Chapter 8, Problem 1P

To determine

Find the horizontal and vertical displacement of joint B.

Answer to Problem 1P

The horizontal deflection at joint B is $\underset{\_}{0.70\text{in}\text{.}\to }$.

The vertical deflection at joint B is $\underset{\_}{0.28\text{in}\text{.}\uparrow }$.

Explanation of Solution

Given information:

Area of all bars is $4\text{in}{\text{.}}^2$ and value of E is $24,000\text{kips/in}{.}^2$.

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 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:

1. 1. 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.
2. 2. 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 $D_x$ and $D_y$ be the horizontal and vertical reactions at the hinged support D.

Let $E_y$ be the vertical reaction at the roller support E.

Find the reactions at the supports using equilibrium equations:

Summation of moments about E is equal to 0.

$\begin{array}{l}\sum M_E=0\\ D_y\left(20\right)-100\left(30\right)=0\\ D_y=150\text{kips}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ D_y-E_y=0\\ E_y=150\text{kips}\downarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -D_x+100=0\\ D_x=100\text{kips}\leftarrow \end{array}$

Find the member forces using method of joints:

The force in the member AB, AE, and DE are zero as it satisfies zero force member condition.

Apply equilibrium equation to the joint C:

$\begin{array}{l}+\to \sum F_x=0\\ 100-F_{BC}\cos 36.87°=0\\ F_{BC}=125\text{kips}\end{array}$

$\begin{array}{l}+\uparrow \sum F_y=0\\ -F_{CD}-F_{BC}\sin 36.87°=0\\ -F_{CD}-\left(125\right)\sin 36.87°=0\\ F_{CD}=-75\text{kips}\end{array}$

Apply equilibrium equation to the joint D:

$\begin{array}{l}+\to \sum F_x=0\\ -100+F_{BD}\cos 36.87°=0\\ F_{BD}=-125\text{kips}\end{array}$

Apply equilibrium equation to the joint E:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -150+F_{BE}=0\\ F_{BE}=150\text{kips}\end{array}$

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

Consider a dummy load of 1 kip directed vertically at joint B with the bar forces $F_{Q_1}$.

The vertical reaction at E is $E_y=1\text{kip}$.

The force in the member AB, AE, DE, BD, BC, and CD are zero as it satisfies zero force member condition.

Apply equilibrium equation to the joint E:

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

Sketch the bar forces $F_{Q_1}$ and reactions produced by the vertical dummy load as shown in Figure 2.

Consider a dummy load of 1 kip directed horizontally at joint B with the bar forces $F_{Q_2}$.

Sketch the bar forces $F_{Q_2}$ and reactions produced by the horizontal dummy load as shown in Figure 3.

Summation of moments about E is equal to 0.

$\begin{array}{l}\sum M_E=0\\ D_y\left(20\right)-1\left(15\right)=0\\ D_y=0.75\text{kips}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ D_y-E_y=0\\ E_y=0.75\text{kips}\downarrow \end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ -D_x+1=0\\ D_x=1\text{kips}\leftarrow \end{array}$

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

Apply equilibrium equation to the joint D:

$\begin{array}{l}+\to \sum F_x=0\\ -1+F_{BD}\cos 36.87°=0\\ F_{BC}=-1.25\text{kips}\end{array}$

Apply equilibrium equation to the joint E:

$\begin{array}{l}+\uparrow \sum F_y=0\\ -0.75+F_{BE}=0\\ F_{BE}=0.75\text{kips}\end{array}$

Find the product of $F_P{F}_{Q_1}L$ and $F_P{F}_{Q2}L$ for each member as shown in Table 1.

Bar	L $\left(\text{ft}\right)$	$\begin{array}{l}{F}_P\\ \left(\text{kips}\right)\end{array}$	$\begin{array}{l}{F}_{Q1}\\ \left(\text{kips}\right)\end{array}$	$\begin{array}{l}{F}_{Q2}\\ \left(\text{kips}\right)\end{array}$	$\begin{array}{l}{F}_P{F}_{Q_1}L\\ \left({\text{kips}}^2\cdot \text{ft}\right)\end{array}$	$\begin{array}{l}{F}_P{F}_{Q_2}L\\ \left({\text{kips}}^2\cdot \text{ft}\right)\end{array}$
AB	25	0	0	0	0	0
BC	25	125	0	0	0	0
CD	30	$-75$	0	0	0	0
DE	20	0	0	0	0	0
AE	20	0	0	0	0	0
BE	15	150	$-1$	0.75	$-2,250$	$1,687.5$
BD	25	$-125$	0	$-1.25$	0	$3,906.25$
				$\sum$	$-2,250$	$5,593.75$

Find the vertical deflection at joint B $\left({\delta }_{By}\right)$:

$\begin{array}{c}\left(1\text{kip}\right)\left({\delta }_{By}\right)=\frac{\sum F_P{F}_{Q_1}L}{AE}\\ {\delta }_{By}=\frac{-2,250{\text{kips}}^2\cdot \text{ft}}{\left(4\text{in}{.}^2\right)\left(24,000\text{kips/in}{.}^2\right)}\\ =\frac{-2,250{\text{kips}}^2\cdot \text{ft}\left(\frac{12\text{in}\text{.}}{1\text{ft}}\right)}{\left(4\text{in}{.}^2\right)\left(24,000\text{kips/in}{.}^2\right)}\\ =-0.28\text{in}\text{.}\\ =0.28\text{in}\text{.}\uparrow \end{array}$

Therefore, the vertical deflection at joint B is $\underset{\_}{0.28\text{in}\text{.}\uparrow }$.

Find the horizontal deflection at joint B $\left({\delta }_{Bx}\right)$:

$\begin{array}{c}\left(1\text{kip}\right)\left({\delta }_{Bx}\right)=\frac{\sum F_P{F}_{Q2}L}{AE}\\ {\delta }_{Bx}=\frac{5,593.75{\text{kips}}^2\cdot \text{ft}}{\left(4\text{in}{.}^2\right)\left(24,000\text{kips/in}{.}^2\right)}\\ =\frac{5,593.75{\text{kips}}^2\cdot \text{ft}\left(\frac{12\text{in}\text{.}}{1\text{ft}}\right)}{\left(4\text{in}{.}^2\right)\left(24,000\text{kips/in}{.}^2\right)}\\ =0.70\text{in}\text{.}\to \end{array}$

Therefore, the horizontal deflection at joint B is $\underset{\_}{0.70\text{in}\text{.}\to }$.