Chapter 12, Problem 53P

(a)

To determine

Draw the influence line for the positive moment at section B using moment distribution method.

Explanation of Solution

Influence line for the moment at support B.

Influence lines will be constructed by placing the unit load at every 20 ft interval from A along the span of the beam. The points are indicated by the circled numbers.

To establish the influence line ordinate at the left end (point 1), the unit load is placed on the beam directly over support A. Since the entire load passes directly into the support, the beam is unstressed. Therefore,

$R_A=1$, $R_C=0$, and $R_D=0$,  and $M_B=0$.

Similarly, if the unit load is moved at C and D, $R_A=0$, $R_C=1$, $R_D=0$, and $M_B=0$ and $R_A=0$, $R_C=0$, $R_D=1$, and $M_B=0$

Case 1:

Apply unit load at 20 ft from A. (point 2).

Draw the beam with points of applying unit load at B (point 2) as in Figure 1.

Refer Figure 1.

Find the fixed end moment at each end of the member CD as follows;

$\begin{array}{c}{\text{FEM}}_{CD}=\frac{-Pa{b}^2}{L^2}\\ =\frac{-\left(1\right)\left(20\right){\left(40\right)}^2}{{60}^2}\\ =-8.89\text{k-ft}\end{array}$

Find the fixed end moment at each end of the member CA as follows;

$\begin{array}{c}{\text{FEM}}_{DC}=\frac{P{a}^{2}b}{L^2}\\ =\frac{\left(1\right){\left(20\right)}^2\left(40\right)}{{60}^2}\\ =4.44\text{k-ft}\end{array}$

Find the distribution factor.

Show the computation of distribution factor as in Table 1.

Joints	Member	Stiffness, K	$\Sigma K$	$DF=\frac{K}{\Sigma K}$
C	CA	$\begin{array}{c}\frac{3}{4}\left(\frac{I}{L_{CA}}\right)=\frac{3}{4}\left(\frac{I}{60}\right)\\ =0.0125I\end{array}$	$\begin{array}{c}\Sigma K=0.0125I+0.0125I\\ =0.025I\end{array}$	$\begin{array}{c}D{F}_{CA}=\frac{0.025I}{0.0125I}\\ =0.5\end{array}$
	CD	$\begin{array}{c}\frac{3}{4}\left(\frac{I}{L_{CD}}\right)=\frac{3}{4}\left(\frac{I}{60}\right)\\ =0.0125I\end{array}$		$\begin{array}{c}D{F}_{CD}=\frac{0.025I}{0.0125I}\\ =0.5\end{array}$

Show the moment distribution computations as in Table 2.

Joint	A	C		D
Member	AC	CA	CD	DC
DF		0.5	0.5
FEM	‑8.89	4.44	0	0
Released moment	+8.89	0	0	0
COM	0	4.44	0
Total	0	8.88	0	0
DEM		‑4.44	‑4.44
COM
Final	0	4.44	‑4.44	0

Find the reaction at A using section AC.

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

Refer Figure 1.

Find the reaction at A.

Consider moment at C

$\begin{array}{l}\Sigma M_C=0\\ R_A\left(60\right)-1\left(40\right)+4.44=0\\ 60R_A=35.56\\ R_A=0.593\text{k}\end{array}$

Find the moment at B.

Consider moment at B

$\begin{array}{c}\Sigma M_B=0\\ M_B=R_A\left(20\right)\\ =0.593\left(20\right)\\ M_B=11.86\text{k-ft}\end{array}$

Thus, the influence line ordinate of moment at B when 1 kip applied at B is 11.86.

Case 2:

Apply unit load at 20 ft from C. (point 5).

Draw the beam with points of applying at point 5 as in Figure 3.

Refer Figure 3.

Find the fixed end moment at each end of the member CD as follows;

$\begin{array}{c}{\text{FEM}}_{CD}=\frac{-Pa{b}^2}{L^2}\\ =\frac{-\left(1\right)\left(20\right){\left(40\right)}^2}{{60}^2}\\ =-8.89\text{k-ft}\end{array}$

Find the fixed end moment at each end of the member DC as follows;

$\begin{array}{c}{\text{FEM}}_{DC}=\frac{P{a}^{2}b}{L^2}\\ =\frac{\left(1\right){\left(20\right)}^2\left(40\right)}{{60}^2}\\ =4.44\text{k-ft}\end{array}$

Show the moment distribution computations as in Table 3.

Joint	A	C		D
Member	AC	CA	CD	DC
DF		0.5	0.5
FEM	0	0	‑8.89	4.440
Released moment	0	0	0	‑4.44
COM	0	0	‑2.22
Total	0	0	‑11.11	0
DEM		+5.56	+5.56
COM
Final	0	5.56	‑5.56	0

Find the reaction at A using section AC.

Draw the free body diagram of section AC as in Figure 4.

Refer Figure 4.

Find the reaction at A.

Consider moment at C

$\begin{array}{l}\Sigma M_C=0\\ R_A\left(32\right)+3=0\\ 32R_A=-3\\ R_A=-0.938\text{k}\end{array}$

Find the moment at B.

