ch14
Short Description
timothy pilophot mechanics of materials 3rd edition solution manual...
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P14.1 Determine the normal stress in a ball (Figure P14.1), which has an outside diameter of 185 mm and a wall thickness of 3 mm, when the ball is inflated to a gage pressure of 80 kPa.
FIGURE P14.1
Solution D 185 mm t 3 mm d 185 mm 2(3 mm) 179 mm
a
pd 4t
(0.110 MPa)(179 mm) 4(3 mm)
1.193 MPa
Ans.
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P14.2 A spherical gas-storage tank with an inside diameter of 2 1 ft is being constructed to store gas under an internal pressure of 160 psi. The Th e tank will be constructed from steel that has a yield strength of 50 ksi. If a factor of safety of 3.0 with respect to the yield strength is required, determine the minimum wall thickness required for the spherical tank.
Solution allow a
Y
FS pd 4t
50 ksi 3.0
t
16.667 ksi pd 4 allow
(160 psi)(21 ft)(12 in./ft) 4(16, 667 psi)
0.605 in.
Ans.
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P14.3 A spherical gas-storage tank with an inside diameter of 9 m is being constructed to store gas under an internal pressure of 1.60 MPa. The tank will be constructed from steel that ha s a yield strength of 340 MPa. If a factor of safety of 3.0 with respect to the yield strength is required, determine the minimum wall thickness required for the spherical tank.
Solution allow a
Y
FS pd 4t
340 MPa 3.0
t
113.333 MPa
pd 4 allow
(1.60 MPa)(9 m)(1,000 mm/m) 4(113.333 MPa)
31.8 mm
Ans.
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P14.4 A spherical pressure vessel has an inside diameter of 6 m and a wall thickness of 15 mm. The vessel will be constructed from steel [ E = 200 GPa; = 0.29] that has a yield strength of 340 MPa. If the internal pressure in the vessel is 1,750 kPa, determine (a) the normal stress in the vessel wall, (b) the factor of safety with respect to the yield strength, (c) the normal strain in the sphere and (d) the increase in the outside diameter of the vessel.
Solution (a) Normal stress in the vessel wall pd (1.750 MPa)(6,000 mm) 175.0 MPa a 4t 4(15 mm) (b) Factor of safety with respect to the yield strength 340 MPa 1.943 FS Y a 175 MPa (c) Normal strain in the sphere 1 x ( x y ) E 1 [175 MPa (0.29)(175 MPa)] 621.25 10 6 mm/mm 621 με 200,000 MPa
Ans.
Ans.
Ans.
(d) Increase in outside diameter
D D (621.25 106 mm/mm) 6,000 mm 2(15 mm) 3.75 mm
Ans.
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P14.5 The normal strain measured on the outside surface of a spherical pressure vessel is 515 . The sphere has an outside diameter of 72 in. and a wall thickness of 0.50 in., and it will be fabricated from an aluminum alloy [ E = 10,000 ksi; = 0.33]. Determine (a) the normal stress in the vessel wall and (b) the internal pressure in the vessel.
Solution (a) Normal stress in the vessel wall E x ( x y ) 1 2 10,000 ksi [(515 10 6 in./in.) (0.33)(515 10 6 in. /in.)] 2 1 (0.33)
7.687 ksi 7.69 ksi
Ans.
(b) Internal pressure D 72 in.
t 0.50 in. d 72 in. 2(0.50 in.) 71 in.
a
pd 4t
p
4 at d
4(7.687 ksi)(0.50 in.) 71 in.
0.216523 ksi 217 psi
Ans.
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P14.6 A typical aluminum-alloy scuba diving tank is shown in Figure P14.6. The outside diameter of the tank is 175 mm and the wall thickness is 12 mm. If the air in the tank is pressurized to 18 MPa, determine: (a) the longitudinal and hoop stresses in the wall of the tank. (b) the maximum shear stress in the plane of the cylinder wall. (c) the absolute maximum shear stress on the outer surface of the cylinder wall. FIGURE P14.6
Solution (a) Longitudinal and hoop stresses D 175 mm
t 12 mm d 175 mm 2(12 mm) 151 mm
long
pd
hoop
pd
4t 2t
(18 MPa)(151 mm)
(18 MPa)(151 mm)
4(12 mm) 2(12 mm)
56.625 MPa 56.6 MPa
Ans.
113.250 MPa 113.3 MPa
Ans.
