Chapter 6, Problem 33P

To determine

Find the slope ${\theta }_B\&{\theta }_D$ and deflection ${\Delta }_B\&{\Delta }_D$ at point B and D of the given beam using the moment-area method.

Answer to Problem 33P

The slope ${\theta }_B$  at point B of the given beam using the moment-area method is $\underset{\_}{0.0099\text{rad}\left(\text{Clockwise}\right)}$.

The deflection ${\Delta }_B$ at point B of the given beam using the moment-area method is $\underset{\_}{0.86\text{in}\text{.}(\downarrow )}$.

The slope ${\theta }_D$ at point D of the given beam using the moment-area method is $\underset{\_}{0.0084\text{rad}\left(\text{Counterclockwise}\right)}$.

The deflection ${\Delta }_D$ at point D of the given beam using the moment-area method is $\underset{\_}{1.44\text{in}\text{.}(\downarrow )}$.

Explanation of Solution

Given information:

The Young’s modulus (E) is 29,000 ksi.

The moment of inertia (I) is $6,000\text{in}{\text{.}}^{\text{4}}$.

Calculation:

Consider flexural rigidity EI of the beam is constant.

Show the free body diagram of the given beam as in Figure (1).

Refer Figure (1),

Consider upward is force is positive and downward force is negative.

Consider clockwise moment is negative and counterclockwise moment is positive.

Split the given beam into two sections such as AC and CE.

Consider the portion CE:

Draw the free body diagram of the portion CE as in Figure (2).

Refer Figure (2),

Consider a reaction at C and take moment about point C.

Determine the reaction at E;

$\begin{array}{l}{R}_E\times \left(24\right)-\left(40\times 12\right)+250=0\\ R_E=\frac{230}{24}\\ R_E=9.58\text{kips}\end{array}$

Determine the reaction at support A;

$\begin{array}{l}{\displaystyle \sum V=0}\\ R_A+R_E-75-40=0\\ R_A=115-9.58\\ R_A=105.42\text{kips}\end{array}$

Show the reaction of the given beam as in Figure (3).

Refer Figure (3),

Determine the moment at A:

$\begin{array}{c}{M}_A=\left(9.58\times 48\right)+250-\left(40\times 36\right)-\left(75\times 12\right)\\ =709.84-2,340\\ \simeq -1,630\text{kips-ft}\end{array}$

Determine the bending moment at B;

$\begin{array}{c}{M}_B=\left(9.58\times 36\right)+250-\left(40\times 24\right)\\ =344.88+250-960\\ \simeq -365\text{kips-ft}\end{array}$

Determine the bending moment at C;

$\begin{array}{c}{M}_C=\left(9.58\times 24\right)+250-\left(40\times 12\right)\\ =479.92-480\\ \simeq 0\end{array}$

Determine the bending moment at D;

$\begin{array}{c}{M}_D=\left(9.58\times 12\right)+250\\ \simeq 365\text{kips-ft}\end{array}$

Determine the bending moment at E;

$M_E=250\text{kips-ft}$

Show the $M/EI$ diagram for the given beam as in Figure (4).

Show the elastic curve diagram as in Figure (5).

Refer Figure (4),

Determine the slope at B;

$\begin{array}{c}{\theta }_B={\theta }_{BA}=\text{Area}\text{of}\text{the}\frac{M}{EI}\text{diagram}\text{between}A\text{and}B\\ =-\left(\frac{1}{2}\times b_1\times h_1+\frac{1}{2}\times b_2\times h_2\right)\end{array}$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b_1$, $\frac{1,630}{EI}$ for $h_1$, 12 ft for $b_2$, and $\frac{365}{EI}$ for $h_2$.

$\begin{array}{c}{\theta }_B={\theta }_{BA}=-\left[\left(\frac{1}{2}\times 12\times \frac{1,630}{EI}\right)+\left(\frac{1}{2}\times 12\times \frac{365}{EI}\right)\right]\\ =-\frac{1}{EI}\left(\frac{1,630+365}{2}\right)12\\ =-\frac{11,970{\text{kips-ft}}^{\text{2}}}{EI}\end{array}$

Substitute 29,000 ksi for E and $6,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\theta }_B=-\frac{11,970{\text{kips-ft}}^{\text{2}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^{\text{2}}}{29,000\times 6,000}\\ =-0.0099\text{rad}\\ =\text{0}\text{.0099}\left(\text{Clockwise}\right)\end{array}$

Hence, the slope at B is $\underset{\_}{0.0099\text{rad}\left(\text{Clockwise}\right)}$.

Determine the deflection between A and B using the relation;

$\begin{array}{c}{\Delta }_B={\Delta }_{BA}=\text{Moment}\text{of}\text{the}\text{area}\text{of}\text{the}M/EI\text{diagram}\text{between}B\text{and}A\text{about}A\\ =-\left[\left(b_1{h}_1\right)\left(\frac{b_1}{2}\right)+\left(\frac{1}{2}{b}_2{h}_2\right)\left(\frac{2}{3}\times b_2\right)\right]\end{array}$

Here, $b_1$ is the width of rectangle, $h_1$ is the height of the rectangle, $b_2$ is the width of the triangle, and $h_2$ is the height of the triangle.

