Chapter 6, Problem 34P

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 34P

The slope ${\theta }_B$ at point B (left) of the given beam using the moment area method is $\underset{\_}{0.0033\text{rad}}$.

The deflection ${\Delta }_B$ at point B (left) of the given beam using the moment area method is $\underset{\_}{0.39\text{in}\text{.}(\uparrow )}$.

The slope ${\theta }_B$ at point B (right) of the given beam using the moment area method is $\underset{\_}{-0.0035\text{rad}}$.

The slope ${\theta }_D$ at point D of the given beam using the moment area method is $\underset{\_}{0.0071\text{rad}}$.

The deflection ${\Delta }_D$ at point D of the given beam using the moment area method is $\underset{\_}{0.62\text{in}\text{.}(\downarrow )}$.

Explanation of Solution

Given information:

The Young’s modulus (E) is 30,000 ksi.

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

The moment of inertia of the section BD is (I) is $3,000\text{in}{\text{.}}^{\text{4}}$.

Calculation:

Consider Young’s modulus (E) of the beam is constant.

To draw a $M/EI$ diagram, the reactions are calculated by considering all the loads acting in the beam. The calculation of positive moment at point A is calculated by simply considering the point load of 35 kips and the reactions calculated by considering all the loads acting in the beam.

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

Refer Figure (1),

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclowise is positive.

Refer Figure (1),

Consider reaction at A and C as $R_A$ and $R_C$.

Take moment about point B.

Determine the reaction at C;

$\begin{array}{l}{R}_C\times \left(8\right)-\left(35\times 16\right)=0\\ R_C=\frac{560}{8}\\ R_C=70\text{kips}\end{array}$

Determine the reaction at support A;

$\begin{array}{l}{\displaystyle \sum V=0}\\ R_A+R_C-\left(2.5\times 16\right)-35=0\\ R_A=75-70\\ R_A=5\text{kips}\end{array}$

Determine the moment at A:

$\begin{array}{c}{M}_A=-\left(35\times 32\right)+\left(70\times 24\right)-\left(25\times 16\times \frac{16}{2}\right)\\ =-240\text{kips-ft}\\ =240\text{kips-ft}\left(\text{Clockwise}\right)\end{array}$

Show the reactions of the given beam as in Figure (2).

Determine the bending moment at B;

$\begin{array}{c}{M}_B=70\times 8-35\times 16\\ =560-560\\ =0\end{array}$

Determine the bending moment at C;

$\begin{array}{c}{M}_C=-\left(35\times 8\right)\\ =-280\text{kips-ft}\end{array}$

Determine the bending moment at D;

$\begin{array}{c}{M}_D=-\left(5\times 32\right)-\left(70\times 8\right)-240+\left(2.5\times 16\times \left(\frac{16}{2}+16\right)\right)\\ =-720+240+960\\ =0\end{array}$

Determine the positive bending moment at A;

$\begin{array}{c}{M}_A=-\left(35\times 32\right)+70\times 24\\ =-1,120+1,680\\ =560\text{kips-ft}\end{array}$

Show the reaction and point load of the beam as in Figure (3).

Determine the value of $M/EI$;

$\frac{M}{EI}=\frac{560\text{kips-ft}}{EI}$

Substitute $\frac{4I}{3}$ for I.

$\begin{array}{c}\frac{M}{EI}=\frac{560\text{kips-ft}}{E\left(\frac{4I}{3}\right)}\\ =\frac{1,680\text{kips-ft}}{4EI}\\ =\frac{420\text{kips-ft}}{EI}\end{array}$

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

Show the elastic curve as in Figure (5).

The slope at point B can be calculated by evaluating the change in slope between A and B.

Express the change in slope using the first moment-area theorem as follows:

$\begin{array}{c}{\theta }_{B\left(\text{left}\right)}={\theta }_{AB}=\text{Area}\text{of}\text{the}M/EI\text{between}A\text{and}B\\ =\text{Area}\text{of}\text{triangle}+\text{Area}\text{of}\text{parabola}\\ =\left[\frac{1}{2}\times b_1\times h_1-\frac{1}{3}\times b_2\times h_2\right]\end{array}$

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

Substitute 16 ft for $b_1$, $\frac{420}{EI}$ for $h_1$, 16 ft for $b_2$, and $-\frac{240}{EI}$ for $h_2$.

