Chapter 9, Problem 157RP

For the loading shown, determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, (c) the slope at the free end.

Fig. P9.157

To determine

The equation of elastic curve (y) for the cantilever beam AB.

Answer to Problem 157RP

The equation of the elastic curve for the cantilever beam AB is $\underset{\_}{-\frac{w_0}{EIL}\left(\frac{L^3{x}^2}{6}-\frac{L^2{x}^3}{12}+\frac{x^5}{120}\right)}$.

Explanation of Solution

Calculation:

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

Calculate the reaction $\left(R_A\right)$ at point A by resolving the vertical component of force.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ R_A-\frac{w_{0}L}{2}=0\\ R_A=\frac{w_{0}L}{2}\end{array}$

Calculate the moment $M_A$ at point A as below:

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -M_A-\frac{w_{0}L}{2}\times \frac{2L}{3}=0\\ -M_A=\frac{w_0{L}^2}{3}\\ M_A=-\frac{w_0{L}^2}{3}\end{array}$

Show the free body diagram of beam section AE as in Figure 2.

Calculate the moment (M) by taking moment about the point E.

$\begin{array}{l}{\displaystyle \sum M_E=0}\\ \frac{w_0{L}^2}{3}-\frac{w_{0}Lx}{2}+\frac{w_0{x}^2}{2L}\times \left(\frac{x}{3}\right)+M=0\\ M=-\frac{w_0{L}^2}{3}+\frac{w_{0}Lx}{2}-\frac{w_0{x}^2}{2L}\times \left(\frac{x}{3}\right)\\ M=-\frac{w_0{L}^2}{3}+\frac{w_{0}Lx}{2}-\frac{w_0{x}^3}{6L}\\ M=-\frac{w_0{L}^2}{3}+\frac{w_{0}Lx}{2}-\frac{w_0{x}^3}{6L}\end{array}$ (1).

Calculate the equation of the elastic curve (y) by integrating the Equation (1).

$\begin{array}{l}EI\frac{d^{2}y}{d{x}^2}=-\frac{w_0{L}^2}{3}+\frac{w_{0}Lx}{2}-\frac{w_0{x}^3}{6L}\\ EI\frac{dy}{dx}=-\frac{w_0{L}^2}{3}x+\frac{w_{0}L}{2}\frac{x^2}{2}-\frac{w_0}{6L}\frac{x^4}{4}+C_1\\ EI\frac{dy}{dx}=-\frac{w_0{L}^2}{3}x+\frac{w_{0}L{x}^2}{4}-\frac{w_0{x}^4}{24L}+C_1\end{array}$ (2).

Substitute 0 for x and 0 for $\frac{dy}{dx}$ in Equation (2).

$\begin{array}{l}EI\left(0\right)=-\frac{w_0{L}^2}{3}\left(0\right)+\frac{w_{0}L{\left(0\right)}^2}{4}-\frac{w_0{\left(0\right)}^4}{24L}+C_1\\ 0=-0+0-0+C_1\\ C_1=0\end{array}$

Integrate equation (2) and Substitute 0 for $C_1$.

$\begin{array}{l}EI\frac{dy}{dx}=-\frac{w_0{L}^2}{3}x+\frac{w_{0}L{x}^2}{4}-\frac{w_0{x}^4}{24L}+\left(0\right)\\ EI\frac{dy}{dx}=-\frac{w_0{L}^2}{3}\frac{x^2}{2}+\frac{w_{0}L}{4}\frac{x^3}{3}-\frac{w_0}{24L}\frac{x^5}{5}+C_2\\ EIy=-\frac{w_0{L}^2{x}^2}{6}+\frac{w_{0}L{x}^3}{12}-\frac{w_0{x}^5}{120L}+C_2\end{array}$ (3)

Substitute 0 for x and 0 for y in Equation (3).

