Textbook Problem

Determine the reactions and draw the shear and bending moment diagrams for the beams shown in Figs. P16.1–P16.5 by using the moment-distribution method.

FIG. P16.1

To determine

Find the reaction and plot the shear and bending moment diagram.

Explanation of Solution

Fixed end moment:

Formula to calculate the relative stiffness for fixed support $\frac{I}{L}$ and for roller support $\left(\frac{3}{4}\right)\left(\frac{I}{L}\right)$.

Formula to calculate the fixed moment for point load with equal length are $\frac{PL}{8}$.

Formula to calculate the fixed moment for point load with equal length are $\frac{PL}{8}$.

Formula to calculate the fixed moment for point load with unequal length are $\frac{Pa{b}^2}{L^2}$ and $\frac{P{a}^{2}b}{L^2}$.

Formula to calculate the fixed moment for UDL is $\frac{W{L}^2}{12}$.

Calculation:

Consider the flexural rigidity EI of the beam is constant.

Show the free body diagram of the entire beam as in Figure 1.

Refer Figure 1,

Calculate the relative stiffness $K_{CA}$ for part CA of the beam:

$\begin{array}{c}{K}_{CA}=\frac{I}{20\text{ft}+\text{10}\text{ft}}\\ =\frac{I}{30}\end{array}$

Calculate the relative stiffness $K_{CE}$ for part CE of the beam:

$\begin{array}{c}{K}_{CE}=\frac{I}{15\text{ft+15}\text{ft}}\\ =\frac{I}{30}\end{array}$

In the above beam, only joint C is free to rotate. Hence, calculate the distribution factor at joint C.

Calculate the distribution factor ${\text{DF}}_{CA}$ for member AC of the beam.

${\text{DF}}_{CA}=\frac{K_{CA}}{K_{CA}+K_{CE}}$

Substitute $\frac{I}{30}$ for $K_{CA}$ and $\frac{I}{30}$ for $K_{CE}$.

$\begin{array}{c}{\text{DF}}_{CA}=\frac{\frac{I}{30}}{\frac{I}{30}+\frac{I}{30}}\\ =0.5\end{array}$

Calculate the distribution factor ${\text{DF}}_{CA}$ for part CE of the beam.

${\text{DF}}_{CE}=\frac{K_{CE}}{K_{CA}+K_{CE}}$

Substitute $\frac{I}{30}$ for $K_{CA}$ and $\frac{I}{30}$ for $K_{CE}$.

$\begin{array}{c}{\text{DF}}_{CE}=\frac{\frac{I}{30}}{\frac{I}{30}+\frac{I}{30}}\\ =0.5\end{array}$

Check for sum of distribution factor:

${\text{DF}}_{CE}+{\text{DF}}_{CA}=1$

Substitute 0.5 for ${\text{DF}}_{CE}$ and 0.5 for ${\text{DF}}_{CA}$.

$0.5+0.5=1$

Hence, OK.

Calculate the fixed end moment for AC.

$\begin{array}{c}{\text{FEM}}_{AC}=\frac{18\times 20\times {10}^2}{{\left(30\right)}^2}\\ =40\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment for CA.

$\begin{array}{c}{\text{FEM}}_{CA}=-\frac{18\times {20}^2\times 10}{{\left(30\right)}^2}\\ =-80\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment for CE.

$\begin{array}{c}{\text{FEM}}_{CE}=\frac{10\times 30}{8}\\ =37.5\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment for EC.

$\begin{array}{c}{\text{FEM}}_{EC}=-\frac{10\times 30}{8}\\ =-37.5\text{k}\cdot \text{ft}\end{array}$

Show the calculation of final moments using moment distribution method as in Table 1.

Consider the member AC of the beam:

Show the free body diagram of the member AC as in Figure 2.

Calculate the vertical reaction at the left end of the joint C by taking moment about point A.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_A=0}\\ C_{y,L}\left(30\right)-18\times \left(20\right)+50.625-58.75=0\\ C_{y,L}\left(30\right)=368.125\\ C_{y,L}=\frac{368.125}{30}\\ C_{y,L}=12.271\text{k}\end{array}$

Calculate the horizontal reaction at point A by resolving the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ A_x=0\end{array}$

Calculate the vertical reaction at point A by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ A_y-18+C_{y,L}=0\\ A_y-18+12.271=0\\ A_y=5.729\text{k}\uparrow \end{array}$

Consider the member CE of the beam:

Show the free body diagram of the member CE as in Figure 3.

Calculate the vertical reaction at the right end of the joint C by taking moment about point E.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_E=0}\\ C_{y,R}\left(30\right)+10\times \left(15\right)-26.875+58.75=0\\ C_{y,R}\left(30\right)=181.875\\ C_{y,R}=\frac{181.875}{30}\\ C_{y,R}=6.063\text{k}\uparrow \end{array}$

Calculate the horizontal reaction at point E by resolving the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ E_x=0\end{array}$

Calculate the vertical reaction at point E by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ E_y-10+C_{y,R}=0\\ E_y-10+6.063=0\\ E_y=3.937\text{k}\uparrow \end{array}$

Calculate the total reaction at point C.

$C_y=C_{y,R}+C_{y,L}$

Substitute $6.063\text{k}$ for $C_{y,R}$ and $12.271\text{k}$ for $C_{y,L}$.

$\begin{array}{c}{C}_y=6.063+12.271\\ =18.334\text{k}\uparrow \end{array}$

Show the reaction of the beam in Figure 4.

Refer Figure 4,

Shear diagram:

Point A:

$\begin{array}{l}{S}_{A,L}=0\\ S_{A,R}=5.729\text{k}\end{array}$

Point B:

$\begin{array}{c}{S}_B=5.729-18\\ =-12.271\text{k}\end{array}$

Point C:

$\begin{array}{c}{S}_{C,L}=-12.271\text{k}\\ S_{C,R}=-12.271+18.334\\ =6.063\text{k}\end{array}$

Point D:

$\begin{array}{c}{S}_{D,L}=6.063\text{k}\\ S_{D,R}=6.063-10\\ =-3.937\text{k}\end{array}$

Point E:

$\begin{array}{c}{S}_{E,L}=-3.937\text{k}\\ S_{E,R}=-3.937\text{k+3}\text{.937}\\ =0\text{k}\end{array}$

Plot the shear force diagram of the beam as in Figure 5.

Refer Figure 4,

Bending moment diagram:

Point A:

$M_A=-50.625\text{k}\cdot \text{ft}$

Point B:

$\begin{array}{c}{M}_B=-50.625+\left(5.729\times 20\right)\\ =63.955\text{k}\cdot \text{ft}\end{array}$

Point C:

$M_C=-58.75\text{k}\cdot \text{ft}\simeq 58.8\text{k}\cdot \text{ft}$

Point D:

$\begin{array}{c}{M}_D=-58.8+\left(6.063\times 15\right)\\ =32.145\text{k}\end{array}$

Point E:

$\begin{array}{c}{M}_E=-26.875\text{k}\cdot \text{ft}\\ \simeq 26.9\text{k}\cdot \text{ft}\end{array}$

Plot the bending moment diagram of the beam as in Figure 6.