8
Applications of Plane Stress
(Pressure Vessels, Beams,
and Combined Loadings)
Spherical Pressure Vessels
When solving the problems for Section 8.2, assume that the given radius or
diameter is an inside dimension and that all internal pressures are gage pressures.
Problem 8.2-1 A large spherical tank (see figure) contains gas at a
pressure of 450 psi. The tank is 42 ft in diameter and is constructed of high-strength
steel having a yield stress in tension of 80 ksi.
Determine the required thickness (to the nearest 1/4 inch) of the wall of the tank
if a factor of safety of 3.5 with respect to yielding is required.
Probs. 8.2-1 and 8.2-2
Solution 8.2-1
Radius:
r5
1
42 * 12
2
r 5 252 in.
Internal Pressure: p 5 450 psi
Yield stress:
t5
prn
2sY
t 5 2.481 in.
to nearest 1/4 inch, t min 5 2.5 in.
;
sY 5 80 ksi (steel)
Factor of safety: n 5 3.5
MINIMUM WALL THICKNESS tmin
From Eq. (8-1): smax 5
pr
pr
sY
5
or
2t
n
2t
649
650
CHAPTER 8
Applications of Plane Stress
Problem 8.2-2 Solve the preceding problem if the internal pressure is 3.75 MPa, the diameter is 19 m, the yield stress is
570 MPa, and the factor of safety is 3.0.
Determine the required thickness to the nearest millimeter.
Solution 8.2-2
Radius:
r5
1
(19 m)
2
r 5 9.5 3 103 mm
Internal Pressure: p 5 3.75 MPa
Yield stress:
t5
prn
2sY
t 5 93.8 mm
Use the next higher millimeter t min 5 94 mm
sY 5 570 MPa
Factor of safety: n 5 3
MINIMUM WALL THICKNESS tmin
From Eq. (8-1): smax 5
pr
pr
sY
5
or
2t
n
2t
Problem 8.2-3 A hemispherical window (or viewport) in a decompression chamber
(see figure) is subjected to an internal air pressure of 80 psi. The port is attached to
the wall of the chamber by 18 bolts.
Find the tensile force F in each bolt and the tensile stress s in the viewport if the
radius of the hemisphere is 7.0 in. and its thickness is 1.0 in.
Solution 8.2-3
Hemispherical viewport
FREE-BODY DIAGRAM
T 5 total tensile force in 18 bolts
a FHORIZ 5 T 2 pA 5 0
T 5 pA 5 p(pr 2)
F 5 force in one bolt
F5
Radius:
r 5 7.0 in.
18 bolts
t 5 1.0 in.
;
TENSILE STRESS IN VIEWPORT (EQ. 8-1)
Internal pressure: p 5 80 psi
Wall thickness:
T
1
5
(ppr 2) 5 684 lb
18
18
s5
pr
5 280 psi
2t
;
;
SECTION 8.2
651
Spherical Pressure Vessels
Problem 8.2-4 A rubber ball (see figure) is inflated to a pressure of 60 kPa. At that pressure
the diameter of the ball is 230 mm and the wall thickness is 1.2 mm. The rubber has modulus of
elasticity E 5 3.5 MPa and Poisson’s ratio v 5 0.45.
Determine the maximum stress and strain in the ball.
Prob. 8.2-4, 8.2-5
Solution 8.2-4 Rubber ball
CROSS-SECTION
MAXIMUM STRESS (EQ. 8-1)
smax 5
(60 kPa)(115 mm)
pr
5
2t
2(1.2 mm)
5 2.88 MPa
;
MAXIMUM STRAIN (EQ. 8-4)
Radius:
r 5 (230 mm)/2 5 115 mm
Internal pressure:
p 5 60 kPa
Wall thickness:
t 5 1.2 mm
âmax 5
(60 kPa)(115 mm)
pr
(1 2 v) 5
(0.55)
2tE
2(1.2 mm)(3.5 MPa)
5 0.452
;
Modulus of elasticity: E 5 3.5 MPa (rubber)
Poisson’s ratio:
v 5 0.45 (rubber)
Problem 8.2-5 Solve the preceding problem if the pressure is 9.0 psi, the diameter is 9.0 in.,
the wall thickness is 0.05 in., the modulus of elasticity is 500 psi, and Poisson’s ratio is 0.45.
