Chapter 9, Problem 6P

To determine

Find the maximum positive and negative shears and the maximum positive and negative bending moments at point C.

Answer to Problem 6P

The maximum positive shear at point C is $\underset{\_}{295.9\text{kN}}$.

The maximum negative shear at point C is $\underset{\_}{-154.2\text{kN}}$.

The maximum positive moment at point C is $\underset{\_}{1,316.7\text{kN-m}}$.

The maximum negative moment at point C is $\underset{\_}{-850\text{kN-m}}$.

Explanation of Solution

Given Information:

The concentrated live load (P) is 150 kN.

The uniformly distributed live load $\left(w_L\right)$ is 50 kN/m.

The uniformly distributed dead load $\left(w_D\right)$ is 25 kN/m.

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 $\left(B_y\right)$ at B using equilibrium equation:

Take moment about point D.

Consider moment equilibrium at point D.

Consider clockwise moment as positive and anticlockwise moment as negative.

Sum of moment at point D is zero.

$\begin{array}{l}\Sigma M_D=0\\ B_y\left(12\right)-1\left(16-x\right)=0\\ 12B_y=16-x\\ B_y=\frac{4}{3}-\frac{x}{12}\end{array}$        (1)

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

Apply vertical equilibrium equation of forces.

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

$B_y+D_y=1$

Substitute $\frac{4}{3}-\frac{x}{12}$ for $B_y$.

$\begin{array}{l}\frac{4}{3}-\frac{x}{12}+D_y=1\\ D_y=1-\frac{4}{3}+\frac{x}{12}\\ D_y=\frac{x}{12}-\frac{1}{3}\end{array}$        (2)

Influence line for the shear at point C.

Apply 1 kN load at just left of C.

Find the equation of shear force at C of portion AB $\left(0\le x\le 8\text{m}\right)$.

Sketch the free body diagram of the section AC 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 $\left(\downarrow -\right)$.

$\Sigma F_y=0$

$\begin{array}{l}{B}_y-S_C-1=0\\ S_C=B_y-1\end{array}$

Substitute $\frac{4}{3}-\frac{x}{12}$ for $B_y$.

$\begin{array}{c}{S}_C=\left(\frac{4}{3}-\frac{x}{12}\right)-1\\ =\frac{1}{3}-\frac{x}{12}\end{array}$

Apply 1 kN load at just right of C.

Find the equation of shear force at C of portion CF $\left(8\text{m}\le x\le 22\text{m}\right)$.

Sketch the free body diagram of the section AC 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 $\left(\downarrow -\right)$.

$\Sigma F_y=0$

$\begin{array}{l}{B}_y-S_C=0\\ S_C=B_y\end{array}$

Substitute $\frac{4}{3}-\frac{x}{12}$ for $B_y$.

$\begin{array}{c}{S}_C=\left(\frac{4}{3}-\frac{x}{12}\right)\\ =\frac{4}{3}-\frac{x}{12}\end{array}$

Thus, the equations of the influence line for $S_C$ as follows

$S_C=\frac{1}{3}-\frac{x}{12},0\le x\le 8\text{m}$        (3)

$S_C=\frac{4}{3}-\frac{x}{12},8\text{m}\le x\le 22\text{m}$        (4)

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

x $\left(\text{m}\right)$	Position	Influence line ordinate of $S_C\left(\text{kN/kN}\right)$
0	A	$\frac{1}{3}$
4	B	$0$
8	$C^{-}$	$-\frac{1}{3}$
8	$C^{+}$	$\frac{2}{3}$
16	D	0
19	E	$-\frac{1}{4}$
22	F	$-\frac{1}{2}$

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

Refer Figure 4.

The maximum positive ILD ordinate at point C is $\frac{2}{3}\text{kN/kN}$.

The maximum negative ILD ordinate at point C is $-\frac{1}{3}\text{kN/kN}$.

Find the positive area $\left(A_1\right)$ of the influence line diagram of shear force at point C.