$\begin{array}{l}EI\left(0\right)=-\frac{w_0{L}^2{\left(0\right)}^2}{6}+\frac{w_{0}L{\left(0\right)}^3}{12}-\frac{w_0{\left(0\right)}^5}{120L}+C_2\\ 0=-0+0-0+C_2\\ C_2=0\end{array}$

Integrate Equation (3) and Substitute 0 for $C_2$.

$EIy=-\frac{w_0{L}^2{x}^2}{6}+\frac{w_{0}L{x}^3}{12}-\frac{w_0{x}^5}{120L}+0$

Multiply the Equation by $\left(\frac{L}{L}\right)$.

$\begin{array}{l}EIy=-\frac{w_0{L}^2{x}^2}{6}\frac{L}{L}+\frac{w_{0}L{x}^3}{12}\frac{L}{L}-\frac{w_0{x}^5}{120L}\frac{L}{L}\\ EIy=-\frac{w_0{L}^3{x}^2}{6L}+\frac{w_0{L}^2{x}^3}{12L}-\frac{w_0{x}^5}{120L}\\ y=-\frac{w_0}{EIL}\left(\frac{L^3{x}^2}{6}-\frac{L^2{x}^3}{12}+\frac{x^5}{120}\right)\end{array}$

Thus, the equation of the elastic curve for the cantilever beam AB is $\underset{\_}{-\frac{w_0}{EIL}\left(\frac{L^3{x}^2}{6}-\frac{L^2{x}^3}{12}+\frac{x^5}{120}\right)}$.

To determine

The deflection $\left(y_B\right)$ the free end at point B.

Answer to Problem 157RP

The deflection $\left(y_B\right)$ the free end at point B is $\underset{\_}{\frac{11}{12}\frac{w_0{L}^4}{EI}\left(\text{downward}\right)}$.

Explanation of Solution

Calculation:

Calculate the deflection $\left(y_B\right)$ using the Equation (3).

Substitute L for x.

$\begin{array}{l}EIy=-\frac{w_0{L}^2{\left(L\right)}^2}{6}+\frac{w_{0}L{\left(L\right)}^3}{12}-\frac{w_0{\left(L\right)}^5}{120L}\\ EIy=-\frac{w_0{L}^4}{6}+\frac{w_0{L}^4}{12}-\frac{w_0{\left(L\right)}^5}{120L}\\ y_B=-\frac{w_0{L}^4}{EI}\left(\frac{1}{6}-\frac{1}{12}+\frac{1}{120}\right)\\ y_B=-\frac{11}{120}\frac{w_0{L}^4}{EI}\end{array}$

Thus, the deflection $\left(y_B\right)$ the free end at point B is $\underset{\_}{\frac{11}{12}\frac{w_0{L}^4}{EI}\left(\text{downward}\right)}$.

To determine

The slope $\left({\theta }_B\right)$ the free end at point B.

Answer to Problem 157RP

The slope $\left({\theta }_B\right)$ the free end at point B is $\underset{\_}{\frac{1}{8}\frac{w_0{L}^3}{EI}}$.

Explanation of Solution

Calculation:

Calculate the slope $\left({\theta }_B\right)$ using the Equation (2).

Substitute L for x and 0 for $C_1$

$\begin{array}{l}EI\frac{dy}{dx}=-\frac{w_0{L}^2}{3}\left(L\right)+\frac{w_{0}L{\left(L\right)}^2}{4}-\frac{w_0{\left(L\right)}^4}{24L}+0\\ EI\frac{dy}{dx}=-\frac{w_0{L}^3}{3}+\frac{w_0{L}^3}{4}-\frac{w_0{L}^3}{24}\\ \frac{dy}{dx}=-\frac{w_0{L}^3}{EI}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{24}\right)\\ \frac{dy}{dx}=-\frac{1}{8}\frac{w_0{L}^3}{EI}\end{array}$

Thus, the slope $\left({\theta }_B\right)$ the free end at point B is $\underset{\_}{\frac{1}{8}\frac{w_0{L}^3}{EI}}$.