solution manual - mgere p1

Spherical Pressure Vessels

When solving the problems for Section 8.2, assume that the given radius ordiameter 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 apressure of 450 psi. The tank is 42 ft in diameter and is constructed of high-strengthsteel having a yield stress in tension of 80 ksi.

Determine the required thickness (to the nearest 1/4 inch) of the wall of the tankif a factor of safety of 3.5 with respect to yielding is required.

Probs. 8.2-1 and 8.2-2

8Applications of Plane Stress

(Pressure Vessels, Beams,

and Combined Loadings)

649

Solution 8.2-1

Radius: r 5 252 in.

Internal Pressure: p 5 450 psiYield stress: sY 5 80 ksi (steel)Factor of safety: n 5 3.5MINIMUM WALL THICKNESS tmin

From Eq. (8-1): or sYn

5pr2 t

smax 5pr2 t

r 512

42 * 12 t 5 2.481 in.

to nearest 1/4 inch, tmin 5 2.5 in. ;

t 5prn2sY

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 is570 MPa, and the factor of safety is 3.0.

Determine the required thickness to the nearest millimeter.

Solution 8.2-2

Radius: r 5 9.5 3 103 mm

Internal Pressure: p 5 3.75 MPaYield stress: sY 5 570 MPaFactor of safety: n 5 3MINIMUM WALL THICKNESS tmin

From Eq. (8-1): or sYn

5pr2 t

smax 5pr2 t

r 512

(19 m) t 5 93.8 mm

Use the next higher millimeter tmin 5 94 mm ;

t 5prn2sY

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 tothe wall of the chamber by 18 bolts.

Find the tensile force F in each bolt and the tensile stress s in the viewport if theradius 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

FHORIZ 5 T 2 pA 5 0 T 5 pA 5 p(pr 2)

F 5 force in one bolt

TENSILE STRESS IN VIEWPORT (EQ. 8-1)

s 5pr2t

5 280 psi ;

F 5T18

5118

(ppr 2) 5 684 lb ;

a

Radius: r 5 7.0 in.Internal pressure: p 5 80 psiWall thickness: t 5 1.0 in.18 bolts

SECTION 8.2 Spherical Pressure Vessels 651

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 ofelasticity E 5 3.5 MPa and Poissons 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)

MAXIMUM STRAIN (EQ. 8-4)

5 0.452 ;

max 5pr2tE

(1 2 v) 5 (60 kPa)(115 mm)2(1.2 mm)(3.5 MPa) (0.55)

5 2.88 MPa ;

smax 5pr2t

5(60 kPa)(115 mm)

2(1.2 mm)

Radius: r 5 (230 mm)/2 5 115 mmInternal pressure: p 5 60 kPaWall thickness: t 5 1.2 mmModulus of elasticity: E 5 3.5 MPa (rubber)Poissons 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 Poissons ratio is 0.45.

Solution 8.2-5 Rubber ball

CROSS-SECTION Modulus of elasticity: E 5 500 psi (rubber)Poissons ratio: v 5 0.45 (rubber)

MAXIMUM STRESS (EQ. 8-1)

MAXIMUM STRAIN (EQ. 8-4)

5 0.446 ;

max 5pr2tE

(1 2 v) 5(9.0 psi)(4.5 in.)

2(0.05 in.)(500 psi) (0.55)

5 405 psi ;smax 5pr2t

5(9.0 psi)(4.5 in.)

2(0.05 in.)Radius:

Internal pressure: p 5 9.0 psiWall thickness: t 5 0.05 in.

r 512

(9.0 in.) 5 4.5 in.


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.)

Cracks incoating

Solution 8.2-6 Spherical vessel with brittle coating

CROSS-SECTION Cracks occur when max 5 150 3 1026

From Eq. (8-4):

5 2.93 MPa ;

P 52(8.0 mm)(205 GPa)(150 * 1026)

(240 mm)(0.70)

P 52tE maxr(12v)

max 5pr2tE

(1 2 v)

r 5 240 mm E 5 205 GPa (steel)t 5 8.0 mm v 5 0.30

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).

(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

Weld

Solution 8.2-7

r 5 24 in. E 5 30 3 106 psit 5 1.75 in. v 5 0.29 (steel)

(a) TENSILE LOAD CARRIED BY WELDT 5 Total load f 5 load per inchT 5 pA 5 ppr 2 c 5 Circumference of tank 5 2pr

5 26.4 k/in. ;

5(2200 psi)(24 in.)

