Question 7.S-P.4: The state of plane stress shown occurs at a critical point o...
The state of plane stress shown occurs at a critical point of a steel machine component. As a result of several tensile tests, it has been found that the tensile yield strength is s_Y = 250 MPa for the grade of steel used. Determine the factor of safety with respect to yield, using (a) the maximum-shearing-stress criterion, and (b) the maximum-distortion-energy criterion.
Mohr’s Circle. We construct Mohr’s circle for the given state of stress and find
s _{\text {ave }}=O C=\frac{1}{2}\left( s _{x}+ s _{y}\right)=\frac{1}{2}(80-40)=20 MPat _{m}=R=\sqrt{(C F)^{2}+(F X)^{2}}=\sqrt{(60)^{2}+(25)^{2}}=65 MPa
Principal Stresses
s _{a}=O C+C A=20+65=+85 MPas _{b}=O C-B C=20-65=-45 MPa
a. Maximum-Shearing-Stress Criterion. Since for the grade of steel used the tensile strength is s _{Y}=250 MPa, the corresponding shearing stress at yield is
t _{Y}=\frac{1}{2} s _{Y}=\frac{1}{2}(250 MPa )=125 MPa
For t _{m}=65 MPa: F . S .=\frac{ t _{Y}}{ t _{m}}=\frac{125 MPa }{65 MPa } F . S .=1.92
b. Maximum-Distortion-Energy Criterion. Introducing a factor of safety into Eq. (7.26), we write
s_{a}^{2}- s _{a} s _{b}+ s _{b}^{2}< s _{Y}^{2} (7.26)
s _{a}^{2}- s _{a} s _{b}+ s _{b}^{2}=\left(\frac{ s _{Y}}{F \cdot S .}\right)^{2}
For s _{a}=+85 MPa , s _{b}=-45 MPa, and s_{Y}=250 MPa , we have
(85)^{2}-(85)(-45)+(45)^{2}=\left(\frac{250}{F \cdot S .}\right)^{2}
114.3=\frac{250}{F . S} F . S .=2.19
Comment. For a ductile material with s _{Y}=250 MPa, we have drawn the hexagon associated with the maximum-shearing-stress criterion and the ellipse associated with the maximum-distortion-energy criterion. The given state of plane stress is represented by point H of coordinates s_{a}=85 MPa and s _{b}=-45 MPa. We note that the straight line drawn through points O and H intersects the hexagon at point T and the ellipse at point M. For each criterion, the value obtained for F.S. can be verified by measuring the line segments indicated and computing their ratios:
(a) F . S .=\frac{O T}{O H}=1.92 (b) F . S .=\frac{O M}{O H}=2.19
