Find the vertical displacement of joint C .

Question

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Answer 1

Question

Chapter 8, Problem 4P

(a)

To determine

Find the vertical displacement of joint C.

(a)

Expert Solution

Answer to Problem 4P

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

The value of E is $10,000 ksi$ .

Procedure to find the deflection of truss by virtual work method is shown below.

For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.

For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces $(Fv)$ in all the member of the truss.

Finally use the desired deflection equation.

Apply the sign conventions for calculating reactions, 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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.

If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the bar forces $FP$ produced by the P-system as follows:

Let $Ax$ and $Ay$ be the horizontal and vertical reactions at the hinged support A.

Let $Dy$ be the vertical reaction at the roller support D.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−15(11)+10(8)=0Dy=3.86 kips↑$

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

$+↑∑Fy=0Dy+Ay−15=03.86+Ay−15=0Ay=11.14 kips↑$

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

$+→∑Fx=0Ax−10=0Ax=10 kips→$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=011.14+FABsin53.13°=0FAB=−13.92 kips$

$+→∑Fx=010+FAE+FABcos53.13°=010+FAE+(−13.92)cos53.13°=0FAE=−1.648 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=03.86+FCDsin53.13°=0FCD=−4.83 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−4.83)cos53.13°=0FED=2.898 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(−13.92)sin53.13°−FBEsin57.99°=0FBE=13.133 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°−10+FBC=0−(−13.92)cos53.13°+(13.133)cos57.99°−10+FBC=0FBC=−5.312 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−4.830)sin53.13°−FCEsin57.99°=0FCE=4.556 kips$

Sketch the bar forces produced by the P-system as shown in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 1

Consider a dummy load of 1 kN directed vertically at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0−Dy(22)+1(16)=0Dy=0.727 kips↓$

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

$+↑∑Fy=0−Dy+Ay+1=0−0.727+Ay+1=0Ay=0.273 kips↓$

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

$+→∑Fx=0Ax=0$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.273+FABsin53.13°=0FAB=0.341 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.341)cos53.13°=0FAE=−0.205 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.727+FCDsin53.13°=0FCD=0.909 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(0.909)cos53.13°=0FED=−0.545 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.341)sin53.13°−FBEsin57.99°=0FBE=−0.322 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.341)cos53.13°+(−0.322)cos57.99°+FBC=0FBC=0.375 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°+1=0−(0.909)sin53.13°−FCEsin57.99°+1=0FCE=0.322 kips$

Sketch the bar forces $FQ$ produced by the vertical dummy load $Cy$ as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 2

Refer Table 1 for vertical displacement of joint C.

(b)

To determine

Find the horizontal displacement of joint C.

(b)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−1(8)=0Dy=0.363 kips↑$

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

$+↑∑Fy=0Dy+Ay=00.363+Ay=0Ay=0.363 kips↓$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.363+FABsin53.13°=0FAB=0.455 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.455)cos53.13°−1=0FAE=0.727 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.363+FCDsin53.13°=0FCD=−0.455 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−0.455)cos53.13°=0FED=0.273 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.455)sin53.13°−FBEsin57.99°=0FBE=−0.429 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.455)cos53.13°+(−0.429)cos57.99°+FBC=0FBC=0.5 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−0.455)sin53.13°−FCEsin57.99°=0FCE=0.492 kips$

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Cx$ as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 3

Refer Table 1 for horizontal displacement of joint C.

(c)

To determine

Find the horizontal displacement of joint D.

(c)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint D with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy=0$

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

$+↑∑Fy=0Ay=0$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−1+FAB=0FAB=1 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=01−FED=0FED=1 kips$

The force in the member AB, BE, EC, BC, and CD are zero as it satisfies zero force member condition.

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Dx$ as shown in Figure 4.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 4

Find the vertical deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCy)=∑FPFQLAE$

Find the horizontal deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCx)=∑FPFQLAE$

Find the product of $FQFPLAE$ for each member as shown in Table 1.

Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$FQ(kips)$			$FQFPLAE(in.)$
Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$Cx$	$Cy$	$Dx$	$Cx$	$Cy$		$Dx$
AB	1	10	$−13.920$	0.455	0.341	0	$−0.076$	$−0.057$	0
BC	1	10	$−5.312$	0.500	0.375	0	$−0.032$	$−0.024$	0
CD	1	10	$−4.830$	$−0.455$	0.909	0	0.026	$−0.053$	0
DE	0.25	11	2.898	0.273	$−0.545$	1	0.042	$−0.083$	0.153
EA	0.25	11	$−1.648$	0.727	$−0.205$	1	$−0.063$	0.018	$−0.087$
EB	0.25	9	13.133	$−0.429$	$−0.322$	0	$−0.255$	$−0.191$	0
EC	0.25	9	4.556	0.429	$0.322$	0	$0.089$	0.066	0
						$δP=∑FQFPLAE$	$−0.269$	$−0.364$	$0.066$

Refer Table 1.