2f 5 T

c5

p1pr 222pr

5pr2

(b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3)

(c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4)

5 3.57 * 1024 ;

max 5pr (1 2 v)

2 tE5

(2200 psi)(24 in.)(0.71)2(1.75 in.)130106 psi2

5 7543 psitmax 5pr4t

5(2200 psi)(24 in.)

4 (1.75 in.)

SECTION 8.2 Spherical Pressure Vessels 653

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 Poissons ratio 0.29.

Solution 8.2-8

r 5 0.5 m E 5 210 GPat 5 48 mm v 5 0.29 (steel)

(a) TENSILE LOAD CARRIED BY WELDT 5 Total load f 5 load per inchT 5 pA 5 ppr2 c 5 Circumference of tank 5 2pr

5 5.5 MN/m ;

f 5 Tc

5p1pr 22

2pr5

pr2

5(22 MPa)(0.5 m)

2

(b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3)

(c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4)

5 3.87 * 1024 ;

max 5pr (1 2 v)

2 tE5

(22 MPa)(0.5 m) 0.712(48 mm)(210 GPa)

5 57.3 MPa ;

tmax 5pr4t

5(22 MPa)(0.5 m)

4(48 mm)

Problem 8.2-9 A spherical stainless-steel tank having a diameter of 22 in. is used to store propane gas at a pressure of2450 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 Poissons 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

r 5 11 in. E 5 30 3 106 psip 5 2450 psi v 5 0.28 (steel)sY 5 140000 psi n 5 2.8tY 5 65000 psi max 5 1100 3 1026

MIMIMUM WALL THICKNESS t

(1) TENSION (EQ. 8-1)

5(2450 psi) (11 in.)

2 140000 psi

2.8

5 0.269 in.

t1 5pr

2smax5

pr

2asYnb

smax 5pr2 t1

(2) SHEAR (EQ. 8-3)

(3) STRAIN (EQ. 8-4)

t3 . t2 . t1 Thus, tmin 5 0.294 in. ;

5 0.294 in.

5(2450 psi)(11 in.)

2 11100 * 10262130 * 106 psi2 0.72

t3 5pr

2max E (1 2 v)

max 5pr

2t3 E (1 2 v)

t2 5pr

4tYn

5(2450 psi)(11 in.)

4 65000 psi

2.8

5 0.29 in.

tmax 5pr4 t2


Problem 8.2-10 Solve the preceding problem if the diameter is 500 mm, the pressure is 18 MPa, the yield stress in tensionis 975 MPa, the yield stress in shear is 460 MPa, the factor of safety is 2.5, the modulus of elasticity is 200 GPa, Poissonsratio is 0.28, and the normal strain must not exceed 1210 3 1026.

Solution 8.2-10

r 5 250 mm E 5 200 GPap 5 18 MPa v 5 0.28 (steel)sY 5 975 MPa n 5 2.5tY 5 460 MPa max 5 1210 3 1026

MINIMUM WALL THICKNESS t

(1) TENSION (EQ. 8-1)

5(18 MPa)(250 mm)

2 975 MPa

2.5

5 5.769 mm

t1 5pr

2smax5

pr

2 asYnb

smax 5pr2t1

(2) SHEAR (EQ. 8-3)

(3) STRAIN (EQ. 8-4)

t3 . t2 . t1 Thus, tmin 5 6.69 mm ;

5 6.694 mm

5(18 MPa)(250 mm)

211210 * 10262(200 GPa) 0.72

t3 5pr

2max E (1 2 v)

max 5pr

2 t3 E (1 2 v)

t2 5pr

4 tY

n

5(18 MPa)(250 mm)

4 460 MPa

2.5

5 6.114 mm

tmax 5pr4t2

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 compressedair in the tank is at a pressure of 24 psi (gage pressure when the tank is outof the water).

At what depth D0 will the wall of the tank be subjected to a compressivestress of 90 psi?

D0

SECTION 8.3 Cylindrical Pressure Vessels 655

Solution 8.2-11 Pressurized sphere under water

CROSS-SECTIONr 5 4.8 in. p1 5 24 psit 5 0.4 in. g 5 density of water 5 62.4 lb/ft3

(1) IN AIR: p1 5 24 psi

D0 5 depth of water (in.)

Compressive stress in tank wall equals 90 psi.(Note: s is positive in tension.)

