Chapter 7, Problem 1P

To determine

Calculate the horizontal deflection and vertical deflection at joint B.

Answer to Problem 1P

The horizontal deflection at joint B is $\underset{\_}{0.567\text{in}\text{.}\leftarrow }$.

The vertical deflection at joint B is $\underset{\_}{0.126\text{in}\text{.}\uparrow }$.

Explanation of Solution

Given information:

The truss is given in the Figure.

The value of E is 29,000 ksi and the value of A is $5.25\text{in}{.}^2$

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 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 member:

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:

Consider the real system.

Find the member axial force (F) for the real system using method of joints:

Let $A_y$ be the vertical reaction at the roller support A.

Let $C_x$ and $C_y$ be the horizontal and vertical reactions at the hinged support C.

Sketch the free body diagram of the truss as shown in Figure 1.

Find the reactions at the supports using equilibrium equations:

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

$\begin{array}{l}+\to \sum F_x=0\\ C_x-75=0\\ C_x=75\text{k}\to \end{array}$

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ -A_y\left(12\right)+75\left(16\right)=0\\ A_y=100\text{k}\uparrow \end{array}$

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

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

Substitute $100\text{k}$ for $A_y$.

$\begin{array}{c}{C}_y=-100\text{k}\\ =100\text{k}\downarrow \end{array}$

Sketch the free body diagram of joint A as shown in Figure 2.

Apply equilibrium equation to the free body diagram of joint A:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 100+F_{AB}\sin 53.13°=0\\ F_{AB}=-125\text{k}\end{array}$

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

$\begin{array}{l}+\to \sum F_x=0\\ F_{AC}+F_{AB}\cos 53.13°=0\end{array}$

Substitute $-125\text{k}$ for $F_{AB}$.

$\begin{array}{l}{F}_{AC}+\left(-125\right)\cos 53.13°=0\\ F_{AC}=75\text{k}\end{array}$

Sketch the free body diagram of joint C as shown in Figure 3.

Apply equilibrium equation to the free body diagram of joint C:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -100+F_{BC}=0\\ F_{BC}=100\text{k}\end{array}$

Sketch the resultant diagram of the real forces as shown in Figure 4.

Consider the virtual system:

For horizontal deflection apply l k at joint B in horizontal direction.

Find the member axial force ($F_{v1}$ ) due to virtual load using method of joints:

Sketch the free body diagram of the truss as shown in Figure 5.

Refer Figure 1 and Figure 5.

The real system is similar to the virtual system except the load of 75k acting in the horizontal direction. So divide the real force (F) by 75 to obtain the virtual force $\left(F_{v1}\right)$.

The virtual force in the member AB is ${\left(F_{v1}\right)}_{AB}=\frac{F_{AB}}{75}$.

Substitute $-125\text{k}$ for $F_{AB}$.

$\begin{array}{c}{\left(F_{v1}\right)}_{AB}=\frac{-125}{75}\\ =-\frac{5}{3}\text{k}\end{array}$

The virtual force in the member AC is ${\left(F_{v1}\right)}_{AC}=\frac{F_{AC}}{75}$.

Substitute $75\text{k}$ for $F_{AC}$.

$\begin{array}{c}{\left(F_{v1}\right)}_{AC}=\frac{75\text{k}}{75}\\ =1\text{k}\end{array}$

The virtual force in the member BC is ${\left(F_{v1}\right)}_{BC}=\frac{F_{BC}}{75}$.

Substitute $100\text{k}$ for $F_{AC}$.

$\begin{array}{c}{\left(F_{v1}\right)}_{BC}=\frac{100\text{k}}{75}\\ =\frac{4}{3}\text{k}\end{array}$

Sketch the resultant diagram of the virtual system $\left(F_{v1}\right)$ as shown in Figure 6.

Consider the virtual system:

For vertical deflection apply l k at joint B in vertical direction.

Find the member axial force $\left(F_{v2}\right)$ due to virtual load using method of joints:

Sketch the free body diagram of the truss as shown in Figure 7.

Find the reactions at the supports using equilibrium equations:

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

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

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ A_y=0\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+C_y-1=0\end{array}$

Substitute $0$ for $A_y$.

$\begin{array}{c}0+C_y-1=0\\ C_y=1\text{k}\uparrow \end{array}$

The reaction at A is zero. So the member AB and AC satisfies the zero force member condition.

Therefore, ${\left(F_{v2}\right)}_{AB}=0$ and ${\left(F_{v2}\right)}_{BC}=0$.

