Chapter 16, Problem 30P

To determine

Find the member end moments and reactions for the frames.

Answer to Problem 30P

The reaction at point A $\left(A_x\right)$, $\left(A_y\right)$ and B $\left(B_x\right)$, $\left(B_y\right)$ are $\underset{\_}{0.04\text{k}\leftarrow }$, $\underset{\_}{16.36\text{k}\uparrow }$, $\underset{\_}{\text{19}\text{.96}\text{k}\leftarrow }$, and $\underset{\_}{31.64\text{k}\uparrow }$ respectively.

The end moment at the member $\left(M_{AC}\right)$, $\left(M_{CA}\right)$ $\left(M_{CD}\right)$, $\left(M_{DC}\right)$, $\left(M_{DB}\right)$, and $\left(M_{BD}\right)$ are $\underset{\_}{47.1\text{k}\cdot \text{ft}}$, $\underset{\_}{19\text{k}\cdot \text{ft}}$, $\underset{\_}{-19\text{k}\cdot \text{ft}}$, $\underset{\_}{-103.3\text{k}\cdot \text{ft}}$, $\underset{\_}{103.3\text{k}\cdot \text{ft}}$, and $\underset{\_}{89.5\text{k}\cdot \text{ft}}$ 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 flexural rigidity EI of the frame is constant.

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

Refer Figure 1,

Calculate the length of the member AC and BD:

$\begin{array}{c}{L}_{AC}=L_{BD}=\sqrt{{16}^2+4^2}\\ =\sqrt{256+16}\\ =16.49\text{ft}\end{array}$

Calculate the relative stiffness $K_{CA}$ and $K_{DB}$ for member AC and BD of the frame as below:

$K_{CA}=K_{DB}=\frac{I}{16.49}$

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

$K_{CD}=\frac{I}{16}$

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

$K_{DC}=\frac{I}{16}$

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

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

Substitute $\frac{I}{16.49}$ for $K_{CA}$ and $\frac{I}{16}$ for $K_{CD}$.

$\begin{array}{c}{\text{DF}}_{CA}=\frac{\frac{I}{16.49}}{\frac{I}{16.49}+\frac{I}{16}}\\ =0.492\end{array}$

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

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

Substitute $\frac{I}{16.49}$ for $K_{CA}$ and $\frac{I}{16}$ for $K_{CD}$.

$\begin{array}{c}{\text{DF}}_{CD}=\frac{\frac{I}{16}}{\frac{I}{16.49}+\frac{I}{16}}\\ =0.508\end{array}$

Check for sum of distribution factor as below:

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

Substitute 0.492 for ${\text{DF}}_{CA}$ and 0.508 for ${\text{DF}}_{CD}$.

$0.492+0.508=1$

Hence, OK.

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

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

Substitute $\frac{I}{16}$ for $K_{DC}$ and $\frac{I}{16.49}$ for $K_{DB}$.

$\begin{array}{c}{\text{DF}}_{DC}=\frac{\frac{I}{16}}{\frac{I}{16}+\frac{I}{16.49}}\\ =0.508\end{array}$

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

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

Substitute $\frac{I}{16}$ for $K_{DC}$ and $\frac{I}{16.49}$ for $K_{DB}$.

$\begin{array}{c}{\text{DF}}_{DB}=\frac{\frac{I}{16.49}}{\frac{I}{16}+\frac{I}{16.49}}\\ =0.492\end{array}$

Check for sum of distribution factor as below:

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

Substitute 0.508 for ${\text{DF}}_{DC}$ and 0.492 for ${\text{DF}}_{DB}$.

$0.508+0.492=1$

Hence, OK.

Calculate the fixed end moment for AC and CA.

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

Calculate the fixed end moment for CD.

$\begin{array}{c}{\text{FEM}}_{CD}=\frac{3\times {\left(16\right)}^2}{12}\\ =64\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment for DC.

$\begin{array}{c}{\text{FEM}}_{DC}=-\frac{3\times {\left(16\right)}^2}{12}\\ =-64\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment for DB and BD.

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

Show the calculation of $M_0$ moments using moment distribution method for side-sway prevented as in Table 1.

Show the arbitrary translation as in Figure 2.

Calculate the relative translation ${\Delta }_{AC}$ between the ends of member AC:

$\begin{array}{c}{\Delta }_{AC}=CC\text{'}=\frac{\sqrt{17}}{4}{\Delta }^{\prime }\\ =1.031{\Delta }^{\prime }\end{array}$

Calculate the relative translation ${\Delta }_{CD}$ between the ends of member CD:

$\begin{array}{c}{\Delta }_{CD}=D{D}^{\prime }=2\left(\frac{1}{4}\right){\Delta }^{\prime }\\ =0.5{\Delta }^{\prime }\end{array}$

Calculate the relative translation ${\Delta }_{BD}$ between the ends of member BD:

$\begin{array}{c}{\Delta }_{BD}=D{D}^{\prime }=\frac{\sqrt{17}}{4}{\Delta }^{\prime }\\ =1.031{\Delta }^{\prime }\end{array}$

Calculate the fixed end moment for AC and CA.

