Chapter 8, Problem 54P

To determine

Draw the influence lines for the force in member AB, BG, DF, and FG.

Explanation of Solution

Calculation:

Find the support reactions.

Apply 1 k moving load from E to G in the top chord member.

Draw the free body diagram of the member as in Figure 1.

Find the reaction at A and B when 1 k load placed from E to G. $\left(0\le x\le 36\text{ft}\right)$.

Apply moment equilibrium at A.

$\begin{array}{l}\Sigma M_A=0\\ -1\left(6-x\right)=B_y\left(24\right)\\ 24B_y=-\left(6-x\right)\\ B_y=\frac{x}{24}-\frac{1}{4}\end{array}$

Apply force equilibrium equation along vertical.

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

$\begin{array}{l}\Sigma F_y=0\\ A_y+B_y-1=0\\ A_y+\frac{x}{24}-\frac{1}{4}=1\\ A_y=1-\frac{x}{24}+\frac{1}{4}\\ A_y=\frac{5}{4}-\frac{x}{24}\end{array}$

Influence line for the force in member AB.

The expressions for the member force $F_{AB}$ can be determined by passing an imaginary section a-a through joint F and then apply a vertical force equilibrium.

Draw the free body diagram of member with section aa as shown in Figure 2.

Refer Figure 2.

Find the equation of member force AB.

Apply a 1 k load at just left of F $\left(0\le x\le 18\text{ft}\right)$.

Consider the right hand portion to section a-a.

Apply moment equilibrium equation at F.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_F=0\\ -B_y\left(18\right)+F_{AB}\left(16\right)=0\\ 16F_{AB}=12B_y\\ F_{AB}=0.75B_y\end{array}$

Substitute $\frac{x}{24}-\frac{1}{4}$ for $B_y$.

$\begin{array}{c}{F}_{AB}=0.75\left(\frac{x}{24}-\frac{1}{4}\right)\\ =\frac{x}{32}-\frac{3}{16}\end{array}$

Apply a 1 k load at just right of F $\left(18\text{ft}\le x\le 36\text{ft}\right)$.

Consider the left hand portion to section a-a.

Apply moment equilibrium equation at F.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_F=0\\ A_y\left(12\right)-F_{AB}\left(16\right)=0\\ 16F_{AB}=12A_y\\ F_{AB}=0.75A_y\end{array}$

Substitute $\frac{5}{4}-\frac{x}{24}$ for $A_y$.

$\begin{array}{c}{F}_{AB}=0.75\left(\frac{5}{4}-\frac{x}{24}\right)\\ =\frac{15}{16}-\frac{x}{32}\end{array}$

Thus, the equation of force in the member AB,

$F_{AB}=\frac{x}{32}-\frac{3}{16},0\le x\le 18\text{ft}$ (1)

$F_{AB}=\frac{15}{16}-\frac{x}{32},18\text{ft}\le x\le 36\text{ft}$ (2)

Find the force in member AB using the Equation (1) and (2) and then summarize the value in Table 1.

x (ft)	Apply 1 k load	Force in member AB (k)	Influence line ordinate for the force in member AB (k/k)
0	E	$\frac{0}{32}-\frac{3}{16}=-0.1875$	$-0.1875$
18	F	$\frac{18}{32}-\frac{3}{16}=0.375$	0.375
36	G	$\frac{15}{16}-\frac{36}{32}=-0.1875$	$-0.1875$

Sketch the influence line diagram for ordinate for the force in member AB using Table 2 as shown in Figure 3.

Influence line for the force in member BG.

The expressions for the member force $F_{BG}$ can be determined by passing an imaginary section a-a through the members BG, BD, and AB and then apply a moment equilibrium at F.

Draw the free body diagram of section a-a as shown in Figure 4.

Refer Figure 4.

Find the force in member BG.

Apply 1 k load just left of F $\left(0\le x\le 18\text{ft}\right)$.

Consider the section EF.

The member force of EF not affected when 1 k load applied from E to F. Therefore, the influence line ordinate of member force BG is 0 k/k from E to F.

Apply a 1 k load just the right of F $\left(18\text{ft}\le x\le 36\text{ft}\right)$.

Apply moment equilibrium at F.

Consider the section right of line a-a.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_F=0\\ -B_y\left(12\right)+F_{AB}\left(16\right)-\frac{16}{\sqrt{6^2+{16}^2}}{F}_{BG}\left(12\right)-\frac{6}{\sqrt{6^2+{16}^2}}{F}_{BG}\left(16\right)=0\\ -12B_y+16F_{AB}-11.24F_{BG}-5.62F_{BG}=0\\ 16.86F_{BG}=16F_{AB}-12B_y\\ F_{BG}=0.95F_{AB}-0.712B_y\end{array}$

Substitute $\frac{15}{16}-\frac{x}{32}$ for $F_{AB}$ and $\frac{x}{24}-\frac{1}{4}$ for $B_y$.

$\begin{array}{c}{F}_{BG}=0.95\left(\frac{15}{16}-\frac{x}{32}\right)-0.712\left(\frac{x}{24}-\frac{1}{4}\right)\\ =0.89-0.0297x-0.0297x+0.178\\ =1.068-0.0594x\end{array}$

Thus, the equation of force in the member BG,

$F_{BG}=0,0\le x\le 18\text{ft}$ (3)

$F_{BG}=1.068-0.0594x,18\text{ft}\le x\le 36\text{ft}$ (4)

Find the force in member BG using the Equation (3) and (4) and then summarize the value in Table 2.

x (ft)	Apply 1 k load	Force in member BG (k)	Influence line ordinate for the force in member BG (k/k)
0	E	$0$	0
18	F	$0$	0
36	G	$1.068-0.0594\left(36\right)=-1.07$	‑1.07

Sketch the influence line diagram for ordinate for the force in member BG using Table 2 as shown in Figure 5.

