Chapter 15, Problem 1P

To determine

Find the reaction and plot the shear and bending moment diagram.

Explanation of Solution

Fixed end moment:

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 $\frac{W{L}^2}{12}$.

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the entire beam as in Figure 1.

Refer Figure 1,

Calculate the fixed end moment for AC.

$\begin{array}{c}{\text{FEM}}_{AC}=\frac{18\times 20\times {10}^2}{{\left(30\right)}^2}\\ =40\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment for CA.

$\begin{array}{c}{\text{FEM}}_{CA}=-\frac{18\times {20}^2\times 10}{{\left(30\right)}^2}\\ =-80\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment for CE.

$\begin{array}{c}{\text{FEM}}_{CE}=\frac{10\times 30}{8}\\ =37.5\text{k}\cdot \text{ft}\end{array}$

Calculate the fixed end moment for EC.

$\begin{array}{c}{\text{FEM}}_{EC}=-\frac{10\times 30}{8}\\ =-37.5\text{k}\cdot \text{ft}\end{array}$

Calculate the slope deflection equation for the member AC.

$M_{AC}=\frac{2EI}{L}\left(2{\theta }_A+{\theta }_C-3\psi \right){\text{+FEM}}_{AC}$

Here, $\psi$ is the chord rotation, ${\theta }_A$ is the slope at the point A, and ${\theta }_C$ is the slope at the point C.

Substitute 0 for $\psi$, 0 for ${\theta }_A$, 30ft for L and $40\text{k}\cdot \text{ft}$ for ${\text{FEM}}_{AC}$.

$\begin{array}{c}{M}_{AC}=\frac{2EI}{30}\left(2\left(0\right)+{\theta }_C-3\left(0\right)\right)\text{+}40\\ =0.0667EI{\theta }_C+40\end{array}$        (1)

Calculate the slope deflection equation for the member CA.

$M_{CA}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_A-3\psi \right){\text{+FEM}}_{CA}$

Substitute 0 for $\psi$, 0 for ${\theta }_A$, 30ft for L and $-80\text{k}\cdot \text{ft}$ for ${\text{FEM}}_{CA}$.

$\begin{array}{c}{M}_{CA}=\frac{2EI}{30}\left(2{\theta }_C+0-3\left(0\right)\right)-80\\ =0.133EI{\theta }_C-80\end{array}$        (2)

Calculate the slope deflection equation for the member CE.

$M_{CE}=\frac{2EI}{L}\left(2{\theta }_C+{\theta }_E-3\psi \right){\text{+FEM}}_{CE}$

Substitute 0 for $\psi$, 0 for ${\theta }_E$, 30ft for L, and $37.5\text{k}\cdot \text{ft}$ for ${\text{FEM}}_{CE}$.

$\begin{array}{c}{M}_{CE}=\frac{2EI}{30}\left(2{\theta }_C+0-3\left(0\right)\right)+37.5\\ =0.133EI{\theta }_C+37.5\end{array}$        (3)

Calculate the slope deflection equation for the member EC.

$M_{EC}=\frac{2EI}{L}\left(2{\theta }_E+{\theta }_C-3\psi \right){\text{+FEM}}_{EC}$

Substitute 0 for $\psi$, 0 for ${\theta }_E$, 30ft for L and $-37.5\text{k}\cdot \text{ft}$ for ${\text{FEM}}_{EC}$.

$\begin{array}{c}{M}_{EC}=\frac{2EI}{30}\left(2\left(0\right)+{\theta }_C-3\left(0\right)\right)-37.5\\ =0.0667EI{\theta }_C-37.5\end{array}$        (4)

Write the equilibrium equation as below.

$M_{CA}+M_{CE}=0$

Substitute equation (2) and equation (3) in above equation.

$\begin{array}{l}0.133EI{\theta }_C-80+0.133EI{\theta }_C+37.5=0\\ 0.267EI{\theta }_C-42.5=0\\ {\theta }_C=\frac{42.5}{0.267EI}\\ {\theta }_C=\frac{159.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}\end{array}$

Calculate the moment about AC.

Substitute $\frac{159.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$ in equation (1).

$\begin{array}{c}{M}_{AC}=0.0667EI\left(\frac{159.2}{EI}\right)+40\\ =50.625\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about CA.

Substitute $\frac{159.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$ in equation (2).