(b) Maximum shear stress in the plane of the cylinder wall If the longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: x 56.625 MPa, y 113.250 MPa, xy 0 MPa
Since the shear stresses on the x and y faces (i.e., in the plane of the cylinder wall) are zero, these normal stresses are, by definition, principal stresses. Therefore,
p1 113.3 MPa and p 2 56.6 MPa The maximum in-plane shear stress can be computed from Eq. (12.16): p1 p 2 113.250 MPa 56.625 MPa 28.313 MPa 28.3 MPa max 2 2
Ans.
(c) Absolute maximum shear stress on the outer surface of the cylinder wall The outer surface of the cylinder wall is in a state of plane stress since the pressure acting on the outer surface of the cylinder is simply atmospheric pressure (i.e., gage pressure = 0). Therefore, z = p3 = 0. Since p1 and p2 are both positive, p1 113.250 MPa Ans. abs max 56.6 MPa 2 2
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P14.7 A cylindrical boiler with an outside diameter of 2.75 m and a wall thickness of 32 mm is made of a steel alloy that has a yield stress of 340 MPa. Determine: (a) the maximum normal stress produced by an internal pressure of 2.3 MPa. (b) the maximum allowable pressure if a factor of safety of 2 .5 with respect to yield is required.
Solution (a) Maximum normal stress D 2,750 mm
hoop
pd 2t
t 32 mm
(2.30 MPa)(2,686 mm) 2(32 mm)
d 2,750 mm 2(32 mm) 2,686 mm
96.5 MPa
(b) Maximum allowable pressure if FS = 2.5 340 MPa allow Y 136.0 MPa FS 2.5 pd 2 t 2(136.0 MPa)(32 mm) allow hoop p allow 3.24 MPa 2t d 2,686 mm
Ans.
Ans.
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P14.8 When filled to capacity, the unpressurized storage tank shown in Figure P14.8 contains water to a height of h = 30 ft. The outside diameter of the tank is 12 ft and the wall thickness is 0.375 in. Determine the maximum normal stress and the absolute maximum shear stress on the outer surface of the tank at its base. 3 (Weight density of water = 62.4 lb/ft .)
FIGURE P14.8
Solution Water pressure p h (62.4 lb/ft 3 )(30 ft) 1,872.0 lb/ft 2 13.00 psi Hoop stress D (12 ft)(12 in./ft) 144 in.
hoop
pd 2t
(13.00 psi)(143.25 in.) 2(0.375 in.)
t 0.375 in.
d 144 in. 2(0.375 in.) 143.25 in.
2,482.992 psi 2,480 psi
Ans.
Principal stresses p1 hoop 2,482.992 psi
p2 long 0 psi
(since the tank is unpressurized)
Maximum shear stress The outer surface of the tank wall is in a state of plane stress since the pressure acting on the outer surface is simply atmospheric pressure (i.e., gage pressure = 0). Therefore, z = p3 = 0. Therefore, p1 2,482.992 psi Ans. 1,241 psi abs max 2 2
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P14.9 A tall open-topped standpipe (Figure P14.9) has an inside diameter of 2,750 mm and a wall thickness of 6 mm. The standpipe contains water, 3 which has a mass density of 1,000 kg/m . (a) What height h of water will produce a circumferential stress of 16 MPa in the wall of the standpipe? (b) What is the axial stress in the wall of the standpipe due to the water pressure?
FIGURE P14.9
Solution Longitudinal and hoop stresses pd p(2,750 mm) hoop 16 MPa 2t 2(6 mm)
p 69.818 103 MPa (a) Height h of water p gh 69.818 10 3 MPa
h
69.818 103 N/m 2 (1,000 kg/m3 )(9.81 m/s 2 )
7.122684 m 7.12 m
Ans.
(b) Axial stress in the wall of the standpipe due to water pressure Since the standpipe is open to the atmosphere at its upper end, the fluid pressure will not create stress in the longitudinal direction of the standpipe; therefore, long 0 Ans.
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P14.10 The pressure tank in Figure P14.10/11 is fabricated from spirally wrapped metal plates that are welded at the seams in the orientation shown where = 40°. The tank has an inside diameter of 480 mm and a wall thickness of 8 mm. Determine the largest gage pressure that can be used inside the tank if the allowable normal stress perpendicular to the weld is 100 MPa and the allowable shear stress parallel to the weld is 25 MPa.
