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 2, Problem 2.5.25P

A capped cast-iron pipe is compressed by a brass rod, as shown. The mil is turned until it is just snug, then add an additional quarter turn to pre-compress the cast-iron pipe. The pitch of the threads of the bolt ap = 52 mils (a mil is one-thousandth of an inch). Use the numerical properties provided.

(a) What stresses a and arwill be produced in the cast-iron pipe and brass rod. respectively, by the additional quarter turn of the nut?

(b) Find the bearing stress ahbeneath the washer and the shear stress t(in the steel cap.

  Chapter 2, Problem 2.5.25P, A capped cast-iron pipe is compressed by a brass rod, as shown. The mil is turned until it is just

(a)

Expert Solution
Check Mark
To determine

The stress produced in pipe and bar.

Answer to Problem 2.5.25P

The stress produced in pipe is= 0.196ksi.

The stress produced in bar is = 3.4204ksi.

Explanation of Solution

Given information:

The pitch of the bolt is 2mins, diameter of washer is 34in, thickness of cap is 1in, outer diameter of cast iron pipe is 6in, inner diameter of cast iron pipe is 5.625in, length of cast iron is 4ft, diameter of brass rod is 12in, modulus of elasticity of steel is 30000ksi, modulus of elasticity of brass is 14000ksi, modulus of elasticity of cast iron is 12000ksiand the number of turns are 14.

Write the expression for area of rod.

  Ar=π4dr2........ (I)

Here, area of rod is Ar, diameter of rod is dr.

Write the expression for area of pipe.

  Ap=π4(do2di2)....... (II)

Here, area of pipe is Ap, outer diameter is doand the inner diameter is di.

Write the expression for displacement between cut ends.

  δnet1=np....... (III)

Here, displacement between cut ends of the rod is δnet1, number of turns is nand the pitch is p.

Write the expression for relative displacement between cut ends.

  (δnet)2=F(Lci+2tcEbAr+LciECAp)....... (IV)

Here, relative displacement between cut ends is (δnet)2, length of cat iron is Lci, thickness of cast iron is tc.

Write the compatibility Equation for displacement.

  δnet1+δnet2=0....... (V)

Write the expression for force on rod.

  Fr=F........ (VI)

Here, force on rod is Fr.

Write the expression for force on pipe.

  Fp=F....... (VII)

Here, force on pipe is Fp.

Write the expression for stress on cast iron pipe.

  σp=FpAP....... (VIII)

Here, stress on cast iron pipe is σp.

Write the expression for stress in rod.

  σr=FrAr....... (IX)

Here, stress on rod is σr.

Calculation:

Substitute 12infor drin Equation (I).

  Ar=π4(12in)2=π4(14in2)=0.19635in2

Substitute 6infor doand 5.625infor diin Equation (II).

  Ap=π4((6in)2(5.625in)2)=π4(36in231.645in2)=3.42in2

Substitute 14for nand 2minsfor pin Equation (III).

  δnet1=14(2mins)=14(2mins)(0.026in)1m=13×103in

Substitute 4ftfor Lci, 1infor tc, 14000ksifor Eb, 12000ksifor Ec, 3.42in2for Ap, 0.19635in2for Arin Equation (IV).

  (δnet)2=F(4ft+2×1in14000ksi×0.19635in2+4ft12000ksi×3.42in2)=F((4ft)(12in1ft)+2×1in14000ksi×0.19635in2+(4ft)(12in1ft)12000ksi×3.42in2)=1.9357×105F

Substitute 1.9357×105Ffor (δnet)2and 13×103infor δnet1in Equation (V).

  13×103in+1.9357×105F=0F=13×1031.9357×105F=671.6lb

Substitute 671.6lbfor Fin Equation (VI).

  Fr=671.6lb

Substitute 671.6lbfor Fin Equation (VII).

  Fp=671.6lb

Substitute 671.6lbfor Fpand 3.42in2for Apin Equation (VIII).

  σp=671.6lb3.42in2=(196.145lb/in2)(1ksi1000lb/in2)=0.196ksi

Substitute 671.6lbfor Frand 0.19635in2for Arin Equation (IX).

  σr=671.6lb0.19635in2=(3420.4227lb/in2)(1ksi1000lb/in2)=3.4204ksi

Conclusion:

The stress produced in pipe is = 0.196ksi.

The stress produced in bar is = 3.4204ksi.

(b)

Expert Solution
Check Mark
To determine

The bearing stress in cap.

The shearing stress in the cap.

Answer to Problem 2.5.25P

The bearing stress in cap is = 2.736ksi.

The shearing stress in the cap is = 0.285ksi.

Explanation of Solution

Given information:

The pitch of the bolt is 2mins, diameter of washer is 34in, thickness of cap is 1in, outer diameter of cast iron pipe is 6in, inner diameter of cast iron pipe is 5.625in, length of cast iron is 4ft, diameter of brass rod is 12in, modulus of elasticity of steel is 30000ksi, modulus of elasticity of brass is 14000ksi, modulus of elasticity of cast iron is 12000ksiand the number of turns are 14.

Write the expression for area of cap.

  Acap=π4(dw2dr2)........ (X)

Here, area of cap is Acap, diameter of washer is dwand the diameter of the rod is dr.

Write the expression for bearing stress on cap.

  σcap=FrAcap......... (XI)

Here, bearing stress on cap is σcap.

Write the expression for shearing stress on the cap.

  τcap=Frπdwtc......... (XII)

Here, shearing stress on the cap is τcap.

Calculation:

Substitute 34infor dwand 12infor drin Equation (X).

  Acap=π4((34in)2(12in)2)=π4(0.3125in2)=0.2454in2

Substitute 0.2454in2for Acapand 671.6lbfor Frin Equation (XI).

  σcap=671.6lb0.2454in2=(2736.75lb/in2)(1ksi1000lb/in2)=2.736ksi

Substitute 671.6lbfor Fr, 34infor dwand 1infor tcin Equation (XII).

  τcap=671.6lbπ(34in)(1in)=671.6lb2.3561in2=(285.036lb/in2)(1ksi1000lb/in2)=0.285ksi

Conclusion:

The bearing stress in cap is = 2.736ksi.

The shearing stress in the cap is = 0.285ksi.

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Chapter 2 Solutions

Mechanics of Materials (MindTap Course List)

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