Chapter 15, Problem 31P

To determine

Find the member end moments and reaction for the frames.

Answer to Problem 31P

The end moments at the member AC $\left(M_{AC}\right)$, CA $\left(M_{CA}\right)$, CD $\left(M_{CD}\right)$, DC $\left(M_{DC}\right)$, CE $\left(M_{CE}\right)$, EC $\left(M_{EC}\right)$, EF $\left(M_{EF}\right)$, FE $\left(M_{FE}\right)$, DF $\left(M_{DF}\right)$, FD $\left(M_{FD}\right)$, BD $\left(M_{BD}\right)$ and DB $\left(M_{DB}\right)$ are $\underset{\_}{119\text{k}\cdot \text{ft}}$, $\underset{\_}{83.4\text{k}\cdot \text{ft}}$, $\underset{\_}{-106.8\text{k}\cdot \text{ft}}$, $\underset{\_}{-106.8\text{k}\cdot \text{ft}}$, $\underset{\_}{23.3\text{k}\cdot \text{ft}}$, $\underset{\_}{44.2\text{k}\cdot \text{ft}}$, $\underset{\_}{-44.2\text{k}\cdot \text{ft}}$, $\underset{\_}{-44.2\text{k}\cdot \text{ft}}$, $\underset{\_}{23.3\text{k}\cdot \text{ft}}$, $\underset{\_}{44.2\text{k}\cdot \text{ft}}$, $\underset{\_}{119\text{k}\cdot \text{ft}}$ and $\underset{\_}{83.4\text{k}\cdot \text{ft}}$ respectively and the reactions at the point A $\left(A_x\right)\left(A_y\right)$ and B $\left(B_x\right)\left(B_y\right)$ are $\underset{\_}{\left(13.5\text{k}\leftarrow \right),\left(10.1\text{k}\uparrow \right)}$ and $\underset{\_}{\left(13.5\text{k}\leftarrow \right),\left(10.1\text{k}\uparrow \right)}$ respectively.

Explanation of Solution

Calculation:

Consider the elastic modulus E of the frame is constant.

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

Refer Figure 1,

Calculate the fixed end moment for AC.

${\text{FEM}}_{AC}=0$

Calculate the fixed end moment for CA.

${\text{FEM}}_{CA}=0$

Calculate the fixed end moment for CD.

${\text{FEM}}_{CD}=0$

Calculate the fixed end moment for DC.

${\text{FEM}}_{DC}=0$

Calculate the fixed end moment for DB.

${\text{FEM}}_{DB}=0$

Calculate the fixed end moment for BD.

${\text{FEM}}_{BD}=0$

Calculate the fixed end moment for CE.

${\text{FEM}}_{CE}=0$

Calculate the fixed end moment for EC.

${\text{FEM}}_{EC}=0$

Calculate the fixed end moment for EF.

${\text{FEM}}_{EF}=0$

Calculate the fixed end moment for FE.

${\text{FEM}}_{FE}=0$

Calculate the fixed end moment for FD.

${\text{FEM}}_{FD}=0$

Calculate the fixed end moment for DF.

${\text{FEM}}_{DF}=0$

Chord rotations:

Show the free body diagram of the chord rotation of the frame as in Figure 2.

Refer Figure 2,

Calculate the chord rotation of the frame AC and BD.

${\psi }_{AC}={\psi }_{BD}=-\frac{{\Delta }_1}{15}$

Calculate the chord rotation of the frame CE and DF.

${\psi }_{CE}={\psi }_{DF}=-\frac{{\Delta }_2}{15}$

Calculate the chord rotation of the frame CD and EF.

${\psi }_{CD}={\psi }_{EF}=0$

Calculate the slope deflection equation for the member AC.

$M_{AC}=\frac{2EI}{L}\left(2{\theta }_A+{\theta }_C-3{\psi }_{AC}\right)+{\text{FEM}}_{AC}$

Substitute 15 ft for L, 0 for ${\theta }_A$, $-\frac{{\Delta }_1}{15}$ for ${\psi }_{AC}$, and 0 for ${\text{FEM}}_{AC}$.