FIGURE P14.10/11
Solution The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: pd pd x , y , xy 0 4t 2t The weld is oriented at 40° as shown; however, the angle required for the stress transformation equations is the angle normal to the weld. Thus, = 40° + 90° = 130° (or = 40° − 90° = −50°). Using this value of , the normal stress transformation equation [Eq. (12-3)] can be used to compute the normal stress perpendicular to the weld: n x cos 2 y sin 2 2 xy sin cos
pd
pd
4t 4t
cos 2 (130)
pd
cos 2 (130)
pd
2t 2t
sin 2 (130) 2(0 MPa)sin(130) cos(130) sin 2 (130)
The normal stress magnitude perpendicular to the weld n must not exceed 100 MPa; thus, 100 MPa
pd 4t
cos 2 (130)
pd 2t
sin 2 (130)
pd 1
cos 2 (130) sin 2 (130) 2t 2
p(480 mm) 0.413176 2(8 mm)
2
0.586824
(23.802361) p Based on the allowable normal stress, p 4.2013 MPa
(a)
Similarly, the shear stress transformation equation [Eq. (12-4)] can be used to compute the shear stress parallel to the weld:
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nt ( x y )sin cos xy (cos 2 sin 2 )
pd pd sin(130) cos(130) (0 MPa )[cos 2 (130 ) sin 2 (130 )] 2t 4t pd pd sin(130)cos(130) 4 t 2 t The shear stress parallel to the weld nt must not exceed a magnitude of 25 MPa; thus,
pd
25 MPa
4t
pd
sin(130) cos(130)
2t
pd 1
1 sin(130)cos(130) 2t 2
p(480 mm) 1 (0.492404) 2(8 mm) 2
7.386058 p Based on the allowable shear stress, p 3.3848 MPa
(b)
Compare the results in Eqs. (a) and (b) to find that the maximum allowable gage pressure is
pallow 3.38 MPa
Ans.
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P14.11 The pressure tank in Figure P14.10/11 is fabricated from spirally wrapped metal plates that are welded at the seams in the orientation shown where = 40°. The tank has an inside diameter of 720 mm and a wall thickness of 8 mm. For a gage pressure of 2.15 MPa, determine (a) the normal stress perpendicular to the weld and (b) the shear stress parallel to the weld.
FIGURE P14.10/11
Solution (a) Normal stress perpendicular to the weld pd (2.15 MPa)(720 mm) long 48.375 MPa 4t 4(8 mm)
hoop
pd 2t
(2.15 MPa)(720 mm) 2(8 mm)
96.750 MPa
The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: x 48.375 MPa, y 96.750 MPa, xy 0 MPa The weld is oriented at 40° as shown; however, the angle required for the stress transformation equations is the angle normal to the weld. Thus, = 40° + 90° = 130° (or = 40° − 90° = −50°). Using this value of , the normal stress transformation equation [Eq. (12-3)] can be used to compute the normal stress perpendicular to the weld: n x cos 2 y sin 2 2 xy sin cos
(48.375 MPa)cos 2 (130 ) (96.750 MPa)sin 2 (130 ) 2(0 MPa)sin(130 )cos(130 ) 76.763 MPa 76.8 MPa (T)
Ans.
(b) Shear stress parallel to the weld Similarly, the shear stress transformation equation [Eq. (12-4)] gives nt : nt ( x y )sin cos xy (cos 2 sin 2 )
[(48.375 MPa) (96.750 MPa)]sin(130 )cos(130 ) (0 MPa)[cos 2(130 ) sin 2(13 0)] 23.820 MPa 23.8 MPa
Ans.
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P14.12 The pressure tank in Figure P14.12/13 is fabricated from spirally wrapped metal plates that are welded at the seams in the orientation shown where = 40°. The tank has an inside diameter o 1,800 mm and a wall thickness of 12 mm. For a gage pressure of 1.75 MPa, determine (a) the normal stress perpendicular to the weld and (b) the shear stress parallel to the weld.
FIGURE P14.12/13
Solution (a) Normal stress perpendicular to the weld pd (1.75 MPa)(1,800 mm) long 65.625 MPa 4t 4(12 mm)
hoop
pd 2t
(1.75 MPa)(1,800 mm) 2(12 mm)
131.250 MPa
The longitudinal axis of the cylinder is defined as the y axis and the circumferential direction is defined as the x axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: x 131.250 MPa, y 65.625 MPa, xy 0 MPa The weld is oriented at 40° as shown. Relative to the positive x axis, this orientation is defined by an angle of = 180° − 40° = 140°. Using this value of , the normal stress transformation equation [Eq. (12-3)] can be used to compute the normal stress perpendicular to the weld: n x cos 2 y sin 2 2 xy sin cos
(131.250 MPa)cos 2 (140 ) (65.625 MPa)sin 2(140 ) 2(0 MPa)sin(140 )cos(140 ) 104.135 MPa 104.1 MPa (T)
Ans.