Sketch the free body diagram of joint C as shown in Figure 8.

Apply equilibrium equation to the free body diagram of joint C:

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ 1+F_{BC}=0\\ {\left(F_{v2}\right)}_{BC}=-1\text{k}\end{array}$

Sketch the resultant diagram of the virtual system $\left(F_{v2}\right)$ as shown in Figure 9.

The expression to find the deflection $1\left(\Delta \right)$ is shown below:

$1\left(\Delta \right)=\sum F_v\left(\frac{FL}{AE}\right)$

Here, L is the length of the member, A is the area of the member, and E is the young’s modulus of the member.

For E and A is constant, the expression becomes,

$\left(1\text{k}\right)\Delta =\frac{1}{AE}\sum F_{v}FL$ (1)

Length of the member AC is,

$\begin{array}{c}{L}_{AC}=12\text{ft}\\ \text{=}12\text{ft}\left(\frac{12\text{in}\text{.}}{1\text{ft}}\right)\\ =144\text{in}\text{.}\end{array}$

Similarly convert the length of the remaining member and show in column 2 of Table 1.

Find the product of $F_{v1}FL$ for the member AC:

Substitute 1 k for $F_{v1}$ , 75 k for F, and 144 in. for L.

$\begin{array}{c}{F}_{v1}FL=\left(1\right)\left(75\right)\left(144\right)\\ =10,800{\text{k}}^2\text{-in}\text{.}\end{array}$

Similarly calculate for the remaining member and show in column 5 of Table 1.

Find the product of $F_{v2}FL$ for the member AC:

Substitute 0 for $F_{v2}$ , 75 k for F, and 144 in. for L.

$\begin{array}{c}{F}_{v1}FL=\left(0\right)\left(75\right)\left(144\right)\\ =0\end{array}$

Similarly calculate for the remaining member and show in column 7 of Table 1.

Frame a table for the above calculated values.

Summarize the calculated values as shown in Table 1.

Member	L $\left(\text{in}\text{.}\right)$	$\begin{array}{l}F\\ \left(\text{k}\right)\end{array}$	$\begin{array}{l}{F}_{v1}\\ \left(\text{k}\right)\end{array}$	$\begin{array}{l}{F}_{v1}FL\\ \left({\text{k}}^2\text{-in}\text{.}\right)\end{array}$	$\begin{array}{l}{F}_{v2}\\ \left(\text{k}\right)\end{array}$	$\begin{array}{l}{F}_{v2}FL\\ \left({\text{k}}^2\text{-in}\text{.}\right)\end{array}$
AC	144	75	1	10,800	0	0
AB	240	$-125$	$-\frac{5}{3}$	50,000	0	0
BC	192	100	$\frac{4}{3}$	25,600	$-1$	$-19,200$
			$\sum$	86,400		$-19,200$

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

Substitute $\text{86,400}{\text{k}}^2\text{-in}\text{.}$ for $\sum F_{v1}FL$, $5.25\text{in}{\text{.}}^{\text{2}}$ for A, and $29,000\text{ksi}$ for E in Equation (1).

$\begin{array}{c}\left(1\text{k}\right){\Delta }_{BH}=\frac{\text{86,400}{\text{k}}^2\text{-in}\text{.}}{\left(5.25\text{in}{\text{.}}^{\text{2}}\right)\left(29,000\text{ksi}\right)}\\ \left(1\text{k}\right){\Delta }_{BH}=0.567\text{k-in}\text{.}\\ {\Delta }_{BH}=0.567\text{in}\text{.}\end{array}$

Therefore, the horizontal deflection at joint B is $\underset{\_}{0.567\text{in}\text{.}\leftarrow }$.

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

Substitute $-19,200{\text{k}}^2\text{-in}\text{.}$ for $\sum F_{v2}FL$, $5.25\text{in}{\text{.}}^{\text{2}}$ for A, and $29,000\text{ksi}$ for E in Equation (1).

$\begin{array}{c}\left(1\text{k}\right){\Delta }_{BV}=\frac{-19,200{\text{k}}^2\text{-in}\text{.}}{\left(5.25\text{in}{\text{.}}^{\text{2}}\right)\left(29,000\text{ksi}\right)}\\ \left(1\text{k}\right){\Delta }_{BV}=-0.126\text{k-in}\text{.}\\ {\Delta }_{BV}=0.126\text{in}\text{.}\uparrow \end{array}$

Therefore, the vertical deflection at joint B is $\underset{\_}{0.126\text{in}\text{.}\uparrow }$.