Chapter 8, Problem 51P

To determine

Draw the influence lines for the force in member CD, CI, DI, and DJ.

Explanation of Solution

Calculation:

Find the support reactions.

Apply 1 k moving load from A to G in the bottom chord member.

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

Refer Figure 1,

Find the reaction at C and E when 1 k load placed from A to G. $\left(0\le x\le 96\text{ft}\right)$.

Apply moment equilibrium at C.

$\begin{array}{l}\Sigma M_C=0\\ -1\left(32-x\right)-E_y\left(32\right)=0\\ 32E_y=-\left(32-x\right)\\ 32E_y=x-32\\ E_y=\frac{x}{32}-1\end{array}$

Apply force equilibrium equation along vertical.

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

$\begin{array}{l}\Sigma F_y=0\\ C_y+E_y-1=0\\ C_y+\frac{x}{32}-1=1\\ C_y=2-\frac{x}{32}\end{array}$

Influence line for the force in member CD.

The expressions for the member force $F_{CD}$ can be determined by passing an imaginary section aa through the members HI, CI, and CD and then apply a moment equilibrium at I.

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

Refer Figure 2.

Apply 1 k load just the left of C $\left(0\le x\le 32\text{ft}\right)$

Find the equation of member force CD from A to C.

Consider the section DG.

Apply moment equilibrium equation at I.

Consider clockwise moment as negative and anticlockwise moment as positive.

$\Sigma M_I=0$

$\begin{array}{l}-F_{CD}\left(20\right)+E_y\left(32\right)=0\\ 20F_{CD}=32E_y\\ F_{CD}=\frac{8}{5}{E}_y\end{array}$

Substitute $\frac{x}{32}-1$ for $E_y$.

$\begin{array}{c}{F}_{CD}=\frac{8}{5}\left(\frac{x}{32}-1\right)\\ =\frac{x}{20}-\frac{8}{5}\end{array}$

Apply 1 k load just the right of C $\left(32\text{ft}\le x\le 96\text{ft}\right)$.

Find the equation of member force CD from C to G.

Consider the section AC.

Apply moment equilibrium equation at I.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\Sigma M_I=0$

$\begin{array}{l}{F}_{CD}\left(20\right)=0\\ F_{CD}=0\end{array}$

Thus, the equation of force in the member CD,

$F_{CD}=\frac{x}{20}-\frac{8}{5},0\le x\le 32\text{ft}$ (1)

$F_{CD}=0,32\text{ft}\le x\le 96\text{ft}$ (2)

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

x (ft)	Apply 1 k load	Force in member CD (k)	Influence line ordinate for the force in member CD (k/k)
0	A	$\begin{array}{c}\frac{x}{20}-\frac{8}{5}=\frac{0}{20}-\frac{8}{5}\\ =-1.6\end{array}$	$-1.6$
16	B	$\begin{array}{c}\frac{x}{20}-\frac{8}{5}=\frac{16}{20}-\frac{8}{5}\\ =-0.8\end{array}$	$-0.8$
32	C	0	0
48	D	0	0
64	E	0	0
80	F	0	0
96	G	0	0

Sketch the influence line diagram for ordinate for the force in member CD using Table 1 as shown in Figure 3.

Influence line for the force in member CI.

Refer Figure 2.

Apply 1 k load just the left of C $\left(0\le x\le 32\text{ft}\right)$.

Find the equation of member force CI from A to C.

Consider the section AC.

Apply moment equilibrium equation at H.

Consider clockwise moment as negative and anticlockwise moment as positive.

$\begin{array}{l}\Sigma M_H=0\\ C_y\left(16\right)+F_{CD}\left(12\right)+F_{CI}\left(16\right)+1\left(16-x\right)=0\\ 16F_{CI}=-16C_y-12F_{CD}-\left(16-x\right)\\ F_{CI}=-\frac{16}{16}{C}_y-\frac{12}{16}{F}_{CD}-\frac{16}{16}+\frac{x}{16}\\ F_{CI}=-C_y-\frac{3}{4}{F}_{CD}-1+\frac{x}{16}\end{array}$

Substitute $2-\frac{x}{32}$ for $C_y$ and $\frac{x}{20}-\frac{8}{5}$ for $F_{CD}$.

$\begin{array}{c}{F}_{CI}=-\left(2-\frac{x}{32}\right)-\frac{3}{4}\left(\frac{x}{20}-\frac{8}{5}\right)-1+\frac{x}{16}\\ =-2+\frac{x}{32}-\frac{3x}{80}+\frac{6}{5}-1+\frac{x}{16}\\ =\frac{9x}{160}-1.8\end{array}$

Apply 1 k load just the right of C $\left(32\text{ft}\le x\le 96\text{ft}\right)$

Find the equation of member force CI from C to G.

Consider the section AC.

