Chapter 8, Problem 21P

To determine

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

Draw the influence lines for the shear at internal hinge D.

Explanation of Solution

Calculation:

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.

$\begin{array}{l}\Sigma M_G=0\\ B_y\left(60\right)+E_y\left(24\right)-1\left(72-x\right)=0\\ 60B_y+24E_y-72+x=0\\ 24E_y=72-x-60B_y\\ E_y=3-\frac{x}{24}-2.5B_y\end{array}$        (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 C.

Find the equation of shear at C of portion AC $\left(0\le x<24\text{ft}\right)$.

Apply 1 k load just left of C.

Consider the portion AC.

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

Refer Figure 2.

Apply 1 k load just right of C.

Apply equilibrium equation of forces.

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

$\Sigma F_y=0$

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

Substitute $1.5-\frac{x}{24}$ for $B_y$.

$\begin{array}{c}{S}_C=\left(1.5-\frac{x}{24}\right)-1\\ =0.5-\frac{x}{24}\end{array}$

Find the equation of shear at C of portion CG $\left(24\text{ft}<x\le 72\text{ft}\right)$.

Consider the portion AC.

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

$\Sigma F_y=0$

$\begin{array}{l}-S_C+B_y=0\\ S_C=B_y\end{array}$        (9)

Find the influence line ordinate of shear at C$\left(S_C\right)$ for portion CD $\left(24\text{ft}\le x\le 36\text{ft}\right)$.

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

$S_C=1.5-\frac{x}{24}$

Find the influence line ordinate of shear at C$\left(S_C\right)$ for portion DG $\left(36\text{ft}\le x\le 72\text{ft}\right)$.

Substitute 0 for $B_y$ in Equation (9).

$S_C=0$

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

$S_C=0.5-\frac{x}{24},0\le x\le 24\text{ft}$        (10)

$S_C=\frac{x}{24}-1.5,24\text{ft}\le x\le 36\text{ft}$        (11)

$S_C=0,36\text{ft}\le x\le 72\text{ft}$        (12)

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

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

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

Influence line for moment at point B:

Refer Figure 2.

Find the equation of bending moment at B of portion AC$\left(0\le x<24\text{ft}\right)$.

Take moment at C.

Consider clockwise moment as positive and anticlockwise moment as negative.

$M_C=B_y\left(12\right)-1\left(24-x\right)$

Substitute $1.5-\frac{x}{24}$ for $B_y$.

$\begin{array}{c}{M}_C=\left(1.5-\frac{x}{24}\right)\left(12\right)-1\left(24-x\right)\\ =18-\frac{x}{2}-24+x\\ =\frac{x}{2}-6\end{array}$

Refer Figure 3.

Find the equation of bending moment at C of portion CG.

Take moment at C.

Consider clockwise moment as negative and anticlockwise moment as positive.

$M_C=B_y\left(12\right)$        (13)

Find the influence line ordinate of moment at C$\left(M_C\right)$ for portion CD$\left(24\text{ft}\le x\le 36\text{ft}\right)$.

Substitute $1.5-\frac{x}{24}$ for $B_y$.

$\begin{array}{c}{M}_C=\left(1.5-\frac{x}{24}\right)\left(12\right)\\ =18-0.5x\end{array}$

Find the influence line ordinate of moment at C $\left(M_C\right)$ for portion DG$\left(36\text{ft}\le x\le 72\text{ft}\right)$.

Substitute 0 for $B_y$ in Equation (13).

$\begin{array}{c}{M}_C=0\left(12\right)\\ =0\end{array}$

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

$M_C=\frac{x}{2}-6,0\le x<24\text{ft}$        (14)

$M_C=18-0.5x,24\text{ft}\le x\le 36\text{ft}$        (15)

$M_C=0,36\text{ft}\le x\le 72\text{ft}$        (16)

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

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

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

Influence line for shear at hinge D:

Find the equation of shear at D of portion AC$\left(0\le x<36\text{m}\right)$.

Sketch the free body diagram of the section AD as shown in Figure 6.

Refer Figure 6.

Apply equilibrium equation of forces.

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

$\Sigma F_y=0$

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

Substitute $1.5-\frac{x}{24}$ for $B_y$.

$\begin{array}{c}{S}_D=\left(1.5-\frac{x}{24}\right)-1\\ =0.5-\frac{x}{24}\end{array}$

Find the equation of shear at C of portion DG $\left(36\text{ft}<x\le 72\text{ft}\right)$.

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

Refer Figure 7.

Apply equilibrium equation of forces.

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

$\Sigma F_y=0$

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

Substitute 0 for $B_y$.

$S_D=0$

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

$S_D=0.5-\frac{x}{24},0\le x<36\text{ft}$        (17)

$S_D=0,36\text{ft}<x\le 72\text{ft}$        (18)

Find the influence line ordinate of $S_D$ using Equation (17) and (18) and summarize the values in Table 3.

x (ft)	Points	Influence line ordinate of $S_D$$\left(\text{k/k}\right)$
0	A	$0.5$
12	$B$	0
24	$C$	$-0.5$
36	D	$-1$
48	E	0
60	F	0
72	G	0

Sketch the influence line diagram for the shear at point D as shown in Figure 8.