Chapter 12, Problem 1P

To determine

Draw the influence lines for the reaction at A.

Draw the influence lines for the shear and moment at points B and C.

Explanation of Solution

Determine the general expression of reaction $R_A$.

Apply a 1 kip unit load between supports A and D 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 at A using equilibrium equation:

Take moment about point D.

Consider moment equilibrium at point D.

Consider clockwise moment as negative and anticlockwise moment as positive

Sum of moment at point D is zero.

$\begin{array}{l}\Sigma M_D=0\\ R_A\left(20\right)-1\left(20-x\right)=0\\ 20R_A=20-x\\ R_A=1-\frac{x}{20}\end{array}$        (1)

Find the value of influence line ordinate at $x=0$ of reaction $R_A$ at support A.

Substitute 0 for x in Equation (1).

$\begin{array}{c}{R}_A=1-\frac{0}{20}\\ =1\end{array}$

Similarly calculate the influence line ordinate of reaction $R_A$ for $x=5,10,\text{and}\text{20}\text{m}$ and summarize the result in Table 1.

x $\left(\text{ft}\right)$	$R_A$
0	1
5	$\frac{3}{4}$
10	$\frac{1}{2}$
20	0

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

Find the equation of shear force at B of portion AB $\left(0\le x<5\text{ft}\right)$.

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

Refer Figure 3.

Apply vertical equilibrium equation of forces.

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

$\Sigma F_y=0$

$\begin{array}{l}{R}_A-V_B-1=0\\ V_B=R_A-1\\ V_B=1-\frac{x}{20}-1\\ V_B=-\frac{x}{20}\end{array}$

Find the equation of shear force at B of portion BD $\left(5\text{ft}<x\le 20\text{ft}\right)$.

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

Refer Figure 4.

Apply vertical equilibrium equation of forces.

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

$\Sigma F_y=0$

$\begin{array}{l}{R}_A-V_B=0\\ 1-\frac{x}{20}-V_B=0\\ V_B=1-\frac{x}{20}\end{array}$

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

$V_B=-\frac{x}{20}0\le x<5\text{ft}$        (2)

$V_B=1-\frac{x}{20}5\text{ft}<x\le 20\text{ft}$        (3)

Find the value of influence line ordinate of shear force at B various points of x using the Equations (2) and (3) and summarize the value as in Table 2.

x $\left(\text{ft}\right)$	$V_B$
0	0
$5^{-}$	$-\frac{1}{4}$
$5^{+}$	$\frac{3}{4}$
10	$\frac{1}{2}$
20	0

Draw the influence lines for the shear force at point B using Table 2 as shown in Figure 5.

Refer Figure 3.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at B of portion AB $\left(0\le x<5\text{ft}\right)$.

$\begin{array}{c}{M}_B=R_A\left(5\right)-\left(1\right)\left(5-x\right)\\ =\left(1-\frac{x}{20}\right)\left(5\right)-5+x\\ =5-\frac{x}{4}-5+x\\ =\frac{3x}{4}\end{array}$

Refer Figure 4.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at B of portion BD $\left(5\text{ft}<x\le 20\text{ft}\right)$.

$\begin{array}{c}{M}_B=R_A\left(5\right)\\ =\left(1-\frac{x}{20}\right)\left(5\right)\\ =5-\frac{x}{4}\end{array}$

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

$M_B=\frac{3x}{4}0\le x<5\text{ft}$        (4)

$M_B=5-\frac{x}{4}5\text{ft}<x\le 20\text{ft}$        (5)

Find the value of influence line ordinate of moment at B various points of x using the Equations (4) and (5) and summarize the value as in Table 3.

x $\left(\text{ft}\right)$	$M_B$
0	0
$5$	$\frac{15}{4}$
20	0

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

Find the equation of shear force at C of portion AC $\left(0\le x<10\text{ft}\right)$.

Sketch the free body diagram of the section AC as shown in Figure 7.

Refer Figure 7.

Apply vertical equilibrium equation of forces.

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

$\Sigma F_y=0$

$\begin{array}{l}{R}_A-V_C-1=0\\ V_B=R_A-1\\ V_C=1-\frac{x}{20}-1\\ V_C=-\frac{x}{20}\end{array}$

Find the equation of shear force at C of portion CD $\left(10\text{ft}<x\le 20\text{ft}\right)$.

Sketch the free body diagram of the section CD as shown in Figure 8.

Refer Figure 8.

Apply equilibrium equation of forces.

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

$\Sigma F_y=0$

$\begin{array}{l}{R}_A-V_C=0\\ 1-\frac{x}{20}-V_C=0\\ V_C=1-\frac{x}{20}\end{array}$

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

$V_C=-\frac{x}{20}0\le x<10\text{ft}$        (6)

$V_C=1-\frac{x}{20}10\text{ft}<x\le 20\text{ft}$        (7)

Find the value of influence line ordinate of shear force at C various points of x using the Equations (6) and (7) and summarize the value as in Table 4.

x $\left(\text{ft}\right)$	$V_C$
0	0
5	$-\frac{1}{4}$
${10}^{-}$	$-\frac{1}{2}$
${10}^{+}$	$\frac{1}{2}$
20	0

Draw the influence lines for the shear force at point C using Table 4 as shown in Figure 9.

Refer Figure 7.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at C of portion AC $\left(0\le x<10\text{ft}\right)$.

$\begin{array}{c}{M}_C=R_A\left(10\right)-\left(1\right)\left(10-x\right)\\ =\left(1-\frac{x}{20}\right)\left(10\right)-10+x\\ =10-\frac{x}{2}-10+x\\ =\frac{x}{2}\end{array}$

Refer Figure 4.

Consider clockwise moment as negative and anticlockwise moment as positive.

Find the equation of moment at C of portion CD $\left(10\text{ft}<x\le 20\text{ft}\right)$.

$\begin{array}{c}{M}_B=R_A\left(10\right)\\ =\left(1-\frac{x}{20}\right)\left(10\right)\\ =10-\frac{x}{2}\end{array}$

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

$M_C=\frac{x}{2}0\le x<10\text{ft}$        (8)

$M_C=10-\frac{x}{2}10\text{ft}<x\le 20\text{ft}$        (9)

Find the value of influence line ordinate of moment at C various points of x using the Equations (8) and (9) and summarize the value as in Table 5.

x $\left(\text{ft}\right)$	$M_C$ $\left(\text{k-ft/k}\right)$
0	0
5	$2.5$
10	5
20	0

Draw the influence lines for the moment at point C using Table 5 as shown in Figure 10.