Chapter 9.3, Problem 61P

To determine

Find the magnitude and location of the largest downward deflection of the beam.

Answer to Problem 61P

The location of the largest downward deflection is $\underset{\_}{x_m=8.40\text{ft}}$.

The largest downward deflection of the beam is $\underset{\_}{y_m=0.281\text{in}\text{.}(\downarrow )}$.

Explanation of Solution

Given information:

The modulus of elasticity of the material is $E=29\times {10}^6\text{psi}$.

Calculation:

Show the free-body diagram of the beam AD as in Figure 1.

Write the singularity equation for load intensity as follows;

$\frac{dV}{dx}=-w\left(x\right)=-3{\langle x-5\rangle }^0$

Integrate the equation to find the shear force.

$V=A_y-3{\langle x-5\rangle }^1-20{\langle x-10\rangle }^0$

By definition, the change in bending moment with respect to change in distance is shear force.

$\frac{dM}{dx}=V=A_y-3{\langle x-5\rangle }^1-20{\langle x-10\rangle }^0$

Integrate the equation to find the bending moment.

$M=A_{y}x-\frac{3}{2}{\langle x-5\rangle }^2-20{\langle x-10\rangle }^1$ (1)

Write the second order differential equation as follows;

$\frac{d^{2}y}{d{x}^2}=\frac{M\left(x\right)}{EI}$

Here, the moment at the corresponding section is $M\left(x\right)$, the modulus of elasticity of the material is E, and the moment of inertia of the section is I.

Substitute $\left(A_{y}x-\frac{3}{2}{\langle x-5\rangle }^2-20{\langle x-10\rangle }^1\right)$ for $M\left(x\right)$ in Equation (1).

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{A_{y}x-\frac{3}{2}{\langle x-5\rangle }^2-20{\langle x-10\rangle }^1}{EI}\\ EI\frac{d^{2}y}{d{x}^2}=A_{y}x-\frac{3}{2}{\langle x-5\rangle }^2+-20{\langle x-10\rangle }^1\end{array}$

Integrate the equation with respect to x;

$EI\frac{dy}{dx}=\frac{A_y{x}^2}{2}-\frac{3}{6}{\langle x-5\rangle }^3-10{\langle x-10\rangle }^2+C_1$ (2)

Integrate the Equation (2) with respect to x.

$EIy=\frac{A_y{x}^3}{6}-\frac{3}{24}{\langle x-5\rangle }^4-\frac{10}{3}{\langle x-10\rangle }^3+C_{1}x+C_2$ (3)

Boundary condition 1:

At the point D; $x=16\text{ft};M=0$.

Substitute 16 ft for x and 0 for M in Equation (1).

$\begin{array}{l}0=A_y\left(16\right)-\frac{3}{2}{\langle 16-5\rangle }^2-20{\langle 16-10\rangle }^1\\ 0=16A_y-181.5-120\\ A_y=18.84375\text{kips}(\uparrow )\end{array}$

Boundary condition 2:

At the point A; $x=0;y=0$.

Substitute 18.84375 kips for $A_y$, 0 for x and 0 for y in Equation (3).

$\begin{array}{l}EI\left(0\right)=\frac{A_y{\left(0\right)}^3}{6}-\frac{3}{24}{\langle 0-5\rangle }^4-\frac{10}{3}{\langle 0-10\rangle }^3+C_1\left(0\right)+C_2\\ 0=0-0+0-0+C_2\\ C_2=0\end{array}$

Boundary condition 3:

At the point D; $x=16\text{ft};y=0$.

Substitute 18.84375 kips for $A_y$, 16 ft for x, 0 for y, and 0 for $C_2$ in Equation (3).

$\begin{array}{l}EI\left(0\right)=\frac{18.84375{\left(16\right)}^3}{6}-\frac{3}{24}{\langle 16-5\rangle }^4-\frac{10}{3}{\langle 16-10\rangle }^3+C_1\left(16\right)+0\\ 0=12,864-1,830.125-720+16C_1\\ C_1=-644.61719{\text{kip-ft}}^2\end{array}$

Refer to Appendix C “The properties of the Rolled-Steel Shapes” in the textbook.

The moment of inertia of the $\text{W}16\times 57$ section is $I=758\text{in}{\text{.}}^4$.

At point A; $x=0;\frac{dy}{dx}={\theta }_A$.

Substitute $29\times {16}^6\text{psi}$ for E, $758\text{in}{\text{.}}^4$ for I, ${\theta }_A$ for $\frac{dy}{dx}$, 18.84375 kips for $A_y$, 0 for x, and $-644.61719{\text{kip-ft}}^2$ for $C_1$ in Equation (2).

$\begin{array}{c}\left(\begin{array}{l}29\times {10}^6\text{psi}\times \frac{\text{1}\text{kip}}{\text{1,000}\text{lb}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2\times 758\text{in}{.}^4\\ \times {\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^4\times {\theta }_A\end{array}\right)=\left\{\begin{array}{l}\frac{18.84375\times {\left(0\right)}^2}{2}-\frac{3}{6}{\langle 0-5\rangle }^3\\ -10{\langle 0-10\rangle }^2-644.61719\end{array}\right\}\\ \left(\frac{29\times {10}^6\times 758}{1,000\times {12}^2}\right){\theta }_A=0-0-0-644.61719\\ {\theta }_A=-4.22\times {10}^{-3}\text{rad}\end{array}$

At point B; $x=5\text{ft};\frac{dy}{dx}={\theta }_B$.