$\begin{array}{c}{M}_{CA}=0.133EI\left(\frac{159.2}{EI}\right)-80\\ =-58.8\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about CE.

Substitute $\frac{159.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$ in equation (3).

$\begin{array}{c}{M}_{CE}=0.133EI\left(\frac{159.2}{EI}\right)+37.5\\ =58.7\text{k}\cdot \text{ft}\end{array}$

Calculate the moment about EC.

Substitute $\frac{159.2}{EI}\text{k}\cdot {\text{ft}}^{\text{2}}$ for ${\theta }_C$ in equation (4).

$\begin{array}{c}{M}_{EC}=0.0667EI\left(\frac{159.2}{EI}\right)-37.5\\ =-26.9\text{k}\cdot \text{ft}\end{array}$

Consider the member AC of the beam:

Show the free body diagram of the member AC as in Figure 2.

Calculate the vertical reaction at the left end of the joint C by taking moment about point A.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_A=0}\\ C_{y,L}\left(30\right)-18\times \left(20\right)+50.625-58.8=0\\ C_{y,L}\left(30\right)=368.175\\ C_{y,L}=\frac{368.175}{30}\\ C_{y,L}=12.27\text{k}\end{array}$

Calculate the horizontal reaction at point A by resolving the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ A_x=0\end{array}$

Calculate the vertical reaction at point A by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ A_y-18+C_{y,L}=0\\ A_y-18+12.27=0\\ A_y=5.73\text{k}\uparrow \end{array}$

Consider the member CE of the beam:

Show the free body diagram of the member CE as in Figure 2.

Calculate the vertical reaction at the right end of the joint C by taking moment about point E.

$\begin{array}{l}+\circlearrowleft {\displaystyle \sum M_E=0}\\ C_{y,R}\left(30\right)+10\times \left(15\right)-26.9+58.7=0\\ C_{y,R}\left(30\right)=181.8\\ C_{y,R}=\frac{181.8}{30}\\ C_{y,R}=6.06\text{k}\uparrow \end{array}$

Calculate the horizontal reaction at point E by resolving the horizontal equilibrium.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ E_x=0\end{array}$

Calculate the vertical reaction at point E by resolving the vertical equilibrium.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ E_y-10+C_{y,R}=0\\ E_y-10+6.06=0\\ E_y=3.94\text{k}\uparrow \end{array}$

Calculate the total reaction at point C .

$C_y=C_{y,R}+C_{y,L}$

Substitute $6.06\text{k}$ for $C_{y,R}$ and $12.27\text{k}$ for $C_{y,L}$.

$\begin{array}{c}{C}_y=6.06+12.27\\ =18.33\text{k}\uparrow \end{array}$

Show the reaction of the beam in Figure 4.

Refer Figure 4,

Shear diagram:

Point A:

$\begin{array}{l}{S}_{A,L}=0\\ S_{A,R}=5.73\text{k}\end{array}$

Point B:

$\begin{array}{c}{S}_B=5.73-18\\ =-12.27\text{k}\end{array}$

Point C:

$\begin{array}{c}{S}_{C,L}=-12.27\text{k}\\ S_{C,R}=-12.27+18.33\\ =6.06\text{k}\end{array}$

Point D:

$\begin{array}{c}{S}_{D,L}=6.06\text{k}\\ S_{D,R}=6.06-10\\ =-3.94\text{k}\end{array}$

Point E:

$\begin{array}{c}{S}_{E,L}=-3.94\text{k}\\ S_{E,R}=-3.94\text{k+3}\text{.94}\\ =0\text{k}\end{array}$

Plot the shear force diagram of the beam as in Figure 5.

Refer Figure 4,

Bending moment diagram:

Point A:

$M_A=-50.625\text{k}\cdot \text{ft}$

Point B:

$\begin{array}{c}{M}_B=-50.625+\left(5.73\times 20\right)\\ =64\text{k}\cdot \text{ft}\end{array}$

Point C:

$M_C=-58.7\text{k}\cdot \text{ft}$

Point D:

$\begin{array}{c}{M}_D=-58.7+\left(6.06\times 15\right)\\ =32.2\text{k}\end{array}$

Point E:

$M_E=-26.9\text{k}\cdot \text{ft}$

Plot the bending moment diagram of the beam as in Figure 6.