# Design Factor and Design Uncertainty Calculation Question

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?

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 $$L_{allowable}\cdot 1.2 = L_{min,f}$$:

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

• 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. Jul 30, 2022 at 13:48
• How do you think the calculation should look like according to the original definition? Jul 30, 2022 at 15:30
• 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. Jul 30, 2022 at 17:08
• 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. Jul 30, 2022 at 18:27
• In my view, the original definition is not very useful, I can understand it only from the example and intent of safe design. Jul 30, 2022 at 18:29