Chapter 6, Problem 1P

To determine

Find the equations for slope and deflection of the beam using direct integration method.

Answer to Problem 1P

The equation for slope is $\underset{\_}{-\frac{M}{6EIL}\left(3x^2-6Lx+2L^2\right)}$.

The equation for deflection is $\underset{\_}{-\frac{M}{6EIL}\left[x^3-3L{x}^2+2L^{2}x\right]}$.

Explanation of Solution

Calculation:

Consider the flexural rigidity EI of the beam is constant.

Draw the free body diagram of the beam as in Figure (1).

Refer Figure (1),

Find the reaction at support B.

$\begin{array}{l}{\displaystyle \sum M_A}=0\\ B_y\times L-M=0\\ B_y=\frac{M}{L}(\uparrow )\end{array}$

Find the reaction at support A.

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ A_y\times L+M=0\\ A_y=-\frac{M}{L}\\ A_y=\frac{M}{L}(\downarrow )\end{array}$

Draw the section at x distance from support A as in Figure (2).

Refer Figure (2),

Write the equation for bending moment at x distance.

$\begin{array}{c}{M}_x=M-A_y\left(x\right)\\ =M-\frac{M}{L}\left(x\right)\\ =\frac{M}{L}\left(L-x\right)\end{array}$

Write the equation for $\frac{M}{EI}$.

$\frac{d^{2}y}{d{x}^2}=\frac{M}{EI}=\frac{M}{EIL}\left(L-x\right)$        (1)

Integrate Equation (1) to find the equation of the slope.

$\begin{array}{c}\frac{dy}{dx}=\theta ={\displaystyle \int \frac{M}{EIL}\left(L-x\right)dx}\\ =\frac{M}{EIL}\left(Lx-\frac{x^2}{2}\right)+C_1\end{array}$        (2)

Integrate Equation (2) to find the equation of the deflection.

$\begin{array}{c}y={\displaystyle \int \left[\frac{M}{EIL}\left(Lx-\frac{x^2}{2}\right)+C_1\right]}dx\\ =\left[\frac{M}{EIL}\left(\frac{L{x}^2}{2}-\frac{x^3}{6}\right)+C_{1}x+C_2\right]\end{array}$        (3)

Find the integration constants $C_1\text{and}{C}_2$:

Apply boundary conditions in Equation (3):

At x=0 and y=0.

$\begin{array}{l}0=\left[\frac{M}{EIL}\left(\frac{L\times 0}{2}-\frac{0}{6}\right)+C_1\times 0+C_2\right]\\ C_2=0\end{array}$

At x=L and y=0.

$\begin{array}{l}0=\left[\frac{M}{EIL}\left(\frac{L\times L^2}{2}-\frac{L^3}{6}\right)+C_1\times L+0\right]\\ 0=\frac{M}{EIL}\left(\frac{L^3}{2}-\frac{L^3}{6}\right)+\left(C_1\times L\right)\\ 0=\frac{M{L}^3}{EIL}\left(\frac{6-2}{12}\right)+\left(C_1\times L\right)\\ C_{1}L=-\frac{M{L}^2}{3EI}\\ C_1=-\frac{ML}{3EI}\end{array}$

Find the equation of the slope.

Substitute $-\frac{ML}{3EI}$ for $C_1$ in Equation (2).

$\begin{array}{c}\theta =\frac{M}{EIL}\left(Lx-\frac{x^2}{2}\right)-\frac{ML}{3EI}\\ =\frac{M}{6EIL}\left(6Lx-\frac{6x^2}{2}\right)-2L\\ =\frac{M}{6EIL}\left(6Lx-3x^2-2L^2\right)\\ =-\frac{M}{6EIL}\left(3x^2-6Lx+2L^2\right)\end{array}$

Thus, the equation of the slope is $\underset{\_}{-\frac{M}{6EIL}\left(3x^2-6Lx+2L^2\right)}$.

Find the equation of the deflection.

Substitute $-\frac{ML}{3EI}$ for $C_1$ and 0 for $C_2$ in Equation (3).

$\begin{array}{c}y=\frac{M}{EIL}\left(\frac{L{x}^2}{2}-\frac{x^3}{6}\right)-\frac{ML}{3EI}x+0\\ =\frac{M}{6EIL}\left[\left(\frac{6L{x}^2}{2}-\frac{6x^3}{6}\right)-2Lx\right]\\ =-\frac{M}{6EIL}\left[x^3-3L{x}^2+2L^{2}x\right]\end{array}$

Thus, the equation of the deflection is $\underset{\_}{-\frac{M}{6EIL}\left[x^3-3L{x}^2+2L^{2}x\right]}$.