$A_1=\frac{1}{2}\left(L_{AB}\right)\left(\frac{1}{3}\right)+\frac{1}{2}\left(L_{CD}\right)\left(\frac{2}{3}\right)$

Here, $L_{AB}$ is the length of beam between A to B and $L_{CD}$ is the length of beam between C to D.

Substitute 4 m for $L_{AB}$ and 8 m for $L_{CD}$.

$\begin{array}{c}{A}_1=\frac{1}{2}\left(4\right)\left(\frac{1}{3}\right)+\frac{1}{2}\left(8\right)\left(\frac{2}{3}\right)\\ =3.333\text{m}\end{array}$

Find the negative area $\left(A_2\right)$ of the influence line diagram of shear force at point C.

$A_2=\frac{1}{2}\left(L_{BC}\right)\left(-\frac{1}{3}\right)+\frac{1}{2}\left(L_{DF}\right)\left(-\frac{1}{2}\right)$

Here, $L_{BC}$ is the length of beam between B to C and $L_{DF}$ is the length of beam between E to F.

Substitute 4 m for $L_{BC}$ and 3 m for $L_{DF}$.

$\begin{array}{c}{A}_2=\frac{1}{2}\left(4\right)\left(-\frac{1}{3}\right)+\frac{1}{2}\left(6\right)\left(-\frac{1}{2}\right)\\ =-0.667-1.5\\ =-2.167\text{m}\end{array}$

Find the maximum positive shear at point C using the equation.

$\text{Maximum positive }{S}_C=P\left(\text{maximum positive ILD ordinate at }C\right)+w_L\left(A_1\right)+w_D\left(A_1+A_2\right)$

Substitute 150 kN for P, $\frac{2}{3}\text{kN/kN}$ for $\text{maximum positive ILD ordinate at }C$, 50 kN/m for $w_L$, $3.333\text{m}$ for $A_1$, 25 kN/m for $w_D$, and $-2.167\text{m}$ for $A_2$.

$\begin{array}{c}\text{Maximum positive }{S}_C=150\left(\frac{2}{3}\right)+50\left(3.333\right)+25\left(3.333-2.167\right)\\ =100+166.67+29.15\\ =295.9\text{kN}\end{array}$

Therefore, the maximum positive shear at point C is $\underset{\_}{295.9\text{kN}}$.

Find the maximum negative shear at point C using the equation.

$\text{Maximum negative }{S}_C=P\left(\text{maximum negative ILD ordinate at }C\right)+w_L\left(A_2\right)+w_D\left(A_1+A_2\right)$

Substitute 150 kN for P, $-\frac{1}{2}\text{kN/kN}$ for $\text{maximum positive ILD ordinate at }C$, 50 kN/m for $w_L$, $3.333\text{m}$ for $A_1$, 25 kN/m for $w_D$, and $-2.167\text{m}$ for $A_2$.

$\begin{array}{c}\text{Maximum negative }{S}_C=150\left(-\frac{1}{2}\right)+50\left(-2.167\right)+25\left(3.333-2.167\right)\\ =-75-108.35+29.15\\ =-154.2\text{kN}\end{array}$

Therefore, the maximum negative shear at point C is $\underset{\_}{-154.2\text{kN}}$.

Influence line for moment at point C.

Refer Figure 2.

Consider clockwise moment as positive and anticlockwise moment as negative.

Find the equation of moment at C of portion AC $\left(0\le x\le 8\text{m}\right)$.

$M_C=B_y\left(4\right)-\left(1\right)\left(8-x\right)$

Substitute $\frac{4}{3}-\frac{x}{12}$ for $B_y$.

$\begin{array}{c}{M}_C=\left(\frac{4}{3}-\frac{x}{12}\right)\left(4\right)-\left(1\right)\left(8-x\right)\\ =\frac{16}{3}-\frac{x}{3}-8+x\\ =\frac{2x}{3}-\frac{8}{3}\end{array}$

Refer Figure 3.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at C of portion CF $\left(8\text{m}\le x\le 22\text{m}\right)$.

$M_C=B_y\left(4\right)$

Substitute $\frac{4}{3}-\frac{x}{12}$ for $B_y$.

