Chapter I
STRESS
2.7 Thin-Walled Pressure Vessels
A pressure vessel is a pressurized
container, often cylindrical or spherical. The
pressure acting on the inner surface is resisted by
tensile stresses in the walls of the vessel. If the
wall thickness t is sufficiently small compared to
the radius r of the vessel, these stresses are
almost uniform throughout the wall thickness. It
can be shown that if 𝑟/𝑡 ≥ 10, the stresses
between the inner and outer surfaces of the wall
vary by less than 5%. In this section we consider
only vessels for which this inequality applies.
a. Cylindrical Vessels
Consider the cylindrical tank of inner
radius r and wall thickness t shown in Fig. 1.9(a).
The tank contains a fluid (or gas) under pressure
𝑝. In this simplified analysis, we assume that the
weights of the fluid and the vessel can be
neglected compared to the other forces that
act on the vessel. The tensile stresses in the wall
that resist the internal pressure are the
longitudinal stress 𝜎𝑙 and the circumferential
stress 𝜎𝑐 (also known as the hoop stress), as
shown in Fig. 2.12(a).
The circumferential stress can be
obtained from the free-body diagram in Fig.
2.12(b). This free body is obtained by taking the
slice of infinitesimal length 𝑑𝑥 shown in Fig.
2.12(a) and cutting it in half along a diametral
plane.
The fluid isolated by the cuts is
considered to be part of the free-body diagram.
The resultant force due to the pressure acting on
the diametral plane is 𝑝(2𝑟𝑑𝑥), where 2𝑟𝑑𝑥 is the
area of the plane.
If we assume the circumferential stress 𝜎𝑐
in the wall of the cylinder is constant throughout
the thickness, then its resultant force is 2(𝜎𝑐 𝑡𝑑𝑥).
Neglecting the weight of the fluid and the
vessel, we find that the equilibrium of vertical
forces becomes
Σ𝐹𝑣 = 0 ↑ +
2(𝜎𝑐 𝑡𝑑𝑥) − 𝑝(2𝑟𝑑𝑥) = 0
which yields for the circumferential stress
𝜎𝑐 =
𝑝𝑟 𝑝𝐷
=
𝑡
2𝑡
Equation (2-17)
To obtain the longitudinal stress s l , we
cut the cylinder into two parts along a crosssectional plane. Isolating the cylinder and the
fluid to the left of the cut gives the free-body
diagram in Fig. 2.12(c).
For thin-walled cylinders, the crosssectional area of the wall can be approximated
by (mean circumference) × (thickness) = (2𝜋𝑟̅)𝑡,
where 𝑟̅ = 𝑟 + 𝑡/2 is the mean radius of the
vessel. Therefore, the resultant of the longitudinal
stress is 𝜎𝑙 (2𝜋𝑟̅𝑡).
The resultant of the pressure acting on
the cross section is 𝑝(𝜋𝑟 2 ). From the equilibrium
of axial forces, we get
Σ𝐹𝐻 = 0 ⟶ +
𝜎𝑙 (2𝜋𝑟̅𝑡) − 𝑝(𝜋𝑟 2 ) = 0
Therefore, the longitudinal stress is 𝜎𝑙 = 𝑝𝑟 2 /2𝑟̅𝑡.
For thin-walled vessels, we can use the
approximation 𝑟̅ ≈ 𝑟, which results in
𝜎𝑙 =
Figure 2.12 (a) Cylindrical pressure vessel; (b) free-body
diagram for computing the circumferential stress 𝝈𝒄 ; (c) freebody diagram for computing the longitudinal stress 𝝈𝒍 .
𝑝𝑟 𝑝𝐷
=
2𝑡
4𝑡
Equation (2-18)
Comparing Eqs. (2-17) and (2-18), we see
that the circumferential stress is twice as large as
the longitudinal stress. It follows that if the
pressure in a cylinder is raised to the bursting
point, the vessel will split along a longitudinal
line.
