Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
9th Edition
ISBN: 9781337594301
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Chapter 8, Problem 8.5.7P

A cylindrical pressure vessel having a radius r = 14 in. and wall thickness t = 0,5 in, is subjected to internal pressure p = 375 psi, In addition, a torque T = 90 kip-ft acts at each end of the cylinder (see figure),

(a) Determine the maximum tensile stress ctniXand the maximum in-plane shear stress Tmjv in the wall of the cylinder.

(b) If the allowable in-plane shear stress is 4.5 ksi, what is the maximum allowable torque T\

(c) If 7 = 150 kip-ft and allowable in-plane shear and allowable normal stresses are 4.5 ksi and 11.5 ksi, respectively, what is the minimum required wall thicknessChapter 8, Problem 8.5.7P, A cylindrical pressure vessel having a radius r = 14 in. and wall thickness t = 0,5 in, is subjected

(a)

Expert Solution
Check Mark
To determine

The maximum tensile stress σmax and the maximum in plane shear stress τmax in the wall of the cylinder.

Answer to Problem 8.5.7P

The maximum tensile stress is 11.09ksi and the maximum shear stress is 3.21ksi

Explanation of Solution

Given Information:

Type of vessel = cylindrical

Radius = 14in

Internal pressure = 375psi

Wall thickness = 0.5in

Torque = 90kipft at each end of cylinder

Concept used:

  Longitudinal stress σx=pr2t

  hoop stress = prt

Where p: Pressure, r=radius , t=thickness

Calculation:

Use the following relation to find the longitudinal stress.

  σx=pr2t

Here, wall thickness is t , radius is r , and internal pressure is ρ .

Substitute 375psi for ρ,14 in for r and 0.5 in. for t .

  σx=375×142×0.5=5,250psi

Use the following relation to find the hoop stress.

  σy=Prt

Substitute 375psifor ρ,14

  r,and 0.5 in. for t

  σy=375×140.5=10,500psi

Use the following relation to find the shear stress.

  τxy=TrIp.............(1)

Here, torque is T and polar moment of inertia is IP .

Express the value of IP .

  IP=π2[r4(rt)4]

Substitute 14 in. for r, and 0.5 in. for t

  Ip=π2[144( 13.5)4]=8,169.63 in4

From equation (1) , substitute 8,169.63 in4 for IP , 14 in. for r , and 90 k-ft for T .

  τxy=90( kft)×14in8,169.63  in4=90( kft)× 10 3 k× 12in 1ft×14in8,169.63in1,851psi

Find the maximum stress (σmax) and the maximum in plane shear stress (τmax) in the wall of the cylinder.

Use the following relation to find the principal stresses.

  σ12=σx+σy2±( σ x σ y 2 )2+τxy2

Substitute 1,851psifor τxy,10,500psiforσy , and 5,250psifor σx

  σ1,2=( 5,250+10,5002)± ( 5,25010,500 2 )2+ ( 1,851 )2=( 5,250+10,5002)±6,890,625+3,426,201=7,875±3,212

Calculate the principal stress, σ2

  σ2=7,8753,212=4,663psi×ksi1000ksi=4,663ksi

As σ1 is greater than σ2 , the maximum stress σmax is given by σ1 is 11.09ksi .

Use the following relation to find the maximum in plane shear stress

  τmax=(σ1σ22)

Substitute 11,087psifor σ1and 4,663psi for σ2

  τmax=11,0874,6632=3,212psi×ksi1000psi=3.21ksi

Hence, the maximum shear stress is 3.21ksi .

Conclusion

The values are found by the concept of longitudinal stress and hoop stress.

(b)

Expert Solution
Check Mark
To determine

Maximum torque.

Answer to Problem 8.5.7P

The maximum torque is 178 kip-ft

Explanation of Solution

Given Information:

  τmax=4.5ksiσx=5250psiσy=10500psiIp=8169.63in4r=14in

Concept used:

  τmax=( σ x σ y 2 )2+( T max ×r I p )2

Calculation:

Find the maximum torque (Tmax) if maximum in plane shear stress is τmax=4.5ksi .

  τmax=( σ x σ y 2 )2+( T max ×r I p )2

Substitute 10,500psifor σy,5,250psifor σx,8,169.63in4 for Ip,4.5ksifor τmax . Are 14 in. for r .