(b) Shear stress parallel to the weld Similarly, the shear stress transformation equation [Eq. (12-4)] gives nt : nt ( x y )sin cos xy (cos 2 sin 2 )
[(131.250 MPa) (65.625 MPa)]sin(140 )cos(140 ) (0 MPa)[cos 2(140 ) sin 2(140 )] 32.314 MPa 32.3 MPa
Ans.
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P14.13 The pressure tank in Figure P14.12/13 is fabricated from spirally wrapped metal plates that are welded at the seams in the orientation shown where = 55°. The tank has an inside diameter of 60 in. and a wall thickness of 0.25 in. Determine the largest allowable gage pressure if the allowable normal stress perpendicular to the weld is 12 ksi and the allowable shear stress parallel to the weld is 7 ksi.
FIGURE P14.12/13
Solution The longitudinal axis of the cylinder is defined as the y axis and the circumferential direction is defined as the x axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: pd pd x , y , xy 0 2t 4t The weld is oriented at 55° as shown. Relative to the positive x axis, this orientation is defined by an angle of = 180° − 55° = 125°. Using this value of , the normal stress transformation equation [Eq. (12-3)] can be used to compute the normal stress perpendicular to the weld: n x cos 2 y sin 2 2 xy sin cos
pd 2t
pd 2t
cos 2 (125)
pd
cos 2 (125)
pd
4t 4t
sin 2 (125) 2(0 MPa)sin(125) cos(125) sin 2 (125)
The normal stress magnitude perpendicular to the weld n must not exceed 12 ksi; thus, 12 ksi
pd 2t
cos 2 (125)
pd 4t
sin 2 (125)
pd
1 cos 2 (125) sin 2 (125) 2t 2
p(60 in.) 0.671010 0.328990 2(0.25 in.) 2
(79.739396) p Based on the allowable normal stress, p 0.150490 ksi
(a)
Similarly, the shear stress transformation equation [Eq. (12-4)] can be used to compute the shear stress parallel to the weld: nt ( x y )sin cos xy (cos 2 sin 2 )
pd pd 2 2 sin(125) cos(125) (0 MPa )[cos (125 ) sin (125 )] 4t 2t pd pd sin(125)cos(125) 4t 2t Excerpts from this work may be reproduced by instructors for distribution on a not -for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work b eyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
The shear stress parallel to the weld nt must not exceed a magnitude of 7 ksi; thus,
pd
7 ksi
2t
pd
sin(125) cos(125)
4t
pd
1 1 sin(125)cos(125) 2t 2
p(60 in.) 1 ( 0.469846) 2(0.25 in.) 2
28.190779 p Based on the allowable shear stress, p 0.248308 ksi
(b)
Compare the results in Eqs. (a) and (b) to find that the maximum allowable gage pressure is
pallow 150.5 psi
Ans.
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P14.14 A strain gage is mounted to the outer surface of a thin-walled boiler as shown in Figure P14.14. The boiler has an inside diameter of 1,800 mm and a wall thickness of 20 mm, and it is made of stainless steel [ E = 193 GPa; = 0.27]. Determine: (a) the internal pressure in the boiler when the strain gage reads 190 . (b) the maximum shear strain in the plane of the boiler wall. (c) the absolute maximum shear strain on the outer surface of the boiler.
FIGURE P14.14
Solution (a) Internal pressure in the boiler The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: pd pd x , y , xy 0 4t 2t
From the generalized Hooke’s Law equations for plane stress, the normal strains on the outer surface of the boiler can be computed from Eqs. (13.21): x
1 E
( x y )
1 pd
pd E 4t 2t
Thus,
d
d 2t 4t
E x p
(193,000 MPa)(190 10 6 in./in.) 3.543 MPa 3.54 MPa 1,800 mm 1 0.27 2t 2 2(20 mm) 2
E x p d 1
Ans.
(b) Maximum shear strain in the plane of the boiler wall −6 The strain in the longitudinal direction is given as x = 190×10 in./in. The strain in the circumferential direction (i.e., the y direction) can be expressed with the generalized Hoo ke’s Law equations as:
y
1 E
( y x )
1 pd
pd E 2t 4t
pd 1 2tE 2
From the pressure computed in part (a), the strain in the y direction is: (3.543 MPa)(1,800 mm) 0.27 1 714.565 106 mm/mm y 2(20 mm)(193, 000 MPa) 2 Since the longitudinal and hoop stresses are principal stresses, the corresponding strains are also principal strains. p1 y 714.565 10 6 mm/mm p 2 x 190 10 6 mm/mm Excerpts from this work may be reproduced by instructors for distribution on a not -for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work b eyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
The maximum shear strain in the plane of the boiler wall can be calculated from Eq. (13.12): max p1 p 2 714.565 10 6 mm/mm 190 10 6 mm/mm
524.565 106 r ad 525 μrad
Ans.
