Chapter 10, Problem 29P

To determine

Analyze the frame and compute all the reactions.

Explanation of Solution

Given information;

The moment of inertia of the member BC is $200\text{in}{\text{.}}^{\text{4}}$.

The moment of inertial of the member AB and CD is $150\text{in}{\text{.}}^{\text{4}}$.

The moment acting at B is $100\text{kips-ft}$.

Apply the sign conventions for calculating reactions using the three equations of equilibrium as shown below.

• For summation of forces along x-direction is equal to zero $\left(\sum F_x=0\right)$, consider the forces acting towards right side as positive $\left(\to +\right)$ and the forces acting towards left side as negative $\left(\leftarrow -\right)$.
• For summation of forces along y-direction is equal to zero $\left(\sum F_y=0\right)$, consider the upward force as positive $\left(\uparrow +\right)$ and the downward force as negative $\left(\downarrow -\right)$.
• For summation of moment about a point is equal to zero $\left(\sum M_{\text{at}\text{a}\text{point}}=0\right)$, consider the clockwise moment as positive and the counter clockwise moment as negative.

Determine the deflection position AB, BC, and CD as follows:

$\begin{array}{l}{\psi }_{AB}=-\frac{\Delta }{25}=-\psi \\ {\psi }_{BC}=\frac{3}{5}\frac{{\Delta }_V}{25}=\frac{3}{4}\psi \\ {\psi }_{CD}=\frac{3}{5}\frac{{\Delta }_H}{20}=\frac{\Delta }{25}=\psi \end{array}$

Set the relative stiffness to $20E=K$.

Determine the stiffness factor for each member as shown below;

$\begin{array}{c}{K}_{AB}=\frac{EI}{L}\\ =\frac{\left(\frac{K}{10}\right)\left(150\right)}{25}\\ =\frac{3}{5}K\end{array}$

$\begin{array}{c}{K}_{BC}=\frac{EI}{L}\\ =\frac{\left(\frac{K}{10}\right)\left(200\right)}{20}\\ =K\end{array}$

$\begin{array}{c}{K}_{CD}=\frac{EI}{L}\\ =\frac{\left(\frac{K}{10}\right)\left(150\right)}{20}\\ =\frac{3}{4}K\end{array}$

Determine the end moment moments of each member shown below;

$\begin{array}{c}{M}_{AB}=\frac{2EI}{L}\left(2{\theta }_A+{\theta }_B-3\psi \right)\\ =2\left(\frac{3}{5}K\right)\left(2{\theta }_A+{\theta }_B-3\left(-\psi \right)\right)\\ =2.4K{\theta }_A+1.2K{\theta }_B+3.6K\psi \\ M_{BA}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_A-3\psi \right)\\ =2\left(\frac{3}{5}K\right)\left(2{\theta }_B+{\theta }_A-3\left(-\psi \right)\right)\\ =2.4K{\theta }_B+1.2K{\theta }_A+3.6K\psi \end{array}$

Determine the end moment moments of each member shown below;

$\begin{array}{c}{M}_{BC}=\frac{2EI}{L}\left(2{\theta }_B+{\theta }_C-3{\psi }_{BC}\right)\\ =2\left(K\right)\left(2{\theta }_B+{\theta }_C-3\left(\frac{3}{4}\psi \right)\right)\\ =4K{\theta }_B+2K{\theta }_C-1.5K\psi \\ M_{CB}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_B-3{\psi }_{BC}\right)\\ =2\left(K\right)\left(2{\theta }_C+{\theta }_B-3\left(\frac{3}{4}\psi \right)\right)\\ =4K{\theta }_C+2K{\theta }_B-4.5K\psi \end{array}$

$\begin{array}{c}{M}_{CD}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_D-3{\psi }_{CD}\right)\\ =2\left(\frac{3}{4}K\right)\left(2{\theta }_C+0-3\psi \right)\\ =3K{\theta }_C-4.5K\psi \\ M_{DC}=\frac{2EI}{L}\left(2{\theta }_D+{\theta }_C-3\psi \right)\\ =2\left(\frac{3}{4}K\right)\left(2\left(0\right)+{\theta }_C-3\psi \right)\\ =1.5K{\theta }_C-4.5K\psi \end{array}$

Show the free body diagram of the frame and point O as in Figure (1).

