CE Board Exam Randomizer

Area Moment Method

Consider 2 points A and B on the beam, as shown in the figure below:

Theorem 1: The change in slope between the tangents drawn to the elastic curve at any 2 points A and B, $\theta_{BA}$, is equal to the product $1/EI$ multiplied by the area of the moment diagram between A and B. That is,

Theorem 2: The deviation of any point B relative to a tangent drawn to the elastic curve at any other point A, $t_{B/A}$, is equal to the product $1/EI$ multiplied by the moment of an area about B of that part of the moment diagram between points A and B. That is,

Sign Convention:
The rotation, $\theta$, is positive if counterclockwise from the left tangent and negative if clockwise from the left tangent. The deviation, $t$, is positive if the tangent is above the point, and negative if the tangent is below the point.

Moment Diagram by Parts

Here, we verify that the effect caused by the actual moment diagram and the moment diagram by parts are the same for the area and area-moment computations.

Problem:

Using the Area Moment Method, determine the following:
a. The slope and deflection at 3m from the left end.
b. The deflection at the free ends, A and D.
c. The deflection and slope at the midspan.
d. The location and value of the maximum deflection between BC.

See images:

Problem:

Refer to the image shown:

See images:

Problem: Simple Beam with Uniform Load and Midspan Load

A 12 m simple beam is loaded with a uniform load w = 40 kN/m over the entire span and a concentrated load R = 200 kN at the midspan. EI is constant.

1. Which of the following most nearly gives the slope at the support due to the concentrated load only?
   1. 4800/EI
   2. 3600/EI
   3. 7200/EI
   4. 1800/EI
2. Which of the following most nearly gives the maximum slope due to the uniform load only?
   1. 9600/EI
   2. 6400/EI
   3. 2880/EI
   4. 10800/EI
3. Which of the following most nearly gives the maximum deflection due to the given loads?
   1. 14400/EI
   2. 10000/EI
   3. 10800/EI
   4. 18000/EI

For a simply supported beam with a concentrated load at midspan, the support slope is:

$$ \theta_P=\frac{PL^2}{16EI} $$ $$ \theta_P=\frac{200(12)^2}{16EI}=\frac{1800}{EI} $$

For a simply supported beam with a uniform load over the entire span, the maximum slope occurs at either support:

$$ \theta_w=\frac{wL^3}{24EI} $$ $$ \theta_w=\frac{40(12)^3}{24EI}=\frac{2880}{EI} $$

The value 4680/EI is the sum of the two slope contributions:

$$ \theta_{\max}=\theta_P+\theta_w $$ $$ \theta_{\max}=\frac{1800}{EI}+\frac{2880}{EI}=\frac{4680}{EI} $$

For maximum deflection at midspan, add the deflection caused by the concentrated load and the deflection caused by the uniform load.

$$ \Delta_P=\frac{PL^3}{48EI}=\frac{200(12)^3}{48EI}=\frac{7200}{EI} $$ $$ \Delta_w=\frac{5wL^4}{384EI}=\frac{5(40)(12)^4}{384EI}=\frac{10800}{EI} $$ $$ \Delta_{\max}=\frac{7200}{EI}+\frac{10800}{EI}=\frac{18000}{EI} $$

Note: The choice shown as 4680/EI in the source item corresponds to combined maximum slope, not maximum deflection. For part c, the corrected maximum deflection is 18000/EI.

Final answers: a. D. 1800/EI; b. C. 2880/EI; c. D. 18000/EI

Problem: Propped Beam with Moment at Simple End

A propped beam with span L carries a concentrated moment M at the simple end. EI is constant.

1. What is the reaction at the simple end?
   1. 2M/3L
   2. M/3L
   3. 3M/2L
   4. 2M/L
2. Compute the location of the maximum deflection from the fixed end.
   1. 2L/3
   2. L/3
   3. L/4
   4. L/2
3. Compute the slope at the simple end if EI is constant.
   1. 4ML/EI
   2. 2ML/EI
   3. ML/2EI
   4. ML/4EI

See images:

Problem:

Refer to the image shown:

See images:

Problem:

Refer to the image shown:

See images:

Problem:

Refer to the image shown:

See images:

Problem:

Refer to the image shown:

See images: