Chapter 9.3, Problem 60P

9.59 through 9.62 For the beam and loading indicated, determine the magnitude and location of the largest downward deflection.

9.60 Beam and loading of Prob. 9.46.

9.46 For the beam and loading shown, determine (a) the slope at end A, (b) the deflection at point B. Use E = 29 × 10⁶ psi.

Fig. P9.46

To determine

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

Answer to Problem 60P

The location of the largest downward deflection is $\underset{\_}{x_m=26.4\text{in}\text{.}}$.

The largest downward deflection of the beam is $\underset{\_}{y_m=\text{0}\text{.1520}\text{in}\text{.}(\downarrow )}$.

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;

$\frac{dV}{dx}=-w\left(x\right)=-10{\langle x-24\rangle }^0+10{\langle x-40\rangle }^0$

Integrate the equation to find the shear force.

$V=A_y-10{\langle x-24\rangle }^1+10{\langle x-40\rangle }^1$

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

$\frac{dM}{dx}=V=A_y-10{\langle x-24\rangle }^1+10{\langle x-40\rangle }^1-200{\langle x-40\rangle }^0$

Integrate the equation to find the bending moment.

$M=A_{y}x-5{\langle x-24\rangle }^2+5{\langle x-40\rangle }^2-200{\langle x-40\rangle }^1$ (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 $\left(A_{y}x-5{\langle x-24\rangle }^2+5{\langle x-40\rangle }^2-200{\langle x-40\rangle }^1\right)$ for $M\left(x\right)$.

$\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{A_{y}x-5{\langle x-24\rangle }^2+5{\langle x-40\rangle }^2-200{\langle x-40\rangle }^1}{EI}\\ EI\frac{d^{2}y}{d{x}^2}=A_{y}x-5{\langle x-24\rangle }^2+5{\langle x-40\rangle }^2-200{\langle x-40\rangle }^1\end{array}$

Integrate the equation with respect to x;

$EI\frac{dy}{dx}=\frac{A_y{x}^2}{2}-\frac{5}{3}{\langle x-24\rangle }^3+\frac{5}{3}{\langle x-40\rangle }^3-100{\langle x-40\rangle }^2+C_1$ (2)

Integrate the Equation (2) with respect to x.

$EIy=\frac{A_y{x}^3}{6}-\frac{5}{12}{\langle x-24\rangle }^4+\frac{5}{12}{\langle x-40\rangle }^4-\frac{100}{3}{\langle x-40\rangle }^3+C_{1}x+C_2$ (3)

Boundary condition 1:

At the point D; $x=48\text{in}\text{.};M=0$.

Substitute 48 in. for x and 0 for M in Equation (1).

$\begin{array}{l}0=A_y\left(48\right)-5{\langle 48-24\rangle }^2+5{\langle 48-40\rangle }^2-200{\langle 48-40\rangle }^1\\ 0=48A_y-2,880+320-1,600\\ A_y=86.667\text{lb}\end{array}$

Boundary condition 2:

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

Substitute 86.667 lb for $A_y$, 0 for x and 0 for y in Equation (3).

$\begin{array}{l}EI\left(0\right)=\frac{86.667{\left(0\right)}^3}{6}-\frac{5}{12}{\langle 0-24\rangle }^4+\frac{5}{12}{\langle 0-40\rangle }^4-\frac{100}{3}{\langle 0-40\rangle }^3+C_1\left(0\right)+C_2\\ 0=0-0+0-0+0+C_2\\ C_2=0\end{array}$

Boundary condition 3:

At the point D; $x=48\text{in}\text{.};y=0$.

Substitute 86.667 lb for $A_y$, 48 in. for x, 0 for y, and 0 for $C_2$ in Equation (3).

$\begin{array}{l}EI\left(0\right)=\frac{86.667{\left(48\right)}^3}{6}-\frac{5}{12}{\langle 48-24\rangle }^4+\frac{5}{12}{\langle 48-40\rangle }^4-\frac{100}{3}{\langle 48-40\rangle }^3+C_1\left(48\right)+0\\ 0=1,597,446.144-138,240+1,706.66667-17,066.66667+48C_1\\ C_1=-30,080.128\text{lb-in}{\text{.}}^2\end{array}$

Determine the moment of inertia (I) of the circular cross section using the equation.

$I=\frac{\pi d^4}{64}$

Here, the diameter of the circular cross section is d.

Substitute 1.25 in. for d.

$\begin{array}{c}I=\frac{\pi \times {1.25}^4}{64}\\ =0.119842\text{in}{\text{.}}^4\end{array}$

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

Substitute $29\times {10}^6\text{psi}$ for E, $0.119842\text{in}{\text{.}}^4$ for I, ${\theta }_A$ for $\frac{dy}{dx}$, 86.667 lb for $A_y$, 0 for x, and $-30,080.128\text{lb-in}{\text{.}}^2$ for $C_1$ in Equation (2).

$\begin{array}{c}29\times {10}^6\times 0.119842\times {\theta }_A=\frac{86.667\times {\left(0\right)}^2}{2}-\frac{5}{3}{\langle 0-24\rangle }^3+\frac{5}{3}{\langle 0-40\rangle }^3-100{\langle 0-40\rangle }^2-30,080.128\\ 29\times {10}^6\times 0.119842\times {\theta }_A=0-0+0-0-30,080.128\\ {\theta }_A=-8.66\times {10}^{-3}\text{rad}\end{array}$

At point B; $x=24\text{in}\text{.};\frac{dy}{dx}={\theta }_B$.

