Acylindrical 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 ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.

1.505

Explanation:

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 ±11 percent, the diameter is known within ±4 percent (tolerances), and the stress that causes failure (strength) is known within ±20 percent, determine the minimum design factor that will guarantee that the part will not fail.

stress is force per unit area

stress=P/A

A = πd^2/4.

uncertainty of axial force P= +/-.11

s=+/-.20, strength

d=+/-.04 diameter

fail load/max allowed

minimum design=fail load/max allowed

minimum design =s/(P/A)

sA/P

A=(.96d^2)/4, so Amin=

(because the diameter  at minimum is (1-0.04=0.96)

minimum design=Pmax/(sminxAmin)

1.11/(.80*.96^2)=

1.505

answer:

i dont know the answer but good luck

explanation:

answer: i have no clue tbh

explanation:

it happens to me to tho