A cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where 2 A d   / 4. If the load is known with an uncertainty of ±10 percent, the diameter is known within ±5 percent (tolerances), and the stress that causes failure (strength) is known within ±15 percent, determine the minimum design factor that will guarantee that the part will not fail.

1.435

Explanation:

The stress (σ) on an object is equivalent to the ratio of pressure applied (P) and cross sectional area (A). Thus:

σ = P/A = P/(πd²/4) = 4P/πd²

The highest stress on the cylindrical part is:

σ = 4(1+0.1)P/π[(1-0.05)d]² = 1.22*4P/πd² = 1.22σ

Similarly, the minimum strength can be estimated as:

= 0.85*S

Failure occurs when the maximum stress is equivalent to the minimum strength. Therefore:

0.85*S = 1.22σ

The design factor = S/σ = 1.22/0.85 = 1.435

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