Breaking stress is the stress at which a material is expected to fail in tension. In practical design, the working
load is kept well below the breaking load by using a safety factor.
1. Stress from axial load
Normal stress is force divided by cross-sectional area:
\[
\boxed{
\sigma=\frac{F}{A}
}.
\]
For a circular cable or rod of diameter \(d\), the cross-sectional area is
\[
\boxed{
A=\frac{\pi d^2}{4}
}.
\]
2. Breaking stress or ultimate tensile strength
The ultimate tensile strength, often abbreviated UTS, is the approximate tensile stress at failure:
\[
\sigma_{\mathrm{UTS}}
=
\frac{F_{\mathrm{break}}}{A}.
\]
Therefore, the estimated breaking load is
\[
\boxed{
F_{\mathrm{break}}=\sigma_{\mathrm{UTS}}A
}.
\]
3. Safety factor
A safety factor \(n\) reduces the allowable working load:
\[
\boxed{
F_{\mathrm{safe}}
=
\frac{F_{\mathrm{break}}}{n}
}.
\]
Equivalently, the allowable working stress is
\[
\boxed{
\sigma_{\mathrm{allow}}
=
\frac{\sigma_{\mathrm{UTS}}}{n}
}.
\]
4. Checking a working load
For an applied working load \(F_{\mathrm{work}}\), the working stress is
\[
\sigma_{\mathrm{work}}
=
\frac{F_{\mathrm{work}}}{A}.
\]
The load is below the selected safe working limit if
\[
F_{\mathrm{work}}\le F_{\mathrm{safe}}.
\]
The utilization of the safe load is
\[
\mathrm{utilization}
=
\frac{F_{\mathrm{work}}}{F_{\mathrm{safe}}}.
\]
5. Actual safety factor
If the working load is known, the actual safety factor is
\[
\boxed{
n_{\mathrm{actual}}
=
\frac{F_{\mathrm{break}}}{F_{\mathrm{work}}}
}.
\]
If \(n_{\mathrm{actual}}\) is smaller than the required safety factor, the design is not acceptable.
6. Required area or diameter
If a cable must carry \(F_{\mathrm{work}}\) with safety factor \(n\), the required area is
\[
\boxed{
A_{\mathrm{req}}
=
\frac{nF_{\mathrm{work}}}{\sigma_{\mathrm{UTS}}}
}.
\]
For a circular cable, the required diameter is
\[
\boxed{
d_{\mathrm{req}}
=
\sqrt{\frac{4A_{\mathrm{req}}}{\pi}}
}.
\]
7. Typical ultimate tensile strengths
| Material |
Approximate UTS |
Comment |
| Structural steel |
\(\approx500\ \mathrm{MPa}\) |
Common cable and structural estimate |
| Mild steel |
\(\approx400\ \mathrm{MPa}\) |
Lower-strength steel |
| High-strength steel |
\(\approx1000\ \mathrm{MPa}\) |
Depends strongly on grade and heat treatment |
| Aluminum alloy |
\(\approx300\ \mathrm{MPa}\) |
Varies widely by alloy |
| Copper |
\(\approx210\ \mathrm{MPa}\) |
Ductile metal |
| Nylon rope |
\(\approx75\ \mathrm{MPa}\) |
Approximate; rope construction matters |
| Kevlar fiber |
\(\approx3600\ \mathrm{MPa}\) |
Very high tensile strength fiber |
8. Why use a safety factor?
Safety factors account for uncertainty, such as:
- material variability,
- wear, corrosion, or fatigue,
- shock loading or dynamic effects,
- manufacturing defects,
- uncertainty in the actual load.
A larger safety factor gives a lower safe working load.
9. Worked example
A steel cable has
\[
\sigma_{\mathrm{UTS}}=500\ \mathrm{MPa},
\qquad
d=12\ \mathrm{mm},
\qquad
n=4.
\]
Convert the diameter:
\[
d=12\ \mathrm{mm}=0.012\ \mathrm{m}.
\]
Calculate area:
\[
A=\frac{\pi d^2}{4}
=
\frac{\pi(0.012)^2}{4}.
\]
\[
A\approx1.131\times10^{-4}\ \mathrm{m^2}.
\]
Convert UTS:
\[
500\ \mathrm{MPa}=500\times10^6\ \mathrm{Pa}.
\]
Breaking load:
\[
F_{\mathrm{break}}
=
\sigma_{\mathrm{UTS}}A
=
(500\times10^6)(1.131\times10^{-4}).
\]
\[
F_{\mathrm{break}}\approx5.65\times10^4\ \mathrm{N}
=
56.5\ \mathrm{kN}.
\]
Safe working load:
\[
F_{\mathrm{safe}}
=
\frac{F_{\mathrm{break}}}{4}
=
\frac{56.5\ \mathrm{kN}}{4}.
\]
\[
\boxed{
F_{\mathrm{safe}}\approx14.1\ \mathrm{kN}
}.
\]
Key idea: safe working load is not the breaking load. It is the breaking load divided by the selected safety factor.
