Chapter 8, Problem 38P

To determine

Draw the influence lines for the reaction moment at support A, the vertical reactions at supports A and B and the shear at the internal hinge C.

Explanation of Solution

Calculation:

Influence line for vertical reaction at support B.

Apply a 1 k unit moving load at a distance of x from left end B.

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

Refer Figure 1.

Apply 1 k load just left of C $\left(0\le x\le 10\text{ft}\right)$.

Consider section BC.

Consider moment equilibrium at C.

Take moment at C from B.

Consider clockwise moment as positive and anticlockwise moment as negative.

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

Apply 1 k load just right of C $\left(10\text{ft}\le x\le 40\text{ft}\right)$.

Consider section BC.

Consider moment equilibrium at C.

Take moment at C from B.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{l}\Sigma M_C=0\\ B_y\left(10\right)=0\\ B_y=0\end{array}$

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

$B_y=1-\frac{x}{10},\left(0\le x\le 10\text{ft}\right)$        (1)

$B_y=0,\left(10\text{ft}\le x\le 40\text{ft}\right)$        (2)

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

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

$\begin{array}{c}{B}_y=1-\frac{0}{10}\\ =1\text{k}\end{array}$

Thus, the influence line ordinate of $B_y$ at B is $1\text{k/k}$.

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

x (ft)	Points	Influence line ordinate of $B_y$$\left(\text{k/k}\right)$
0	B	1
10	C	0
20	D	0
30	E	0
40	F	0

Sketch the influence line diagram for vertical support reaction at B using Table 1 as shown in Figure 2.

Influence line for vertical reaction at support A.

Apply a 1 k 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 k load just left of E $\left(0\le x\le 30\text{ft}\right)$.

Consider section EF.

Consider moment equilibrium at point E.

Consider clockwise moment as positive and anticlockwise moment as negative

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

Apply 1 k load just right of E $\left(30\text{ft}\le x\le 40\text{ft}\right)$.

Consider section EF.

Consider moment equilibrium at point E.

Consider clockwise moment as positive and anticlockwise moment as negative

$\begin{array}{l}\Sigma M_E=0\\ F_y\left(10\right)+1\left(x-30\right)=0\\ 10F_y=-x+30\\ F_y=3-\frac{x}{10}\end{array}$

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

$F_y=0,\left(30\text{ft}\le x\le 40\text{ft}\right)$        (3)

$F_y=3-\frac{x}{10},\left(30\text{ft}\le x\le 40\text{ft}\right)$        (4)

Apply a 1 k unit moving load at a distance of x from left end B.

Refer Figure 1.

Apply vertical equilibrium in the system.

Consider upward force as positive and downward force as negative.

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

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

Substitute $1-\frac{x}{10}$ for $B_y$ and 0 for $F_y$ in Equation (5).

$\begin{array}{c}{A}_y=1-\left(1-\frac{x}{10}\right)+0\\ =1-1+\frac{x}{10}\\ =\frac{x}{10}\end{array}$

Find the equation of vertical support reaction $\left(A_y\right)$ from C to E $\left(10\text{ft}\le x\le 30\text{ft}\right)$ using Equation (5).

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

$\begin{array}{c}{A}_y=1-0-0\\ =1\text{k}\end{array}$

Find the equation of vertical support reaction $\left(A_y\right)$ from E to F $\left(30\text{ft}\le x\le 40\text{ft}\right)$ using Equation (5).

Substitute $0$ for $B_y$ and $3-\frac{x}{10}$ for $F_y$ in Equation (5).

$\begin{array}{c}{A}_y=1-0+\left(3-\frac{x}{10}\right)\\ =1+3-\frac{x}{10}\\ =4-\frac{x}{10}\end{array}$

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

$A_y=\frac{x}{10},\left(0\le x\le 10\text{ft}\right)$        (6)

$A_y=1\text{k},\left(10\text{ft}\le x\le 30\text{ft}\right)$        (7)

$A_y=4-\frac{x}{10},\left(30\text{ft}\le x\le \text{40}\text{ft}\right)$        (8)

Find the influence line ordinate of $A_y$ at F using Equation (8).

Substitute 40 ft for $x$ in Equation (8).

$\begin{array}{c}{A}_y=4-\frac{40}{10}\\ =4-4\\ =0\end{array}$

Thus, the influence line ordinate of $A_y$ at F is $0\text{k/k}$.

