A cylindrical part of diameter d is loaded by an axial force P. This causes a stress of P/A, where A = πd2/4. If the load is known with an uncertainty of ±10 percent, the diameter is known within ±3 percent (tolerances), and the stress that causes failure (strength) is known within ±16 percent, determine the minimum design factor that will guarantee that the part will not fail. The design factor that guarantee the part will not fail is.

1.4

Explanation:

The relative uncertainties known are:

eP = 0.1

eD = 0.03

eσ = 0.16

The equation for the stress is:

σ = P / A

The load in function of the strength and section is:

P = σ * A

P = π/4 * D^2 * σ

Now we add a safety factor:

P / f = π/4 * D^2 * σ

This safety factor will make P smaller.

We think of the worst condition that could happen, which is if the load is greatest and the diameter and strength smallest:

1 + 10% = 1.1

1 - 3% = 0.97

1 - 16% = 0.84

1.1*P / f = π/4 * (0.97*D) ^2 * 0.84*σ

We forget about the actual variables and focus only on the numbers that multiply them:

1.1 / f = 0.97^2 * 0.84

1.1 / (0.97^2 * 0.84) = f

f = 1.4

With a factor of safety of 1.4 the largest load wont cause a failure even in the case of the smallest diameter and the smallest material strength.