Chapter 9, Problem 13P

To determine

Find the maximum positive shear and bending moment at point B.

Answer to Problem 13P

The maximum positive shear at point B is $\underset{\_}{61.67\text{kN}}$.

The maximum positive bending moment at point B is $\underset{\_}{733.33\text{kN-m}}$.

Explanation of Solution

Calculation:

Apply a 1 kN 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.

$\begin{array}{l}\Sigma M_C=0\\ A_y\left(15\right)-1\left(15-x\right)=0\\ A_y\left(15\right)-15+x=0\\ 15A_y=15-x\end{array}$

$A_y=1-\frac{x}{15}$        (1)

Find the equation of support reaction $\left(B_y\right)$ at B 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 $1-\frac{x}{15}$ for $A_y$.

$\begin{array}{l}1-\frac{x}{15}+C_y=1\\ C_y=1-1+\frac{x}{15}\\ C_y=\frac{x}{15}\end{array}$        (2)

Influence line for shear at point B.

Find the equation of shear force at B of portion AB $\left(0\le x<10\text{m}\right)$.

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 $1-\frac{x}{15}$ for $A_y$.

$\begin{array}{c}{S}_B=1-\frac{x}{15}-1\\ =-\frac{x}{15}\end{array}$

Find the equation of shear force at B of portion BC $\left(10\text{m}<x\le 15\text{m}\right)$.

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 $\frac{x}{15}$ for $C_y$.

$S_B=1-\frac{x}{15}$

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

$S_B=-\frac{x}{15}0\le x<10\text{m}$        (3)

$S_B=1-\frac{x}{15}10\text{m}<x\le 15\text{m}$        (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 $\left(\text{m}\right)$	$S_B$ $\left(\text{kN/kN}\right)$
0	0
${10}^{-}$	$-\frac{2}{3}$
${10}^{+}$	$\frac{1}{3}$
16	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.

${\theta }_{AB}=\frac{-\frac{2}{3}}{L_{AB}}$

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

Substitute 10 m for $L_{AB}$.

$\begin{array}{c}{\theta }_{AB}=\frac{-\frac{2}{3}}{10}\\ =-\frac{1}{15}\end{array}$

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

${\theta }_{BC}=\frac{\frac{1}{3}}{L_{BC}}$

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

Substitute 5 m for $L_{BC}$.

$\begin{array}{c}{\theta }_{BC}=\frac{\frac{1}{3}}{5}\\ =\frac{1}{15}\end{array}$

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.

${\left(S_B\right)}_1=125\left(L_{BC}\right)\left({\theta }_{BC}\right)+100\left(L_{BC}-2\right)\left({\theta }_{BC}\right)+50\left(0\right)$

Substitute 5 m for $L_{BC}$ and $\frac{1}{15}$ for ${\theta }_{BC}$.

$\begin{array}{c}{\left(S_B\right)}_1=125\left(5\right)\left(\frac{1}{15}\right)+100\left(5-2\right)\left(\frac{1}{15}\right)+50\left(0\right)\\ =41.67+20+0\\ =61.67\text{kN}\end{array}$

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.

${\left(S_B\right)}_2=125\left(L_{AB}-2\right)\left({\theta }_{AB}\right)+100\left(L_{BC}\right)\left({\theta }_{BC}\right)+50\left(L_{BC}-3\right)\left({\theta }_{BC}\right)$

Substitute 10 m for $L_{AB}$, $-\frac{1}{15}$ for ${\theta }_{AB}$, 5 m for $L_{BC}$, and $\frac{1}{15}$ for ${\theta }_{BC}$.

$\begin{array}{c}{\left(S_B\right)}_2=125\left(10-2\right)\left(-\frac{1}{15}\right)+100\left(5\right)\left(\frac{1}{15}\right)+50\left(5-3\right)\left(\frac{1}{15}\right)\\ =-66.67+33.33+6.67\\ =-26.67\text{k}\end{array}$

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.

${\left(S_B\right)}_3=125\left(L_{AB}-5\right)\left({\theta }_{AB}\right)+100\left(L_{AB}-3\right)\left({\theta }_{AB}\right)+50\left(L_{BC}\right)\left({\theta }_{BC}\right)$

Substitute 10 m for $L_{AB}$, $-\frac{1}{15}$ for ${\theta }_{AB}$, 5 m for $L_{BC}$, and $\frac{1}{15}$ for ${\theta }_{BC}$.

$\begin{array}{c}{\left(S_B\right)}_3=125\left(10-5\right)\left(-\frac{1}{15}\right)+100\left(10-3\right)\left(-\frac{1}{15}\right)+50\left(5\right)\left(\frac{1}{15}\right)\\ =-41.67-46.67+16.67\\ =-71.67\text{kN}\end{array}$

Maximum positive shear force at B as follows.

