Chapter 7, Problem 2P

To determine

Find the equations for slope and deflection for the beam by the double integration method and calculate the deflection at B and C.

Answer to Problem 2P

The equation for slope is $\underset{\_}{\frac{dy}{dx}=\frac{1}{EI}\left(\frac{3wL{x}^2}{40}-\frac{w{L}^{2}x}{30}-\frac{w{x}^4}{24L}\right)}$.

The equation for deflection is $\underset{\_}{y=\frac{1}{EI}\left(\frac{wL{x}^3}{40}-\frac{w{L}^2{x}^2}{60}-\frac{w{x}^5}{120L}\right)}$.

The deflection of the beam at B is $\underset{\_}{{\Delta }_B=-0.000659\frac{w{L}^4}{EI}}$.

The deflection of the beam at C is $\underset{\_}{{\Delta }_C=-0.001302\frac{w{L}^4}{EI}}$.

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

Refer to Figure 1.

Use equilibrium equations:

Summation of moments about D is equal to 0.

$\begin{array}{l}\sum M_D=0\\ R_A\times L-\frac{w{L}^2}{30}-\frac{1}{2}\times L\times w\times \frac{1}{3}L+\frac{w{L}^2}{20}=0\\ R_A-\frac{wL}{30}-\frac{wL}{6}+\frac{wL}{20}=0\\ R_A=\frac{9wL}{60}\end{array}$

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

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_D+\frac{9wL}{60}-\frac{1}{2}wL=0\\ R_D=\frac{7wL}{20}\end{array}$

Consider a section at a distance x from A.

Sketch the cross section as shown in Figure 2.

Refer to Figure 1.

Calculate the load intensity at a distance x from A.

$\begin{array}{l}\frac{w}{L}=\frac{w_x}{x}\\ w_x=\frac{w}{L}x\end{array}$

Calculate the moment at section x from B.

$\begin{array}{c}M\left(x\right)=-\frac{w{L}^2}{30}+\frac{9wL}{60}\times x-\frac{1}{2}\left(\frac{w}{L}x\right)x\left(\frac{1}{3}x\right)\\ =\frac{3wLx}{20}-\frac{w{L}^2}{30}-\frac{w{x}^3}{6L}\end{array}$

Apply double integration method as shown below.

$\begin{array}{l}EI\frac{d^{2}y}{d{x}^2}=M\left(x\right)\\ EI\frac{d^{2}y}{d{x}^2}=\frac{3wLx}{20}-\frac{w{L}^2}{30}-\frac{w{x}^3}{6L}\end{array}$

Integrate with respect to x.

$\begin{array}{c}EI\frac{dy}{dx}=\frac{3wL}{20}\frac{x^2}{2}-\frac{w{L}^2}{30}x-\frac{w}{6L}\frac{x^4}{4}+C_1\\ =\frac{3wL{x}^2}{40}-\frac{w{L}^{2}x}{30}-\frac{w{x}^4}{24L}+C_1\end{array}$        (1)

Integrate with respect to x.

$\begin{array}{c}EIy=\frac{3wL}{40}\frac{x^3}{3}-\frac{w{L}^2}{30}\frac{x^2}{2}-\frac{w}{24L}\frac{x^5}{5}+C_{1}x+C_2\\ =\frac{wL{x}^3}{40}-\frac{w{L}^2{x}^2}{60}-\frac{w{x}^5}{120L}+C_{1}x+C_2\end{array}$        (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).

$\begin{array}{l}EI\left(0\right)=\frac{3wL{\left(0\right)}^2}{40}-\frac{w{L}^2\left(0\right)}{30}-\frac{w{\left(0\right)}^4}{24L}+C_1\\ C_1=0\end{array}$

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

$\begin{array}{l}EI\left(0\right)=\frac{wL{\left(0\right)}^3}{40}-\frac{w{L}^2{\left(0\right)}^2}{60}-\frac{w{\left(0\right)}^5}{120L}+C_1\left(0\right)+C_2\\ C_2=0\end{array}$

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

$\begin{array}{l}EI\frac{dy}{dx}=\frac{3wL{x}^2}{40}-\frac{w{L}^{2}x}{30}-\frac{w{x}^4}{24L}+0\\ \frac{dy}{dx}=\frac{1}{EI}\left(\frac{3wL{x}^2}{40}-\frac{w{L}^{2}x}{30}-\frac{w{x}^4}{24L}\right)\end{array}$        (3)

Hence, the equation for slope is $\underset{\_}{\frac{dy}{dx}=\frac{1}{EI}\left(\frac{3wL{x}^2}{40}-\frac{w{L}^{2}x}{30}-\frac{w{x}^4}{24L}\right)}$.

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

$\begin{array}{l}EIy=\frac{wL{x}^3}{40}-\frac{w{L}^2{x}^2}{60}-\frac{w{x}^5}{120L}+\left(0\right)x+\left(0\right)\\ y=\frac{1}{EI}\left(\frac{wL{x}^3}{40}-\frac{w{L}^2{x}^2}{60}-\frac{w{x}^5}{120L}\right)\end{array}$        (4)

Hence, the equation for deflection is $\underset{\_}{y=\frac{1}{EI}\left(\frac{wL{x}^3}{40}-\frac{w{L}^2{x}^2}{60}-\frac{w{x}^5}{120L}\right)}$.

Calculate the deflection at B as shown below.

Substitute $\frac{L}{4}$ for x in Equation (3).

$\begin{array}{l}{y}_B=\frac{1}{EI}\left(\frac{wL}{40}{\left(\frac{L}{4}\right)}^3-\frac{w{L}^2}{60}{\left(\frac{L}{4}\right)}^2-\frac{w}{120L}{\left(\frac{L}{4}\right)}^5\right)\\ {\Delta }_B=\frac{w{L}^4}{120EI}\left(\frac{3}{64}-\frac{2}{16}-\frac{1}{1024}\right)\\ =\frac{w{L}^4}{120EI}\left(\frac{48-128-1}{1024}\right)\\ =-0.000659\frac{w{L}^4}{EI}\end{array}$

Hence, the deflection of the beam at B is $\underset{\_}{{\Delta }_B=-0.000659\frac{w{L}^4}{EI}}$.

Calculate the deflection at C as shown below.

Substitute $\frac{L}{2}$ for x in Equation (3).

$\begin{array}{l}{y}_C=\frac{1}{EI}\left(\frac{wL}{40}{\left(\frac{L}{2}\right)}^3-\frac{w{L}^2}{60}{\left(\frac{L}{2}\right)}^2-\frac{w}{120L}{\left(\frac{L}{2}\right)}^5\right)\\ {\Delta }_C=\frac{w{L}^4}{120EI}\left(\frac{3}{8}-\frac{2}{4}-\frac{1}{32}\right)\\ =\frac{w{L}^4}{120EI}\left(\frac{12-16-1}{32}\right)\\ =-0.001302\frac{w{L}^4}{EI}\end{array}$

Therefore, the deflection of the beam at C is $\underset{\_}{{\Delta }_C=-0.001302\frac{w{L}^4}{EI}}$.