Apply moment equilibrium equation at H.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_H=0\\ C_y\left(16\right)+F_{CD}\left(12\right)+F_{CI}\left(16\right)=0\\ 16F_{CI}=-16C_y-12F_{CD}\\ F_{CI}=-\frac{16}{16}{C}_y-\frac{12}{16}{F}_{CD}\\ F_{CI}=-C_y-\frac{3}{4}{F}_{CD}\end{array}$

Substitute $2-\frac{x}{32}$ for $C_y$ and $0$ for $F_{CD}$.

$\begin{array}{c}{F}_{CI}=-\left(2-\frac{x}{32}\right)-\frac{3}{4}\left(0\right)\\ =-2+\frac{x}{32}\\ =\frac{x}{32}-2\end{array}$

Thus, the equation of force in the member CI,

$F_{CI}=\frac{9x}{160}-1.8,0\le x\le 32\text{ft}$ (3)

$F_{CI}=\frac{x}{32}-2,32\text{ft}<x\le 96\text{ft}$ (4)

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

x (ft	Apply 1 k load	Force in member CI (k)	Influence line ordinate for the force in member CI (k/k)
0	A	$\begin{array}{c}\frac{9x}{160}-1.8=\frac{9\left(0\right)}{160}-1.8\\ =-1.8\end{array}$	$-1.8$
16	B	$\begin{array}{c}\frac{9x}{160}-1.8=\frac{9\left(16\right)}{160}-1.8\\ =-0.9\end{array}$	$-0.9$
32	C	$\begin{array}{c}\frac{9x}{160}-1.8=\frac{9\left(32\right)}{160}-1.8\\ =0\end{array}$	0
48	D	$\begin{array}{c}\frac{x}{32}-2=\frac{48}{32}-2\\ =-0.5\end{array}$	$-0.5$
64	E	$\begin{array}{c}\frac{x}{32}-2=\frac{64}{32}-2\\ =0\end{array}$	0
80	F	$\begin{array}{c}\frac{x}{32}-2=\frac{80}{32}-2\\ =0.5\end{array}$0	0.5
96	G	$\begin{array}{c}\frac{x}{32}-2=\frac{96}{32}-2\\ =1\end{array}$	1

Sketch the influence line diagram for ordinate for the force in member CI using Table 2 as shown in Figure 4.

Influence line for the force in member DI.

The expressions for the member force $F_{DI}$ can be determined by passing an imaginary section bb through the members IJ, DI and CD and then apply a moment equilibrium at J.

Draw the free body diagram of the section bb as shown in Figure 5.

Refer Figure 5.

Apply 1 k load just the left of C $\left(0\le x\le 32\text{ft}\right)$.

Find the equation of member force DI from A to C.

Consider the section DG.

Apply moment equilibrium equation at J.

Consider clockwise moment as negative and anticlockwise moment as positive.

$\begin{array}{l}\Sigma M_J=0\\ E_y\left(16\right)-F_{CD}\left(24\right)-\left(\frac{16}{\sqrt{{16}^2+{20}^2}}\right)F_{DI}\left(24\right)=0\\ 15F_{DI}=16E_y-24F_{CD}\\ F_{DI}=\frac{16}{15}{E}_y-1.6F_{CD}\end{array}$

Substitute $\frac{x}{20}-\frac{8}{5}$ for $F_{CD}$ and $\frac{x}{32}-1$ for $E_y$.

$\begin{array}{c}{F}_{DI}=\frac{16}{15}\left(\frac{x}{32}-1\right)-1.6\left(\frac{x}{20}-\frac{8}{5}\right)\\ =\frac{x}{30}-\frac{16}{15}-\frac{2x}{25}+\frac{64}{25}\\ =\frac{112}{75}-\frac{7x}{150}\end{array}$

Apply 1 k load just the right of C $\left(32\text{ft}\le x\le 96\text{ft}\right)$.

Find the equation of member force DI from C to G.

Consider the section AC.

Apply moment equilibrium equation at J.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_J=0\\ C_y\left(16\right)-F_{CD}\left(24\right)-\frac{20}{\sqrt{{16}^2+{20}^2}}{F}_{DI}\left(16\right)-\frac{16}{\sqrt{{16}^2+{20}^2}}{F}_{DI}\left(4\right)=0\\ 12.5F_{DI}+2.5F_{DI}=16C_y-24F_{CD}\\ 15F_{DI}=16C_y-24F_{CD}\\ F_{DI}=\frac{16}{15}{C}_y-\frac{24}{15}{F}_{CD}\end{array}$

Substitute $2-\frac{x}{32}$ for $C_y$ and 0 for $F_{CD}$.

