Chapter 10, Problem 21P

To determine

Analyze and find all the reactions of the frame.

Explanation of Solution

Determine the deflection position of AB and BA using the relation;

$\begin{array}{c}{\psi }_{AB}={\psi }_{BA}=\frac{\Delta }{L}\\ =\frac{\text{1}\text{in}\text{.}}{\left(\text{10}\text{ft×}\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}\\ =\frac{1}{120}\end{array}$

Determine the deflection position of BC and CB using the relation;

$\begin{array}{c}{\psi }_{BC}={\psi }_{CB}=-\frac{\Delta }{L}\\ =-\frac{\text{1}\text{in}\text{.}}{\left(\text{10}\text{ft×}\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}\\ =-\frac{1}{120}\end{array}$

Determine the end moment of each member as shown below;

$\begin{array}{c}{M}_{AB}=\frac{2EI}{L}\left({\theta }_B+2{\theta }_A-3{\psi }_{AB}\right)\\ =\frac{2EI}{10}\left({\theta }_B+0-3\frac{1}{120}\right)\\ =\frac{2EI}{10}\left({\theta }_B-\frac{3}{120}\right)\\ M_{BA}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_A-3{\psi }_{BA}\right)\\ =\frac{2EI}{10}\left(2{\theta }_B+0-3\frac{1}{120}\right)\\ =\frac{2EI}{10}\left(2{\theta }_B-\frac{3}{120}\right)\end{array}$

$\begin{array}{c}{M}_{BC}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_C-3{\psi }_{BC}\right)\\ =\frac{2EI}{10}\left(2{\theta }_B+{\theta }_C-3\left(-\frac{1}{120}\right)\right)\\ =\frac{2EI}{10}\left(2{\theta }_B+{\theta }_C+\frac{3}{120}\right)\\ M_{CB}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_B-3{\psi }_{CB}\right)\\ =\frac{2EI}{10}\left(2{\theta }_C+{\theta }_B-3\left(-\frac{1}{120}\right)\right)\\ =\frac{2EI}{10}\left(2{\theta }_C+{\theta }_B+\frac{3}{120}\right)\end{array}$

$\begin{array}{c}{M}_{BE}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_E\right)\\ =\frac{2EI}{10}\left(2{\theta }_B+0\right)\\ =\frac{2EI}{10}\left(2{\theta }_B\right)\\ M_{EB}=\frac{2EI}{L}\left(2{\theta }_E+{\theta }_B\right)\\ =\frac{2EI}{10}\left(2\left(0\right)+{\theta }_B\right)\\ =\frac{2EI}{10}\left({\theta }_B\right)\end{array}$

$\begin{array}{c}{M}_{CD}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_D\right)\\ =\frac{2EI}{15}\left(2{\theta }_C+0\right)\\ =\frac{2EI}{15}\left(2{\theta }_C\right)\\ M_{DC}=\frac{2EI}{L}\left(2{\theta }_D+{\theta }_C\right)\\ =\frac{2EI}{15}\left(2\left(0\right)+{\theta }_C\right)\\ =\frac{2EI}{15}\left({\theta }_C\right)\end{array}$

Apply Equilibrium at joint B;

$\begin{array}{l}{M}_{BA}+M_{BE}+M_{BC}=0\\ \frac{2EI}{10}\left(2{\theta }_B-\frac{3}{120}\right)+\frac{2EI}{10}\left(2{\theta }_B\right)+\frac{2EI}{10}\left(2{\theta }_B+{\theta }_C+\frac{3}{120}\right)=0\\ 0.4EI{\theta }_B-0.005EI+0.4EI{\theta }_B+0.4EI{\theta }_B+0.2EI{\theta }_C+0.005EI=0\\ 1.2EI{\theta }_B-0.2EI{\theta }_C=0\end{array}$        (1)

Apply Equilibrium at joint C;

$\begin{array}{l}{M}_{CB}+M_{CD}=0\\ \frac{2EI}{10}\left(2{\theta }_C+{\theta }_B+\frac{3}{120}\right)+\frac{2EI}{15}\left(2{\theta }_C\right)=0\\ 0.4EI{\theta }_C+0.2EI{\theta }_B+0.005EI+0.267EI{\theta }_C=0\\ 0.667EI{\theta }_C+0.2EI{\theta }_B+0.005EI=0\end{array}$        (2)

Solve Equation (1) and (2).

$\begin{array}{l}{\theta }_B=\frac{1}{760}\text{rad}\\ {\theta }_C=-\frac{3}{380}\text{rad}\end{array}$

Calculation of end moment of each member as shown below;

$\begin{array}{c}{M}_{AB}=\frac{2\left(29,000\right)\left(100\right)}{\text{10}}\left(\frac{1}{760}-\frac{3}{120}\right){\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^2\\ =-95.4\text{kips-ft}\\ M_{BA}=\frac{2EI}{10}\left(2\left(\frac{1}{760}\right)-\frac{3}{120}\right)\\ =\frac{2\left(29,000\right)\left(100\right)}{\text{10}}\left(2\left(\frac{1}{760}\right)-\frac{3}{120}\right){\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^2\\ =-90.1\text{kips-ft}\end{array}$

Hence, the end moment of member AB is $\underset{\_}{-95.4\text{kips-ft}}$.

Hence, the end moment of member BA is $\underset{\_}{-90.1\text{kips-ft}}$.

