SME
Materials Science & Material Properties
Mechanical properties
Use this for “A 12.8 mm bar with a 50 mm gauge length peaks at 50 kN and breaks at 62 mm long with a 9 mm neck. Find the UTS, %EL and %RA.”
Diagram
The engineering curve: elastic, yielding, hardening to the UTS, necking, fracture (the knee is a sketch: the page gives no σY).
This is an example. Type your own values to start a new problem.
UTS (MPa)
=
Fₘₐₓ (kN)
A₀ (mm²)
%EL (%)
=
Lf (mm)
−
L₀ (mm)
L₀ (mm)
×
100
%RA (%)
=
(1
−
df (mm)
d₀ (mm)
2
)
×
100
d₀ and L₀ are the bar’s diameter and gauge length before the test; df is the neck’s diameter and Lf the gauge length after it breaks. A₀ = πd₀² ÷ 4.
Formulas
A₀
=
π
×
d₀
2
4
128.6796
=
π
×
12.8
2
4
UTS
=
Fₘₐₓ
A₀
388.5619
=
50,000
128.6796
%EL
=
Lf −
L₀
L₀
×
100
24
=
62 − 50
50
×
100
%RA
=
(1
−
(
df
d₀
)
2
)
×
100
50.5615
=
(1
−
(
9
12.8
)
2
)
×
100
Formulas are worked in N. The step-by-step shows the conversions.
Assumptions
•
Engineering stress: the load over the starting area, not the shrinking one.
•
%EL is measured after fracture, once the elastic stretch has sprung back.
•
The steps work in N, mm and MPa: 1 MPa is 1 N/mm².
Step-by-step
WE KNOW
d₀ = 12.8 mm, Fₘₐₓ = 50 kN, L₀ = 50 mm, Lf = 62 mm, df = 9 mm
FIND
starting area (A₀), tensile strength (UTS), elongation at fracture (%EL), reduction of area (%RA)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N).
Fₘₐₓ = 50 kN = 50,000 N (1 kN = 1000 N)
Step 2 · Find starting area (A₀)
Use A₀ = π × d₀² ÷ 4
The cross-section of a round bar is a circle: π times the diameter squared, over 4.
A₀
=
π
×
12.8
2
4
A₀
=
π
×
163.84
4
A₀
=
163.84π
4
A₀
=
128.67964
mm²
Step 3 · Find tensile strength (UTS)
Use UTS = Fₘₐₓ ÷ A₀
The
tensile
strength
is
the
largest
load
over
the
starting
area:
N
mm²
gives
MPa.
UTS
=
50,000
128.67964
UTS = 388.5619 MPa
Step 4 · Find elongation at fracture (%EL)
Use %EL = (Lf − L₀) ÷ L₀ × 100
The gain in length over the gauge length, as a percent.
%EL
=
62 − 50
50
×
100
%EL
=
12
50
×
100
%EL = 0.24 × 100
%EL = 24%
Step 5 · Find reduction of area (%RA)
Use %RA = (1 − (df ÷ d₀)²) × 100
The area lost at the neck, as a percent: areas go as the diameter squared.
%RA
=
(1
−
(
9
12.8
)
2
)
×
100
%RA = (1 − 0.49438) × 100
%RA = 50.5615%
Check
Put the numbers back in. Both sides should match.
128.67964
=
π
×
12.8
2
4
✓
388.5619
=
50,000
128.67964
✓
24
=
62 − 50
50
×
100
✓
50.5615
=
(1
−
(
9
12.8
)
2
)
×
100
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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