$\begin{array}{c}{M}_C=\left(\frac{4}{3}-\frac{x}{12}\right)\left(4\right)\\ =\frac{16}{3}-\frac{x}{3}\\ =\frac{16-x}{3}\end{array}$

Thus, the equations of the influence line for $M_C$ as follows,

$M_C=\frac{2x}{3}-\frac{8}{3},0\le x\le 8\text{m}$        (5)

$M_C=\frac{16-x}{3},8\text{m}\le x\le 22\text{m}$        (6)

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

x $\left(\text{m}\right)$	Position	Influence line ordinate of $M_C\left(\text{kN-m/kN}\right)$
0	A	$-\frac{8}{3}$
4	B	0
8	$C$	$\frac{8}{3}$
16	D	0
19	E	$-1$
22	F	$-2$

Draw the influence lines for the moment at point C using Table 2 as shown in Figure 5.

Refer Figure 5.

The maximum positive ILD ordinate of moment at C is $\frac{\text{8}}{3}\text{kN-m/kN}$.

The maximum negative ILD ordinate of moment at C is $-\frac{\text{8}}{3}\text{kN-m/kN}$.

Find the positive area $\left(A_3\right)$ of ILD ordinate of moment at C.

$A_3=\frac{1}{2}\left(\frac{8}{3}\right)L_{BD}$

Here, $L_{BD}$ is the length of beam between B and D.

Substitute 12 m for $L_{BD}$.

$\begin{array}{c}{A}_3=\frac{1}{2}\left(\frac{8}{3}\right)\left(12\right)\\ =16{\text{m}}^2\end{array}$

Find the negative area $\left(A_4\right)$ of ILD ordinate of moment at C.

$A_4=\frac{1}{2}\left(-\frac{8}{3}\right)L_{AB}+\frac{1}{2}\left(-2\right)L_{DF}$

Substitute 4 m for $L_{AB}$ and 6 m for $L_{DF}$.

$\begin{array}{c}{A}_4=\frac{1}{2}\left(-\frac{8}{3}\right)\left(4\right)+\frac{1}{2}\left(-2\right)\left(6\right)\\ =-11.333{\text{m}}^2\end{array}$

Find the maximum positive moment at point C using the equation.

$\text{Maximum positive }{M}_C=P\left(\text{maximum positive ILD ordinate of moment}\right)+w_L\left(A_3\right)+w_D\left(A_3+A_4\right)$

Substitute 150 kN for P, $\frac{8}{3}\text{kN-m/kN}$ for $\text{maximum positive ILD ordinate of moment}$, $50\text{kN/m}$ for $w_L$, $16{\text{m}}^2$ for $A_3$, 25 kN/m for $w_D$, and $-11.333{\text{m}}^2$ for $A_4$.

$\begin{array}{c}\text{Maximum positive }{M}_C=150\left(\frac{8}{3}\right)+50\left(16\right)+25\left(16-11.333\right)\\ =400+800+116.675\\ =1,316.7\text{kN-m}\end{array}$

Therefore, the maximum positive moment at point C is $\underset{\_}{1,316.7\text{kN-m}}$

Find the maximum negative moment at point C using the equation.

$\text{Maximum positive }{M}_C=P\left(\text{maximum negative ILD ordinate of moment}\right)+w_L\left(A_4\right)+w_D\left(A_3+A_4\right)$

Substitute 150 kN for P, $-\frac{\text{8}}{3}\text{kN-m/kN}$ for $\text{maximum positive ILD ordinate of moment}$, $50\text{kN/m}$ for $w_L$, $16{\text{m}}^2$ for $A_3$, 25 kN/m for $w_D$, and $-11.333{\text{m}}^2$ for $A_4$.

$\begin{array}{c}\text{Maximum negative }{M}_C=150\left(-\frac{8}{3}\right)+50\left(-11.333\right)+25\left(16-11.333\right)\\ =-400-566.65+116.675\\ =-850\text{kN-m}\end{array}$

Therefore, the maximum negative moment at point C is $\underset{\_}{-850\text{kN-m}}$.