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

x (ft)	Points	Influence line ordinate of $A_y$$\left(\text{k/k}\right)$
0	B	0
10	C	1
20	D	1
30	E	1
40	F	0

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

Influence line for moment at support A.

Apply a 1 k unit moving load at a distance of x from left end B.

Refer Figure 1.

Apply 1 k load just left of C $\left(0\le x\le 10\text{ft}\right)$.

Take moment at A from B.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{c}{M}_A=B_y\left(20\right)-1\left(20-x\right)\\ =20B_y-20+x\end{array}$

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

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

Apply 1 k load just right of C to just left of D $\left(10\text{ft}\le x\le 20\text{ft}\right)$.

Take moment at A from B.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{c}{M}_A=B_y\left(20\right)-1\left(20-x\right)\\ =20B_y-20+x\end{array}$

Substitute $0$ for $B_y$.

$\begin{array}{c}{M}_A=20\left(0\right)-20+x\\ =x-20\end{array}$

Apply 1 k load just right of D to just left of E $\left(20\text{ft}\le x\le 30\text{ft}\right)$.

Take moment at A from F.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{c}{M}_A=F_y\left(20\right)-1\left(20-x\right)\\ =20F_y-20+x\end{array}$

Substitute $0$ for $F_y$.

$\begin{array}{c}{M}_A=20\left(0\right)-20+x\\ =x-20\end{array}$

Apply 1 k load just right of E $\left(30\text{ft}\le x\le 40\text{ft}\right)$.

Take moment at A from F.

Consider clockwise moment as positive and anticlockwise moment as negative.

$\begin{array}{c}{M}_A=F_y\left(20\right)-1\left(20-x\right)\\ =20F_y-20+x\end{array}$

Substitute $0$ for $F_y$.

$\begin{array}{c}{M}_A=20\left(3-\frac{x}{10}\right)-20+x\\ =60-2x-20+x\\ =40-x\end{array}$

Thus, the equation of moment at A as follows,

$M_A=-x,\left(0\le x\le 10\text{ft}\right)$        (9)

$M_A=x-20,\left(10\text{ft}\le x\le 30\text{ft}\right)$        (10)

$M_A=40-x,\left(30\text{ft}\le x\le \text{40}\text{ft}\right)$        (11)

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

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

$\begin{array}{c}{M}_A=-0\\ =0\text{k-ft}\end{array}$

Thus, the influence line ordinate of $M_A$ at B is $0\text{k-ft/k}$.

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

x (ft)	Points	Influence line ordinate of $M_A$$\left(\text{k-ft/k}\right)$
0	B	0
10	C	‑10
20	D	0
30	E	10
40	F	0

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

Influence line for shear at point C.

Find the equation of shear force at C of portion BC $\left(0\le x\le 10\text{ft}\right)$.

Sketch the free body diagram of the section BC when 1 k load placed between BC 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 $\left(\downarrow -\right)$.

$\begin{array}{l}\Sigma F_y=0\\ B_y-S_C-1=0\\ S_C=1-B_y\end{array}$

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

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

Find the equation of shear force at C of portion CF $\left(10\text{ft}\le x\le 40\text{ft}\right)$.

Sketch the free body diagram of the section BC when 1 k load placed between CF as shown in Figure 6.

Refer Figure 5.

Apply equilibrium equation of forces.

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

$\begin{array}{l}\Sigma F_y=0\\ B_y-S_C=0\\ S_C=B_y\end{array}$

Substitute $0$ for $B_y$.

$S_C=0$

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

$S_C=-\frac{x}{10},0\text{ft}\le x<10\text{ft}$        (12)

$S_C=0,10\text{ft}\le x<40\text{ft}$        (13)

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

Substitute 10 m for $x$ in Equation (1).

$\begin{array}{c}{S}_C=-\frac{10}{10}\\ =-1\text{k}\end{array}$

Thus, the influence line ordinate of $S_C$ at just left of C is $-1\text{k/k}$.

Find the shear force of $S_C$ at various points of x using the Equations (12) and (13) and summarize the values as in Table 4.

x (ft)	Points	Influence line ordinate of $S_C$$\left(\text{k/k}\right)$
0	B	0
10	$C$	‑1
20	$D$	0
30	E	0
40	F	0

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

Therefore, the influence lines for the moment at support A and the vertical reactions at supports A and B and the influence lines for the shear hinge C are drawn.