Plates (continued)22

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

This preview shows pages 1–4. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online