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29 April, 19:19

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.

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  1. 29 April, 19:20
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    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.
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