Answered

• 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 :

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

[tex]\sigma a[/tex] = [tex]\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}[/tex]    ...........1

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

put here value and we get

[tex]\sigma a[/tex] =  [tex]\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}[/tex]  

[tex]\sigma a[/tex] = 12 kpsi

and

[tex]\sigma m[/tex] = [tex]\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}[/tex]     ...........2

put here value and we get

[tex]\sigma m[/tex] = [tex]\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}[/tex]  

[tex]\sigma m[/tex] = 17.34 kpsi

now we apply here goodman line equation here that is

[tex]\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}[/tex]     ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

[tex]\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}[/tex]  

solve it we get

FOS = 1.5432

${teks-lihat-gambar} DeniceSandidge

Other Questions