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 3.8.18P

Two pipes {L, = 2.5 m and L, = 1.5 m) are joined al B by flange plales (thickness (, = 14 mm) with five bolts [dlt, = 13 mm] arranged in a circular pal tor n (see figure). Also, each pipe segment is atlaehed to a wall (at .1 and ( '. see figure! using a base plate Uh = 15 mm) and four bolts (dM, = 16 mm). All bolts are tightened until just snug. Assume £, = 110 GPa,E2 = 73 GPa,», = 0.33,andv, = 0.25. Neglect the self-weight of the pipes, and assume the pipes are in a stress-free stale initially. The cross-sectional areas of the pipes are At = 1500 mm: and A2 = (3/5)4. The ollter diameter of Pipe 1 is 60 mm. The outer diameter of Pipe 2 is equal to the inner diameter of Pipe 1. The bolt radius r = 64 mm for both base and flange plates.

(a)

If torque '/'is applied at .v = Lt. find an expression for reactive torques Iit and IL in terms of T.

(b)

Find the maximum load variable /'(i.e., Tmal) if allowable torsional stress in the two pipes is

Tall0* = 65 MPa-id Draw torsional moment iTMD i and torsional displacement (TDD) diagrams. Label all key ordinales. What is '/>.ll('.'

(d) Find mail, if allowable shear and bearing stresses in the base plate and flange bolts cannot be exceeded. Assume allowable stresses in shear an.: :vari:'.g I all bolls are r |Nill, = 45 MPa andtr MaK =90 MPa.

(e) Remove torque Tat x — L,. Now assume the flange-plate bolt holes are misaligned by some angle ß (see figure). Find the expressions for reactive torques Rx and R2 if the pipes are twisted to align the flange-plate bolt holes, bolts are then inserted, and the pipes released.

(f) What is the maximum permissible misalignment angle ß mix if allowable stresses in shear and bearing for all bolts [from part (d)] are not to be exceeded?

  Chapter 3, Problem 3.8.18P, Two pipes {L, = 2.5 m and L, = 1.5 m) are joined al B by flange plales (thickness (, = 14 mm) with

(a)

Expert Solution
Check Mark
To determine

The reactive torque in terms of torque.

Answer to Problem 3.8.18P

The reactive torque at point (A) is = 0.77T.

The reactive torque at point (B) is = 0.23T.

Explanation of Solution

Given information:

       Figure (1) shows the systematic diagram of the system:

  Mechanics of Materials (MindTap Course List), Chapter 3, Problem 3.8.18P , additional homework tip  1

       Figure-(1)

The length of first pipe is 2.5m, the length of second pipe is 1.5m, thickness of flange is 14mm, diameter of the flange bolt is 13mm, thickness of the base plate attached to the wall is 15mm, the diameter of the base plate bolt is 16mm, young’s modulus of first pipe is 110GPa, young’s modulus of second pipe is 73GPa, poison ratio of first pipe is 0.33, poison ratio of second pipe is 0.25, cross sectional area of the first pipe is 1500mm2, cross sectional area of second pipe is (3/5)A1, the outer diameter of first pipe is 60mm, the outer diameter of second pipe is equal to inner diameter of first pipe and the bolt radius is 64mm.

Write the expression for cross sectional area of the pipe (1).

  A1=π4(d12d22)...... (I)

Here, the cross sectional area of the pipe (1) is A1, outside diameter of the pipe is d1and inner diameter is d2.

Write the expression for cross sectional area of the pipe (2).

  A2=π4(d22d32)...... (II)

Here, the cross sectional area of the pipe (2) is A2, outside diameter of the pipe is d2and inner diameter is d3.

Write the expression for the thickness of the pipe (1).

  t1=d1d22...... (III)

Here, the thickness of the pipe (1) is t1.

Write the expression for the thickness of the pipe (2).

  t2=d2d32...... (III)

Here, the thickness of the pipe (2) is t2.

Write the expression for polar moment of inertia for pipe (1).

