RayLeigh Rittz Method

Using Rayleigh-Ritz method to solve plate Probllems ü Assuming the deflection In[1]:= b = a; In[2]:= w = A11 ∗ H1 − 4 ∗ x^2 ê a^2L ∗ H1 − 4 ∗ y^2 ê b^2L Out[2]= A11 J1 − 4x 2 ccccccccc a 2 NJ1 − 4y 2 ccccccccc a 2 N Calculating the strain Energy In[4]:= uo = Simplify@ HD@w, 8x, 2<D + D@w, 8y, 2<DL ^2 − 2 H1 −νL ∗ HD@w, 8x, 2<D ∗ D@w, 8y, 2<D − D@w, x, yD ^2LD Out[4]= 1 cccccc a 8 H128 A11 2 H8 Hx 4 + y 4 + 2x 2 y 2 H4 − 3 νLL + a 4 H1 +νL − 4a 2 Hx 2 + y 2 LH1 +νLLL In[5]:= U = Dp ê 2 ∗ ‡ −aê2 aê2 ‡ −bê2 bê2 uo Å x Å y Out[5]= 2816 A11 2 Dp cccccccccccccccc ccccccccccccccc 45 a 2 Calculating the Potential Energy In[6]:= V =−qo ∗ ‡ −aê2 aê2 ‡ −bê2 bê2 w Å x Å y Out[6]= − 4 cccc 9 a 2 A11 qo Total Potential Energy In[7]:= TPE = U + V Out[7]= 2816 A11 2 Dp cccccccccccccccc ccccccccccccccc 45 a 2 − 4 cccc 9 a 2 A11 qo ü Differentiate the TPE and solve for the unknowns In[8]:= E1 = D@TPE, A11D m 0 Out[8]= 5632 A11 Dp cccccccccccccccc cccccccccccc 45 a 2 − 4a 2 qo cccccccccccccccc 9 m 0 In[9]:= S1 = Solve@E1, A11D Out[9]= 99A11 → 5a 4 qo cccccccccccccccccc 1408 Dp == RayLeigh-Ritz Method.nb 1

RayLeigh Rittz Method

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