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• Diego Bonfim

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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

;