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 33P
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.

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Answer to Problem 33P

The slope θB  at point B of the given beam using the moment-area method is 0.0099rad(Clockwise)_.

The deflection ΔB at point B of the given beam using the moment-area method is 0.86in.()_.

The slope θD at point D of the given beam using the moment-area method is 0.0084rad(Counterclockwise)_.

The deflection ΔD at point D of the given beam using the moment-area method is 1.44in.()_.

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).

Structural Analysis, 5th Edition, Chapter 6, Problem 33P , additional homework tip  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).

Structural Analysis, 5th Edition, Chapter 6, Problem 33P , additional homework tip  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).

Structural Analysis, 5th Edition, Chapter 6, Problem 33P , additional homework tip  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).

Structural Analysis, 5th Edition, Chapter 6, Problem 33P , additional homework tip  4

Show the elastic curve diagram as in Figure (5).

Structural Analysis, 5th Edition, Chapter 6, Problem 33P , additional homework tip  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.

θB=11,970kips-ft2×(12in.1ft)229,000×6,000=0.0099rad=0.0099(Clockwise)

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

Determine the deflection between A and B using the relation;

ΔB=ΔBA=MomentoftheareaoftheM/EIdiagrambetweenBandAaboutA=[(b1h1)(b12)+(12b2h2)(23×b2)]

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;

ΔC=ΔCA=MomentoftheareaoftheM/EIdiagrambetweenCandAaboutA=[(b1h1)(12+b12)+(12b2h2)(12+23×b2)+(12b3h3)(23×b3)]

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;

ΔCE=MomentoftheareaoftheM/EIdiagrambetweenCandEaboutE=[(12b1h1)(23×b1)+(b2h2)(12+b22)+(12b3h3)(12+13×b3)]

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

ΔCE=[(12×12×365EI)(23×12)+(12×250EI)(12+122)+(12×12×(365EI250EI))(12+13×12)]=82,560kips-ft3EI

Determine the slope at E using the relation;

θE=ΔC+ΔCELCE

Here, LCE is the length between C and E.

Substitute 248,160kips-ft3EI for ΔC, 82,560kips-ft3EI for ΔCE, and 24 ft for LCE.

θE=248,160kips-ft3EI+(82,560kips-ft3EI)24=13,780kips-ft2EI

Determine the slope between D and E using the relation;

θDE=AreaoftheMEIdiagrambetweenDandE=(12×b1×h1+12×b2×h2)

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

Substitute 12 ft for b1, 365EI for h1, 12 ft for b2, and 250EI for h2.

θDE=[(12×12×365EI)+(12×12×250EI)]=1EI(365+2502)12=3,690kips-ft2EI

Determine the slope at D using the relation;

θD=θEθDE

Substitute 3,690kips-ft2EI for θE and 3,690kips-ft2EI for θDE.

θD=13,780kips-ft2EI(3,690kips-ft2EI)=10,090kips-ft2EI

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

θD=10,090kips-ft2×(12in.1ft)229,000×6,000=0.00835rad0.0084rad(Counterclockwise)

Hence, the slope at D is 0.0084rad(Counterclockwise)_.

Determine the deflection between D and E using the relation;

ΔDE=MomentoftheareaoftheM/EIdiagrambetweenDandEaboutE=[(b1h1)(b12)+(12b2h2)(13×b2)]

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

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

ΔDE=[(12×250EI)(122)+(12×12×(365EI250EI))(13×12)]=20,760kips-ft3EI

Determine the deflection at D using the relation;

ΔD=LDEθEΔDE

Substitute 12 ft for LDE, 13,780kips-ft2EI for θE, and 20,760kips-ft3EI for ΔDE.

ΔD=12(13,780kips-ft2EI)(20,760kips-ft3EI)=165,360+20,760EI=144,600kips-ft3EI

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

ΔD=144,600kips-ft3×(12in.1ft)329,000×6,000=1.44in.=1.44in.()

Hence, the deflection at point D is 1.44in.()_.

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