$\begin{array}{c}{M}_{BC}=\frac{2\left(29,000\right)\left(100\right)}{10}\left(2\left(\frac{1}{760}\right)-\frac{3}{380}+\frac{3}{120}\right){\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^2\\ =79.5\text{kips-ft}\\ M_{CB}=\frac{2\left(29,000\right)\left(100\right)}{10}\left(2\left(-\frac{3}{380}\right)+\left(\frac{1}{760}\right)+\frac{3}{120}\right){\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^2\\ =42.4\text{kips-ft}\end{array}$

Hence, the end moment of member BC is $\underset{\_}{79.5\text{kips-ft}}$.

Hence, the end moment of member CB is $\underset{\_}{42.4\text{kips-ft}}$.

$\begin{array}{c}{M}_{BE}=\frac{2\left(29,000\right)\left(100\right)}{10}\left(2\left(\frac{1}{760}\right)\right){\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^{\text{2}}\\ =10.6\text{kips-ft}\\ M_{EB}=\frac{2\left(29,000\right)\left(100\right)}{10}\left(\frac{1}{760}\right){\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^{\text{2}}\\ =5.3\text{kips-ft}\end{array}$

Hence, the end moment of member BE is $\underset{\_}{10.6\text{kips-ft}}$.

Hence, the end moment of member EB is $\underset{\_}{5.3\text{kips-ft}}$.

$\begin{array}{c}{M}_{CD}=\frac{2\left(29,000\right)\left(100\right)}{15}\left(2\left(-\frac{3}{380}\right)\right){\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^2\\ =-42.4\text{kips-ft}\\ M_{DC}=\frac{2\left(29,000\right)\left(100\right)}{15}\left(-\frac{3}{380}\right){\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^2\\ =21.2\text{kips-ft}\end{array}$

Hence, the end moment of member CD is $\underset{\_}{-42.4\text{kips-ft}}$.

Hence, the end moment of member DC is $\underset{\_}{21.2\text{kips-ft}}$.

Show the free body diagram of support A, span AB, span BC, span EB, and span DC as in Figure (1).

Consider span AB;

Consider clockwise moment is positive and counterclockwise moment is negative.

Determine the vertical reaction at support B;

Take moment about point A;

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -R_B\times 10-95.4-90.1=0\\ R_B=-\frac{185.5}{10}\\ R_B=-18.6\text{kips}\\ R_B=18.6\text{kips}(\downarrow )\end{array}$

Consider upward is positive and downward is negative.

Determine the vertical reaction at A;

$\begin{array}{l}{\displaystyle \sum V=0}\\ R_A+=0\\ R_A=-R_B\\ R_A=-\left(-18.6\right)\\ R_A=18.6\text{kips}\end{array}$

Hence, the vertical reaction at A is $\underset{\_}{18.6\text{kips}}$.

Consider span BC;

Determine the vertical reaction at support C;

Take moment about point B;

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -R_C\times 10+42.4+79.5=0\\ R_C=\frac{121.9}{10}\\ R_C=12.2\text{kips}\end{array}$

Determine the vertical reaction at B;

$\begin{array}{l}{\displaystyle \sum V=0}\\ R_B+R_C=0\\ R_B=-R_C\\ R_B=-12.2\text{kips}\\ R_B=12.2\text{kips}(\downarrow )\end{array}$

Hence, the total reaction at B is $\underset{\_}{30.8\text{kips}}$.

Consider span BE;

Refer Figure (1),

The vertical reaction of 30.8 kips acts as upward reaction at B.

Determine the vertical reaction at E;

$\begin{array}{l}{\displaystyle \sum V=0}\\ R_B+R_E=0\\ 30.8+R_E=0\\ R_E=-30.8\text{kips}\\ R_E=30.8\text{kips}(\downarrow )\end{array}$

Determine the horizontal reaction E;

Take moment about B;

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ \left(H_E\times 10\right)+10.6+5.3=0\\ H_E=-\frac{15.9}{10}\\ H_E=-1.6\text{kips}\\ H_E\simeq 1.60\text{kips}(\to )\end{array}$

Determine the horizontal reaction at B;

$\begin{array}{l}{\displaystyle \sum H=0}\\ H_B+H_E=0\\ H_B=-\left(1.60\text{kips}\right)\\ H_B=1.60\text{kips}(\leftarrow )\end{array}$

Consider span CD;

Refer Figure (1),

The vertical reaction 12.2 kips at point C will acts downward reaction at C.

Determine the vertical reaction at E;

$\begin{array}{l}{\displaystyle \sum V=0}\\ R_C+R_D=0\\ -12.2+R_D=0\\ R_D=12.2\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at D is $\underset{\_}{12.2\text{kips}}$.

Determine the horizontal reaction D;

Take moment about C;

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ \left(H_D\times 15\right)+42.4+21.2=0\\ H_D=-\frac{63.6}{15}\\ H_D=-4.2\text{kips}\\ H_D=4.2\text{kips}(\leftarrow )\end{array}$

Determine the horizontal reaction at C;

$\begin{array}{l}{\displaystyle \sum H=0}\\ H_D+H_C=0\\ H_C=-\left(2.8\text{kips}\right)\\ H_C=2.8\text{kips}(\to )\end{array}$

Determine the horizontal force at A and B for span AB;

$\begin{array}{c}{H}_A=4.2-1.6\\ =2.6\text{kips}(\to )\\ H_B=1.6-4.2\\ =-2.6\text{kips}\\ =2.6\text{kips}(\leftarrow )\end{array}$