5 1080 in. 5 90 ft ;

Solve for D0: D0 5234

0.21667

5 144 2 0.21667 D0

290 psi 5(24 psi 2 0.03611 D0)(4.8 in.)

2(0.4 in.)

s 5pr2t

5( p1 2 p2)r

2t s 5 290 psi

p2 5 gD0 5 a 62.4 lb/ft3

1728 in.3/ ft3bD0 5 0.036111 D0 ( psi)

(2) UNDER WATER: p1 5 24 psi

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 is35,000 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?

(1) IN AIR

(2) UNDER WATER


Solution 8.3-1 Scuba tank

Cylindrical pressure vesselp 5 1600 psi n 5 2.0 d 5 7.0 in.r 5 3.5 in. sY 5 35,000 psi tY 5 16,000 psi

Find required wall thickness t.

(1) BASED ON TENSION (EQ. 8-5)

(2) BASED ON SHEAR (EQ. 8-10)

Shear governs since t2 . t1[ tmin 5 0.350 in. ;

t2 5pr

2tallow5

(1600 psi)(3.5 in.)2(8,000 psi) 5 0.350 in.

tmax 5pr2t

t1 5pr

sallow5

(1600 psi)(3.5 in.)17,500 psi 5 0.320 in.

smax 5prt

sallow 5sY

n5 17,500 psi tallow 5

tY

n5 8,000 psi

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.

(a) What height h of water will produce a circumferential stress of 12 MPa in the wall ofthe standpipe?

(b) What is the axial stress in the wall of the tank due to the water pressure?

d

h

Solution 8.3-2

d 5 2.2 m r 5 1.1 m t 5 20 mmweight density of water height of water hwater pressure p 5 gh

(a) HEIGHT OF WATER

s1 5prt

5 12 MPa 50.00981h (1.1 m)

20 mm

g 5 9.81 kN/m3 (b) AXIAL STRESS IN THE WALL DUE TO WATERPRESSURE ALONE

Because the top of the tank is open, the internalpressure of the water produces no axial (longitu-dinal) stresses in the wall of the tank. Axial stressequals zero. ;

h 512 (20)

0.00981 (1.1) 5 22.2 m


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 asmall blower and has a radius of 40 ft when fully inflated. Alongitudinal seam runs the entire length of the ridge of thestructure.

If the longitudinal seam along the ridge tears open when itis subjected to a tensile load of 540 pounds per inch of seam,what is the factor of safety n against tearing when the internalpressure is 0.5 psi and the structure is fully inflated?

Longitudinal seam

Solution 8.3-3 Inflatable structure

Half-circular cylinderr 5 40 ft 5 480 in.Internal pressure p 5 0.5 psi

T 5 tensile force per unit length of longitudinal seamSeam tears when T 5 Tmax 5 540 lb/in.Find factor of safety against tearing.

CIRCUMFERENTIAL STRESS (EQ. 8-5)

where t 5 thickness of fabric

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

n 5Tmax

T5

540 lb/in.240 lb/in.

5 2.25 ;

s1 5prt

Problem 8.3-4 A thin-walled cylindrical pressure vessel of radius r is subjectedsimultaneously to internal gas pressure p and a compressive force F acting at theends (see figure).

What should be the magnitude of the force F in order to produce pure shearin the wall of the cylinder?

FF

Solution 8.3-4 Cylindrical pressure vessel

r 5 Radiusp 5 Internal pressure

STRESSES (SEE EQ. 8-5 AND 8-6):

s2 5pr2t

2FA

5pr2t

2F

2prt

s1 5prt


FOR PURE SHEAR, the stresses s1 and s2 must be equal inmagnitude and opposite in sign (see, e.g., Fig. 7-11 inSection 7.3).[ s1 5 2s2

OR

Solve for F: F 5 3ppr2 ;

prt

5 2 apr2t

2F

2prtb

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 is200. 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

E 5 10 3 106 psi v 5 0.33

0 5 change in strain when pressure is released5 170 3 1026

Find internal pressure p.

r

t5 200

STRAIN IN LONGITUDINAL DIRECTION (EQ. 8-11a)

Substitute numerical values:

p 52(10 * 106 psi)(170 * 1026)

(200)(1 2 0.66) 5 50 psi ;

2 5 0 p 52tE0

(r)(1 2 2v) 52E0

(r/t) (1 2 2v)

2 5pr2tE

(1 2 2v) or p 5 2tE2r(1 2 2v)


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 controlroom. 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 reducethe pressure in the tank? (Data for the steel are as follows: modulus ofelasticity E 5 205 GPa and Poissons ratio v 5 0.30.)