Substitute 12 ft for $b_1$, $\frac{365}{EI}$ for $h_1$, 12 ft for $b_2$, and $\left(\frac{1,630}{EI}-\frac{365}{EI}\right)$ for $h_2$.

$\begin{array}{c}{\Delta }_B={\Delta }_{BA}=-\left[\left(12\times \frac{365}{EI}\right)\left(\frac{12}{2}\right)+\frac{1}{2}\times 12\times \left(\frac{1,630}{EI}-\frac{365}{EI}\right)\left(\frac{2}{3}\times 12\right)\right]\\ =-\frac{87,000\text{kips}\cdot {\text{ft}}^{\text{3}}}{EI}\end{array}$

Substitute 29,000 ksi for E and $6,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\Delta }_B={\Delta }_{BA}=-\frac{87,000\text{kips}\cdot {\text{ft}}^{\text{3}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^3}{29,000\times 6,000}\\ =-0.864\text{in}\text{.}\\ \simeq 0.86\text{in}\text{.}(\downarrow )\end{array}$

Hence, the deflection at B is $\underset{\_}{0.86\text{in}\text{.}(\downarrow )}$.

To determine the slope at point E, it is necessary to determine the deflection at point C and the deflection between C and E.

Determine the deflection at C and A using the relation;

$\begin{array}{c}{\Delta }_C={\Delta }_{CA}=\text{Moment}\text{of}\text{the}\text{area}\text{of}\text{the}M/EI\text{diagram}\text{between}C\text{and}A\text{about}A\\ =-\left[\left(b_1{h}_1\right)\left(12+\frac{b_1}{2}\right)+\left(\frac{1}{2}{b}_2{h}_2\right)\left(12+\frac{2}{3}\times b_2\right)+\left(\frac{1}{2}{b}_3{h}_3\right)\left(\frac{2}{3}\times b_3\right)\right]\end{array}$

Substitute 12 ft for $b_1$, $\frac{365}{EI}$ for $h_1$, 12 ft for $b_2$, $\left(\frac{1,630}{EI}-\frac{365}{EI}\right)$ for $h_2$, 12 for $b_3$, and $\frac{365}{EI}$ for $h_3$.

$\begin{array}{c}{\Delta }_C={\Delta }_{CA}=-\left[\begin{array}{l}\left(12\times \frac{365}{EI}\right)\left(12+\frac{12}{2}\right)+\left(\frac{1}{2}\times 12\times \left(\frac{1,630}{EI}-\frac{365}{EI}\right)\right)\left(12+\frac{2}{3}\times 12\right)\\ +\left(\frac{1}{2}\times 12\times \frac{365}{EI}\right)\left(\frac{2}{3}\times 12\right)\end{array}\right]\\ =-\frac{248,160{\text{kips-ft}}^{\text{3}}}{EI}\end{array}$

Determine the deflection between C and E using the relation;

$\begin{array}{c}{\Delta }_{CE}=\text{Moment}\text{of}\text{the}\text{area}\text{of}\text{the}M/EI\text{diagram}\text{between}C\text{and}E\text{about}E\\ =-\left[\left(\frac{1}{2}{b}_1{h}_1\right)\left(\frac{2}{3}\times b_1\right)+\left(b_2{h}_2\right)\left(12+\frac{b_2}{2}\right)+\left(\frac{1}{2}{b}_3{h}_3\right)\left(12+\frac{1}{3}\times b_3\right)\right]\end{array}$

Substitute 12 ft for $b_1$, $\frac{365}{EI}$ for $h_1$, 12 ft for $b_2$, $\frac{250}{EI}$ for $h_2$, 12 ft for $b_3$, and $\left(\frac{365}{EI}-\frac{250}{EI}\right)$ for $h_3$.

$\begin{array}{c}{\Delta }_{CE}=-\left[\begin{array}{l}\left(\frac{1}{2}\times 12\times \frac{365}{EI}\right)\left(\frac{2}{3}\times 12\right)+\left(12\times \frac{250}{EI}\right)\left(12+\frac{12}{2}\right)+\\ \left(\frac{1}{2}\times 12\times \left(\frac{365}{EI}-\frac{250}{EI}\right)\right)\left(12+\frac{1}{3}\times 12\right)\end{array}\right]\\ =-\frac{82,560{\text{kips-ft}}^{\text{3}}}{EI}\end{array}$

Determine the slope at E using the relation;

${\theta }_E=\frac{{\Delta }_C+{\Delta }_{CE}}{L_{CE}}$

Here, $L_{CE}$ is the length between C and E.

Substitute $-\frac{248,160{\text{kips-ft}}^{\text{3}}}{EI}$ for ${\Delta }_C$, $-\frac{82,560{\text{kips-ft}}^{\text{3}}}{EI}$ for ${\Delta }_{CE}$, and 24 ft for $L_{CE}$.

