University:
Massachusetts Institute of Technology
Course:
12.520 | Geodynamics
Academic year:

2023
Views:

471

Pages:

5
Author:

compressqk4

Lecture Notes 19 Plates (continued) Flexural equation: D where D = d 4w d 2w + P 2 + ∆ρ gw = q(x) dx 4 dx Eh 3 . 12(1 − ν 2 ) t T �c w �m Figure 19.1 T = T0 cos kx = T0 cos 2π x λ “Harmonic” load t = t 0 cos kx , w = w0 cos kx , t 0 = T0 − w0 h D d 4w + ρ m gw = t 0 ρ c g cos kx when P = 0 dx 4 k 4 Dw0 cos kx + ρ m gw0 cos kx = t 0 ρ c g cos kx w0 = t0 Dk 4 ρm −1+ ρc ρc g ⎛ D ⎞ Call 2 π ⎜ ⎟ ⎝ ρc g ⎠ 1/ 4 ≡ λl flexural wavelength For λ ? λl , (λl k)4 = 1 , w0 ; t0 ρm −1 ρc ; isostacy For λ = λl , w0 ; 0 ⇒ uncompensated L t x Figure 19.2 ∞ ⎡ 2 π nx s 2 π nx ⎤ t(x) = ∑ ⎢t nc cos + t n sin ⎥ ⎣ L ⎦ L n=0 Find t nc , t ns Assume D, calculate wns , wnc Synthesize w(x) , compare to observations. Plate subject to an end load y L x Figure 19.3 Shear force: dV = −q dx Since q = 0, V = const. = Va Bending moment: dM dw =V +P dx dx Since P = 0, dM = V ⇒ M = Va x + const ⇒ 0 at x = L dx M = Va ( x − L ) Displacement: d 4w =0 dx 4 ⇒ d 3w = const dx 3 But M = −D d 2w =0 , dx 2 d 3w dM = −D 3 = Va dx dx V d 3w =− a 3 dx D V d 2w = − a (x − L) 2 dx D Subject to w, w= dw = 0 at x = 0 dx Va x 2 x (L − ) 2D 3 ⇒ cubic displacement y Va L h x W Figure 19.4 w= Va 2 x x (L − ) 2D 3 Assumption σ xy = σ xx σ xx = E ε xx 1− ν2 ε xx = −y d 2w dx 2 M = −D ε xx = d 2w dx 2 y M D σ xxmax = E h 1 6V L 6V ⎛ L ⎞ Va L = a2 = a ⎜ ⎟ 2 h h ⎝h⎠ 1− ν 2 D σ xy = Va 1 h max = σ xx h 6L