462
Transformations of Stress and Strain
7.9. STRESSES IN THIN-WALLED PRESSURE VESSELS
Fig. 7.48
Thin-walled pressure vessels provide an important application of the analysis of plane stress. Since their walls offer little resistance to bending, it can be assumed that the internal forces exerted on a given portion of wall are tangent to the surface of the vessel (Fig. 7.48). The resulting stresses on an element of wall will thus be contained in a plane tangent to the surface of the vessel. Our analysis of stresses in thin-walled pressure vessels will be limited to the two types of vessels most frequently encountered: cylindrical pressure vessels and spherical pressure vessels (Figs. 7.49 and 7.50).
Fig. 7.50
Fig. 7.49
y
1 2 1
t
2
r
z
x
Fig. 7.51
y
x 1 dA
t r
z
p dA
1 dA Fig. 7.52
r t
x
Consider a cylindrical vessel of inner radius r and wall thickness t containing a fluid under pressure (Fig. 7.51). We propose to determine the stresses exerted on a small element of wall with sides respectively parallel and perpendicular to the axis of the cylinder. Because of the axisymmetry of the vessel and its contents, it is clear that no shearing stress is exerted on the element. The normal stresses s1 and s2 shown in Fig. 7.51 are therefore principal stresses. The stress s1 is known as the hoop stress, because it is the type of stress found in hoops used to hold together the various slats of a wooden barrel, and the stress s2 is called the longitudinal stress. In order to determine the hoop stress s1, we detach a portion of the vessel and its contents bounded by the xy plane and by two planes parallel to the yz plane at a distance ¢x from each other (Fig. 7.52). The forces parallel to the z axis acting on the free body defined in this fashion consist of the elementary internal forces s1 dA on the wall sections, and of the elementary pressure forces p dA exerted on the portion of fluid included in the free body. Note that p denotes the gage pressure of the fluid, i.e., the excess of the inside pressure over the outside atmospheric pressure. The resultant of the internal forces s1 dA is equal to the product of s1 and of the cross-sectional area 2t ¢x of the wall, while the resultant of the pressure forces p dA is equal to the product of p and of the area 2r ¢x. Writing the equilibrium equation ©Fz ⫽ 0, we have
©Fz ⫽ 0:
s1 12t ¢x2 ⫺ p12r ¢x2 ⫽ 0
7.9. Stresses in Thin-Walled Pressure Vessels
and, solving for the hoop stress s1, s1 ⫽
pr t
(7.30)
To determine the longitudinal stress s2, we now pass a section perpendicular to the x axis and consider the free body consisting of the portion of the vessel and its contents located to the left of the section y
2 dA
t
r x
z p dA Fig. 7.53
(Fig. 7.53). The forces acting on this free body are the elementary internal forces s2 dA on the wall section and the elementary pressure forces p dA exerted on the portion of fluid included in the free body. Noting that the area of the fluid section is pr 2 and that the area of the wall section can be obtained by multiplying the circumference 2pr of the cylinder by its wall thickness t, we write the equilibrium equation:† 兺Fx ⫽ 0:
s2 12prt2 ⫺ p1pr 2 2 ⫽ 0
and, solving for the longitudinal stress s2, s2 ⫽
pr 2t
(7.31)
We note from Eqs. (7.30) and (7.31) that the hoop stress s1 is twice as large as the longitudinal stress s2: s1 ⫽ 2s2
(7.32)
†Using the mean radius of the wall section, rm ⫽ r ⫹ 12 t, in computing the resultant of the forces on that section, we would obtain a more accurate value of the longitudinal stress, namely, s2 ⫽
pr
1
2t
1⫹
t 2r
17.31¿ 2
However, for a thin-walled pressure vessel, the term t Ⲑ2r is sufficiently small to allow the use of Eq. (7.31) for engineering design and analysis. If a pressure vessel is not thin-walled (i.e., if tⲐ2r is not small), the stresses s1 and s2 vary across the wall and must be determined by the methods of the theory of elasticity.
