Chapter 8, Problem 37P

To determine

Draw the influence lines for the reaction moment at support A, the vertical reactions at supports A and F and the shear and bending moment at point E.

Explanation of Solution

Calculation:

Influence line for moment at support A.

Apply a 1 kN unit moving load at a distance of x from left end C.

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

Refer Figure 1.

Apply 1 kN load just left of C $\left(0\le x\le 5\text{m}\right)$.

Take moment at A from B.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{c}{M}_A=-1\left(5-x\right)\\ =x-5\end{array}$

Apply 1 kN load just right of C and just left of D $\left(5\text{m}\le x\le 10\text{m}\right)$.

Take moment at A from D.

$\begin{array}{c}{M}_A=1\left(x-5\right)\\ =x-5\end{array}$

Apply 1 kN load just right of D and just right of F $\left(10\text{m}\le x\le 20\text{m}\right)$.

Take moment at A from F.

$\begin{array}{c}{M}_A=1\left(x-5\right)-F_y\left(15\right)\\ =x-5-15F_y\end{array}$

Thus, the equation of moment at A as follows,

$M_A=x-5,\left(0\le x\le 10\text{m}\right)$ (1)

$M_A=x-5-15F_y,\left(10\text{m}\le x\le \text{20}\text{m}\right)$ (2)

Find the influence line ordinate of $M_A$ at B using Equation (1).

Substitute 0 for $x$ in Equation (1).

$\begin{array}{c}{M}_A=0-5\\ =-5\text{kN-m}\end{array}$

Find the influence line ordinate of $M_A$ at F using Equation (2).

The vertical reaction at F is 1 kN when 1 kN applied at F.

Substitute 20 m for $x$ and 1 kN for $F_y$ in Equation (2).

$\begin{array}{c}{M}_A=20-5-15\left(1\right)\\ =15-15\\ =0\end{array}$

Thus, the influence line ordinate of $M_A$ at F is $0\text{kN-m/kN}$.

Similarly calculate the influence line ordinate of $M_A$ at various points on the frame and summarize the values in Table 1.

x (m)	Points	Influence line ordinate of $M_A$$\left(\text{kN-m/kN}\right)$
0	B	‑5
5	C	0
10	D	5
20	F	0

Sketch the influence line diagram for the moment at support A using Table 1 as shown in Figure 2.

Influence line for vertical reaction at support F.

Apply a 1 kN unit moving load at a distance of x from left end C.

Refer Figure 1.

Find the vertical support reaction $\left(F_y\right)$ at F using equilibrium equation:

Apply 1 kN load just left of D $\left(0\le x\le 10\text{m}\right)$.

Consider section DF.

Consider moment equilibrium at point D.

Consider clockwise moment as positive and anticlockwise moment as negative

$\begin{array}{l}\Sigma M_D=0\\ -F_y\left(10\right)=0\\ F_y=0\end{array}$

Apply 1 kN load just right of D $\left(10\text{m}\le x\le 20\text{m}\right)$.

Consider section DF.

Consider moment equilibrium at point D.

Consider clockwise moment as positive and anticlockwise moment as negative

$\begin{array}{l}\Sigma M_D=0\\ -F_y\left(10\right)+1\left(x-10\right)=0\\ 10F_y=x-10\\ F_y=\frac{x}{10}-1\end{array}$

Thus, the equation of vertical support reaction at F as follows,

$F_y=0,\left(0\le x\le 10\text{m}\right)$ (3)

$F_y=\frac{x}{10}-1,\left(10\text{m}\le x\le \text{20}\text{m}\right)$ (4)

Find the influence line ordinate of $F_y$ at F using Equation (2).

Substitute 20 for $x$ in Equation (1).

$\begin{array}{c}{F}_y=\frac{20}{10}-1\\ =2-1\\ =1\text{kN}\end{array}$

Thus, the influence line ordinate of $F_y$ at F is $1\text{kN/kN}$.

Similarly calculate the influence line ordinate of $F_y$ at various points on the frame and summarize the values in Table 2.

x (m)	Points	Influence line ordinate of $F_y$$\left(\text{kN/kN}\right)$
0	B	0
5	C	0
10	D	0
15	E	0.5
20	F	1

Sketch the influence line diagram for the vertical reaction at support F using Table 2 as shown in Figure 3.

