SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Size a rod for a factor of safety
Use this for “A steel rod with σY = 250 MPa must carry 80 kN with a factor of safety of 2. Find its diameter.”
Diagram
σallow
=
σY
n
=
250
MPa
2
=
125
MPa.
A
=
P
σallow
=
640
mm²,
spread
evenly
across
the
section.
d
=
√(
4A
π
)
=
28.55
mm.
This is an example. Type your own values to start a new problem.
σallow (MPa)
=
σY (MPa)
n
=
P (kN)
A (mm²)
A (mm²)
=
π
×
(
d (mm)
)
2
4
Formulas
σallow
=
σY
n
125
=
250
2
σallow
=
P
A
125
=
80,000
640
A
=
π
×
d
2
4
640
=
π
×
28.546
2
4
Formulas are worked in N. The step-by-step shows the conversions.
Assumptions
•
The rod is in axial tension, so the stress P ÷ A is even across it.
•
The allowed stress is the yield strength over the factor of safety: σallow = σY ÷ n.
•
Round the diameter up to a stock size; a larger n covers uncertain loads.
Step-by-step
WE KNOW
P = 80 kN, σY = 250 MPa, n = 2
FIND
allowed stress (σallow), area needed (A), diameter needed (d)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N).
P = 80 kN = 80,000 N (1 kN = 1000 N)
Step 2 · Find allowed stress (σallow)
Use σallow = σY ÷ n
Keep
the
working
stress
n
times
below
yield:
divide
the
yield
strength
by
n
.
σallow
=
250
2
σallow = 125 MPa
Step 3 · Find area needed (A)
Use σallow = P ÷ A
The smallest area keeps the stress at the allowed value: divide the load by it.
A
=
P
σallow
A
=
80,000
125
A
=
640
mm²
Step 4 · Find diameter needed (d)
Use A = π × d² ÷ 4
Multiply the area by 4, divide by π, then take the square root.
d
=
√(4
×
A
π
)
d
=
√(4
×
640
π
)
d
=
√(
2,560
π
)
d
=
√
814.9
d
=
28.546
mm
Check
Put the numbers back in. Both sides should match.
125
=
250
2
✓
125
=
80,000
640
✓
640
=
π
×
28.546
2
4
✓
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Diameter change by Poisson’s ratio
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Principal stresses and Mohr’s circle
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