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

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Answer 2

Question

Chapter 8, Problem 4P

(a)

To determine

Find the vertical displacement of joint C.

(a)

Expert Solution

Answer to Problem 4P

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

The value of E is $10,000 ksi$ .

Procedure to find the deflection of truss by virtual work method is shown below.

For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.

For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces $(Fv)$ in all the member of the truss.

Finally use the desired deflection equation.

Apply the sign conventions for calculating reactions, 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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.

If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the bar forces $FP$ produced by the P-system as follows:

Let $Ax$ and $Ay$ be the horizontal and vertical reactions at the hinged support A.

Let $Dy$ be the vertical reaction at the roller support D.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−15(11)+10(8)=0Dy=3.86 kips↑$

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

$+↑∑Fy=0Dy+Ay−15=03.86+Ay−15=0Ay=11.14 kips↑$

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

$+→∑Fx=0Ax−10=0Ax=10 kips→$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=011.14+FABsin53.13°=0FAB=−13.92 kips$

$+→∑Fx=010+FAE+FABcos53.13°=010+FAE+(−13.92)cos53.13°=0FAE=−1.648 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=03.86+FCDsin53.13°=0FCD=−4.83 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−4.83)cos53.13°=0FED=2.898 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(−13.92)sin53.13°−FBEsin57.99°=0FBE=13.133 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°−10+FBC=0−(−13.92)cos53.13°+(13.133)cos57.99°−10+FBC=0FBC=−5.312 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−4.830)sin53.13°−FCEsin57.99°=0FCE=4.556 kips$

Sketch the bar forces produced by the P-system as shown in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 1

Consider a dummy load of 1 kN directed vertically at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0−Dy(22)+1(16)=0Dy=0.727 kips↓$

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

$+↑∑Fy=0−Dy+Ay+1=0−0.727+Ay+1=0Ay=0.273 kips↓$

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

$+→∑Fx=0Ax=0$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.273+FABsin53.13°=0FAB=0.341 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.341)cos53.13°=0FAE=−0.205 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.727+FCDsin53.13°=0FCD=0.909 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(0.909)cos53.13°=0FED=−0.545 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.341)sin53.13°−FBEsin57.99°=0FBE=−0.322 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.341)cos53.13°+(−0.322)cos57.99°+FBC=0FBC=0.375 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°+1=0−(0.909)sin53.13°−FCEsin57.99°+1=0FCE=0.322 kips$

Sketch the bar forces $FQ$ produced by the vertical dummy load $Cy$ as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 2

Refer Table 1 for vertical displacement of joint C.

(b)

To determine

Find the horizontal displacement of joint C.

(b)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−1(8)=0Dy=0.363 kips↑$

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

$+↑∑Fy=0Dy+Ay=00.363+Ay=0Ay=0.363 kips↓$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.363+FABsin53.13°=0FAB=0.455 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.455)cos53.13°−1=0FAE=0.727 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.363+FCDsin53.13°=0FCD=−0.455 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−0.455)cos53.13°=0FED=0.273 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.455)sin53.13°−FBEsin57.99°=0FBE=−0.429 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.455)cos53.13°+(−0.429)cos57.99°+FBC=0FBC=0.5 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−0.455)sin53.13°−FCEsin57.99°=0FCE=0.492 kips$

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Cx$ as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 3

Refer Table 1 for horizontal displacement of joint C.

(c)

To determine

Find the horizontal displacement of joint D.

(c)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint D with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy=0$

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

$+↑∑Fy=0Ay=0$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−1+FAB=0FAB=1 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=01−FED=0FED=1 kips$

The force in the member AB, BE, EC, BC, and CD are zero as it satisfies zero force member condition.

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Dx$ as shown in Figure 4.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 4

Find the vertical deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCy)=∑FPFQLAE$

Find the horizontal deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCx)=∑FPFQLAE$

Find the product of $FQFPLAE$ for each member as shown in Table 1.

Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$FQ(kips)$			$FQFPLAE(in.)$
Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$Cx$	$Cy$	$Dx$	$Cx$	$Cy$		$Dx$
AB	1	10	$−13.920$	0.455	0.341	0	$−0.076$	$−0.057$	0
BC	1	10	$−5.312$	0.500	0.375	0	$−0.032$	$−0.024$	0
CD	1	10	$−4.830$	$−0.455$	0.909	0	0.026	$−0.053$	0
DE	0.25	11	2.898	0.273	$−0.545$	1	0.042	$−0.083$	0.153
EA	0.25	11	$−1.648$	0.727	$−0.205$	1	$−0.063$	0.018	$−0.087$
EB	0.25	9	13.133	$−0.429$	$−0.322$	0	$−0.255$	$−0.191$	0
EC	0.25	9	4.556	0.429	$0.322$	0	$0.089$	0.066	0
						$δP=∑FQFPLAE$	$−0.269$	$−0.364$	$0.066$

Refer Table 1.

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

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Answer 3

Question

Chapter 8, Problem 4P

(a)

To determine

Find the vertical displacement of joint C.

(a)

Expert Solution

Answer to Problem 4P

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

The value of E is $10,000 ksi$ .

Procedure to find the deflection of truss by virtual work method is shown below.

For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.

For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces $(Fv)$ in all the member of the truss.

Finally use the desired deflection equation.

Apply the sign conventions for calculating reactions, 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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.

If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the bar forces $FP$ produced by the P-system as follows:

Let $Ax$ and $Ay$ be the horizontal and vertical reactions at the hinged support A.

Let $Dy$ be the vertical reaction at the roller support D.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−15(11)+10(8)=0Dy=3.86 kips↑$

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

$+↑∑Fy=0Dy+Ay−15=03.86+Ay−15=0Ay=11.14 kips↑$

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

$+→∑Fx=0Ax−10=0Ax=10 kips→$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=011.14+FABsin53.13°=0FAB=−13.92 kips$

$+→∑Fx=010+FAE+FABcos53.13°=010+FAE+(−13.92)cos53.13°=0FAE=−1.648 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=03.86+FCDsin53.13°=0FCD=−4.83 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−4.83)cos53.13°=0FED=2.898 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(−13.92)sin53.13°−FBEsin57.99°=0FBE=13.133 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°−10+FBC=0−(−13.92)cos53.13°+(13.133)cos57.99°−10+FBC=0FBC=−5.312 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−4.830)sin53.13°−FCEsin57.99°=0FCE=4.556 kips$

Sketch the bar forces produced by the P-system as shown in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 1

Consider a dummy load of 1 kN directed vertically at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0−Dy(22)+1(16)=0Dy=0.727 kips↓$

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

$+↑∑Fy=0−Dy+Ay+1=0−0.727+Ay+1=0Ay=0.273 kips↓$

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

$+→∑Fx=0Ax=0$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.273+FABsin53.13°=0FAB=0.341 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.341)cos53.13°=0FAE=−0.205 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.727+FCDsin53.13°=0FCD=0.909 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(0.909)cos53.13°=0FED=−0.545 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.341)sin53.13°−FBEsin57.99°=0FBE=−0.322 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.341)cos53.13°+(−0.322)cos57.99°+FBC=0FBC=0.375 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°+1=0−(0.909)sin53.13°−FCEsin57.99°+1=0FCE=0.322 kips$

Sketch the bar forces $FQ$ produced by the vertical dummy load $Cy$ as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 2

Refer Table 1 for vertical displacement of joint C.

(b)

To determine

Find the horizontal displacement of joint C.

(b)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−1(8)=0Dy=0.363 kips↑$

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

$+↑∑Fy=0Dy+Ay=00.363+Ay=0Ay=0.363 kips↓$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.363+FABsin53.13°=0FAB=0.455 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.455)cos53.13°−1=0FAE=0.727 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.363+FCDsin53.13°=0FCD=−0.455 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−0.455)cos53.13°=0FED=0.273 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.455)sin53.13°−FBEsin57.99°=0FBE=−0.429 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.455)cos53.13°+(−0.429)cos57.99°+FBC=0FBC=0.5 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−0.455)sin53.13°−FCEsin57.99°=0FCE=0.492 kips$

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Cx$ as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 3

Refer Table 1 for horizontal displacement of joint C.

(c)

To determine

Find the horizontal displacement of joint D.

(c)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint D with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy=0$

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

$+↑∑Fy=0Ay=0$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−1+FAB=0FAB=1 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=01−FED=0FED=1 kips$

The force in the member AB, BE, EC, BC, and CD are zero as it satisfies zero force member condition.

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Dx$ as shown in Figure 4.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 4

Find the vertical deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCy)=∑FPFQLAE$

Find the horizontal deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCx)=∑FPFQLAE$

Find the product of $FQFPLAE$ for each member as shown in Table 1.

Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$FQ(kips)$			$FQFPLAE(in.)$
Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$Cx$	$Cy$	$Dx$	$Cx$	$Cy$		$Dx$
AB	1	10	$−13.920$	0.455	0.341	0	$−0.076$	$−0.057$	0
BC	1	10	$−5.312$	0.500	0.375	0	$−0.032$	$−0.024$	0
CD	1	10	$−4.830$	$−0.455$	0.909	0	0.026	$−0.053$	0
DE	0.25	11	2.898	0.273	$−0.545$	1	0.042	$−0.083$	0.153
EA	0.25	11	$−1.648$	0.727	$−0.205$	1	$−0.063$	0.018	$−0.087$
EB	0.25	9	13.133	$−0.429$	$−0.322$	0	$−0.255$	$−0.191$	0
EC	0.25	9	4.556	0.429	$0.322$	0	$0.089$	0.066	0
						$δP=∑FQFPLAE$	$−0.269$	$−0.364$	$0.066$

Refer Table 1.

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

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Answer 4

Question

Chapter 8, Problem 4P

(a)

To determine

Find the vertical displacement of joint C.

(a)

Expert Solution

Answer to Problem 4P

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

The value of E is $10,000 ksi$ .

Procedure to find the deflection of truss by virtual work method is shown below.

For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.

For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces $(Fv)$ in all the member of the truss.

Finally use the desired deflection equation.

Apply the sign conventions for calculating reactions, 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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.

If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the bar forces $FP$ produced by the P-system as follows:

Let $Ax$ and $Ay$ be the horizontal and vertical reactions at the hinged support A.

Let $Dy$ be the vertical reaction at the roller support D.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−15(11)+10(8)=0Dy=3.86 kips↑$

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

$+↑∑Fy=0Dy+Ay−15=03.86+Ay−15=0Ay=11.14 kips↑$

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

$+→∑Fx=0Ax−10=0Ax=10 kips→$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=011.14+FABsin53.13°=0FAB=−13.92 kips$

$+→∑Fx=010+FAE+FABcos53.13°=010+FAE+(−13.92)cos53.13°=0FAE=−1.648 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=03.86+FCDsin53.13°=0FCD=−4.83 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−4.83)cos53.13°=0FED=2.898 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(−13.92)sin53.13°−FBEsin57.99°=0FBE=13.133 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°−10+FBC=0−(−13.92)cos53.13°+(13.133)cos57.99°−10+FBC=0FBC=−5.312 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−4.830)sin53.13°−FCEsin57.99°=0FCE=4.556 kips$

Sketch the bar forces produced by the P-system as shown in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 1

Consider a dummy load of 1 kN directed vertically at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0−Dy(22)+1(16)=0Dy=0.727 kips↓$

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

$+↑∑Fy=0−Dy+Ay+1=0−0.727+Ay+1=0Ay=0.273 kips↓$

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

$+→∑Fx=0Ax=0$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.273+FABsin53.13°=0FAB=0.341 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.341)cos53.13°=0FAE=−0.205 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.727+FCDsin53.13°=0FCD=0.909 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(0.909)cos53.13°=0FED=−0.545 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.341)sin53.13°−FBEsin57.99°=0FBE=−0.322 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.341)cos53.13°+(−0.322)cos57.99°+FBC=0FBC=0.375 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°+1=0−(0.909)sin53.13°−FCEsin57.99°+1=0FCE=0.322 kips$

Sketch the bar forces $FQ$ produced by the vertical dummy load $Cy$ as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 2

Refer Table 1 for vertical displacement of joint C.

(b)

To determine

Find the horizontal displacement of joint C.

(b)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−1(8)=0Dy=0.363 kips↑$

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

$+↑∑Fy=0Dy+Ay=00.363+Ay=0Ay=0.363 kips↓$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.363+FABsin53.13°=0FAB=0.455 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.455)cos53.13°−1=0FAE=0.727 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.363+FCDsin53.13°=0FCD=−0.455 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−0.455)cos53.13°=0FED=0.273 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.455)sin53.13°−FBEsin57.99°=0FBE=−0.429 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.455)cos53.13°+(−0.429)cos57.99°+FBC=0FBC=0.5 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−0.455)sin53.13°−FCEsin57.99°=0FCE=0.492 kips$

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Cx$ as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 3

Refer Table 1 for horizontal displacement of joint C.

(c)

To determine

Find the horizontal displacement of joint D.