Influence line for vertical reaction at support A.

Apply a 1 kN unit moving load at a distance of x from left end C.

Refer Figure 1.

Apply vertical equilibrium in the system.

Consider upward force as positive and downward force as negative.

$\begin{array}{l}{A}_y+F_y-1=0\\ A_y=1-F_y\end{array}$ (5)

Find the equation of vertical support reaction $\left(A_y\right)$ from B to C $\left(0\le x\le 10\text{m}\right)$ using Equation (5).

Substitute 0 for $F_y$ in Equation (5).

$\begin{array}{c}{A}_y=1-\left(0\right)\\ =1\text{kN}\end{array}$

Find the equation of vertical support reaction $\left(A_y\right)$ from C to F $\left(10\text{m}\le x\le 20\text{m}\right)$ using Equation (5).

Substitute $\frac{x}{10}-1$ for $F_y$ in Equation (5).

$\begin{array}{c}{A}_y=1-\left(\frac{x}{10}-1\right)\\ =1-\frac{x}{10}+1\\ =2-\frac{x}{10}\end{array}$

Thus, the equation of vertical support reaction at A as follows,

$A_y=1\text{kN},\left(0\le x\le 10\text{m}\right)$ (6)

$A_y=2-\frac{x}{10},\left(10\text{m}\le x\le \text{20}\text{m}\right)$ (7)

Find the influence line ordinate of $A_y$ at F using Equation (7).

Substitute 20 m for $x$ in Equation (7).

$\begin{array}{c}{A}_y=2-\frac{20}{10}\\ =2-2\\ =0\end{array}$

Thus, the influence line ordinate of $F_y$ at F is $1\text{kN/kN}$.

Similarly calculate the influence line ordinate of $A_y$ at various points on the beam and summarize the values in Table 3.

x (m)	Points	Influence line ordinate of $A_y$$\left(\text{kN/kN}\right)$
0	B	1
5	C	1
10	D	1
15	E	0.5
20	F	0

Sketch the influence line diagram for the vertical reaction at support A using Table 3 as shown in Figure 4.

Influence line for shear at point E.

Sketch the free body diagram of the section BD as shown in Figure 5.

Refer Figure 5.

Apply equilibrium equation of forces.

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

$\Sigma F_y=0$

$\begin{array}{l}{A}_y-S_E-1=0\\ S_E=A_y-1\end{array}$ (8)

Find the equation of shear force at E of portion BD $\left(0\le x\le 10\text{m}\right)$.

Substitute $1\text{kN}$ for $A_y$ in Equation (8).

$\begin{array}{c}{S}_E=1-1\\ =0\end{array}$

Find the equation of shear force at E of portion DE $\left(10\text{m}\le x\le 15\text{m}\right)$.

Substitute $2-\frac{x}{10}$ for $A_y$ in Equation (8).

$\begin{array}{c}{S}_E=\left(2-\frac{x}{10}\right)-1\\ =2-\frac{x}{10}-1\\ =1-\frac{x}{10}\end{array}$

Find the equation of shear force at E of portion EF $\left(10\text{m}<x\le 20\text{m}\right)$.

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

$\Sigma F_y=0$

$\begin{array}{l}{S}_E-1+F_y=0\\ S_E=1-F_y\end{array}$

Substitute $\frac{x}{10}-1$ for $F_y$.

$\begin{array}{c}{S}_E=1-\left(\frac{x}{10}-1\right)\\ =1-\frac{x}{10}+1\\ =2-\frac{x}{10}\end{array}$

Thus, the equations of the influence line for $S_E$ as follows,

$S_E=0,0\le x<10\text{m}$ (9)

$S_E=1-\frac{x}{10},10\text{m}\le x<15\text{m}$ (10)

$S_E=2-\frac{x}{10},15\text{m}<x\le 20\text{m}$ (11)

Find the influence line ordinate of $S_E$ at just left of E using Equation (10).

Substitute 15 m for $x$ in Equation (1).

