SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Diameter change by Poisson’s ratio
Use this for “A 20 mm steel rod is stretched to a strain of 7.96 × 10⁻⁴, with ν = 0.3. How much does its diameter change?”
Diagram
εₗₐₜ = −νε = −0.0002388: a bar that stretches gets thinner. Δd = εₗₐₜ × d = −0.004776 mm.
Both changes are drawn 200 times larger to be seen.
This is an example. Type your own values to start a new problem.
εₗₐₜ
=
−
ν
×
ε
Δd (mm)
=
εₗₐₜ
×
d (mm)
Formulas
εₗₐₜ
=
−
ν
×
ε
−2.388 ×
10
-4
=
−0.3
×
(7.96
×
10
-4
)
Δd
=
εₗₐₜ
×
d
−0.004776
=
−2.388 ×
10
-4
×
20
Assumptions
•
Linear elastic and isotropic: εₗₐₜ = −νε, the same in every sideways direction.
•
Tension is +: a bar that stretches (ε > 0) gets thinner (εₗₐₜ < 0, Δd < 0).
•
Steel by default, ν = 0.3; ν lies between 0 and 0.5.
Step-by-step
WE KNOW
ε = +7.96 × 10⁻⁴, ν = 0.3, d = 20 mm
FIND
lateral strain (εₗₐₜ), diameter change (Δd)
Step 1 · Find lateral strain (εₗₐₜ)
Use εₗₐₜ = −ν × ε
Poisson’s ratio is the sideways strain over the axial strain, with the sign flipped.
εₗₐₜ
=
−0.3
×
(7.96
×
10
-4
)
εₗₐₜ
=
−2.388 ×
10
-4
Step 2 · Find diameter change (Δd)
Use Δd = εₗₐₜ × d
Strain
is
change
over
original
size,
so
the
diameter
changes
by
εₗₐₜ
times
d
.
Δd
=
−2.388 ×
10
-4
×
20
Δd = −0.004776 mm
Check
Put the numbers back in. Both sides should match.
−2.388 ×
10
-4
=
−0.3
×
(7.96
×
10
-4
)
✓
−0.004776
=
−2.388 ×
10
-4
×
20
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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