# Use the moment-area method to determine the slopes and deflections at points B and D of the beam shown. FIG. P6.33, P6.59

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931

Chapter
Section

### Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 6, Problem 33P
Textbook Problem
70 views

## Use the moment-area method to determine the slopes and deflections at points B and D of the beam shown.FIG. P6.33, P6.59

To determine

Find the slope θB&θD and deflection ΔB&ΔD at point B and D of the given beam using the moment-area method.

### Explanation of Solution

Given information:

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

The moment of inertia (I) is 6,000in.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;

RE×(24)(40×12)+250=0RE=23024RE=9.58kips

Determine the reaction at support A;

V=0RA+RE7540=0RA=1159.58RA=105.42kips

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

Refer Figure (3),

Determine the moment at A:

MA=(9.58×48)+250(40×36)(75×12)=709.842,3401,630kips-ft

Determine the bending moment at B;

MB=(9.58×36)+250(40×24)=344.88+250960365kips-ft

Determine the bending moment at C;

MC=(9.58×24)+250(40×12)=479.924800

Determine the bending moment at D;

MD=(9.58×12)+250365kips-ft

Determine the bending moment at E;

ME=250kips-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;

θB=θBA=AreaoftheMEIdiagrambetweenAandB=(12×b1×h1+12×b2×h2)

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

Substitute 12 ft for b1, 1,630EI for h1, 12 ft for b2, and 365EI for h2.

θB=θBA=[(12×12×1,630EI)+(12×12×365EI)]=1EI(1,630+3652)12=11,970kips-ft2EI

Substitute 29,000 ksi for E and 6,000in.4 for I.

Hence, the slope at B is 0.0099rad(Clockwise)_.

Determine the deflection between A and B using the relation;

Here, b1 is the width of rectangle, h1 is the height of the rectangle, b2 is the width of the triangle, and h2 is the height of the triangle.

Substitute 12 ft for b1, 365EI for h1, 12 ft for b2, and (1,630EI365EI) for h2.

ΔB=ΔBA=[(12×365EI)(122)+12×12×(1,630EI365EI)(23×12)]=87,000kipsft3EI

Substitute 29,000 ksi for E and 6,000in.4 for I.

ΔB=ΔBA=87,000kipsft3×(12in.1ft)329,000×6,000=0.864in.0.86in.()

Hence, the deflection at B is 0.86in.()_.

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;

Substitute 12 ft for b1, 365EI for h1, 12 ft for b2, (1,630EI365EI) for h2, 12 for b3, and 365EI for h3.

ΔC=ΔCA=[(12×365EI)(12+122)+(12×12×(1,630EI365EI))(12+23×12)+(12×12×365EI)(23×12)]=248,160kips-ft3EI

Determine the deflection between C and E using the relation;

Substitute 12 ft for b1, 365EI for h1, 12 ft for b2, 250EI for h2, 12 ft for b3, and (365EI250EI) for h3

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