Chapter 9, Problem 14P

To determine

Find maximum positive shear and bending moment at point B due to the series of four moving concentrated loads

Explanation of Solution

Calculation:

Apply a 1 k unit moving load at a distance of x from left end A.

Sketch the free body diagram of beam as shown in Figure 1.

Refer Figure 1.

Find the equation of support reaction $\left(A_y\right)$ at A using equilibrium equation:

Take moment about point C.

Consider moment equilibrium at point C.

Consider clockwise moment as positive and anticlockwise moment as negative.

Sum of moment at point C is zero.

$\begin{array}{l}\Sigma M_C=0\\ A_y\left(30\right)-1\left(30-x\right)=0\\ A_y\left(30\right)-30+x=0\\ 30A_y=30-x\end{array}$

$A_y=1-\frac{x}{30}$        (1)

Find the equation of support reaction $\left(C_y\right)$ at C using equilibrium equation:

Apply vertical equilibrium equation of forces.

Consider upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.

$A_y+C_y=1$

Substitute $1-\frac{x}{30}$ for $A_y$.

$\begin{array}{l}1-\frac{x}{30}+C_y=1\\ C_y=1-1+\frac{x}{30}\\ C_y=\frac{x}{30}\end{array}$        (2)

Influence line for shear at B.

Find the equation of shear force at B of portion AB $\left(0\le x<15\text{ft}\right)$.

Sketch the free body diagram of the section AB as shown in Figure 2.

Refer Figure 2.

Apply equilibrium equation of forces.

Consider upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.

$\begin{array}{l}\Sigma F_y=0\\ A_y-S_B-1=0\\ S_B=A_y-1\end{array}$

Substitute $1-\frac{x}{30}$ for $A_y$.

$\begin{array}{c}{S}_B=1-\frac{x}{30}-1\\ =-\frac{x}{30}\end{array}$

Find the equation of shear force at B of portion BC $\left(15\text{ft}<x\le 30\text{ft}\right)$.

Sketch the free body diagram of the section BC as shown in Figure 3.

Refer Figure 3.

Apply equilibrium equation of forces.

Consider upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.

$\Sigma F_y=0$

$\begin{array}{l}{S}_B-1+C_y=0\\ S_B=1-C_y\end{array}$

Substitute $\frac{x}{30}$ for $C_y$.

$S_B=1-\frac{x}{30}$

Thus, the equations of the influence line for $S_B$ are,

$S_B=-\frac{x}{30}0\le x<15\text{ft}$        (3)

$S_B=1-\frac{x}{30}15\text{ft}<x\le 30\text{ft}$        (4)

Find the value of influence line ordinate of shear force at various points of x using the Equations (3) and (4) and summarize the value as in Table 1.

x $\left(\text{ft}\right)$	$S_B\left(\text{k/k}\right)$
0	1
${15}^{-}$	$-\frac{1}{2}$
${15}^{+}$	$\frac{1}{2}$
30	0

Draw the influence lines for the shear force at point B using Table 1 as shown in Figure 4.

Refer Figure 4.

Find the slope $\left({\theta }_{AB}\right)$ of influence line diagram of shear at B in the portion AB.

${\theta }_{AB}=\frac{-\frac{1}{2}}{L_{AB}}$

Here, $L_{AB}$ is the length of the beam from A to B.

Substitute 15 ft for $L_{AB}$.

$\begin{array}{c}{\theta }_{AB}=\frac{-\frac{1}{2}}{15}\\ =-\frac{1}{30}\end{array}$

Find the slope $\left({\theta }_{BC}\right)$ of influence line diagram of shear at B in the portion BC.

${\theta }_{BC}=\frac{\frac{1}{2}}{L_{BC}}$

Here, $L_{BC}$ is the length of the beam from B to C.

Substitute 15 ft for $L_{BC}$.

$\begin{array}{c}{\theta }_{BC}=\frac{\frac{1}{2}}{15}\\ =\frac{1}{30}\end{array}$

Sketch the loading position as shown in Figure 5.

Find the maximum positive shear force at B.

Sketch the loading position on the beam when the load 1 placed at just right of B as shown in Figure 6.

Refer Figure 6.

Find the shear force at B when the load 1 placed at just right of B.

${\left(S_B\right)}_1=10\left(L_{BC}\right)\left({\theta }_{BC}\right)+20\left(0\right)+20\left(0\right)+5\left(0\right)$

Substitute 15 ft for $L_{BC}$ and $\frac{1}{30}$ for ${\theta }_{BC}$.

