Chapter 16, Problem 29P

To determine

Find the member end moments and reactions for the frames.

Answer to Problem 29P

The reaction at point A $\left(A_x\right)$, $\left(A_y\right)$ and B $\left(B_x\right)$, $\left(B_y\right)$ are $\underset{\_}{17\text{kN}\leftarrow }$, $\underset{\_}{50.8\text{kN}\uparrow }$, $\underset{\_}{33\text{kN}\leftarrow }$, and $\underset{\_}{39.2\text{kN}\uparrow }$ 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 $\underset{\_}{11.7\text{kN}\cdot \text{m}}$, $\underset{\_}{-43.9\text{kN}\cdot \text{m}}$, $\underset{\_}{43.9\text{kN}\cdot \text{m}}$, $\underset{\_}{-14.7\text{kN}\cdot \text{m}}$, $\underset{\_}{-14.7\text{kN}\cdot \text{m}}$, and $\underset{\_}{0\text{kN}\cdot \text{m}}$ 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:

$\begin{array}{c}{L}_{DB}=\sqrt{3^2+4^2}\\ =\sqrt{9+16}\\ =5\text{m}\end{array}$

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

$K_{CA}=\frac{I}{4}$

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

$K_{CD}=\frac{I}{5}$

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

$K_{DC}=\frac{I}{5}$

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

$\begin{array}{c}{K}_{DB}=\left(\frac{3}{4}\right)\frac{I}{5}\\ =\frac{3I}{20}\end{array}$

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

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

Substitute $\frac{I}{4}$ for $K_{CA}$ and $\frac{I}{5}$ for $K_{CD}$.

$\begin{array}{c}{\text{DF}}_{CA}=\frac{\frac{I}{4}}{\frac{I}{4}+\frac{I}{5}}\\ =0.556\end{array}$

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

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

Substitute $\frac{I}{4}$ for $K_{CA}$ and $\frac{I}{5}$ for $K_{CD}$.

$\begin{array}{c}{\text{DF}}_{CD}=\frac{\frac{I}{5}}{\frac{I}{4}+\frac{I}{5}}\\ =0.444\end{array}$

Check for sum of distribution factor as below:

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

Substitute 0.556 for ${\text{DF}}_{CA}$ and 0.444 for ${\text{DF}}_{CD}$.

$0.556+0.444=1$

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 $\frac{I}{5}$ for $K_{DC}$ and $\frac{3I}{20}$ for $K_{DB}$.

$\begin{array}{c}{\text{DF}}_{DC}=\frac{\frac{I}{5}}{\frac{I}{5}+\frac{3I}{20}}\\ =0.571\end{array}$

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

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

Substitute $\frac{I}{5}$ for $K_{DC}$ and $\frac{3I}{20}$ for $K_{DB}$.

$\begin{array}{c}{\text{DF}}_{DB}=\frac{\frac{3I}{20}}{\frac{I}{5}+\frac{3I}{20}}\\ =0.429\end{array}$

Check for sum of distribution factor as below:

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

Substitute 0.571 for ${\text{DF}}_{DC}$ and 0.429 for ${\text{DF}}_{DB}$.

$0.571+0.429=1$

Hence, OK.

Calculate the fixed end moment for AC.

$\begin{array}{c}{\text{FEM}}_{AC}=\frac{50\times \left(4\right)}{8}\\ =25\text{kN}\cdot \text{m}\end{array}$

Calculate the fixed end moment for CA.

$\begin{array}{c}{\text{FEM}}_{CA}=-\frac{50\times \left(4\right)}{8}\\ =-25\text{kN}\cdot \text{m}\end{array}$

Calculate the fixed end moment for CD.

$\begin{array}{c}{\text{FEM}}_{CD}=\frac{18\times {\left(5\right)}^2}{12}\\ =37.5\text{kN}\cdot \text{m}\end{array}$

Calculate the fixed end moment for DC.