463
464
Drawing Mohr’s circle through the points A and B that correspond respectively to the principal stresses s1 and s2 (Fig. 7.54), and recalling that the maximum in-plane shearing stress is equal to the radius of this circle, we have
Transformations of Stress and Strain
D'
max 2
D 1 2 2
O
B E E'
2
2
1 2 2
2 2 1
s1 ⫽ s2
2 dA t r C
D'
B A
(7.35)
1 2
pr 2t
(7.36)
Since the principal stresses s1 and s2 are equal, Mohr’s circle for transformations of stress within the plane tangent to the surface of the vessel reduces to a point (Fig. 7.57); we conclude that the in-plane normal stress is constant and that the in-plane maximum shearing stress is zero. The maximum shearing stress in the wall of the vessel, however, is not zero; it is equal to the radius of the circle of diameter OA and corresponds to a rotation of 45° out of the plane of stress. We have
p dA
1 2
s1 ⫽ s2 ⫽
x
Fig. 7.56
O
(7.34)
To determine the value of the stress, we pass a section through the center C of the vessel and consider the free body consisting of the portion of the vessel and its contents located to the left of the section (Fig. 7.56). The equation of equilibrium for this free body is the same as for the free body of Fig. 7.53. We thus conclude that, for a spherical vessel,
Fig. 7.55
max
(7.33)
We now consider a spherical vessel of inner radius r and wall thickness t, containing a fluid under a gage pressure p. For reasons of symmetry, the stresses exerted on the four faces of a small element of wall must be equal (Fig. 7.55). We have
1
Fig. 7.57
pr 2t
tmax ⫽ s2 ⫽
Fig. 7.54
1
pr 4t
This stress corresponds to points D and E and is exerted on an element obtained by rotating the original element of Fig. 7.51 through 45° within the plane tangent to the surface of the vessel. The maximum shearing stress in the wall of the vessel, however, is larger. It is equal to the radius of the circle of diameter OA and corresponds to a rotation of 45° about a longitudinal axis and out of the plane of stress.† We have
A
tmax 1in plane2 ⫽ 12 s2 ⫽
1
tmax ⫽ 12 s1 ⫽
pr 4t
(7.37)
†It should be observed that, while the third principal stress is zero on the outer surface of the vessel, it is equal to ⫺p on the inner surface, and is represented by a point C1⫺p, 02 on a Mohr-circle diagram. Thus, close to the inside surface of the vessel, the maximum shearing stress is equal to the radius of a circle of diameter CA, and we have tmax ⫽
pr 1 t 1s ⫹ p2 ⫽ a1 ⫹ b 2 1 2t r
For a thin-walled vessel, however, the term t/r is small, and we can neglect the variation of tmax across the wall section. This remark also applies to spherical pressure vessels.
SAMPLE PROBLEM 7.5 A compressed-air tank is supported by two cradles as shown; one of the cradles is designed so that it does not exert any longitudinal force on the tank. The cylindrical body of the tank has a 30-in. outer diameter and is fabricated from a 38-in. steel plate by butt welding along a helix that forms an angle of 25° with a transverse plane. The end caps are spherical and have a uniform wall thickness of 165 in. For an internal gage pressure of 180 psi, determine (a) the normal stress and the maximum shearing stress in the spherical caps, (b) the stresses in directions perpendicular and parallel to the helical weld.
8 ft
30 in. 25°
SOLUTION
a
a. Spherical Cap.
1
2
p ⫽ 180 psi, t ⫽
in. ⫽ 0.3125 in., r ⫽ 15 ⫺ 0.3125 ⫽ 14.688 in. 1180 psi2 114.688 in.2 pr ⫽ s1 ⫽ s2 ⫽ s ⫽ 4230 psi 䉳 2t 210.3125 in.2
0
We note that for stresses in a plane tangent to the cap, Mohr’s circle reduces to a point (A, B) on the horizontal axis and that all in-plane shearing stresses are zero. On the surface of the cap the third principal stress is zero and corresponds to point O. On a Mohr’s circle of diameter AO, point D¿ represents the maximum shearing stress; it occurs on planes at 45° to the plane tangent to the cap.
b
4230 psi 1 2 D'
max O A, B
C
tmax ⫽ 12 14230 psi2
p ⫽ 180 psi, t ⫽ 38 in. ⫽ 0.375 in., r ⫽ 15 ⫺ 0.375 ⫽ 14.625 in. 1180 psi2 114.625 in.2 pr s 2 ⫽ 12s 1 ⫽ 3510 psi ⫽ ⫽ 7020 psi s1 ⫽ t 0.375 in. save ⫽ 12 1s1 ⫹ s2 2 ⫽ 5265 psi R ⫽ 12 1s1 ⫺ s2 2 ⫽ 1755 psi
1 7020 psi
O
Stresses at the Weld. Noting that both the hoop stress and the longitudinal stress are principal stresses, we draw Mohr’s circle as shown. An element having a face parallel to the weld is obtained by rotating the face perpendicular to the axis Ob counterclockwise through 25°. Therefore, on Mohr’s circle we locate the point X¿ corresponding to the stress components on the weld by rotating radius CB counterclockwise through 2u ⫽ 50°.
b
2 3510 psi
1
sw ⫽ save ⫺ R cos 50° ⫽ 5265 ⫺ 1755 cos 50° tw ⫽ R sin 50° ⫽ 1755 sin 50°
1 7020 psi ave 5265 psi
C
B 2 50°
R X'
w
sw ⫽ ⫹4140 psi 䉳 tw ⫽ 1344 psi 䉳
Since X¿ is below the horizontal axis, tw tends to rotate the element counterclockwise.