$\begin{array}{c}{\theta }_E=\frac{-\frac{248,160{\text{kips-ft}}^{\text{3}}}{EI}+\left(-\frac{82,560{\text{kips-ft}}^{\text{3}}}{EI}\right)}{24}\\ =-\frac{13,780{\text{kips-ft}}^2}{EI}\end{array}$

Determine the slope between D and E using the relation;

$\begin{array}{c}{\theta }_{DE}=\text{Area}\text{of}\text{the}\frac{M}{EI}\text{diagram}\text{between}D\text{and}E\\ =\left(\frac{1}{2}\times b_1\times h_1+\frac{1}{2}\times b_2\times h_2\right)\end{array}$

Here, b is the width and h is the height of respective triangle.

Substitute 12 ft for $b_1$, $\frac{365}{EI}$ for $h_1$, 12 ft for $b_2$, and $\frac{250}{EI}$ for $h_2$.

$\begin{array}{c}{\theta }_{DE}=-\left[\left(\frac{1}{2}\times 12\times \frac{365}{EI}\right)+\left(\frac{1}{2}\times 12\times \frac{250}{EI}\right)\right]\\ =-\frac{1}{EI}\left(\frac{365+250}{2}\right)12\\ =-\frac{3,690{\text{kips-ft}}^{\text{2}}}{EI}\end{array}$

Determine the slope at D using the relation;

${\theta }_D={\theta }_E-{\theta }_{DE}$

Substitute $-\frac{3,690{\text{kips-ft}}^{\text{2}}}{EI}$ for ${\theta }_E$ and $-\frac{3,690{\text{kips-ft}}^{\text{2}}}{EI}$ for ${\theta }_{DE}$.

$\begin{array}{c}{\theta }_D=-\frac{13,780{\text{kips-ft}}^2}{EI}-\left(-\frac{3,690{\text{kips-ft}}^{\text{2}}}{EI}\right)\\ =-\frac{10,090{\text{kips-ft}}^{\text{2}}}{EI}\end{array}$

Substitute 29,000 ksi for E and $6,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\theta }_D=-\frac{10,090{\text{kips-ft}}^{\text{2}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^{\text{2}}}{29,000\times 6,000}\\ =-0.00835\text{rad}\\ \simeq \text{0}\text{.0084}\text{rad}\left(\text{Counterclockwise}\right)\end{array}$

Hence, the slope at D is $\underset{\_}{\text{0}\text{.0084}\text{rad}\left(\text{Counterclockwise}\right)}$.

Determine the deflection between D and E using the relation;

$\begin{array}{c}{\Delta }_{DE}=\text{Moment}\text{of}\text{the}\text{area}\text{of}\text{the}M/EI\text{diagram}\text{between}D\text{and}E\text{about}E\\ =-\left[\left(b_1{h}_1\right)\left(\frac{b_1}{2}\right)+\left(\frac{1}{2}{b}_2{h}_2\right)\left(\frac{1}{3}\times b_2\right)\right]\end{array}$

Here, b is the width and h is the height of respective rectangle and triangle.

Substitute 12 ft for $b_1$, $\frac{250}{EI}$ for $h_1$, 12 for $b_2$, and $\left(\frac{365}{EI}-\frac{250}{EI}\right)$ for $h_2$.

$\begin{array}{c}{\Delta }_{DE}=-\left[\left(12\times \frac{250}{EI}\right)\left(\frac{12}{2}\right)+\left(\frac{1}{2}\times 12\times \left(\frac{365}{EI}-\frac{250}{EI}\right)\right)\left(\frac{1}{3}\times 12\right)\right]\\ =-\frac{20,760{\text{kips-ft}}^{\text{3}}}{EI}\end{array}$

Determine the deflection at D using the relation;

${\Delta }_D=L_{DE}{\theta }_E-{\Delta }_{DE}$

Substitute 12 ft for $L_{DE}$, $-\frac{13,780{\text{kips-ft}}^2}{EI}$ for ${\theta }_E$, and $-\frac{20,760{\text{kips-ft}}^{\text{3}}}{EI}$ for ${\Delta }_{DE}$.

$\begin{array}{c}{\Delta }_D=12\left(-\frac{13,780{\text{kips-ft}}^2}{EI}\right)-\left(-\frac{20,760{\text{kips-ft}}^{\text{3}}}{EI}\right)\\ =\frac{-165,360+20,760}{EI}\\ =-\frac{144,600{\text{kips-ft}}^{\text{3}}}{EI}\end{array}$

Substitute 29,000 ksi for E and $6,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\Delta }_D=-\frac{144,600{\text{kips-ft}}^{\text{3}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^{\text{3}}}{29,000\times 6,000}\\ =-1.44\text{in}\text{.}\\ \text{=1}\text{.44}\text{in}\text{.}(\downarrow )\end{array}$

Hence, the deflection at point D is $\underset{\_}{1.44\text{in}\text{.}(\downarrow )}$.