SME
Materials Science & Material Properties · Problem type
Modulus of resilience
Use this for “A steel yields at 250 MPa and has E = 200 GPa. How much energy per cubic meter can it store elastically?”
Diagram
The elastic line σ = Eε; the 0.2% offset line meets the curve at σY.
Uᵣ
=
σY
2
2E
=
250
2
2 × 200,000
=
0.156
MJ/
m
3
,
the
shaded
triangle.
This is an example. Type your own values to start a new problem.
Uᵣ (MJ/m³)
=
(
σY (MPa)
)
2
2
×
E (GPa)
εY
Strain at yield
calculated
Formulas
εY
=
σY
E
0.00125
=
250
200,000
Uᵣ
=
σY
2
2 ×
E
0.1563
=
250
2
2 × 200,000
Formulas are worked in MPa. The step-by-step shows the conversions.
Assumptions
•
Linear elastic up to σY: the energy stored is the triangle under the line, ½σYεY.
•
Uᵣ is energy per volume: 1 MPa is 1 MJ/m³, and 1 psi is 1 in·lbf/in³.
•
σY stands in for the proportional limit, as the texts do.
Step-by-step
WE KNOW
σY = 250 MPa, E = 200 GPa
FIND
strain at yield (εY), modulus of resilience (Uᵣ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to MPa).
E = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
Step 2 · Find strain at yield (εY)
Use εY = σY ÷ E
Hooke’s law up to yield: the strain there is the yield strength over the modulus.
εY
=
250
200,000
εY = 0.00125
Step 3 · Find modulus of resilience (Uᵣ)
Use Uᵣ = σY² ÷ (2 × E)
The
triangle’s
area
is
1
2
σYεY,
and
εY
=
σY
E
:
so
σY
2
over
2
E
.
Uᵣ
=
250
2
2 × 200,000
Uᵣ
=
250
2
400,000
Uᵣ
=
62,500
400,000
Uᵣ
=
0.1563
MJ/
m
3
Check
Put the numbers back in. Both sides should match.
0.00125
=
250
200,000
✓
0.1563
=
250
2
2 × 200,000
✓
Topic check
See what you know: 5 questions on this topic
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