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

Answer:

minimum factor of safety for fatigue is = 1.5432

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

$\sigma a$ = $\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}$    ...........1

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

put here value and we get

$\sigma a$ =  $\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}$  

$\sigma a$ = 12 kpsi

and

$\sigma m$ = $\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}$     ...........2

put here value and we get

$\sigma m$ = $\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}$  

$\sigma m$ = 17.34 kpsi

now we apply here goodman line equation here that is

$\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}$     ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

$\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}$  

solve it we get

FOS = 1.5432