(c)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint D with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy=0$

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

$+↑∑Fy=0Ay=0$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−1+FAB=0FAB=1 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=01−FED=0FED=1 kips$

The force in the member AB, BE, EC, BC, and CD are zero as it satisfies zero force member condition.

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Dx$ as shown in Figure 4.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 4

Find the vertical deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCy)=∑FPFQLAE$

Find the horizontal deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCx)=∑FPFQLAE$

Find the product of $FQFPLAE$ for each member as shown in Table 1.

Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$FQ(kips)$			$FQFPLAE(in.)$
Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$Cx$	$Cy$	$Dx$	$Cx$	$Cy$		$Dx$
AB	1	10	$−13.920$	0.455	0.341	0	$−0.076$	$−0.057$	0
BC	1	10	$−5.312$	0.500	0.375	0	$−0.032$	$−0.024$	0
CD	1	10	$−4.830$	$−0.455$	0.909	0	0.026	$−0.053$	0
DE	0.25	11	2.898	0.273	$−0.545$	1	0.042	$−0.083$	0.153
EA	0.25	11	$−1.648$	0.727	$−0.205$	1	$−0.063$	0.018	$−0.087$
EB	0.25	9	13.133	$−0.429$	$−0.322$	0	$−0.255$	$−0.191$	0
EC	0.25	9	4.556	0.429	$0.322$	0	$0.089$	0.066	0
						$δP=∑FQFPLAE$	$−0.269$	$−0.364$	$0.066$

Refer Table 1.

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

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Answer 5

Chapter 8, Problem 4P

(a)

To determine

Find the vertical displacement of joint C.

(a)

Expert Solution

Answer to Problem 4P

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

The value of E is $10,000 ksi$ .

Procedure to find the deflection of truss by virtual work method is shown below.

For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.

For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces $(Fv)$ in all the member of the truss.

Finally use the desired deflection equation.

Apply the sign conventions for calculating reactions, 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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.

If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the bar forces $FP$ produced by the P-system as follows:

Let $Ax$ and $Ay$ be the horizontal and vertical reactions at the hinged support A.

Let $Dy$ be the vertical reaction at the roller support D.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−15(11)+10(8)=0Dy=3.86 kips↑$

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

$+↑∑Fy=0Dy+Ay−15=03.86+Ay−15=0Ay=11.14 kips↑$

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

$+→∑Fx=0Ax−10=0Ax=10 kips→$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=011.14+FABsin53.13°=0FAB=−13.92 kips$

$+→∑Fx=010+FAE+FABcos53.13°=010+FAE+(−13.92)cos53.13°=0FAE=−1.648 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=03.86+FCDsin53.13°=0FCD=−4.83 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−4.83)cos53.13°=0FED=2.898 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(−13.92)sin53.13°−FBEsin57.99°=0FBE=13.133 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°−10+FBC=0−(−13.92)cos53.13°+(13.133)cos57.99°−10+FBC=0FBC=−5.312 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−4.830)sin53.13°−FCEsin57.99°=0FCE=4.556 kips$

Sketch the bar forces produced by the P-system as shown in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 1

Consider a dummy load of 1 kN directed vertically at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0−Dy(22)+1(16)=0Dy=0.727 kips↓$

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

$+↑∑Fy=0−Dy+Ay+1=0−0.727+Ay+1=0Ay=0.273 kips↓$

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

$+→∑Fx=0Ax=0$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.273+FABsin53.13°=0FAB=0.341 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.341)cos53.13°=0FAE=−0.205 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.727+FCDsin53.13°=0FCD=0.909 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(0.909)cos53.13°=0FED=−0.545 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.341)sin53.13°−FBEsin57.99°=0FBE=−0.322 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.341)cos53.13°+(−0.322)cos57.99°+FBC=0FBC=0.375 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°+1=0−(0.909)sin53.13°−FCEsin57.99°+1=0FCE=0.322 kips$

Sketch the bar forces $FQ$ produced by the vertical dummy load $Cy$ as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 2

Refer Table 1 for vertical displacement of joint C.

(b)

To determine

Find the horizontal displacement of joint C.

(b)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−1(8)=0Dy=0.363 kips↑$

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

$+↑∑Fy=0Dy+Ay=00.363+Ay=0Ay=0.363 kips↓$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.363+FABsin53.13°=0FAB=0.455 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.455)cos53.13°−1=0FAE=0.727 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.363+FCDsin53.13°=0FCD=−0.455 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−0.455)cos53.13°=0FED=0.273 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.455)sin53.13°−FBEsin57.99°=0FBE=−0.429 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.455)cos53.13°+(−0.429)cos57.99°+FBC=0FBC=0.5 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−0.455)sin53.13°−FCEsin57.99°=0FCE=0.492 kips$

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Cx$ as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 3

Refer Table 1 for horizontal displacement of joint C.

