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
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
Solution:
𝑆𝑇 =
Tangential Stress:
𝑆𝑇 =
𝑝𝐷
2𝑡
𝐾𝑁
=
(9.81 3 )(50 𝑚)(0.3 𝑚)
𝑚
= 𝟔𝟏𝟑𝟏. 𝟐𝟓
𝒐𝒓 𝑲𝑷𝒂
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
=
=
=
2𝑡
=
1.269 𝑀𝑃𝑎 (0.3 𝑚)
2(0.010 𝑚)
= 𝟏𝟐. 𝟔𝟗 𝑴𝑷𝒂
PROBLEM # 4:
2(0.012 𝑚)
𝑲𝑵
𝒎𝟐
𝑝𝐷
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:
𝑝𝐷
𝑆𝑇 =
2𝑡
124 𝑀𝑃𝑎 (
1000 𝑘𝑃𝑎
1 𝑀𝑃𝑎
)=
(5500 𝑘𝑃𝑎)(0.45 𝑚)
2𝑡
𝒕 = 𝟎. 𝟎𝟎𝟗𝟗𝟖 𝒎 𝒐𝒓 𝟗. 𝟗𝟖 𝒎𝒎 𝒔𝒂𝒚 𝟏𝟎 𝒎𝒎
Allowable tensile stress of hoops, 𝑆𝑡 = 110 𝑀𝑃𝑎
Cross-sectional area of hoops,
𝐴ℎ = 50(6) = 300 𝑚𝑚2
Pipe diameter,
𝐷 = 6𝑚 = 6000 𝑚𝑚
Maximum pressure of the tank (at bottom):
𝐾𝑁
𝑝 = (9.81 3 ) (0.8)(7 𝑚) = 54.936 𝑘𝑃𝑎
𝑚
Spacing of Hoops, S:
𝑆=
2𝑆𝑡 𝐴ℎ
𝑝𝐷
=
2(110000 𝐾𝑃𝑎)(300 𝑚𝑚2 )
(54.936 𝑘𝑃𝑎)(6000 𝑚𝑚)
= 𝟐𝟎𝟎. 𝟐𝟑 𝒎𝒎 𝑠𝑎𝑦
= 𝟐𝟎𝟎 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:
𝐹 = γhA
KN
8m
= (9.81 3 ) ( ) (8 m)(3 m)
m
2
= 941.76 𝑘𝑁
Wall stress
pD
St =
4t
(1700 kPa)(3500 mm)
60000 kPa =
t = 24.79 mm
∑ 𝑀𝑡𝑜𝑝 ℎ𝑜𝑜𝑝 = 0
4t
2T2 (6 m) = F (
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.
∑ MTOP = 0
2T2 (3 m) = F(2.3 m)
T2 = 0.3833F → E1
F = γhA
KN 2.1 m
)(
) (2 m)(2.1 m)
m3
2
= 43.26 kN
= (9.81
3
m)
13
2T2 (6 m) = (941.76 𝑘𝑁) ( m)
3
𝐓𝟐 = 𝟑𝟒𝟎. 𝟎𝟖 𝐊𝐍
PROBLEM # 6:
Solution:
13
∑ 𝑀𝑡𝑜𝑝 ℎ𝑜𝑜𝑝 = 0
5
2T1 (6 m) = F ( m)
3
Substitute F to Eq. 1:
T2 = 0.3833(43.26 kN)
= 𝟏𝟔. 𝟓𝟖 𝒌𝑵 (tension at the bottom hoop)
∑ 𝐹𝐻 = 0
2T2 + 2T1 − F = 0
2(16.58 𝑘𝑁) + 2T1 − 43.26 kN = 0
𝐓𝟏 = 𝟓. 𝟎𝟓 𝐊𝐍
5
