Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Chapter 6, Problem 34P
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 34P

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 0.0035rad_.

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 Young’s modulus (E) 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, Chapter 6, Problem 34P , 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 C;

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 reactions of the given beam as in Figure (2).

Structural Analysis, Chapter 6, Problem 34P , 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;

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, Chapter 6, Problem 34P , 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, Chapter 6, Problem 34P , additional homework tip  4

Show the elastic curve as in Figure (5).

Structural Analysis, Chapter 6, Problem 34P , additional homework tip  5

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(left)=θAB=AreaoftheM/EIbetweenAandB=Areaoftriangle+Areaofparabola=[12×b1×h113×b2×h2]

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

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

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

Determine the slope B (left) using the relation;

θB(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 deflection at B using the relation;

ΔB=ΔBA=MomentoftheareaoftheM/EIdiagrambetweenBandAaboutA=[MEI(Areaoftriangle)+MEI(Areaofparabola)]

Substitute 420EI for MEI, (12×16×(23×16)) for area of triangle, (240EI) for MEI, and (13×16×(34×16)) for area of triangle.

ΔBA=[420EI(12×16×(23×16))+(240EI)(13×16×(34×16))]=1EI(35,84015,360)=20,480kips-ft3EI

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

ΔBA=20,480kips-ft3EI

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

ΔBA=20,480kips-ft3×(12in.1ft)330,0000×3,000=0.39in.()

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

Determine the deflection between B and C using the relation;

ΔBC=MomentoftheareaoftheM/EIdiagrambetweenBandCaboutC=[MEI(Areaoftriangle)×(23×b)]=[12×b×h×(23×b)]

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

Substitute 8 ft for b and 280EI for h.

ΔBC=[12×8×(280EI)×(23×8)]=5,973.33kips-ft3EI

Express the relationship between the deflection and slope of span BC as follows:

ΔBC+ΔB=LBCθCθC=ΔBC+ΔBLBC

Here, LBC is the length between point B and C.

Substitute 5,973.33kips-ft3EI for ΔBC, 20,480kips-ft3EI for ΔB, and 8 ft for LBC.

θC=5,973.33EI+20,480EI8=3,306.67kips-ft2EI

Determine the slope between B and C using the relation;

θBC=AreaoftheM/EIbetweenBandC=Areaoftriangle=[12×b×h]

Substitute 8 ft for b and 280EI for h.

θBC=[12×8×280EI]=1,120kips-ft2EI

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

θB(right)=θCθBC

Substitute 3,306.67kips-ft2EI for θC and 1,120kips-ft2EI for θBC.

θB(right)=3,306.67kips-ft2EI1,120kips-ft2EI=2,186.67kips-ft2EI

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

θB(right)=2,186.67kips-ft2×(12in.1ft)230,000×3,000=0.0035rad

Hence, the slope at B (right) is 0.0035rad_.

Determine the slope between C and D using the relation;

θCD=AreaoftheM/EIbetweenAandB=Areaoftriangle=[12×b×h]

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

Substitute 8 ft for b and 280EI for h.

θCD=[12×b×h]=[12×8×280EI]=1,120kips-ft2EI

Determine the slope at D using the relation;

θD=θC+θCD

Substitute  3,306.67kips-ft2EI for θC and 1,120kips-ft2EI for θCD.

θD=3,306.67kips-ft2EI+1,120kips-ft2EI=4,426.67kips-ft3EI

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

θD=4,426.67kips-ft3×(12in.1ft)230,000×3,000=0.0071rad

Hence, the slope at point D is 0.0071rad_.

Determine the deflection between C and D using the relation;

ΔDC=θCD×(13×16)

Substitute 1,120kips-ft2EI for θCD.

ΔDC=1,120kips-ft2EI×(13×16)=5,973.33kips-ft3EI

Determine the deflection at point D using the relation;

ΔD=LCDθC+ΔDC

Substitute 8 ft for LCD, 3,306.67kips-ft2EI for θC, and 5,973.33kips-ft3EI for ΔDC.

ΔD=8(3,306.67kips-ft2EI)+5,973.33kips-ft3EI=32,426.69kips-ft3EI

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

ΔD=32,426.69kips-ft3×(12in.1ft)330,000×3,000=0.62in.()

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

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