$\begin{array}{c}{M}_{AC}=\frac{2EI}{15}\left(2\left(0\right)+{\theta }_C-3\left(-\frac{{\Delta }_1}{15}\right)\right)+0\\ =0.133EI{\theta }_C+0.0267EI{\Delta }_1\end{array}$        (1)

Calculate the slope deflection equation for the member CA.

$M_{CA}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_A-3{\psi }_{CA}\right)+{\text{FEM}}_{CA}$

Substitute 15 ft for L, 0 for ${\theta }_A$, $-\frac{{\Delta }_1}{15}$ for ${\psi }_{CA}$, and 0 for ${\text{FEM}}_{CA}$.

$\begin{array}{c}{M}_{CA}=\frac{2EI}{15}\left(2{\theta }_C+0-3\left(-\frac{{\Delta }_1}{15}\right)\right)-0\\ =0.2667EI{\theta }_C+0.0267EI{\Delta }_1\end{array}$        (2)

Calculate the slope deflection equation for the member CD.

$M_{CD}=\frac{2E\left(2I\right)}{L}\left(2{\theta }_C+{\theta }_D-{\psi }_{CD}\right){\text{+FEM}}_{CD}$

Substitute 30 ft for L, 0 for ${\psi }_{CD}$, and 0 for ${\text{FEM}}_{CD}$.

$\begin{array}{c}{M}_{CD}=\frac{2E\left(2I\right)}{30}\left(2{\theta }_C+{\theta }_D-3\left(0\right)\right)+0\\ =0.2667EI{\theta }_C+0.133EI{\theta }_D\end{array}$        (3)

Calculate the slope deflection equation for the member DC.

$M_{DC}=\frac{2E\left(2I\right)}{L}\left(2{\theta }_D+{\theta }_C-{\psi }_{DC}\right){\text{+FEM}}_{DC}$

Substitute 30 ft for L, 0 for ${\psi }_{DC}$, and 0 for ${\text{FEM}}_{DC}$.

$\begin{array}{c}{M}_{DC}=\frac{2E\left(2I\right)}{30}\left(2{\theta }_D+{\theta }_C-3\left(0\right)\right)-0\\ =0.133EI{\theta }_C+0.2667EI{\theta }_D\end{array}$        (4)

Calculate the slope deflection equation for the member DB.

$M_{DB}=\frac{2EI}{L}\left(2{\theta }_D+{\theta }_B-3{\psi }_{DB}\right){\text{+FEM}}_{DB}$

Substitute 15 ft for L, 0 for ${\theta }_B$, $-\frac{{\Delta }_1}{15}$ for ${\psi }_{DB}$, and $0\text{k}\cdot \text{ft}$ for ${\text{FEM}}_{DB}$.

$\begin{array}{c}{M}_{DB}=\frac{2EI}{15}\left(2{\theta }_D+\left(0\right)-3\left(-\frac{{\Delta }_1}{15}\right)\right)+0\\ =0.2667EI{\theta }_D+0.0267EI{\Delta }_1\end{array}$        (5)

Calculate the slope deflection equation for the member BD.

$M_{BD}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_D-3{\psi }_{BD}\right){\text{+FEM}}_{BD}$

Substitute 15 ft for L, 0 for ${\theta }_B$, $-\frac{{\Delta }_1}{15}$ for ${\psi }_{BD}$, and $0\text{k}\cdot \text{ft}$ for ${\text{FEM}}_{BD}$.

$\begin{array}{c}{M}_{BD}=\frac{2EI}{15}\left({\theta }_D+2\left(0\right)-3\left(-\frac{{\Delta }_1}{15}\right)\right)+0\\ =0.133EI{\theta }_D+0.0267EI{\Delta }_1\end{array}$        (6)

Calculate the slope deflection equation for the member CE.

$M_{CE}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_E-3{\psi }_{CE}\right)+{\text{FEM}}_{CE}$

Substitute 15 ft for L, $-\frac{{\Delta }_2}{15}$ for ${\psi }_{CE}$, and 0 for ${\text{FEM}}_{CE}$.

