Chapter 9, Problem 1P

To determine

Find the maximum negative bending moment at point B.

Answer to Problem 1P

The maximum negative bending moment at point B is $\underset{\_}{-150\text{kN-m}}$.

Explanation of Solution

Given Information:

The concentrated live load (P) is 75 kN.

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 $\left(C_y\right)$ at C using equilibrium equation:

Take moment about point A.

Consider moment equilibrium at point A.

Consider clockwise moment as positive and anticlockwise moment as negative.

Sum of moment at point A is zero.

$\begin{array}{l}\Sigma M_A=0\\ C_y\left(14\right)-1\left(x\right)=0\\ 14C_y=x\\ C_y=\frac{x}{14}\end{array}$        (1)

Find the equation of support reaction $\left(A_y\right)$ at A 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 $\frac{x}{14}$ for $C_y$.

$\begin{array}{l}{A}_y+\frac{x}{14}=1\\ A_y=1-\frac{x}{14}\end{array}$        (2)

Find the equation of moment at B.

Apply 1 kN at just left of B $\left(0\le x\le 7\text{m}\right)$.

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

Refer Figure 2.

Consider moment at B.

Consider clockwise moment as positive and anticlockwise moment as negative.

$M_B=A_y\left(7\right)-\left(1\right)\left(7-x\right)$

Substitute $1-\frac{x}{14}$ for $A_y$.

$\begin{array}{c}{M}_B=\left(1-\frac{x}{14}\right)\left(7\right)-\left(1\right)\left(7-x\right)\\ =7-\frac{x}{2}-7+x\\ =\frac{x}{2}\end{array}$

Apply 1 kN at just right of B $\left(7\text{m}<x\le 18\text{m}\right)$.

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

Refer Figure 3.

Consider moment at B.

Consider clockwise moment as positive and anticlockwise moment as negative.

Find the equation of moment at B of portion BC $\left(7\text{m}<x\le 18\text{m}\right)$.

$\begin{array}{c}{M}_B=C_y\left(7\right)-\left(1\right)\left[7-\left(14-x\right)\right]\\ =7C_y-7+\left(14-x\right)\\ =7C_y+7-x\end{array}$

Substitute $\frac{x}{14}$ for $C_y$.

$\begin{array}{c}{M}_B=7\left(\frac{x}{14}\right)+7-x\\ =\frac{x}{2}+7-x\\ =7-\frac{x}{2}\end{array}$

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

$M_B=\frac{x}{2},0\le x\le 7\text{m}$        (3)

$M_B=7-\frac{x}{2},7\text{m}\le x\le 18\text{m}$        (4)

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

x $\left(\text{m}\right)$	$M_B$ $\left(\text{kN-m/kN}\right)$
0	0
7	$\frac{7}{2}$
14	0
28	–2

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

Refer Figure 4,

The maximum negative influence line ordinate of bending moment at B is $-2\text{kN-m/kN}$.

Find the maximum negative bending moment at point B using the equation.

$\text{Maximum Negative }{M}_B=P\times \left(\text{maximum negative ILD of }{M}_B\right)$

Substitute 75 kN for P and $-2\text{kN-m/kN}$ for $\text{maximum negative ILD of }{M}_B$.

$\begin{array}{c}\text{Maximum Negative }{M}_B=75\times \left(-2\right)\\ =-150\text{kN-m}\end{array}$

Therefore, the maximum negative bending moment at point B is $\underset{\_}{-150\text{kN-m}}$.