$\begin{array}{c}{\theta }_{B\left(\text{left}\right)}=\left[\frac{1}{2}\times 16\times \frac{420}{EI}-\frac{1}{3}\times 16\times \frac{240}{EI}\right]\\ =\frac{2,080{\text{kips-ft}}^{\text{2}}}{EI}\end{array}$

Determine the slope B (left) using the relation;

${\theta }_{B\left(\text{left}\right)}=\frac{2,080{\text{kips-ft}}^{\text{2}}}{EI}$

Substitute 30,000 ksi for E and $4,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\theta }_{B\left(\text{left}\right)}=\frac{2,080{\text{kips-ft}}^{\text{2}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}{30,000\times 3,000}\\ =0.0033\text{rad}\end{array}$

Hence, the slope at B (left) is $\underset{\_}{0.0033\text{rad}}$.

Determine the deflection at B using the relation;

$\begin{array}{c}{\Delta }_B={\Delta }_{BA}=\text{Moment}\text{of}\text{the}\text{area}\text{of}\text{the}M/EI\text{diagram}\text{between}B\text{and}A\text{about}A\\ =\left[\frac{M}{EI}\left(\text{Area}\text{of}\text{triangle}\right)+\frac{M}{EI}\left(\text{Area}\text{of}\text{parabola}\right)\right]\end{array}$

Substitute $\frac{420}{EI}$ for $\frac{M}{EI}$, $\left(\frac{1}{2}\times 16\times \left(\frac{2}{3}\times 16\right)\right)$ for area of triangle, $\left(-\frac{240}{EI}\right)$ for $\frac{M}{EI}$, and $\left(\frac{1}{3}\times 16\times \left(\frac{3}{4}\times 16\right)\right)$ for area of triangle.

$\begin{array}{c}{\Delta }_{BA}=\left[\frac{420}{EI}\left(\frac{1}{2}\times 16\times \left(\frac{2}{3}\times 16\right)\right)+\left(-\frac{240}{EI}\right)\left(\frac{1}{3}\times 16\times \left(\frac{3}{4}\times 16\right)\right)\right]\\ =\frac{1}{EI}\left(35,840-15,360\right)\\ =\frac{20,480{\text{kips-ft}}^3}{EI}\end{array}$

Determine the deflection at B (left) using the relation;

${\Delta }_{BA}=\frac{20,480{\text{kips-ft}}^3}{EI}$

Substitute 30,000 ksi for E and $3,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\Delta }_{BA}=\frac{20,480{\text{kips-ft}}^3\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^3}{30,0000\times 3,000}\\ =0.39\text{in}\text{.}(\uparrow )\end{array}$

Hence, the deflection at B (left) is $\underset{\_}{0.39\text{in}\text{.}(\uparrow )}$.

Determine the deflection between B and C using the relation;

$\begin{array}{c}{\Delta }_{BC}=\text{Moment}\text{of}\text{the}\text{area}\text{of}\text{the}M/EI\text{diagram}\text{between}B\text{and}C\text{about}C\\ =\left[\frac{M}{EI}\left(\text{Area}\text{of}\text{triangle}\right)\times \left(\frac{2}{3}\times b\right)\right]\\ =\left[\frac{1}{2}\times b\times h\times \left(\frac{2}{3}\times b\right)\right]\end{array}$

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

Substitute 8 ft for b and $\frac{280}{EI}$ for h.

$\begin{array}{c}{\Delta }_{BC}=\left[\frac{1}{2}\times 8\times \left(\frac{280}{EI}\right)\times \left(\frac{2}{3}\times 8\right)\right]\\ =\frac{5,973.33{\text{kips-ft}}^{\text{3}}}{EI}\end{array}$

Express the relationship between the deflection and slope of span BC as follows:

$\begin{array}{l}{\Delta }_{BC}+{\Delta }_B=L_{BC}{\theta }_C\\ {\theta }_C=\frac{{\Delta }_{BC}+{\Delta }_B}{L_{BC}}\end{array}$

Here, $L_{BC}$ is the length between point B and C.

Substitute $\frac{5,973.33{\text{kips-ft}}^{\text{3}}}{EI}$ for ${\Delta }_{BC}$, $\frac{20,480{\text{kips-ft}}^3}{EI}$ for ${\Delta }_B$, and 8 ft for $L_{BC}$.

$\begin{array}{c}{\theta }_C=\frac{\frac{5,973.33}{EI}+\frac{20,480}{EI}}{8}\\ =\frac{3,306.67{\text{kips-ft}}^2}{EI}\end{array}$

Determine the slope between B and C using the relation;

$\begin{array}{c}{\theta }_{BC}=\text{Area}\text{of}\text{the}M/EI\text{between}B\text{and}C\\ =\text{Area}\text{of}\text{triangle}\\ =\left[\frac{1}{2}\times b\times h\right]\end{array}$

Substitute 8 ft for b and $\frac{280}{EI}$ for h.

