Chapter 14, Problem 6P

To determine

Calculate all the reactions and draw the moment diagrams.

Answer to Problem 6P

The vertical reaction at A is $\underset{\_}{57.42\text{kips}}$.

The vertical reaction at B is $\underset{\_}{67.5\text{kips}}$.

The vertical reaction at C is $\underset{\_}{13.08\text{kips}}$.

The bending moment at A is $\underset{\_}{-182.57\text{kips}\cdot \text{ft}}$.

The bending moment at B is $\underset{\_}{-120.85\text{kips}\cdot \text{ft}}$.

The bending moment at C is $\underset{\_}{-74.57\text{kips}\cdot \text{ft}}$.

Explanation of Solution

Restrained structure: Clamp joint 2:

Show the free body diagram of the restrained structure as in Figure (1).

Refer Figure (1),

Find the fixed end moment of the member 1-2 $\left({\text{FEM}}_{12}\right)$ using the relation;

$\begin{array}{c}\left({\text{FEM}}_{12}\right)=\pm \frac{w{L}^2}{12}\\ =\pm \frac{6\times {18}^2}{12}\\ =\pm 162\text{kips}\cdot \text{ft}\end{array}$

Find the fixed end moment of the member 2-3 $\left({\text{FEM}}_{23}\right)$ using the relation;

$\begin{array}{c}\left({\text{FEM}}_{23}\right)=\pm \frac{PL}{8}\\ =\pm \frac{30\times 24}{8}\\ =\pm 90\text{kips}\cdot \text{ft}\end{array}$

Find the moment in clamp using the relation;

$\begin{array}{c}\text{Moment in clamp}=\left({\text{FEM}}_{21}\right)+\left({\text{FEM}}_{23}\right)\\ =162-90\\ =72\text{kips}\cdot \text{ft}\end{array}$

Show the free body diagram of the restrained structure with FEM and moment as in Figure (2).

Show the free body diagram of the structure with unit rotation at 2 Figure (3).

Moments due to unit rotation at 2:

Find the moment at the member 1-2 using the relation;

$\begin{array}{c}\left(M_{12}^{JD}\right)=\frac{2EI}{L}\left(\theta \right)\\ =\frac{2EI}{18}\left(1\right)\\ =\frac{EI}{9}\end{array}$

Find the moment at the member 2-1 using the relation;

$\begin{array}{c}\left(M_{21}^{JD}\right)=\frac{2EI}{L}\left(2\theta \right)\\ =\frac{2EI}{18}\left(2\times 1\right)\\ =\frac{2EI}{9}\end{array}$

Find the moment at the member 2-3 using the relation;

$\begin{array}{c}\left(M_{23}^{JD}\right)=\frac{2EI}{L}\left(2\theta \right)\\ =\frac{2EI}{24}\left(2\times 1\right)\\ =\frac{EI}{6}\end{array}$

Find the moment at the member 3-2 using the relation;

$\begin{array}{c}\left(M_{32}^{JD}\right)=\frac{2EI}{L}\left(\theta \right)\\ =\frac{2EI}{24}\left(1\right)\\ =\frac{EI}{12}\end{array}$

Find the stiffness coefficient $K_2$ using the relation;

$\begin{array}{c}{K}_2=\left(M_{21}^{JD}\right)+\left(M_{23}^{JD}\right)\\ =\left(\frac{2EI}{2}\right)+\left(\frac{EI}{6}\right)\\ =\frac{7EI}{18}\end{array}$

Find the rotation at joint 2;

Take summation moment about joint 2.

$\begin{array}{l}{\displaystyle \sum M_2}=0\\ M_{\text{clamp}}+K_2{\theta }_2=0\\ 72+\frac{7EI}{18}\times {\theta }_2=0\\ {\theta }_2=-\frac{185.14}{EI}\end{array}$

Find the moment at the member 1-2 using the relation;

$\begin{array}{c}{M}_{12}=-{\text{FEM}}_{12}+{\theta }_2{M}_{12}^{JD}\\ =-162-\frac{185.14}{EI}\times \left(\frac{EI}{9}\right)\\ =-182.57\text{kips}\cdot \text{ft}\end{array}$

Find the moment at the member 2-1 using the relation;

$\begin{array}{c}{M}_{21}={\text{FEM}}_{12}+{\theta }_2{M}_{21}^{JD}\\ =162-\frac{185.14}{EI}\times \left(\frac{2EI}{9}\right)\\ =120.85\text{kips}\cdot \text{ft}\end{array}$

