Chapter 7, Problem 6P

To determine

Find the equations for slope and deflection for the beam by the double integration method and calculate the slope at each support and the deflection at midspan.

Answer to Problem 6P

The equation for slope is $\underset{\_}{\frac{dy}{dx}=\frac{1}{EI}\left(\frac{P{x}_1^2}{6}-0.0493P{L}^2\right)}$ for $0\le x_1\le \frac{2}{3}L$ and $\underset{\_}{\frac{dy}{dx}=\frac{1}{EI}\left(-\frac{P{x}_2^2}{3}+\frac{2}{9}PL{x}_2+0.0247P{L}^2\right)}$ for $0\le x_2\le \frac{L}{3}$.

The equation for deflection is $\underset{\_}{y=\frac{1}{EI}\left(\frac{P{x}_1^3}{18}-0.0493P{L}^2{x}_1\right)}$ for $0\le x_1\le \frac{2}{3}L$ and $\underset{\_}{y=\frac{1}{EI}\left(-\frac{P{x}_2^3}{9}+\frac{PL{x}_2^2}{9}+0.0247P{L}^2{x}_2-0.01646P{L}^3\right)}$ for $0\le x_2\le \frac{1}{3}L$.

The slope of the beam at support A is $\underset{\_}{{\theta }_A=-\frac{0.0493P{L}^2}{EI}}$.

The deflection of the beam at midspan is $\underset{\_}{{\Delta }_{\frac{L}{2}}=-0.0177\frac{P{L}^3}{EI}}$.

The slope of the beam at support B is $\underset{\_}{{\theta }_B=0.0617\frac{P{L}^2}{EI}}$.

Explanation of Solution

Given information:

Maximum deflection occurs at $x=0.544L$.

Apply the sign conventions for calculating reactions 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 positive and the counter clockwise moment as negative.

Sketch the Free Body Diagram of the beam as shown in Figure 1.

Refer to Figure 1.

Use equilibrium equations:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_A=0\\ -R_{B}L+P\times \frac{2L}{3}=0\\ R_B=\frac{2}{3}P\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_A+R_B-P=0\\ R_A-P+\frac{2}{3}P=0\\ R_A=\frac{1}{3}P\end{array}$

Refer to Figure 1.

Consider a section at a distance $x_1$ from A for $0\le x_1\le \frac{2}{3}L$.

Calculate the moment at section $x_2$ from B.

$M\left(x_1\right)=\frac{1}{3}P{x}_1$

Apply double integration method as shown below.

$\begin{array}{l}EI\frac{d^{2}y}{d{x}^2}=M\left(x_1\right)\\ EI\frac{d^{2}y}{d{x}^2}=\frac{1}{3}P{x}_1\end{array}$

Integrate with respect to x.

$\begin{array}{c}EI\frac{dy}{dx}=\frac{1}{3}P\frac{x_1^2}{2}+C_1\\ =\frac{P{x}_1^2}{6}+C_1\end{array}$        (1)

Integrate with respect to x.

$\begin{array}{c}EIy=\frac{P}{6}\frac{x_1^3}{3}+C_1{x}_1+C_2\\ =\frac{P{x}_1^3}{18}+C_1{x}_1+C_2\end{array}$        (2)

Apply the boundary conditions as shown below.

i) At a distance $x_1=0$ the deflection $y=0$.

ii) At a distance $x_1=0.544L$ the deflection $\frac{dy}{dx}=0$.

Apply boundary condition (i) in Equation (2).

$\begin{array}{l}EI\left(0\right)=\frac{P{\left(0\right)}^3}{18}+C_1\left(0\right)+C_2\\ C_2=0\end{array}$

Apply boundary condition (ii) in Equation (2).

$\begin{array}{l}EI\left(0\right)=\frac{P{\left(0.544L\right)}^2}{6}+C_1\\ C_1=-0.0493P{L}^2\end{array}$

Substitute $-0.0493P{L}^2$ for $C_1$ in Equation (1).

$\begin{array}{l}EI\frac{dy}{dx}=\frac{P{x}_1^2}{6}-0.0493P{L}^2\\ \frac{dy}{dx}=\frac{1}{EI}\left(\frac{P{x}_1^2}{6}-0.0493P{L}^2\right)\end{array}$        (3)

Hence, the equation for slope is $\underset{\_}{\frac{dy}{dx}=\frac{1}{EI}\left(\frac{P{x}_1^2}{6}-0.0493P{L}^2\right)}$ for $0\le x_1\le \frac{2}{3}L$.

Substitute $-0.0493P{L}^2$ for $C_1$ and 0 for $C_2$ in Equation (2).

$\begin{array}{l}EIy=\frac{P{x}_1^3}{18}-0.0493P{L}^2{x}_1+0\\ y=\frac{1}{EI}\left(\frac{P{x}_1^3}{18}-0.0493P{L}^2{x}_1\right)\end{array}$        (4)

Hence, the equation for deflection is $\underset{\_}{y=\frac{1}{EI}\left(\frac{P{x}_1^3}{18}-0.0493P{L}^2{x}_1\right)}$ for $0\le x_1\le \frac{2}{3}L$.

Calculate the slope at support A as shown below.

$\begin{array}{l}{\left(\frac{dy}{dx}\right)}_{x=0}=\frac{1}{EI}\left(\frac{P{\left(0\right)}^2}{6}-0.0493P{L}^2\right)\\ {\theta }_A=-\frac{0.0493P{L}^2}{EI}\end{array}$

Hence, the slope of the beam at support A is $\underset{\_}{{\theta }_A=-\frac{0.0493P{L}^2}{EI}}$.

Calculate the slope at $x=\frac{2L}{3}$ as shown below.

