Textbook Problem

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7.24 through 7.27 Use the virtual work method to determine the deflection at point C of the beam shown.

FIG. P7.27, P7.62

To determine

Find the deflection at point C of the beam using virtual work method.

Explanation of Solution

Given information:

The beam is given in the Figure.

The value of E is 29,000 ksi and I is $350\text{in}{\text{.}}^4$.

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.

Calculation:

Consider the real system.

Sketch the real system of the beam as shown in Figure 1.

Find the reactions at the supports A and B:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ B_y\left(15\right)-2\left(21\right)\left(\frac{21}{2}\right)=0\\ B_y=29.4\text{k}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+B_y-2\left(21\right)=0\\ A_y+29.4-2\left(21\right)=0\\ A_y=12.6\text{k}\uparrow \end{array}$

Consider the virtual system.

Remove all the real loads and apply unit load where the point to find the deflection.

Let the bending moment due to virtual load be $M_v$.

Sketch the virtual system of the beam with unit load at point C as shown in Figure 2.

Find the reactions at the supports A and B:

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ B_y\left(15\right)-1\left(21\right)=0\\ B_y=1.4\text{k}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+B_y-1=0\\ -A_y+1.4-1=0\\ A_y=0.4\text{k}\downarrow \end{array}$

Find the equations for M and $M_v$ for the 2 segments of the beam as shown in Table 1.

Segment	x-coordinate		M $\left(\text{k-ft}\right)$	$M_v$ $\left(\text{k-ft}\right)$
	Origin	Limits (ft)
AB	A	$0-15$	$12.6x-x^2$	$-0.4x$
CB	C	$0-6$	$-x^2$	$-x$

Find the deflection at C using the virtual work expression:

$1\left({\Delta }_C\right)={\displaystyle {\int }_0^L\frac{M_{v}M}{EI}dx}$ (1)

Rearrange Equation (1) for the limits $0-15$ and $0-6$ as follows