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STRESS ON THIN-WALLED VESSEL (Sample Problem) PROBLEM # 1:
PROBLEM # 3:
A 300 mm diameter steel pipe 12 mm thick carries water under a head of 50 m of water. Determine the stress in the steel.
Given:
Determine the stress at the walls of a 300 mm diameter pipe, 10 mm thick, under a pressure of 150 m of water and submerged to a depth of 20 m in salt water.
Solution:
inside dia. of pipe, D thickness of pipewall, t pressure head, h
= = =
300 mm or 0.3 m 12 mm or 0.012 m 50 m
Solution:
p= pinside − poutside KN KN ¿ 9.81 3 ( 150 m )− 9.81 3 ( 1.03 ) ( 20 m) m m ¿ 1269.4 KPa or 1.2694 MPa
(
)
Tangential Stress:
ST =
pD = 2t
(
9.81
KN ( 50 m ) ( 0.3 m) 3 m 2 ( 0.012 m )
)
KN ¿ 6131. 25 2 ∨KPa m PROBLEM # 2: Determine the required thickness of a 450 mm diameter steel pipe to carry a maximum pressure of 5500 kPa if the allowable working stress of steel is 124 MPa. Given: inside dia. of pipe, D Allowable stress of steel, St maximum pressure, p
= = =
ST =
(
)
pD 1.269 MPa ( 0.3 m ) = =12.69 MPa 2t 2(0.010 m)
PROBLEM # 4: A wooden storage vat is 6 m in diameter and is filled with 7 m of oil, s = 0.8. The wood staves are bound by flat steel bands, 50 mm wide by 6 mm thick, whose allowable tensile stress is 110 MPa. What is the required spacing of the bands near the bottom of the vat, neglecting any initial stress? Solution:
450 mm or 0.45 m 124 MPa 5500 kPa
Solution: Tangential Stress:
ST =
pD 2t
124 MPa
kPa ( 5500 kPa ) ( 0.45 m ) = ( 1000 1 M Pa ) 2t
Allowable tensile stress of hoops,
S t =110 MPa A h=50 ( 6 )=300 mm2 D=6 m=6000 mm
Cross-sectional area of hoops, Pipe diameter, Maximum pressure of the tank (at bottom):
(
p= 9.81
KN ( 0.8 ) ( 7 m )=54.936 kPa m3
)
t=0 . 00998 m∨9 . 98 mm say 10 mm
PROBLEM # 7:
PROBLEM # 5: A thin-walled hollow sphere 3.5 m in diameter holds helium gas at1700 kPa. Determine the minimum wall thickness of the sphere if its allowable stress is 60 MPa.
A cylindrical container 8 m high and 3 m diameter is reinforced with two hoops 1 meter from each end. When it is filled with water, what is the tension in each hoop due to water?
Solution: Solution: Wall stress
St =
F=γ h A KN 8 m ¿ 9.81 3 ( 8 m ) ( 3 m) 2 m ¿ 941.76 kN
60000 kPa=
)( )
(
pD 4t
( 1700 kPa ) ( 3500mm ) 4t
∑ M top hoop=0
t=24.79 mm
2T 2 ( 6 m) =F
( 133 m)
PROBLEM # 6:
2T 2 ( 6 m) =( 941.76 kN )
A vertical cylindrical tank is 2 meters in diameter and 3 meters high. Its sides are held in position by means of two steel hoops, one at the top and the other at the bottom. If the tank is filled with water to a depth of 2.1 m, determine the tensile stress in each hoop.
T 2 =340.08 KN
∑ M top hoop=0 2T 1 ( 6 m) =F
( 53 m)
2T 1 ( 6 m) =( 941.76 kN ) T 1 =130.8 kN
Substitute F to Eq. 1:
Solution:
∑ M TOP=0 2T 2 (3 m )=F ( 2.3 m)
( 133 m)
T 2 =0.3833 ( 43.26 kN ) ¿ 16.58 kN (tension at the bottom hoop)
T 2 =0.3833 F → E1 F=γ h A KN ¿ 9.81 3 m ¿ 43.26 kN
(
( 2 m) ( 2.1 m) () 2.1m 2 )
PROBLEM # 8: A cylindrical tank with its axis vertical is 1 meter in diameter and 6 m high. It is held together by two steel hoops, one at the top and the other at the bottom. Three liquids A, B, and C having sp. Gravity of 1.0, 2.0, and 3.0 respectively fills this tank each having a depth of 1.20 m. On the surface of A there is atmospheric pressure. Find the tensile stress in each hoop if each has a cross-sectional area of 1250 m2 . Solution:
Moment Arm:
Pressure:
p1=0 p2= p1 +γ A h A
(
¿ 0+ ( 1.0 ) 9.81
KN ( 1.2 m ) m3
)
¿ 11.77 kPa p3= p2 +γ B h B
(
¿ 11.77kPa+ ( 2.0 ) 9.81
KN ( 1.2m ) 3 m
)
¿ 35.31 kPa
2 2 4 y 1= ( AB )= ( 1.2 )= ∨0.8 m 3 3 5 1 1 y 2= ( BC ) + AB= ( 1.2m ) +1.2 m 2 2 ¿ 1.8 m 2 2 y 3= ( BC ) + AB= ( 1.2 m )+1.2 m 3 3 ¿2m 1 1 y 4 = (CD ) + BC + AB= ( 1.2 m) +1.2 m+ 1.2m 2 2 ¿3m 2 2 y 5= ( CD ) +BC + AB= ( 1.2m ) +1.2 m+1.2 3 3 ¿ 3.2 m
( 2 T 2 ) ( 3.6 m )=F 1 ( y 1 ) + F 2 ( y 2 ) + F 3 ( y 3 )+ F 4 ( y 4 ) + F 5 ( y 5 ) ( 7.2 m) T 2 =7.06 kN ( 0.8 m )+14.12 kN ( 1.8 m ) +14.12 kN ( 2m ) +42.37 kN ( 3 m) +21.19 kN ( 3.2 m ) T 2 =35.31kN Stress in the bottom hoop:
S 2=
T2 = A2
35.31 kN =28248 kPa o r 28.25 Mpa 2 1m 2 1250 mm 1000 mm
∑ F H =0
(
)
2T 1 +2T 2=F 1+ F 2 + F 3+ F 4 + F 5 2T 1=7.06 kN +14.12 kN +14.12 kN +42.37 kN
¿ 7.06 kN F2 =p 2 ( BC ) ( dia. ) ¿ ( 11.77 kPa ) ( 1.2 m) ( 1 m) ¿ 14.12 kN 1 F3 = ( p 3− p2 ) ( BC )( dia . ) 2 1 ¿ ( 35.31kPa−11.77 kPa ) ( 1.2 m) ( 1 m) 2 ¿ 14.12 kN F 4= p3 ( CD ) ( dia.) ¿ ( 35.31 kPa )( 1.2 m )( 1 m ) ¿ 42.37 kN 1 F5 = ( p 4− p3 ) ( CD ) (dia .) 2 1 ¿ (70.63 kPa−35.31 kPa )( 1.2 m )( 1 m ) 2 ¿ 21.19 kN
