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Structural Analysis

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

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Section
BuyFindarrow_forward

Structural Analysis

6th Edition
KASSIMALI + 1 other
Publisher: Cengage,
ISBN: 9781337630931
Chapter 6, Problem 15P
Textbook Problem
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6.14 through 6.17 Use the moment-area method to determine the slopes and deflections at points B and C of the beam shown.

FIG. P6.15, P6.41

Chapter 6, Problem 15P, 6.14 through 6.17 Use the moment-area method to determine the slopes and deflections at points B and

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

Given information:

The Young’s modulus (E) is 70 GPa.

The moment of inertia (I) is 500×106mm4.

Calculation:

Consider elastic modulus E of the beam is constant.

Show the given beam as in Figure (1).

Refer Figure (1),

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.

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclockwise is positive.

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

V=0RA+RC100=0RA+0100=0RA=100kN

Determine the bending moment at C;

MC=300kNm

Determine the bending moment at B;

MB=300kNm

Determine the bending moment at A;

MA=100×6+300=900kNm

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×h1+(b×h2)

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

Substitute 6 m for b, 300EI for h1, and 150EI for h2.

θB=θBA=(12×300EI(6)+150EI(6))=1EI(1,800)=1,800kNm2EI

Determine the slope at B using the relation;

θB=1,800kNm2EI

Substitute 70 GPa for E and 500×106mm4 for I.

θB=1,800kNm2(70GPa×106kN/m21GPa)(500×106mm4×(1m1,000mm)4)=0.0514rad

Hence, the slope at point B is 0.0514rad(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=MomentoftheareaoftheM/EIdiagrambetweenAandBaboutB=12×b×h1×(23×b)+(b×h2)(b2)

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

Substitute 6 for b, 300EI for h1, and 150EI for h2.

ΔB=ΔBA=(12×300EI(6)(23×6)+150EI(6)×62)=6,300kNm3EI

Determine the deflection at B using the relation;

ΔB=ΔBA=6,300kNm3EI

Substitute 70 GPa for E and 500×106mm4 for I.

ΔB=ΔBA=6,300kNm3(70GPa×106kN/m21GPa)(500×106mm4×(1m1,000mm)4)=0

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Chapter 6 Solutions

Structural Analysis
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