$\begin{array}{c}{M}_{CE}=\frac{2EI}{15}\left(2{\theta }_C+{\theta }_E-3\left(-\frac{{\Delta }_2}{15}\right)\right)+0\\ =0.2667EI{\theta }_C+0.133EI{\theta }_E+0.0267EI{\Delta }_2\end{array}$        (7)

Calculate the slope deflection equation for the member EC.

$M_{EC}=\frac{2EI}{L}\left(2{\theta }_E+{\theta }_C-3{\psi }_{EC}\right)+{\text{FEM}}_{EC}$

Substitute 15 ft for L, $-\frac{{\Delta }_2}{15}$ for ${\psi }_{EC}$ and 0 for ${\text{FEM}}_{EC}$.

$\begin{array}{c}{M}_{EC}=\frac{2EI}{15}\left(2{\theta }_E+{\theta }_C-3\left(-\frac{{\Delta }_2}{15}\right)\right)+0\\ =0.133EI{\theta }_C+0.2667EI{\theta }_E+0.0267EI{\Delta }_2\end{array}$        (8)

Calculate the slope deflection equation for the member EF.

$M_{EF}=\frac{2E\left(2I\right)}{L}\left(2{\theta }_E+{\theta }_F-{\psi }_{EF}\right){\text{+FEM}}_{EF}$

Substitute 30 ft for L, 0 for ${\psi }_{EF}$ and 0 for ${\text{FEM}}_{EF}$.

$\begin{array}{c}{M}_{EF}=\frac{2E\left(2I\right)}{30}\left(2{\theta }_E+{\theta }_F-3\left(0\right)\right)+0\\ =0.2667EI{\theta }_E+0.133EI{\theta }_F\end{array}$        (9)

Calculate the slope deflection equation for the member FE.

$M_{FE}=\frac{2E\left(2I\right)}{L}\left(2{\theta }_F+{\theta }_E-{\psi }_{FE}\right){\text{+FEM}}_{FE}$

Substitute 30 ft for L, 0 for ${\psi }_{FE}$ and 0 for ${\text{FEM}}_{FE}$.

$\begin{array}{c}{M}_{FE}=\frac{2E\left(2I\right)}{30}\left(2{\theta }_F+{\theta }_E-3\left(0\right)\right)+0\\ =0.133EI{\theta }_E+0.2667EI{\theta }_F\end{array}$        (10)

Calculate the slope deflection equation for the member FD.

$M_{FD}=\frac{2EI}{L}\left(2{\theta }_F+{\theta }_D-3{\psi }_{FD}\right){\text{+FEM}}_{FD}$

Substitute 15 ft for L, $-\frac{{\Delta }_2}{15}$ for ${\psi }_{FD}$ and $0\text{k}\cdot \text{ft}$ for ${\text{FEM}}_{FD}$.

$\begin{array}{c}{M}_{FD}=\frac{2EI}{15}\left(2{\theta }_F+{\theta }_D-3\left(-\frac{{\Delta }_2}{15}\right)\right)+0\\ =0.2667EI{\theta }_F+0.133EI{\theta }_D+0.0267EI{\Delta }_2\end{array}$        (11)

Calculate the slope deflection equation for the member DF.

$M_{DF}=\frac{2EI}{L}\left(2{\theta }_D+{\theta }_F-3{\psi }_{DF}\right){\text{+FEM}}_{DF}$

Substitute 15 ft for L, $-\frac{{\Delta }_2}{15}$ for ${\psi }_{DF}$, and $0\text{k}\cdot \text{ft}$ for ${\text{FEM}}_{DF}$.

$\begin{array}{c}{M}_{DF}=\frac{2EI}{15}\left(2{\theta }_D+{\theta }_F-3\left(-\frac{{\Delta }_2}{15}\right)\right)+0\\ =0.2667EI{\theta }_D+0.133EI{\theta }_F+0.0267EI{\Delta }_2\end{array}$        (12)

Write the equilibrium equation as below.

$M_{CA}+M_{CD}+M_{CE}=0$

Substitute equation (2), equation (3), and equation (7) in above equation.