${\text{FEM}}_{AC}={\text{FEM}}_{CA}=\frac{6EI\left({\Delta }_{AC}\right)}{{\left(16.49\right)}^2}$

Substitute $1.031{\Delta }^{\prime }$ for ${\Delta }_{AC}$.

${\text{FEM}}_{AC}={\text{FEM}}_{CA}=\frac{6EI\left(1.031{\Delta }^{\prime }\right)}{{\left(16.49\right)}^2}$ (1).

Calculate the fixed end moment for CD and DC.

${\text{FEM}}_{CD}={\text{FEM}}_{DC}=-\frac{6EI\left({\Delta }_{CD}\right)}{{\left(16\right)}^2}$

Substitute $0.5{\Delta }^{\prime }$ for ${\Delta }_{CD}$.

${\text{FEM}}_{CD}={\text{FEM}}_{DC}=-\frac{6EI\left(0.5{\Delta }^{\prime }\right)}{{\left(16\right)}^2}$        (2)

Calculate the fixed end moment for BD and DB.

${\text{FEM}}_{BD}={\text{FEM}}_{DB}=\frac{6EI\left({\Delta }_{BD}\right)}{{\left(16.49\right)}^2}$

Substitute $1.031{\Delta }^{\prime }$ for ${\Delta }_{BD}$.

${\text{FEM}}_{BD}={\text{FEM}}_{DB}=\frac{6EI\left(1.031{\Delta }^{\prime }\right)}{{\left(16.49\right)}^2}$        (3)

Assume the Fixed-end moment at AC and CA as $-100\text{k}\cdot \text{ft}$.

$\begin{array}{c}FE{M}_{AC}=FE{M}_{CA}\\ =100\text{k}\cdot \text{ft}\end{array}$

Calculate the value of $EI{\Delta }^{\prime }$ using the equation (1).

Substitute $-100\text{k}\cdot \text{ft}$ for $FE{M}_{AC}$.

$\begin{array}{c}-100=\frac{6EI\left(1.031{\Delta }^{\prime }\right)}{{\left(16.49\right)}^2}\\ EI{\Delta }^{\prime }=\frac{100\times {\left(16.49\right)}^2}{6\times 1.031}\\ EI{\Delta }^{\prime }=4,395.7\end{array}$

Calculate the fixed end moment of CD and DC.

Substitute 4,395.7 for $EI{\Delta }^{\prime }$ in equation (2).

$\begin{array}{c}{\text{FEM}}_{CD}={\text{FEM}}_{DC}=-\frac{6\times 0.5\times 4,395.7}{{\left(16\right)}^2}\\ =-51.5\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment of BD and DB.

Substitute 4,395.7 for $EI{\Delta }^{\prime }$ in equation (3).

$\begin{array}{c}{\text{FEM}}_{BD}={\text{FEM}}_{DB}=\frac{6\times 1.031\left(4,395.7\right)}{{\left(16.49\right)}^2}\\ =100\text{k}\cdot \text{ft}\end{array}$

Show the calculation of $M_Q$ moments using moment distribution method for side-sway permitted as in Table 2.

Show the free body diagram of the member AC, CD and DB for side-sway permitted as in Figure 3.

Consider member CD:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -C_{y1}\left(16\right)-81-81=0\\ C_{y1}\left(16\right)=-162\\ C_{y1}=-\frac{162}{16}\\ C_{y1}=10.13\text{k}\downarrow \end{array}$

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

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ C_{y1}+D_{y1}=0\\ -10.13+D_{y1}=0\\ D_{y1}=10.13\text{k}\uparrow \end{array}$

Consider member AC

Calculate vertical reaction at joint A using the relation:

$\begin{array}{c}{A}_{y1}=C_{y1}\\ =10.13\text{k}\downarrow \end{array}$

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ -A_{x1}\left(16\right)+\left(10.13\times 4\right)+90.5+81=0\\ -A_{x1}\left(16\right)=212.02\\ A_{x1}=-\frac{212.02}{16}\\ A_{x1}=13.25\text{k}\leftarrow \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ A_{x1}+C_{x1}=0\\ -13.25+C_{x1}=0\\ C_{x1}=13.25\text{k}\to \end{array}$

Consider member DB:

Calculate vertical reaction at joint B using the relation:

$\begin{array}{c}{B}_{y1}=D_{y1}\\ =10.13\text{k}\uparrow \end{array}$

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -B_{x1}\left(16\right)+\left(10.13\times 4\right)+81+90.5=0\\ B_{x1}\left(16\right)=-212.02\\ B_{x1}=-\frac{212.02}{16}\\ B_{x1}=13.25\text{k}\leftarrow \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ B_{x1}+D_{x1}=0\\ -13.25+D_{x1}=0\\ D_{x1}=13.25\text{k}\to \end{array}$

Show the unknown load Q as in Figure 4.

