Chapter 6, Problem 3P

To determine

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

Answer to Problem 3P

The equation for slope for segment AB is $\underset{\_}{\frac{wx}{2EI}\left(a^2-L^2+\left(L-a\right)x\right)}$.

The equation for deflection for segment AB is $\underset{\_}{\frac{w{x}^2}{2EI}\left(\frac{a^2-L^2}{2}+\frac{\left(L-a\right)x}{3}\right)}$.

The equation for slope for segment BC is $\underset{\_}{\frac{w}{2EI}\left(xL\left(x-L\right)-\frac{x^3}{3}+\frac{a^3}{3}\right)}$.

The equation for deflection for segment BC is $\underset{\_}{\frac{w}{2EI}\left(x^{2}L\left(\frac{x}{3}-\frac{L}{2}\right)-\frac{x^4}{12}-\frac{a^4}{12}+\frac{a^{3}x}{3}\right)}$.

Explanation of Solution

Calculation:

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

Refer Figure (1),

Find the reaction at support A.

Apply vertical equilibrium along y-axis.

Consider upward force as positive.

$\begin{array}{l}\Sigma F_y=0\\ A_y-w\left(L-a\right)=0\\ A_y=w\left(L-a\right)\end{array}$

Find the moment at A.

Consider anticlockwise moment as positive.

$\begin{array}{l}0=M_A-w\times \left(L-a\right)\times \left[\frac{\left(L-a\right)}{2}+a\right]\\ M_A=w\left(L-a\right)\left[\left(\frac{L-a+2a}{2}\right)\right]\\ M_A=w\left(L-a\right)\left(\frac{L+a}{2}\right)\\ M_A=\frac{w}{2}\left(L^2-a^2\right)\end{array}$

Segment AB: $0\le x\le a$.

Consider a section $X_1{X}_1$ in the segment AB at a distance of x from A.

Sketch the free body diagram when section $X_1{X}_1$ consider in the segment AB as shown in Figure 2.

Refer Figure (2),

Take the moment at section $X_1{X}_1$ .

$M=-\frac{w}{2}\left(L^2-a^2\right)+w\times \left(L-a\right)\left(x\right)$

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

$\frac{d^{2}y}{d{x}^2}=\frac{M}{EI}=\frac{1}{EI}\left[-\frac{w}{2}\left(L^2-a^2\right)+w\left(L-a\right)x\right]$        (1)

Find the equation for slope $\left(\theta \right)$.

Integrate Equation (1) with respect to x.

$\begin{array}{l}\frac{dy}{dx}=\theta ={\displaystyle \int \frac{M}{EI}dx}\\ \theta ={\displaystyle \int \frac{-\frac{w}{2}\left(L^2-a^2\right)+w\times \left(L-a\right)\left(x\right)}{EI}dx}\\ EI\theta =-\frac{w}{2}\left(L^2-a^2\right)x+w\times \left(L-a\right)\left(\frac{x^2}{2}\right)+C_1\end{array}$        (2)

Find the equation for deflection $\left(\theta \right)$.

Integrate again Equation (2) with respect to x.

$EIy=-\frac{w}{2}\left(L^2-a^2\right)\frac{x^2}{2}+w\times \left(L-a\right)\left(\frac{x^3}{6}\right)+C_{1}x+C_2$        (3)

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

Apply boundary conditions in Equation (2):

At $x=0$ and $y=0$.

$\begin{array}{l}0=\frac{M\times 0^2}{2EI}+C_1\times 0+C_2\\ C_2=0\end{array}$

Apply boundary conditions in Equation (1):

At $x=0$ and $\theta =0$.

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

Find the equation for slope of segment AB.

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

$\begin{array}{l}EI\theta =-\frac{w}{2}\left(L^2-a^2\right)x+w\times \left(L-a\right)\left(\frac{x^2}{2}\right)+0\\ EI\theta =-\frac{wx}{2}\left(L^2-a^2\right)+\frac{w{x}^2}{2}\left(L-a\right)\\ \theta =\frac{wx}{2EI}\left[a^2-L^2+\left(L-a\right)x\right]\end{array}$

Thus, the equation for slope of segment AB is $\underset{\_}{\frac{wx}{2EI}\left[a^2-L^2+\left(L-a\right)x\right]}$.

Find the equation for deflection of segment AB.

Substitute 0 for $C_1$ and 0 for $C_2$ in Equation (3).