$\begin{array}{c}{F}_{DI}=\frac{16}{15}\left(2-\frac{x}{32}\right)-\frac{24}{15}\left(0\right)\\ =\frac{32}{15}-\frac{x}{30}\end{array}$

Thus, the equation of force in the member DI,

$F_{DI}=\frac{112}{75}-\frac{7x}{150},0\le x\le 32\text{ft}$ (5)

$F_{DI}=\frac{32}{15}-\frac{x}{30},32\text{ft}\le x\le 96\text{ft}$ (6)

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

x (ft)	Apply 1 k load	Force in member DI (k)	Influence line ordinate for the force in member DI (k/k)
0	A	$\begin{array}{c}\frac{112}{75}-\frac{7x}{150}=\frac{112}{75}-\frac{7\left(0\right)}{150}\\ =1.494\end{array}$	1.494
16	B	$\begin{array}{c}\frac{112}{75}-\frac{7x}{150}=\frac{112}{75}-\frac{7\left(16\right)}{150}\\ =0.747\end{array}$	0.747
32	C	$\begin{array}{c}\frac{112}{75}-\frac{7x}{150}=\frac{112}{75}-\frac{7\left(32\right)}{150}\\ =0\end{array}$	0
48	D	$\begin{array}{c}\frac{32}{15}-\frac{x}{30}=\frac{32}{15}-\frac{48}{30}\\ =0.534\end{array}$	0.534
64	E	0	0
80	F	$\begin{array}{c}\frac{32}{15}-\frac{x}{30}=\frac{32}{15}-\frac{80}{30}\\ =-0.534\end{array}$0	$-0.534$
96	G	$\begin{array}{c}\frac{32}{15}-\frac{x}{30}=\frac{32}{15}-\frac{96}{30}\\ =-1.067\end{array}$	$-1.067$

Sketch the influence line diagram for ordinate for the force in member DI using Table 3 as shown in Figure 6.

Influence line for the force in member DJ.

The expressions for the member force $F_{DJ}$ can be determined by passing an imaginary section cc through the members JK, DJ, DI, and CD and then apply a moment equilibrium at K.

Draw the free body diagram of the section cc as shown in Figure 7.

Refer Figure 7.

Apply 1 k load just the left of C $\left(0\le x\le 32\text{ft}\right)$

Find the equation of member force DJ from A to C.

Consider the section DG.

Apply moment equilibrium equation at C.

The member force DI is resolved in horizontal and vertical.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_K=0\\ F_{CD}\left(20\right)+\frac{16}{\sqrt{{16}^2+{20}^2}}{F}_{DI}\left(20\right)+\frac{20}{\sqrt{{16}^2+{20}^2}}{F}_{DI}\left(16\right)+F_{DJ}\left(16\right)=0\\ 20F_{CD}+12.494F_{DI}+12.494F_{DI}+16F_{DJ}=0\\ 16F_{DJ}=-20F_{CD}-25F_{DI}\\ F_{DJ}=-1.25F_{CD}-1.5625F_{DI}\end{array}$

Substitute $\frac{x}{20}-\frac{8}{5}$ for $F_{CD}$ and $\frac{112}{75}-\frac{7x}{150}$ for $F_{DI}$.

$\begin{array}{c}{F}_{DJ}=-1.25\left(\frac{x}{20}-\frac{8}{5}\right)-1.5625\left(\frac{112}{75}-\frac{7x}{150}\right)\\ =-0.0625x+2-2.333+0.07292x\\ =0.0104x-0.333\end{array}$

Apply 1 k load just the right of C $\left(32\text{ft}\le x\le 96\text{ft}\right)$.

Find the equation of member force DJ from C to G.

Consider the section DG.

Apply moment equilibrium equation at K.

The member force DI is resolved in horizontal and vertical.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_K=0\\ -F_{CD}\left(20\right)-\frac{20}{\sqrt{{16}^2+{20}^2}}{F}_{DI}\left(32\right)-F_{DJ}\left(16\right)+C_y\left(32\right)=0\\ -20F_{CD}-25F_{DI}-16F_{DJ}+32C_y=0\end{array}$

Substitute 0 for $F_{CD}$, $\frac{32}{15}-\frac{x}{30}$ for $F_{DI}$, and $2-\frac{x}{32}$ for $C_y$.

$\begin{array}{l}-20\left(0\right)-25\left(\frac{32}{15}-\frac{x}{30}\right)-16F_{DJ}+32\left(2-\frac{x}{32}\right)=0\\ 0-53.333+0.833x-16F_{DJ}+64-x=0\\ 10.667-0.167x-16F_{DJ}=0\\ 16F_{DJ}=10.667-0.167x\\ F_{DJ}=0.667-0.0104x\end{array}$

Thus, the equation of force in the member DJ,

$F_{DJ}=0.0104x-0.333,0\le x\le 32\text{ft}$ (7)

$F_{DJ}=0.0104x-0.667,32\text{ft}\le x\le 96\text{ft}$ (8)

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

x (ft)	Apply 1 k load	Force in member DJ (k)	Influence line ordinate for the force in member DJ (k/k)
0	A	$0.0104\left(0\right)-0.333=-0.333$	$-0.333$
16	B	$0.0104\left(16\right)-0.333=-0.167$	$-0.167$
32	C	$0.0104\left(32\right)-0.333=0$	0
48	D	$0.0104\left(48\right)-0.667=0.167$	0.167
64	E	$0.667-0.0104\left(64\right)=0$	0
80	F	$0.667-0.0104\left(80\right)=-0.167$	$-0.167$
96	G	$0.667-0.0104\left(96\right)=-0.333$	$-0.333$

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