Chapter 13, Problem 1P

To determine

Find the moment at the support B of the beam using approximate analysis.

Sketch the shear and moment diagrams of the beam.

Answer to Problem 1P

The moment at the support B of the beam using approximate analysis is $\underset{\_}{-360\text{kip}\cdot \text{ft}}$.

Explanation of Solution

Given information:

The value of EI is constant.

Apply the sign conventions for calculating reactions, forces and moments 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 negative and the counter clockwise moment as positive.

Calculation:

Let consider the location of a point of inflection at $0.25L$ to the left of support B which means 6 ft to the left of support B or 18 ft from the support A to the right.

Show the location of a point of inflection as in Figure 1.

Find the reaction $R_A$ at support A:

Summation of moments about P.I. is equal to 0.

$\begin{array}{l}\sum M_{P.I}=0\\ R_A\left(18\right)-5\left(18\right)\left(\frac{18}{2}\right)=0\\ R_A=45\text{kips}\uparrow \end{array}$

Let $R_B$ and $R_C$ be the reactions at the supports B and C.

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_C=0\\ R_A\left(44\right)+R_B\left(20\right)-5\left(44\right)\left(\frac{44}{2}\right)=0\\ 45\left(44\right)+R_B\left(20\right)-5\left(44\right)\left(\frac{44}{2}\right)=0\\ R_B=143\text{kips}\uparrow \end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_A+R_B+R_C-5\left(44\right)=0\\ 45+143+R_C-5\left(44\right)=0\\ R_C=32\text{kips}\uparrow \end{array}$

Find the shear values of the beam:

$F_A=45\text{kips}$

$\begin{array}{c}{F}_{B,JL}=45-5\left(24\right)\\ =-75\text{kips}\end{array}$

$\begin{array}{c}{F}_{B,JR}=45-5\left(24\right)+143\\ =68\text{kips}\end{array}$

$F_C=-32\text{kips}$

Find the point of zero shear force for span AB,

$\begin{array}{l}\sum F_{xx}=0\\ 45-5x=0\\ x=9\text{ft}\text{from}\text{point}\text{A}\end{array}$

For span BC,

$\begin{array}{l}\sum F_{xx}=0\\ 32-5x=0\\ x=6.4\text{ft}\text{from}\text{point}\text{C}\end{array}$

Find the moment values of the beam:

$M_A=0$

$\begin{array}{c}{M}_B=45\left(24\right)-5\left(24\right)\left(\frac{24}{2}\right)\\ =-360\text{kip}\cdot \text{ft}\end{array}$

$M_C=0$

$\begin{array}{c}{M}_{x=9\text{ft}\text{from}\text{point}\text{A}}=45\left(9\right)-5\left(9\right)\left(\frac{9}{2}\right)\\ =202.5\text{kip}\cdot \text{ft}\end{array}$

$\begin{array}{c}{M}_{x=6.4\text{ft}\text{from}\text{point}\text{C}}=32\left(6.4\right)-5\left(6.4\right)\left(\frac{6.4}{2}\right)\\ =102.4\text{kip}\cdot \text{ft}\end{array}$

Sketch the shear and moment diagram of the beam using approximate analysis as shown in Figure 2.

Check the shear and moment values using moment distribution method:

Find the fixed end moments in the span AB and BC:

$\begin{array}{c}{\text{FEM}}_{AB}=-\frac{w{L}^2}{12}\\ =-\frac{5{\left(24\right)}^2}{12}\\ =-240\text{kip}\cdot \text{ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BA}=-{\text{FEM}}_{AB}\\ =240\text{kip}\cdot \text{ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{BC}=-\frac{w{L}^2}{12}\\ =-\frac{5{\left(20\right)}^2}{12}\\ =-166.7\text{kip}\cdot \text{ft}\end{array}$

$\begin{array}{c}{\text{FEM}}_{CB}=-{\text{FEM}}_{BC}\\ =166.7\text{kip}\cdot \text{ft}\end{array}$

Find the distribution factor at joint B:

$\begin{array}{c}D{F}_{BA}=\frac{20}{44}\\ =\frac{5}{11}\end{array}$

$\begin{array}{c}D{F}_{BC}=\frac{24}{44}\\ =\frac{6}{11}\end{array}$

Show the moment distribution table as in Figure 3.

Consider span AB, find the reactions at the supports:

Summation of moments about B is equal to 0.

$\begin{array}{l}\sum M_B=0\\ R_A\left(24\right)-5\left(24\right)\left(\frac{24}{2}\right)+310=0\\ R_A=47.1\text{kips}\uparrow \end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_A+R_B-5\left(24\right)=0\\ 47.1+R_B-5\left(24\right)=0\\ R_B=72.9\text{kips}\uparrow \end{array}$

Consider span BC, find the reactions at the supports:

Summation of moments about C is equal to 0.

$\begin{array}{l}\sum M_B=0\\ R_C\left(20\right)-5\left(20\right)\left(\frac{20}{2}\right)+310=0\\ R_C=34.5\text{kips}\uparrow \end{array}$

Summation of forces along y-direction is equal to 0.

$\begin{array}{l}+\uparrow \sum F_y=0\\ R_B+R_C-5\left(20\right)=0\\ R_B+34.5-5\left(20\right)=0\\ R_B=65.5\text{kips}\uparrow \end{array}$

Therefore, the total $R_B$ is $72.9+65.5=138.4\text{kips}$

Find the shear values of the beam:

$F_A=47.1\text{kips}$

$\begin{array}{c}{F}_{B,JL}=47.1-5\left(24\right)\\ =-72.9\text{kips}\end{array}$

$\begin{array}{c}{F}_{B,JR}=47.1-5\left(24\right)+138.4\\ =65.5\text{kips}\end{array}$

$F_C=-34.5\text{kips}$

Find the point of zero shear force for span AB,

$\begin{array}{l}\sum F_{xx}=0\\ 47.1-5x=0\\ x=9.42\text{ft}\text{from}\text{point}\text{A}\end{array}$

For span BC,

$\begin{array}{l}\sum F_{xx}=0\\ 34.5-5x=0\\ x=6.9\text{ft}\text{from}\text{point}\text{C}\end{array}$

Find the moment values of the beam:

$M_A=0$

$\begin{array}{c}{M}_B=47.1\left(24\right)-5\left(24\right)\left(\frac{24}{2}\right)\\ =-310\text{kip}\cdot \text{ft}\end{array}$

$M_C=0$

$\begin{array}{c}{M}_{x=9.42\text{ft}\text{from}\text{point}\text{A}}=47.1\left(9.42\right)-5\left(9.42\right)\left(\frac{9.42}{2}\right)\\ =221.8\text{kip}\cdot \text{ft}\end{array}$

$\begin{array}{c}{M}_{x=6.9\text{ft}\text{from}\text{point}\text{C}}=34.5\left(6.9\right)-5\left(6.9\right)\left(\frac{6.9}{2}\right)\\ =119.025\text{kip}\cdot \text{ft}\end{array}$

Sketch the moment distribution results, shear, and moment diagrams of the beam using moment distribution method as shown in Figure 3.

Therefore, the moment at the support B of the beam using approximate analysis is $\underset{\_}{-360\text{kip}\cdot \text{ft}}$.