Chapter 16, Problem 31P

To determine

Find the member end moments and reactions for the frames.

Answer to Problem 31P

The reaction at point A $\left(A_x\right)$, $\left(A_y\right)$ and B $\left(B_x\right)$, $\left(B_y\right)$ are $\underset{\_}{13.5\text{k}\leftarrow }$, $\underset{\_}{10.1\text{k}\downarrow }$, $\underset{\_}{13.5\text{k}\leftarrow }$, and $\underset{\_}{10.1\text{k}\uparrow }$ respectively.

The end moment at the member $\left(M_{AC,BD}\right)$, $\left(M_{CA,DB}\right)$ $\left(M_{CD,DC}\right)$, $\left(M_{CE,DF}\right)$, $\left(M_{EC,FD}\right)$, and $\left(M_{EF,FE}\right)$ are $\underset{\_}{119\text{k}\cdot \text{ft}}$, $\underset{\_}{83.5\text{k}\cdot \text{ft}}$, $\underset{\_}{-106.9\text{k}\cdot \text{ft}}$, $\underset{\_}{-23.3\text{k}\cdot \text{ft}}$, $\underset{\_}{44.2\text{k}\cdot \text{ft}}$, and $\underset{\_}{-44.2\text{k}\cdot \text{ft}}$ respectively.

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 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 $\frac{W{L}^2}{12}$.

Formula to calculate the fixed moment for UVL are $\frac{W{L}^2}{30}$ and $\frac{W{L}^2}{20}$.

Formula to calculate the fixed moment for deflection is $\frac{6EI\Delta }{L^2}$

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 relative stiffness $K_{CA}$, $K_{DB}$, $K_{CE}$, and $K_{DF}$ for member CA, DB, CE and DF of the frame as below:

$K_{CA}=K_{DB}=K_{CE}=K_{DF}=\frac{I}{15}$

Calculate the relative stiffness $K_{CD}$ and $K_{EF}$ for member CD and EF of the frame as below:

$\begin{array}{c}{K}_{CD}=K_{EF}=\frac{2I}{30}\\ =\frac{I}{15}\end{array}$

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

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

Substitute $\frac{I}{15}$ for $K_{CA}$, $\frac{I}{15}$ for $K_{CD}$, and $\frac{I}{15}$ for $K_{CE}$.

$\begin{array}{c}{\text{DF}}_{CA}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{3}\end{array}$

Calculate the distribution factor ${\text{DF}}_{CD}$ for member CD of the frame.

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

Substitute $\frac{I}{15}$ for $K_{CA}$, $\frac{I}{15}$ for $K_{CD}$, and $\frac{I}{15}$ for $K_{CE}$.

$\begin{array}{c}{\text{DF}}_{CD}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{3}\end{array}$

Calculate the distribution factor ${\text{DF}}_{CE}$ for member CE of the frame.

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

Substitute $\frac{I}{15}$ for $K_{CA}$, $\frac{I}{15}$ for $K_{CD}$, and $\frac{I}{15}$ for $K_{CE}$.

$\begin{array}{c}{\text{DF}}_{CE}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{3}\end{array}$

Check for sum of distribution factor as below:

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

Substitute $\frac{1}{3}$ for ${\text{DF}}_{CA}$, ${\text{DF}}_{CD}$ , and ${\text{DF}}_{CE}$.

$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$

Hence, OK.

Calculate the distribution factor ${\text{DF}}_{EC}$ for member EC of the frame.

${\text{DF}}_{EC}=\frac{K_{EC}}{K_{EC}+K_{EF}}$

Substitute $\frac{I}{15}$ for $K_{EC}$ and $\frac{I}{15}$ for $K_{EF}$.

$\begin{array}{c}{\text{DF}}_{EC}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{2}\end{array}$

Calculate the distribution factor ${\text{DF}}_{EF}$ for member EF of the frame.

${\text{DF}}_{EF}=\frac{K_{EF}}{K_{EC}+K_{EF}}$

Substitute $\frac{I}{15}$ for $K_{EC}$ and $\frac{I}{15}$ for $K_{EF}$.

$\begin{array}{c}{\text{DF}}_{EF}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{2}\end{array}$

Check for sum of distribution factor as below:

${\text{DF}}_{EC}+{\text{DF}}_{EF}=1$

Substitute $\frac{1}{2}$ for ${\text{DF}}_{EC}$ and $\frac{1}{2}$ for ${\text{DF}}_{EF}$.

$\frac{1}{2}+\frac{1}{2}=1$

Hence, OK.

Calculate the distribution factor ${\text{DF}}_{FE}$ for member EC of the frame.

${\text{DF}}_{FE}=\frac{K_{FE}}{K_{FD}+K_{FE}}$

Substitute $\frac{I}{15}$ for $K_{FD}$ and $\frac{I}{15}$ for $K_{FE}$.