(c) Absolute maximum shear strain on the outer surface of the boiler The strain in the radial direction (i.e., the out-of-plane direction) can be expressed with the generalized Hooke’s Law equations as:
z
E
( x y )
pd
E 4t
pd
pd 1 1 2t 2tE 2
From the pressure computed in part (a), the strain in the z direction is: (0.27)(3.543 MPa)(1,800 mm) 1 1 334.565 106 mm/mm z 2(20 mm)(193,000 MPa) 2 The strain in the z direction is also a principal strain; therefore, p 3 z 334.565 106 mm/mm Since p1 is positive and p3 is negative, the absolute maximum shear strain is abs max 714.565 10 6 mm/mm ( 334.565 10 6 mm/mm)
1, 049.130 106 rad 1,049 μrad
Ans.
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P14.15 A closed cylindrical tank containing a pressurized fluid has an inside diameter of 830 mm and a wall thickness of 10 mm. The stresses in the wall of the tank acting on a rotated element have the values shown in Figure P14.15. What is the fluid pressure in the tank?
FIGURE P14.15
Solution Let the given stresses be designated as: x 51 MPa, y 66 MPa, xy 18 MPa The principal stress magnitudes can be computed from Eq. (12-12): p1, p 2
x y 2
2
x y 2xy 2
(51 MPa) (66 MPa) 2
2
(51 MPa) (66 MPa) (18 MPa) 2 2
58.50 MPa 19.50 MPa p1 78.0 MPa and p 2 39.0 MPa Since this is a cylindrical pressure vessel subjected to internal pressure only, we know that the principal stresses occur in the hoop and longitudinal directions. Thus, we can assert that: pd pd p1 hoop and p 2 long 2t 4t The internal pressure can be calculated from either expression: pd 78 MPa 2t
p
2(10 mm)(78 MPa) 830 mm
1.880 MPa
Ans.
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P14.16 A closed cylindrical vessel (Figure P14.16) contains a fluid at a pressure of 5.0 MPa. The cylinder, which has an outside diameter of 2,500 mm and a wall thickness of 20 mm, is fabricated from stainless steel [ E = 193 GPa; = 0.27]. Determine the increase in both the diameter and the length of the cylinder.
FIGURE P14.16
Solution D 2,500 mm long
pd
hoop
pd
4t
2t
t 20 mm
(5 MPa)(2,460 mm)
(5 MPa)(2,460 mm)
4(20 mm)
2(20 mm)
d 2,500 mm 2(20 mm) 2,460 mm
153.750 MPa 307.500 MPa
From the generalized Hooke’s law, the strain in the longitudinal direction is: 1 x ( x y ) E 1 ( long hoop ) E 1 153.750 MPa (0.27)(307.500 MPa) 193,000 MPa
366.451 10 6 mm/mm Therefore, the change in length of the cylinder is:
L x L (366.451 106 mm/mm)(6,000 mm) 2.20 mm
Ans.
Similarly, the strain in the circumferential direction is: 1 y ( y x ) E 1 ( hoop long ) E 1 307.500 MPa (0.27)(153.750 MPa) 193,000 MPa
1,378.174 106 mm/mm The change in diameter of the cylinder is:
D y D (1,378.174 106 mm/mm)(2,500 mm) 3.45 mm
Ans.
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P14.17 A strain gage is mounted at an angle of = 20° with respect to the longitudinal axis of the cylindrical pressure vessel shown in Figure P14.17/18. Th e pressure vessel is fabricated from aluminum [ E = 10,000 ksi; = 0.33], and it has an inside diameter of 48 in. and a wall thickness of 0.25 in. If the strain gage measures a normal strain of 470 , determine: (a) the internal pressure in the cylinder. (b) the absolute maximum shear stress on the outer surface of the cylinder. (c) the absolute maximum shear stress on the inner surface of the cylinder.