$\begin{array}{c}EIy=-\frac{w}{2}\left(L^2-a^2\right)\frac{x^2}{2}+w\times \left(L-a\right)\left(\frac{x^3}{6}\right)+0\times x+0\\ EIy=-\frac{w}{2}\left(L^2-a^2\right)\frac{x^2}{2}+w\times \left(L-a\right)\left(\frac{x^3}{6}\right)\\ EIy=-\frac{w{x}^2}{2}\left(\frac{L^2-a^2}{2}\right)+\frac{w{x}^3}{6}\left(L-a\right)\\ y=\frac{w{x}^2}{2EI}\left(\frac{a^2-L^2}{2}+\frac{\left(L-a\right)x}{3}\right)\end{array}$

Thus, the equation for deflection is $\underset{\_}{\frac{w{x}^2}{2EI}\left(\frac{a^2-L^2}{2}+\frac{\left(L-a\right)x}{3}\right)}$.

Segment BC: $a\le x\le L$.

Consider a section $X_2{X}_2$ in the segment BC at a distance of x from A.

Sketch the free body diagram when section $X_2{X}_2$ consider segment BC as shown in Figure 3.

Refer Figure 3.

Write the equation for bending moment at section $X_2{X}_2$.

$M=-\frac{w}{2}\left(L^2-a^2\right)+w\times \left(L-a\right)\left(x\right)-\frac{w}{2}{\left(x-a\right)}^2$

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

$\frac{d^{2}y}{d{x}^2}=\frac{M}{EI}=\frac{1}{EI}\left[-\frac{w}{2}\left(L^2-a^2\right)+w\times \left(L-a\right)\left(x\right)-\frac{w}{2}{\left(x-a\right)}^2\right]$        (4)

Write the equation for slope.

Integrate Equation (4) with respect to x.

$\begin{array}{c}\frac{dy}{dx}=\theta ={\displaystyle \int \frac{M}{EI}dx}\\ \theta ={\displaystyle \int \frac{\left[-\frac{w}{2}\left(L^2-a^2\right)+w\times \left(L-a\right)\left(x\right)-\frac{w}{2}{\left(x-a\right)}^2\right]}{EI}dx}\\ EI\theta =-\frac{w}{2}\left(L^2-a^2\right)x+w\times \left(L-a\right)\left(\frac{x^2}{2}\right)-\frac{w}{2}\left(\frac{x^3}{3}+a^{2}x-a{x}^2\right)+C_3\end{array}$        (5)

Write the equation for deflection.

Integrate Equation (5) with respect to x.

$EIy=-\frac{w}{2}\left(L^2-a^2\right)\frac{x^2}{2}+w\times \left(L-a\right)\left(\frac{x^3}{6}\right)-\frac{w}{2}\left(\frac{x^4}{12}+\frac{a^2{x}^2}{2}-\frac{a{x}^3}{3}\right)+C_{3}x+C_4$        (6)

Find the integration constants $C_3\text{and}{C}_4$:

Apply boundary conditions in Equation (3):

At $x=0$, $\left(L-a\right)=0$, and ${\theta }_{B,\text{Left}}={\theta }_{B,Right}$.

$\begin{array}{l}EI\theta =-\frac{w}{2}\left(\frac{a^3}{3}+a^2\times a-a\times a^2\right)+C_3\\ 0=-\frac{w}{2}\left(\frac{a^3}{3}+a^3-a^3\right)+C_3\\ 0=-\frac{w{a}^3}{6}+C_3\\ C_3=\frac{w{a}^3}{6}\end{array}$

Apply boundary conditions in Equation (4):

At $x=0$, $\left(L-a\right)=0$, and $y_{B,\text{Left}}=y_{B,Right}$.

$\begin{array}{l}0=-\frac{w}{2}\left(\frac{a^4}{12}+\frac{a^2{a}^2}{2}-\frac{a\times a^3}{3}\right)+\frac{w{a}^3}{6}\times a+C_4\\ 0=-\frac{w}{2}\left(\frac{a^4+6a^4-4a^4}{12}\right)+\frac{w{a}^4}{6}+C_4\\ 0=-\frac{w}{2}\left(\frac{3a^4}{12}\right)+\frac{w{a}^4}{6}+C_4\\ 0=-\frac{w{a}^4}{8}+\frac{w{a}^4}{6}+C_4\end{array}$

$\begin{array}{l}0=\frac{-6w{a}^4+8w{a}^4}{48}+C_4\\ 0=\frac{w{a}^4}{24}+C_4\\ C_4=-\frac{w{a}^4}{24}\end{array}$

Find the equation for slope of segment BC.