$\begin{array}{c}{\text{FEM}}_{DC}=-\frac{18\times {\left(5\right)}^2}{12}\\ =-37.5\text{kN}\cdot \text{m}\end{array}$

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 free body diagram of the member AC, CD and DB for side-sway prevented as in Figure 2.

Consider member CD:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -C_{y1}\left(5\right)+\left(18\times 5\times \frac{5}{2}\right)+38.7-18.5=0\\ C_{y1}\left(5\right)=245.2\\ C_{y1}=\frac{245.2}{5}\\ C_{y1}=49.04\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ C_{y1}+D_{y1}-\left(18\times 5\right)=0\\ 49.04+D_{y1}-90=0\\ D_{y1}=40.96\text{kN}\uparrow \end{array}$

Consider member AC

Calculate vertical reaction at joint A using the relation:

$\begin{array}{c}{A}_{y1}=C_{y1}\\ =49.04\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ -A_{x1}\left(4\right)+\left(50\times 2\right)+18.1-38.7=0\\ A_{x1}\left(4\right)=79.4\\ A_{x1}=\frac{79.4}{4}\\ A_{x1}=19.85\text{kN}\leftarrow \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ A_{x1}+C_{x1}+50=0\\ -19.85+C_{x1}+50=0\\ C_{x1}=30.15\text{kN}\leftarrow \end{array}$

Consider member DB:

Calculate vertical reaction at joint B:

$\begin{array}{c}{B}_{y1}=D_{y1}\\ =40.96\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -B_{x1}\left(4\right)+\left(40.96\times 3\right)+18.5+0=0\\ B_{x1}\left(4\right)=141.38\\ B_{x1}=\frac{141.38}{4}\\ B_{x1}=35.35\text{kN}\leftarrow \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ B_{x1}+D_{x1}=0\\ -35.35+D_{x1}=0\\ D_{x1}=35.35\text{kN}\to \end{array}$

Show the unknown load R as in Figure 3.

Calculate the reaction R:

$\begin{array}{c}R=C_{x1}+D_{x1}\\ =-30.15+\left(35.35\right)\\ =5.2\text{kN}\to \end{array}$

Show the arbitrary translation as in Figure 4.

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

${\Delta }_{AC}={\Delta }^{\prime }$

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

$\begin{array}{c}{\Delta }_{CD}=\left(\frac{3}{4}\right){\Delta }^{\prime }\\ =0.75{\Delta }^{\prime }\end{array}$

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

$\begin{array}{c}{\Delta }_{BD}=\left(\frac{5}{4}\right){\Delta }^{\prime }\\ =1.25{\Delta }^{\prime }\end{array}$

Calculate the fixed end moment for AC and CA.

${\text{FEM}}_{AC}={\text{FEM}}_{CA}=-\frac{6EI\left({\Delta }_{AC}\right)}{{\left(4\right)}^2}$

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

${\text{FEM}}_{AC}={\text{FEM}}_{CA}=-\frac{6EI\left({\Delta }^{\prime }\right)}{{\left(4\right)}^2}$ (1).

Calculate the fixed end moment for CD and DC.

${\text{FEM}}_{CD}={\text{FEM}}_{DC}=\frac{6EI\left({\Delta }_{CD}\right)}{{\left(5\right)}^2}$

Substitute $0.75{\Delta }^{\prime }$ for ${\Delta }_{CD}$.

${\text{FEM}}_{CD}={\text{FEM}}_{DC}=\frac{6EI\left(0.75{\Delta }^{\prime }\right)}{{\left(5\right)}^2}$        (2)

Calculate the fixed end moment for BD and DB.

${\text{FEM}}_{BD}={\text{FEM}}_{DB}=\frac{6EI\left({\Delta }_{BD}\right)}{{\left(5\right)}^2}$

Substitute $1.25{\Delta }^{\prime }$ for ${\Delta }_{BD}$.

