Chapter 8, Problem 4P

(a)

To determine

Find the vertical displacement of joint C.

Answer to Problem 4P

The vertical deflection at joint C $\left({\delta }_{Cy}\right)$ is $\underset{\_}{-0.364\text{in}\text{.}\downarrow }$.

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $\text{1}\text{in}{\text{.}}^{\text{2}}$ and area of members AE, ED, BE, and CE are $\text{0}\text{.25}\text{in}{\text{.}}^{\text{2}}$.

The value of E is $10,000\text{ksi}$.

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 $\left(F_v\right)$ 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 $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.

For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.

For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, 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 $F_P$ produced by the P-system as follows:

Let $A_x$ and $A_y$ be the horizontal and vertical reactions at the hinged support A.

Let $D_y$ 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.

$\begin{array}{l}\sum M_A=0\\ D_y\left(22\right)-15\left(11\right)+10\left(8\right)=0\\ D_y=3.86\text{kips}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ D_y+A_y-15=0\\ 3.86+A_y-15=0\\ A_y=11.14\text{kips}\uparrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x-10=0\\ A_x=10\text{kips}\to \end{array}$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$\begin{array}{l}+\uparrow \sum F_y=0\\ 11.14+F_{AB}\sin 53.13°=0\\ F_{AB}=-13.92\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ 10+F_{AE}+F_{AB}\cos 53.13°=0\\ 10+F_{AE}+\left(-13.92\right)\cos 53.13°=0\\ F_{AE}=-1.648\text{kips}\end{array}$

Apply equilibrium equation to the joint D:

$\begin{array}{l}+\uparrow \sum F_y=0\\ 3.86+F_{CD}\sin 53.13°=0\\ F_{CD}=-4.83\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ -F_{ED}-F_{CD}\cos 53.13°=0\\ -F_{ED}-\left(-4.83\right)\cos 53.13°=0\\ F_{ED}=2.898\text{kips}\end{array}$

Apply equilibrium equation to the joint B:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{AB}\sin 53.13°-F_{BE}\sin 57.99°=0\\ -\left(-13.92\right)\sin 53.13°-F_{BE}\sin 57.99°=0\\ F_{BE}=13.133\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ -F_{AB}\cos 53.13°+F_{BE}\cos 57.99°-10+F_{BC}=0\\ -\left(-13.92\right)\cos 53.13°+\left(13.133\right)\cos 57.99°-10+F_{BC}=0\\ F_{BC}=-5.312\text{kips}\end{array}$

Apply equilibrium equation to the joint C:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{CD}\sin 53.13°-F_{CE}\sin 57.99°=0\\ -\left(-4.830\right)\sin 53.13°-F_{CE}\sin 57.99°=0\\ F_{CE}=4.556\text{kips}\end{array}$

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

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

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -D_y\left(22\right)+1\left(16\right)=0\\ D_y=0.727\text{kips}\downarrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -D_y+A_y+1=0\\ -0.727+A_y+1=0\\ A_y=0.273\text{kips}\downarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x=0\end{array}$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -0.273+F_{AB}\sin 53.13°=0\\ F_{AB}=0.341\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ F_{AE}+F_{AB}\cos 53.13°=0\\ F_{AE}+\left(0.341\right)\cos 53.13°=0\\ F_{AE}=-0.205\text{kips}\end{array}$

Apply equilibrium equation to the joint D:

$\begin{array}{l}+\uparrow \sum F_y=0\\ 0.727+F_{CD}\sin 53.13°=0\\ F_{CD}=0.909\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ -F_{ED}-F_{CD}\cos 53.13°=0\\ -F_{ED}-\left(0.909\right)\cos 53.13°=0\\ F_{ED}=-0.545\text{kips}\end{array}$

Apply equilibrium equation to the joint B:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{AB}\sin 53.13°-F_{BE}\sin 57.99°=0\\ -\left(0.341\right)\sin 53.13°-F_{BE}\sin 57.99°=0\\ F_{BE}=-0.322\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ -F_{AB}\cos 53.13°+F_{BE}\cos 57.99°+F_{BC}=0\\ -\left(0.341\right)\cos 53.13°+\left(-0.322\right)\cos 57.99°+F_{BC}=0\\ F_{BC}=0.375\text{kips}\end{array}$

Apply equilibrium equation to the joint C:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{CD}\sin 53.13°-F_{CE}\sin 57.99°+1=0\\ -\left(0.909\right)\sin 53.13°-F_{CE}\sin 57.99°+1=0\\ F_{CE}=0.322\text{kips}\end{array}$

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

Refer Table 1 for vertical displacement of joint C.

(b)

To determine

Find the horizontal displacement of joint C.

Answer to Problem 4P

The horizontal deflection at joint C $\left({\delta }_{Cx}\right)$ is $\underset{\_}{-0.269\text{in}\text{.}\leftarrow }$.

