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 2.5.21P

Pipe 2 has been inserted snugly into Pipe I. but the holes Tor a connecting pin do not line up; there is a gap s. The user decides to apply either force P:lo Pipe I or force P-, to Pipe 2, whichever is smaller. Determine the following using the numerical properties in the box.

(a) If only P{is applied, find Pt{tips} required to close gap s; if a pin is then inserted and Ptremoved, what are reaction forces RAand RBfor this load case?

(b) If only P2is applied, find P2{kips) required to close gap a; if a pin is inserted and P2removed, what are reaction forces R^ and RBfor this load case?

(c) What is the maximum shear stress in the pipes, for the loads in parts (a) and (b)?

(d) If a temperature increase IT is to be applied to the entire structure to close gaps{instead of applying forces Ptand P2), find the AT required to close the gap. If a pin is inserted after the gaphas closed, what are reaction forces .''.', and RBfor this case? (e) Finally, if the structure (with pin inserted) then cools to the original ambient temperature, what are reaction forces Rtand P Chapter 2, Problem 2.5.21P, Pipe 2 has been inserted snugly into Pipe I. but the holes Tor a connecting pin do not line up;

(a)

Expert Solution
Check Mark
To determine

The reactions at A and B.

Answer to Problem 2.5.21P

The reaction at A is = 55.24kips.

The reaction at B is = 55.24kips.

Explanation of Solution

Given information:

The gap between the pipes is sand the force P1is applied, modulus of elasticity of pipe 1 is 30000ksi, modulus of elasticity of pipe 2 is 14000ksi, the gap is 0.05in, length of pipe 1 is 56in, length of pipe 2 is 36in, diameter of pipe 1 is 6in, diameter of pipe 2 is 5in, thickness of pipe 1 is 0.5inthickness of pipe 2 is 0.25in, area of pipe 1 is 8.64in2and the area of pipe 2 is 3.73in2.

Write the expression for elongation in pipe 1.

  s=P1L1E1A1....... (I)

Here, length of pipe 1 is L1, modulus of elasticity of pipe 1 is E1and the area is A1.

Write the expression for elongation at point B.

  (δB)2=RB(L1A1E1+L2A2E2)........ (II)

Here, elongation at point B is (δB)2, length of pipe 2 is L2, modulus of elasticity of pipe 2 is E2, and area of pipe 2 is A2and the reaction at B is RB.

Write the expression for net elongation at B.

  (δB)1+(δB)2=0........ (III)

Here, elongation at point 1 is (δB)1.

Substitute sfor (δB)1and RB(L1A1E1+L2A2E2)for (δB)2in Equation (III).

  s+RB(L1A1E1+L2A2E2)=0RB=sL1A1E1+L2A2E2....... (IV)

Write the reaction at point A.

  RA=RB........ (V)

Here, reaction at A is RA.

Calculation:

Substitute 0.05infor s, 56infor L1, 30000ksifor E1and 8.64in2for A1in Equation (I).

  0.05in=P1(56in)(30000ksi)(8.64in2)P1=0.05in×30000ksi×8.64in256inP1=231.43kips

The force required to close the gap is 231.43kips.

Substitute 0.05infor s, 56infor L1, 30000ksifor E1and 8.64in2for A1, 36infor L2, 3.73in2for A2, 14000ksifor E2in and 231.43kipsfor P1Equation (IV).

  RB=0.05in(56in)(30000ksi)(1000psi1ksi)(8.64in2)+(36in)(14000ksi)(1000psi1ksi)(3.73in2)=0.05in2.16×107psiin2+6.89×107psiin2=(55.24×103psiin2)(1kips1000psiin2)=55.24kips

Substitute 55.24kipsfor RBin Equation (V).

  RA=55.24kips

Conclusion:

The reaction at A is 55.24kips.

The reaction at B is 55.24kips.

(b)

Expert Solution
Check Mark
To determine

The reaction at A is 55.24kips.

The reaction at B is 55.24kips.

Answer to Problem 2.5.21P

The reaction at A is 55.24kips.

The reaction at B is 55.24kips.

Explanation of Solution

Given information:

The gap between the pipes is s, modulus of elasticity of pipe 1 is 30000ksi, modulus of elasticity of pipe 2 is 14000ksi, the gap is 0.05in, length of pipe 1 is 56in, length of pipe 2 is 36in, diameter of pipe 1 is 6in, diameter of pipe 2 is 5in, thickness of pipe 1 is 0.5in, thickness of pipe 2 is 0.25in, area of pipe 1 is 8.64in2and the area of pipe 2 is 3.73in2.

Write the expression for force applied at pipe 2.

  P2=2(E2A2L2)s…... (VI)

Here, force applied at pipe 2 is P2.

Calculation:

Substitute 36infor L2, 3.73in2for A2, 14000ksifor E2and 0.05infor sin Equation (VI).

  P2=2((14000ksi)(3.73in2)36in)0.05in=(0.1in)(14000ksi)(3.73in2)36in=145.05kips

Substitute 0.05infor s, 56infor L1, 30000ksifor E1and 8.64in2for A1, 36infor L2, 3.73in2for A2, 14000ksifor E2in Equation (IV).