(c)

To determine

Find the horizontal displacement of joint D.

(c)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint D with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy=0$

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

$+↑∑Fy=0Ay=0$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−1+FAB=0FAB=1 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=01−FED=0FED=1 kips$

The force in the member AB, BE, EC, BC, and CD are zero as it satisfies zero force member condition.

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Dx$ as shown in Figure 4.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 4

Find the vertical deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCy)=∑FPFQLAE$

Find the horizontal deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCx)=∑FPFQLAE$

Find the product of $FQFPLAE$ for each member as shown in Table 1.

Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$FQ(kips)$			$FQFPLAE(in.)$
Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$Cx$	$Cy$	$Dx$	$Cx$	$Cy$		$Dx$
AB	1	10	$−13.920$	0.455	0.341	0	$−0.076$	$−0.057$	0
BC	1	10	$−5.312$	0.500	0.375	0	$−0.032$	$−0.024$	0
CD	1	10	$−4.830$	$−0.455$	0.909	0	0.026	$−0.053$	0
DE	0.25	11	2.898	0.273	$−0.545$	1	0.042	$−0.083$	0.153
EA	0.25	11	$−1.648$	0.727	$−0.205$	1	$−0.063$	0.018	$−0.087$
EB	0.25	9	13.133	$−0.429$	$−0.322$	0	$−0.255$	$−0.191$	0
EC	0.25	9	4.556	0.429	$0.322$	0	$0.089$	0.066	0
						$δP=∑FQFPLAE$	$−0.269$	$−0.364$	$0.066$

Refer Table 1.

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Answer 6

Chapter 8, Problem 4P

(a)

To determine

Find the vertical displacement of joint C.

(a)

Expert Solution

Answer to Problem 4P

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

The value of E is $10,000 ksi$ .

Procedure to find the deflection of truss by virtual work method is shown below.

For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.

For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces $(Fv)$ in all the member of the truss.

Finally use the desired deflection equation.

Apply the sign conventions for calculating reactions, 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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.

If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the bar forces $FP$ produced by the P-system as follows:

Let $Ax$ and $Ay$ be the horizontal and vertical reactions at the hinged support A.

Let $Dy$ be the vertical reaction at the roller support D.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−15(11)+10(8)=0Dy=3.86 kips↑$

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

$+↑∑Fy=0Dy+Ay−15=03.86+Ay−15=0Ay=11.14 kips↑$

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

$+→∑Fx=0Ax−10=0Ax=10 kips→$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=011.14+FABsin53.13°=0FAB=−13.92 kips$

$+→∑Fx=010+FAE+FABcos53.13°=010+FAE+(−13.92)cos53.13°=0FAE=−1.648 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=03.86+FCDsin53.13°=0FCD=−4.83 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−4.83)cos53.13°=0FED=2.898 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(−13.92)sin53.13°−FBEsin57.99°=0FBE=13.133 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°−10+FBC=0−(−13.92)cos53.13°+(13.133)cos57.99°−10+FBC=0FBC=−5.312 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−4.830)sin53.13°−FCEsin57.99°=0FCE=4.556 kips$

Sketch the bar forces produced by the P-system as shown in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 1

Consider a dummy load of 1 kN directed vertically at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0−Dy(22)+1(16)=0Dy=0.727 kips↓$

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

$+↑∑Fy=0−Dy+Ay+1=0−0.727+Ay+1=0Ay=0.273 kips↓$

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

$+→∑Fx=0Ax=0$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.273+FABsin53.13°=0FAB=0.341 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.341)cos53.13°=0FAE=−0.205 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.727+FCDsin53.13°=0FCD=0.909 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(0.909)cos53.13°=0FED=−0.545 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.341)sin53.13°−FBEsin57.99°=0FBE=−0.322 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.341)cos53.13°+(−0.322)cos57.99°+FBC=0FBC=0.375 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°+1=0−(0.909)sin53.13°−FCEsin57.99°+1=0FCE=0.322 kips$

Sketch the bar forces $FQ$ produced by the vertical dummy load $Cy$ as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 2

Refer Table 1 for vertical displacement of joint C.

Answer 7

Chapter 8, Problem 4P

(a)

To determine

Find the vertical displacement of joint C.

Answer 8

(a)

Expert Solution

Answer to Problem 4P

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

The value of E is $10,000 ksi$ .

Procedure to find the deflection of truss by virtual work method is shown below.

