Complete Question:
An AISI 1018 steel has a yield strength, [tex]S_{y} = 295 MPa[/tex]. Using the distortion-energy theory for the given state of plane stress,
[tex]\sigma_{x} = 31 MPa[/tex], [tex]\sigma_{y} = 41 MPa[/tex] , [tex]T_{xy} = 45 MPa[/tex]
(a) determine the factor of safety,
(b) plot the failure locus, the load line, and estimate the factor of safety by graphical measurement.
Answer:
a) Factor of safety = 3.42
b) See the plot failure locus in the file attached
Explanation:
According to the maximum distortion energy theorem:
[tex]\sigma_{1} ^{2} - \sigma_{1} \sigma_{2} + \sigma_{2} ^{2} \leq (\frac{S_{y} }{n}) ^{2}[/tex]..............(1)
n = Factor of safety
We have to calculate [tex]\sigma_{1} and \sigma_{2}[/tex]
[tex]\sigma_{1} = \frac{\sigma_{x}+ \sigma_{y} }{2} + \sqrt{(\frac{\sigma_{x}-\sigma_{y} }{2}) ^{2} + T_{xy} ^{2} }[/tex]
[tex]\sigma_{1} = \frac{\sigma_{x}+ \sigma_{y} }{2} - \sqrt{(\frac{\sigma_{x}-\sigma_{y} }{2}) ^{2} + T_{xy} ^{2} }[/tex]
[tex]\sigma_{1} = \frac{31+ 41 }{2} + \sqrt{(\frac{31-41 }{2}) ^{2} +45 ^{2} }[/tex]
[tex]\sigma_{1} = 81.28 MPa[/tex]
[tex]\sigma_{2} = \frac{31+ 41 }{2} - \sqrt{(\frac{31-41 }{2}) ^{2} +45 ^{2} }[/tex]
[tex]\sigma_{2} = -9.28 MPa[/tex]
Substituting these values into equation (1)
[tex]81.28 ^{2} +(81.28*9.28) + (-9.28) ^{2} \leq (\frac{295 }{n}) ^{2}[/tex]
[tex]n^{2} \geq 11.7[/tex]
[tex]n \geq \sqrt{11.7} \\n \geq 3.42[/tex]
