Textbook Problem

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8.53 through 8.57 Draw the influence lines for the forces in the members identified by an “×” of the trusses shown in Figs. P8.53–P8.57. Live loads are transmitted to the top chords of the trusses.

FIG. P8.55

To determine

Draw the influence lines for the force in member BC, BF, BG, and FG.

Explanation of Solution

Calculation:

Find the support reactions.

Apply 1 kN moving load from E to D 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 kN load placed from E to G. $\left(0\le x\le 12\text{m}\right)$.

Apply moment equilibrium at A.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_A=0\\ E_x\left(3\right)+1\left(x\right)=0\\ 3E_x=-x\\ E_x=-\frac{x}{3}\end{array}$

Apply force equilibrium equation along horizontal.

Consider the force acting towards right side as positive $\left(\to +\right)$ and the force acting towards left side as negative $\left(\leftarrow -\right)$.

$\begin{array}{l}\Sigma F_x=0\\ A_x+E_x=0\\ A_x-\frac{x}{3}=0\\ A_x=\frac{x}{3}\end{array}$

Consider joint E.

Sketch the free body diagram of the joint E as shown in Figure 2.

Refer Figure 2.

Apply vertical equilibrium at     E.

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

$\begin{array}{l}\Sigma F_y=0\\ E_y=0\end{array}$

Refer Figure 1.

Apply vertical equilibrium at     E.

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

$\begin{array}{l}\Sigma F_y=0\\ A_y+E_y=1\\ A_y+0=1\\ A_y=1\text{kN}\end{array}$

Influence line for the force in member AB.

The expressions for the member force $F_{BC}$ can be determined by passing an imaginary section aa through member FG, BG, and BC and then apply a moment equilibrium at G.

Draw the free body diagram of the section aa as shown in Figure 3.

Refer Figure 3.

Slope of line BC is 4 horizontal and 1 vertical.

Find the equation of member force BG.

Apply a 1 kN load at just left of G $\left(0\le x\le 8\text{m}\right)$.

Consider the section GD.

Apply moment equilibrium equation at G.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_G=0\\ \left(\frac{4}{\sqrt{4^2+1^2}}\right)F_{BC}\left(1\right)=0\\ F_{BC}=0\end{array}$

Apply a 1 kN load at just right of G $\left(8\text{m}\le x\le 12\text{m}\right)$.

Consider the section GD

Apply moment equilibrium equation at G.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_G=0\\ \left(\frac{4}{\sqrt{4^2+1^2}}\right)F_{BC}\left(1\right)+1\left(x-8\right)=0\\ \frac{4}{\sqrt{17}}{F}_{BC}=-x+8\\ F_{BC}=-\frac{\sqrt{17}}{4}x+2\sqrt{17}\end{array}$

Thus, the equation of force in the member BC,

$F_{BC}=0,0\le x\le 8\text{m}$ (1)

$F_{BC}=-\frac{\sqrt{17}}{4}x+2\sqrt{17},8\text{m}\le x\le 12\text{m}$ (2)

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

x (m)	Apply 1 kN load	Force in member BC (kN)	Influence line ordinate for the force in member BC (kN/kN)
0	E	$0$	$0$
4	F	$0$	0
8	G	$0$	$0$
12	D	$-\frac{\sqrt{17}}{4}\left(12\right)+2\sqrt{17}=4.123$	4.123

Sketch the influence line diagram for ordinate for the force in member BC using Table 1 as shown in Figure 4.

Influence line for the force in member FG.

Refer Figure 4.

Find the force in member FG.

Apply 1 kN load just left of F $\left(0\le x\le 4\text{m}\right)$.

Consider the section GD.

Apply moment equilibrium equation at B.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_B=0\\ F_{FG}\left(2\right)=0\\ F_{FG}=0\end{array}$

Apply a 1 kN load at just right of F $\left(4\text{m}\le x\le 12\text{m}\right)$.

Consider the section FG.

Apply moment equilibrium equation at B.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_B=0\\ -F_{FG}\left(2\right)+1\left(x-4\right)=0\\ 2F_{FG}=x-4\\ F_{FG}=\frac{x}{2}-2\end{array}$

Thus, the equation of force in the member FG,

$F_{FG}=0,0\le x\le 4\text{m}$ (3)

$F_{FG}=\frac{x}{2}-2,4\text{m}\le x\le 12\text{m}$ (4)

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

x (m)	Apply 1 kN load	Force in member FG (kN)	Influence line ordinate for the force in member FG (kN/kN)
0	E	$0$	$0$
4	F	$0$	0
8	G	2	2
12	D	$\frac{12}{2}-2=4$	4

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

Influence line for the force in member BG.

Refer Figure 4.

Slope of member is 4 horizontal and 2 vertical.

Find the force in member BG.

Apply 1 k load just left of F $\left(0\le x\le 8\text{m}\right)$

Consider the section GD.

Apply vertical force equilibrium equation.

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