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $\text{1}\text{in}{\text{.}}^{\text{2}}$ and area of members AE, ED, BE, and CE are $\text{0}\text{.25}\text{in}{\text{.}}^{\text{2}}$.

Calculation:

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

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ D_y\left(22\right)-1\left(8\right)=0\\ D_y=0.363\text{kips}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ D_y+A_y=0\\ 0.363+A_y=0\\ A_y=0.363\text{kips}\downarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x=1\text{kip}\leftarrow \end{array}$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -0.363+F_{AB}\sin 53.13°=0\\ F_{AB}=0.455\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ F_{AE}+F_{AB}\cos 53.13°=0\\ F_{AE}+\left(0.455\right)\cos 53.13°-1=0\\ F_{AE}=0.727\text{kips}\end{array}$

Apply equilibrium equation to the joint D:

$\begin{array}{l}+\uparrow \sum F_y=0\\ 0.363+F_{CD}\sin 53.13°=0\\ F_{CD}=-0.455\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ -F_{ED}-F_{CD}\cos 53.13°=0\\ -F_{ED}-\left(-0.455\right)\cos 53.13°=0\\ F_{ED}=0.273\text{kips}\end{array}$

Apply equilibrium equation to the joint B:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{AB}\sin 53.13°-F_{BE}\sin 57.99°=0\\ -\left(0.455\right)\sin 53.13°-F_{BE}\sin 57.99°=0\\ F_{BE}=-0.429\text{kips}\end{array}$

$\begin{array}{l}+\to \sum F_x=0\\ -F_{AB}\cos 53.13°+F_{BE}\cos 57.99°+F_{BC}=0\\ -\left(0.455\right)\cos 53.13°+\left(-0.429\right)\cos 57.99°+F_{BC}=0\\ F_{BC}=0.5\text{kips}\end{array}$

Apply equilibrium equation to the joint C:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{CD}\sin 53.13°-F_{CE}\sin 57.99°=0\\ -\left(-0.455\right)\sin 53.13°-F_{CE}\sin 57.99°=0\\ F_{CE}=0.492\text{kips}\end{array}$

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

Refer Table 1 for horizontal displacement of joint C.

(c)

To determine

Find the horizontal displacement of joint D.

Answer to Problem 4P

The horizontal deflection at joint D $\left({\delta }_{Dx}\right)$ is $\underset{\_}{0.066\text{in}\text{.}\to }$.

Explanation of Solution

Given information:

Area of members AB, BC, and CD are $\text{1}\text{in}{\text{.}}^{\text{2}}$ and area of members AE, ED, BE, and CE are $\text{0}\text{.25}\text{in}{\text{.}}^{\text{2}}$.

Calculation:

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

Find the reactions at the supports using equilibrium equations:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ D_y=0\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y=0\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ A_x=1\text{kip}\leftarrow \end{array}$

Find the member forces using method of joints:

Apply equilibrium equation to the joint A:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -1+F_{AB}=0\\ F_{AB}=1\text{kips}\end{array}$

Apply equilibrium equation to the joint D:

$\begin{array}{l}+\uparrow \sum F_y=0\\ 1-F_{ED}=0\\ F_{ED}=1\text{kips}\end{array}$

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

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

Find the vertical deflection at joint C $\left({\delta }_{Cy}\right)$ using the relation:

$\left(1\text{kips}\right)\left({\delta }_{Cy}\right)=\frac{\sum F_P{F}_{Q}L}{AE}$

Find the horizontal deflection at joint C $\left({\delta }_{Cy}\right)$ using the relation:

$\left(1\text{kips}\right)\left({\delta }_{Cx}\right)=\frac{\sum F_P{F}_{Q}L}{AE}$

Find the product of $F_Q\frac{F_{P}L}{AE}$ for each member as shown in Table 1.

Bar	A $\left(\text{in}{\text{.}}^{\text{2}}\right)$	L $\left(\text{ft}\right)$	$\begin{array}{l}{F}_P\\ \left(\text{kips}\right)\end{array}$	$F_Q\left(\text{kips}\right)$			$\begin{array}{l}{F}_Q\frac{F_{P}L}{AE}\\ \left(\text{in}\text{.}\right)\end{array}$
				$C_x$	$C_y$	$D_x$	$C_x$	$C_y$		$D_x$
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
						$\begin{array}{l}{\delta }_P=\\ \sum F_Q\frac{F_{P}L}{AE}\end{array}$	$-0.269$	$-0.364$	$0.066$

Refer Table 1.

The vertical deflection at joint C $\left({\delta }_{Cy}\right)$ is $\underset{\_}{-0.364\text{in}\text{.}\downarrow }$.

The horizontal deflection at joint C $\left({\delta }_{Cx}\right)$ is $\underset{\_}{-0.269\text{in}\text{.}\leftarrow }$.

The horizontal deflection at joint D $\left({\delta }_{Dx}\right)$ is $\underset{\_}{0.066\text{in}\text{.}\to }$.