Question

A cylindrical part of diameter d is loaded by an axial force p. this causes a stress of pya, where a 5 pd2y4. if the load is known with an uncertainty of 610 percent, the diameter is known within 65 percent (tolerances), and the stress that causes failure (strength) is known within 615 percent, determine the minimum design factor that will guarantee that the part will not fail.

Answer 1

Here is the correct question.

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 ±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.

Answer:

the minimum design factor that will guarantee that the part will not fail. = 1.434

Explanation:

Looking at the uncertainty; loss of strength must be raised to $\dfrac{1}{0.85}$ due to the stress that causes the failure (strength)  is known within ±15% uncertainty.

Looking at the uncertainty; the maximum allowable load  must be reduced to $\dfrac{1}{1.1}$ because the load is known with an uncertainty of ±10.

Looking at the uncertainty; the diameter must be raised to $\dfrac{1}{0.95}$  because the diameter is known within an uncertainty of ±5.

The decrease in the maximum allowable stress can be estimated as:

$\sigma' = \dfrac{P'}{A'}$

where,

$\sigma =$ stress

P = load

A = cross-sectional area of the cylinder

∴

$\sigma' = \dfrac{P'}{\dfrac{\pi}{4}(d')^2}$

replacing P' with $\dfrac{1}{1.1}P$   and d' with $\dfrac{1}{0.95}d$, we have:

$\sigma' = \dfrac{(\dfrac{1}{1.1})\times p }{\dfrac{\pi}{4}(\dfrac{1}{0.95 } d)^2 }$

$\sigma' =\dfrac{P}{\dfrac{\pi}{4}d^2} (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }$

$\sigma' =\sigma \times (\dfrac{\dfrac{1}{1.1} }{(\dfrac{1}{0.95})^2 }) }$

$\sigma' =\sigma \times 0.82045$

$\dfrac{\sigma' }{\sigma } =0.82045$

Thus, the uncertainty in diameter and the load of the allowable stress needs to decrease to 0.82045

Now, the minimum design factor that will ascertain that the part will not fail can be computed as:

$n_d = \dfrac{loss \ of \ function \ parameter }{maximum \ allowable \ parameter}$

where;

the design factor = $n_d$

$n_d =\dfrac{\dfrac{1}{0.85} }{0.82045}$

the design factor  $n_d$ = 1.434.

Thus,  the minimum design factor that will guarantee that the part will not fail. = 1.434