$\begin{array}{l}\left\{\begin{array}{l}0.2667EI{\theta }_C+0.0267EI{\Delta }_1+0.2667EI{\theta }_C+0.133EI{\theta }_D+\\ 0.2667EI{\theta }_C+0.133EI{\theta }_E+0.0267EI{\Delta }_2\end{array}\right\}=0\\ 0.8EI{\theta }_C+0.133EI{\theta }_D+0.133EI{\theta }_E+0.0267EI{\Delta }_1+0.0267EI{\Delta }_2=0\end{array}$        (13)

Write the equilibrium equation as below.

$M_{DC}+M_{DB}+M_{DF}=0$

Substitute equation (4), equation (5) and equation (12) in above equation.

$\begin{array}{l}\left\{\begin{array}{l}0.133EI{\theta }_C+0.2667EI{\theta }_D+0.2667EI{\theta }_D+0.0267EI{\Delta }_1\\ +0.2667EI{\theta }_D+0.133EI{\theta }_F+0.0267EI{\Delta }_2\end{array}\right\}=0\\ 0.133EI{\theta }_C+0.8EI{\theta }_D++0.133EI{\theta }_F+0.0267EI{\Delta }_1+0.0267EI{\Delta }_2=0\end{array}$        (14)

Write the equilibrium equation as below.

$M_{EC}+M_{EF}=0$

Substitute equation (8) and equation (9) in above equation.

$\begin{array}{l}\left\{0.133EI{\theta }_C+0.2667EI{\theta }_E+0.0267EI{\Delta }_2+0.2667EI{\theta }_E+0.133EI{\theta }_F\right\}=0\\ 0.133EI{\theta }_C+0.533EI{\theta }_E+0.133EI{\theta }_F+0.0267EI{\Delta }_2=0\end{array}$        (15)

Write the equilibrium equation as below.

$M_{FD}+M_{FE}=0$

Substitute equation (10) and equation (11) in above equation.

$\begin{array}{l}\left\{0.133EI{\theta }_E+0.2667EI{\theta }_F+0.2667EI{\theta }_F+0.133EI{\theta }_D+0.0267EI{\Delta }_2\right\}=0\\ 0.133EI{\theta }_E+0.533EI{\theta }_F+0.133EI{\theta }_D+0.0267EI{\Delta }_2=0\end{array}$        (16)

Show the free body diagram of the joint E and F due to sway force as in Figure 3.

Calculate the horizontal reaction at the member CE due to sway force by taking moment about point C.

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ \left(-S_{CE}\times 15\right)+M_{CE}+M_{EC}=0\\ S_{CE}=\frac{M_{CE}+M_{EC}}{15}\end{array}$

Calculate the horizontal reaction at the member DF due to sway force by taking moment about point D.

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ \left(-S_{DF}\times 15\right)+M_{DF}+M_{FD}=0\\ S_{DF}=\frac{M_{DF}+M_{FD}}{15}\end{array}$

Calculate the reaction of the support E and support F due to sway force by considering horizontal equilibrium.

$\begin{array}{l}+\leftarrow {\displaystyle \sum F_x=0}\\ S_{CE}+S_{DF}=9\\ \left(\frac{M_{CE}+M_{EC}}{15}\right)+\left(\frac{M_{DF}+M_{FD}}{15}\right)=9\\ \frac{M_{CE}+M_{EC}+M_{DF}+M_{FD}}{15}=9\\ M_{CE}+M_{EC}+M_{DF}+M_{FD}=135\end{array}$

Substitute equation (7), (8), (11) and (12).

$\begin{array}{l}\left\{\begin{array}{l}0.2667EI{\theta }_C+0.133EI{\theta }_E+0.0267EI{\Delta }_2+\\ 0.133EI{\theta }_C+0.2667EI{\theta }_E+0.0267EI{\Delta }_2\\ +0.2667EI{\theta }_F+0.133EI{\theta }_D+0.0267EI{\Delta }_2+\\ 0.2667EI{\theta }_D+0.133EI{\theta }_F+0.0267EI{\Delta }_2\end{array}\right\}=135\\ 0.4EI{\theta }_C+0.4EI{\theta }_E+0.4EI{\theta }_F+0.4EI{\theta }_D+0.1068EI{\Delta }_2=135\end{array}$        (17)

Show the free body diagram of the joint C and D due to sway force as in Figure 4.