$\begin{array}{c}{\text{DF}}_{FE}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{2}\end{array}$

Calculate the distribution factor ${\text{DF}}_{FD}$ for member FD of the frame.

${\text{DF}}_{FD}=\frac{K_{FD}}{K_{FD}+K_{FE}}$

Substitute $\frac{I}{15}$ for $K_{FD}$ and $\frac{I}{15}$ for $K_{FE}$.

$\begin{array}{c}{\text{DF}}_{FD}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{2}\end{array}$

Check for sum of distribution factor as below:

${\text{DF}}_{FE}+{\text{DF}}_{FD}=1$

Substitute $\frac{1}{2}$ for ${\text{DF}}_{FE}$ and $\frac{1}{2}$ for ${\text{DF}}_{FD}$.

$\frac{1}{2}+\frac{1}{2}=1$

Hence, OK.

Calculate the distribution factor ${\text{DF}}_{DF}$ for member of the frame.

${\text{DF}}_{DF}=\frac{K_{DF}}{K_{DF}+K_{DC}+K_{DB}}$

Substitute $\frac{I}{15}$ for $K_{DF}$, $\frac{I}{15}$ for $K_{DC}$, and $\frac{I}{15}$ for $K_{DB}$.

$\begin{array}{c}{\text{DF}}_{DF}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{3}\end{array}$

Calculate the distribution factor ${\text{DF}}_{DC}$ for member DC of the frame.

${\text{DF}}_{DC}=\frac{K_{DC}}{K_{DF}+K_{DC}+K_{DB}}$

Substitute $\frac{I}{15}$ for $K_{DF}$, $\frac{I}{15}$ for $K_{DC}$, and $\frac{I}{15}$ for $K_{DB}$.

$\begin{array}{c}{\text{DF}}_{DC}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{3}\end{array}$

Calculate the distribution factor ${\text{DF}}_{DB}$ for member DB of the frame.

${\text{DF}}_{DB}=\frac{K_{DB}}{K_{DF}+K_{DC}+K_{DB}}$

Substitute $\frac{I}{15}$ for $K_{DF}$, $\frac{I}{15}$ for $K_{DC}$, and $\frac{I}{15}$ for $K_{DB}$.

$\begin{array}{c}{\text{DF}}_{DB}=\frac{\frac{I}{15}}{\frac{I}{15}+\frac{I}{15}+\frac{I}{15}}\\ =\frac{1}{3}\end{array}$

Check for sum of distribution factor as below:

${\text{DF}}_{DF}+{\text{DF}}_{DC}+{\text{DF}}_{DB}=1$

Substitute $\frac{1}{3}$ for ${\text{DF}}_{DF}$, $\frac{1}{3}$ for ${\text{DF}}_{DC}$, and $\frac{1}{3}$ for ${\text{DF}}_{DB}$.

$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$

Hence, OK.

Show the translation ${{\Delta }^{\prime }}_1$ of the frame as in figure 2.

Write the expression to calculate the Fixed-end moment of the member.

$FE{M}_{AC}=FE{M}_{CA}=FE{M}_{BD}=FE{M}_{DB}=\frac{6EI{{\Delta }^{\prime }}_1}{{\left(15\right)}^2}$

Assume the Fixed-end moment of the members AC, CA, BD and DB as $100\text{k}\cdot \text{ft}$.

$FE{M}_{AC}=FE{M}_{CA}=FE{M}_{BD}=FE{M}_{DB}=100\text{k}\cdot \text{ft}$

Write the expression to calculate the Fixed-end moment of the member.

$FE{M}_{CE}=FE{M}_{EC}=FE{M}_{DF}=FE{M}_{FD}=-\frac{6EI{{\Delta }^{\prime }}_1}{{\left(15\right)}^2}$

Assume the Fixed-end moment of the members CE, EC, DF and FD as $100\text{k}\cdot \text{ft}$.

$FE{M}_{CE}=FE{M}_{EC}=FE{M}_{DF}=FE{M}_{FD}=-100\text{k}\cdot \text{ft}$

Show the calculation of $M_{Q1}$ moments using moment distribution method as in Table 1.

Show the free body diagram of the frame with unknown reaction $Q_{21}$ as in Figure 3.