For Real system: If the deflection of truss is determined by the external loads, then apply method of joints or method of sections to find the real axial forces (F) in all the members of the truss.

For virtual system: Remove all given real loads, apply a unit load at the joint where is deflection is required and also in the direction of desired deflection. Use method of joints or method of sections to find the virtual axial forces $(Fv)$ in all the member of the truss.

Finally use the desired deflection equation.

Apply the sign conventions for calculating reactions, 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.

Method of joints:

The negative value of force in any member indicates compression (C) and the positive value of force in any member indicates tension (T).

Condition for zero force members:

If only two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both the members is zero.

If three members, two of which are collinear are connected to a joint that has no external loads or reactions applied to it, then the force in non-collinear member is zero.

Calculation:

Find the bar forces $FP$ produced by the P-system as follows:

Let $Ax$ and $Ay$ be the horizontal and vertical reactions at the hinged support A.

Let $Dy$ be the vertical reaction at the roller support D.

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−15(11)+10(8)=0Dy=3.86 kips↑$

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

$+↑∑Fy=0Dy+Ay−15=03.86+Ay−15=0Ay=11.14 kips↑$

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

$+→∑Fx=0Ax−10=0Ax=10 kips→$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=011.14+FABsin53.13°=0FAB=−13.92 kips$

$+→∑Fx=010+FAE+FABcos53.13°=010+FAE+(−13.92)cos53.13°=0FAE=−1.648 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=03.86+FCDsin53.13°=0FCD=−4.83 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−4.83)cos53.13°=0FED=2.898 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(−13.92)sin53.13°−FBEsin57.99°=0FBE=13.133 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°−10+FBC=0−(−13.92)cos53.13°+(13.133)cos57.99°−10+FBC=0FBC=−5.312 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−4.830)sin53.13°−FCEsin57.99°=0FCE=4.556 kips$

Sketch the bar forces produced by the P-system as shown in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 1

Consider a dummy load of 1 kN directed vertically at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0−Dy(22)+1(16)=0Dy=0.727 kips↓$

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

$+↑∑Fy=0−Dy+Ay+1=0−0.727+Ay+1=0Ay=0.273 kips↓$

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

$+→∑Fx=0Ax=0$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.273+FABsin53.13°=0FAB=0.341 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.341)cos53.13°=0FAE=−0.205 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.727+FCDsin53.13°=0FCD=0.909 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(0.909)cos53.13°=0FED=−0.545 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.341)sin53.13°−FBEsin57.99°=0FBE=−0.322 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.341)cos53.13°+(−0.322)cos57.99°+FBC=0FBC=0.375 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°+1=0−(0.909)sin53.13°−FCEsin57.99°+1=0FCE=0.322 kips$

Sketch the bar forces $FQ$ produced by the vertical dummy load $Cy$ as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 2

Refer Table 1 for vertical displacement of joint C.

Answer 13

(b)

To determine

Find the horizontal displacement of joint C.

(b)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−1(8)=0Dy=0.363 kips↑$

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

$+↑∑Fy=0Dy+Ay=00.363+Ay=0Ay=0.363 kips↓$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.363+FABsin53.13°=0FAB=0.455 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.455)cos53.13°−1=0FAE=0.727 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.363+FCDsin53.13°=0FCD=−0.455 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−0.455)cos53.13°=0FED=0.273 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.455)sin53.13°−FBEsin57.99°=0FBE=−0.429 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.455)cos53.13°+(−0.429)cos57.99°+FBC=0FBC=0.5 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−0.455)sin53.13°−FCEsin57.99°=0FCE=0.492 kips$

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Cx$ as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 3

Refer Table 1 for horizontal displacement of joint C.

Answer 14

(b)

To determine

Find the horizontal displacement of joint C.

Answer 15

(b)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint C with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy(22)−1(8)=0Dy=0.363 kips↑$

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

$+↑∑Fy=0Dy+Ay=00.363+Ay=0Ay=0.363 kips↓$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−0.363+FABsin53.13°=0FAB=0.455 kips$

$+→∑Fx=0FAE+FABcos53.13°=0FAE+(0.455)cos53.13°−1=0FAE=0.727 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=00.363+FCDsin53.13°=0FCD=−0.455 kips$

$+→∑Fx=0−FED−FCDcos53.13°=0−FED−(−0.455)cos53.13°=0FED=0.273 kips$

Apply equilibrium equation to the joint B:

$+↑∑Fy=0−FABsin53.13°−FBEsin57.99°=0−(0.455)sin53.13°−FBEsin57.99°=0FBE=−0.429 kips$

$+→∑Fx=0−FABcos53.13°+FBEcos57.99°+FBC=0−(0.455)cos53.13°+(−0.429)cos57.99°+FBC=0FBC=0.5 kips$

Apply equilibrium equation to the joint C:

$+↑∑Fy=0−FCDsin53.13°−FCEsin57.99°=0−(−0.455)sin53.13°−FCEsin57.99°=0FCE=0.492 kips$

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Cx$ as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 3

Refer Table 1 for horizontal displacement of joint C.

