Chapter 4, Problem 18P

To determine

The reaction force equation at point $A$.

The reaction force equation at point $C$.

The shear force equation for section $AB$.

The shear force equation for section $BC$.

The bending moment equation for section $AB$.

The bending moment equation for section $BC$.

The deflection equations for section $AB$.

The deflection equations for section $BC$.

Answer to Problem 18P

The reaction force equation at point $A$ is $R_1=\frac{wa}{2l}\left(2l-a\right)$.

The reaction force equation at point $C$ is $R_2=\frac{w{a}^2}{2l}$.

The shear force equation for section $AB$ is $V_{AB}=\frac{w}{2l}\left[2l\left(a-x\right)-a^2\right]$.

The shear force equation for section $BC$ is $V_{BC}=-\frac{w{a}^2}{2l}$.

The bending moment equation for section $AB$ is $M_{AB}=\frac{wx}{2l}\left(2al-a^2-lx\right)$.

The bending moment equation for section $BC$ is $M_{BC}=\frac{w{a}^2}{2l}\left(l-x\right)$.

The deflection equations for section $AB$ is $y_{AB}=\frac{wx}{24EIl}\left[2a{x}^2\left(2l-a\right)-l{x}^3-a^2{\left(2l-a\right)}^2\right]$.

The deflection equations for section $BC$ is $y_{BC}=y_{AB}+\frac{w}{24EI}{\left(x-a\right)}^4$.

Explanation of Solution

Write the balanced force equation in vertical direction.

    $R_1+R_2=wa$                                                                      (I)

Here, the reaction at point $A$ is $R_1$, the load on the beam is $w$, the length of the section $BC$ is $a$ and the reaction at point $B$ is $R_2$.

Take the net moment about point $C$.

    $\begin{array}{l}{R}_{1}l-wa\left(l-\frac{a}{2}\right)=0\\ R_{1}l=\frac{wa\left(2l-a\right)}{2}\\ R_1=\frac{wa}{2l}\left(2l-a\right)\end{array}$

Thus, the reaction force at point A is $\frac{wa}{2l}\left(2l-a\right)$.

Substitute $\frac{wa}{2l}\left(2l-a\right)$ for $R_1$ in Equation (I).

    $\begin{array}{l}\frac{wa}{2l}\left(2l-a\right)+R_2=wa\\ wa-\frac{w{a}^2}{2l}+R_2=wa\\ R_2=\frac{w{a}^2}{2l}\end{array}$

Thus, the reaction force at point $C$ is $\frac{w{a}^2}{2l}$.

Take a section at a distance $x$ from left end as shown in below figure.

Figure (1)

Write the shear force equation for part AB.

    $V_{AB}=R_1-wx$                                                                          (II)

Here, the shear force for the section $AB$ is $V_{AB}$.

Substitute $\frac{wa}{2l}\left(2l-a\right)$ for $R_1$ in Equation (II).

    $\begin{array}{c}{V}_{AB}=\frac{wa}{2l}\left(2l-a\right)-wx\\ =\frac{w}{2l}\left[\left(2la-a^2\right)-2lx\right]\\ =\frac{w}{2l}\left[2l\left(a-x\right)-a^2\right]\end{array}$

Thus, the shear force equation for region $AB$ is $\frac{w}{2l}\left[2l\left(a-x\right)-a^2\right]$.

Take a section at a distance $x$ from left end as shown in below figure.

Figure (2)

Write the shear force equation for part $BC$.

    $V_{BC}=R_1-wa$                                                                (III)

Here, the shear force for the section $BC$ is $V_{BC}$.

Substitute $\frac{wa}{2l}\left(2l-a\right)$ for $R_1$ in Equation (III).

    $\begin{array}{c}{V}_{BC}=\frac{wa}{2l}\left(2l-a\right)-wa\\ =wa-\frac{w{a}^2}{2l}-wa\\ =-\frac{w{a}^2}{2l}\end{array}$

Thus, the shear force equation for region $BC$ is $-\frac{w{a}^2}{2l}$.

Write the moment equation for section $AB$ using figure (1).

    $M_{AB}=R_{1}x-wx\cdot \frac{x}{2}$                                                           (IV)

Here, the moment for the section $AB$ is $M_{AB}$.

Substitute $\frac{wa}{2l}\left(2l-a\right)$ for $R_1$ in Equation (IV).