Consider member EC:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_E=0}\\ C_{x1}\left(15\right)-85.2-61.8=0\\ C_{x1}\left(15\right)=147\\ C_{x1}=\frac{147}{15}\\ C_{x1}=9.8\text{k}\to \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ E_{x1}+C_{x1}=0\\ E_{x1}+9.8=0\\ E_{x1}=9.8\text{k}\leftarrow \end{array}$

Consider member FD:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_F=0}\\ D_{x1}\left(15\right)-85.2-61.8=0\\ D_{x1}\left(15\right)=147\\ D_{x1}=\frac{147}{15}\\ D_{x1}=9.8\text{k}\to \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ F_{x1}+D_{x1}=0\\ F_{x1}+9.8=0\\ F_{x1}=9.8\text{k}\leftarrow \end{array}$

Calculate the reaction $Q_{21}$ using the relation:

$\begin{array}{c}{Q}_{21}=E_{x1}+F_{x1}\\ =9.8+9.8\\ =19.6\text{k}\leftarrow \end{array}$

Show the free body diagram of the frame with unknown reaction $Q_{11}$ as in Figure 4.

Consider member AC:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ -A_{x1}\left(15\right)+97.1+94.2=0\\ A_{x1}\left(15\right)=191.3\\ A_{x1}=\frac{191.3}{15}\\ A_{x1}=12.75\text{k}\leftarrow \end{array}$

Calculate the horizontal reaction at joint C by resolving the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ A_{x1}+C_{x11}=0\\ C_{x11}-12.75=0\\ C_{x11}=12.75\text{k}\to \end{array}$

Consider member BD:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -B_{x1}\left(15\right)+94.2+97.1=0\\ B_{x1}\left(15\right)=191.3\\ B_{x1}=\frac{191.3}{15}\\ B_{x1}=12.75\text{k}\leftarrow \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ B_{x1}+D_{x11}=0\\ -12.75+D_{x11}=0\\ D_{x11}=12.75\text{k}\to \end{array}$

Calculate the reaction $Q_{11}$ using the relation:

$\begin{array}{c}{Q}_{11}=C_{x11}+D_{x11}+E_{x1}+F_{x1}\\ =12.75+12.75+9.8+9.8\\ =45.1\text{k}\to \end{array}$

Show the translation ${{\Delta }^{\prime }}_2$ of the frame as in Figure 5.

Write the expression to calculate the Fixed-end moment of the member.

$FE{M}_{CE}=FE{M}_{EC}=FE{M}_{DF}=FE{M}_{FD}=\frac{6EI{{\Delta }^{\prime }}_1}{{\left(15\right)}^2}$

Assume the Fixed-end moment of the members CE, EC, DF and FD as $100\text{k}\cdot \text{ft}$.

$FE{M}_{CE}=FE{M}_{EC}=FE{M}_{DF}=FE{M}_{FD}=100\text{k}\cdot \text{ft}$

Show the calculation of $M_{Q2}$ moments using moment distribution method as in Table 2.

Show the free body diagram of the frame with unknown reaction $Q_{22}$ as in Figure 6.

Consider member EC:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_E=0}\\ -C_{x2}\left(15\right)+58.9+53=0\\ C_{x2}\left(15\right)=111.9\\ C_{x2}=\frac{111.9}{15}\\ C_{x2}=7.46\text{k}\leftarrow \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ E_{x2}+C_{x2}=0\\ E_{x2}-7.46=0\\ E_{x2}=7.46\text{k}\to \end{array}$

Consider member FD:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_F=0}\\ -D_{x2}\left(15\right)+53+58.9=0\\ D_{x2}\left(15\right)=111.9\\ D_{x2}=\frac{111.9}{15}\\ D_{x2}=7.46\text{k}\to \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ F_{x2}+D_{x2}=0\\ F_{x2}-7.46=0\\ F_{x2}=7.46\text{k}\to \end{array}$

Calculate the reaction $Q_{21}$:

$\begin{array}{c}{Q}_{21}=E_{x2}+F_{x2}\\ =7.46+7.46\\ =14.9\text{k}\to \end{array}$

Show the free body diagram of the frame with unknown reaction $Q_{12}$ as in Figure 7.

Consider member AC:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ A_{x2}\left(15\right)-11.8-23.5=0\\ A_{x2}\left(15\right)=35.3\\ A_{x2}=\frac{35.3}{15}\\ A_{x2}=2.35\text{k}\to \end{array}$

Calculate the horizontal reaction at joint C by resolving the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ A_{x2}+C_{x22}=0\\ C_{x22}+2.35=0\\ C_{x22}=2.35\text{k}\leftarrow \end{array}$

Consider member BD:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ B_{x2}\left(15\right)-23.5-11.8=0\\ B_{x2}\left(15\right)=35.3\\ B_{x2}=\frac{35.3}{15}\\ B_{x2}=2.35\text{k}\to \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ B_{x2}+D_{x22}=0\\ 2.35+D_{x22}=0\\ D_{x11}=2.35\text{k}\leftarrow \end{array}$

Calculate the reaction $Q_{11}$ using the relation:

$\begin{array}{c}{Q}_{11}=C_{x22}+D_{x22}+E_{x2}+F_{x2}\\ =2.35+2.35+7.46+7.46\\ =19.6\text{k}\leftarrow \end{array}$

Write the equation by superimposing the horizontal forces at joints C,

$45.1c_1-19.6c_2=18$ (1).