A

Cylindrical tank

Pressure reliefvalve

Solution 8.3-6

tULT 5 84 MPa E 5 205 GPa v 5 0.3

n 5 2.5 tmax 5 33.6 MPa

Find maximum allowable strain reading at the gage

From Eq. (8-10)

Pmax 52ttmax

rtmax 5

s12

5pr2t

s2 5pr2t

s1 5prt

tmax 5tULT

n

From Eq. (8-11a)

max 5 6.56 3 1025max 5tmax

E (1 2 2v)

2max 5pmax r2tE

(1 2 2v) 5 tmaxE

(1 2 2v)

2 5pr

2tE (1 2 2v)

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 shearstress tallow in the wall of the cylinder is 5500 psi.

What is the minimum permissible thickness tmin of the cylinderwall? (See the figure on the next page.)

Probs. 8.3-7 and 8.3-8

Cylinder

F

Piston

p

Solution 8.3-7 Cylinder with internal pressure

d 5 1.80 in. r 5 0.90 in.F 5 3500 lb tallow 5 5500 psiFind minimum thickness tmin.

Pressure in cylinder: p 5FA

5F

pr 2

Maximum shear stress (Eq. 8-10):

Minimum thickness:


tmin 53500 lb

2p(0.90 in.)(5500 psi) 5 0.113 in. ;

tmin 5F

2prtallow

tmax 5pr2t

5F

2prt


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

d 5 90 mm r 5 45 mmF 5 42.0 kN tallow 5 40 MPaFind minimum thickness tmin.

Pressure in cylinder: p 5FA

5F

pr 2

Maximum shear stress (Eq. 8-10):

Minimum thickness:


tmin 542.0 kN

2p(45 mm)(40 MPa) 5 3.71 mm ;

tmin 5F

2pr tallow

tmax 5pr2t

5F

2prt

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 thebottom 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

d 5 12 ft 5 144 in. r 5 72 in. t 5 6 in.

s1 5 hoop stress at bottom of standpipe 5 130 psi

g 5 62.4 lb/ft3 562.41728

lb/in.3

(a) FIND HEIGHT h OF WATER IN THE STANDPIPEp 5 pressure at bottom of standpipe 5 gh

From Eq. (8-5):


5 25 ft ;

h 5(130 psi)(6 in.)

a 62.41728

lb/in.3b (72 in.)5 300 in.

s1 5prt

5ghr

t or h 5

s1t

gr


HORIZONTAL PIPES

d1 5 2 ft 5 24 in. r1 5 12 in. t1 5 1.0 in.

(b) FIND HOOP STRESS s1 IN THE PIPESSince the pipes are 2 ft in diameter, the depth ofwater to the center of the pipes is about 24 ft.h1 < 24 ft 5 288 in. p1 5 gh1

Based on the average pressure in the pipes:s1 < 125 psi ;

5 125 psi

5

a 62.41728

lb/in.3b (288 in.)(12 in.)1.0 in.

s1 5p1r1t1

5gh1r1

t1

Problem 8.3-10 A cylindrical tank with hemispherical heads is constructed of steel sections that are welded circumferentially (see figure). The tank diameter is1.25 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. Probs. 8.3-10 and 8.3-11(e) Determine the maximum shear stress tc in the cylindrical part of the tank.

Welded seams

Solution 8.3-10

d 5 1.25 m t 5 22 mm p 5 1750 kPa

(a) MAXIMUM TENSILE STRESS IN HEMISPHERES (EQ. 8-1)

sh 5 24.9 MPa

(b) MAXIMUM STRESS IN CYLINDER (EQ. 8-5)

sc 5 49.7 MPa ;sc 5prt

;sh 5pr2 t

r 5d2

(c) TENSILE STRESS IN WELDS (EQ. 8-6)

sw 5 24.9 MPa

(d) MAXIMUM SHEAR STRESS IN HEMISPHERES (EQ. 8-3)

th 5 12.43 MPa

(e) MAXIMUM SHEAR STRESS IN CYLINDER (EQ. 8-10)

tc 5 24.9 MPa ;tc 5pr2 t

;th 5pr4 t

;sw 5pr2t

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 isconstructed of steel sections that are welded circumferentially (see figure). The heads of the tank are hemispherical. Theallowable tensile and shear stresses are 8200 psi and 3000 psi, respectively. Also, the allowable tensile stress perpendicularto a weld is 6250 psi.

