Chapter 7, Problem 64P

To determine

Find the slope and deflection at point D of the beam using Castigliano’s second theorem.

Answer to Problem 64P

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

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:

Let apply a load P and couple $\overline{M}$ at point D in the desired direction to find the deflection and slope.

The value of load P is 35 k and couple $\overline{M}$ is zero.

Sketch the beam with load P and couple $\overline{M}$ as shown in Figure 1.

Let the equation for bending moment at distance x in terms of load P be $M$, the derivative of M with respect to P is $\frac{\partial M}{\partial P}$.

The derivative of M with respect to $\overline{M}$ is $\frac{\partial M}{\partial \overline{M}}$.

Find the reactions and moment at the supports:

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

$\begin{array}{l}\sum M_B=0\\ 8C_y-P\left(16\right)-\overline{M}=0\\ 8C_y=16P+\overline{M}\\ C_y=2P+\frac{\overline{M}}{8}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ A_y+C_y-2.5\left(16\right)-P=0\\ A_y+2P+\frac{\overline{M}}{8}-2.5\left(16\right)-P=0\\ A_y=40-P-\frac{\overline{M}}{8}\end{array}$

Summation of moments about A is equal to 0.

$\begin{array}{l}\sum M_A=0\\ M_A-P\left(32\right)-\overline{M}+C_y\left(24\right)-2.5\left(16\right)\left(\frac{16}{2}\right)=0\\ M_A-32P-\overline{M}+48P+3\overline{M}-320=0\\ M_A=320-16P-2\overline{M}\end{array}$

Find the equations for M, $\frac{\partial M}{\partial \overline{M}}$, and $\frac{\partial M}{\partial P}$ for the 3 segments of the beam as shown in Table 1.

Segment	x-coordinate		M	$\frac{\partial M}{\partial \overline{M}}$	$\frac{\partial M}{\partial P}$
	Origin	Limits (ft)
DC	D	$0-8$	$-Px-\overline{M}$	$-1$	$-x$
CB	D	$8-16$	$-Px-\overline{M}+\left(2P+\frac{\overline{M}}{8}\right)\left(x-8\right)$	$-1+\frac{1}{8}\left(x-8\right)$	$x-16$
AB	A	$0-16$	$\begin{array}{l}-\left(320-16P-2\overline{M}\right)\\ +\left(40-P-\frac{\overline{M}}{8}\right)x-1.25x^2\end{array}$	$2-\frac{x}{8}$	$16-x$

The expression for slope at D using Castigliano’s second theorem $\left({\theta }_D\right)$ is shown as follows:

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

Here, L is the length of the beam.

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

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

Substitute the value of load P as 35 k and couple $\overline{M}$ as 0 in the column 4 of Table 1.

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

$\begin{array}{c}{\theta }_D=\frac{1}{E}\left[\begin{array}{l}\frac{1}{I}{\displaystyle {\int }_0^8\left(-1\right)\left(-35x\right)dx}+\frac{1}{I}{\displaystyle {\int }_8^{16}\left(-1+\frac{1}{8}\left(x-8\right)\right)\left(-35x+70\left(x-8\right)\right)dx}\\ +\frac{1}{\frac{4}{3}I}{\displaystyle {\int }_0^{16}\left(2-\frac{x}{8}\right)\left(240+5x-1.25x^2\right)dx}\end{array}\right]\\ =\frac{1}{E}\left[\begin{array}{l}\frac{1}{I}{\displaystyle {\int }_0^8\left(35x\right)dx}+\frac{1}{I}{\displaystyle {\int }_8^{16}\left(70x-140x+1,120-\frac{35x^2}{8}+\frac{70x^2}{8}-\frac{560x}{8}\right)dx}\\ +\frac{3}{4I}{\displaystyle {\int }_0^{16}\left(480+10x-2.5x^2-\frac{240x}{8}-\frac{5x^2}{8}+\frac{1.25x^3}{8}\right)dx}\end{array}\right]\\ =\frac{1}{EI}\left[\begin{array}{l}{\left(\frac{35x^2}{2}\right)}_0^8+{\left(-\frac{140x^2}{2}+1,120x+\frac{4.375x^3}{3}\right)}_8^{16}\\ +\frac{3}{4}{\left(480x-\frac{20x^2}{2}-\frac{3.125x^3}{3}+\frac{1.25x^4}{32}\right)}_0^8\end{array}\right]\\ =\frac{4,426.67{\text{k-ft}}^{\text{2}}}{EI}\end{array}$

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

$\begin{array}{c}{\theta }_D=\frac{4,426.67{\text{k-ft}}^{\text{2}}}{\left(30,000\text{ksi}\right)\left(3,000\text{in}{\text{.}}^4\right)}\\ =\frac{4,426.67{\text{k-ft}}^{\text{2}}\times \frac{{12}^2{\text{in}}^{\text{2}}}{1{\text{ft}}^{\text{2}}}}{\left(30,000\text{ksi}\right)\left(3,000\text{in}{\text{.}}^4\right)}\\ =0.0071\text{rad}\end{array}$

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

The expression for deflection at D using Castigliano’s second theorem $\left({\Delta }_D\right)$ is shown as follows:

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

Here, L is the length of the beam.

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

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

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

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

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

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 $\underset{\_}{0.62\text{in}\text{.}\downarrow }$.