I recently had a discussion about design factors and design uncertainty. Looking in Shigley's Mechanical Engineering Design, pg. 17, ed. 3, there is a detailed discussion of this topic. The authors define the design factor as follows.

$$ n_d = \frac{\text{loss of function parameter}}{\text{maximum allowable parameter}} $$

Where $n_d$ is the design factor.

This is as I would expect. However, they then present the following example calculation.

Consider that the maximum load on a structure is known with an uncertainty of $\pm 20\%$, and the load causing failure is known within $\pm 15\%$. If the load causing failure is nominally $10\text{kN}$, determine the design factor and the maximum allowable load that will offset the absolute uncertainties.

They state that, in this instance, the design factor can be calculated as follows.

$$ n_d = \frac{1/0.85}{1/{1.2}} $$

As I understand it, this is inconsistent with their original definition. The original definition is a ratio between the absolute values of the given parameter for failure and use (adjusted for uncertainty). This isn't the same as the second definition, which is a ratio between the percentage uncertainties themselves.

Is this a mistake, or do I misunderstand something?


1 Answer 1


I don't understand the definition, but the example seems to be correct. If they state:

the design factor can be calculated as follows

I would agree, since the nominal values just cancel out: $$n_d = \frac{10kN/0.85}{10kN/1.2} = \frac{1/0.85}{1/1.2}$$

Just checking: If nominal failure load is $L_{nom} = 10 kN$, then the minimum failure load is $15\%$ less $L_{min,f} = 0.85\cdot L_{nom}$. Now the allowable load should be such, that if you add $20\%$ to it, it will be at the minimum failure load $L_{allowable}\cdot 1.2 = L_{min,f}$:

$$L_{allowable} = \frac{0.85}{1.2}\cdot L_{nom} = \frac{L_{nom}}{n_d}$$

  • $\begingroup$ I think that's the problem, the nominal values aren't the same. If they were both 10kN, it would work, but they aren't; only the load causing failure is 10kN. I assume the maximum allowable parameter is lower since this is the load you would expect through normal use. $\endgroup$ Commented Jul 30, 2022 at 13:48
  • $\begingroup$ How do you think the calculation should look like according to the original definition? $\endgroup$ Commented Jul 30, 2022 at 15:30
  • $\begingroup$ I don't see how there is enough information in the question to calculate the design factor according to the original definition. I think you need either the design factor or the maximum allowable load. The way I understand it, you have two unknowns in the question and only one equation. In my experience, a design factor would be specified by a design standards code, and then you would calculate the maximum allowable load. $\endgroup$ Commented Jul 30, 2022 at 17:08
  • $\begingroup$ You have everything you need. You are supposed to find maximum allowable load, knowing the actual load value may be $20\%$ higher and this has to be at most $15\%$ lower than the failure load. $n_d$ just comes out from appropriately combining the 2 uncertainties. $\endgroup$ Commented Jul 30, 2022 at 18:27
  • $\begingroup$ In my view, the original definition is not very useful, I can understand it only from the example and intent of safe design. $\endgroup$ Commented Jul 30, 2022 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.