Solution 8.2-5 Rubber ball
CROSS-SECTION
Modulus of elasticity: E 5 500 psi (rubber)
Poisson’s ratio:
v 5 0.45 (rubber)
MAXIMUM STRESS (EQ. 8-1)
smax 5
1
(9.0 in.) 5 4.5 in.
2
Radius:
r5
Internal pressure:
p 5 9.0 psi
Wall thickness:
t 5 0.05 in.
(9.0 psi)(4.5 in.)
pr
5 405 psi
5
2t
2(0.05 in.)
;
MAXIMUM STRAIN (EQ. 8-4)
âmax 5
(9.0 psi)(4.5 in.)
pr
(1 2 v) 5
(0.55)
2tE
2(0.05 in.)(500 psi)
5 0.446
;
652
CHAPTER 8
Applications of Plane Stress
Problem 8.2-6 A spherical steel pressure vessel (diameter 480 mm,
thickness 8.0 mm) is coated with brittle lacquer that cracks when the strain
reaches 150 3 1026 (see figure).
What internal pressure p will cause the lacquer to develop cracks? (Assume
E 5 205 GPa and v 5 0.30.)
Solution 8.2-6
Spherical vessel with brittle coating
Cracks occur when «max 5 150 3 1026
pr
(1 2 v)
From Eq. (8-4): âmax 5
2tE
CROSS-SECTION
‹ P5
r 5 240 mm
t 5 8.0 mm
Cracks in
coating
E 5 205 GPa (steel)
v 5 0.30
P5
2tE âmax
r(1 2v)
2(8.0 mm)(205 GPa)(150 * 1026)
(240 mm)(0.70)
5 2.93 MPa
;
Problem 8.2-7 A spherical tank of diameter 48 in. and wall thickness 1.75 in. contains
compressed air at a pressure of 2200 psi. The tank is constructed of two hemispheres
joined by a welded seam (see figure).
Weld
(a) What is the tensile load f (lb per in. of length of weld) carried by the weld?
(b) What is the maximum shear stress tmax in the wall of the tank?
(c) What is the maximum normal strain « in the wall? (For steel, assume
E 5 30 3 106 psi and v 5 0.29.)
Probs. 8.2-7 and 8.2-8
Solution 8.2-7
r 5 24 in.
t 5 1.75 in.
E 5 30 3 106 psi
v 5 0.29 (steel)
(b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3)
tmax 5
(2200 psi)(24 in.)
pr
5 7543 psi
5
4t
4 (1.75 in.)
(a) TENSILE LOAD CARRIED BY WELD
T 5 Total load
f 5 load per inch
T 5 pA 5 ppr 2
c 5 Circumference of tank 5 2pr
T p1pr 2 pr (2200 psi)(24 in.)
f5 5
5 5
c
2pr
2
2
2
5 26.4 k/in.
;
(c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4)
âmax 5
(2200 psi)(24 in.)(0.71)
pr (1 2 v)
5
2 tE
2(1.75 in.)130106 psi2
5 3.57 * 1024
;
SECTION 8.2
653
Spherical Pressure Vessels
Problem 8.2-8 Solve the preceding problem for the following data: diameter 1.0 m, thickness 48 mm, pressure 22 MPa,
modulus 210 GPa, and Poisson’s ratio 0.29.
Solution 8.2-8
r 5 0.5 m
(b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3)
E 5 210 GPa
t 5 48 mm
v 5 0.29 (steel)
tmax 5
(a) TENSILE LOAD CARRIED BY WELD
T 5 Total load
5 57.3 MPa
;
f 5 load per inch
T 5 pA 5 ppr2
c 5 Circumference of tank 5 2pr
p1pr 2
pr
(22 MPa)(0.5 m)
T
5
5
5
c
2pr
2
2
(c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4)
2
f5
(22 MPa)(0.5 m)
pr
5
4t
4 (48 mm)
5 5.5 MN/m
âmax 5
(22 MPa)(0.5 m) 0.71
pr (1 2 v)
5
2 tE
2(48 mm)(210 GPa)
5 3.87 * 1024
;
;
Problem 8.2-9 A spherical stainless-steel tank having a diameter of 22 in. is used to store propane gas at a pressure of
2450 psi. The properties of the steel are as follows: yield stress in tension, 140,000 psi; yield stress in shear, 65,000 psi;
modulus of elasticity, 30 3 106 psi; and Poisson’s ratio, 0.28. The desired factor of safety with respect to yielding is 2.8.