Calculate the reaction R using the relation:

$\begin{array}{c}Q=C_{x1}+D_{x1}\\ =13.25+13.25\\ =26.5\text{k}\to \end{array}$

Calculate the actual member end moments of the member AC:

$M_{AC}={\left(M_0\right)}_{AC}+\left(\frac{20}{Q}\right){\left(M_Q\right)}_{AC}$

Substitute $-21.2\text{k}\cdot \text{ft}$ for ${\left(M_0\right)}_{AC}$, $90.5\text{k}\cdot \text{ft}$ for ${\left(M_Q\right)}_{AC}$, and $26.5\text{k}$ for Q.

$\begin{array}{c}{M}_{AC}=-21.2+\left(\frac{20}{26.5}\right)\times \left(90.5\right)\\ =-21.2+68.30\\ =47.1\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member CA:

$M_{CA}={\left(M_0\right)}_{CA}+\left(\frac{20}{Q}\right){\left(M_Q\right)}_{CA}$

Substitute $-42.2\text{k}\cdot \text{ft}$ for ${\left(M_0\right)}_{CA}$, $81\text{k}\cdot \text{ft}$ for ${\left(M_Q\right)}_{CA}$, and $26.5\text{k}$ for Q.

$\begin{array}{c}{M}_{CA}=-42.2+\left(\frac{20}{26.5}\right)\times \left(81\right)\\ =-42.2+61.13\\ =18.93\text{k}\cdot \text{ft}\\ \simeq 19\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member CD:

$M_{CD}={\left(M_0\right)}_{CD}+\left(\frac{20}{Q}\right){\left(M_Q\right)}_{CD}$

Substitute $42.2\text{k}\cdot \text{ft}$ for ${\left(M_0\right)}_{CD}$, $-81\text{k}\cdot \text{ft}$ for ${\left(M_Q\right)}_{CD}$, and $26.5\text{k}$ for Q.

$\begin{array}{c}{M}_{CD}=42.2+\left(\frac{20}{26.5}\right)\times \left(-81\right)\\ =42.2-61.13\\ =-18.93\text{k}\cdot \text{ft}\\ \simeq -19\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member DC:

$M_{DC}={\left(M_0\right)}_{DC}+\left(\frac{20}{Q}\right){\left(M_Q\right)}_{DC}$

Substitute $-42.2\text{k}\cdot \text{ft}$ for ${\left(M_0\right)}_{DC}$, $-81\text{k}\cdot \text{ft}$ for ${\left(M_Q\right)}_{DC}$, and $26.5\text{k}$ for Q.

$\begin{array}{c}{M}_{DC}=-42.2+\left(\frac{20}{26.5}\right)\times \left(-81\right)\\ =-42.2-61.13\\ =-103.3\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member DB:

$M_{DB}={\left(M_0\right)}_{DB}+\left(\frac{20}{Q}\right){\left(M_Q\right)}_{DB}$

Substitute $42.2\text{k}\cdot \text{ft}$ for ${\left(M_0\right)}_{DB}$, $81\text{k}\cdot \text{ft}$ for ${\left(M_Q\right)}_{DB}$, and $26.5\text{k}$ for Q.

$\begin{array}{c}{M}_{DB}=42.2+\left(\frac{20}{26.5}\right)\times \left(81\right)\\ =42.2+61.13\\ =103.3\text{k}\cdot \text{ft}\end{array}$

Calculate the actual member end moments of the member BD:

$M_{BD}={\left(M_0\right)}_{BD}+\left(\frac{20}{Q}\right){\left(M_Q\right)}_{BD}$

Substitute $21.2\text{k}\cdot \text{ft}$ for ${\left(M_0\right)}_{BD}$, $90.5\text{k}\cdot \text{ft}$ for ${\left(M_Q\right)}_{BD}$, and $26.5\text{k}$ for Q.

$\begin{array}{c}{M}_{BD}=21.2+\left(\frac{20}{26.5}\right)\times \left(90.5\right)\\ =21.2+68.30\\ =89.5\text{k}\cdot \text{ft}\end{array}$

Show the section free body diagram of the member AC, CD and DB as in Figure 5.

Consider the member CD.

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ D_y\left(16\right)-\left(3\times 16\times \frac{16}{2}\right)-19-103.4=0\\ D_y\left(16\right)=506.4\\ D_y=\frac{506.4}{16}\\ D_y=31.65\text{k}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ D_y+C_y=\left(3\times 16\right)\\ C_y+31.65=48\\ C_y=16.34\text{k}\uparrow \end{array}$

Consider the member AC.

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

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ C_y+A_y=0\\ -16.34+A_y=0\\ A_y=16.34\text{k}\uparrow \end{array}$

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ -A_x\left(16\right)-\left(16.35\times 4\right)+47.3+19=0\\ A_x\left(16\right)=0.9\\ A_x=\frac{0.9}{16}\\ A_x=0.05\text{k}\leftarrow \end{array}$

Consider the member BD.

Calculate the vertical reaction at joint B by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ B_y+D_y=0\\ -31.65+B_y=0\\ B_y=31.65\text{k}\uparrow \end{array}$

Consider the entire frame.

Calculate the horizontal reaction at the joint B by considering the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ B_x+A_x=-20\\ B_x-0.05=-20\\ B_x=19.95\text{k}\leftarrow \end{array}$

Show the reactions of the frame as in Figure 6.