$\begin{array}{c}{S}_E=1-\frac{15}{10}\\ =1-\frac{3}{2}\\ =-\frac{1}{2}\text{kN}\end{array}$

Thus, the influence line ordinate of $S_E$ at just left of E is $-0.5\text{kN-m/kN}$.

Find the shear force of $S_E$ at various points of x using the Equations (8) and (9) and summarize the value as in Table 4.

x (m)	Points	Influence line ordinate of $S_E$$\left(\text{kN/kN}\right)$
0	B	0
5	$C$	0
10	$D$	0
15	$E^{-}$	$-\frac{1}{2}$
15	$E^{+}$	$\frac{1}{2}$
20	F	0

Draw the influence lines for the shear force at point E using Table 4 as shown in Figure 7.

Influence line for moment at point E.

Refer Figure 5.

Consider section BE.

Consider clockwise moment as positive and anticlockwise moment as negative.

Take moment at E.

Find the equation of moment at E of portion BE.

$M_E=-A_y\left(10\right)+\left(1\right)\left(15-x\right)-M_A$ (12)

Find the equation of moment at E of portion BD $\left(0\le x<10\text{m}\right)$.

Substitute $1\text{kN}$ for $A_y$ and $x-5$ for $M_A$ in Equation (12)

$\begin{array}{c}{M}_E=\left(1\right)\left(10\right)-1\left(15-x\right)-\left(x-5\right)\\ =10-15+x-x+5\\ =0\end{array}$

Find the equation of moment at E of portion DE $\left(10\text{m}\le x<15\text{m}\right)$.

Substitute $2-\frac{x}{10}$ for $A_y$ and $x-5-15F_y$ for $M_A$ in Equation (12).

$\begin{array}{c}{M}_E=\left(2-\frac{x}{10}\right)\left(10\right)-1\left(15-x\right)-\left(x-5-15F_y\right)\\ =20-x-15+x-x+5+15F_y\\ M_E=10-x+15F_y\end{array}$

Substitute $\frac{x}{10}-1$ for $F_y$.

$\begin{array}{c}{M}_E=10-x+15\left(\frac{x}{10}-1\right)\\ =10-x+\frac{3x}{2}-15\\ =\frac{x}{2}-5\end{array}$

Refer Figure 6.

Consider section EF.

Find the equation of moment at E of portion EF $\left(15\text{m}\le x<20\text{m}\right)$.

Consider clockwise moment as positive and anticlockwise moment as negative.

Take moment at E.

Find the equation of moment at E of portion EF.

$M_E=-F_y\left(5\right)-1\left[5-\left(20-x\right)\right]$

Substitute $\frac{x}{10}-1$ for $F_y$.

$\begin{array}{c}{M}_E=\left(\frac{x}{10}-1\right)\left(5\right)-5+20-x\\ =\frac{x}{2}-5-5+20-x\\ =10-\frac{x}{2}\end{array}$

Thus, the equations of the influence line for $M_E$ as follows,

$M_E=0,0\le x<10\text{m}$ (13)

$M_E=\frac{x}{2}-5,10\text{m}\le x<15\text{m}$ (14)

$M_E=10-\frac{x}{2},15\text{m}<x\le 20\text{m}$ (15)

Find the influence line ordinate of $M_E$ at E using Equation (14).

Substitute 15 m for $x$ in Equation (1).

$\begin{array}{c}{M}_E=\frac{15}{2}-5\\ =\frac{5}{2}\text{kN-m}\end{array}$

Thus, the influence line ordinate of $M_E$ at E is $\frac{5}{2}\text{kN-m/kN}$.

Find the moment at various points of x using the Equations (5) and (6) and summarize the value as in Table 5.

x (m)	Points	Influence line ordinate of $M_E$$\left(\text{kN-m/kN}\right)$
0	B	0
5	C	0
10	D	0
15	E	$\frac{5}{2}$
20	F	0

Draw the influence lines for the moment at point E using Table 5 as shown in Figure 8.

Therefore, the influence lines for the vertical reactions at supports A and F and the influence lines for the shear and bending moment at point E are drawn.