FIGURE P14.17/18
Solution (a) Internal pressure in the cylinder A strain transformation equation [Eq. (13.3)] n x cos2 y sin 2 xy sin cos
can be written for the normal strain in the direction of the strain gage: 470 με x cos 2 (20) y sin 2 (20) xy sin(20) cos(20) Since this is a cylindrical pressure vessel, the shear stress xy must equal zero, and hence, the shear strain xy must also equal zero. The strain transformation equation reduces to: 470 με 470 10 6in./in. x cos 2 (20 ) y sin 2 (20 ) Substitute Eqs. (13.21) for x and y to obtain an expression in terms of x and y: 470 10 6 x cos 2 (20) y sin 2(20)
1 E 1 E
( x y )cos 2 (20)
1 E
( y x )sin 2 (20)
[ x cos 2 (20) x sin 2(20)]
x
pd
[cos 2 (20) sin 2 (20)]
y
1
[ y sin 2(20) y cos 2(20)] E
[sin 2 (20) cos 2 (20)]
E E The normal stress x is the longitudinal stress caused by the internal pressure, and y is the hoop stress. Substitute expressions for long and hoop to obtain: long hoop 470 10 6 [cos 2 (20) sin 2 (20)] [sin 2 (20) cos 2(20)] E E pd pd [cos 2 (20) sin 2 (20)] [sin 2 (20) cos 2 (20)] 4tE 2tE
cos 2 (20) sin 2 (20) 2sin 2 (20) 2 cos 2 (20) 4tE
Thus, the pressure p can be expressed as: 4tE (470 106 ) p d cos2 (20) sin2 (20) 2sin2 (20 ) 2 cos2 (20 )
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Compute the internal pressure: 4(0.25 in.)(10,000 ksi)(470 106 in./in.) p (48 in.) cos 2 (20) (0.33)sin 2 (20) 2sin 2 (20) 2(0.33) cos 2 (20)
4(0.25 in.)(10,000 ksi)(470 106 in./in.) (48 in.)(0.495580)
0.197580 ksi 197.6 psi
Ans.
(b) Absolute maximum shear stress on the outer surface of the cylinder The principal stresses are: pd (0.197580 ksi)(48 in.) 9.484 ksi p 2 long 4t 4(0.25 in.)
hoop
pd
(0.197580 ksi)(48 in.)
18.968 ksi p1 2t 2(0.25 in.) The outer surface of the cylinder is in plane stress; therefore, the absolute maximum shear stress is: p1 p3 18.968 ksi 0 ksi Ans. abs max 9.48 ksi 2 2 (c) Absolute maximum shear stress on the inner surface of the cylinder Inside the cylinder, the pressure creates a stress in the radial d irection; therefore, p 3 radial p 0.197580 ksi
The absolute maximum shear stress inside the cylinder is p1 p3 18.968 ksi ( 0.197580 ksi) 9.58 k si abs max 2 2
Ans.
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P14.18 A strain gage is mounted at an angle of = 20° with respect to the longitudinal axis of the cylindrical pressure shown in Figure P14.17/18. The pressure vessel is fabricated from aluminum [ E = 10,000 ksi; = 0.33], and it has an inside diameter of 48 in. and a wall thickness of 0.50 in. If the internal pressure in the cylinder is 350 psi, determine: (a) the expected strain gage reading (in ). (b) the principal strains, the maximum shear strain, and the absolute maximum shear strain on the outer surface of the cylinder.
FIGURE P14.17/18
Solution (a) Expected strain gage reading A strain transformation equation [Eq. (13.3)] n x cos2 y sin 2 xy sin cos
can be written for the normal strain in the direction of the strain gage: n x cos 2 (20) y sin 2 (20) xy sin(20 )cos(20 ) Since this is a cylindrical pressure vessel, the shear stress xy must equal zero, and hence, the shear strain xy must also equal zero. The strain transformation equation reduces to: n x cos2 (20) y sin2 (20) Substitute Eqs. (13.21) for x and y to obtain an expression in terms of x and y: n x cos 2 (20) y sin 2 (20)
1 E 1 E
( x y )cos 2 (20)
1 E
( y x )sin 2 (20)
[ x cos 2 (20) x sin 2 (20)]
x
pd
[cos 2 (20) sin 2 (20)]
y
1
[ y sin 2(20) y cos 2(20)] E
[sin 2 (20) cos 2 (20)]
E E The normal stress x is the longitudinal stress caused by the internal pressure, and y is the hoop stress. Substitute expressions for long and hoop to obtain: long hoop n [cos 2 (20) sin 2 (20)] [sin 2 (20) cos 2 (20)] E E pd pd [cos 2 (20) sin 2 (20)] [sin 2 (20) cos 2 (20)] 4tE 2tE
cos 2 (20) sin 2 (20) 2sin 2 (20) 2 cos 2 (20) 4tE
The expected strain gage reading is thus: (350 psi)(48 in.) cos 2 (20) (0.33)sin 2 (20 ) 2sin 2 (20 ) 2(0.33)cos 2(20 ) n 4(0.50 in.)(10,000,000 psi)
416.288 10 6 in./in. 416 με
Ans.
