Chapter 5, Problem 155RP

For the beam and loading shown, determine the equations of the shear and bending-moment curves and the maximum absolute value of the bending moment in the beam, knowing that (a) k = 1, (b) k = 0.5.

Fig. P5.155

To determine

The equations of the shear and bending moment curves and the maximum absolute value of the bending moment when $k=1$.

Answer to Problem 155RP

The equation of the shear force curve is $\underset{\_}{V=-\frac{w_0{x}^2}{L}+w_{0}x}$.

The equation of the bending moment curve is $\underset{\_}{M=-\frac{w_0{x}^3}{3L}+\frac{w_0{x}^2}{2}}$.

The maximum absolute bending moment in the beam is $\underset{\_}{M_{\max }=\frac{w_0{L}^2}{6}}$.

Explanation of Solution

Find the load function $w\left(x\right)$ as follows;

$\begin{array}{c}w\left(x\right)=\frac{w_{0}x}{L}-\frac{k{w}_0\left(L-x\right)}{L}\\ =\frac{w_{0}x}{L}-k{w}_0+\frac{k{w}_{0}x}{L}\end{array}$

By definition;

$\frac{dV}{dx}=-w\left(x\right)$

Substitute $\left(\frac{w_{0}x}{L}-k{w}_0+\frac{k{w}_{0}x}{L}\right)$ for $w\left(x\right)$.

$\frac{dV}{dx}=-\frac{w_{0}x}{L}+k{w}_0-\frac{k{w}_{0}x}{L}$

Integrate the equation to find V.

$V=-\frac{w_0{x}^2}{2L}+k{w}_{0}x-\frac{k{w}_0{x}^2}{2L}+C_1$ (1)

By definition;

$\frac{dM}{dx}=V$

Substitute $\left(-\frac{w_0{x}^2}{2L}+k{w}_{0}x-\frac{k{w}_0{x}^2}{2L}+C_1\right)$ for V.

$\frac{dM}{dx}=-\frac{w_0{x}^2}{2L}+k{w}_{0}x-\frac{k{w}_0{x}^2}{2L}+C_1$

Integrate the equation to find M.

$M=-\frac{w_0{x}^3}{6L}+\frac{k{w}_0{x}^2}{2}-\frac{k{w}_0{x}^3}{6L}+C_{1}x+C_2$ (2)

Apply the boundary condition; $M=0\text{at}x=0$.

Substitute 0 for M and 0 for x in Equation (2).

$\begin{array}{l}0=-\frac{w_0{\left(0\right)}^3}{6L}+\frac{k{w}_0{\left(0\right)}^2}{2}-\frac{k{w}_0{\left(0\right)}^3}{6L}+C_1\left(0\right)+C_2\\ 0=-0+0-0+0+C_2\\ C_2=0\end{array}$

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

$\begin{array}{c}M=-\frac{w_0{x}^3}{6L}+\frac{k{w}_0{x}^2}{2}-\frac{k{w}_0{x}^3}{6L}+C_{1}x+0\\ =-\frac{w_0{x}^3}{6L}+\frac{k{w}_0{x}^2}{2}-\frac{k{w}_0{x}^3}{6L}+C_{1}x\end{array}$ (3)

Apply the boundary condition; $V=0\text{at}x=0$.

Substitute 0 for V and 0 for x in Equation (1).

$\begin{array}{l}0=-\frac{w_0{\left(0\right)}^2}{2L}+k{w}_0\left(0\right)-\frac{k{w}_0{\left(0\right)}^2}{2L}+C_1\\ 0=-0+0-0+C_1\\ C_1=0\end{array}$

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

$\begin{array}{c}M=-\frac{w_0{x}^3}{6L}+\frac{k{w}_0{x}^2}{2}-\frac{k{w}_0{x}^3}{6L}+\left(0\right)x\\ =-\frac{w_0{x}^3}{6L}+\frac{k{w}_0{x}^2}{2}-\frac{k{w}_0{x}^3}{6L}\end{array}$ (4)

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

$\begin{array}{c}V=-\frac{w_0{x}^2}{2L}+k{w}_{0}x-\frac{k{w}_0{x}^2}{2L}+\left(0\right)\\ =-\frac{w_0{x}^2}{2L}+k{w}_{0}x-\frac{k{w}_0{x}^2}{2L}\end{array}$ (5)

When $k=1$;

Substitute 1 for k in Equation (4).

