Chapter 7, Problem 1P

To determine

Find the equations for slope and deflection for the beam by the double integration method and compare the deflection at B with the deflection at midspan.

Answer to Problem 1P

The equation for slope of the beam is $\underset{\_}{\frac{dy}{dx}=-\frac{1}{6EI}\left(3w{L}^{2}x-3wL{x}^2+w{x}^3\right)}$.

The equation for deflection of the beam is $\underset{\_}{y=-\frac{1}{6EI}\left(\frac{3}{2}w{L}^2{x}^2-wL{x}^3+\frac{w{x}^4}{4}\right)}$.

The deflection at midspan is equal to 0.35 times the deflection at B.

Explanation of Solution

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 deflected shape of the beam as shown in Figure 1.

Refer to Figure 1.

Consider a section at a distance x from A.

Sketch the cross section from B as shown in Figure 2.

Calculate the moment at the section $\left(L-x\right)$ from B.

$\begin{array}{l}{\displaystyle \sum M_{L-x}=0}\\ M\left(x\right)-w\times \left(L-x\right)\times \frac{\left(L-x\right)}{2}=0\\ M\left(x\right)=\frac{w{\left(L-x\right)}^2}{2}\end{array}$

Apply double integration method as shown below.

$\begin{array}{c}EI\frac{d^{2}y}{d{x}^2}=-M\left(x\right)\\ =-\frac{w{\left(L-x\right)}^2}{2}\\ =-\frac{w}{2}\left(L^2-2Lx+x^2\right)\end{array}$

Integrate with respect to x.

$\begin{array}{c}EI\frac{dy}{dx}=-\frac{w}{2}\left(L^{2}x-2L\frac{x^2}{2}+\frac{x^3}{3}\right)+C_1\\ =-\frac{w}{6}\left(3L^{2}x-3L{x}^2+x^3\right)+C_1\end{array}$        (1)

Integrate with respect to x.

$\begin{array}{c}EIy=-\frac{w}{6}\left(3L^2\frac{x^2}{2}-3L\frac{x^3}{3}+\frac{x^4}{4}\right)+C_{1}x+C_2\\ =-\frac{w}{6}\left(\frac{3}{2}{L}^2{x}^2-L{x}^3+\frac{x^4}{4}\right)+C_{1}x+C_2\end{array}$        (2)

Apply the boundary conditions as shown below.

i) At a distance $x=0$ the slope $\frac{dy}{dx}=0$.

ii) At a distance $x=0$ the deflection $y=0$.

Apply boundary condition (i) in Equation (1).

$\begin{array}{l}EI\left(0\right)=-\frac{w}{6}\left(3L^2\left(0\right)-3L{\left(0\right)}^2+{\left(0\right)}^3\right)+C_1\\ C_1=0\end{array}$

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

$\begin{array}{l}EI\left(0\right)=-\frac{w}{6}\left(\frac{3}{2}{L}^2{\left(0\right)}^2-L{\left(0\right)}^3+\frac{{\left(0\right)}^4}{4}\right)+C_1\left(0\right)+C_2\\ C_2=0\end{array}$

Substitute 0 for $C_1$ in Equation (1).

$\begin{array}{l}EI\frac{dy}{dx}=-\frac{w}{6}\left(3L^{2}x-3L{x}^2+x^3\right)+0\\ \frac{dy}{dx}=-\frac{1}{6EI}\left(3w{L}^{2}x-3wL{x}^2+w{x}^3\right)\end{array}$        (3)

Hence, the equation for slope of the beam is $\underset{\_}{\frac{dy}{dx}=-\frac{1}{6EI}\left(3w{L}^{2}x-3wL{x}^2+w{x}^3\right)}$.

Substitute 0 for $C_1$ and 0 for $C_2$ in Equation (2).

$\begin{array}{l}EIy=-\frac{w}{6}\left(\frac{3}{2}w{L}^2{x}^2-wL{x}^3+\frac{w{x}^4}{4}\right)+\left(0\right)x+\left(0\right)\\ y=-\frac{1}{6EI}\left(\frac{3}{2}w{L}^2{x}^2-wL{x}^3+\frac{w{x}^4}{4}\right)\end{array}$        (4)

Hence, the equation for deflection of the beam is $\underset{\_}{y=-\frac{1}{6EI}\left(\frac{3}{2}w{L}^2{x}^2-wL{x}^3+\frac{w{x}^4}{4}\right)}$.

Calculate the deflection at B as shown below.

Substitute L for x in Equation (4).

$\begin{array}{l}{y}_B=-\frac{1}{6EI}\left(\frac{3}{2}w{L}^2\times L^2-wL\times L^3+\frac{w\times L^4}{4}\right)\\ {\Delta }_B=-\frac{w{L}^4}{6EI}\left(\frac{6-4+1}{4}\right)\\ =-\frac{w{L}^4}{8EI}\end{array}$

Calculate the deflection at midspan as shown below.

Substitute $\frac{L}{2}$ for x in Equation (4).

$\begin{array}{l}{y}_{\frac{L}{2}}=-\frac{1}{6EI}\left(\frac{3}{2}w{L}^2\times {\left(\frac{L}{2}\right)}^2-wL\times {\left(\frac{L}{2}\right)}^3+\frac{w}{4}{\left(\frac{L}{2}\right)}^4\right)\\ {\Delta }_{\frac{L}{2}}=-\frac{w{L}^4}{6EI}\left(\frac{3}{2}\times \frac{1}{4}-\frac{1}{8}+\frac{1}{4}\times \frac{1}{16}\right)\\ =-\frac{w{L}^4}{6EI}\left(\frac{16+1}{64}\right)\\ =-\frac{17w{L}^4}{384EI}\end{array}$

Compare the deflection at B with the deflection at midspan as shown below.

$\begin{array}{c}\frac{{\Delta }_{\frac{L}{2}}}{{\Delta }_B}=\frac{-\frac{17w{L}^4}{384EI}}{-\frac{w{L}^4}{8EI}}\\ =0.35\end{array}$

Therefore, the deflection at midspan is equal to 0.35 times the deflection at B.