Chapter 11, Problem 27P

To determine

Find the reactions and moments of the members in the frame.

Find the horizontal displacement of joint B.

Answer to Problem 27P

Therefore, the horizontal displacement at point B is $\underset{\_}{0.71\text{in}\text{.}}$.

Explanation of Solution

Given information:

The value of Young’s modulus (E) is $3,000\text{kips}/\text{in}{\text{.}}^2$.

The moment of inertia (I) of the material is $1,500\text{in}{\text{.}}^4$.

Calculation:

Show the free-body diagram of the sway as in Figure 1.

Find the stiffness of each member connected to joint B as follows;

$\begin{array}{c}{K}_{BA}=\frac{I}{L_{BA}}\\ =\frac{I}{12}\\ K_{BC}=\frac{I}{L_{BC}}\\ =\frac{I}{16}\end{array}$

Find the total stiffness at joint B as follows;

$\begin{array}{c}{\displaystyle \sum K_B}=K_{BA}+K_{BC}\\ =\frac{I}{12}+\frac{I}{16}\\ =\frac{7I}{48}\end{array}$

Find the distribution factors at joint B using the relation;

$\begin{array}{c}D{F}_{BA}=\frac{K_{BA}}{{\displaystyle \sum K_B}}\\ =\frac{\frac{I}{12}}{\frac{7I}{48}}\\ =\frac{4}{7}\end{array}$

$\begin{array}{c}D{F}_{BC}=\frac{K_{BC}}{{\displaystyle \sum K_B}}\\ =\frac{\frac{I}{16}}{\frac{7I}{48}}\\ =\frac{3}{7}\end{array}$

Find the stiffness of each member connected to joint C as follows;

$\begin{array}{c}{K}_{CB}=\frac{I}{L_{CB}}\\ =\frac{I}{16}\\ K_{CD}=\frac{I}{L_{CD}}\\ =\frac{I}{16}\\ K_{CF}=\frac{I}{L_{CF}}\\ =\frac{1.5I}{12}\end{array}$

Find the total stiffness at joint C as follows;

$\begin{array}{c}{\displaystyle \sum K_C}=K_{CB}+K_{CD}+K_{CF}\\ =\frac{I}{16}+\frac{I}{16}+\frac{1.5I}{12}\\ =\frac{4I}{16}\end{array}$

Find the distribution factors at joint C using the relation;

$\begin{array}{c}D{F}_{CB}=\frac{K_{CB}}{{\displaystyle \sum K_C}}\\ =\frac{\frac{I}{16}}{\frac{4I}{16}}\\ =\frac{1}{4}\end{array}$

$\begin{array}{c}D{F}_{CD}=\frac{K_{CD}}{{\displaystyle \sum K_C}}\\ =\frac{\frac{I}{16}}{\frac{4I}{16}}\\ =\frac{1}{4}\end{array}$

$\begin{array}{c}D{F}_{CF}=\frac{K_{CF}}{{\displaystyle \sum K_C}}\\ =\frac{\frac{1.5I}{12}}{\frac{4I}{16}}\\ =\frac{1}{2}\end{array}$

Find the stiffness of each member connected to joint D as follows;

$\begin{array}{c}{K}_{DC}=\frac{I}{L_{DC}}\\ =\frac{I}{16}\\ K_{DE}=\frac{I}{L_{DE}}\\ =\frac{I}{12}\end{array}$

Find the total stiffness at joint D as follows;

$\begin{array}{c}{\displaystyle \sum K_D}=K_{DC}+K_{DE}\\ =\frac{I}{16}+\frac{I}{12}\\ =\frac{7I}{48}\end{array}$

Find the distribution factors at joint D using the relation;

$\begin{array}{c}D{F}_{DC}=\frac{K_{DC}}{{\displaystyle \sum K_D}}\\ =\frac{\frac{I}{16}}{\frac{7I}{48}}\\ =\frac{3}{7}\end{array}$

$\begin{array}{c}D{F}_{DE}=\frac{K_{DE}}{{\displaystyle \sum K_D}}\\ =\frac{\frac{I}{12}}{\frac{7I}{48}}\\ =\frac{4}{7}\end{array}$

