SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Thin-walled pressure vessel
Use this for “A tank 500 mm in radius with a 10 mm wall holds 2 MPa. Find the hoop and axial stresses,” or the stress in a sphere (σₐ).
Diagram
Hoop
stress
σₕ
=
pr
t
round
the
wall
is
twice
the
axial
σₐ
=
pr
2t
.
This is an example. Type your own values to start a new problem.
σₕ (MPa)
=
p (MPa)
×
r (mm)
t (mm)
σₐ (MPa)
=
p (MPa)
×
r (mm)
2
×
t (mm)
Formulas
σₕ
=
p
×
r
t
100
=
2
×
500
10
σₐ
=
p
×
r
2 ×
t
50
=
2
×
500
2 × 10
Assumptions
•
Thin wall: r ÷ t is 10 or more, so the stress is even through the wall.
•
r is the inside radius; the closed ends carry the pressure, which makes σₐ.
•
A sphere has σ = pr ÷ (2t) in every direction: the axial value σₐ here.
•
MPa × mm ÷ mm stays MPa, so the steps need no conversion factor.
Step-by-step
WE KNOW
p = 2 MPa, r = 500 mm, t = 10 mm
FIND
hoop stress (σₕ), axial stress (σₐ)
Step 1 · Find hoop stress (σₕ)
Use σₕ = p × r ÷ t
Cut
the
cylinder
along
its
length:
the
pressure
on
2
r
is
held
by
two
walls
t
thick.
σₕ
=
2
×
500
10
σₕ
=
1,000
10
σₕ = 100 MPa
Step 2 · Find axial stress (σₐ)
Use σₐ = p × r ÷ (2 × t)
Cut
across
the
cylinder:
the
pressure
on
the
end,
πr²
p
,
is
held
by
the
ring
of
wall,
2πrt.
σₐ
=
2
×
500
2 × 10
σₐ
=
2
×
500
20
σₐ
=
1,000
20
σₐ = 50 MPa
Check
Put the numbers back in. Both sides should match.
100
=
2
×
500
10
✓
50
=
2
×
500
2 × 10
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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