${\text{FEM}}_{BD}={\text{FEM}}_{DB}=-\frac{6EI\left(1.25{\Delta }^{\prime }\right)}{{\left(5\right)}^2}$        (3)

Assume the Fixed-end moment at AC, and CA as $-100\text{kN}\cdot \text{m}$

$\begin{array}{c}FE{M}_{AC}=FE{M}_{CA}\\ =-100\text{kN}\cdot \text{m}\end{array}$

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

Substitute $-100\text{kN}\cdot \text{m}$ for $FE{M}_{AB}$.

$\begin{array}{c}-100=-\frac{6EI\left({\Delta }^{\prime }\right)}{{\left(4\right)}^2}\\ EI{\Delta }^{\prime }=\frac{100\times {\left(4\right)}^2}{6}\\ EI{\Delta }^{\prime }=266.7\end{array}$

Calculate the fixed end moment of CD and DC.

Substitute 266.7 for $EI{\Delta }^{\prime }$ in equation (2).

$\begin{array}{c}{\text{FEM}}_{CD}={\text{FEM}}_{DC}=\frac{6\times 0.75\times 266.7}{{\left(5\right)}^2}\\ =48\text{kN}\cdot \text{m}\end{array}$

Calculate the fixed end moment of BD and DB.

Substitute 266.7 for $EI{\Delta }^{\prime }$ in equation (3).

$\begin{array}{c}{\text{FEM}}_{BD}={\text{FEM}}_{DB}=-\frac{6\times 1.25\times 266.7}{{\left(5\right)}^2}\\ =-80\text{kN}\cdot \text{m}\end{array}$

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 5.

Consider member CD:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -C_{y1}\left(5\right)+67.7+48.9=0\\ C_{y1}\left(5\right)=116.6\\ C_{y1}=\frac{116.6}{5}\\ C_{y1}=23.32\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ C_{y1}+D_{y1}=0\\ 23.32+D_{y1}=0\\ D_{y1}=23.32\text{kN}\downarrow \end{array}$

Consider member AC

Calculate vertical reaction at joint A using the relation:

$\begin{array}{c}{A}_{y1}=C_{y1}\\ =23.32\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ A_{x1}\left(4\right)-83.9-67.7=0\\ A_{x1}\left(4\right)=151.6\\ A_{x1}=\frac{151.6}{4}\\ A_{x1}=37.9\text{kN}\to \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ A_{x1}+C_{x1}=0\\ 37.9+C_{x1}=0\\ C_{x1}=37.9\text{kN}\leftarrow \end{array}$

Consider member DB:

Calculate vertical reaction at joint B:

$\begin{array}{c}{B}_{y1}=D_{y1}\\ =23.32\text{kN}\downarrow \end{array}$

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ B_{x1}\left(4\right)-\left(23.32\times 3\right)-48.9+0=0\\ B_{x1}\left(4\right)=118.86\\ B_{x1}=\frac{118.86}{4}\\ B_{x1}=29.72\text{kN}\to \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ B_{x1}+D_{x1}=0\\ 29.72+D_{x1}=0\\ D_{x1}=29.72\text{kN}\leftarrow \end{array}$

Show the unknown load Q as in Figure 6.

Calculate the reaction Q:

$\begin{array}{c}Q=C_{x1}+D_{x1}\\ =37.9+29.72\\ =67.62\text{kN}\leftarrow \end{array}$

Calculate the actual member end moments of the member AC:

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

Substitute $18.1\text{kN}\cdot \text{m}$ for ${\left(M_0\right)}_{AC}$, $-83.9\text{kN}\cdot \text{m}$ for ${\left(M_Q\right)}_{AC}$, $67.62\text{kN}$ for Q, and $5.2\text{kN}$ for R.

$\begin{array}{c}{M}_{AC}=18.1+\left(\frac{5.2}{67.62}\right)\times \left(-83.9\right)\\ =18.1-6.45\\ =11.7\text{kN}\cdot \text{m}\end{array}$

Calculate the actual member end moments of the member CA:

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

Substitute $-38.7\text{kN}\cdot \text{m}$ for ${\left(M_0\right)}_{CA}$, $-67.7\text{kN}\cdot \text{m}$ for ${\left(M_Q\right)}_{CA}$, $67.62\text{kN}$ for Q, and $5.2\text{kN}$ for R.

