Answer:
a) MSS: n = 3.5 , DE: n = 3.5
b) MSS: n = 3.5 , DE: n = 4.0414
c) MSS: n = 2 , DE: n = 2.3015
Explanation:
Given:
- Minimum yield strength in tension and compression S_y = 350 MPa
Solution:
- The Factor of safety n for MSS is given as follows:
n = S_y / ( σ_1 - σ_3 )
- The Factor of safety n for distortion-energy is given as follows:
n = S_y / σ'
σ' = ( σ_x^2 - σ_x*σ_y + σ_y^2 + 3*τ_xy )^0.5
Case a: σx=100 MPa, σy=100 MPa
- MSS: σ_1 = 100 MPa, σ_2 = 100 MPa , σ_3 = 0
n = 350 / ( 100 - 0 )
n = 3.5
- DE:
σ' = ( σ_x^2 - σ_x*σ_y + σ_y^2 )^0.5
σ' = ( 100^2 - 100*100 + 100^2 )^0.5
σ' = 100 MPa
n = 350 / 100
n = 3.5
Case b: σx=100 MPa, σy=50 MPa
- MSS: σ_1 = 100 MPa, σ_2 = 50 MPa , σ_3 = 0
n = 350 / ( 100 - 0 )
n = 3.5
- DE:
σ' = ( σ_x^2 - σ_x*σ_y + σ_y^2 )^0.5
σ' = ( 100^2 - 100*50 + 50^2 )^0.5
σ' = 86.603 MPa
n = 350 / 86.603
n = 4.0414
Case c: σx=100 MPa, σy=-75 MPa
- MSS: σ_1 = 100 MPa, σ_2 = 0 MPa , σ_3 = -75
n = 350 / ( 100 + 75 )
n = 2
- DE:
σ' = ( σ_x^2 - σ_x*σ_y + σ_y^2 )^0.5
σ' = ( 100^2 + 100*75 + 75^2 )^0.5
σ' = 152.07 MPa
n = 350 / 86.603
n = 2.3016
