PL cantilever crude attempt

This is the Title PhL 3.26.05 I am going to try this problem before reading about it in Lai. The famous cantilever beam. Assume 2 is down and 1 is to the right. The Navier equation is local to a point as says ρg is the gravity force so 0 = 0 + ∂jT1j // the 1 direction equation 0 = ρg + ∂jT2j // the 2 direction equation We assume things stay in a vertical plane so u3 = 0 and also T3i = 0 for all i. Then the equations above are 0 = ∂1T11 + ∂2T12 0 = ρg + ∂1T21+ ∂2T22 Assume T symmetric. And assume isoptropic elastic solid so that Tij = λeδij + 2μEij Eij= (1/2)(∂iuj+∂jui) e = div u = ∂1u1+ ∂2u2 So where does this lead us? T11 = λe + 2μE11 = λe + 2μ∂1u1 = λ∂1u1+ λ∂2u2 + 2μ∂1u1 = (λ+2μ) ∂1u1 + λ∂2u2 T22 = λe + 2μE22 = λe + 2μ∂2u2 = λ∂1u1+ λ∂2u2 +2μ∂2u2 = λ ∂1u1 + (λ+2μ)∂2u2 T12 = 2μE12= μ(∂1u2+∂2u1) Now we need derivatives: ∂1T11 = ∂1[(λ+2μ) ∂1u1 + λ∂2u2] = (λ+2μ) ∂12u1 + λ ∂1∂2u2 ∂2T12 = ∂2[μ(∂1u2+∂2u1)] = μ∂1∂2u2 +μ∂22u1 So insisting these add to zero requires that (λ+2μ) ∂12u1 + λ ∂1∂2u2 + μ∂1∂2u2 +μ∂22u1 = 0 (λ+2μ) ∂12u1 + (λ+μ) ∂1∂2u2 + ∂22u1 = 0 (1) which is a strange 2D PDE. The other equation needs ∂2T22 = λ ∂1∂2u1 + (λ+2μ)∂22u2 and then it says -ρg = ∂1T21+ ∂2T22 = [μ(∂12u2+∂1∂2u1) + λ ∂1∂2u1 + (λ+2μ)∂22u2 -ρg = μ∂12u2 + (μ+λ) ∂1∂2u1 + (λ+2μ)∂22u2 (2) So there are our two equations. Write again as (λ+2μ) ∂12u1 + (λ+μ) ∂1∂2u2 + ∂22u1 = 0 (1) μ∂12u2 + (λ+μ) ∂1∂2u1 + (λ+2μ)∂22u2 = -ρg (2) In the small motion limit, I have done everything exactly. Now, suppose we assume that u1 << u2 . As our cantilever bends down, the up down motion is larger than the left right motion of a particle. Then we could just set perhaps u1 = 0 and then have (λ+μ) ∂1∂2u2 = 0 μ∂12u2 + (λ+2μ)∂22u2 = -ρg This is all for a particle in our little beam, it is just self loading. Do I know the solution of this: A∂x2f + B∂y2f = C This is a 2nd order linear PDE, so I presume there are two solutions. Try f = a + bx + cy + αxy + β x2 + γ y2 ∂x2f = 2β ∂y2f = 2γ A2β + B2γ = C So OK, we assume then that u2 = a + bx1 + cx2 + α x1x2 + β x12 + γ x22 Then our first two equations say (λ+μ) ∂1[c + αx1+2γx2] = 0 (λ+μ)α = 0 => α = 0 which I guess is OK by me The second equation says μ2β + (λ+2μ)2γ = -ρg and this just makes a connection between β and γ. β + [λ+2μ)/μ]γ = -ρg/2μ => β = -ρg/2μ – [λ+2μ)/μ]γ I would also require that u2(0,0) = 0 and so then a = 0. Our solution might then be u2(x1,y1) = bx1 + cx2 + β x12 + γ x22 OK, stop. I am just goofing around, let's now see how it's done right then come back here and make repairs. I know I omitted the fact that the total Mg of the beam equals the total shear force at the connection. I think there are approximations I need to make beyond those above.