Also, the normal strain must not exceed 1100 3 1026.
Determine the minimum permissible thickness tmin of the tank.
Solution 8.2-9
E 5 30 3 106 psi
r 5 11 in.
p 5 2450 psi
(2) SHEAR (EQ. 8-3)
v 5 0.28 (steel)
sY 5 140000 psi
t2 5
n 5 2.8
«max 5 1100 3 1026
tY 5 65000 psi
MIMIMUM WALL THICKNESS t
(1) TENSION (EQ. 8-1)
smax 5
pr
sY
2a b
n
(2450 psi) (11 in.)
pr
2 t1
2
140000 psi
2.8
t3 5
pr
4t2
(2450 psi)(11 in.)
pr
5 0.29 in.
5
65000 psi
tY
4n
4
2.8
(3) STRAIN (EQ. 8-4)
pr
t1 5
5
2smax
5
tmax 5
pr
2âmax E
âmax 5
pr
(1 2 v)
2t 3 E
(1 2 v)
(2450 psi)(11 in.)
5
5 0.269 in.
211100 * 10262130 * 106 psi2
0.72
5 0.294 in.
t3 . t2 . t1
Thus, tmin 5 0.294 in.
;
654
CHAPTER 8
Applications of Plane Stress
Problem 8.2-10 Solve the preceding problem if the diameter is 500 mm, the pressure is 18 MPa, the yield stress in tension
is 975 MPa, the yield stress in shear is 460 MPa, the factor of safety is 2.5, the modulus of elasticity is 200 GPa, Poisson’s
ratio is 0.28, and the normal strain must not exceed 1210 3 1026.
Solution 8.2-10
r 5 250 mm
E 5 200 GPa
p 5 18 MPa
v 5 0.28 (steel)
(2) SHEAR (EQ. 8-3)
n 5 2.5
tY 5 460 MPa
«max 5 1210 3 1026
MINIMUM WALL THICKNESS t
t1 5
pr
5
2smax
smax 5
pr
4t 2
pr
(18 MPa)(250 mm)
5
5 6.114 mm
tY
460 MPa
4
4
n
2.5
pr
(3) STRAIN (EQ. 8-4) âmax 5
(1 2 v)
2t 3E
t2 5
sY 5 975 MPa
(1) TENSION (EQ. 8-1)
tmax 5
pr
2t 1
pr
sY
2a b
n
(18 MPa)(250 mm)
5
5 5.769 mm
975 MPa
2
2.5
t3 5
5
pr
(1 2 v)
2âmax E
(18 MPa) (250 mm)
211210 * 10262(200 GPa)
0.72
5 6.694 mm
Thus, tmin 5 6.69 mm
t3 . t2 . t1
Problem 8.2-11 A hollow pressurized sphere having radius r 5 4.8 in. and
wall thickness t 5 0.4 in. is lowered into a lake (see figure). The compressed
air in the tank is at a pressure of 24 psi (gage pressure when the tank is out
of the water).
At what depth D0 will the wall of the tank be subjected to a compressive
stress of 90 psi?
D0
;
SECTION 8.3
Solution 8.2-11
CROSS-SECTION
r 5 4.8 in.
t 5 0.4 in.
Cylindrical Pressure Vessels
655
Pressurized sphere under water
D0 5 depth of water (in.)
p1 5 24 psi
3
g 5 density of water 5 62.4 lb/ft
(1) IN AIR: p1 5 24 psi
p2 5 gD0 5 a
pr
( p1 2 p2)r
5
2t
2t
290 psi 5
(2) UNDER WATER: p1 5 24 psi
1728 in.3/ ft3
bD0 5 0.036111 D0 ( psi)
Compressive stress in tank wall equals 90 psi.
(Note: s is positive in tension.)
s5
(1) IN AIR
62.4 lb/ft3
s 5 290 psi
(24 psi 2 0.03611 D0)(4.8 in.)
2(0.4 in.)
5 144 2 0.21667 D0
Solve for D0: D0 5
234
0.21667
5 1080 in. 5 90 ft
(2) UNDER WATER
Cylindrical Pressure Vessels
When solving the problems for Section 8.3, assume that the given radius or diameter is an
inside dimension and that all internal pressures are gage pressures.