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

Therefore, the maximum positive shear at point B is $\underset{\_}{61.67\text{kN}}$.

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 $\left(0\le x<10\text{m}\right)$.

$M_B=A_y\left(10\right)-\left(1\right)\left(10-x\right)$

Substitute $1-\frac{x}{15}$ for $A_y$.

$\begin{array}{c}{M}_B=\left(1-\frac{x}{15}\right)\left(10\right)-\left(1\right)\left(10-x\right)\\ =10-\frac{2}{3}x-10+x\\ =\frac{x}{3}\end{array}$

Refer Figure 3.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at B of portion BC $\left(10\text{m}<x\le 15\text{m}\right)$.

$\begin{array}{c}{M}_B=C_y\left(5\right)-\left(1\right)\left[5-\left(15-x\right)\right]\\ =5C_y-5+\left(15-x\right)\\ =5C_y+10-x\end{array}$

Substitute $\frac{x}{15}$ for $C_y$.

$\begin{array}{c}{M}_B=5\left(\frac{x}{15}\right)+10-x\\ =\frac{x}{3}+10-x\\ =10-\frac{2}{3}x\end{array}$

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

$M_B=\frac{x}{3},0\le x<10\text{m}$        (5)

$M_B=10-\frac{2}{3}x,10\text{m}<x\le 15\text{m}$        (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 $\left(\text{m}\right)$	$M_B$ $\left(\text{kN-m/kN}\right)$
0	0
10	$\frac{10}{3}$
15	0

Draw the influence lines for the moment at point B using Table 2 as shown in Figure 9.

Refer Figure 9.

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

${\varphi }_{AB}=\frac{\frac{10}{3}}{L_{AB}}$

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

Substitute 10 m for $L_{AB}$.

$\begin{array}{c}{\varphi }_{AB}=\frac{\frac{10}{3}}{10}\\ =\frac{1}{3}\end{array}$

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

${\varphi }_{BC}=\frac{\frac{10}{3}}{L_{BC}}$

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

Substitute 5 m for $L_{BC}$.

$\begin{array}{c}{\varphi }_{BC}=\frac{\frac{10}{3}}{5}\\ =\frac{2}{3}\end{array}$

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.

${\left(M_B\right)}_1=125\left(L_{BC}\right)\left({\varphi }_{BC}\right)+100\left(L_{BC}-2\right)\left({\varphi }_{BC}\right)+50\left(0\right)$

Substitute 5 m for $L_{BC}$ and $\frac{2}{3}$ for ${\varphi }_{BC}$.

$\begin{array}{c}{\left(M_B\right)}_1=125\left(5\right)\left(\frac{2}{3}\right)+100\left(5-2\right)\left(\frac{2}{3}\right)+50\left(0\right)\\ =416.67+200+0\\ =616.67\text{kN-m}\end{array}$

Refer Figure 7.

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

${\left(M_B\right)}_2=125\left(L_{AB}-2\right)\left({\varphi }_{AB}\right)+100\left(L_{BC}\right)\left({\varphi }_{BC}\right)+50\left(L_{BC}-3\right)\left({\varphi }_{BC}\right)$

Substitute 10 m for $L_{AB}$, $\frac{1}{3}$ for ${\varphi }_{AB}$, 5 m for $L_{BC}$, and $\frac{2}{3}$ for ${\varphi }_{BC}$.

$\begin{array}{c}{\left(M_B\right)}_2=125\left(10-2\right)\left(\frac{1}{3}\right)+100\left(5\right)\left(\frac{2}{3}\right)+50\left(5-3\right)\left(\frac{2}{3}\right)\\ =333.33+333.33+66.67\\ =733.33\text{kN-m}\end{array}$

Refer Figure 8.

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

${\left(M_B\right)}_3=125\left(L_{AB}-5\right)\left({\varphi }_{AB}\right)+100\left(L_{AB}-3\right)\left({\varphi }_{AB}\right)+50\left(L_{BC}\right)\left({\varphi }_{BC}\right)$

Substitute 10 m for $L_{AB}$, $\frac{1}{3}$ for ${\varphi }_{AB}$, 5 m for $L_{BC}$, and $\frac{2}{3}$ for ${\varphi }_{BC}$.

$\begin{array}{c}{\left(M_B\right)}_3=125\left(10-5\right)\left(\frac{1}{3}\right)+100\left(10-3\right)\left(\frac{1}{3}\right)+50\left(5\right)\left(\frac{2}{3}\right)\\ =208.33+233.33+166.67\\ =608.33\text{kN-m}\end{array}$

Maximum positive bending moment at B. as follows.

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

Therefore, the maximum positive bending moment at point B is $\underset{\_}{733.33\text{kN-m}}$.