  4.5ksi= ( 5,25010,500 2 )2+( ( T max ) 2 × ( 14 ) 2 ( 8,169.63 ) 2 )(4.5ksi× 10 3 psi ksi)2=( 5,25010,500 2)2+( ( T max ) 2 × ( 14 ) 2 ( 8,169.63 ) 2 )20.25×106=(6.89× 106)+(T max2×2.937× 10 6)Tmax2=( 20.256.89)× 1062.937× 10 6=4.549×1012Tmax=2.133×106lbin

Conclusion:

The maximum torque is calculated by equating 1p , Tmax, r, etc.

(C)

Expert Solution
Check Mark
To determine

What is the minimum required wall thickness.

Answer to Problem 8.5.7P

The minimum required wall thickness is 0.519psi

Explanation of Solution

Given Information:

  τall=4.5ksiσall=11.5ksihoop stress = prtLongitudinalstress=pr2t

Concept used:

  σa=pr2t+prt2+( pr 2t + pr t 2 )2+[ T.r π 2 [ r 4 ( rt ) 4 ]]2

Calculation:

  σa=pr2t+prt2+( pr 2t + pr t 2 )2+[ T.r π 2 [ r 4 ( rt ) 4 ]]2

Here, allowable in plane shear stress is τa and allowable normal stress is σa .

Substitute 11.5ksifor σa,14 in. for r,0.5 in. for t,375psifor p , and 150kipft for T .

  σa= 375×14 2t+ 375×14t2+( 375×14 2t + 375×14 t 2 )+ ( 1.8× 10 6 ×14×2 π[ ( 14 4 ( 14t ) 4 )] )2= 2,625t+ 5,250t2+ ( 2,6255,250 2t )2+ ( A )2

Express the value of A form equation (2) .

  A=BC..............(3)

Here, assumed

Variable are A,B and C respectively.

Evaluate the value of B from equation (2) .

  B=1.8×106×14×2π=16.0428×106

Evaluate the value of C from Equation (2)

  C=144(14t)4=( 14 2)2( [ 14t] 2)2=( 142+ ( 14t )2)( 142 ( 14t )2)=(196+196+t228t)(196196t2+28t)=(392+t228t)(28tt2)=10,976t392t2+28t3t4784t2+28t3=t4+56t31,176t2+1,0976t

Substitute t4+56t31,176t2+1,0976t for C and 16.0428×106 for B .

  A=16.0428×106(t4+56t31,176t2+1,0976t)

Substitute the value of A and σa in equation (2) .

  11.5×103= 2,625t+ 5,250t2+ ( 2,6255,250 2t )2+ ( 16.0428× 10 6 ( t 4 +56 t 3 1,176 t 2 +1,0976t ) )2=3,937.5t+ ( 3,312.5 t )2+ [ 16.0428× 10 6 t( t 3 +56 t 2 1,176t+10,976 ) ]2

Use the trial and error method to find the thickness t value.

Trial 1

Substitute t=0.50in . In Equation (2) .

  σallow=7,875+6,890,625+9,514,743.15=7,875+4,050.35=11,925.35psi

Trial ll

Substitute t=0.55in . In Equation (2)

  σallow=7,159.09+5,694,731.405+7,948,628.98=7,159.09+3,693.69=10,852.78psi

Trial lll

Substitute t=0.515in . In equation (2)

  σallow=7,645.63+6,495,074.94+8,997,596.67=7,645.63+3,936.07=11,581.7psi

Substitute t=0.519in . In Equation (2) .

  σallow=7,586.71+6,395,343.981+8,867,080.985=7,586.71+3,906.71=1,1493.4psi

Express the allowable in plane shear stress.

  τallow=( pr 2t pr t 2 )2+[ T.r π 2 [ r 4 ( rt ) 4 ]]2

Substitute 14 in for r,0.5 . For t,375psifor p , and 150kipft for T .

  τallow=( 375×14 2t 375×14 t 2 )+ ( 1.8× 10 6 ×14×2 π[ ( 14 2 ( 14t ) 4 )] )24,500= [ 1312.5 t ]2+ [ 16.0428× 10 6 t 4 +56 t 3 117 t 2 +10,976t ]2

Substitute 0.519in . For t in equation (4) .

  4,500= [ 1312.5 0.519in ]2+[ 16.0428× 10 6 ( 0.519in ) 4 +56 ( 0.519in ) 3 1176 ( 0.519in ) 2 +10,976( 0.519in )]τallow=3,906.71psi

As τallow1t the thickness value is to be increased to get τallow=4,500psi .

Hence, from the trial and error method the minimum wall thickness tmin is 0.519in

Conclusion:

The minimum thickness is calculated by trial and error method.

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Chapter 8 Solutions

Bundle: Mechanics Of Materials, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card

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