Answer 20

(c)

To determine

Find the horizontal displacement of joint D.

(c)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint D with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy=0$

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

$+↑∑Fy=0Ay=0$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−1+FAB=0FAB=1 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=01−FED=0FED=1 kips$

The force in the member AB, BE, EC, BC, and CD are zero as it satisfies zero force member condition.

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Dx$ as shown in Figure 4.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 4

Find the vertical deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCy)=∑FPFQLAE$

Find the horizontal deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCx)=∑FPFQLAE$

Find the product of $FQFPLAE$ for each member as shown in Table 1.

Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$FQ(kips)$			$FQFPLAE(in.)$
Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$Cx$	$Cy$	$Dx$	$Cx$	$Cy$		$Dx$
AB	1	10	$−13.920$	0.455	0.341	0	$−0.076$	$−0.057$	0
BC	1	10	$−5.312$	0.500	0.375	0	$−0.032$	$−0.024$	0
CD	1	10	$−4.830$	$−0.455$	0.909	0	0.026	$−0.053$	0
DE	0.25	11	2.898	0.273	$−0.545$	1	0.042	$−0.083$	0.153
EA	0.25	11	$−1.648$	0.727	$−0.205$	1	$−0.063$	0.018	$−0.087$
EB	0.25	9	13.133	$−0.429$	$−0.322$	0	$−0.255$	$−0.191$	0
EC	0.25	9	4.556	0.429	$0.322$	0	$0.089$	0.066	0
						$δP=∑FQFPLAE$	$−0.269$	$−0.364$	$0.066$

Refer Table 1.

The vertical deflection at joint C $(δCy)$ is $−0.364 in.↓_$ .

The horizontal deflection at joint C $(δCx)$ is $−0.269 in.←_$ .

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Answer 21

(c)

To determine

Find the horizontal displacement of joint D.

Answer 22

(c)

Expert Solution

Answer to Problem 4P

The horizontal deflection at joint D $(δDx)$ is $0.066 in.→_$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $1 in.2$ and area of members AE, ED, BE, and CE are $0.25 in.2$ .

Calculation:

Consider a dummy load of 1 kN directed horizontally at joint D with the bar forces $FQ$ .

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$∑MA=0Dy=0$

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

$+↑∑Fy=0Ay=0$

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

$+→∑Fx=0Ax=1 kip←$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$+↑∑Fy=0−1+FAB=0FAB=1 kips$

Apply equilibrium equation to the joint D:

$+↑∑Fy=01−FED=0FED=1 kips$

The force in the member AB, BE, EC, BC, and CD are zero as it satisfies zero force member condition.

Sketch the bar forces $FQ$ produced by the horizontal dummy load $Dx$ as shown in Figure 4.

Fundamentals Of Structural Analysis:, Chapter 8, Problem 4P , additional homework tip 4

Find the vertical deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCy)=∑FPFQLAE$

Find the horizontal deflection at joint C $(δCy)$ using the relation:

$(1 kips)(δCx)=∑FPFQLAE$

Find the product of $FQFPLAE$ for each member as shown in Table 1.

Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$FQ(kips)$			$FQFPLAE(in.)$
Bar	A $(in.2)$	L $(ft)$	$FP(kips)$	$Cx$	$Cy$	$Dx$	$Cx$	$Cy$		$Dx$
AB	1	10	$−13.920$	0.455	0.341	0	$−0.076$	$−0.057$	0
BC	1	10	$−5.312$	0.500	0.375	0	$−0.032$	$−0.024$	0
CD	1	10	$−4.830$	$−0.455$	0.909	0	0.026	$−0.053$	0
DE	0.25	11	2.898	0.273	$−0.545$	1	0.042	$−0.083$	0.153
EA	0.25	11	$−1.648$	0.727	$−0.205$	1	$−0.063$	0.018	$−0.087$
EB	0.25	9	13.133	$−0.429$	$−0.322$	0	$−0.255$	$−0.191$	0
EC	0.25	9	4.556	0.429	$0.322$	0	$0.089$	0.066	0
						$δP=∑FQFPLAE$	$−0.269$	$−0.364$	$0.066$