Consider moment at B

$\begin{array}{c}\Sigma M_B=0\\ M_B=R_A\left(20\right)\\ =-0.0938\left(20\right)\\ M_B=-1.876\text{k-ft}\end{array}$

Thus, the influence line ordinate of moment at B when 1 kip applied at point 5 is ‑1.876.

Similarly calculate the influence line ordinate by applying the points 3 and 5 and tabulate the values as in Table 4.

Apply 1 k at point	Influence line ordinate of moment at B.
1	0
2	11.86
4	0
5	‑1.876
6	‑1.876
7	0

Draw the influence line diagram of influence line ordinate of moment at B using Table 4 as in Figure 5.

(b)

To determine

Find the maximum moment at B for the given loading condition.

Answer to Problem 53P

The maximum moment at B for the given loading condition is $\underset{\_}{887\text{k-ft}}$.

Explanation of Solution

Consider the uniformly distributed load of 2 kips/ft act on the span AC only and the concentrated live load of 20 kips act at B.

Draw the free body diagram of the beam for the given loading as in Figure 6.

Refer Figure 6.

Find the fixed end moment at each end of the member AC as follows;

$\begin{array}{c}{\text{FEM}}_{AC}=\frac{-w{L}^2}{12}-\frac{Pa{b}^2}{L^2}\\ =\frac{-2{\left(60\right)}^2}{12}-\frac{20\left(20\right){\left(40\right)}^2}{{60}^2}\\ =-600-177.78\\ =-777.78\text{k-ft}\end{array}$

Find the fixed end moment at each end of the member CA as follows;

$\begin{array}{c}{\text{FEM}}_{CA}=\frac{w{L}^2}{12}+\frac{P{a}^{2}b}{L^2}\\ =\frac{2{\left(60\right)}^2}{12}-\frac{20{\left(20\right)}^2\left(40\right)}{{60}^2}\\ =600+88.89\\ =688.89\text{k-ft}\end{array}$

Show the moment distribution computations as in Table 5.

Joint	A	C		D
Member	AC	CA	CD	DC
DF		0.5	0.5
FEM	‑777.78	688.89	0	0
Released moment	+777.78	00	0	0
COM		388.89
Total	0	1,077.78	0	0
DEM		‑538.89	‑538.89
COM
Final	0	538.89	‑538.89	0

Find the reaction at A using section AC.

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

Refer Figure 7.

Find the reaction at A.

Consider moment at C

$\begin{array}{l}\Sigma M_C=0\\ R_A\left(60\right)-20\left(40\right)-2\left(60\right)\left(\frac{60}{2}\right)+538.89=0\\ 60R_A=800+3,600-538.89\\ R_A=64.35\text{k}\end{array}$

Find the moment at B.

Consider moment at B

$\begin{array}{c}\Sigma M_B=0\\ M_B=R_A\left(20\right)-2\left(20\right)\left(\frac{20}{2}\right)\\ =64.35\left(20\right)-400\\ M_B=887\text{k-ft}\end{array}$

Therefore, the maximum moment at B for the given loading condition is $\underset{\_}{887\text{k-ft}}$.

(c)

To determine

Find the maximum moment at B due to wheel loads using moment distribution method.

Answer to Problem 53P

The maximum moment at B due to wheel loads using moment distribution method is $\underset{\_}{179.8\text{k-ft}}$.

Explanation of Solution

The maximum moment at B is produced when wheel load is placed at B.

Draw the free body diagram of the beam for the wheel loading as in Figure 8.

Refer Figure 8.

Find the fixed end moment at each end of the member AC as follows;

$\begin{array}{c}{\text{FEM}}_{AC}=-\frac{Pa{b}^2}{L^2}-\frac{PL}{8}\\ =-\frac{9\left(20\right){\left(40\right)}^2}{{60}^2}-\frac{9\left(60\right)}{8}\\ =-80-67.5\\ =-147.5\text{k-ft}\end{array}$

Find the fixed end moment at each end of the member CA as follows;

$\begin{array}{c}{\text{FEM}}_{CA}=\frac{P{a}^{2}b}{L^2}\frac{PL}{8}\\ =\frac{9{\left(20\right)}^2\left(40\right)}{{60}^2}+\frac{9\left(60\right)}{8}\\ =40+67.5\\ =107.5\text{k-ft}\end{array}$

Show the moment distribution computations as in Table 6.

Joint	A	C		D
Member	AC	CA	CD	DC
DF		0.5	0.5
FEM	‑147.5	107.5	0	0
Released moment	+147.5	0	0	0
COM		73.75
Total	0	181.25	0	0
DEM		‑90.63	90.63
COM
Final	0	‑90.63	90.63	0

Find the reaction at A using section AC.

Draw the free body diagram of section AC as in Figure 9.

Refer Figure 7.

Find the reaction at A.

Consider moment at C

$\begin{array}{l}\Sigma M_C=0\\ R_A\left(60\right)-9\left(40\right)-9\left(30\right)+90.63=0\\ 60R_A=360+270-90.63\\ R_A=8.99\text{k}\end{array}$

Find the moment at B.

Consider moment at B

$\begin{array}{c}\Sigma M_B=0\\ M_B=R_A\left(20\right)\\ =8.99\left(20\right)\\ M_B=179.8\text{k-ft}\end{array}$

Therefore, the maximum moment at B for the given loading condition is $\underset{\_}{179.8\text{k-ft}}$.