Influence line for the force in member DF.

The expressions for the member force $F_{DF}$ can be determined by passing an imaginary section bb through the members FG, DF, and AB and then apply moment equilibrium at F

Draw the free body diagram of section a-a as shown in Figure 6.

Refer Figure 6.

Find the force in member DF.

Apply 1 k load just left of F $\left(0\le x\le 18\text{ft}\right)$.

Consider the section right of line bb.

Apply moment equilibrium at G.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_G=0\\ F_{AB}\left(16\right)+B_y\left(6\right)+\frac{4}{5}{F}_{DF}\left(12\right)+\frac{3}{5}{F}_{DF}\left(8\right)=0\\ 16F_{AB}+6B_y+9.6F_{DF}+4.8F_{DF}=0\\ 14.4F_{DF}=-16F_{AB}-6B_y\end{array}$

Substitute $\frac{x}{32}-\frac{3}{16}$ for $F_{AB}$ and $\frac{x}{24}-\frac{1}{4}$ for $B_y$.

$\begin{array}{l}14.4F_{DF}=-16\left(\frac{x}{32}-\frac{3}{16}\right)-6\left(\frac{x}{24}-\frac{1}{4}\right)\\ 14.4F_{DF}=-0.5x+3-0.25x+1.5\\ 14.4F_{DF}=-0.75x+4.5\\ F_{DF}=\frac{-0.75x+4.5}{14.4}\end{array}$

Apply 1 k load just right of F $\left(18\text{ft}\le x\le 36\text{ft}\right)$.

Consider the section left of line bb.

Apply moment equilibrium at E.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_E=0\\ -F_{AB}\left(16\right)-A_y\left(6\right)+\frac{4}{5}{F}_{DF}\left(18\right)+\frac{3}{5}{F}_{DF}\left(0\right)=0\\ -16F_{AB}-6A_y+14.4F_{DF}=0\\ 14.4F_{DF}=16F_{AB}+6A_y\end{array}$

Substitute $\frac{15}{16}-\frac{x}{32}$ for $F_{AB}$ and $\frac{5}{4}-\frac{x}{24}$ for $A_y$.

$\begin{array}{l}14.4F_{DF}=16\left(\frac{15}{16}-\frac{x}{32}\right)+6\left(\frac{5}{4}-\frac{x}{24}\right)\\ 14.4F_{DF}=15-0.5x+7.5-0.25x\\ 14.4F_{DF}=-0.75x+22.5\\ F_{DF}=\frac{-0.75x+22.5}{14.4}\end{array}$

Thus, the equation of force in the member DF,

$F_{DF}=\frac{-0.75x+4.5}{14.4},0\le x\le 18\text{ft}$ (5)

$F_{DF}=\frac{-0.75x+22.5}{14.4},18\text{ft}\le x\le 36\text{ft}$ (6)

Find the force in member DF using the Equation (5) and (6) and then summarize the value in Table 3.

x (ft)	Apply 1 k load	Force in member DF (k)	Influence line ordinate for the force in member DF (k/k)
0	E	$0$.3125	0.3125
18	F	‑0.625	‑0.625
36	G	‑0.3125	‑0.3125

Sketch the influence line diagram for ordinate for the force in member DF using Table 3 as shown in Figure 7.

Influence line for the force in member FG.

Refer Figure 6.

Find the force in member FG.

Apply 1 k load just left of F $\left(0\le x\le 18\text{ft}\right)$.

Consider the section right of line bb.

Apply moment equilibrium at B.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_B=0\\ -F_{FG}\left(16\right)=0\\ F_{FG}=0\end{array}$

Apply 1 k load just right of F $\left(18\text{ft}\le x\le 36\text{ft}\right)$.

Consider the section left of line bb.

Apply moment equilibrium at A.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_A=0\\ F_{FG}\left(16\right)+\frac{4}{5}{F}_{DF}\left(12\right)+\frac{3}{5}{F}_{DF}\left(16\right)=0\\ 16F_{FG}+9.6F_{DF}+9.6F_{DF}=0\\ 16F_{FG}=-19.2F_{DF}\\ F_{FG}=-1.2F_{DF}\end{array}$

Substitute $\frac{-0.75x+22.5}{14.4}$ for $F_{DF}$.

$\begin{array}{c}{F}_{FG}=-1.2\left(\frac{-0.75x+22.5}{14.4}\right)\\ F_{FG}=\frac{0.9x-27}{14.4}\end{array}$

Thus, the equation of force in the member FG,

$F_{GF}=0,0\le x\le 18\text{ft}$ (7)

$F_{FG}=\frac{0.9x-27}{14.4},18\text{ft}\le x\le 36\text{ft}$ (8)

Find the force in member FG using the Equation (7) and (8) and then summarize the value in Table 4.

x (ft)	Apply 1 k load	Force in member FG (k)	Influence line ordinate for the force in member FG (k/k)
0	E	$0$.3125	0.3125
18	F	‑0.625	‑0.625
36	G	‑0.3125	‑0.3125

Sketch the influence line diagram for ordinate for the force in member FG using Table 4 as shown in Figure 8.