Plates (continued)22

Plates (continued)22 - 12.005 Lecture Notes 22 Plates...

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12.005 Lecture Notes 22 Plates (continued) Figure 22.1 w = V a 2 D x 2 ( L x 3 ) Assumption σ xy = xx xx = E 1 ν 2 ε xx xx =− y d 2 w dx 2 M D d 2 w dx 2 xx = y D M xx max = E 1 2 h 2 1 D V a L = 6 V a L h 2 = 6 V a h L h L V a W h x y Figure by MIT OCW.
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σ xy = V a h = 1 6 h L xx max Response to a line load (e.g. volcanic chain under water) Figure 22.2 D d 4 w dx 4 + ( ρ m w ) gw = 0a t x 0 V 0 at x = 0 Solution to homogeneous equation w = e x / α ( C 1 cos x + C 2 sin x ) + e x / ( C 3 cos x + C 4 sin x ) with = 4 D ( m w ) g 1/4 is flexural parameter. Invoke symmetry, boundedness. Determine solution only for x 0. dw dx = 0 at x = 0 C 3 = C 4 V a ρ w ρ c ρ m h x { Figure by MIT OCW.
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w 0 x →∞ C 1 = C 2 = 0 w = C 3 e x / α (cos x + sin x ) Now to evaluate , go back to the end load problem (or original definition) C 3 dM dx = V + P dw dx C 3 depends on V . 0 d 3 w dx 3 =− 1 2 V D 1 2 V 0 = D d 3 w dx 3 x = 0 = 4 DC 3 3 V 0 is negative load, half supported by each side.
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Plates (continued)22 - 12.005 Lecture Notes 22 Plates...

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