  RB=0.05in(56in)(30000ksi)(1000psi1ksi)(8.64in2)+(36in)(14000ksi)(1000psi1ksi)(3.73in2)=0.05in2.16×107psiin2+6.89×107psiin2=(55.24×103psiin2)(1kips1000psiin2)=55.24kips

Substitute 55.24kipsfor RBin Equation (V).

  RA=55.24kips

Conclusion:

The reaction at A is 55.24kips.

The reaction at B is 55.24kips.

(b)

Expert Solution
Check Mark
To determine

The maximum shear stress in pipe 1 and pipe 2.

.

Answer to Problem 2.5.21P

The maximum shear stress in pipe 1 is 13.9ksi.

The maximum shear stress in pipe 2 is 19.44ksi.

Explanation of Solution

Given information:

The gap between the pipes is sand the force P1is applied, modulus of elasticity of pipe 1 is 30000ksi, modulus of elasticity of pipe 2 is 14000ksi, the gap is 0.05in, length of pipe 1 is 56in, length of pipe 2 is 36in, diameter of pipe 1 is 6in, diameter of pipe 2 is 5in, thickness of pipe 1 is 0.5in, thickness of pipe 2 is 0.25in, area of pipe 1 is 8.64in2and the area of pipe 2 is 3.73in2.

Write the expression for maximum shear stress in pipe 1.

  τmax1=P1A12........ (VII)

Here, maximum shear stress in pipe 1 is τmax1.

Write the expression for maximum shear stress in pipe.

  τmax2=P2A22........ (VIII)

Here, maximum shear stress in pipe 2 is τmax2.

Calculation:

Substitute 231.43kipsfor P1and 8.64in2for A1in Equation (VII).

  τmax1=231.43kips8.64in22=26.7858kips/in22=(13.39kips/in2)(1ksi1kips/in2)=1.39ksi

The maximum shear stress in pipe 1 is = 1.39ksi.

Substitute 145.05kipsfor P2

  3.73in2for A2in Equation (VIII).

  τmax2=145.05kips3.73in22=38.8873kips/in22=(19.44kips/in2)(1ksi1kips/in2)=19.44ksi

Conclusion:

The maximum shear stress in pipe 1 is = 13.9ksi.

The maximum shear stress in pipe 2 is = 19.44ksi.

(d)

Expert Solution
Check Mark
To determine

The rise in temperature required to close the gap.

The reactions.

Answer to Problem 2.5.21P

The rise in temperature required to close the gap is 65.8°F.

The reactions are 0.

Explanation of Solution

Given information:

The gap between the pipes is sand the force P1is applied, modulus of elasticity of pipe 1 is 30000ksi, modulus of elasticity of pipe 2 is 14000ksi, the gap is 0.05in, length of pipe 1 is 56in, length of pipe 2 is 36in, diameter of pipe 1 is 6in, diameter of pipe 2 is 5in, thickness of pipe 1 is 0.5in, thickness of pipe 2 is 0.25in, area of pipe 1 is 8.64in2and the area of pipe 2 is 3.73in2.

Write the expression for temperature raise.

  ΔT=sα1L1+α2L2........ (IX)

Here, raise in temperature is ΔT, coefficient of thermal expansion for pipe 1 is α1and the coefficient of thermal expansion for pipe 2 is α2.

Calculation:

Substitute 0.05infor s, 56infor L1, 36infor L2, 6.56×106/°Ffor α1and 11×106/°Ffor α2in Equation (IX).

  ΔT=0.05in(6.5×106/°F)(56in)+(11×106/°F)(36in)=0.05in760×106in/°F=65.8°F

Since the temperature remains constant, so the reactions are zero.

Conclusion:

The temperature raise required to close the gap is 65.8°F.

The reactions are 0.

(e)

Expert Solution
Check Mark
To determine

The reaction at A.

The reaction at B.

Answer to Problem 2.5.21P

The reaction at A is 55.24kips.

The reaction at B is 55.24kips.

Explanation of Solution

Given information:

The gap between the pipes is sand the force P1is applied, modulus of elasticity of pipe 1 is 30000ksi, modulus of elasticity of pipe 2 is 14000ksi, the gap is 0.05in, length of pipe 1 is 56in, length of pipe 2 is 36in, diameter of pipe 1 is 6in, diameter of pipe 2 is 5in, thickness of pipe 1 is 0.5in, thickness of pipe 2 is 0.25in, area of pipe 1 is 8.64in2and the area of pipe 2 is 3.73in2.

Calculation:

Substitute 0.05infor s, 56infor L1, 30000ksifor E1and 8.64in2for A1, 36infor L2, 3.73in2for A2, 14000ksifor E2in and 231.43kipsfor P1Equation (IV).

  RB=0.05in(56in)(30000ksi)(1000psi1ksi)(8.64in2)+(36in)(14000ksi)(1000psi1ksi)(3.73in2)=0.05in2.16×107psiin2+6.89×107psiin2=(55.24×103psiin2)(1kips1000psiin2)=55.24kips

Substitute 55.24kipsfor RBin Equation (V).

  RA=55.24kips

Conclusion:

The reaction at A is = 55.24kips.

The reaction at B is = 55.24kips.

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Chapter 2 Solutions

Mechanics of Materials (MindTap Course List)

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