$\begin{array}{c}{\left(S_B\right)}_1=10\left(15\right)\left(\frac{1}{30}\right)+20\left(0\right)+20\left(0\right)+5\left(0\right)\\ =5\text{k}\end{array}$

Sketch the loading position on the beam when the load 2 placed at just right of B as shown in Figure 7.

Refer Figure 7.

Find the shear force at B when the load 2 placed at just right of B.

${\left(S_B\right)}_2=10\left(0\right)+20\left(L_{BC}\right)\left({\theta }_{BC}\right)+20\left(L_{BC}-10\right)\left({\theta }_{BC}\right)+5\left(0\right)$

Substitute 15 ft for $L_{BC}$ and $\frac{1}{30}$ for ${\theta }_{BC}$.

$\begin{array}{c}{\left(S_B\right)}_2=10\left(0\right)+20\left(15\right)\left(\frac{1}{30}\right)+20\left(15-10\right)\left(\frac{1}{30}\right)+5\left(0\right)\\ =0+10+3.33+0\\ =13.33\text{k}\end{array}$

Sketch the loading position on the beam when the load 3 placed at just right of B as shown in Figure 8.

Refer Figure 8.

Find the shear force at B when the load 3 placed at just right of B.

${\left(S_B\right)}_3=10\left(0\right)+20\left(L_{AB}-10\right)\left({\theta }_{AB}\right)+20\left(L_{BC}\right)\left({\theta }_{BC}\right)+5\left(L_{BC}-10\right)\left({\theta }_{BC}\right)$

Substitute 15 ft for $L_{AB}$, $-\frac{1}{30}$ for ${\theta }_{AB}$, 15 ft for $L_{BC}$, and $\frac{1}{30}$ for ${\theta }_{BC}$.

$\begin{array}{c}{\left(S_B\right)}_3=10\left(0\right)+20\left(15-10\right)\left(-\frac{1}{30}\right)+20\left(15\right)\left(\frac{1}{30}\right)+5\left(15-10\right)\left(\frac{1}{30}\right)\\ =0-3.33+10+0.83\\ =7.5\text{k}\end{array}$

Sketch the loading position on the beam when the load 4 placed at just right of B as shown in Figure 9.

Refer Figure 9.

Find the shear force at B when the load 3 placed at just right of B.

${\left(S_B\right)}_4=10\left(0\right)+20\left(0\right)+20\left(L_{AB}-10\right)\left({\theta }_{AB}\right)+5\left(L_{BC}\right)\left({\theta }_{BC}\right)$

Substitute 15 ft for $L_{AB}$, $-\frac{1}{30}$ for ${\theta }_{AB}$, 15 ft for $L_{BC}$, and $\frac{1}{30}$ for ${\theta }_{BC}$.

$\begin{array}{c}{\left(S_B\right)}_4=10\left(0\right)+20\left(0\right)+20\left(15-10\right)\left(-\frac{1}{30}\right)+5\left(15\right)\left(\frac{1}{30}\right)\\ =0+0-3.33+2.5\\ =-0.83\text{k}\end{array}$

Maximum positive shear force at B as follows.

The maximum positive shear at B is the maximum of ${\left(S_B\right)}_1$, ${\left(S_B\right)}_2$, ${\left(S_B\right)}_3$, and ${\left(S_B\right)}_4$.

Therefore, the maximum positive shear at point B is $\underset{\_}{13.33\text{k}}$.

Influence line for moment at B.

Refer Figure 2.

Consider clockwise moment as positive and anticlockwise moment as negative.

Find the equation of moment at B of portion AB $\left(0\le x<15\text{ft}\right)$.

$M_B=A_y\left(15\right)-\left(1\right)\left(15-x\right)$

Substitute $1-\frac{x}{30}$ for $A_y$.

$\begin{array}{c}{M}_B=\left(1-\frac{x}{30}\right)\left(15\right)-\left(1\right)\left(15-x\right)\\ =15-\frac{x}{2}-15+x\\ =\frac{x}{2}\end{array}$

Refer Figure 3.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at B of portion BC $\left(15\text{ft}<x\le 30\text{ft}\right)$.

$\begin{array}{c}{M}_B=C_y\left(15\right)-\left(1\right)\left[15-\left(30-x\right)\right]\\ =15C_y-15+\left(30-x\right)\\ =15C_y+15-x\end{array}$

Substitute $\frac{x}{30}$ for $C_y$.