Substitute $\frac{w{a}^3}{6}$ for $C_3$ in Equation (5).

$\begin{array}{c}EI\theta =-\frac{w}{2}\left(L^2-a^2\right)x+w\times \left(L-a\right)\left(\frac{x^2}{2}\right)-\frac{w}{2}\left(\frac{x^3}{3}+a^{2}x-a{x}^2\right)+\frac{w{a}^3}{6}\\ EI\theta =-\frac{w}{2}\left(L^{2}x-a^{2}x\right)+\frac{w}{2}\left(L{x}^2-a{x}^2\right)-\frac{w}{2}\left(\frac{x^3}{3}+a^{2}x-a{x}^2\right)+\frac{w{a}^3}{6}\\ EI\theta =-\frac{w}{2}\left(L^{2}x\right)+\frac{w}{2}\left(L{x}^2\right)-\frac{w}{2}\left(\frac{x^3}{3}\right)+\frac{w{a}^3}{6}\\ EI\theta =\frac{w}{2}\left[L^{2}x+L{x}^2-\frac{x^3}{3}+\frac{a^3}{3}\right]\end{array}$

$\theta =\frac{w}{2EI}\left[xL\left(x-L\right)-\frac{x^3}{3}+\frac{a^3}{3}\right]$

Thus, the equation for slope of segment BC is $\underset{\_}{\frac{w}{2EI}\left[xL\left(x-L\right)-\frac{x^3}{3}+\frac{a^3}{3}\right]}$.

Find the equation for deflection of segment BC.

Substitute $\frac{w{a}^3}{6}$ for $C_3$ and $-\frac{w{a}^4}{24}$ for $C_4$ in Equation (6).

$\begin{array}{l}EIy=-\frac{w}{2}\left(L^2-a^2\right)\frac{x^2}{2}+w\times \left(L-a\right)\left(\frac{x^3}{6}\right)-\frac{w}{2}\left(\frac{x^4}{12}+\frac{a^2{x}^2}{2}-\frac{a{x}^3}{3}\right)+\frac{w{a}^3}{6}x-\frac{w{a}^4}{24}\\ EIy=-\frac{w}{2}\left(\frac{L^2{x}^2}{2}-\frac{a^2{x}^2}{2}\right)+w\left(\frac{L{x}^3}{6}-\frac{a{x}^3}{6}\right)-\frac{w}{2}\left(\frac{x^4}{12}+\frac{a^2{x}^2}{2}-\frac{a{x}^3}{3}\right)+\frac{wx{a}^3}{6}-\frac{w{a}^4}{24}\\ EIy=\left\{\begin{array}{l}\frac{w}{2}\left(-\frac{L^2{x}^2}{2}\right)+\frac{w}{2}\left(\frac{a^2{x}^2}{2}\right)+\frac{w}{2}\left(\frac{L{x}^3}{3}-\frac{a{x}^3}{3}\right)+\frac{w}{2}\left(-\frac{x^4}{12}+\frac{a{x}^3}{3}\right)+\\ \frac{w}{2}\left(-\frac{a^2{x}^2}{2}\right)+\frac{wx{a}^3}{6}-\frac{w{a}^4}{24}\end{array}\right\}\\ y=\frac{w}{2}\left(-\frac{L^2{x}^2}{2}+\frac{L{x}^3}{3}-\frac{a{x}^3}{3}-\frac{x^4}{12}+\frac{a{x}^3}{3}+\frac{x{a}^3}{3}-\frac{a^4}{12}\right)\end{array}$$\begin{array}{c}y=\frac{w}{2}\left[x^{2}L\left(\frac{x}{3}-\frac{L}{2}\right)-\frac{x^4}{12}-\frac{a^4}{12}+\frac{x{a}^3}{3}\right]\\ =\frac{w}{2EI}\left[x^{2}L\left(\frac{x}{3}-\frac{L}{2}\right)-\frac{x^4}{12}-\frac{a^4}{12}+\frac{x{a}^3}{3}\right]\end{array}$

Thus, the equation for deflection of segment BC is $\underset{\_}{\frac{w}{2EI}\left[x^{2}L\left(\frac{x}{3}-\frac{L}{2}\right)-\frac{x^4}{12}-\frac{a^4}{12}+\frac{x{a}^3}{3}\right]}$.