Calculate the horizontal reaction at the member AC due to sway force by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ \left(-S_{AC}\times 15\right)+M_{AC}+M_{CA}=0\\ S_{AC}=\frac{M_{AC}+M_{CA}}{15}\end{array}$

Calculate the horizontal reaction at the member BD due to sway force by taking moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ \left(-S_{BD}\times 15\right)+M_{BD}+M_{DB}=0\\ S_{BD}=\frac{M_{BD}+M_{DB}}{15}\end{array}$

Calculate the reaction of the support C and support D due to sway force by considering horizontal equilibrium.

$\begin{array}{l}+\leftarrow {\displaystyle \sum F_x=0}\\ S_{AC}+S_{BD}=27\\ \left(\frac{M_{AC}+M_{CA}}{15}\right)+\left(\frac{M_{BD}+M_{DB}}{15}\right)=27\\ \frac{M_{AC}+M_{CA}+M_{BD}+M_{DB}}{15}=27\\ M_{AC}+M_{CA}+M_{BD}+M_{DB}=405\end{array}$

Substitute equation (1), equation (2), equation (5), and equation (6).

$\begin{array}{l}\left\{\begin{array}{l}0.133EI{\theta }_C+0.0267EI{\Delta }_1+0.2667EI{\theta }_C+0.0267EI{\Delta }_1\\ +0.2667EI{\theta }_D+0.0267EI{\Delta }_1+0.133EI{\theta }_D+0.0267EI{\Delta }_1\end{array}\right\}=405\\ 0.4EI{\theta }_C+0.4EI{\theta }_D+0.1068EI{\Delta }_1=405\end{array}$        (18)

Solve the equation (13), equation (14), equation (15), equation (16), equation (17) and equation (18).

$\begin{array}{l}{\theta }_C={\theta }_D=\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}\\ {\theta }_E={\theta }_F=\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}\\ {\Delta }_1=\frac{5793.7}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}\\ {\Delta }_2=\frac{4095.9}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}\end{array}$

Calculate the moment about AC.

Substitute $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$ and $\frac{5793.7}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}$ for ${\Delta }_1$ in equation (1).

$\begin{array}{c}{M}_{AC}=0.133EI\left(\frac{-267.2}{EI}\right)+0.0267EI\left(\frac{5793.7}{EI}\right)\\ =119\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about CA.

Substitute $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$ and $\frac{5793.7}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}$ for ${\Delta }_1$ in equation (2).

$\begin{array}{c}{M}_{CA}=0.2667EI\left(\frac{-267.2}{EI}\right)+0.0267EI\left(\frac{5793.7}{EI}\right)\\ =83.4\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about CD.

Substitute $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$ and $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_D$ in equation (3).

$\begin{array}{c}{M}_{CD}=0.2667EI\left(\frac{-267.2}{EI}\right)+0.133EI\left(\frac{-267.2}{EI}\right)\\ =-106.8\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about DC.

Substitute $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$ and $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_D$ in equation (4).

$\begin{array}{c}{M}_{DC}=0.133EI\left(\frac{-267.2}{EI}\right)+0.2667EI\left(\frac{-267.2}{EI}\right)\\ =-106.8\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about DB.

Substitute $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_D$ and $\frac{5793.7}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}$ for ${\Delta }_1$ in equation (5).

$\begin{array}{c}{M}_{DB}=0.2667EI\left(\frac{-267.2}{EI}\right)+0.0267EI\left(\frac{5793.7}{EI}\right)\\ =83.4\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about BD.

Substitute $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_D$ and $\frac{5793.7}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}$ for ${\Delta }_1$ in equation (6).

$\begin{array}{c}{M}_{BD}=0.133EI\left(\frac{-267.2}{EI}\right)+0.0267EI\left(\frac{5793.7}{EI}\right)\\ =119\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about CE.

Substitute $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$, $\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_E$ and $\frac{4095.9}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}$ for ${\Delta }_2$ in equation (7).

$\begin{array}{c}{M}_{CE}=0.2667EI\left(\frac{-267.2}{EI}\right)+0.133EI\left(\frac{-110.8}{EI}\right)+0.0267EI\left(\frac{4095.9}{EI}\right)\\ =23.3\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about EC.

