will be same as calculated at joint B.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 5

Figure-(5)

Write the Equation for sum of vertical forces.

$ΣFy=0FCD+FECsinθ−6.5 k=0$

Here, force in the member CD is $FCD$ .

Substitute $35$ for $sinθ$ , and $2.08 k$ for $FEC$ .

$FCD+(2.08 k×35)−6.5 k=0FCD=5.25 k(Compression)$

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Answer 7

Chapter 12, Problem 12.1P

To determine

The forces in each member of the truss.

Answer 8

Expert Solution & Answer

Answer to Problem 12.1P

The forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Check the determinacy of the truss as shown below.

$m+r−2j=0$

Here, number of members are $m$ , number of reactions are $r$ and number of joints are $j$ .

Substitute $11$ for $m$ , $3$ for $r$ and $6$ for $j$ .

$11+3−2×6>02>0$

Hence, the given truss is statically indeterminate by $2$ degree.

The following diagram shows the free body diagram of the truss.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 1

Figure-(1)

Write the equation for moment about A.

$ΣMA=0(Cy×16 ft)−(4 k×16 ft)−(5 k×8 ft)=0Cy×16 ft=104 k⋅ftCy=6.5 k$

Here, vertical reaction at C is $Cy$ .

Write the equation for sum of vertical forces.

$ΣFy=0Ay+Cy−3 k−4 k−5 k=0Ay+Cy=12 k$ ...... (I)

Here, vertical reaction at A is $Ay$ .

Substitute $6.5 k$ for $Cy$ in Equation (I).

$6.5 k+Ay=12 kAy=(12−6.5)kAy=5.5 k$

Consider the force in diagonal members FB is in tension and AE is in compression.

$FFB=FAE$

Here, force in member AE and FB are $FAE$ and $FFB$ .

Use the method of section and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 2

Figure-(2)

Calculate the angles as shown below.

$cosθ=35sinθ=45$

Here, the angle between the members FA and FB is $θ$ .

Consider the joint F.

Write the Equation for sum of vertical forces.

$ΣFy=05.5 k−3 k−FFBcosθ−FAEcosθ=0$

Substitute $35$ for $cosθ$ .

$5.5 k−3 k−2FFB×35=02FFB×35=2.5 kFFB=2.08 k(Tension)FAE=2.08 k(Compression)$

Write the equation for the sum of horizontal forces.

$ΣFx=0FFE+FFBsinθ=0$

Here, force in member FE is $FFE$ .

Substitute $2.08 k$ for $FFB$ and $45$ for $sinθ$ .

$FFE+(2.08 k×45)=0FFE=1.67 k(Compression)$

Consider joint A.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 3

Figure-(3)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the equation for sum of vertical forces.

$ΣFy=0FFA+FAEsinθ−5.5 k=0$

Substitute $2.08 k$ for $FAE$ and $35$ for $sinθ$ .

$FFA+(2.08 k×35)−5.5 k=0FFA=4.25 k(Compression)$

Here, force in member FA is $FFA$ .

Write the Equation for sum of horizontal forces.

$ΣFx=0FAB−FAEcosθ=0$

Substitute $2.08 k$ for $FAE$ and $45$ for $cosθ$ .

$FAB−2.08 k×45=0FAB=1.67 k(Tension)$

Here, force in member AB is $FAB$ .

Consider the force in diagonal members BD is in tension and EC is in compression.

$FBD=FEC$

Here, force in member BD and EC are $FBD$ and $FEC$ .

Use the method of section method and cut the section as shown in figure below and calculate the forces in the members.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 4

Figure-(4)

Calculate the angles as shown below.

$cosθ=45sinθ=35$

Write the Equation for sum of vertical forces.

$ΣFy=05.5−3−5+FBDsinθ+FECsinθ=02FBDsinθ=2.5$

Substitute $35$ for $sinθ$ .

$2FBD×35=2.5FBD=2.08 k(Tension)FCE=2.08(Compression)$

Here, force in the member BD and CE are $FBD$ and $FCE$ .

Write the equation for moment about B.

$(FED×6+FECcosθ×6+4×8−6.5×8)k⋅ft=0$

Here, force in the member ED is $FED$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FEC$ .

$(FED×6+2.08×45×6+4×8−6.5×8) k⋅ft=0FED×6 ft=10.01 k⋅ftFED=1.67 k(Compression)$

Write the Equation for moment about E.

$(FBC×6+FBDcosθ×6−6.5×8+4×8)=0$

Here, force in the member BC is $FBC$ .

Substitute $45$ for $cosθ$ and $2.08 k$ for $FBD$ .

$(FBC×6+2.08 k×45×6−6.5×8+4×8) k⋅ft=0FBC×6 ft=10.01 k⋅ftFBC=1.67 k(Tension)$

Write the Equation for sum of vertical forces.

$ΣFy=0FBE+FFBsinθ+FBDsinθ=0$

Substitute $35$ for $sinθ$ , $2.08 k$ for $FFB$ and $2.08 k$ for $FBD$ .

$FBE+2.08 k×35+2.08 k×35=0FBE+1.25 k+1.25 k=0FBE=2.5 k(Compression)$

Here, force in the member BE is $FBE$ .

Consider the joint C.

Angles will be same as calculated at joint B.

Structural Analysis (10th Edition), Chapter 12, Problem 12.1P , additional homework tip 5

Figure-(5)

Write the Equation for sum of vertical forces.

$ΣFy=0FCD+FECsinθ−6.5 k=0$

Here, force in the member CD is $FCD$ .

Substitute $35$ for $sinθ$ , and $2.08 k$ for $FEC$ .

$FCD+(2.08 k×35)−6.5 k=0FCD=5.25 k(Compression)$

Conclusion:

Therefore, the forces in each member of the truss are shown below.

The force in member AE is, $2.08 k(Compression)$ .

The force in member BF is, $2.08 k(Tension)$ .

The force in member EF is, $1.67 k(Compression)$ .

The force in member AB is, $1.67 k(Tension)$ .

The force in member CE is, $2.08(Compression)$ .

The force in member BD is, $2.08 k(Tension)$ .

The force in member DE is, $1.67 k(Compression)$ .

The force in member BC is, $1.67 k(Tension)$ .

The force in member AF is, $4.25 k(Compression)$ .

The force in member BE is, $2.5 k(Compression)$ .

The force in member CD is, $5.25 k(Compression)$ .

Answer 12

Ch. 12 - Prob. 12.1P Ch. 12 - Prob. 12.2P Ch. 12 - Prob. 12.3P Ch. 12 - Prob. 12.4P Ch. 12 - Prob. 12.5P Ch. 12 - Prob. 12.6P Ch. 12 - Prob. 12.7P Ch. 12 - Prob. 12.8P Ch. 12 - Prob. 12.9P Ch. 12 - Prob. 12.10P

The forces in each member of the truss.

The forces in each member of the truss.

Answer to Problem 12.1P

Explanation of Solution

Want to see more full solutions like this?

Chapter 12 Solutions