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(b) Principal strains on outer surface of cylinder The principal stresses are: pd (350 psi)(48 in.) 8,400 psi p 2 long 4t 4(0.50 in.)
pd
hoop
(350 psi)(48 in.)
16,800 psi p1 2t 2(0.50 in.) From the generalized Hooke’s Law equations for plane stress, the normal strains produced in the plate can be computed from Eqs. (13.21): 1 1 x ( long hoop ) [8, 400 psi (0.33)(16,800 psi)] 285.600 10 6 in./in. 6 E 10 10 psi y
1 E
z
( hoop long )
E Therefore:
1 10 10 psi 6
( long hoop )
p1 1,403 με
[16,800 psi (0.33)(8, 400 psi)] 1, 402.800 10 6 in./in.
0.33 10 10 psi 6
[8, 400 psi 16, 800 psi] 831.600 10 6 in./in.
p 2 286 με
p 3 832 με
Ans.
Maximum shear strain
max p1 p 2 1,402.800 106 285.600 106 1,117 μrad
Ans.
Absolute maximum shear strain
abs max p1 p3 1, 402.800 106 (831.600 106 ) 2,230 μr ad
Ans.
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P14.19 The pressure vessel in Figure P14.19 consists of spirally wrapped steel plates that are welded at the seams in the orientation shown where = 35°. The cylinder has an inside diameter of 540 mm and a wall thickness of 10 mm. The ends of the cylinder are capped by two rigid end plates. The gage pressure inside the cylinder is 4.25 MPa, and compressive axial loads of P = 215 kN are applied to the rigid end caps. Determine: (a) the normal stress perpendicular to the weld seams. (b) the shear stress parallel to the weld seams. (c) the absolute maximum shear stress in the cylinder. FIGURE P14.19
Solution (a) Normal stress perpendicular to the weld pd (4.25 MPa)(540 mm) long 57.375 MPa 4t 4(10 mm)
hoop
pd
(4.25 MPa)(540 mm)
114.750 MPa 2t 2(10 mm) The compressive axial load also creates a normal stress in the x direction. A D 2 d 2 (560 mm) 2 (540 mm) 2 17,278.760 mm 2 4 4 P 215,000 N axial 12.443 MPa A 17,278.760 mm 2 The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore, the normal and shear stresses on longitudinal and circumferential faces of a stress element are: x 44.932 MPa, y 114.750 MPa, xy 0 MPa The weld is oriented at 35° as shown. The angle required for the stress transformation equations is the angle normal to the weld, which is also 35°. Using this value of , the normal stress transformation equation [Eq. (12-3)] can be used to compute the normal stress perpendicular to the weld: n x cos 2 y sin 2 2 xy sin cos
(44.932 MPa)cos 2 (35) (114.750 MPa)sin 2(35 ) 2(0 MPa)sin(35 )cos(35 ) 67.901 MPa 67.9 MPa (T)
Ans.
(b) Shear stress parallel to the weld Similarly, the shear stress transformation equation [Eq. (12-4)] gives nt : nt ( x y )sin cos xy (cos 2 sin 2 )
[(44.932 MPa) (114.750 MPa)]sin(35 )cos(35 ) (0 MPa)[cos 2(35 ) sin 2 (35)] 32.804 MPa 32.8 MPa
Ans.
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(c) Absolute maximum shear stress in the cylinder The principal stresses on the outside of the c ylinder are: p1 hoop 114.750 MPa p 2 long axial 44.932 MPa
p3 0
On the outside surface of the cylinder: p1 0 114.750 MPa abs max 57.375 MPa 2 2 Inside the cylinder, the third principal stress is equal in magnitude to the internal pressure: p 3 radial p 4.25 MPa On the inside surface of the cylinder: p1 p 3 114.750 MPa (4.25 MPa) 59.500 MPa abs max 2 2 Thus, the absolute maximum shear stress in the cylinder is
abs max 59.5 MPa
Ans.
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P14.20 The cylindrical pressure vessel shown in Figure P14.20/21 has an inside diameter of 610 mm and a wall thickness of 3 mm. The cylinder is made of an aluminum alloy that has an elastic modulus of E = 70 GPa and a shear modulus of G = 26.3 GPa. Two strain gages are mounted on the exterior surface of the cylinder at right angles to each other; however, the angle is not known. If the strains measured by the two gages are a = 360 and b = 975 , what is the pressure in the vessel? Notice that when two orthogonal strains are measured, the angle is not needed to determine the normal stresses.
