Plates (continued)21

# Plates (continued)21 - dM dx = V P dw dx Since P = 0 dM dx...

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12.005 Lecture Notes 21 Plates (continued) Flexural equation: D d 4 w dx 4 + P d 2 w dx 2 + ∆ ρ gw = q ( x ) where D = Eh 3 12(1 ν 2 ) . Figure 21.1 T = T 0 cos kx = T 0 cos 2 π x λ “Harmonic” load t = t 0 cos kx , , w = w 0 cos kx t 0 = T 0 w 0 D d 4 w dx 4 + ρ m gw = t 0 ρ c g cos kx when P = 0 h T t w ρ c ρ m Figure by MIT OCW.

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k 4 Dw 0 cos kx + ρ m gw 0 cos kx = t 0 ρ c g cos kx w 0 = t 0 ρ m ρ c 1 + Dk 4 ρ c g Call 2 π D ρ c g 1/4 λ l flexural wavelength For λ ? λ l , , ( λ l k ) 4 = 1 w 0 ; t 0 ρ m ρ c 1 ; isostacy For λ = λ l , uncompensated w 0 ; 0 Figure 21.2 t ( x ) = t n c cos 2 π nx L + t n s sin 2 π nx L n = 0 Find t Assume D, calculate n c , t n s w n s , w n c Synthesize w ( x ), compare to observations. x t L Figure by MIT OCW.
Plate subject to an end load Figure 21.3 Shear force: dV dx = − q Since

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Unformatted text preview: dM dx = V + P dw dx Since P = 0, dM dx = V ⇒ M = V a x + const ⇒ 0 at x = L M = V a ( x − L ) Displacement: d 4 w dx 4 = 0 ⇒ d 3 w dx 3 = const L x y Figure by MIT OCW. But M = − D d 2 w dx 2 = 0 , dM dx = − D d 3 w dx 3 = V a d 3 w dx 3 = − V a D d 2 w dx 2 = − V a D ( x − L ) Subject to w , dw dx = 0 at x = 0 w = V a x 2 2 D ( L − x 3 ) ⇒ cubic displacement...
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