$\begin{array}{l}{\left(\frac{dy}{dx}\right)}_{x=\frac{2L}{3}}=\frac{1}{EI}\left(\frac{P{\left(\frac{2L}{3}\right)}^2}{6}-0.0493P{L}^2\right)\\ {\theta }_C=\frac{1}{EI}\left(0.0740P{L}^2-0.0493P{L}^2\right)\\ =0.0247\frac{P{L}^2}{EI}\end{array}$

Calculate the deflection at mid span as shown below.

$\begin{array}{l}{\left(y\right)}_{x=\frac{L}{2}}=\frac{1}{EI}\left(\frac{P{\left(\frac{L}{2}\right)}^3}{18}-0.0493P{L}^2\left(\frac{L}{2}\right)\right)\\ {\Delta }_{\frac{L}{2}}=-0.0177\frac{P{L}^3}{EI}\end{array}$

Hence, the deflection of the beam at midspan is $\underset{\_}{{\Delta }_{\frac{L}{2}}=-0.0177\frac{P{L}^3}{EI}}$.

Consider a section at a distance $x_2$ from C for $0\le x_2\le \frac{1}{3}L$.

Calculate the moment at section $x_2$ from C.

$M\left(x_2\right)=-\frac{2}{3}P{x}_2+\frac{2}{9}PL$

Apply double integration method as shown below.

$\begin{array}{l}EI\frac{d^{2}y}{d{x}^2}=M\left(x_2\right)\\ EI\frac{d^{2}y}{d{x}^2}=-\frac{2}{3}P{x}_2+\frac{2}{9}PL\end{array}$

Integrate with respect to x.

$\begin{array}{c}EI\frac{dy}{dx}=-\frac{2}{3}P\frac{x_2^2}{2}+\frac{2}{9}PL{x}_2+C_3\\ =-\frac{P{x}_2^2}{3}+\frac{2}{9}PL{x}_2+C_3\end{array}$        (5)

Integrate with respect to x.

$\begin{array}{c}EIy=-\frac{P}{3}\frac{x_2^3}{3}+\frac{2}{9}PL\frac{x_2^2}{2}+C_3{x}_2+C_4\\ =-\frac{P{x}_2^3}{9}+\frac{PL{x}_2^2}{9}+C_3{x}_2+C_4\end{array}$        (6)

Apply the boundary conditions as shown below.

iii) At a distance $x_2=0$ the deflection $\frac{dy}{dx}=0.0247\frac{P{L}^2}{EI}$.

iv) At a distance $x_2=\frac{L}{3}$ the deflection $y=0$.

Apply boundary condition (iii) in Equation (5).

$\begin{array}{l}EI\left(0.0247\frac{P{L}^2}{EI}\right)=-\frac{2}{3}P\frac{{\left(0\right)}^2}{2}+\frac{2}{9}PL{\left(0\right)}^2+C_3\\ C_3=0.0247P{L}^2\end{array}$

Apply boundary condition (iv) in Equation (6).

$\begin{array}{l}EI\left(0\right)=-\frac{P}{9}{\left(\frac{L}{3}\right)}^3+\frac{PL}{9}{\left(\frac{L}{3}\right)}^2+\left(0.0247P{L}^2\right)\left(\frac{L}{3}\right)+C_4\\ 0=-\frac{P{L}^3}{243}+\frac{P{L}^3}{81}+0.0247\frac{P{L}^3}{3}+C_4\\ C_4=-0.01646P{L}^3\end{array}$

Substitute $0.0247P{L}^2$ for $C_3$ in Equation (5).

$\begin{array}{l}EI\frac{dy}{dx}=-\frac{P{x}_2^2}{3}+\frac{2}{9}PL{x}_2+0.0247P{L}^2\\ \frac{dy}{dx}=\frac{1}{EI}\left(-\frac{P{x}_2^2}{3}+\frac{2}{9}PL{x}_2+0.0247P{L}^2\right)\end{array}$        (7)

Hence, the equation for slope is $\underset{\_}{\frac{dy}{dx}=\frac{1}{EI}\left(-\frac{P{x}_2^2}{3}+\frac{2}{9}PL{x}_2+0.0247P{L}^2\right)}$ for $0\le x_2\le \frac{L}{3}$.

Substitute $0.0247P{L}^2$ for $C_3$ and $-0.01646P{L}^3$ for $C_4$ in Equation (2).

$\begin{array}{l}EIy=-\frac{P{x}_2^3}{9}+\frac{PL{x}_2^2}{9}+0.0247P{L}^2{x}_2-0.01646P{L}^3\\ y=\frac{1}{EI}\left(-\frac{P{x}_2^3}{9}+\frac{PL{x}_2^2}{9}+0.0247P{L}^2{x}_2-0.01646P{L}^3\right)\end{array}$        (4)

Hence, the equation for deflection is $\underset{\_}{y=\frac{1}{EI}\left(-\frac{P{x}_2^3}{9}+\frac{PL{x}_2^2}{9}+0.0247P{L}^2{x}_2-0.01646P{L}^3\right)}$ for $0\le x_2\le \frac{1}{3}L$.

Calculate the slope at support B as shown below.

$\begin{array}{l}{\left(\frac{dy}{dx}\right)}_{x_2=\frac{L}{3}}=\frac{1}{EI}\left(-\frac{P{\left(\frac{L}{3}\right)}^2}{3}+\frac{2}{9}PL\left(\frac{L}{3}\right)+0.0247P{L}^2\right)\\ {\theta }_B=\frac{1}{EI}\left(-\frac{P{L}^2}{27}+\frac{2}{27}P{L}^2+0.0247P{L}^2\right)\\ =0.0617\frac{P{L}^2}{EI}\end{array}$

Therefore, the slope of the beam at support B is $\underset{\_}{{\theta }_B=0.0617\frac{P{L}^2}{EI}}$.