Chapter 6, Problem 16P

To determine

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

Answer to Problem 16P

The slope at point B of the given beam using the direct moment-area method is $\underset{\_}{-0.0112\text{rad}}$.

The deflection at point B of the given beam using the direct moment-area method is $\underset{\_}{0.782\text{in}\text{.}(\downarrow )}$.

The slope at point C of the given beam using the direct moment-area method is $\underset{\_}{-0.0155\text{rad}}$.

The deflection at point C of the given beam using the direct moment-area method is $\underset{\_}{4.227\text{in}\text{.}(\downarrow )}$.

Explanation of Solution

Given information:

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

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

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 positive and downward is negative.

Consider clockwise is negative and counterclowise is positive.

Since support C is a free end there is no reaction.

$\begin{array}{l}{\displaystyle \sum V=0}\\ R_A-\left(60+\left(3\times 10\right)\right)=0\\ R_A=90\text{kips}\end{array}$

Determine the bending moment at A;

$\begin{array}{l}{M}_A-\left(3\times 10\times \frac{10}{2}+20\right)-\left(60\times 10\right)=0\\ M_A=1,350\text{kips}\cdot \text{ft}\end{array}$

Determine the bending moment at B;

$\begin{array}{l}{M}_B-1,350+\left(90\times 10\right)=0\\ M_B=1,350-900\\ M_B=450\text{kips}\cdot \text{ft}\end{array}$

Determine the moment at D;

$\begin{array}{l}{M}_D+90\times 20-\left(60\times 10\right)-1,350=0\\ M_B=-1,800+1,950\\ M_B=150\text{kips}\cdot \text{ft}\end{array}$

Determine the bending moment at C;

$\begin{array}{l}{M}_C-1,350+\left(90\times 30\right)-\left(60\times 20\right)-\left(3\times 10\times \frac{10}{2}\right)=0\\ M_C=-2,700+2,700\\ M_C=0\end{array}$

Show the $M/EI$ diagram for the given beam as in Figure (2).

Elastic curve:

The sign of $M/EI$ diagram is negative; therefore, the beam bends downward. The support A of the given beam is fixed and the slope at A is zero. Therefore, the tangent to the elastic curve at A is horizontal.

Show the elastic curve diagram as in Figure (3).

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:

$\begin{array}{c}{\theta }_B={\theta }_{BA}=\text{Area}\text{of}\text{the}M/EI\text{between}A\text{and}B\\ =\text{Area}\text{of}\text{triangle}+\text{Area}\text{of}\text{rectangle}\\ =\frac{1}{2}\times b\times h+\left(b_1\times h_1\right)\end{array}$

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

Substitute 10 ft for b, $\frac{900}{EI}$ for h, 10 for $b_1$, and $\frac{450}{EI}$ for $h_1$.

$\begin{array}{c}{\theta }_B={\theta }_{BA}=\left(\frac{1}{2}\times \frac{900}{EI}\left(10\right)+\frac{450}{EI}\left(10\right)\right)\\ =\frac{1}{EI}\left(9,000\right)\\ =\frac{9,000\text{kips}\cdot {\text{ft}}^{\text{2}}}{EI}\end{array}$

Determine the slope at B using the relation;

${\theta }_B=\frac{9,000\text{kips}\cdot {\text{ft}}^{\text{2}}}{EI}$

Substitute $29,000\text{ksi×}{\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^{\text{2}}$ for E and $4,000\text{in}{\text{.}}^{\text{4}}\times {\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^{\text{4}}$ for I.

$\begin{array}{c}{\theta }_B=\frac{9,000\text{kips}\cdot {\text{ft}}^{\text{2}}}{\left(29,000\text{ksi×}{\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^{\text{2}}\right)\left(4,000\text{in}{\text{.}}^{\text{4}}\times {\left(\frac{\text{1}\text{ft}}{\text{12}\text{in}\text{.}}\right)}^{\text{4}}\right)}\\ =-0.0112\text{rad}\end{array}$

Hence, the slope at point B is $\underset{\_}{-0.0112\text{rad}\left(\text{Clockwise}\right)}$.

The deflection of B with respect to the undeforemd axis of the beam is equal to the tangential deviation of B from the tangent at A.

Express the deflection at B using the second moment-area theorem as follows:

$\begin{array}{c}{\Delta }_B={\Delta }_{BA}=\text{Moment}\text{of}\text{the}\text{area}\text{of}\text{the}M/EI\text{diagram}\text{between}A\text{and}B\text{about}B\\ =\frac{1}{2}\times b\times h\times \left(\frac{2}{3}\times b\right)+\left(b\times h_1\right)\left(\frac{b}{2}\right)\end{array}$

Substitute 10 ft for b, $\frac{900}{EI}$ for h, and $\frac{450}{EI}$ for $h_1$.

$\begin{array}{c}{\Delta }_B={\Delta }_{BA}=\left(\frac{1}{2}\times \frac{900}{EI}\left(10\right)\left(\frac{2}{3}\times 10\right)+\frac{450}{EI}\left(10\right)\times \frac{10}{2}\right)\\ =\frac{52,500\text{kips}\cdot {\text{ft}}^{\text{3}}}{EI}\end{array}$

Determine the deflection at B using the relation;

${\Delta }_B={\Delta }_{BA}=\frac{52,500\text{kips}\cdot {\text{ft}}^{\text{3}}}{EI}$

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

$\begin{array}{c}{\Delta }_B={\Delta }_{BA}=\frac{52,500\text{kips}\cdot {\text{ft}}^{\text{3}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^3}{\left(29,000\text{ksi}\right)\left(4,000\text{in}{\text{.}}^{\text{4}}\right)}\\ =0.782\text{in}\text{.}(\downarrow )\end{array}$

Hence, the deflection at B is $\underset{\_}{0.782\text{in}\text{.}(\downarrow )}$.

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

$\begin{array}{c}{\theta }_C={\theta }_{CA}={\theta }_B\text{+Area}\text{of}\text{the}M/EI\text{between}A\text{and}C\\ ={\theta }_B\text{+Area}\text{of}\text{rectangle}+\text{Area}\text{of}\text{triangle}+\text{Area}\text{of}\text{parabola}\\ ={\theta }_B\text{+}\left(b\times h\right)+\left(\frac{1}{2}\times b\times h\right)+\left(\frac{1}{3}\times b\times h\right)\end{array}$

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

Substitute $\frac{9,000}{EI}$ for ${\theta }_B$, 10 ft for b, and $\frac{150}{EI}$, $\frac{300}{EI}$, and $\frac{150}{EI}$ for h.

$\begin{array}{c}{\theta }_C={\theta }_{CA}=\frac{9,000}{EI}+\left(\left(10\times \frac{150}{EI}\right)+\left(\frac{1}{2}\times \frac{300}{EI}\times 10\right)+\left(\frac{1}{3}\times \frac{150}{EI}\times 10\right)\right)\\ =\frac{12,500\text{kips}\cdot {\text{ft}}^2}{EI}\end{array}$

Determine the slope at C using the relation;

${\theta }_C=\frac{12,500\text{kips}\cdot {\text{ft}}^2}{EI}$

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

$\begin{array}{c}{\theta }_B=\frac{12,500\text{kips}\cdot {\text{ft}}^{\text{2}}{\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^2}{\left(29,000\text{ksi}\right)\left(4,000\text{in}{\text{.}}^{\text{4}}\right)}\\ =0.0155\text{rad}\end{array}$

Hence, the slope at point C is $\underset{\_}{-0.0155\text{rad}}$.

The deflection of C with respect to the undeforemd axis of the beam is equal to the tangential deviation of C from the tangent at A.

Express the deflection at C using the second moment-area theorem as follows:

$\begin{array}{c}{\Delta }_C={\Delta }_{CA}=\text{Moment}\text{of}\text{the}\text{area}\text{of}\text{the}M/EI\text{diagram}\text{between}A\text{and}C\text{about}C\\ =\left[\begin{array}{c}\left(lb\left(20+\frac{b}{2}\right)\right)+\left(\frac{1}{2}bh\times \left(20+\frac{2}{3}b\right)\right)\\ +\left(lb\left(10+\frac{b}{2}\right)\right)+\left(\frac{1}{2}bh\times \left(10+\frac{2}{3}b\right)\right)\\ +\left(\frac{1}{3}bh\times \left(\frac{3}{4}b\right)\right)\end{array}\right]\\ =\left[\begin{array}{c}\left(\frac{450}{EI}\left(10\right)\left(20+\frac{b}{2}\right)\right)+\left(\frac{1}{2}\left(10\right)\left(\frac{900}{EI}\right)\times \left(20+\frac{2}{3}\times 10\right)\right)\\ +\left(10\left(\frac{150}{EI}\right)\left(10+\frac{10}{2}\right)\right)+\left(\frac{1}{2}\left(10\right)\left(\frac{300}{EI}\right)\times \left(10+\frac{2}{3}\times 10\right)\right)\\ +\left(\frac{1}{3}\left(10\right)\left(\frac{150}{EI}\right)\times \left(\frac{3}{4}\times 10\right)\right)\end{array}\right]\end{array}$

$\begin{array}{c}{\Delta }_C={\Delta }_{CA}=\frac{1}{EI}\left(112,500+120,000+22,500+25,000+3,750\right)\\ =\frac{283,750\text{kips}\cdot {\text{ft}}^{\text{3}}}{EI}\end{array}$

Determine the deflection at C using the relation;

${\Delta }_C=\frac{283,750\text{kips}\cdot {\text{ft}}^{\text{3}}}{EI}$

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

$\begin{array}{c}{\Delta }_C={\Delta }_{CA}=\frac{283,750\text{kips}\cdot {\text{ft}}^{\text{3}}\times {\left(\frac{\text{12}\text{in}\text{.}}{\text{1}\text{ft}}\right)}^3}{\left(29,000\text{ksi}\right)\left(4,000\text{in}{\text{.}}^{\text{4}}\right)}\\ =4.227\text{in}\text{.}(\downarrow )\end{array}$

Hence, the deflection at C is $\underset{\_}{4.227\text{in}\text{.}(\downarrow )}$.