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I am trying to evaluate the expression Subscript[u, θ]. However, the results of the integration show another integral. I want to find the final equation. with all integrals evaluated.

enter image description here

(* Stress functions *)
Subscript[σ, rr] = ((-a^2 + r^2)*S*(r^2 + (-3*a^2 + r^2)*Cos[2* θ]))/(2*r^4)

Subscript[σ, rθ] = ((a - r)*(a + r)*(3 a^2 + r^2)*S*Cos[θ]*Sin[θ])/r^4

Subscript[σ, θθ] = (S (r^2 *(a^2 + r^2) - (3 a^4 + r^4)* Cos[2* θ]))/(2*r^4)

(* Constitutive equilibruim equations for plain stress *)
Subscript[ϵ, rr] = Subscript[σ, rr]/Ε - (ν*Subscript[σ, θθ])/Ε

Subscript[ϵ, rθ] = (1 + ν)/Ε*Subscript[σ, rθ]

Subscript[ϵ, θθ] = Subscript[σ, θθ]/Ε - (ν*Subscript[σ, rr])/Ε

(* Finding Subscript[u, r] and Subscript[u, θ] *)
Subscript[u, r] = ∫Subscript[σ, rr] \[DifferentialD]r + f[θ] // FullSimplify

Subscript[u, θ] = ∫(r*Subscript[ϵ, θθ] - Subscript[u, r]) \[DifferentialD]θ + g[r]
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    $\begingroup$ Welcome to Mathematica SE. This issue is hard to find without having the complete definitions of the used symbols. Can you please provide the code (and not the screenshot), so we can work with the code directly. $\endgroup$ – mgamer Dec 14 '20 at 13:46
  • $\begingroup$ Thank you so much for your reply. How could I attach my mathematica code? I am so sorry that I am new into Mathematica. I have put my code now. $\endgroup$ – Ahmed Moussa Dec 14 '20 at 13:57
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    $\begingroup$ There has to be an integral since f[theta] is not defined. You can get further along if you integrate the individual terms of the last integral, i.e., Subscript[u, \[Theta]] = FullSimplify[(Integrate[#, \[Theta]] & /@ (r*Subscript[\[Epsilon], \[Theta]\[Theta]] - Subscript[u, r])) + g[r]] $\endgroup$ – Bob Hanlon Dec 14 '20 at 14:20
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Try the following: in addition to your definitions write two last expressions as follows

  Subscript[u, r] = 
     f[\[Theta]] + \[Integral]Subscript[\[Sigma], rr] \[DifferentialD]r //
       Simplify

(*  (S (r^2 (a^2 + r^2) + (-a^4 + 4 a^2 r^2 + r^4) Cos[2 \[Theta]]))/(
 2 r^3) + f[\[Theta]]   *)

and

Subscript[u, \[Theta]] = 
 g[r] + Distribute[\[Integral](r*
         Subscript[\[Epsilon], \[Theta]\[Theta]] - 
        Subscript[u, r]) \[DifferentialD]\[Theta]] // Simplify

(*  (S \[Theta] (a^2 (1 - \[CapitalEpsilon] + \[Nu]) - 
    r^2 (-1 + \[CapitalEpsilon] + \[Nu])))/(2 r \[CapitalEpsilon]) + 
 g[r] - \[Integral]f[\[Theta]] \[DifferentialD]\[Theta] + (
 S (4 a^2 r^2 (-\[CapitalEpsilon] + \[Nu]) - 
    r^4 (1 + \[CapitalEpsilon] + \[Nu]) + 
    a^4 (\[CapitalEpsilon] - 
       3 (1 + \[Nu]))) Cos[\[Theta]] Sin[\[Theta]])/(
 2 r^3 \[CapitalEpsilon])    *)

which integrates the expression to the end, but leaves only the integral of f[theta] unevaluated.

Have fun!

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  • $\begingroup$ Thank you so much. I highly appreciate your help. $\endgroup$ – Ahmed Moussa Dec 15 '20 at 8:21

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