When a cylindrical tank is manufactured
from curved sheets that are riveted together, as
in Fig. 2.13, the strength of longitudinal joint
should be twice the strength of girth joints.
denotes the mean radius of the vessel and 𝑡 is
the wall thickness. Therefore, the resultant force
due to 𝜎 is 𝜎(2𝜋𝑟̅𝑡). The equilibrium equation
Σ𝐹𝑣 = 0 ↑ +
𝜎(2𝜋𝑟̅𝑡) = 𝑝(𝜋𝑟 2 )
yields 𝜎 = 𝑝𝑟 2 /2𝑟̅ 𝑡. If we again neglect the small
difference between 𝑟̅ and 𝑟, the stress becomes
𝜎=
𝑝𝑟 𝑝𝐷
=
2𝑡
4𝑡
Equation (2-19)
As pointed out before, the di¤erence between the
inner radius 𝑟 and the mean radius 𝑟̅ of a thin-walled
vessel (𝑟/𝑡 ≥ 10) is insignificant, so that either radius
may be substituted for 𝑟 in Eqs. (2-17)–(2-19). The
stresses computed using 𝑟̅ rather than 𝑟 would be
different, of course, but the discrepancy is at most a few
percent.
Figure 2.13 Cylindrical pressure vessel made of curved sheets.
b. Spherical vessels
Using an analysis similar to that used for
cylinders, we can derive the expression for the
tensile stress s in the wall of the thin-walled,
spherical pressure vessel in Fig. 2.14(a).
ILLUSTRATIVE PROBLEMS
1.5 A cylindrical steel pressure vessel 400 mm in
diameter with a wall thickness of 20 mm, is
subjected to an internal pressure of 4.5 MN/m2.
(a) Calculate the circumferential and
longitudinal stresses in the steel.
(b) To what value may the internal pressure be
increased if the stress in the steel is limited to 120
MN/m2?
Solution:
Figure 2.14 (a) Spherical pressure vessel; (b) free-body
diagram for computing the stress 𝝈.
Because of symmetry, di¤erent directions
on the surface of the sphere are
indistinguishable. Therefore, the stress is constant
throughout the vessel.
As shown in Fig. 2.14(b), we use half of
the vessel as the free-body diagram. The fluid is
included in the free-body diagram, but its
weight is neglected together with the weight of
the vessel. The resultant force due to the
pressure acting on the circular surface of the
fluid is 𝑝(𝜋𝑟 2 ), where 𝑟 is the inner radius of the
vessel. We use again the approximation 2𝜋𝑟̅𝑡 for
the cross-sectional area of the wall, where 𝑟̅
a. Circumferential stress (longitudinal section)
𝜎𝑐 =
𝑝𝑟 𝑝𝐷
=
𝑡
2𝑡
𝜎𝑐 =
4.5(400)
2(20)
𝝈𝒄 = 𝟒𝟓 𝑴𝑷𝒂
1.6 A cylindrical steel pressure vessel has
hemispherical end-caps. The inner radius of the
vessel is 24 in. and the wall thickness is constant
at 0.25 in. When the vessel is pressurized to 125
psi, determine the stresses and the change in
the radius of (1) the cylinder; and (2) the endcaps. Use E = 29 x 106 psi and 𝑣 = 0.28 for steel.
Longitudinal stress (Transverse section)
Solution:
Part 1
The circumferential and longitudinal stresses in
the cylinder are
𝑝𝑟 𝑝𝐷 125(24)
=
=
𝑡
2𝑡
0.25
𝜎𝑙 =
𝑝𝑟 𝑝𝐷
=
2𝑡
4𝑡
𝜎𝑐 =
𝜎𝑙 =
4.5(400)
4(20)
𝝈𝒄 = 𝟏𝟐𝟎𝟎𝟎 𝒑𝒔𝒊
𝝈𝒍 = 𝟐𝟐. 𝟓 𝑴𝑷𝒂
b. Substitute the stress for circumferential or
longutidanl stress:
In circumferential stress:
𝑝1 𝑟 𝑝1 𝐷
=
𝑡
2𝑡
𝑝1 (400)
120 =
2(20)
𝜎𝑐 =
𝑝1 = 12 𝑀𝑃𝑎
In longitudinal stress:
𝑝2 𝑟 𝑝2 𝐷
=
2𝑡
4𝑡
𝑝2 (400)
120 =
4(20)
𝜎𝑙 =
𝑝2 = 24 𝑀𝑃𝑎
Since the pressure in longitudinal section is less
than the stress in the transverse section,
therefore, the masimum internal pressure that
can be applied until the vessel burst is 12 MPa.