  Ip1=π32(d14d24)..... (IV)

Here, the polar moment of inertia of pipe (1) is IP1.

Write the expression for polar moment of inertia for pipe (2).

  Ip2=π32(d24d34)..... (V)

Here, the polar moment of inertia of pipe (2) is IP2.

Write the expression for modulus of rigidity for pipe (1).

  G1=E12(1+v1)...... (VI)

Here, modulus of rigidity for pipe (1) is G1, young’s modulus of pipe is E1and poison ratio is v1.

Write the expression for modulus of rigidity for pipe (2).

  G2=E22(1+v2)...... (VII)

Here, modulus of rigidity for pipe (2) is G2, young’s modulus of pipe is E2and poison ratio is v2.

       Figure-(2) shows free body diagram of the system with internal torque and reactions at the end:

  Mechanics of Materials (MindTap Course List), Chapter 3, Problem 3.8.18P , additional homework tip  2

       Figure-(2)

Write the expression of equilibrium condition for figure (2)

  R1+R2=T...... (VIII)

Here, the reaction torque acting at the point (A) is R1, reaction torque acting at the point (B) is R2and the torque applied at x=Lis T.

Write the expression for torsional flexibility of pipe (1)

  fT1=L1G1Ip1...... (IX)

Here, the torsional rigidity of pipe (1) is fT1.

Write the expression for torsional flexibility of pipe (2)

  fT2=L2G2Ip2...... (X)

Here, the torsional rigidity of pipe (1) is fT2.

Write the expression for angle of twist for pipe (1)

  ϕ1=T×fT1

Here, the angle of twist for pipe (1) is ϕ1.

Write the expression for angle of twist for pipe (2)

  ϕ2=R2(fT1+fT2)fT2

Here, the angle of twist for pipe (1) is ϕ2.

Since, the angle of twist in pipe (1) and pipe (2) are opposite in direction

  ϕ1=ϕ2..... (XI)

Substitute, T×fT1for ϕ1and R2(fT1+fT2)fT2for ϕ2in Equation (XI)

  T×fT1=(R2(fT1+fT2)fT2)R2=TfT1fT1+fT2...... (XII)

Calculation:

Substitute 1500mm2for A1and 60mmfor d1in Equation (I).

  1500mm2=π4((60mm)2d22)1909.85mm2=((60mm)2d22)1909.85mm2(60mm)2=d2241.11mm=d2

Substitute (3/5)A1for A2and 41.11mmfor d2in Equation (II).

  (35)1500mm2=π4((41.11mm)2d32)1145.9mm2=((41.11mm)2d32)1145.9mm2(41.11mm)2=d3223.33mm=d3

Substitute 60mmfor d1and 41.11mmfor d2in Equation (III).

  t1=(60mm)(41.11mm)2=9.945mm

Substitute 41.11mmfor d2and 23.33mmfor d3in Equation (IV).

  t1=(41.11mm)(23.33mm)2=8.89mm

Substitute 60mmfor d1, 41.11mmfor d2in Equation (V)

  Ip1=π32((60mm)4(41.11mm)4)=0.098((12960000mm4)(2856208.4mm4))=991.91×103mm4

Substitute 41.11mmfor d2, 23.33mmfor d3in Equation (V)

  Ip2=π32((41.11mm)4(23.33mm)4)=0.098((2856208.4mm4)(296250.4mm4))=251.36×103mm4

Substitute 110GPafor E1, 0.33for v1in Equation (VI)

  G1=110GPa2(1+0.33)=110GPa(109Pa1GPa)2.66=0.414×105N/mm2

Substitute 73GPafor E2, 0.25for v2in Equation (VII)

  G2=73GPa2(1+0.25)=73GPa(109Pa1GPa)2.5=0.292×105N/mm2

Substitute 2500mmfor L1, 0.414×105N/mm2for G1and 991.91×103mm4for IP1in Equation (IX)

  fT1=2500mm0.414×105N/mm2(991.91×103mm4)=2500mm4.106×1010=6.088×108

Substitute 1500mmfor L1, 0.292×105N/mm2for G1and 251.36×103mm4for IP2in Equation (X)

  fT2=1500mm0.292×105N/mm2(251.36×103mm4)=2500mm7.339×1010=20.437×108

Substitute 6.088×108for fT1and 20.437×108for fT2in equation (XII).