Problem 8.3-1 A scuba tank (see figure) is being designed for an internal pressure of
1600 psi with a factor of safety of 2.0 with respect to yielding. The yield stress of the steel is
35,000 psi in tension and 16,000 psi in shear.
If the diameter of the tank is 7.0 in., what is the minimum required wall thickness?
;
656
CHAPTER 8
Applications of Plane Stress
Solution 8.3-1 Scuba tank
sallow 5
sY
5 17,500 psi
n
t allow 5
tY
5 8,000 psi
n
Find required wall thickness t.
(1) BASED ON TENSION (EQ. 8-5)
t1 5
pr
sallow
(1600 psi)(3.5 in.)
5
17,500 psi
(2) BASED ON SHEAR (EQ. 8-10)
t2 5
Cylindrical pressure vessel
p 5 1600 psi
r 5 3.5 in.
n 5 2.0
d 5 7.0 in.
sY 5 35,000 psi
tY 5 16,000 psi
smax 5
pr
t
5 0.320 in.
tmax 5
pr
2t
(1600 psi)(3.5 in.)
pr
5 0.350 in.
5
2tallow
2(8,000 psi)
Shear governs since t2 . t1
[ tmin 5 0.350 in.
;
Problem 8.3-2 A tall standpipe with an open top (see figure) has diameter d 5 2.2 m
and wall thickness t 5 20 mm.
d
(a) What height h of water will produce a circumferential stress of 12 MPa in the wall of
the standpipe?
(b) What is the axial stress in the wall of the tank due to the water pressure?
h
Solution 8.3-2
d 5 2.2 m
r 5 1.1 m
t 5 20 mm
weight density of water g 5 9.81 kN/m3
h5
12 (20)
5 22.2 m
0.00981 (1.1)
(b) AXIAL
STRESS
IN
THE
WALL
DUE
TO
WATER
height of water
h
PRESSURE ALONE
water pressure
p 5 gh
Because the top of the tank is open, the internal
pressure of the water produces no axial (longitudinal) stresses in the wall of the tank. Axial stress
equals zero.
;
(a) HEIGHT OF WATER
s1 5
0.00981h (1.1 m)
pr
5 12 MPa 5
t
20 mm
SECTION 8.3
Problem 8.3-3 An inflatable structure used by a traveling
circus has the shape of a half-circular cylinder with closed ends
(see figure). The fabric and plastic structure is inflated by a
small blower and has a radius of 40 ft when fully inflated. A
longitudinal seam runs the entire length of the “ridge” of the
structure.
If the longitudinal seam along the ridge tears open when it
is subjected to a tensile load of 540 pounds per inch of seam,
what is the factor of safety n against tearing when the internal
pressure is 0.5 psi and the structure is fully inflated?
Solution 8.3-3
Cylindrical Pressure Vessels
657
Longitudinal seam
Inflatable structure
Half-circular cylinder
r 5 40 ft 5 480 in.
Internal pressure p 5 0.5 psi
T 5 tensile force per unit length of longitudinal seam
Seam tears when T 5 Tmax 5 540 lb/in.
Find factor of safety against tearing.
CIRCUMFERENTIAL STRESS (EQ. 8-5)
s1 5
pr
where t 5 thickness of fabric
t
Actual value of T due to internal pressure 5 s1t
[ T 5 s1t 5 pr 5 (0.5 psi)(480 in.) 5 240 lb/in.
FACTOR OF SAFETY
n5
Tmax
540 lb/in.
5
5 2.25
T
240 lb/in.
Problem 8.3-4 A thin-walled cylindrical pressure vessel of radius r is subjected
simultaneously to internal gas pressure p and a compressive force F acting at the
ends (see figure).
What should be the magnitude of the force F in order to produce pure shear
in the wall of the cylinder?
Solution 8.3-4
F
Cylindrical pressure vessel
STRESSES (SEE EQ. 8-5 AND 8-6):
r 5 Radius
p 5 Internal pressure
s1 5
pr
t
s2 5
pr
pr
F
F
2 5
2
2t
A
2t
2prt
;
F
658
CHAPTER 8
Applications of Plane Stress
FOR PURE SHEAR, the stresses s1 and s2 must be equal in
magnitude and opposite in sign (see, e.g., Fig. 7-11 in
Section 7.3).
