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Mohr's Circle 
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#1
Aug2509, 01:50 PM

P: 19

Hey. I've been having some trouble getting through Mohr's circle. The googles that I've found haven't been too useful. I've seen several questions of the same calibre and what I generally need is how to derive the magnitude/direction of the principal strains as well as the principle stresses when given a set up like a rosette gage with 3 strains at 60degrees, with for example 100 epsilon  200 epsilon 300 epsilon on each axis. If you can show me an example solution or the general procedure, preferrably illustrated, I'd be that much wiser.
Thanks in advance. 


#2
Aug2509, 02:11 PM

P: 34

draw the stresses on a shearnormal coordiate system.
shear: cw+, ccw normal: tension+, compressive. connect two tips of shear stress, angle between the line and normal axis is double of stress angle. when you have mohr's circle done, the maximum coordinates of the circle can reach will be principle stresses. Read your book and do some examples, it's not a hard topic. 


#3
Aug2609, 04:27 AM

P: 19

I don't have acess to any books to refer to. I don't need help with signage. It's not the difficulty, it's the lack of a detailed methodology.
I can find the principal strains for 45 through formulas using: R=[0.5(εAεB)2+0.5(εBεC)2]^1/2 φ=0.5tan^1[(2εBεAεC)/(εAεC)] on the strain system. I have another set for 60. I can then find the stresses through the stress/strain relationship as I have the Young's Modulus and the v value. What I'm looking for however is the purely graphical Pole method which has been poorly described. If anyone can assist with me that, I'd be much obliged. Thanks. 


#4
Aug2609, 11:24 AM

P: 34

Mohr's Circle
Hope this may help.



#5
Aug2709, 12:11 PM

P: 19

Thank you. I certainly appreciate you having taken the time to take those pictures!



#6
Aug2709, 04:15 PM

P: 34

Since you can find [tex]2\varphi[/tex], the angle between principle stress and normal stress is [tex]\varphi[/tex]. If you draw your mohr's cicle, the rest will be only a metter of rotating [tex]\varphi[/tex] degress from horizontal axis to find principle stress direction. In geometry, round angle is half of centre angle; that's where pole comes from. From above picture, you can see the angle between blue line and horizontal axis is [tex]\varphi[/tex] (Orange line is parallel to horizontal axis). For stresses on any other planes, no matter pole or whatever, just use your geometry skills. It won't be more difficult than your middle school geometry class. Good Luck. 


#7
Nov2310, 04:13 AM

P: 2

I am not sure whether you finish your work. If not, send me you data and I will show it to you.
ipichet@yahoo.com 


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