  R2=T(6.088×108)(6.088×108)+(20.437×108)R2=0.23T

Substitute 0.23Tfor R2in Equation (VIII).

  R1+0.23T=TR1=T+0.23TR1=0.77T

Conclusion:

The reactive torque at point (A) is 0.77T.

The reactive torque at point (B) is 0.23T.

(b)

Expert Solution
Check Mark
To determine

The maximum load variable

Answer to Problem 3.8.18P

The maximum load variable

Explanation of Solution

Given information:

Permissible torsional stress in two pipes is 65MPa.

Write the expression for maximum load in pipe (1).

  T1=τaIp1(d12)...... (XIII)

Here, maximum load in the pipe (1) is T1and the permissible stress is τa.

Write the expression for maximum load in pipe (1).

  T2=τaIp2(d22)...... (XIV)

Here, maximum load in the pipe (2) is T2.

Write the expression for maximum load variable in pipe (1).

  Tmax=T1(fT2fT1+fT2)...... (XV)

Here, the maximum load variable is Tmax.

Calculation:

Substitute 65MPafor τa, 991.91×103mm4for IP1, and 60mmfor d1in equation (XIII).

  T1=65MPa(991.91×103mm4)(60mm2)=65MPa(991.91×103mm4)(30mm)=2149×103Nmm(1m1000mm)(1kN1000N)=2.149kNm

Substitute 65MPafor τa, 251.36×103mm4for IP1, and 41.11mmfor d1in equation (XIII).

  T2=65MPa(251.36×103mm4)(41.11mm2)=65MPa(991.91×103mm4)(20.55mm)=795×103Nmm(1m1000mm)(1kN1000N)=0.795kNm

Since, the pipe (1) is greater, the maximum load variable in pipe (1).

Substitute 2.149kNmfor T1, 6.088×108for fT1and 20.437×108for fT2in Equation (XV).

  Tmax=2.149kNm(6.088×1086.088×108+20.437×108)=2.149kNm(0.77)=2.79kNm

Conclusion:

The maximum load variable is = 2.79kNm.

(c)

Expert Solution
Check Mark
To determine

The torsional moment diagrams.

The torsional displacement diagrams.

The maximum angle of twist.

Answer to Problem 3.8.18P

The maximum angle of twist is 7.49°.

Explanation of Solution

Given information:

       Figure (3) shows torsional moment diagram and torsional displacement diagram for the given system.

  Mechanics of Materials (MindTap Course List), Chapter 3, Problem 3.8.18P , additional homework tip  3

       Figure-(3)

Write the expression for maximum angle of twist

  ϕmax=[(fT1×fT2)(fT1+fT2)]×Tmax...... (XVI)

Here, maximum angle of twist is = ϕmax.

Calculation:

Substitute 2.79kNmfor Tmax, 6.088×108for fT1and 20.437×108for fT2in Equation (XVI).

  ϕmax=[((6.088×108)×(20.437×108))((6.088×108)+(20.437×108))]×2.79kNm=[(1.2442×1014)((26.525×108))]×2.79kNm(1000N1kN)=0.1308(180π)=7.49°

Conclusion:

The maximum angle of twist is = 7.49°.

(c)

Expert Solution
Check Mark
To determine

The maximum torque for limiting value shear and bearing stress of base plate and flange bolts.

Answer to Problem 3.8.18P

The maximum torque for limiting value shear and bearing stress of base plate and flange bolts is = 2480727.27N.

Explanation of Solution

Given information:

The permissible shear stress for all bolts is 45MPa, and permissible bearing stress is 90MPa.

Write the expression for cross -sectional area of the base plate bolt.