$\begin{array}{c}{M}_B=15\left(\frac{x}{30}\right)+15-x\\ =\frac{x}{2}+15-x\\ =15-\frac{x}{2}\end{array}$

Thus, the equations of the influence line for $M_B$ are,

$M_B=\frac{x}{2}0\le x<15\text{ft}$        (5)

$M_B=15-\frac{x}{2}15\text{ft}<x\le 30\text{ft}$        (6)

Find the value of influence line ordinate of moment at various points of x using the Equations (5) and (6) and summarize the value as in Table 2.

x $\left(\text{ft}\right)$	$M_B$ $\left(\text{k-ft/k}\right)$
0	0
$15$	$7$.5
30	0

Draw the influence lines for the moment at point B using Table 2 as shown in Figure 10.

Refer Figure 9.

The slope of portion AB and BC is same.

Find the slope $\left(\varphi \right)$ of influence line diagram of moment at B.

$\varphi =\frac{7.5}{L_{AB}}$

Here, $L_{AB}$ is the length of the beam from A to B.

Substitute 15 ft for $L_{AB}$.

$\begin{array}{c}{\varphi }_{AB}=\frac{7.5}{15}\\ =\frac{1}{2}\end{array}$

Find the maximum positive bending moment at B.

Refer Figure 6.

Find the bending moment at B when the load 1 placed at just right of B.

${\left(M_B\right)}_1=10\left(L_{BC}\right)\left(\varphi \right)+20\left(0\right)+20\left(0\right)+5\left(0\right)$

Substitute 15 ft for $L_{BC}$ and $\frac{1}{2}$ for $\varphi$.

$\begin{array}{c}{\left(M_B\right)}_1=10\left(15\right)\left(\frac{1}{2}\right)+20\left(0\right)+20\left(0\right)+5\left(0\right)\\ =75\text{k-ft}\end{array}$

Refer Figure 7.

Find the bending moment at B when the load 2 placed at just right of B.

${\left(M_B\right)}_2=10\left(0\right)+20\left(L_{BC}\right)\left(\varphi \right)+20\left(L_{BC}-10\right)\left(\varphi \right)+5\left(0\right)$

Substitute 15 ft for $L_{BC}$ and $\frac{1}{2}$ for $\varphi$.

$\begin{array}{c}{\left(M_B\right)}_2=10\left(0\right)+20\left(15\right)\left(\frac{1}{2}\right)+20\left(15-10\right)\left(\frac{1}{2}\right)+5\left(0\right)\\ =0+150+50\\ =200\text{k-ft}\end{array}$

Refer Figure 8.

Find the bending moment at B when the load 3 placed at just right of B.

${\left(M_B\right)}_3=10\left(0\right)+20\left(L_{AB}-10\right)\left(\varphi \right)+20\left(L_{BC}\right)\left(\varphi \right)+5\left(L_{BC}-10\right)\left(\varphi \right)$

Substitute 15 ft for $L_{AB}$, $\frac{1}{2}$ for $\varphi$, and 15 ft for $L_{BC}$.

$\begin{array}{c}{\left(M_B\right)}_3=10\left(0\right)+20\left(15-10\right)\left(\frac{1}{2}\right)+20\left(15\right)\left(\frac{1}{2}\right)+5\left(15-10\right)\left(\frac{1}{2}\right)\\ =0+50+150+12.5\\ =212.5\text{k-ft}\end{array}$

Refer Figure 9.

Find the bending moment at B when the load 1 placed at just right of B.

${\left(M_B\right)}_4=10\left(0\right)+20\left(0\right)+20\left(L_{AB}-10\right)\left(\varphi \right)+5\left(L_{BC}\right)\left(\varphi \right)$

Substitute 15 ft for $L_{AB}$, $\frac{1}{2}$ for $\varphi$, and 15 ft for $L_{BC}$.

$\begin{array}{c}{\left(M_B\right)}_4=10\left(0\right)+20\left(0\right)+20\left(15-10\right)\left(\frac{1}{2}\right)+5\left(15\right)\left(\frac{1}{2}\right)\\ =0+0+50+37.5\\ =87.5\text{k-ft}\end{array}$

Maximum positive bending moment at B as follows.

The maximum positive bending moment at B is the maximum of ${\left(M_B\right)}_1$, ${\left(M_B\right)}_2$, ${\left(M_B\right)}_3$, and ${\left(M_B\right)}_4$.

Therefore, the maximum positive bending moment at point B is $\underset{\_}{212.5\text{k-ft}}$.