Find the fixed end moments in the member AB as follows:

$\begin{array}{c}{\text{FEM}}_{AB}=-\frac{6EI\Delta }{L_{AB}^2}\\ =-\frac{6\times 3000\times 1500\times 1}{{\left(\text{12}\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}\\ =1,302.083\text{kip-in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\\ =-108.5\text{kips-ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BA}=-\frac{6EI\Delta }{L_{BA}^2}\\ =-\frac{6\times 3000\times 1500\times 1}{{\left(\text{12}\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}\\ =1,302.083\text{kip-in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\\ =-108.5\text{kips-ft}\end{array}$

Find the fixed end moments in the member DE as follows:

$\begin{array}{c}{\text{FEM}}_{DE}=-\frac{6EI\Delta }{L_{DE}^2}\\ =-\frac{6\times 3000\times 1500\times 1}{{\left(\text{12}\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}\\ =1,302.083\text{kip-in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\\ =-108.5\text{kips-ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{ED}=-\frac{6EI\Delta }{L_{ED}^2}\\ =-\frac{6\times 3000\times 1500\times 1}{{\left(\text{12}\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}\\ =1,302.083\text{kip-in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\\ =-108.5\text{kips-ft}\end{array}$

Find the fixed end moments in the member CF as follows:

$\begin{array}{c}{\text{FEM}}_{CF}=-\frac{6EI\Delta }{L_{CF}^2}\\ =-\frac{6\times 3000\times 1.5\times 1500\times 1}{{\left(\text{12}\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}\\ =1,953.125\text{kip-in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\\ =-162.8\text{kips-ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{FC}=-\frac{6EI\Delta }{L_{FC}^2}\\ =-\frac{6\times 3000\times 1.5\times 1500\times 1}{{\left(\text{12}\text{ft}\times \frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}\\ =1,953.125\text{kip-in}\text{.}\times \frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\\ =-162.8\text{kips-ft}\end{array}$

Show the moment distribution computations as in Figure 2.

Show the free-body diagram of the members split up as in Figure 3.

Due to symmetry the vertical reaction at the joint C will be zero. Therefore consider the member BCD for the calculation purpose.

Consider member BCD:

Take moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B^{BCD}=0}\\ V_D\left(32\right)-55.3-101.3-55.3=0\\ V_D=6.6\text{kips}\end{array}$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ -V_B+V_D=0\\ -V_B+6.6=0\\ V_B=6.6\text{kips}\end{array}$

Consider the member AB:

Take moment about point B;

$\begin{array}{l}{\displaystyle \sum M_B^{BA}=0}\\ -82-55.3+A_x\left(12\right)=0\\ A_x=11.4\text{kips}\\ H_B=A_x=11.4\text{kips}\end{array}$

Also, $A_y=V_B=6.6\text{kips}$.

Consider the member CF:

Take moment about point C;

$\begin{array}{l}{\displaystyle \sum M_C^{CF}=0}\\ -132-101.3+F_x\left(12\right)=0\\ F_x=19.4\text{kips}\\ H_C=F_x=19.4\text{kips}\end{array}$

Consider the member DE:

Take moment about point D;

$\begin{array}{l}{\displaystyle \sum M_D^{DE}=0}\\ -82-55.3+E_x\left(12\right)=0\\ E_x=11.4\text{kips}\\ H_D=E_x=11.4\text{kips}\end{array}$

Also, $E_y=V_D=6.6\text{kips}$.

Find the value of using the relation:

$\begin{array}{c}S=H_B+H_C+H_D\\ =11.4+19.4+11.4\\ =42.2\text{kips}(\to )\end{array}$

Find the horizontal displacement at point B using the relation:

$\begin{array}{c}{\Delta }_{HB}=\frac{30}{42.2}\\ =0.71\text{in}\text{.}\end{array}$

Therefore, the horizontal displacement at point B is $\underset{\_}{0.71\text{in}\text{.}}$.

Multiply all the values in the Figure 3 by 0.71 in. to get the corrected reaction and moment values.

Show the final reaction and moment values as in Figure 4.