Mechanics of Materials (MindTap Course List)
Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 2, Problem 2.3.26P

A uniformly tapered lube AB of circular cross section and length L is shown in the figure. The average diameters at the ends are dAand d£= 2d t. Assume E is constant. Find the elongation S of the tube when it is subjected to loads P acting at the ends. Use the following numerical data:^ = 35 mm, L = WO mm, E = 2.1 GPa. and P = 25 tN. Consider the following cases.

(a) A hole of constant diameter dAis drilled from B toward A to form a hollow section of length x - U2.

(b) A hole of variable diameter a\.x) is drilled, from B toward A to form a hollow section of length x = L/2 and constant thickness t = dA/20.

  Chapter 2, Problem 2.3.26P, A uniformly tapered lube AB of circular cross section and length L is shown in the figure. The

a.

Expert Solution
Check Mark
To determine

The elongation of the tube when a hole of constant diameter dA is drilled from B to A to form a hollow section.

Answer to Problem 2.3.26P

The elongation of the tube when a hole of constant diameter dA is drilled from B to A to form a hollow section is 2.18 mm.

Explanation of Solution

Given:

  P=25 kNL=300 mmdA=35 mmdB=2dAE=2.1 GPa

  d(ζ)=dA(1+ζL)

From the given information, the area is calculated.

For the length Lx

  A(ζ)=πd(ζ)24

For the length x .

  A(ζ)=π4[d(ζ)2dA2]

The elongation is calculated.

  δ=PE[ 1 A( ζ ) dζ]δ=PE[0 Lx 4 πd ( ζ ) 2 dζ+ Lx L 4 π( d ( ζ ) 2 d A 2 ) dζ]δ=PE[0Lx1[ π 4 [ d A ( 1+ ζ L )] 2 ]dζ+LxL1[ π 4[ [ d A ( 1+ ζ L )] 2 d A 2 ]]dζ]δ=PE[4L2(x2)πdA2+(4Lπ d A 2+ LxL 1 [ π 4 [ [ d A ( 1+ ζ L )] 2 d A 2 ]]dζ)]δ=PE[4L2(x2)πdA2+(4Lπ d A 22Lln3π d A 2+2Lln( Lx)+ln( 3Lx)π d A 2)]

Given x=L2

  δ=PE(43LπdA22Lln3πdA2+2Lln( 1 2 L)+ln( 5 2 L)πdA2)

Substituting the values:

  δ=PE(43L π d A 2 2L ln3 π d A 2 +2L ln( 1 2 L )+ln( 5 2 L ) π d A 2 )δ=252.1(43 300 π ( 35 ) 2 2( 300) ln3 π ( 35 ) 2 +2( 300) ln( 1 2 300 )+ln( 5 2 300 ) π ( 35 ) 2 )δ=2.18 mm

Conclusion:

The elongation of the tube when a hole of constant diameter dA is drilled from B to A to form a hollow section is 2.18 mm.

b.

Expert Solution
Check Mark
To determine

The elongation of the tube when a hole of variable diameter d(x) is drilled from B to A to form a hollow section.

Answer to Problem 2.3.26P

The elongation of the tube when a hole of variable diameter d(x) is drilled from B to A to form a hollow section is 6.74 mm.

Explanation of Solution

  d(ζ)=dA(1+ζL)

From the given information, the area is calculated.

For the length Lx

  A(ζ)=πd(ζ)24

For the length x .

  A(ζ)=π4[d(ζ)2(d( ζ)2 d A 20)2]

The elongation is calculated.

  δ=PE[ 1 A( ζ ) dζ]δ=PE[0 Lx 4 πd ( ζ ) 2 dζ+ Lx L 4 π[ d ( ζ ) 2 ( d( ζ )2 d A 20 ) 2 ] dζ]δ=PE[0Lx4[ π [ d A ( 1+ ζ L )] 2 ]dζ+LxL4[π[ [ d A ( 1+ ζ L )] 2 [ d A ( 1+ ζ L )2 d A 20 ] 2 ]]dζ]δ=PE[4L2(x2L)πdA2+4LπdA2+(20Lln3+ln13+2ln d A+lnLπ d A 2)20L2lndA+ln(39L20x)πdA2]

Given x=L2

  δ=PE(43LπdA2+20Lln3+ln13+2lndA+lnLπdA220L2lndA+ln( 29L)πdA2)

Substituting the values:

  δ=PE(43L π d A 2 2L ln3 π d A 2 +2L ln( 1 2 L )+ln( 5 2 L ) π d A 2 )δ=252.1(43 300 π ( 35 ) 2 2( 300) ln3 π ( 35 ) 2 +2( 300) ln( 1 2 300 )+ln( 5 2 300 ) π ( 35 ) 2 )δ=6.74 mm

Conclusion:

The elongation of the tube when a hole of variable diameter d(x) is drilled from B to A to form a hollow section is 6.74 mm.

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Chapter 2 Solutions

Mechanics of Materials (MindTap Course List)

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