FIGURE P14.20/21
Solution Strain invariance: From strain invariance [Eq. (13.8)], we can state a b x y
Let’s first focus on the right-hand side of this equation. From Eqs. (13.21), the sum of the strains in the x and y directions can be expressed as 1 1 x y ( x y ) ( y x ) E E 1 1 x y y x x (1 ) y (1 ) E E 1 x y E The longitudinal axis of the cylinder is defined as the x axis and the circumferential direction is defined as the y axis; therefore: pd pd x long and y hoop 4t 2t Substitution of these expressions gives
1 pd
pd
1 pd 3(1 ) pd 1 2 E 4t 2t E 4t E 4t Now, based on strain invariance, we can equate x y and a b x y
3(1 ) pd E
4t
a b
Before calculating p, we need to derive an expression for from Eq. (13.18): E E G 1 2(1 ) 2G Determine Poisson’s ratio from this expression: E 70 GPa 1 1 0.331 2G 2(26.3 GPa) Having calculated the value of Poisson’s ratio, we can now calculate the int ernal pressure p: 4( a b ) Et 4(360 10 6 975 10 6 )(70,000 MPa)(3 mm) 0.916 MPa p 3(1 )d 3(1 0.331)(610 mm)
Ans.
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P14.21 The cylindrical pressure vessel shown in Figure P14.20/21 has an inside diameter of 900 mm and a wall thickness of 12 mm. The cylinder is made of an aluminum alloy that has an elastic modulus of E = 70 GPa and a shear modulus of G = 26.3 GPa. Two strain gages are mounted on the exterior surface of the cylinder at right an gles to each other. The angle is 25°. If the pressure in the vessel is 1.75 MPa, determine (a) the strains that act in the x and y directions. (b) the strains expected in gages a and b. (c) the normal stresses n and t . (d) the shear stress nt .
FIGURE P14.20/21
Solution (b) Strains in the x and y directions pd (1.75 MPa)(900 mm) 32.813 MPa x long 4t 4(12 mm)
y hoop
pd 2t
(1.75 MPa)(900 mm) 2(12 mm)
65.625 MPa
Before proceeding, we need to derive an expression for from Eq. (13.18): E E G 1 2(1 ) 2G Determine Poisson’s ratio from this expression: E 70 GPa 1 1 0.331 2G 2(26.3 GPa) From the generalized Hooke’s law, the strain in the x direction is: 1 x ( x y ) E 1 32.813 MPa (0.331)(65.625 MPa) 70,000 MPa
158.626 106 mm/mm 158.6 με
Ans.
Similarly, the strain in the y direction is: 1 y ( y x ) E 1 65.625 MPa (0.331)(32.813 MPa) 70,000 MPa
782.438 10 6 mm/mm 782 με
Ans.
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(b) Expected strain gage readings in gages a and b A strain transformation equation [Eq. (13.3)] n x cos2 y sin 2 xy sin cos
can be written for the normal strain in the direction of strain gage a: a x cos 2 (25) y sin 2 (25) xy sin(25 )cos(25 ) Since this is a cylindrical pressure vessel, the shear stress xy must equal zero, and hence, the shear strain xy must also equal zero. The strain transformation equation reduces to: a x cos2 (25) y sin2 (25 ) The expected strain reading in gage a is thus:
a x cos2 (25) y sin2 (25 )
(158.626 με)cos2 (25) (782.438 με)sin 2 (25) 270 με
Ans.
Use a value of = 25° + 90° = 115° to obtain the strain reading expected in gage b: b x cos 2 (115) y sin 2 (115)
(158.626 με)cos2 (115) (782.438 με)sin2 (115) 671 με
Ans.
(c) Normal stresses n and t . Use the normal stress transformation equation [Eq. (12.3)] to calculate n: n x cos 2 y sin 2 2 xy sin cos
(32.813 MPa)cos 2 (25 ) (65.625 MPa)sin 2(25 ) 2(0 MPa)sin(25 )cos(25 ) 38.673 MPa 38.7 MPa (T)
Ans.
and t : t x cos 2 ( 90 ) y sin 2 ( 90 ) 2 xy sin( 90 )cos( 90 )
(32.813 MPa)cos 2 (115 ) (65.625 MPa)sin 2(115 ) 2(0 MPa)sin(115 )cos(115 ) 59.764 MPa 59.8 MPa (T)
Ans.
(d) Shear stress nt . The shear stress transformation equation [Eq. (12.4)] gives nt : nt ( x y )sin cos xy (cos 2 sin 2 )
[(32.813 MPa) (65.625 MPa)]sin(25 )cos(25 ) (0 MPa)[cos 2(25 ) sin 2(25 )] 12.568 MPa 12.57 MPa
Ans.
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