Design Factor and Design Uncertainty Calculation Question

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?

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: $$n_d = \frac{10kN/0.85}{10kN/1.2} = \frac{1/0.85}{1/1.2}$$

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

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