Substitute $29\times {16}^6\text{psi}$ for E, $758\text{in}{\text{.}}^4$ for I, ${\theta }_B$ for $\frac{dy}{dx}$, 18.84375 kips for $A_y$, 5 ft for x, and $-644.61719{\text{kip-ft}}^2$ for $C_1$ in Equation (2).

$\begin{array}{c}\left(\begin{array}{l}29\times {10}^6\text{psi}\times \frac{\text{1}\text{kip}}{\text{1,000}\text{lb}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2\times 758\text{in}{.}^4\\ \times {\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^4\times {\theta }_B\end{array}\right)=\left\{\begin{array}{l}\frac{18.84375\times {\left(5\right)}^2}{2}-\frac{3}{6}{\langle 5-5\rangle }^3\\ -10{\langle 5-10\rangle }^2-644.61719\end{array}\right\}\\ \left(\frac{29\times {10}^6\times 758}{1,000\times {12}^2}\right){\theta }_B=235.546875-0-0-644.61719\\ {\theta }_B=-2.680\times {10}^{-3}\text{rad}\end{array}$

At point C; $x=10\text{ft};\frac{dy}{dx}={\theta }_C$.

Substitute $29\times {16}^6\text{psi}$ for E, $758\text{in}{\text{.}}^4$ for I, ${\theta }_C$ for $\frac{dy}{dx}$, 18.84375 kips for $A_y$, 10 ft for x, and $-644.61719{\text{kip-ft}}^2$ for $C_1$ in Equation (2).

$\begin{array}{c}\left(\begin{array}{l}29\times {10}^6\text{psi}\times \frac{\text{1}\text{kip}}{{\text{10}}^3\text{lb}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2\times 758\text{in}{.}^4\\ \times {\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^4\times {\theta }_C\end{array}\right)=\left\{\begin{array}{l}\frac{18.84375\times {\left(10\right)}^2}{2}-\frac{3}{6}{\langle 10-5\rangle }^3\\ -10{\langle 10-10\rangle }^2-644.61719\end{array}\right\}\\ \left(\frac{29\times {10}^6\times 758}{1,000\times {12}^2}\right){\theta }_C=942.1875-62.5-0-644.61719\\ {\theta }_C=1.540\times {10}^{-3}\text{rad}\end{array}$

At point D; $x=16\text{ft};\frac{dy}{dx}={\theta }_D$.

Substitute $29\times {16}^6\text{psi}$ for E, $758\text{in}{\text{.}}^4$ for I, ${\theta }_D$ for $\frac{dy}{dx}$, 18.84375 kips for $A_y$, 16 ft for x, and $-644.61719{\text{kip-ft}}^2$ for $C_1$ in Equation (2).

$\begin{array}{c}\left(\begin{array}{l}29\times {10}^6\text{psi}\times \frac{\text{1}\text{kip}}{{\text{10}}^3\text{lb}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2\times 758\text{in}{.}^4\\ \times {\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^4\times {\theta }_D\end{array}\right)=\left\{\begin{array}{l}\frac{18.84375\times {\left(16\right)}^2}{2}-\frac{3}{6}{\langle 16-5\rangle }^3\\ -10{\langle 16-10\rangle }^2-644.61719\end{array}\right\}\\ \left(\frac{29\times {10}^6\times 758}{1,000\times {12}^2}\right){\theta }_D=2,412-665.5-360-644.61719\\ {\theta }_D=4.860\times {10}^{-3}\text{rad}\end{array}$

The slope changes from negative to positive in the section BC.

The maximum deflection occurs where the slope changes sign. i.e., $\frac{dy}{dx}=0$.

Substitute 0 for $\frac{dy}{dx}$, 18.84375 kips for $A_y$, and $-644.61719{\text{kip-ft}}^2$ for $C_1$ in Equation (2).

$\begin{array}{l}EI\left(0\right)=\frac{18.84375\times x_m^2}{2}-\frac{3}{6}{\langle x_m-5\rangle }^3-10{\langle x_m-10\rangle }^2-644.61719\\ 0=9.421875x_m^2-0.5{\left(x_m-5\right)}^3-0-644.61719\\ 0=9.421875x_m^2-0.5{\left(x_m-5\right)}^3-644.61719\end{array}$

Solve the equation;

$x_m=8.40\text{ft}$.

Therefore, the location of the largest downward deflection is $\underset{\_}{x_m=8.40\text{ft}}$.

At largest deflection point; $x_m=8.40\text{ft};y=y_m$.

Substitute $29\times {16}^6\text{psi}$ for E, $758\text{in}{\text{.}}^4$ for I, $y_m$ for y, 18.84375 kips for $A_y$, 8.40 ft for x, $-644.61719{\text{kip-ft}}^2$ for $C_1$, and 0 for $C_2$ in Equation (3).

$\begin{array}{c}\left(\begin{array}{l}29\times {10}^6\text{psi}\times \frac{\text{1}\text{kip}}{{\text{10}}^3\text{lb}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2\times 758\text{in}{.}^4\\ \times {\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^4{y}_m\end{array}\right)=\left\{\begin{array}{l}\frac{18.84375\times {8.40}^3}{6}-\frac{3}{24}{\langle 8.40-5\rangle }^4\\ -\frac{10}{3}{\langle 8.40-10\rangle }^3\\ -644.61719\left(8.40\right)+0\end{array}\right\}\\ \left(\frac{29\times {10}^6\times 758}{1,000\times {12}^2}\right)y_m=1,861.461-16.7042-0-5,414.784396\\ y_m=-0.02339\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\\ =0.281\text{in}\text{.}(\downarrow )\end{array}$

Therefore, the largest downward deflection of the beam is $\underset{\_}{y_m=0.281\text{in}\text{.}(\downarrow )}$.