Determine the minimum required thickness tmin of (a) the cylindrical part of the tank and (b) the hemispherical heads.


Solution 8.3-11

d 5 18 in. p 5 450 psi

sallow 5 8200 psi (tension)tallow 5 3000 psi (shear)WELD sa 5 6250 psi (tension)

(a) FIND MINIMUM THICKNESS OF CYLINDER

TENSION

tmin 5 0.494 in.

SHEAR

tmin 5 0.675 in.

WELD s 5pr2t

tmin 5pr

2tallow

tmax 5pr2 t

tmin 5pr

sallow

smax 5prt

r 5d2

tmin 5 0.324 in.

(b) FIND MINIMUM THICKNESS OF HEMISPHERES

TENSION

tmin 5 0.247 in.

SHEAR

tmin 5 0.338 in.

tmin 5 0.338 in. ;

tmin 5pr

4tallow

tmax 5pr4t

tmin 5pr

2sallow

smax 5pr2t

tmin 5 0.675 in. ;

tmin 5pr

2sa

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 tankhas radius r 5 0.6 m, wall thickness t 5 18 mm, and internal pressure p 52.8 MPa. Also, the steel has modulus of elasticity E 5 200 GPa and Poissonsratio v 5 0.30.

Determine the following quantities for the cylindrical part of the tank.(a) The circumferential and longitudinal stresses.(b) The maximum in-plane and out-of-plane shear stresses.(c) The circumferential and longitudinal strains.(d) The normal and shear stresses acting on planes parallel and perpendicular to Probs. 8.3-12 and 8.3-13

the weld (show these stresses on a properly oriented stress element).

Solution 8.3-12

a 5 55 deg r 5 0.6 m t 5 18 mmp 5 2.8 MPa E 5 200 GPa v 5 0.3

(a) CIRCUMFERENTIAL STRESS

s1 5 93.3 MPa

LONGITUDIAL STRESS

s2 5 46.7 MPa ;s2 5pr2t

;s1 5prt

(b) IN-PLANE SHEAR STRESS

t1 5 23.3 MPa

OUT-OF-PLANE SHEAR STRESS

t2 5 46.7 MPa ;t2 5s12

;t1 5s1 2 s2

2

Helical weld

a


(c) CIRCUMFERENTIAL STRAIN

1 5 3.97 3 1024

LONGITUDINAL STRAIN

2 5 9.33 3 1025

(d) STRESS ON WELDu 5 90 deg 2 a u 5 35 degsx 5 s2 sx 5 46.667 MPa sy 5 s1sy 5 93.333 MPa txy 5 0

;2 5s2E

(1 2 2v)

;1 5s12E

(2 2 v)For u 5 35 deg

sx1 5 62.0 MPa

tx1y1 5 21.9 MPasy1 5 sx 1 sy 2 sx1 sy1 5 78.0 MPa ;

;

tx1y1 5 2sx 2 sy

2 sin (2u) + txy cos (2u)

;

+ txy sin (2u)

sx1 5sx + sy

2+

sx 2 sy

2 cos (2u)

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. t 5 0.65 in.p 5 240 psi E 5 30 3 106 psi v 5 0.3

(a) CIRCUMFERENTIAL STRESS

LONGITUDIAL STRESS

(b) IN-PLANE SHEAR STRESS

OUT-OF-PLANE SHEAR STRESS

(c) CIRCUMFERENTIAL STRAIN

1 5 1.988 3 1024 ;1 5s12E

(2 2 v)

;t2 5 3508 psit2 5s12

;t1 5 1754 psit1 5s1 2 s2

2

;s2 5 3508 psis2 5pr2t

;s1 5 7015 psis1 5prt

LONGITUDINAL STRAIN

2 5 4.68 3 1025

(d) STRESS ON WELDu 5 90 deg 2 a u 5 28 degsx 5 s2 sx 5 3.508 3 103 psi sy 5 s1sy 5 7.015 3 103 psi txy 5 0

For u 5 28 deg

sy1 5 sx 1 sy 2 sx1 ;sy1 5 6242 psi

;tx1y1 5 1454 psi

tx1y1 5 2sx 2 sy

2 sin (2u) + txy cos (2u)

;sx1 5 4281 psi

+ txy sin (2u)

sx1 5sx + sy

2+

sx 2 sy

2 cos (2u)

;2 5s2E

(1 2 2v)

solution manual - mgere p1

Documents