    $\begin{array}{c}{M}_{AB}=\frac{wa}{2l}\left(2l-a\right)\cdot x-wx\cdot \frac{x}{2}\\ =\frac{wx}{2l}\left(2al-a^2-lx\right)\end{array}$

Thus, the bending moment equation for region AB is $\frac{wx}{2l}\left(2al-a^2-lx\right)$.

Write the moment equation for section $BC$ using figure (2).

    $M_{BC}=R_{1}x-wa\cdot \left(x-\frac{a}{2}\right)$                                                                  (V)

Here, the moment for the section $BC$ is $M_{BC}$.

Substitute $\frac{wa}{2l}\left(2l-a\right)$ for $R_1$ in Equation (V).

    $\begin{array}{c}{M}_{BC}=\frac{wa}{2l}\left(2l-a\right)\cdot x-wa\cdot \left(x-\frac{a}{2}\right)\\ =\frac{w}{2l}\left(2alx-a^{2}x-2alx+a^{2}l\right)\\ =\frac{w}{2l}\left(-a^{2}x+a^{2}l\right)\\ =\frac{w{a}^2}{2l}\left(l-x\right)\end{array}$

Thus, the bending moment equation for region $BC$ is $\frac{w{a}^2}{2l}\left(l-x\right)$.

Write the bending moment equation for section $AB$ using figure (1).

    $EI\frac{d^2{y}_{AB}}{d{x}^2}=-M_{AB}$                                                                          (VI)

Here, Young’s modulus of the beam is $E$ and the moment of inertia of the beam is $I$.

Substitute $\frac{wx}{2l}\left(2al-a^2-lx\right)$ for $M_{AB}$ in Equation (VI).

    $EI\frac{d^2{y}_{AB}}{d{x}^2}=-\frac{wx}{2l}\left(2al-a^2-lx\right)$

Integrate Equation (VI).

    $EI\frac{d{y}_{AB}}{dx}=-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{x^2}{2}-l\cdot \frac{x^3}{3}\right]+C_1$                                     (VII)

Here, the integration constant is $C_1$.

Integrate Equation (VII).

    $EI{y}_{AB}=-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{x^3}{6}-l\cdot \frac{x^4}{12}\right]+C_{1}x+C_2$                                (VIII)

Here, the second integration constant is $C_2$.

Substitute $0$ for $x$ and $0$ for $y_{AB}$ in Equation (VIII).

    $\begin{array}{c}EI\cdot 0=-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot 0-l\cdot 0\right]+C_1\cdot 0+C_2\\ 0=-0+0+C_2\\ C_2=0\end{array}$

Substitute $0$ for $C_2$ in the Equation (VIII).

    $\begin{array}{c}EI{y}_{AB}=-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{x^3}{6}-l\cdot \frac{x^4}{12}\right]+C_{1}x+0\\ EI{y}_{AB}=-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{x^3}{6}-l\cdot \frac{x^4}{12}\right]+C_{1}x\end{array}$                            (IX)

Here, the deflection for the section $AB$ is $y_{AB}$.

Substitute $\frac{w{a}^2}{24l}\left(a^2-4al-4l^2\right)$ for $C_1$ and $0$ for $C_2$ in Equation (VIII).

    $\begin{array}{c}EI{y}_{AB}=-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{x^3}{6}-l\cdot \frac{x^4}{12}\right]+\frac{w{a}^2}{24l}\left(a^2-4al-4l^2\right)x+0\\ =\frac{w}{24l}\left[-4al{x}^3+2a^2{x}^3+l{x}^4+a^{4}x-4a^{3}lx-4a^2{l}^{2}x\right]\\ y_{AB}=\frac{wx}{24EIl}\left[-4al{x}^2+2a^2{x}^2+l{x}^3+a^4-4a^{3}l-4a^2{l}^2\right]\\ =\frac{wx}{24EIl}\left[2a{x}^2\left(2l-a\right)-l{x}^3-a^2{\left(2l-a\right)}^2\right]\end{array}$

Thus, the beam deflection equation for region AB is $y_{AB}=\frac{wx}{24EIl}\left[2a{x}^2\left(2l-a\right)-l{x}^3-a^2{\left(2l-a\right)}^2\right]$.

Write the bending moment equation for section $BC$ using figure (2).

