notes_24_plate_bendin

# notes_24_plate_bendin - Solution of Plate Bending Equation...

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b Solution of Plate Bending Equation Uniform Load Simply Supported Free to pull in via sinusoidal loading ( loading px , y) := p o sin π⋅ y sin x a pxy = = = w0 m x = m y = 0 for x0 y = 0 xb y = a = = m x = m y = 0 => d d x 2 2 w = d d y 2 2 w = 0 x0 y = 0 y = a 1

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x y D all boundary conditions satisfied if take wx , y) := C sin π⋅ y sin x ( a b substitute in plate equation: d 4 d 2 d 2 d 4 p o sin y sin x a b 4 w(x,y) + 2 2 2 w(x,y) + 4 w(x,y) = D dx dx dy dy d 4 4 , y) + 2 d 2 2 d 2 2 , y) + d 4 4 , y) 4C sin ( y ) sin ( x ) ⋅π 4 ( ( ( dx dx dy dy after collecting terms: ( ( d 4 4 , y) + 2 d 2 2 d 2 2 , y) + d 4 4 , y) = C sin y sin x π 4 + 2 π 4 + π 4 dx ( dx dy dy a b b 4 a 2 b 2 a 4 2 = C sin y sin x π 2 + π 2 a b b 2 a 2 is a solution if 2 p o sin y sin x C 4 sin y sin x 1 + 1 = a b a b b 2 a 2 D is a solution if 2 p o 1 C 4 b 1 2 + a 1 2 = p D o or . ... C = D 4 b 1 2 + a 1 2 2 , y) := p o 1 sin y sin x is the displacement for a sinusoidal loading in x and y ( D 4 1 1 2 a b moments and stresses are determined from: + 2 b 2 a d 2 d 2  d 2 d 2 m x := D 2 w(x,y) + ν 2 w(x,y) m y := D 2 w(x,y) + ν 2 w(x,y) dx d  dy d 2
n n t t t z max m x 3 m x 6 and σ x := I max I := σ x := I 2 σ x := m x 12 2 t this result can first be generalized to: for a loading of a higher order sinusoidal loading ( wx , y) := p o 1 sin m π⋅ y sin n x ( D ⋅π 4 2 2 2 a b px , y) := p o sin

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## This note was uploaded on 02/24/2012 for the course MECHANICAL 2.082 taught by Professor Davidburke during the Spring '03 term at MIT.

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notes_24_plate_bendin - Solution of Plate Bending Equation...

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