𝜎𝑙 = 𝜎𝑐 /2 = 𝟔𝟎𝟎𝟎 𝒑𝒔𝒊
The circumferential strain is obtained from biaxial
Hooke’s law—see Eq. (2.10):
𝜖𝑐 =
1
(𝜎 − 𝑣𝜎𝑙 )
𝐸 𝑐
𝜖𝑐 =
1
[12000 − 0.28(6000)]
29 × 106
𝜖𝑐 = 355.9 × 10−6
Because the radius is proportional to the
circumference, 𝜖𝑐 is also the strain of the radius
(change of radius per unit length); that is,
𝜖𝑐 = 𝛿𝑟/𝑟. Therefore, the change in the radius of
the cylinder is
𝛿𝑟 = 𝜖𝑐 𝑟 = 355.9 × 10−6 (24)
𝜹𝒓 = 𝟎. 𝟎𝟎𝟖𝟒𝟓 𝒊𝒏.
Part 2
The stress in the spherical end-caps is
𝜎=
𝑝𝑟 𝑝𝐷 125(24)
=
=
2𝑡
4𝑡
2(0.25)
𝝈 = 𝟔𝟎𝟎𝟎 𝒑𝒔𝒊
Because 𝜎 acts biaxially, the strain must again
be computed from biaxial Hooke’s law, which
yields
𝜖=
1
1
(𝜎 − 𝑣𝜎) =
[6000 − 0.28(6000)]
𝐸
29 × 106
𝜖 = 148.97 × 10−6
Therefore, the change in the radius of an end-cap is
𝛿𝑟 = 𝜖𝑟 = 148.97 × 10−6 (24)
𝜹𝒓 = 𝟎. 𝟎𝟎𝟑𝟓𝟖 𝒊𝒏.
1.7 The cylindrical pressure vessel with
hemispherical end-caps is made of steel. The
vessel has a uniform thickness of 18 mm and an
outer diameter of 400 mm. When the vessel is
pressurized to 3.6 MPa, determine the change in
length in the cylinder and hemisphere and the
overall length of the vessel. Use E = 200 GPa and
𝑣 = 0.3 for steel. Neglect localized bending.
Because 𝜎 acts biaxially, the strain must again
be computed from biaxial Hooke’s law, which
yields
1
1
[18.2 − 0.30(18.2)]
𝜖 = (𝜎 − 𝑣𝜎) =
𝐸
200 × 103
𝜖 = 0.0000637
Therefore, the change in the radius of an end-cap is
364
𝛿𝑟 = 𝜖𝑟 = 0.0000637 (
)
2
𝜹𝒓 = 𝟎. 𝟎𝟏𝟏𝟔 𝒎𝒎
The overall change in length of the vessel is,
𝛿𝐿 𝑇 = 0.0218 + 2(0.0116
𝜹𝑳𝑻 = 𝟎. 𝟎𝟒𝟓 𝒎𝒎
Solution:
Part 1 Cylindrical part
Solving the inner diameter of the cylinder,
𝐷 = 400 − 2(18) = 364 𝑚𝑚
Solving the circumferential and longitudinal
stress,
𝜎𝑐 =
𝑝𝑟 𝑝𝐷 3.6(364)
=
=
𝑡
2𝑡
2(18)
𝜎𝑐 = 36.4 𝑀𝑃𝑎
𝜎𝑙 = 𝜎𝑐 /2 = 18.2 𝑀𝑃𝑎
The longitudinal strain is obtained from biaxial
Hooke’s law—see Eq. (2.10):
𝜖𝑙 =
1
(𝜎 − 𝑣𝜎𝑐 )
𝐸 𝑙
𝜖𝑙 =
1
[18.2 − 0.30(36.4)]
200 × 103
𝜖𝑙 = 0.0000364
𝛿𝐿 = 𝜖𝑙 𝐿 = 0.0000364(600)
𝜹𝑳 = 𝟎. 𝟎𝟐𝟏𝟖 𝒎𝒎
Part 2 Hemispherical part
Solving the inner diameter of the cylinder,
𝜎=
𝑝𝑟 𝑝𝐷 3.6(364)
=
=
2𝑡
4𝑡
4(18)
𝜎 = 18.20 𝑀𝑃𝑎