$\begin{array}{c}{M}_{CA}=-38.7+\left(\frac{5.2}{67.62}\right)\times \left(-67.7\right)\\ =-38.7-5.2\\ =-43.9\text{kN}\cdot \text{m}\end{array}$

Calculate the actual member end moments of the member CD:

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

Substitute $38.7\text{kN}\cdot \text{m}$ for ${\left(M_0\right)}_{CD}$, $67.7\text{kN}\cdot \text{m}$ for ${\left(M_Q\right)}_{CD}$, $67.62\text{kN}$ for Q, and $5.2\text{kN}$ for R.

$\begin{array}{c}{M}_{CD}=38.7+\left(\frac{5.2}{67.62}\right)\times \left(67.7\right)\\ =+38.7+5.2\\ =43.9\text{kN}\cdot \text{m}\end{array}$

Calculate the actual member end moments of the member DC:

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

Substitute $-18.5\text{kN}\cdot \text{m}$ for ${\left(M_0\right)}_{DC}$, $48.9\text{kN}\cdot \text{m}$ for ${\left(M_Q\right)}_{DC}$, $67.62\text{kN}$ for Q, and $5.2\text{kN}$ for R.

$\begin{array}{c}{M}_{DC}=-18.5+\left(\frac{5.2}{67.62}\right)\times \left(48.9\right)\\ =-18.5+3.76\\ =-14.7\text{kN}\cdot \text{m}\end{array}$

Calculate the actual member end moments of the member DB:

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

Substitute $18.5\text{kN}\cdot \text{m}$ for ${\left(M_0\right)}_{DB}$, $-48.9\text{kN}\cdot \text{m}$ for ${\left(M_Q\right)}_{DB}$, $67.62\text{kN}$ for Q, and $5.2\text{kN}$ for R.

$\begin{array}{c}{M}_{DB}=18.5+\left(\frac{5.2}{67.62}\right)\times \left(-48.9\right)\\ =18.5-3.76\\ =-14.7\text{kN}\cdot \text{m}\end{array}$

Calculate the actual member end moments of the member BD:

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

Substitute $0\text{kN}\cdot \text{m}$ for ${\left(M_0\right)}_{BD}$, $0\text{kN}\cdot \text{m}$ for ${\left(M_Q\right)}_{BD}$, $67.62\text{kN}$ for Q, and $5.2\text{kN}$ for R.

$\begin{array}{c}{M}_{BD}=0+\left(\frac{5.2}{67.62}\right)\times \left(0\right)\\ =0\text{kN}\cdot \text{m}\end{array}$

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

Consider member CD:

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_D=0}\\ -C_y\left(5\right)+\left(18\times 5\times \frac{5}{2}\right)+43.9-14.7=0\\ C_y\left(5\right)=254.2\\ C_y=\frac{254.2}{5}\\ C_y=50.8\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ C_y+D_y-\left(18\times 5\right)=0\\ 50.8+D_y-90=0\\ D_y=39.2\text{kN}\uparrow \end{array}$

Consider member AC

Calculate vertical reaction at joint A:

$\begin{array}{c}{A}_{y1}=C_{y1}\\ =50.8\text{kN}\uparrow \end{array}$

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

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_C=0}\\ A_x\left(4\right)+\left(50\times 2\right)+11.7-43.9=0\\ A_x\left(4\right)=67.8\\ A_x=\frac{67.8}{4}\\ A_x=17\text{kN}\to \end{array}$

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

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ A_x+C_x-50=0\\ 17+C_x-50=0\\ C_x=33\text{kN}\leftarrow \end{array}$

Consider member DB:

Calculate vertical reaction at joint B using the relation:

$\begin{array}{c}{B}_y=D_y\\ =39.2\text{kN}\uparrow \end{array}$

Calculate horizontal reaction at joint B.

$\begin{array}{c}{B}_x=C_x\\ =33\text{kN}\end{array}$

Show the reactions of the frame as in Figure 8.