2T1 (6 m) = (941.76 𝑘𝑁) ( m)
3
𝐓𝟏 = 𝟏𝟑𝟎. 𝟖 𝐤𝐍
(tension in the top hoop)
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:
Pressure:
𝑝1 = 0
𝑝2 = 𝑝1 + 𝛾𝐴 ℎ𝐴
= 0 + (1.0) (9.81
= 11.77 𝑘𝑃𝑎
Moment Arm:
2
2
4
𝑦1 = (𝐴𝐵) = (1.2) = 𝑜𝑟 0.8 𝑚
KN
m3
) (1.2 𝑚)
𝑝3 = 𝑝2 + 𝛾𝐵 ℎ𝐵
= 11.77 𝑘𝑃𝑎 + (2.0) (9.81
= 35.31 𝑘𝑃𝑎
𝑝4 = 𝑝3 + 𝛾𝐶 ℎ𝑐
= 35.31 𝑘𝑃𝑎 + (3.0) (9.81
= 70.63 𝑘𝑃𝑎
KN
m3
KN
m3
) (1.2 𝑚)
) (1.2 𝑚)
Forces acting on the Tank wall:
1
𝐹1 = 𝑝2 (𝐴𝐵)(𝑑𝑖𝑎. )
2
1
= (11.77 𝑘𝑃𝑎)(1.2 𝑚)(1 𝑚)
2
= 7.06 𝑘𝑁
𝐹2 = 𝑝2 (𝐵𝐶)(𝑑𝑖𝑎. )
= (11.77 𝑘𝑃𝑎)(1.2𝑚)(1 𝑚)
= 14.12 𝑘𝑁
1
𝐹3 = (𝑝3 − 𝑝2 )(𝐵𝐶)(𝑑𝑖𝑎. )
2
1
= (35.31 𝑘𝑃𝑎 − 11.77 𝑘𝑃𝑎)(1.2 𝑚)(1 𝑚)
2
= 14.12 𝑘𝑁
𝐹4 = 𝑝3 (𝐶𝐷)(𝑑𝑖𝑎. )
= (35.31 𝑘𝑃𝑎)(1.2𝑚)(1 𝑚)
= 42.37 𝑘𝑁
1
𝐹5 = (𝑝4 − 𝑝3 )(𝐶𝐷)(𝑑𝑖𝑎. )
2
1
= (70.63 𝑘𝑃𝑎 − 35.31 𝑘𝑃𝑎)(1.2𝑚)(1 𝑚)
2
= 21.19 𝑘𝑁
3
1
3
1
5
𝑦2 = (𝐵𝐶) + 𝐴𝐵 = (1.2 𝑚) + 1.2 𝑚
2
2
= 1.8 𝑚
2
2
𝑦3 = (𝐵𝐶) + 𝐴𝐵 = (1.2 𝑚) + 1.2 𝑚
3
3
= 2𝑚
1
1
𝑦4 = (𝐶𝐷) + 𝐵𝐶 + 𝐴𝐵 = (1.2𝑚) + 1.2𝑚 + 1.2𝑚
2
2
= 3𝑚
2
2
𝑦5 = (𝐶𝐷) + 𝐵𝐶 + 𝐴𝐵 = (1.2𝑚) + 1.2𝑚 + 1.2
3
3
= 3.2 𝑚
∑ 𝑀𝐴 = 0
(2𝑇2 )(3.6 𝑚) = 𝐹1 (𝑦1 ) + 𝐹2 (𝑦2 ) + 𝐹3 (𝑦3 ) + 𝐹4 (𝑦4 ) + 𝐹5 (𝑦5 )
(7.2 𝑚)𝑇2 = 7.06 𝑘𝑁(0.8 𝑚) + 14.12 𝑘𝑁(1.8𝑚)
+14.12 𝑘𝑁(2 𝑚) + 42.37 𝑘𝑁(3 𝑚)
+ 21.19 𝑘𝑁(3.2 𝑚)
𝑇2 = 35.31 𝑘𝑁
Stress in the bottom hoop:
𝑇
35.31 𝑘𝑁
𝑆2 = 2 =
1𝑚
𝐴2
1250𝑚𝑚2 (
)
2
= 𝟐𝟖𝟐𝟒𝟖𝒌𝑷𝒂 𝒐𝒓 𝟐𝟖. 𝟐𝟓 𝑴𝒑𝒂
1000 𝑚𝑚
∑ 𝐹𝐻 = 0
2𝑇1 + 2𝑇2 = 𝐹1 + 𝐹2 + 𝐹3 + 𝐹4 + 𝐹5
2𝑇1 = 7.06 𝑘𝑁 + 14.12 𝑘𝑁 + 14.12 𝑘𝑁 + 42.37 𝑘𝑁
+21.19 𝑘𝑁 − 2(35.31 𝑘𝑁)
𝑇1 = 14.12 𝑘𝑁
Stress in the top hoop:
𝑇
14.12 𝑘𝑁
𝑆1 = 1 =
1𝑚
𝐴1
1250𝑚𝑚2 (
1000 𝑚𝑚
2
)
= 𝟏𝟏𝟐𝟗𝟔 𝑲𝑷𝒂 𝒐𝒓 𝟏𝟏. 𝟑 𝑴𝑷𝒂