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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Textbook Question
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Chapter 3, Problem 3.4.15P

A uniformly tapered aluminum-alloy tube AB with a circular cross section and length L is shown in the figure. The outside diameters at the ends arc dAand dB= 2dA. A hollow section of length LB and constant thickness t = dA/10 is cast into the tube and extends from B halfway toward A. (a)

Find the angle of twist é of the tube

when it is subjected to torques T acting at the ends. Use numerical values: dA= 2.5 in., L = 48 in., G = 3.9 × 106 psi, and T =40,000 in.-lb.

(b)

Repeat part (a) if the hollow section has con stant diameter rfj (see figure part b)

  Chapter 3, Problem 3.4.15P, A uniformly tapered aluminum-alloy tube AB with a circular cross section and length L is shown in

(a)

Expert Solution
Check Mark
To determine

The angle of twist of the tube when it is subjected to torques at it ends.

Answer to Problem 3.4.15P

The angle of twist of the tube when it is subjected to torques at it end Ais 2.793° .

The angle of twist of the tube when it is subjected to torques at it end Bis 2.207°

Explanation of Solution

Given information:

The diameter of the smaller section of the tube is 2.5in , the length of the tube is 48in. , the torque acting on the body is 40000inlband the diameter of the larger section of the tube is 5in .

Write the expression for outer diameter.

  d=2dAxdAL

Here, the elemental area of the tube is dA , the length of the elemental area is x,and the length of the tube is L .

Write the expression for inner diameter.

  di=15dA(9L+5x)L

Write the expression for angle of twist of side A .

  ϕA=TG(32π)(0L/21 d 4 d 4 idx+L/2L1d4dx)...... (I)

Substitute 2dAxdALfor dand 15dA(9L+5x)Lfor diin Equation (I).

  ϕA=TG(32π)[ 0 L/2 1 (2dAxdAL )4 (15dA9L+5x2 )4 dx+ L/2 L 1 ( 2dAxdAL)4dx]=[TG(32π)( 125L2( 3ln(2)+2ln(7)ln( 197)dA4 ) 125L2( 2ln( 19)+ln( 181)dA4 )+ 19L 81dA4 )]

  ϕA=16TL81Gπd4A(38+10125ln( 71117 70952))=3.868TLGπd4A...... (II)

Here, the torque at the ends of the tube is T , the length of the tube is L,and the diameter of the smaller side is dA .

Write the expression for angle of twist of side B .

  ϕB=TG(32π)[ 0 L/2 1 (2dAxdAL )4 (dA )4 dx+ L/2 L 1 ( 2dAxdAL)4dx]ϕB=TG(32π)[14Lln(5)+20tan(32)dA414Lln(3)+20tan(32)dA4+19L81dA4]=3.057TLGdA4 ...... (III)

Calculation:

Substitute 40000inlbfor T , 48infor L

  3.9×106psifor G,and 2.5infor dA .in Equation (II).

  ϕA=3.868×40000inlb×48in3.9×106psi×( 2.5in)4=3.868×40000inlb×48in3.9×106psi×( 1lb/ in2 1psi)×( 2.5in)4=3.868×1.92×106in2lb152.343×106in2lb=0.04875radian

  ϕA=0.04875radian=0.04875radian(180°πradian)=2.793°

Substitute 40000inlbfor T , 48infor L

  3.9×106psifor Gand 5infor dB .in Equation (III).

  ϕB=3.057×40000inlb×48in3.9×106psi×( 2.5in)4=3.057×40000inlb×48in3.9×106psi×( 1lb/ in2 1psi)×( 2.5in)4=3.057×1.92×106in2lb152.343×106in2lb=0.03852radian

  ϕB=0.03852radian=0.03852radian(180°πradian)=2.207°

Conclusion:

The angle of twist of the tube when it is subjected to torques at it end Ais 2.793° .

The angle of twist of the tube when it is subjected to torques at it end Bis 2.207°

(b)

Expert Solution
Check Mark
To determine

The angle of twist when the tapered shaft is hollow.

Answer to Problem 3.4.15P

The angle of twist of the tube when it is subjected to torques at it end Ais 22.46° .

The angle of twist of the tube when it is subjected to torques at it end Bis 7.53° .

Explanation of Solution

Write the expression for angle of twist on side A .

  ϕA=32(TTA)πG0L/21d4dx...... (IV)

The figure below shows the torque acting on the tube.

  Mechanics of Materials (MindTap Course List), Chapter 3, Problem 3.4.15P

Figure-(1)

Write the expression for angle of twist on side B .

  ϕB=32(TTB)πGL/2L1d4di4dx ....... (V)

Write the expression for equation of compatibility.

  ϕA+ϕB=0...... (VI)

Substitute 32(TTA)πG0L/21d4dxfor ϕAand 32(TTB)πGL/2L1d4di4dxfor ϕBin Equation (VI).

  32TBπG0L/21d4dx32TAπGL/2L1d4di4dx ..... (VII)

Write the expression for the total torque T .

  T=TA+TB...... (IX)

Here, the torque acting on the side Ais TAand the torque onthe side Bis TB .

Calculation:

Substitute 48infor Lin Equation (VII).

  TBTA= 24 48 ( 48in)4 ( 48in+x)4( 48in)4dx024 ( 48in)4 ( 48in+x)4 dxTBTA=11.25933.1454TB=3.5796TA...... (VIII)

Substitute 3.5796TAfor TBin Equation (IX)

  TA+11.25393.1454TA=40000inlbTA+3.57789TA=40000inlb4.57789TA=40000inlbTA=87376inlb

Substitute 87376inlbfor TAin Equation (VIII).

  TB=3.5796(87376inlb)=312771.12inlb

Substitute 87376inlbfor TAand 312771.12inlbin Equation (IX).

  T=87376inlb+312771.12inlb=400147.1296inlb

Substitute 400147.1296inlbfor Tand 87376inlb for TAin Equation (IV).

  ϕA=32(400147.1296inlb87376inlb)π×(3.9× 10 6psi)0241 (5in) 4=32(3.91× 106inlb)π×(3.9× 106psi×( 1lb/ in 21psi ))×1625in40241dx=10.212in3×1625in4×[240]=0.3921radian

  ϕA=0.3921radian=0.3921radian(180°πradian)=22.46°

Substitute 400147.1296inlbfor Tand 312771.12inlbfor TBin Equation (V).

  ϕB=32(400147.1296inlb312771.12inlb)π(3.2× 10 6psi)24481 (5in) 4 (2.5in) 4=2.79603×106inlbπ(3.9× 106psi×( 1lb/ in 21psi ))×(0.024in4)2448dx=(5.47×103)×(24)=0.13144radian

  ϕA=0.13144radian=0.13144radian(180°πradian)=7.53°

Conclusion:

The angle of twist of the tube when it is subjected to torques at it end Ais 22.46° .

The angle of twist of the tube when it is subjected to torques at it end Bis 7.53° .

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Chapter 3 Solutions

Mechanics of Materials (MindTap Course List)

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