    $EI\frac{d^2{y}_{BC}}{d{x}^2}=-M_{BC}$                                                                            (X)

Here, the moment for the section $BC$ is $M_{BC}$.

Substitute $\frac{w{a}^2}{2l}\left(l-x\right)$ for $M_{BC}$ in Equation (X)

    $EI\frac{d^2{y}_{BC}}{d{x}^2}=-\frac{w{a}^2}{2l}\left(l-x\right)$

Integrate the Equation (X).

    $EI\frac{d{y}_{BC}}{dx}=-\frac{w{a}^2}{2l}\left(lx-\frac{x^2}{2}\right)+C_3$                                                     (XI)

Here, the first integration constant is $C_3$.

Integrate the Equation (XI).

    $EI{y}_{BC}=-\frac{w{a}^2}{2l}\left(l\cdot \frac{x^2}{2}-\frac{x^3}{6}\right)+C_{3}x+C_4$                                             (XII)

Here, the second integration constant is $C_4$.

Substitute $l$ for $x$ and $0$ for $y_{BC}$ in Equation (XII).

    $\begin{array}{l}EI\cdot 0=-\frac{w{a}^2}{2l}\left(l\cdot \frac{l^2}{2}-\frac{l^3}{6}\right)+C_{3}l+C_4\\ C_{3}l+C_4=-\frac{w{a}^2{l}^2}{6}\\ C_4=-\frac{w{a}^2{l}^2}{6}-C_{3}l\end{array}$

Substitute $-\frac{w{a}^2{l}^2}{6}-C_{3}l$ for $C_4$ in the Equation (XII).

    $EI{y}_{BC}=-\frac{w{a}^2}{2l}\left(l\cdot \frac{x^2}{2}-\frac{x^3}{6}\right)+C_{3}x+\left(-\frac{w{a}^2{l}^2}{6}-C_{3}l\right)$                          (XIII)

At $x=a$, $y_{AB}=y_{BC}$ or $EI{y}_{AB}=EI{y}_{BC}$.

Equate the right hand side of equation (XI) and (XIII) and substitute $a$ for $x$.

    $\begin{array}{l}-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{x^3}{6}-l\cdot \frac{x^4}{12}\right]+C_{1}x=\left\{\begin{array}{l}-\frac{w{a}^2}{2l}\left(l\cdot \frac{x^2}{2}-\frac{x^3}{6}\right)+C_{3}x\\ +\left(-\frac{w{a}^2{l}^2}{6}-C_{3}l\right)\end{array}\right\}\\ -\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{a^3}{6}-l\cdot \frac{a^4}{12}\right]+C_{1}a=\left\{\begin{array}{l}-\frac{w{a}^2}{2l}\left(l\cdot \frac{a^2}{2}-\frac{a^3}{6}\right)+C_{3}a\\ +\left(-\frac{w{a}^2{l}^2}{6}-C_{3}l\right)\end{array}\right\}\\ C_{1}a=\left\{\begin{array}{l}\frac{w{a}^4}{6}-\frac{w{a}^5}{12l}-\frac{w{a}^4}{24}-\frac{w{a}^4}{4}\\ +\frac{w{a}^5}{12l}-\frac{w{a}^2{l}^2}{6}+C_3\left(a-l\right)\end{array}\right\}\\ C_{1}a=-\frac{w{a}^4}{8}-\frac{w{a}^2{l}^2}{6}+C_3\left(a-l\right)\end{array}$  (XIV)

At $x=a$,

    $\begin{array}{c}\frac{d{y}_{AB}}{dx}=\frac{d{y}_{BC}}{dx}\\ EI\frac{d{y}_{AB}}{dx}=EI\frac{d{y}_{BC}}{dx}\end{array}$                                                                        (XIV)

Substitute $-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{x^2}{2}-l\cdot \frac{x^3}{3}\right]+C_1$ for $EI\frac{d{y}_{AB}}{dx}$ and $-\frac{w{a}^2}{2l}\left(lx-\frac{x^2}{2}\right)+C_3$ for $EI\frac{d{y}_{BC}}{dx}$ and $a$ for $x$ in Equation (XIV)