Find the moment at the member 2-3 using the relation;

$\begin{array}{c}{M}_{23}=-{\text{FEM}}_{23}+{\theta }_2{M}_{23}^{JD}\\ =-90-\frac{185.14}{EI}\times \left(\frac{EI}{6}\right)\\ =-120.85\text{kips}\cdot \text{ft}\end{array}$

Find the moment at the member 3-2 using the relation;

$\begin{array}{c}{M}_{32}={\text{FEM}}_{23}+{\theta }_2{M}_{32}^{JD}\\ =90-\frac{185.14}{EI}\times \left(\frac{EI}{12}\right)\\ =74.57\text{kips}\cdot \text{ft}\end{array}$

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

Refer Figure (4),

Consider AB:

Find the vertical reaction at A;

Take summation moment about B.

$\begin{array}{l}{\displaystyle \sum M_B}=0\\ 182.57+6\times 18\times \frac{18}{2}-120.85+A_y\times 18=0\\ A_y=\frac{\text{1,033}\text{.72}}{18}\\ A_y=57.42\text{kips}\end{array}$

Thus, the vertical reaction at A is $\underset{\_}{57.42\text{kips}}$.

Find the vertical reaction at B;

Summation of forces about y axis is equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -6\times 18+A_y+B_y=0\\ -108+57.42+B_y=0\\ B_y=50.58\text{kips}\end{array}$

Consider BC:

Find the vertical reaction at B;

Take summation moment about C.

$\begin{array}{l}{\displaystyle \sum M_C}=0\\ 30\times 12+120.85-74.57+B_y\times 24=0\\ B_y=\frac{\text{406}\text{.28}}{24}\\ B_y=16.92\text{kips}\end{array}$

Find the vertical reaction at C;

Summation of forces about y axis is equal to zero.

$\begin{array}{l}{\displaystyle \sum F_y}=0\\ -30+B_y+C_y=0\\ -30+16.92+C_y=0\\ C_y=13.08\text{kips}\end{array}$

Thus, the vertical reaction at C is $\underset{\_}{13.08\text{kips}}$.

Find the total vertical reaction at B;

$\begin{array}{c}{B}_y=50.58+16.92\\ =67.5\text{kips}\end{array}$

Thus, the vertical reaction at B is $\underset{\_}{67.5\text{kips}}$.

Show the free body diagram of the beam as in Figure (5).

Refer Figure (5),

Shear force calculation:

$\begin{array}{c}{V}_A=A_y\\ =57.42\text{kips}\\ V_{BL}=A_y-6\times 18\\ =57.42-6\times 18\\ =-50.58\text{kips}\\ V_{BR}\text{=}-50.58+B_y\\ =-50.58+67.5\\ =16.92\text{kips}\\ V_C=16.92-30\\ =-13.08\text{kips}\end{array}$

Find the distance when shear force is zero (x):

Equate the equation of shear force to zero.

$\begin{array}{l}{\displaystyle \sum V_x}=0\\ 57.42-6x=0\\ x=\frac{57.42}{6}\\ x=9.57\text{ft}\end{array}$

Bending moment calculation:

$\begin{array}{c}{M}_A=-182.57\text{kips}\cdot \text{ft}\\ M_{\max }\text{=}-182.57+6\times x\times \frac{x}{2}\\ =-182.57+6\times 9.57\times \frac{9.57}{2}\\ =92.18\text{kips}\cdot \text{ft}\end{array}$

$\begin{array}{c}{M}_B=-120.85\text{kips}\cdot \text{ft}\\ M_{BC}{}_{\text{center}}\text{=}-182.57+57.42\times 30-6\times 18\times \left(\frac{18}{2}+12\right)+67.5\times 12\\ =82.03\text{kips}\cdot \text{ft}\\ M_C=-74.57\text{kips}\cdot \text{ft}\end{array}$

Thus, the bending moment at A is $\underset{\_}{-182.57\text{kips}\cdot \text{ft}}$.

Thus, the bending moment at B is $\underset{\_}{-120.85\text{kips}\cdot \text{ft}}$.

Thus, the bending moment at C is $\underset{\_}{-74.57\text{kips}\cdot \text{ft}}$.

Show the shear force and bending moment diagram as in Figure (6).