Chapter 7, Problem 35P

To determine

Find the slope and deflection at point D of the beam using virtual work method.

Answer to Problem 35P

The slope at point D of the beam is $\underset{\_}{0.0071\text{rad}}$ in the clockwise direction.

The deflection at point D of the beam is $\underset{\_}{0.62\text{in}\text{.}\downarrow }$.

Explanation of Solution

Given information:

The beam is given in the Figure.

Value of E is 30,000 ksi, I is $4,000\text{in}{\text{.}}^4$ for span AB, and $3,000\text{in}{\text{.}}^4$ for span BD.

Calculation:

Consider the real system.

Draw a diagram showing all the given real loads acting on it.

Let an equation expressing the variation of bending moment due to real loading be M.

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

Find the reactions and moment at the supports:

Consider portion BCD, Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ 8C_y-35\left(16\right)=0\\ C_y=70\text{k}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+C_y-2.5\left(16\right)-35=0\\ A_y+70-2.5\left(16\right)-35=0\\ A_y=5\text{k}\uparrow \end{array}$

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -M_A-35\left(32\right)+70\left(24\right)-2.5\left(16\right)\left(\frac{16}{2}\right)=0\\ M_A=240\text{k-ft}\end{array}$

Consider the virtual system.

Draw a diagram of beam without the given real loads. For deflection apply unit load at the point and in the desired direction.

For slope calculation, apply a unit couple at the point on the beam where the slope is desired.

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

Let an equation expressing the variation of bending moment due to virtual couple be $M_{v1}$.

Consider portion BCD, Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ 8C_y-1=0\\ C_y=\frac{1}{8}\text{k}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+C_y=0\\ -A_y+\frac{1}{8}\\ A_y=\frac{1}{8}\text{k}\downarrow \end{array}$

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -M_A-1+\frac{1}{8}\left(24\right)=0\\ M_A=2\text{k-ft}\end{array}$

Sketch the virtual system of the beam with unit load at point D as shown in Figure 3.

Let an equation expressing the variation of bending moment due to virtual load be $M_{v2}$.

Consider portion BCD, Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ 8C_y-1\left(16\right)=0\\ C_y=2\text{k}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ -A_y+C_y-1=0\\ -A_y+2-1=0\\ A_y=1\text{k}\downarrow \end{array}$

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -M_A-1\left(32\right)+2\left(24\right)=0\\ M_A=16\text{k-ft}\end{array}$

Find the equations for M, $M_{v1}$, and $M_{v2}$ for the 3 segments of the beam as shown in Table 1.

Segment	x-coordinate		M $\left(\text{kN}\cdot \text{m}\right)$	$M_{v1}$ $\left(\text{kN}\cdot \text{m}\right)$	$M_{v2}$ $\left(\text{kN}\cdot \text{m}\right)$
	Origin	Limits (m)
DC	D	$0-8$	$-35x$	$-1$	$-1x$
CB	D	$8-16$	$-35x+70\left(x-8\right)$	$-1+\frac{1}{8}\left(x-8\right)$	$x-16$
AB	A	$0-16$	$240+5x-1.25x^2$	$2-\frac{x}{8}$	$16-x$

Find the slope at D using the virtual work expression:

$1\left({\theta }_D\right)={\displaystyle {\int }_0^L\frac{M_{v1}M}{EI}dx}$ (1)

Here, L is the length of the beam, E is the young’s modulus, and I is the moment of inertia.

Rearrange Equation (1) for the limits $0-8$, $8-16$, and $0-16$ as follows.

${\theta }_D=\frac{1}{E}\left[\frac{1}{I}{\displaystyle {\int }_0^8\left(M_{v1}M\right)dx+\frac{1}{I}{\displaystyle {\int }_8^{16}\left(M_{v1}M\right)dx}}+\frac{1}{I}{\displaystyle {\int }_0^{16}\left(M_{v1}M\right)dx}\right]$

Substitute $-1$ for $M_{v1}$, $-35x$ for $M$ for the limit $0-8$, $-1+\frac{1}{8}\left(x-8\right)$ for $M_{v1}$, $-35x+70\left(x-8\right)$ for $M$ for the limit $8-16$, $2-\frac{x}{8}$ for $M_{v1}$, $\frac{4I}{3}$ for I, and $240+5x-1.25x^2$ for $M$ for the limit $0-16$.

$\begin{array}{c}{\theta }_D=\frac{1}{E}\left[\begin{array}{l}\frac{1}{I}{\displaystyle {\int }_0^8\left(-1\right)\left(-35x\right)dx}+\frac{1}{I}{\displaystyle {\int }_8^{16}\left(-1+\frac{1}{8}\left(x-8\right)\right)\left(-35x+70\left(x-8\right)\right)dx}\\ +\frac{1}{\frac{4}{3}I}{\displaystyle {\int }_0^{16}\left(2-\frac{x}{8}\right)\left(240+5x-1.25x^2\right)dx}\end{array}\right]\\ =\frac{1}{E}\left[\begin{array}{l}\frac{1}{I}{\displaystyle {\int }_0^8\left(35x\right)dx}+\frac{1}{I}{\displaystyle {\int }_8^{16}\left(70x-140x+1,120-\frac{35x^2}{8}+\frac{70x^2}{8}-\frac{560x}{8}\right)dx}\\ +\frac{3}{4I}{\displaystyle {\int }_0^{16}\left(480+10x-2.5x^2-\frac{240x}{8}-\frac{5x^2}{8}+\frac{1.25x^3}{8}\right)dx}\end{array}\right]\\ =\frac{1}{EI}\left[\begin{array}{l}{\left(\frac{35x^2}{2}\right)}_0^8+{\left(-\frac{140x^2}{2}+1,120x+\frac{4.375x^3}{3}\right)}_8^{16}\\ +\frac{3}{4}{\left(480x-\frac{20x^2}{2}-\frac{3.125x^3}{3}+\frac{1.25x^4}{32}\right)}_0^8\end{array}\right]\\ =\frac{4426.67{\text{k-ft}}^{\text{2}}}{EI}\end{array}$

