Question

• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the modified : Goodman criterion. The shaft CD rotates at a constant speed, has a constant diameter of 1.13 in, and is made from cold-drawn AISI 1018 steel. From problem 3-85, the critical stress element in shaft CD experiences a completely reversed bending stress due to the rotation, as well as steady torsional and axial stresses. Thus, a,bend= 12 kpsi, Om, bend= 0 kpsi, Oa,axial= 0 kpsi, om, axial= -0.9 kpsi, Ta = 0 kpsi, and Im = 10 kpsi. The minimum factor of safety for fatigue is

Answer 1

minimum factor of safety for fatigue is = 1.5432

Step-by-step explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

`\sigma a` =
`√((\sigma a * kb)^2+3* (za* kt)^2)` ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

`\sigma a` =
`√((12 * 1)^2+3* (0* 1)^2)`

`\sigma a` = 12 kpsi

and

`\sigma m` =
`√((\sigma m * kb)^2+3* (zm* kt)^2)` ...........2

put here value and we get

`\sigma m` =
`√((-0.9 * 1)^2+3* (10* 1)^2)`

`\sigma m` = 17.34 kpsi

now we apply here goodman line equation here that is

`(\sigma m)/(Sut) + (\sigma a)/(Se) = (1)/(FOS)` ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

`(17.34)/(63.800) + (12)/(31.9) = (1)/(FOS)`

solve it we get

FOS = 1.5432