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# Solving simple trigonometric equation (Mohr's circle)

Posted 10 years ago
 Intro: I am working towards constructing Mohr's circle for stresses at a point. This involves the Plane Stress Transformation equations, one of them being stress=(sigmax+sigmay)/2 + (sigmax-sigmay) Cos[2 phi]/2 + tau Sin[2 phi]. I want to determine the angle (preferably in the form Tan[2 phi]) at which the the stress is maximum, i.e. the principle direction of stress. Using this value I can then determine the principle stresses. Problem: I do the following: sol=Solve[D[stress,phi]==0,phi] hoping it will give me the following:{phi -> (1/2) ArcTan[2 tau / (sigmax-sigmay)]} but to no avail. It gives a screen full of ConditionalExpressions instead which I don't know how to begin to use. Question: How to 'solve' for phi in the above conditions (i.e. when the differential of stress is 0)? Thank you inadvance.
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Posted 10 years ago
 Bill Simpson, your guess was correct.I modified it slightly to the following: sol = Simplify[Solve[D[stress, phi] == 0, phi] /. C[1] -> 0, Element[tau,Reals] && Element[sigmax,Reals] && Element[sigmay,Reals]] To account for the more general case of there being negative stresses and shears (e.g. tension and whatnot)Thank you
Posted 10 years ago
 This is just a guess In[1]:= stress = (sigmax + sigmay)/2 + (sigmax - sigmay) Cos[2 phi]/2 + tau Sin[2 phi]; sol = Simplify[Solve[D[stress, phi] == 0, phi] /. C[1] -> 0, tau > 0 && sigmax > 0 && sigmay > 0] Out[2]= {{phi -> 1/2 ArcTan[sigmax - sigmay, 2 tau]}, {phi -> 1/2 ArcTan[-sigmax + sigmay, -2 tau]}} Note there are two forms of ArcTan, one that accepts a single argument y/x and another that accepts two arguments x,y. The second form, shown in the result above, sometimes let you determine which angle to use when the y/x form may not. Be careful of the order of those arguments. Other than that this looks like it might be the solution you were looking for. You could verify my assumptions given to Solve correctly reflect your problem.
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