Chapter 8, Problem 22P

To determine

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

Explanation of Solution

Equation of influence line ordinate of support B:

Apply a 1 k unit moving load at a distance of x from left end D.

Sketch the free body diagram of frame as shown in Figure 1.

Refer Figure 1.

Consider the unit load at a variable position x to the left hinge D. (placed portion AC of the beam $0\le x<36\text{ft}$).

Find the vertical support reaction $\left(B_y\right)$ at B for portion AD.

Take Moment at hinge D from left end A.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_D^{AD}=0\\ B_y\left(24\right)-1\left(36-x\right)=0\\ 24B_y=36-x\\ B_y=1.5-\frac{x}{24}\end{array}$

Consider the unit load at a variable position x to the right hinge D. (Place portion DG of the beam $36\text{ft}\le x\le 72\text{ft}$).

Find the vertical support reaction $\left(B_y\right)$ at D for portion DG.

Take Moment at hinge D from left end A.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_D^{AD}=0\\ B_y\left(24\right)=0\\ B_y=0\end{array}$

Thus, the equations of the influence line ordinate for $B_y$ are,

$B_y=1.5-\frac{x}{24},0\le x<36\text{ft}$        (1)

$B_y=0,36\text{ft}\le x\le 72\text{ft}$        (2)

Equation of influence line ordinate of support E:

Refer Figure 1.

Find the equation of influence line ordinate for the vertical reaction $\left(E_y\right)$ using equilibrium equation.

Apply moment equilibrium at G.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_G=0\\ B_y\left(60\right)+E_y\left(24\right)-1\left(72-x\right)=0\\ 60B_y+24E_y-72+x=0\\ 24E_y=72-x-60B_y\end{array}$

$E_y=3-\frac{x}{24}-2.5B_y$        (3)

Find the influence line ordinate of vertical reaction $\left(E_y\right)$ for portion AD $\left(0\le x<36\text{ft}\right)$.

Substitute $1.5-\frac{x}{24}$ for $B_y$ in Equation (3).

$\begin{array}{c}{E}_y=3-\frac{x}{24}-2.5\left(1.5-\frac{x}{24}\right)\\ =3-\frac{x}{24}-3.75+\frac{2.5x}{24}\\ =\frac{1.5x}{24}-0.75\end{array}$

Find the influence line ordinate of vertical reaction $\left(E_y\right)$ for portion DG $\left(36\text{ft}\le x\le 72\text{ft}\right)$.

Substitute $0$ for $B_y$ in Equation (3).

$\begin{array}{c}{E}_y=3-\frac{x}{24}-2.5\left(0\right)\\ =3-\frac{x}{24}\end{array}$

Thus, the equations of the influence line ordinate for $E_y$ are,

$E_y=\frac{1.5x}{24}-0.75,0\le x<36\text{ft}$        (4)

$E_y=3-\frac{x}{24},36\text{ft}\le x\le 72\text{ft}$        (5)

Equation of influence line ordinate of support G:

Refer Figure 1.

Find the equation of influence line ordinate for the vertical reaction $\left(G_y\right)$ using force equilibrium equation.

Consider the vertical forces equilibrium condition, take the upward force as positive $\left(\uparrow +\right)$ and downward force as negative $\left(\downarrow -\right)$.

$\begin{array}{l}\Sigma F_y=0\\ B_y+E_y+G_y=1\\ G_y=1-B_y-G_y\end{array}$        (6)

Find the influence line ordinate of vertical reaction $\left(G_y\right)$ for portion AD $\left(0\le x<36\text{ft}\right)$.

Substitute $1.5-\frac{x}{24}$ for $B_y$ and $\frac{1.5x}{24}-0.75$ for $E_y$ in Equation (6).

$\begin{array}{c}{G}_y=1-\left(1.5-\frac{x}{24}\right)-\left(\frac{1.5x}{24}-0.75\right)\\ =1-1.5+\frac{x}{24}-\frac{1.5x}{24}+0.75\\ =0.25-\frac{0.5x}{24}\end{array}$

Find the influence line ordinate of vertical reaction $\left(G_y\right)$ for portion DG $\left(36\text{ft}\le x\le 72\text{ft}\right)$.

Substitute $0$ for $B_y$ and $3-\frac{x}{24}$ for $E_y$ in Equation (6).

$\begin{array}{c}{G}_y=1-\left(0\right)-\left(3-\frac{x}{24}\right)\\ =1-3+\frac{x}{24}\\ =\frac{x}{24}-2\end{array}$

Thus, the equations of the influence line ordinate for $G_y$ are,

$G_y=0.25-\frac{0.5x}{24},0\le x<36\text{ft}$        (7)

$G_y=\frac{x}{24}-2,36\text{ft}\le x\le 72\text{ft}$        (8)

Influence line for shear at point F.

Find the equation of shear at F of portion AG $\left(0\le x<72\text{ft}\right)$.

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

$\Sigma F_y=0$

$\begin{array}{l}{B}_y-1+E_y-S_F=0\\ S_F=B_y+E_y-1\end{array}$        (9)

Find the influence line ordinate of shear at F $\left(S_F\right)$ for portion AD $\left(0\le x\le 36\text{ft}\right)$.