2 3510 psi O
tmax ⫽ 2115 psi 䉳
b. Cylindrical Body of the Tank. We first determine the hoop stress s 1 and the longitudinal stress s 2. Using Eqs. (7.30) and (7.32), we write
a
2
Using Eq. (7.36), we write 5 16
R 1755 psi
A
w
x'
w 4140 psi w 1344 psi Weld
465
PROBLEMS
7.98 A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball is inflated to a 120-kPa gage pressure. 7.99 A spherical pressure vessel of 1.2-m outer diameter is to be fabricated from a steel having an ultimate stress U ⫽ 450 MPa. Knowing that a factor of safety of 4.0 is desired and that the gage pressure can reach 3 MPa, determine the smallest wall thickness that should be used. 7.100 A spherical gas container made of steel has a 20-ft outer diameter and a wall thickness of 167 in. Knowing that the internal pressure is 75 psi, determine the maximum normal stress and the maximum shearing stress in the container. 7.101 A spherical pressure vessel has an outer diameter of 3 m and a wall thickness of 12 mm. Knowing that for the steel used all ⫽ 80 MPa, E ⫽ 200 GPa, and n ⫽ 0.29, determine (a) the allowable gage pressure, (b) the corresponding increase in the diameter of the vessel. 7.102 A spherical gas container having an outer diameter of 15 ft and a wall thickness of 0.90 in. is made of a steel for which E ⫽ 29 ⫻ 106 psi and n ⫽ 0.29. Knowing that the gage pressure in the container is increased from zero to 250 psi, determine (a) the maximum normal stress in the container, (b) the increase in the diameter of the container.
25 ft
48 ft
Fig. P7.104
h
7.103 The maximum gage pressure is known to be 10 MPa in a spherical steel pressure vessel having a 200-mm outer diameter and a 6-mm wall thickness. Knowing that the ultimate stress in the steel used is U ⫽ 400 Mpa, determine the factor of safety with respect to tensile failure. 7.104 The unpressurized cylindrical storage tank shown has a 163 -in. wall thickness and is made of steel having a 60-ksi ultimate strength in tension. Determine the maximum height h to which it can be filled with water if a factor of safety of 4.0 is desired. (Specific weight of water ⫽ 62.4 lb/ft3.) 7.105 For the storage tank of Prob. 7.104, determine the maximum normal stress and the maximum sharing stress in the cylindrical wall when the take is filled to capacity (h ⫽ 48 ft). 7.106 A standard-weight steel pipe of 12-in. nominal diameter carries water under a pressure of 400 psi. (a) Knowing that the outside diameter is 12.75 in. and the wall thickness is 0.375 in., determine the maximum tensile stress in the pipe. (b) Solve part a, assuming an extra-strong pipe is used, of 12.75-in. outside diameter and 0.500-in. wall thickness. 7.107 A storage tank contains liquified propane under a pressure of 210 psi at a temperature of 100⬚F. Knowing that the tank has an outer diameter of 12.6 in. and a wall thickness of 0.11 in., determine the maximum normal stress and the maximum shearing stress in the tank.
466
7.108 The bulk storage tank shown in Fig. 7.49 has an outer diameter of 3.5 m and a wall thickness of 20 mm. At a time when the internal pressure of the tank is 1.2 MPa, determine the maximum normal stress and the maximum shearing stress in the tank.
Problems
7.109 Determine the largest internal pressure that can be applied to a cylindrical tank of 1.75-m outer diameter and 16-mm wall thickness if the ultimate normal stress of the steel used is 450 MPa and a factor of safety of 5.0 is desired. 7.110 A steel penstock has a 750-mm outer diameter, a 12-mm wall thickness, and connects a reservoir at A with a generating station at B. Knowing that the density of water is 1000 kg/m3, determine the maximum normal stress and the maximum shearing stress in the penstock under static conditions.