[ s1 5 2s2
OR
pr
pr
F
5 2a 2
b
t
2t
2prt
Solve for F: F 5 3ppr2
;
Problem 8.3-5 A strain gage is installed in the longitudinal direction on the surface
of an aluminum beverage can (see figure). The radius-to-thickness ratio of the can is
200. When the lid of the can is popped open, the strain changes by [0 5 170 3 1026.
What was the internal pressure p in the can? (Assume E 5 10 3 106 psi
and v 5 0.33.)
12 FL OZ (355 mL)
Solution 8.3-5 Aluminum can
STRAIN IN LONGITUDINAL DIRECTION (EQ. 8-11a)
â2 5
pr
(1 2 2v)
2tE
â2 5 â0
‹
or p 5
p5
2tEâ2
r(1 2 2v)
2Eâ0
2tEâ0
5
(r)(1 2 2v)
(r/t) (1 2 2v)
Substitute numerical values:
r
5 200
t
E 5 10 3 106 psi
p5
v 5 0.33
«0 5 change in strain when pressure is released
5 170 3 1026
Find internal pressure p.
2(10 * 106 psi)(170 * 1026)
(200)(1 2 0.66)
5 50 psi
;
SECTION 8.3
Problem 8.3-6 A circular cylindrical steel tank (see figure) contains
a volatile fuel under pressure. A strain gage at point A records the
longitudinal strain in the tank and transmits this information to a control
room. The ultimate shear stress in the wall of the tank is 84 MPa, and
a factor of safety of 2.5 is required.
At what value of the strain should the operators take action to reduce
the pressure in the tank? (Data for the steel are as follows: modulus of
elasticity E 5 205 GPa and Poisson’s ratio v 5 0.30.)
659
Cylindrical Pressure Vessels
Pressure relief
valve
Cylindrical tank
A
Solution 8.3-6
tULT 5 84 MPa
E 5 205 GPa
n 5 2.5
tULT
n
tmax 5
v 5 0.3
tmax 5 33.6 MPa
Find maximum allowable strain reading at the gage
s1 5
pr
t
s2 5
pr
2t
From Eq. (8-11a)
â2 5
pr
(1 2 2v)
2tE
â2max 5
âmax 5
From Eq. (8-10)
tmax 5
pr
s1
5
2
2t
Pmax 5
pmax r
tmax
(1 2 2v) 5
(1 2 2v)
2tE
E
tmax
(1 2 2v)
E
«max 5 6.56 3 1025
2ttmax
r
Cylinder
Problem 8.3-7 A cylinder filled with oil is under pressure from a
piston, as shown in the figure. The diameter d of the piston is 1.80 in.
and the compressive force F is 3500 lb. The maximum allowable shear
stress tallow in the wall of the cylinder is 5500 psi.
What is the minimum permissible thickness tmin of the cylinder
wall? (See the figure on the next page.)
F
p
Piston
Probs. 8.3-7 and 8.3-8
Solution 8.3-7
Cylinder with internal pressure
Maximum shear stress (Eq. 8-10):
tmax 5
d 5 1.80 in.
r 5 0.90 in.
F 5 3500 lb
tallow 5 5500 psi
Find minimum thickness tmin.
F
F
Pressure in cylinder: p 5 5
A
pr 2
pr
F
5
2t
2prt
Minimum thickness:
t min 5
F
2prtallow
Substitute numerical values:
t min 5
3500 lb
5 0.113 in.
2p(0.90 in.)(5500 psi)
;
660
CHAPTER 8
Applications of Plane Stress
Problem 8.3-8 Solve the preceding problem if d 5 90 mm, F 5 42 kN, and tallow 5 40 MPa.
Solution 8.3-8
Cylinder with internal pressure
Maximum shear stress (Eq. 8-10):
pr
F
5
2t
2prt
tmax 5
Minimum thickness:
d 5 90 mm
F 5 42.0 kN
r 5 45 mm
t min 5
tallow 5 40 MPa
F
2pr tallow
Find minimum thickness tmin.
Substitute numerical values:
F
F
Pressure in cylinder: p 5 5
A
pr 2
t min 5
42.0 kN
5 3.71 mm
2p(45 mm)(40 MPa)
;
Problem 8.3-9 A standpipe in a water-supply system (see figure) is 12 ft in
diameter and 6 inches thick. Two horizontal pipes carry water out of the
standpipe; each is 2 ft in diameter and 1 inch thick. When the system is
shut down and water fills the pipes but is not moving, the hoop stress at the
bottom of the standpipe is 130 psi.