Apply Equation of equilibrium at A;

$\begin{array}{l}{M}_{AB}=0\\ 2.4K{\theta }_A+1.2K{\theta }_B+3.6K\psi =0\\ 2.4{\theta }_A+1.2{\theta }_B+3.6\psi =0\end{array}$        (1)

Apply Equation of equilibrium at joint B;

$\begin{array}{l}{M}_{BA}+M_{BC}-100=0\\ 2.4K{\theta }_B+1.2K{\theta }_A+3.6K\psi +4K{\theta }_B+2K{\theta }_C-1.5K\psi =100\\ \frac{6}{5}{\theta }_A+\frac{32}{5}{\theta }_B+2{\theta }_C-\frac{9}{10}\psi =\frac{100}{K}\end{array}$        (2)

Apply Equation of equilibrium at joint C;

$\begin{array}{l}{M}_{CB}+M_{CD}=0\\ 4K{\theta }_C+2K{\theta }_B-4.5K\psi +3K{\theta }_C-4.5K\psi =0\\ 7K{\theta }_C+2K{\theta }_B-9K\psi =0\end{array}$        (3)

Refer Figure (1).

Take moment about point O;

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ -\frac{M_{BA}}{25}\left(58.33\right)+100+M_{DC}+\frac{M_{CD}+M_{DC}}{20}=0\\ \left[\begin{array}{l}-\frac{\left(2.4K{\theta }_B+1.2K{\theta }_A+3.6K\psi \right)}{25}\left(58.33\right)+100\\ +\left(1.5K{\theta }_C-4.5K\psi \right)+\frac{\left(3K{\theta }_C-4.5K\psi \right)+\left(1.5K{\theta }_C-4.5K\psi \right)}{20}\end{array}\right]=0\\ -2.8K{\theta }_A-5.6K{\theta }_B+3K{\theta }_C-15.9K\psi =-\frac{100}{\psi }\end{array}$        (4)

Solve Equation (1), (2), (3), and (4).

$\begin{array}{l}{\theta }_A=-\frac{10.158}{K}\\ {\theta }_B=\frac{19.136}{K}\\ {\theta }_C=-\frac{4.938}{K}\\ \psi =\frac{0.412}{K}\end{array}$

Hence, the slope at A is $\underset{\_}{-\frac{10.158}{K}}$.

Hence, the slope at B is $\underset{\_}{\frac{19.136}{K}}$.

Hence, the slope at C is $\underset{\_}{-\frac{4.938}{K}}$.

Determine the end moments of each member as shown below;

$\begin{array}{c}{M}_{AB}=2.4K\left(-\frac{10.158}{K}\right)+1.2K\left(\frac{19.136}{K}\right)+3.6K\left(\frac{0.412}{K}\right)\\ \simeq 0\\ M_{BA}=2.4K\left(\frac{19.136}{K}\right)+1.2K\left(-\frac{10.158}{K}\right)+3.6K\left(\frac{0.412}{K}\right)\\ =35.22\text{kips-ft}\end{array}$

$\begin{array}{c}{M}_{BC}=4K\left(\frac{19.136}{K}\right)+2K\left(-\frac{4.938}{K}\right)-1.5K\left(\frac{0.412}{K}\right)\\ =66.05\text{kips-ft}\\ M_{CB}=4K\left(-\frac{4.938}{K}\right)+2K\left(\frac{19.136}{K}\right)-4.5K\left(\frac{0.412}{K}\right)\\ =16.67\text{kips-ft}\end{array}$

$\begin{array}{c}{M}_{CD}=3K\left(-\frac{4.938}{K}\right)-4.5K\left(\frac{0.412}{K}\right)\\ =-16.67\text{kips-ft}\\ M_{DC}=1.5K\left(-\frac{4.938}{K}\right)-4.5K\left(\frac{0.412}{K}\right)\\ =-9.26\text{kips-ft}\end{array}$

Hence, the end moment of member AB is $\underset{\_}{\text{zero}}$.

Hence, the end moment of member BA is $\underset{\_}{35.22\text{kips-ft}}$.

Hence, the end moment of member BC is $\underset{\_}{66.05\text{kips-ft}}$.

Hence, the end moment of member CB is $\underset{\_}{16.67\text{kips-ft}}$.

Hence, the end moment of member CD is $\underset{\_}{-16.67\text{kips-ft}}$.

Hence, the end moment of member DC is $\underset{\_}{-9.26\text{kips-ft}}$.

Consider span AB;

Show the fee body diagram of span AB as in Figure (2).