A
300 m
7.111 A steel penstock has a 750-mm outer diameter and connects a reservoir at A with a generating station at B. Knowing that the density of water is 1000 kg/m3 and that the allowable normal stress in the steel is 85 MPa, determine the smallest thickness that can be used for the penstock. 7.112 The steel pressure tank shown has a 30-in. inner diameter and a 0.375-in. wall thickness. Knowing that the butt welded seams form an angle  ⫽ 50⬚ with the longitudinal axis of the tank and that the gage pressure in the tank is 200 psi, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld. 7.113 The pressurized tank shown was fabricated by welding strips of plate along a helix forming an angle  with a transverse plane. Determine the largest value of  that can be used if the normal stress perpendicular to the weld is not be larger than 85 percent of the maximum stress in the tank. 7.114 The cylindrical portion of the compressed air tank shown is fabricated of 0.25-in.-thick plate welded along a helix forming an angle  ⫽ 30⬚ with the horizontal. Knowing that the allowable stress normal to the weld is 10.5 ksi, determine the largest gage pressure that can be used in the tank.
20 in.
60 in.
Fig. P7.114
7.115 For the compressed-air tank of Prob. 7.114, determine the gage pressure that will cause a shearing stress parallel to the weld of 4 ksi.
B 750 mm Fig. P7.110 and P7.111
Fig. P7.112 and P7.113
467
468
7.116 Square plates, each of 16-mm thickness, can be bent and welded together in either of the two ways shown to form the cylindrical portion of a compressed air tank. Knowing that the allowable normal stress perpendicular to the weld is 65 MPa, determine the largest allowable gage pressure in each case.
Transformations of Stress and Strain
5m
5m
45 8m
(a)
(b)
Fig. P7.116 3m 1.6 m
Fig. P7.117
7.117 The pressure tank shown has an 8-mm wall thickness and butt welded seams forming an angle  ⫽ 20⬚ with a transverse plane. For a gage pressure of 600 kPa, determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld. 7.118 For the tank of Prob. 7.117, determine the largest allowable gage pressure, knowing that the allowable normal stress perpendicular to the weld is 120 MPa and the allowable shearing stress parallel to the weld is 80 MPa. 7.119 For the tank of Prob. 7.117, determine the range of values of  that can be used if the shearing stress parallel to the weld is not to exceed 12 MPa when the gage pressure is 600 kPa. 7.120 A torque of magnitude T ⫽ 12 kN ⴢ m is applied to the end of a tank containing compressed air under a pressure of 8 MPa. Knowing that the tank has a 180-mm inner diameter and a 12-mm wall thickness, determine the maximum normal stress and the maximum shearing stress in the tank.
T
Fig. P7.120 and P7.121
7.121 The tank shown has a 180-mm inner diameter and a 12-mm wall thickness. Knowing that the tank contains compressed air under a pressure of 8 MPa, determine the magnitude T of the applied torque for which the maximum normal stress is 75 MPa.
7.122 A pressure vessel of 10-in. inner diameter and 0.25-in. wall thickness is fabricated from a 4-ft section of spirally welded pipe AB and is equipped with two rigid end plates. The gage pressure inside the vessel is 300 psi and 10-kip centric axial forces P and P¿ are applied to the end plates. Determine (a) the normal stress perpendicular to the weld, (b) the shearing stress parallel to the weld.
Problems
y 150 mm 4 ft P' B
A
P 35
P
B
Fig. P7.122 600 mm
7.123 Solve Prob. 7.122, assuming that the magnitude P of the two forces is increased to 30 kips. 7.124 The compressed-air tank AB has a 250-mm outside diameter and an 8-mm wall thickness. It is fitted with a collar by which a 40-kN force P is applied at B in the horizontal direction. Knowing that the gage pressure inside the tank is 5 MPa, determine the maximum normal stress and the maximum shearing stress at point K. 7.125 In Prob. 7.124, determine the maximum normal stress and the maximum shearing stress at point L. 7.126 A brass ring of 5-in. outer diameter and 0.25-in. thickness fits exactly inside a steel ring of 5-in. inner diameter and 0.125-in. thickness when the temperature of both rings is 50⬚F. Knowing that the temperature of both rings is then raised to 125⬚F, determine (a) the tensile stress in the steel ring, (b) the corresponding pressure exerted by the brass ring on the steel ring.
1.5 in.
STEEL ts 81 in. Es 29 106 psi
ss 6.5 10–6/F 5 in. BRASS tb 14 in. Eb 15 106 psi
bs 11.6 10–6/F Fig. P7.126
7.127 Solve Prob. 7.126, assuming that the brass ring is 0.125 in. thick and the steel ring is 0.25 in. thick.
K
L
A z
150 mm x
Fig. P7.124
469
