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 p x , y) := p o sin π⋅ y sin π⋅ x a pxy = = = w 0 m x = m y = 0 for x 0 y = 0 x b y = a = = m x = m y = 0 => d d x 2 2 w = d d y 2 2 w = 0 x 0 y = 0 x b y = a 1
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x y D all boundary conditions satisfied if take w x , y) := C sin π⋅ y sin π⋅ x ( a b substitute in plate equation: d 4 d 2 d 2 d 4 p o sin π⋅ y sin π⋅ 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 w x , y) + 2 d 2 2 d 2 2 w x , y) + d 4 4 w x , y) 4 C sin ( π⋅ y ) sin ( π⋅ x ) ⋅π 4 ( ( ( dx dx dy dy after collecting terms: ( ( d 4 4 w x , y) + 2 d 2 2 d 2 2 w x , y) + d 4 4 w x , y) = C sin π⋅ y sin π⋅ 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 π⋅ y sin π⋅ x π 2 + π 2 a b b 2 a 2 is a solution if 2 p o sin π⋅ y sin π⋅ x C ⋅π 4 sin π⋅ y sin π⋅ x 1 + 1 = a b a b b 2 a 2 D is a solution if 2 p o 1 C ⋅π 4 b 1 2 + a 1 2 = p D o or .... C = D ⋅π 4 b 1 2 + a 1 2 2 w x , y) := p o 1 sin π⋅ y sin π⋅ x is the displacement for a sinusoidal loading in x and y ( D ⋅π 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
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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 (
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