Aductile 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 = 50 mpa (b) σx = 100 mpa, τxy = –75 mpa (c) σx = –50 mpa, σy = –75 mpa, τxy = –50 mpa (d) σx = 100 mpa, σy = 20 mpa, τxy = –20 mpa

See explanation

Explanation:

a)

MSS: σ1 = 100 MPa ; σ2 = 50 MPa ; σ3 = 0

n =350 /(100 - 0) = 3.5   Ans.

DE: σ' = 100² -100*(50)+50²) ^0.5  = 86.6 MPa

n = 350/86.6 = 4.04    Ans.

b) σA, σB = 100/2 ± √((100/2)² + (-75)²) = 140,

σ1 = 140,  σ2= 0,  σ3= -40 MPa

MSS: n = 350/(140-(-40)) =1.94  Ans.

DE: σ' = [100² + 3 * (-75)²] ^0.5  = 164 MPa

  n= 350/164  = 2.13  Ans.

c) σA,σB = ((-50 -75) / 2)  ± √(((-50 +75)/2)² + (-50)²)

σ1 = 0, σ2 = -11.0,  σ3 = -114.0 MPa

MSS: n = 350/(0-(-1142.0)) = 3.07 Ans.

DE: σ' [(-50)² - (-50)(-75) + (-75)² + 3(-50)²]^0.5 = 105.0 MPa

n = 350/109.0 = 3.21  Ans.

d)  σA,σB = (100 +20)/1  ± √(((100-20)/2)² + (-20)²)

σ1 = 104.7, σ2 = 15.3,  σ3 = 0 MPa

MSS: n= 350/(140.7 - 0) = 3.34 Ans.

DE: σ' = [100² - 100(20) + 20² + 3 (-20)²]^0.5  = 98.0 MPa

n= 350/ 98.0 = 3.57 Ans.

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explanation: