   # Use the moment-area method to determine the slopes and deflections at points Band C of the beam shown. FIG. P6.14, P6.40

#### Solutions

Chapter
Section
Chapter 6, Problem 14P
Textbook Problem
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## Use the moment-area method to determine the slopes and deflections at points Band C of the beam shown. FIG. P6.14, P6.40

To determine

Find the slope θB&θC and deflection ΔB&ΔC at point B and C of the given beam using the moment-area method.

### Explanation of Solution

Calculation:

Consider elastic modulus E of the beam is constant.

Show the given beam as in Figure (1).

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclockwise is positive.

Since the point C is free end there is no support reaction at Point C. Therefore, the reaction at point C is also equal to zero.

Determine the reaction at support A using the Equation of equilibrium;

V=0RA=P

Determine the moment at point A using the relation;

MA(P×(L2+L2))=0MA=PL

Determine the moment at point B using the relation;

MB(P×(L2))=0MB=PL2

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:

θB=θBA=AreaoftheM/EIbetweenAandB=Areaoftriangle+Areaofrectangle=12×b×h+(b×h)

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

Substitute L2 for b and PL4EI for h.

θB=θBA=12×L2×(PL4EI)+(L2)×(PL4EI)=PL216EI+PL28EI=PL2+2PL216EI=3PL216EI

Hence, the slope at point B is 3PL216EI(Clockwise)_.

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:

ΔB=ΔBA=2PL3+3PL396EI=5PL396EI

Hence, the deflection at B is 5PL396EI()_

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