Substitute $29\times {10}^6\text{psi}$ for E, $0.119842\text{in}{\text{.}}^4$ for I, ${\theta }_B$ for $\frac{dy}{dx}$, 86.667 lb for $A_y$, 24 in. for x, and $-30,080.128\text{lb-in}{\text{.}}^2$ for $C_1$ in Equation (2).

$\begin{array}{c}29\times {10}^6\times 0.119842\times {\theta }_B=\left\{\begin{array}{l}\frac{86.667\times {\left(24\right)}^2}{2}-\frac{5}{3}{\langle 24-24\rangle }^3+\frac{5}{3}{\langle 24-40\rangle }^3-100{\langle 24-40\rangle }^2\\ -30,080.128\end{array}\right\}\\ 29\times {10}^6\times 0.119842\times {\theta }_B=24,960.096-0+0-0-30,080.128\\ {\theta }_B=-1.473\times {10}^{-3}\text{rad}\end{array}$

At point C; $x=40\text{in}\text{.};\frac{dy}{dx}={\theta }_C$.

Substitute $29\times {10}^6\text{psi}$ for E, $0.119842\text{in}{\text{.}}^4$ for I, ${\theta }_C$ for $\frac{dy}{dx}$, 86.667 lb for $A_y$, 40 in. for x, and $-30,080.128\text{lb-in}{\text{.}}^2$ for $C_1$ in Equation (2).

$\begin{array}{c}29\times {10}^6\times 0.119842\times {\theta }_C=\left\{\begin{array}{l}\frac{86.667\times {\left(40\right)}^2}{2}-\frac{5}{3}{\langle 40-24\rangle }^3+\frac{5}{3}{\langle 40-40\rangle }^3-100{\langle 40-40\rangle }^2\\ -30,080.128\end{array}\right\}\\ 29\times {10}^6\times 0.119842\times {\theta }_C=69,333.6-6,826.667+0-0-30,080.128\\ {\theta }_C=9.33\times {10}^{-3}\text{rad}\end{array}$

At point D; $x=48\text{in}\text{.};\frac{dy}{dx}={\theta }_D$.

Substitute $29\times {10}^6\text{psi}$ for E, $0.119842\text{in}{\text{.}}^4$ for I, ${\theta }_D$ for $\frac{dy}{dx}$, 86.667 lb for $A_y$, 48 in. for x, and $-30,080.128\text{lb-in}{\text{.}}^2$ for $C_1$ in Equation (2).

$\begin{array}{c}29\times {10}^6\times 0.119842\times {\theta }_D=\left\{\begin{array}{l}\frac{86.667\times {\left(48\right)}^2}{2}-\frac{5}{3}{\langle 48-24\rangle }^3+\frac{5}{3}{\langle 48-40\rangle }^3-100{\langle 48-40\rangle }^2\\ -30,080.128\end{array}\right\}\\ 29\times {10}^6\times 0.119842\times {\theta }_D=99,840.384-23,040+853.333-6,400-30,080.128\\ {\theta }_D=11.847\times {10}^{-3}\text{rad}\end{array}$

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 $\frac{dy}{dx}$, 86.667 lb for $A_y$, and $-30,080.128\text{lb-in}{\text{.}}^2$ for $C_1$ in Equation (2).

$\begin{array}{l}EI\left(0\right)=\frac{86.667x_m^2}{2}-\frac{5}{3}{\langle x_m-24\rangle }^3+\frac{5}{3}{\langle x_m-40\rangle }^3-100{\langle x_m-40\rangle }^2-30,080.128\\ 0=43.333x_m^2-\frac{5}{3}{\left(x_m-24\right)}^3+0-0-30,080.128\\ 0=43.333x_m^2-\frac{5}{3}{\left(x_m-24\right)}^3-30,080.128\end{array}$

Solve the Equation;

$x_m=26.4\text{in}\text{.}$.

Therefore, the location of the largest downward deflection is $\underset{\_}{x_m=26.4\text{in}\text{.}}$.

At largest deflection point; $x_m=26.4\text{in}\text{.};y=y_m$.

Substitute $29\times {10}^6\text{psi}$ for E, $0.119842\text{in}{\text{.}}^4$ for I, $y_B$ for y, 86.667 lb for $A_y$, 26.4 in. for x, $-30,080.128\text{lb-in}{\text{.}}^2$ for $C_1$, and 0 for $C_2$ in Equation (3).

$\begin{array}{c}29\times {10}^6\times 0.119842\times y_m=\left\{\begin{array}{l}\frac{86.667\times {26.4}^3}{6}-\frac{5}{12}{\langle 26.4-24\rangle }^4+\frac{5}{12}{\langle 26.4-40\rangle }^4-\frac{100}{3}{\langle 26.4-40\rangle }^3\\ -30,080.128\left(26.4\right)+0\end{array}\right\}\\ =265,775.1022-13.824+0-0-794,115.3792\\ y_m=-0.1520\text{in}\text{.}\\ =\text{0}\text{.1520}\text{in}\text{.}(\downarrow )\end{array}$

Therefore, the largest downward deflection of the beam is $\underset{\_}{y_m=\text{0}\text{.1520}\text{in}\text{.}(\downarrow )}$.