Write the equation by superimposing the horizontal forces at joints E,

$-19.6c_1+14.9c_2=9$ (2).

Calculate the value of $c_1$ and $c_2$ by solving equation (1) and (2).

$\begin{array}{l}{c}_1=1.545\\ c_2=2.636\end{array}$

Calculate the actual member end moments of the member AC and BD:

$M_{AC,BD}=c_1{\left(M_{Q1}\right)}_{AC,BD}+c_2{\left(M_{Q2}\right)}_{AC,BD}$

Substitute $97.1\text{k}\cdot \text{ft}$ for ${\left(M_{Q1}\right)}_{AC,BD}$, $-11.8\text{k}\cdot \text{ft}$ for ${\left(M_{Q2}\right)}_{AC,BD}$,1.545 for $c_1$ and 2.636 for $c_2$.

$\begin{array}{c}{M}_{AC,BD}=\left(1.545\times 97.1\right)+2.636\times \left(-11.8\right)\\ =150-31\\ =119\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member CA and DB:

$M_{CA,DB}=c_1{\left(M_{Q1}\right)}_{CA,DB}+c_2{\left(M_{Q2}\right)}_{CA,DB}$

Substitute $94.2\text{k}\cdot \text{ft}$ for ${\left(M_{Q1}\right)}_{CA,DB}$, $-23.5\text{k}\cdot \text{ft}$ for ${\left(M_{Q2}\right)}_{CA,DB}$,1.545 for $c_1$ and 2.636 for $c_2$.

$\begin{array}{c}{M}_{CA,DB}=\left(1.545\times 94.2\right)+2.636\times \left(-23.5\right)\\ =145.5-62\\ =83.5\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member CD and DC:

$M_{CD,DC}=c_1{\left(M_{Q1}\right)}_{CD,DC}+c_2{\left(M_{Q2}\right)}_{CD,DC}$

Substitute $-8.8\text{k}\cdot \text{ft}$ for ${\left(M_{Q1}\right)}_{CD,DC}$, $-35.4\text{k}\cdot \text{ft}$ for ${\left(M_{Q2}\right)}_{CD,DC}$,1.545 for $c_1$ and 2.636 for $c_2$.

$\begin{array}{c}{M}_{CD,DC}=\left(1.545\times \left(-8.8\right)\right)+2.636\times \left(-35.4\right)\\ =-13.6-93.3\\ =-106.9\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member CE and DF:

$M_{CE,DF}=c_1{\left(M_{Q1}\right)}_{CE,DF}+c_2{\left(M_{Q2}\right)}_{CE,DF}$

Substitute $-85.2\text{k}\cdot \text{ft}$ for ${\left(M_{Q1}\right)}_{CE,DF}$, $58.9\text{k}\cdot \text{ft}$ for ${\left(M_{Q2}\right)}_{CE,DF}$,1.545 for $c_1$ and 2.636 for $c_2$.

$\begin{array}{c}{M}_{CE,DF}=\left(1.545\times \left(-85.2\right)\right)+2.636\times \left(58.9\right)\\ =-132+155.3\\ =-23.3\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member EC and FD:

$M_{EC,FD}=c_1{\left(M_{Q1}\right)}_{EC,FD}+c_2{\left(M_{Q2}\right)}_{EC,FD}$

Substitute $-61.8\text{k}\cdot \text{ft}$ for ${\left(M_{Q1}\right)}_{EC,FD}$, $53\text{k}\cdot \text{ft}$ for ${\left(M_{Q2}\right)}_{EC,FD}$,1.545 for $c_1$ and 2.636 for $c_2$.

$\begin{array}{c}{M}_{EC,FD}=\left(1.545\times \left(-61.8\right)\right)+2.636\times \left(53\right)\\ =-95.48+139.71\\ =44.2\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member EC and FD:

$M_{EF,FE}=c_1{\left(M_{Q1}\right)}_{EF,FE}+c_2{\left(M_{Q2}\right)}_{EF,FE}$

Substitute $61.7\text{k}\cdot \text{ft}$ for ${\left(M_{Q1}\right)}_{EF,FE}$, $-52.8\text{k}\cdot \text{ft}$ for ${\left(M_{Q2}\right)}_{EF,FE}$,1.545 for $c_1$ and 2.636 for $c_2$.

$\begin{array}{c}{M}_{EF,FE}=\left(1.545\times \left(61.8\right)\right)+2.636\times \left(-53\right)\\ =95.48-139.71\\ =-44.2\text{k}\cdot \text{ft}\end{array}$

Show the section free body diagram of the member as in Figure 8.

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.

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

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.

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

Calculate the reaction at joint A using the relation:

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

Calculate the reaction at joint B using the relation:

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

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 9.