  Acbb=π4dbb2...... (XVII)

Here, the cross-sectional area of the base plate bolt is Acbb, the diameter of the base plate dbb.

Write the expression for cross sectional area of the flange plate bolt.

  Acbf=π4dbf2...... (XVII)

Here, the cross-sectional area of the flange plate bolt is Abf, the diameter of the flange plate dbf.

Write the expression for maximum shear load in base plate bolt.

  Tmax,bb=nb×r(τallow×Acbb)(fT2fT1+fT2)...... (XVIII)

Here, the maximum shear load in base plate is Tmax,bband number of base plate bolts is nbWrite the expression for maximum shear load in flange plate bolt.

  Tmax,bf=nf×r(τallow×Acbf)(fT2fT1+fT2)...... (XIX)

Here, the maximum shear load in flange plate is Tmax,bfand number of base plate bolts is nf.

Write the expression for area of the base plate bolt.

  Abb=dbb×tb...... (XX)

Here, the area of base plate bolt is Abband the thickness of the is tb.

Write the expression for area of the flange plate bolt.

  Abf=dbf×tf...... (XXI)

Here, the area of base plate bolt is Abfand the thickness of the is tf.

Write the expression for bearing load on base plate

  Tmax,bb=nb×r(σallow×Abb)(fT2fT1+fT2)...... (XXII)

Here, the maximum shear load in base plate is Tmax,bband number of base plate bolts is nbWrite the expression for maximum shear load in flange plate bolt.

  Tmax,bb=nf×r(σallow×Abf)(fT2fT1+fT2)...... (XXIII)

Here, the maximum shear load in flange plate is Tmax,bfand number of base plate bolts is = nf.

Calculation:

Substitute 16mmfor dbbin Equation (XVII).

  Abb=π4(16mm)2=(0.785)(256mm2)=200.96mm2

Substitute 13mmfor dbfin Equation (XVIII).

  Abf=π4(13mm)2=(0.785)(169mm2)=132.65mm2

Substitute 200.96mm2for Abb, 64mmfor r, 4for nb, 45MPafor τallow, 6.088×108for fT1and 20.437×108for fT2in Equation (XVIII)

  Tmax,bb=4×64mm((45MPa)×(200.96mm2))(20.437×108(6.088×108)+(20.437×108))=4×64mm((45MPa)×(200.96mm2))0.77=2.315×106N0.77=3006493.5N

Substitute 132.65mm2for Abf, 64mmfor r, 5for nb, 45MPafor τallow, 6.088×108for fT1and 20.437×108for fT2in Equation (XVIII)

  Tmax,bb=5×64mm((45MPa)×(132.65mm2))(20.437×108(6.088×108)+(20.437×108))=5×64mm((45MPa)×(132.65mm2))0.77=1.910×106N0.77=2480727.27N

Substitute 16mmfor dbband 15mmfor tin Equation (XX).

  Abb=16mm×15mm=240mm2

Substitute 13mmfor dbband 14mmfor tin Equation (XXI).

  Abf=13mm×14mm=182mm2

Substitute 240mm2for Abb, 64mmfor r, 4for nb, 90MPafor τallow, 6.088×108for fT1and 20.437×108for fT2in Equation (XVIII)

  Tmax,bb=4×64mm((90MPa)×(240mm2))(20.437×108(6.088×108)+(20.437×108))=4×64mm((90MPa)×(240mm2))0.77=5.529×106N0.77=7181298.701N

Substitute 182mm2for Abf, 64mmfor r, 5for nb, 90MPafor τallow, 6.088×108for fT1and 20.437×108for fT2in Equation (XVIII)

  Tmax,bf=5×64mm((90MPa)×(182mm2))(20.437×108(6.088×108)+(20.437×108))=5×64mm((90MPa)×(182mm2))0.77=5.184×106N0.77=6732467.532N

Conclusion:

The maximum torque for limiting value shear and bearing stress of base plate and flange bolts is = 2480727.27N.

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Chapter 3 Solutions

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