The smallest yield stress for steelbased on the maximum distortion energy theory.

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Chapter 10.7, Problem 84P

To determine

The smallest yield stress for steelbased on the maximum distortion energy theory.

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Determine the resulting maximum value of the normal stress. Specify the orientation of the plane on which these maximum values occur. **The answer is tensile stress is 0 ksi at 90 degrees. **The answer is compressive stress is 7 ksi at 0 degrees. Can you explain how that is? This was my thought process: I know that tensile would be zero because the force P is actually going inwards and not outwards. I know that means that there would be a compressive force. I am confused on the angles, how is a tensile force going 90 degress if there technically is no force in the tensile direction. And how is there a compressive force at 90 degrees if there is a stress? thank you!

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Problem #1: Re-evaluate the beam from the Module 3 homework. Recall that you determine the principal stresses at point A and B on the beam. Note: Assume the yield strength is 248 MPa. Part A: Based on that information, determine if any of the failure criteria (Rankine's theory, Tresca criterion, and Von Mises criterion) predict failure by yielding at either A or B. Present the factor of safety for all three criteria. Part B: Based on the results of Part A and what we know about the failure surfaces of these three criteria, do these computed factors of safety make sense? Where does this data point land on the failure surfaces? 20 kN 100 mm 10 kN B 100 mm 100 mm •B |A 50 mm 50 mm 2 m 2 m