Substitute $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$, $\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_E$ and $\frac{4095.9}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}$ for ${\Delta }_2$ in equation (8).

$\begin{array}{c}{M}_{EC}=0.133EI\left(\frac{-267.2}{EI}\right)+0.2667EI\left(\frac{-110.8}{EI}\right)+0.0267EI\left(\frac{4095.9}{EI}\right)\\ =44.2\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about EF.

Substitute $\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_F$ and $\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_E$ in the equation (9).

$\begin{array}{c}{M}_{EF}=0.2667EI\left(\frac{-110.8}{EI}\right)+0.133EI\left(\frac{-110.8}{EI}\right)\\ =-44.2\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about FE.

Substitute $\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_F$ and $\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_E$ in the equation (10).

$\begin{array}{c}{M}_{FE}=0.133EI\left(\frac{-110.8}{EI}\right)+0.2667EI\left(\frac{-110.8}{EI}\right)\\ =-44.2\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about FD.

Substitute $\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_F$, $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_D$ and $\frac{4095.9}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}$ for ${\Delta }_2$ in equation (11).

$\begin{array}{c}{M}_{FD}=0.2667EI\left(\frac{-110.8}{EI}\right)+0.133EI\left(\frac{-267.2}{EI}\right)+0.0267EI\left(\frac{4095.9}{EI}\right)\\ =44.2\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about DF.

Substitute $\frac{-110.8}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_F$, $\frac{-267.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_D$ and $\frac{4095.9}{EI}\text{k}\cdot {\text{ft}}^{\text{3}}$ for ${\Delta }_2$ in the equation (12).

$\begin{array}{c}{M}_{DF}=0.2667EI\left(\frac{-267.2}{EI}\right)+0.133EI\left(\frac{-110.8}{EI}\right)+0.0267EI\left(\frac{4095.9}{EI}\right)\\ =23.3\text{k}\cdot \text{ft}\end{array}$

Show the section free body diagram of the member EF as in Figure 5.

Consider member EF:

Calculate the vertical reaction at the joint E by taking moment about point F.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_F=0}\\ -E_y\left(30\right)-44.2-44.2=0\\ -E_y\left(30\right)=88.4\\ E_y=-\frac{88.4}{30}\\ E_y=2.95\text{k}\downarrow \end{array}$

Calculate the vertical reaction at joint F by resolving the horizontal equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ E_y+F_y=0\\ -2.95+F_y=0\\ F_y=2.95\text{k}\uparrow \end{array}$

Show the section free body diagram of the member CD as in Figure 6.

Consider member CD:

Calculate the vertical reaction at the joint C by taking moment about point D.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -C_y\left(30\right)-106.9-106.9=0\\ -C_y\left(30\right)=213.8\\ C_y=-\frac{213.8}{30}\\ C_y=7.13\text{k}\downarrow \end{array}$

Calculate the vertical reaction at joint D by resolving the horizontal equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ C_y+D_y=0\\ -7.13+D_y=0\\ D_y=7.13\text{k}\uparrow \end{array}$

Show the section free body diagram of the member AC, CE, DB and FD as in Figure 7.

Calculate the reaction at joint A:

$\begin{array}{c}{A}_y=E_y+C_y\\ =2.95\text{k}\downarrow +7.13\text{k}\downarrow \\ =10.1\text{k}\downarrow \end{array}$

Calculate the reaction at joint B:

$\begin{array}{c}{B}_y=F_y+D_y\\ =2.95\text{k}\uparrow +7.13\text{k}\uparrow \\ =10.1\text{k}\uparrow \end{array}$

Consider member AC:

Calculate the horizontal reaction at the joint A by taking moment about point C.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ -A_x\left(15\right)+119+83.5=0\\ A_x\left(15\right)=202.5\\ A_x=\frac{202.5}{15}\\ A_x=13.5\text{k}\leftarrow \end{array}$

Consider member BD:

Calculate the horizontal reaction at the joint B by taking moment about point D.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -B_x\left(15\right)+119+83.5=0\\ B_x\left(15\right)=202.5\\ B_x=\frac{202.5}{15}\\ B_x=13.5\text{k}\leftarrow \end{array}$

Show the reactions of the frame as in Figure 8.