Determine the shear force $V_{AB}$ using the relation;

Take moment about point B;

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ -V_{AB}\times \left(25\right)+100=0\\ V_{AB}=\frac{100}{25}\\ V_{AB}=4.0\text{kips}(\downarrow )\end{array}$

Hence, the vertical reaction at A is $\underset{\_}{4.0\text{kips}(\downarrow )}$.

Determine the shear force using $V_{BA}$ the relation;

$\begin{array}{l}{\displaystyle \sum V=0}\\ V_{AB}+V_{BA}=0\\ V_{BA}=-4.0\text{kips}\\ V_{BA}=4.0\text{kips}(\leftarrow )\end{array}$

Consider span BC;

Show the free body diagram of span BC as in Figure (3).

Determine the shear force $V_{BC}$ using the relation;

Take moment about point C;

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ V_{BC}\times \left(20\right)+66.05+16.67=0\\ V_{BC}=\frac{82.72}{20}\\ V_{BC}=4.1\text{kips}\end{array}$

Determine the shear force using $V_{CB}$ the relation;

$\begin{array}{l}{\displaystyle \sum V=0}\\ V_{BC}+V_{CB}=0\\ V_{CB}=-4.1\text{kips}\\ V_{CB}=4.1\text{kips}(\leftarrow )\end{array}$

Consider member CD;

Show the free body diagram of span CD as in Figure (4).

Determine the shear force $V_{CD}$ using the relation;

Take moment about point D;

$\begin{array}{l}{\displaystyle \sum M_D=0}\\ V_{CD}\times \left(20\right)-16.67-9.26=0\\ V_{CD}=\frac{25.93}{20}\\ V_{CD}=1.30\text{kips}\end{array}$

Determine the shear force using $V_{DC}$ the relation;

$\begin{array}{l}{\displaystyle \sum V=0}\\ V_{CD}+V_{DC}=0\\ V_{DC}=-1.30\text{kips}\\ V_{DC}=1.30\text{kips}(\leftarrow )\end{array}$

Hence, the horizontal reaction at D is $\underset{\_}{1.3\text{kips}}$.

Determine the horizontal reaction at A;

$\begin{array}{l}{\displaystyle \sum H=0}\\ H_A+H_D=0\\ H_A=-H_D\\ H_A=-1.30\text{kips}\\ H_A=1.30\text{kips}(\leftarrow )\end{array}$

Determine the vertical reaction at D using the relation;

$\begin{array}{l}{\displaystyle \sum V=}0\\ R_A+R_D=0\\ R_D=-R_A\\ R_D=-\left(4\text{kips}\right)\\ R_D=4\text{kips}(\uparrow )\end{array}$

Hence, the vertical reaction at D is $\underset{\_}{4\text{kips}(\uparrow )}$.

Shear force calculation;

$\begin{array}{l}{V}_{AB}=4.0\text{kips}\\ V_{BA}=4.0\text{kips}(\leftarrow )\\ V_{BC}=4.1\text{kips}\\ V_{CB}=4.1\text{kips}(\leftarrow )\\ V_{CD}=1.30\text{kips}\\ V_{DC}=1.30\text{kips}(\leftarrow )\end{array}$

Bending moment calculation;

$\begin{array}{c}{M}_{AB}=-2.90\text{kips-ft}\\ M_{BA}=35.22\text{kips-ft}\\ M_{BC}=66.05\text{kips-ft}\\ M_{CB}=16.67\text{kips-ft}\\ M_{CD}=-16.67\text{kips-ft}\\ M_{DC}=-9.26\text{kips-ft}\end{array}$

Consider span BC;

Consider a section at a distance of x from point C within 20 ft;

Determine the where the moment is equal to zero;

$\begin{array}{l}{M}_x=0\\ 16.7-4.1x=0\\ x=\frac{16.7}{4.1}\\ x=4.073\text{ft}\text{from}\text{support}C\end{array}$

Hence, the bending moment is zero at 4.073 ft from support C.

Consider span CD;

Consider a section at a distance of x from point D within 20 ft;

Determine the where the moment is equal to zero;

$\begin{array}{l}{M}_x=0\\ -9.26+1.30x=0\\ x=\frac{9.26}{1.30}\\ x=7.123\text{ft}\text{from support}D\end{array}$

Hence, the bending moment is zero at 7.123 ft from support D.

Show the shear force and bending moment diagram as in Figure (5).

Show the deflected shape as in Figure (6).