    $\begin{array}{c}-\frac{w}{2l}\left[\left(2al-a^2\right)\cdot \frac{a^2}{2}-l\cdot \frac{a^3}{3}\right]+C_1=-\frac{w{a}^2}{2l}\left(la-\frac{a^2}{2}\right)+C_3\\ -\frac{w}{2l}\left[a^{3}l-\frac{a^4}{2}-\frac{a^{3}l}{3}\right]+C_1=-\frac{w}{2l}\left[a^{3}l-\frac{a^4}{2}\right]+C_3\\ -\frac{w}{2l}\cdot \frac{2a^{3}l}{3}+\frac{w{a}^4}{4l}+C_1=-\frac{w}{2l}\cdot a^{3}l+\frac{w{a}^4}{4l}+C_3\\ C_1+\frac{w{a}^3}{6}=C_3\end{array}$               (XV)

Substitute $C_1+\frac{w{a}^3}{6}$ for $C_3$ in Equation (XV).

    $\begin{array}{c}{C}_{1}a=-\frac{w{a}^4}{8}-\frac{w{a}^2{l}^2}{6}+\left(C_1+\frac{w{a}^3}{6}\right)\left(a-l\right)\\ C_{1}a=-\frac{w{a}^4}{8}-\frac{w{a}^2{l}^2}{6}+C_{1}a-C_{1}l+\frac{w{a}^4}{6}-\frac{w{a}^{3}l}{6}\\ C_{1}l=\frac{w{a}^4}{24}-\frac{w{a}^{3}l}{6}-\frac{w{a}^2{l}^2}{6}\\ C_1=\frac{w{a}^2}{24l}\left(a^2-4al-4l^2\right)\end{array}$                                   (XVI)

Substitute $\frac{w{a}^2}{24l}\left(a^2-4al-4l^2\right)$ for $C_1$ in Equation (XV).

    $\begin{array}{l}\frac{w{a}^2}{24l}\left(a^2-4al-4l^2\right)+\frac{w{a}^3}{6}=C_3\\ \frac{w{a}^4}{24l}-\frac{w{a}^3}{6}-\frac{w{a}^{2}l}{6}+\frac{w{a}^3}{6}=C_3\\ \frac{w{a}^2}{24l}\left(a^2-4l\right)=C_3\end{array}$

Substitute $\frac{w{a}^2}{24l}\left(a^2-4l\right)$ for $C_3$ and $0$ for $C_4$ in Equation (XII).

    $\begin{array}{c}EI{y}_{BC}=-\frac{w{a}^2}{2l}\left(l\cdot \frac{x^2}{2}-\frac{x^3}{6}\right)+C_{3}x+\left(-\frac{w{a}^2{l}^2}{6}-C_{3}l\right)\\ =-\frac{w{a}^2{x}^2}{4}+\frac{w{a}^2{x}^3}{12l}+\frac{w{a}^2}{24l}\left(a^2-4l\right)\left(x-l\right)-\frac{w{a}^2{l}^2}{6}\\ =\frac{w}{24}\left[-6a^2{x}^2+\frac{2a^2{x}^3}{l}+\frac{a^2}{l}\left(a^{2}x-a^{2}l-4lx+4l^2\right)-4a^2{l}^2\right]\\ =\frac{w}{24l}\left[2a^2{x}^3+a^{4}x-a^{4}l-4l{a}^{2}x+4l^2{a}^2\right]+\frac{w}{24}\left[-6a^2{x}^2-4a^2{l}^2\right]\end{array}$

Solve the equation further,

    $\begin{array}{c}{y}_{BC}=\frac{w}{24EIl}\left[2a^2{x}^3+a^{4}x-a^{4}l-4l{a}^{2}x+4l^2{a}^2\right]+\frac{w}{24EI}\left[-6a^2{x}^2-4a^2{l}^2\right]\\ =\frac{wx}{24EIl}\left[2a{x}^2\left(2l-a\right)-l{x}^3-a^2{\left(2l-a\right)}^2\right]+\frac{w}{24EI}{\left(x-a\right)}^4\end{array}$ (XVII)

Substitute $\frac{wx}{24EIl}\left[2a{x}^2\left(2l-a\right)-l{x}^3-a^2{\left(2l-a\right)}^2\right]$ for $y_{AB}$ in (XVII).

    $y_{BC}=y_{AB}+\frac{w}{24EI}{\left(x-a\right)}^4$

Thus, the beam deflection equation for region BC is $y_{BC}=y_{AB}+\frac{w}{24EI}{\left(x-a\right)}^4$.