Mechanics of Materials: Lesson 53 - Mohr’s Circle on Thin Wall Pressure Vessel

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All the calculations are done with 0.5" as the O.D., not I.D. With I.D.=0.5", σx=7.35 ksi, σy=5.00 ksi, and τxy=23.18 ksi. Therefore, σ1=29.38 ksi, σ2= -17.03 ksi, and τmax=23.21 ksi. Alternatively, O.D. can be switched to equal 0.5" to make it easier to follow along.

peter_neff
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Nice to see you back making engineering videos. I'm past this course but still have to do thermo and dynamics so hope you circle back to those in the spring as I'll be pretty much forced to take at least one of them in the summer

LyleH-
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Can't thank you enough for the help. Wish my professor taught like this.

gerrymullins
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right in time for the end of my course too

FunniestFlicks
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Wonderful solids lessons!! really taking the time to explain the problems along with a good sense of humor :))

manuboker
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So finding the angles of these points on the Mohr circle means at what angle/2 we need to rotate our stress element to find the corresponding sigma(x)’ and tau(xy)’ stresses?
Great videos, sir!
Greetings from Norway.

NinjaMartin