Substitute $30,000\text{ksi}$ for $E$ and $3,000\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}{\theta }_D=\frac{4426.67{\text{k-ft}}^{\text{2}}}{\left(30,000\text{ksi}\right)\left(3,000\text{in}{\text{.}}^4\right)}\\ =\frac{4426.67{\text{k-ft}}^{\text{2}}\times \frac{{12}^2{\text{in}}^{\text{2}}}{1{\text{ft}}^{\text{2}}}}{\left(30,000\text{ksi}\right)\left(3,000\text{in}{\text{.}}^4\right)}\\ =0.0071\text{rad}\end{array}$

Therefore, the slope at point D of the beam is $\underset{\_}{0.0071\text{rad}}$ in the clockwise direction.

Find the deflection at D using the virtual work expression:

$1\left({\Delta }_D\right)={\displaystyle {\int }_0^L\frac{M_{v2}M}{EI}dx}$ (2)

Rearrange Equation (2) for the limits $0-8$, $8-16$, and $0-16$ as follows.

${\Delta }_D=\frac{1}{EI}\left[{\displaystyle {\int }_0^8\left(M_{v2}M\right)dx+{\displaystyle {\int }_8^{16}\left(M_{v2}M\right)dx}}+{\displaystyle {\int }_0^{16}\left(M_{v2}M\right)dx}\right]$

Substitute $-1x$ for $M_{v2}$, $-35x$ for $M$ for the limit $0-8$, $x-16$ for $M_{v2}$, $-35x+70\left(x-8\right)$ for $M$ for the limit $8-16$, $16-x$ for $M_{v2}$, $\frac{4I}{3}$ for I, and $240+5x-1.25x^2$ for $M$ for the limit $0-16$.

$\begin{array}{c}{\Delta }_D=\frac{1}{E}\left[\begin{array}{l}\frac{1}{I}{\displaystyle {\int }_0^8\left(-1x\right)\left(-35x\right)dx}+\frac{1}{I}{\displaystyle {\int }_8^{16}\left(x-16\right)\left(-35x+70\left(x-8\right)\right)dx}\\ +\frac{1}{\frac{4}{3}I}{\displaystyle {\int }_0^{16}\left(16-x\right)\left(240+5x-1.25x^2\right)dx}\end{array}\right]\\ =\frac{1}{E}\left[\begin{array}{l}\frac{1}{I}{\displaystyle {\int }_0^8\left(35x^2\right)dx}+\frac{1}{I}{\displaystyle {\int }_8^{16}\left(-35x^2+70x^2-560x+560x-1,120x+8,960\right)dx}\\ +\frac{3}{4I}{\displaystyle {\int }_0^{16}\left(3,840+80x-20x^2-240x-5x^2+1.25x^3\right)dx}\end{array}\right]\\ =\frac{1}{E}\left[\begin{array}{l}\frac{1}{I}{\displaystyle {\int }_0^8\left(35x^2\right)dx}+\frac{1}{I}{\displaystyle {\int }_8^{16}\left(350x^2-1,120x+8,960\right)dx}\\ +\frac{3}{4I}{\displaystyle {\int }_0^{16}\left(3,840-160x-25x^2+1.25x^3\right)dx}\end{array}\right]\\ =\frac{1}{EI}\left[\begin{array}{l}{\left(\frac{35x^3}{3}\right)}_0^8+{\left(\frac{350x^3}{3}-1,120\frac{x^2}{2}+8,960x\right)}_8^{16}\\ +\frac{3}{4}{\left(3,840x-\frac{160x^2}{2}-\frac{25x^3}{3}+\frac{1.25x^4}{4}\right)}_0^8\end{array}\right]\end{array}$

${\Delta }_D=\frac{32,426.67{\text{k-ft}}^{\text{3}}}{EI}$

Substitute $30,000\text{ksi}$ for $E$ and $3,000\text{in}{\text{.}}^4$ for I.

$\begin{array}{c}{\Delta }_D=\frac{32,426.67{\text{k-ft}}^{\text{3}}}{\left(30,000\text{ksi}\right)\left(3,000\text{in}{\text{.}}^4\right)}\\ =\frac{32,426.67{\text{k-ft}}^{\text{3}}\times \frac{{12}^3{\text{in}}^3}{1{\text{ft}}^3}}{\left(30,000\text{ksi}\right)\left(3,000\text{in}{\text{.}}^4\right)}\\ =0.62\text{in}\text{.}\downarrow \end{array}$

Therefore, the deflection at point D of the beam is $\underset{\_}{0.62\text{in}\text{.}\downarrow }$.