Structural Analysis, 5th Edition
Structural Analysis, 5th Edition
5th Edition
ISBN: 9788131520444
Author: Aslam Kassimali
Publisher: Cengage Learning
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Chapter 6, Problem 60P
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.

Expert Solution & Answer
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Answer to Problem 60P

The slope θB at point B (left) of the given beam using the moment area method is 0.0033rad_.

The deflection ΔB at point B (left) of the given beam using the moment area method is 0.39in.()_.

The slope θB at point B (right) of the given beam using the moment area method is 0.00263rad_.

The slope θD at point D of the given beam using the moment area method is 0.0071rad_.

The deflection ΔD at point D of the given beam using the moment area method is 0.62in.()_.

Explanation of Solution

Given information:

The Young’s modulus (E) is 30,000 ksi.

The moment of inertia of the section AB is (I) is 4,000in.4.

The moment of inertia of the section BD is (I) is 3,000in.4.

Calculation:

Consider flexural rigidity EI of the beam is constant.

To draw a M/EI diagram, the reactions are calculated by considering all the loads acting in the beam. The calculation of positive moment at point A is calculated by simply considering the point load of 35 kips and the reactions calculated by considering all the loads acting in the beam.

Show the free body diagram of the given beam as in Figure (1).

Structural Analysis, 5th Edition, Chapter 6, Problem 60P , additional homework tip  1

Refer Figure (1),

Consider upward is positive and downward is negative.

Consider clockwise is negative and counterclowise is positive.

Refer Figure (1),

Consider reaction at A and C as RA and RC.

Take moment about point B.

Determine the reaction at D;

RC×(8)(35×16)=0RC=5608RC=70kips

Determine the reaction at support A;

V=0RA+RC(2.5×16)35=0RA=7570RA=5kips

Determine the moment at A:

MA=(35×32)+(70×24)(25×16×162)=240kips-ft=240kips-ft(Clockwise)

Show the reaction of the given beam as in Figure (2).

Structural Analysis, 5th Edition, Chapter 6, Problem 60P , additional homework tip  2

Determine the bending moment at B;

MB=70×835×16=560560=0

Determine the bending moment at C;

MC=(35×8)=280kips-ft

Determine the bending moment at D;

MD=(5×32)(70×8)240+(2.5×16×(162+16))=720+240+960=0

Determine the positive bending moment at A using the relation;

MA=(35×32)+70×24=1,120+1,680=560kips-ft

Show the reaction and point load of the beam as in Figure (3).

Structural Analysis, 5th Edition, Chapter 6, Problem 60P , additional homework tip  3

Determine the value of M/EI;

MEI=560kips-ftEI

Substitute 4I3 for I.

MEI=560kips-ftE(4I3)=1,680kips-ft4EI=420kips-ftEI

Show the M/EI diagram of the beam as in Figure (4).

Structural Analysis, 5th Edition, Chapter 6, Problem 60P , additional homework tip  4

Show the conjugate beam as in Figure (5).

Structural Analysis, 5th Edition, Chapter 6, Problem 60P , additional homework tip  5

Determine the support reaction at support B;

RB×8+1EI[13×240×16×(8+34×16)12×16×420×(8+23×16)+12×8×280×(13×8)]=0RB=25,60062,731+2,9878EIRB=4,268kips-ft2EIRB=4,268kips-ft2EI()

Determine the shear force at B (left) using the relation;

SB(left)=[12×b1×h113×b2×h2]

Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, and 240EI for h2.

SB(left)=[12×16×420EI13×16×240EI]=2,080kips-ft2EI

Determine the slope at B (left) using the relation;

SB(left)=2,080kips-ft2EI

Substitute 30,000 ksi for E and 4,000in.4 for I.

θB(left)=2,080kips-ft2×(12in.1ft)230,000×3,000=0.0033rad

Hence, the slope at B (left) is 0.0033rad_.

Determine the slope at B (right) using the relation;

θB(right)=θB(left)RB

Substitute 2,080kips-ft2EI for θB(left) and 4,268kips-ft2EI for RB.

θB(right)=2,080kips-ft2EI4,268kips-ft2EI=2,188kips-ft2EI

Substitute 30,000 ksi for E and 4,000in.4 for I.

θB(right)=2,188kips-ft2×(12in.1ft)230,000×4,000=0.00263rad

Hence, the deflection at B (right) is 0.00263rad_.

Determine the bending moment at B using the relation;

M=[12×b1×h1×(23×b1)+13×b2×h2×(34×b2)]

Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, (240EI) for h2.

MB=[12×16×420EI×(23×16)13×16×(240EI)×(34×16)]=35,84015,360EI=20,480kips-ft2EI

Determine the deflection at B using the relation;

MB=20,480kips-ft2EI

Substitute 30,000 ksi for E and 4,000in.4 for I.

MB=20,480kips-ft2×(12in.1ft)330,000×3,000=0.39in.()

Hence, the deflection at B is 0.39in.()_.

Determine the shear force at D using the relation;

SD=[12×b1×h1+13×b2×h2+RB+12×b3×h3]

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

Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, 240EI for h2, 4,268kips-ft2EI for RB, 16 ft for b3, and 280EI for h3.

SD=[12×16×420EI13×16×(240EI)4,26812×16×(280EI)]=3,3601,2804,2682,240EI=4,427kips-ft2EI

Determine the slope at D using the relation;

θD=4,427kips-ft2EI

Substitute 30,000 ksi for E and 3,000in.4 for I.

θD=4,427kips-ft2×(12in.1ft)230,000×3,000=0.0071rad

Hence, the deflection at D is 0.0071rad_

Determine the bending moment at D using the relation;

MD=[12×b1×h1×(16+23×b1)+13×b2×h2×(16+34×b2)+(RB×16)+12×b3×h3×(24×b3)]

Substitute 16 ft for b1, 420EI for h1, 16 ft for b2, 240EI for h2, 4,268kips-ft2EI for RB, 16 ft for b3, and 280EI for h3.

MD=[12×16×(420EI)(16+23×16)13×16×(240EI)(16+34×16)(4,268×16)12×16×(280EI)×(24×16)]=89,61135,84068,28817,920EI=32,437kips-ft3EI

Determine the deflection at D using the relation;

ΔD=32,437kips-ft3EI

Substitute 30,000 ksi for E and 3,000in.4 for I.

ΔD=32,437kips-ft3×(12in.1ft)330,000×3,000=0.62in.=0.62in.()

Hence, the deflection at D is 0.62in.()_

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