Substitute $1.5-\frac{x}{24}$ for $B_y$ and $\frac{1.5x}{24}-0.75$ for $E_y$ in Equation (9).

$\begin{array}{c}{S}_F=1.5-\frac{x}{24}+\frac{1.5x}{24}-0.75-1\\ =-0.25+\frac{0.5x}{24}\end{array}$

Find the influence line ordinate of shear at F $\left(S_F\right)$ for portion DF $\left(36\text{ft}\le x\le 60\text{ft}\right)$.

Substitute $0$ for $B_y$ and $3-\frac{x}{24}$ for $E_y$ in Equation (9).

$\begin{array}{c}{S}_F=0+3-\frac{x}{24}-1\\ =2-\frac{x}{24}\end{array}$

Find the equation of shear at F of portion FG $\left(60\text{ft}<x\le 72\text{ft}\right)$.

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

$\Sigma F_y=0$

$\begin{array}{l}-S_F+B_y+E_y=0\\ S_F=B_y+E_y\end{array}$

Substitute 0 for $B_y$ and $3-\frac{x}{24}$ for $E_y$.

$\begin{array}{c}{S}_F=0+3-\frac{x}{24}\\ =3-\frac{x}{24}\end{array}$

Thus, the equations of the influence line ordinate for $S_F$ are,

$S_F=-0.25+\frac{0.5x}{24},0\le x\le 36\text{ft}$        (10)

$S_F=2-\frac{x}{24},36\text{ft}\le x\le 60\text{ft}$        (11)

$S_F=3-\frac{x}{24},60\text{ft}\le x\le 72\text{ft}$        (12)

Find the influence line ordinate of $S_F$ using Equation (10), (11), and (12) and summarize the values in Table 1.

x (ft)	Points	Influence line ordinate of $S_F$ $\left(\text{k/k}\right)$
0	A	$-0.25$
12	$B$	0
24	C	0.25
36	D	0.5
48	$E$	$0$
60	$F^{-}$	$-0.5$
60	$F^{+}$	0.5
72	G	0

Sketch the influence line diagram for the shear at point F as shown in Figure 4.

Refer Figure 2.

Find the equation of bending moment at E of portion AD $\left(0\le x\le 36\text{ft}\right)$.

Take moment at F.

Consider clockwise moment as positive and anticlockwise moment as negative.

$M_F=B_y\left(48\right)-1\left(60-x\right)+E_y\left(12\right)$        (13)

Find the influence line ordinate of bending moment at F $\left(M_F\right)$ for portion AD $\left(0\le x\le 36\text{ft}\right)$.

Substitute $1.5-\frac{x}{24}$ for $B_y$ and $\frac{1.5x}{24}-0.75$ for $E_y$ in Equation (13).

$\begin{array}{c}{M}_F=\left(1.5-\frac{x}{24}\right)\left(48\right)-1\left(60-x\right)+\left(\frac{1.5x}{24}-0.75\right)\left(12\right)\\ =72-2x-60+x+0.75x-9\\ =-0.25x+3\end{array}$

Find the influence line ordinate of bending moment at F $\left(S_F\right)$ for portion DF $\left(36\text{ft}\le x\le 60\text{ft}\right)$.

Substitute $0$ for $B_y$ and $3-\frac{x}{24}$ for $E_y$ in Equation (13).

$\begin{array}{c}{M}_F=\left(0\right)\left(48\right)-1\left(60-x\right)+\left(3-\frac{x}{24}\right)\left(12\right)\\ =0-60+x+36-0.5x\\ =0.5x-24\end{array}$

Refer Figure 3.

Find the equation of bending moment at F of portion FG $\left(60\text{ft}\le x\le 72\text{ft}\right)$.

Take moment at F.

Consider clockwise moment as negative and anticlockwise moment as positive.

$M_F=B_y\left(48\right)+E_y\left(12\right)$

Substitute $0$ for $B_y$ and $3-\frac{x}{24}$ for $E_y$.

$\begin{array}{c}{M}_F=\left(0\right)\left(48\right)+\left(3-\frac{x}{24}\right)\left(12\right)\\ =0+36-0.5x\\ =36-0.5x\end{array}$

Thus, the equations of the influence line ordinate for $M_F$ are,

$M_F=-0.25x+3,0\le x<36\text{ft}$        (14)

$M_F=0.5x-24,36\text{ft}\le x\le 60\text{ft}$        (15)

$M_F=36-0.5x,60\text{ft}\le x\le 72\text{ft}$        (16)

Find the influence line ordinate of $M_E$ using Equation (14), (15), and (16) and summarize the values in Table 2.

x (ft)	Points	Influence line ordinate of $M_F$ $\left(\text{k-ft/k}\right)$
0	A	3
12	$B$	0
24	C	$-3$
36	D	$-6$
48	E	0
60	F	6
72	G	0

Sketch the influence line diagram for the bending moment at point F as shown in Figure 5.