(a) What is the height h of the water in the standpipe?
(b) If the bottoms of the pipes are at the same elevation as the bottom of the
standpipe, what is the hoop stress in the pipes?
Solution 8.3-9
Vertical standpipe
(a) FIND HEIGHT h OF WATER IN THE STANDPIPE
p 5 pressure at bottom of standpipe 5 gh
From Eq. (8-5): s1 5
pr
ghr
5
t
t
or h 5
Substitute numerical values:
h5
d 5 12 ft 5 144 in.
g 5 62.4 lb/ft3 5
r 5 72 in.
t 5 6 in.
62.4
lb/in.3
1728
s1 5 hoop stress at bottom of standpipe 5 130 psi
(130 psi)(6 in.)
62.4
lb/in.3 b(72 in.)
a
1728
5 25 ft
;
5 300 in.
s1t
gr
SECTION 8.3
HORIZONTAL PIPES
d1 5 2 ft 5 24 in.
r1 5 12 in.
t1 5 1.0 in.
s1 5
Since the pipes are 2 ft in diameter, the depth of
water to the center of the pipes is about 24 ft.
h1 < 24 ft 5 288 in.
661
p1r1
gh 1r1
5
t1
t1
a
(b) FIND HOOP STRESS s1 IN THE PIPES
Cylindrical Pressure Vessels
5
62.4
lb/in.3 b(288 in.)(12 in.)
1728
1.0 in.
5 125 psi
p1 5 gh1
Based on the average pressure in the pipes:
s1 < 125 psi
Problem 8.3-10 A cylindrical tank with hemispherical heads is constructed of
steel sections that are welded circumferentially (see figure). The tank diameter is
1.25 m, the wall thickness is 22 mm, and the internal pressure is 1750 kPa.
(a) Determine the maximum tensile stress sh in the heads of the tank.
(b) Determine the maximum tensile stress sc in the cylindrical part
of the tank.
(c) Determine the tensile stress sw acting perpendicular to the welded joints.
(d) Determine the maximum shear stress th in the heads of the tank.
(e) Determine the maximum shear stress tc in the cylindrical part of the tank.
;
Welded seams
Probs. 8.3-10 and 8.3-11
Solution 8.3-10
d 5 1.25 m
r5
d
2
t 5 22 mm
p 5 1750 kPa
(c) TENSILE STRESS IN WELDS (EQ. 8-6)
sw 5
pr
2t
sw 5 24.9 MPa
;
(a) MAXIMUM TENSILE STRESS IN HEMISPHERES (EQ. 8-1)
sh 5
pr
2t
sh 5 24.9 MPa
(d) MAXIMUM SHEAR STRESS IN HEMISPHERES (EQ. 8-3)
;
th 5
pr
4t
th 5 12.43 MPa
;
(b) MAXIMUM STRESS IN CYLINDER (EQ. 8-5)
sc 5
pr
t
sc 5 49.7 MPa
;
(e) MAXIMUM SHEAR STRESS IN CYLINDER (EQ. 8-10)
tc 5
pr
2t
tc 5 24.9 MPa
;
Problem 8.3-11 A cylindrical tank with diameter d 5 18 in. is subjected to internal gas pressure p 5 450 psi. The tank is
constructed of steel sections that are welded circumferentially (see figure). The heads of the tank are hemispherical. The
allowable tensile and shear stresses are 8200 psi and 3000 psi, respectively. Also, the allowable tensile stress perpendicular
to a weld is 6250 psi.
Determine the minimum required thickness tmin of (a) the cylindrical part of the tank and (b) the hemispherical heads.
662
CHAPTER 8
Applications of Plane Stress
Solution 8.3-11
d 5 18 in.
r5
sallow 5 8200 psi
d
2
t min 5
p 5 450 psi
pr
2sa
tmin 5 0.324 in.
t min 5 0.675 in.
(tension)
;
tallow 5 3000 psi (shear)
WELD
sa 5 6250 psi
(b) FIND MINIMUM THICKNESS OF HEMISPHERES
(tension)
(a) FIND MINIMUM THICKNESS OF CYLINDER
pr
TENSION
smax 5
t
pr
t min 5
tmin 5 0.494 in.
sallow
SHEAR
t min
tmax 5
pr
5
2tallow
WELD
s5
TENSION
t min 5
smax 5
pr
2sallow
SHEAR
tmin 5 0.247 in.
tmax 5
pr
2t
t min 5
tmin 5 0.675 in.
t min 5 0.338 in.
pr
4tallow
pr
4t
tmin 5 0.338 in.
