SME
Mechanics of Materials (Solid Mechanics I)
Stress and strain
Use this for “A steel rod 20 mm across and 2 m long carries 50 kN. Find the stress, the strain and the stretch.”
Diagram
The elastic line σ = Eε.
The
point:
σ
=
Eε
=
200,000
MPa
×
7.96 ×
10
-4
=
159
MPa.
σ
=
F
A
=
50,000
N
314
mm²
=
159
MPa.
ε
=
δ
L
=
1.59
mm
2,000
mm
=
7.96 ×
10
-4
.
δ
is
drawn
200
times
larger
to
be
seen.
This is an example. Type your own values to start a new problem.
A (mm²)
=
π
×
(
d (mm)
)
2
4
σ (MPa)
=
P (kN)
A (mm²)
=
E (GPa)
×
ε
ε
=
δ (mm)
L (m)
Formulas
A
=
π
×
d
2
4
314.1593
=
π
×
20
2
4
σ
=
P
A
159.1549
=
50,000
314.1593
ε
=
δ
L
7.9577 ×
10
-4
=
1.5915
2,000
σ
=
E
×
ε
159.1549
=
200,000
×
(7.9577
×
10
-4
)
Formulas are worked in N, mm, MPa. The step-by-step shows the conversions.
Assumptions
•
The load is axial, through the centroid, and the stress stays below the proportional limit.
•
Linear elastic, small deflections; steel by default, E = 200 GPa. Tension is +.
•
The steps work in N, mm and MPa: 1 MPa is 1 N/mm².
Step-by-step
WE KNOW
P = 50 kN, d = 20 mm, L = 2 m, E = 200 GPa
FIND
cross-section area (A), normal stress (σ), normal strain (ε), elongation (δ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N, mm, MPa).
P = 50 kN = 50,000 N (1 kN = 1000 N)
L = 2 m = 2,000 mm (1 m = 1000 mm)
E = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
Step 2 · Find cross-section area (A)
Use A = π × d² ÷ 4
The cross-section of a round bar is a circle: π times the diameter squared, over 4.
A
=
π
×
20
2
4
A
=
π
×
400
4
A
=
400π
4
A
=
314.1593
mm²
Step 3 · Find normal stress (σ)
Use σ = P ÷ A
Normal
stress
is
the
axial
load
spread
over
the
cross-section:
N
mm²
gives
MPa.
σ
=
50,000
314.1593
σ
=
159.1549
MPa
Step 4 · Find normal strain (ε)
Use σ = E × ε
Divide the stress by the modulus.
ε
=
σ
E
ε
=
159.1549
200,000
ε
=
7.9577 ×
10
-4
Step 5 · Find elongation (δ)
Use ε = δ ÷ L
Multiply the strain by the length for the stretch.
δ
=
ε
×
L
δ
=
7.9577 ×
10
-4
×
2,000
δ
=
1.5915
mm
Check
Put the numbers back in. Both sides should match.
314.1593
=
π
×
20
2
4
✓
159.1549
=
50,000
314.1593
✓
7.9577 ×
10
-4
=
1.59155
2,000
✓
159.1549
=
200,000
×
(7.957747
×
10
-4
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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