$\begin{array}{c}{\theta }_{BC}=\left[\frac{1}{2}\times 8\times \frac{280}{EI}\right]\\ =\frac{1,120{\text{kips-ft}}^{\text{2}}}{EI}\end{array}$

Determine the slope at B (right) using the relation;

${\theta }_{B\left(\text{right}\right)}={\theta }_C-{\theta }_{BC}$

Substitute $\frac{3,306.67{\text{kips-ft}}^2}{EI}$ for ${\theta }_C$ and $\frac{1,120{\text{kips-ft}}^{\text{2}}}{EI}$ for ${\theta }_{BC}$.

$\begin{array}{c}{\theta }_{B\left(\text{right}\right)}=\frac{3,306.67{\text{kips-ft}}^2}{EI}-\frac{1,120{\text{kips-ft}}^{\text{2}}}{EI}\\ =\frac{2,186.67{\text{kips-ft}}^{\text{2}}}{EI}\end{array}$

Substitute 30,000 ksi for E and $3,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\theta }_{B\left(\text{right}\right)}=\frac{2,186.67{\text{kips-ft}}^{\text{2}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}{30,000\times 3,000}\\ =-0.0035\text{rad}\end{array}$

Hence, the slope at B (right) is $\underset{\_}{-0.0035\text{rad}}$.

Determine the slope between C and D using the relation;

$\begin{array}{c}{\theta }_{CD}=\text{Area}\text{of}\text{the}M/EI\text{between}A\text{and}B\\ =\text{Area}\text{of}\text{triangle}\\ =\left[\frac{1}{2}\times b\times h\right]\end{array}$

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

Substitute 8 ft for b and $\frac{280}{EI}$ for h.

$\begin{array}{c}{\theta }_{CD}=\left[\frac{1}{2}\times b\times h\right]=\left[\frac{1}{2}\times 8\times \frac{280}{EI}\right]\\ =\frac{1,120{\text{kips-ft}}^{\text{2}}}{EI}\end{array}$

Determine the slope at D using the relation;

${\theta }_D={\theta }_C+{\theta }_{CD}$

Substitute  $\frac{3,306.67{\text{kips-ft}}^2}{EI}$ for ${\theta }_C$ and $\frac{1,120{\text{kips-ft}}^{\text{2}}}{EI}$ for ${\theta }_{CD}$.

$\begin{array}{c}{\theta }_D=\frac{3,306.67{\text{kips-ft}}^2}{EI}+\frac{1,120{\text{kips-ft}}^{\text{2}}}{EI}\\ =\frac{4,426.67{\text{kips-ft}}^{\text{3}}}{EI}\end{array}$

Substitute 30,000 ksi for E and $3,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\theta }_D=\frac{4,426.67{\text{kips-ft}}^{\text{3}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}{30,000\times 3,000}\\ =0.0071\text{rad}\end{array}$

Hence, the slope at point D is $\underset{\_}{0.0071\text{rad}}$.

Determine the deflection between C and D using the relation;

${\Delta }_{DC}={\theta }_{CD}\times \left(\frac{1}{3}\times 16\right)$

Substitute $\frac{1,120{\text{kips-ft}}^{\text{2}}}{EI}$ for ${\theta }_{CD}$.

$\begin{array}{c}{\Delta }_{DC}=\frac{1,120{\text{kips-ft}}^{\text{2}}}{EI}\times \left(\frac{1}{3}\times 16\right)\\ =\frac{5,973.33{\text{kips-ft}}^3}{EI}\end{array}$

Determine the deflection at point D using the relation;

${\Delta }_D=L_{CD}{\theta }_C+{\Delta }_{DC}$

Substitute 8 ft for $L_{CD}$, $\frac{3,306.67{\text{kips-ft}}^2}{EI}$ for ${\theta }_C$, and $\frac{5,973.33{\text{kips-ft}}^3}{EI}$ for ${\Delta }_{DC}$.

$\begin{array}{c}{\Delta }_D=8\left(\frac{3,306.67{\text{kips-ft}}^2}{EI}\right)+\frac{5,973.33{\text{kips-ft}}^3}{EI}\\ =\frac{32,426.69{\text{kips-ft}}^{\text{3}}}{EI}\end{array}$

Substitute 30,000 ksi for E and $3,000\text{in}{\text{.}}^{\text{4}}$ for I.

$\begin{array}{c}{\Delta }_D=\frac{32,426.69{\text{kips-ft}}^{\text{3}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^3}{30,000\times 3,000}\\ =0.62\text{in}\text{.}(\downarrow )\end{array}$

Hence, the deflection at point D is $\underset{\_}{0.62\text{in}\text{.}(\downarrow )}$.