;
pr
2t
Problem *8.3-12 A pressurized steel tank is constructed with a helical weld
that makes an angle a 5 55° with the longitudinal axis (see figure). The tank
has radius r 5 0.6 m, wall thickness t 5 18 mm, and internal pressure p 5
2.8 MPa. Also, the steel has modulus of elasticity E 5 200 GPa and Poisson’s
ratio v 5 0.30.
Determine the following quantities for the cylindrical part of the tank.
(a)
(b)
(c)
(d)
pr
2t
Helical weld
a
The circumferential and longitudinal stresses.
The maximum in-plane and out-of-plane shear stresses.
The circumferential and longitudinal strains.
The normal and shear stresses acting on planes parallel and perpendicular to
the weld (show these stresses on a properly oriented stress element).
Probs. 8.3-12 and 8.3-13
Solution 8.3-12
a 5 55 deg
p 5 2.8 MPa
r 5 0.6 m
t 5 18 mm
E 5 200 GPa
v 5 0.3
(a) CIRCUMFERENTIAL STRESS
pr
s1 5
t
s1 5 93.3 MPa
pr
2t
s2 5 46.7 MPa
t1 5
s1 2 s2
2
t1 5 23.3 MPa
OUT-OF-PLANE SHEAR STRESS
;
LONGITUDIAL STRESS
s2 5
(b) IN-PLANE SHEAR STRESS
;
t2 5
s1
2
t2 5 46.7 MPa
;
;
SECTION 8.3
(c) CIRCUMFERENTIAL STRAIN
s1
â1 5
(2 2 v)
2E
«1 5 3.97 3 1024
;
For u 5 35 deg
sx + sy
sx 2 sy
sx1 5
+
cos (2u)
2
2
LONGITUDINAL STRAIN
s2
(1 2 2v)
â2 5
E
+ txy sin (2u)
«2 5 9.33 3 1025
;
(d) STRESS ON WELD
u 5 90 deg 2 a
sx 5 s2
sx1 5 62.0 MPa
;
sx 2 sy
tx1y1 5 2
sin (2u) + txy cos (2u)
2
tx1y1 5 21.9 MPa
u 5 35 deg
sx 5 46.667 MPa
sy 5 93.333 MPa
663
Cylindrical Pressure Vessels
;
sy1 5 78.0 MPa
sy1 5 sx 1 sy 2 sx1
sy 5 s1
;
txy 5 0
Problem *8.3-13 Solve the preceding problem for a welded tank with a 5 62°, r 5 19 in., t 5 0.65 in., p 5 240 psi,
E 5 30 3 106 psi, and v 5 0.30.
Solution 8.3-13
a 5 62 deg
r 5 19 in.
p 5 240 psi
E 5 30 3 106 psi
LONGITUDINAL STRAIN
t 5 0.65 in.
v 5 0.3
â2 5
(a) CIRCUMFERENTIAL STRESS
s1 5
pr
t
pr
2t
;
u 5 90 deg 2 a
sx 5 s2
s2 5 3508 psi
s1 2 s2
2
s1
2
t1 5 1754 psi
t2 5 3508 psi
s1
(2 2 v)
2E
sy 5 s1
txy 5 0
For u 5 28 deg
sx 2 sy
sx + sy
+
cos (2u)
sx1 5
2
2
;
+ txy sin (2u)
sx1 5 4281 psi
;
sx 2 sy
sin (2u) + txy cos (2u)
tx1y1 5 2
2
;
tx1y1 5 1454 psi
(c) CIRCUMFERENTIAL STRAIN
â1 5
sx 5 3.508 3 103 psi
sy 5 7.015 3 10 psi
;
OUT-OF-PLANE SHEAR STRESS
t2 5
;
u 5 28 deg
3
(b) IN-PLANE SHEAR STRESS
t1 5
«2 5 4.68 3 1025
(d) STRESS ON WELD
s1 5 7015 psi
LONGITUDIAL STRESS
s2 5
s2
(1 2 2v)
E
«1 5 1.988 3 1024
;
;
sy1 5 sx 1 sy 2 sx1
sy1 5 6242 psi
;