Engineering, 27.02.2020 17:23 bankskry

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following plane stress states:

(a) σx=100 MPa, σy=100 MPa
(b) σx=100 MPa, σy=50 MPa
(c) σx=100 MPa, σy=-75 MPa

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Answer from: alisonlebron15

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

Answer from: Quest

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Answer from: Quest

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A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. U...

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