Chapter 8, Problem 1P

To determine

Draw the influence lines for the vertical reactions at supports A and C.

Draw the influence lines for the shear and bending moment at point B.

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.

$\begin{array}{l}\Sigma M_C=0\\ A_y\left(28\right)-1\left(28-x\right)=0\\ A_y\left(28\right)-28+x=0\\ 28A_y=28-x\\ A_y=1-\frac{x}{28}\end{array}$        (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 $1-\frac{x}{28}$ for $A_y$.

$\begin{array}{l}1-\frac{x}{28}+C_y=1\\ C_y=1-1+\frac{x}{28}\\ C_y=\frac{x}{28}\end{array}$        (2)

Consider Equation (1).

Find the value of influence line ordinate of reaction $A_y$ at support A.

Substitute 0 for x in Equation (1).

$\begin{array}{c}{A}_y=1-\frac{0}{28}\\ =1\text{k/k}\end{array}$

Similarly calculate the influence line ordinate of reaction $A_y$ for different value of x and summarize the result in Table 1.

x $\left(\text{ft}\right)$	$A_y\left(\text{k/k}\right)$
0	1
14	0.5
28	0

Draw the influence line diagram for the vertical reactions at support A using Table 1 as shown in Figure 2.

Consider Equation (2).

Find the influence line ordinate of reaction $C_y$ at support C.

Substitute 28 for x in Equation (2).

$\begin{array}{c}{C}_y=\frac{28}{28}\\ =1\end{array}$

Similarly calculate the influence line ordinate of reaction $C_y$ for different value of x and summarize the result in Table 2.

x $\left(\text{ft}\right)$	$C_y\left(\text{k/k}\right)$
0	0
14	0.5
28	1

Draw the influence line diagram for the vertical reactions at support C using Table 2 as shown in Figure 3.

Find the equation of shear force at B of portion AB $\left(0\le x<14\text{ft}\right)$.

Sketch the free body diagram of the section AB as shown in Figure 4.

Refer Figure 4.

Apply equilibrium equation of forces.

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

$\Sigma F_y=0$

$\begin{array}{l}{A}_y-S_B-1=0\\ S_B=A_y-1\end{array}$

Substitute $1-\frac{x}{28}$ for $A_y$.

$\begin{array}{c}{S}_B=1-\frac{x}{28}-1\\ =-\frac{x}{28}\end{array}$

Find the equation of shear force at B of portion BC $\left(14\text{ft}<x\le 28\text{ft}\right)$.

Sketch the free body diagram of the section BC as shown in Figure 5.

Refer Figure 5.

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 $\frac{x}{28}$ for $C_y$.

$S_B=1-\frac{x}{28}$

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

$S_B=-\frac{x}{28}0\le x<14\text{ft}$        (3)

$S_B=1-\frac{x}{28}14\text{ft}<x\le 28\text{ft}$        (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 3.

x $\left(\text{ft}\right)$	$S_B\left(\text{k/k}\right)$
0	1
${14}^{-}$	$-\frac{1}{2}$
${14}^{+}$	$\frac{1}{2}$
28	0

Draw the influence lines for the shear force at point B using Table 3 as shown in Figure 6.

Refer Figure 4.

Consider clockwise moment as positive and anticlockwise moment as negative.

Find the equation of moment at B of portion AB $\left(0\le x<14\text{ft}\right)$.

$M_B=A_y\left(14\right)-\left(1\right)\left(14-x\right)$

Substitute $1-\frac{x}{28}$ for $A_y$.

$\begin{array}{c}{M}_B=\left(1-\frac{x}{28}\right)\left(14\right)-\left(1\right)\left(14-x\right)\\ =14-\frac{x}{2}-14+x\\ =\frac{x}{2}\end{array}$

Refer Figure 5.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at B of portion BC $\left(14\text{ft}<x\le 28\text{ft}\right)$.

$\begin{array}{c}{M}_B=C_y\left(14\right)-\left(1\right)\left[14-\left(28-x\right)\right]\\ =14C_y-14+\left(28-x\right)\\ =14C_y+14-x\end{array}$

Substitute $\frac{x}{28}$ for $C_y$.

$\begin{array}{c}{M}_B=14\left(\frac{x}{28}\right)+14-x\\ =\frac{x}{2}+14-x\\ =14-\frac{x}{2}\end{array}$

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

$M_B=\frac{x}{2}0\le x<14\text{ft}$        (5)

$M_B=14-\frac{x}{2}14\text{ft}<x\le 28\text{ft}$        (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 4.

x $\left(\text{ft}\right)$	$M_B$ $\left(\text{k-ft/k}\right)$
0	0
14	$7.0$
28	0

Draw the influence lines for the moment at point B using Table 4 as shown in Figure 7.

Therefore, the influence lines for the vertical reactions at supports A and C and the influence lines for the shear and bending moment at point B are drawn.