$\begin{array}{c}M=-\frac{w_0{x}^3}{6L}+\frac{\left(1\right)w_0{x}^2}{2}-\frac{\left(1\right)w_0{x}^3}{6L}\\ =-\frac{w_0{x}^3}{6L}+\frac{w_0{x}^2}{2}-\frac{w_0{x}^3}{6L}\\ =-\frac{w_0{x}^3}{3L}+\frac{w_0{x}^2}{2}\end{array}$ (6)

Substitute 1 for k in Equation (5).

$\begin{array}{c}V=-\frac{w_0{x}^2}{2L}+\left(1\right)w_{0}x-\frac{\left(1\right)w_0{x}^2}{2L}\\ =-\frac{w_0{x}^2}{2L}+w_{0}x-\frac{w_0{x}^2}{2L}\\ =-\frac{w_0{x}^2}{L}+w_{0}x\end{array}$

Maximum bending moment;

The maximum bending moment occurs at the fixed end. $x=L$.

Substitute L for x in Equation (6).

$\begin{array}{c}{M}_{\max }=-\frac{w_0{\left(L\right)}^3}{3L}+\frac{w_0{\left(L\right)}^2}{2}\\ =-\frac{w_0{L}^2}{3}+\frac{w_0{L}^2}{2}\\ =-\frac{2w_0{L}^2}{6}+\frac{3w_0{L}^2}{6}\\ =\frac{w_0{L}^2}{6}\end{array}$

Therefore,

The equation of the shear force curve is $\underset{\_}{V=-\frac{w_0{x}^2}{L}+w_{0}x}$.

The equation of the bending moment curve is $\underset{\_}{M=-\frac{w_0{x}^3}{3L}+\frac{w_0{x}^2}{2}}$.

The maximum absolute bending moment in the beam is $\underset{\_}{M_{\max }=\frac{w_0{L}^2}{6}}$.

To determine

The maximum absolute value of the bending moment when $k=0.5$.

Answer to Problem 155RP

The equation of the shear force curve is $\underset{\_}{V=-\frac{3w_0{x}^2}{4L}+\frac{w_{0}x}{2}}$.

The equation of the bending moment curve is $\underset{\_}{M=-\frac{w_0{x}^3}{4L}+\frac{w_0{x}^2}{4}}$.

The maximum absolute bending moment in the beam is $\underset{\_}{M_{\max }=\frac{w_0{L}^2}{27}}$.

Explanation of Solution

When $k=0.5$;

Substitute 0.5 for k in Equation (4).

$\begin{array}{c}M=-\frac{w_0{x}^3}{6L}+\frac{\left(0.5\right)w_0{x}^2}{2}-\frac{\left(0.5\right)w_0{x}^3}{6L}\\ =-\frac{w_0{x}^3}{6L}+\frac{w_0{x}^2}{4}-\frac{w_0{x}^3}{12L}\\ =-\frac{3w_0{x}^3}{12L}+\frac{w_0{x}^2}{4}\\ =-\frac{w_0{x}^3}{4L}+\frac{w_0{x}^2}{4}\end{array}$ (7)

Substitute 0.5 for k in Equation (5).

$\begin{array}{c}V=-\frac{w_0{x}^2}{2L}+\left(0.5\right)w_{0}x-\frac{\left(0.5\right)w_0{x}^2}{2L}\\ =-\frac{w_0{x}^2}{2L}+\frac{w_{0}x}{2}-\frac{w_0{x}^2}{4L}\\ =-\frac{3w_0{x}^2}{4L}+\frac{w_{0}x}{2}\end{array}$ (8)

The maximum bending moment occurs where the shear force changes sign. i.e., $V=0$.

Substitute 0 for V in Equation (8).

$\begin{array}{l}0=-\frac{3w_0{x}^2}{4L}+\frac{w_{0}x}{2}\\ \frac{3x}{2L}=1\\ x=\frac{2L}{3}\end{array}$

Substitute $\frac{2L}{3}$ for x in Equation (7).

$\begin{array}{c}{M}_{\max }=-\frac{w_0{\left(\frac{2L}{3}\right)}^3}{4L}+\frac{w_0{\left(\frac{2L}{3}\right)}^2}{4}\\ =-\frac{8w_0{L}^3}{108L}+\frac{4w_0{L}^2}{36}\\ =\frac{-8w_0{L}^2+12w_0{L}^2}{108}\\ =\frac{w_0{L}^2}{27}\end{array}$

Therefore,

The equation of the shear force curve is $\underset{\_}{V=-\frac{3w_0{x}^2}{4L}+\frac{w_{0}x}{2}}$.

The equation of the bending moment curve is $\underset{\_}{M=-\frac{w_0{x}^3}{4L}+\frac{w_0{x}^2}{4}}$.

The maximum absolute bending moment in the beam is $\underset{\_}{M_{\max }=\frac{w_0{L}^2}{27}}$.