Description
Problem 1: Vibrations of a plate
material properties are the same as the cantilever beam from the last project. Let’s put our origin at theLet’s analyze the vibrations of a plate. Consider a plate of dimensions L = 1m, W = 1m and H = 0.001m. The
. We are still using the same equation of motion. So we don’t have to non-dimensionalize anything again.x = 0,y =
0,z = 0
- a) direction) force ofLet’s consider the case where the plate is clamped at the left face x = 0 −z
100N 0.98L ≤ x ≤ L,0.49W ≤ y ≤
| together on a single plot.(0.5L,0,H) (0.5L,0.25W,H) (0.5L,0.5W,H) (0.5L,0.75W,H) (0.5L,W,H) | |
| (1b) | Repeat the above problem for L = 2m,W = 2m. How does the frequency and amplitude change? |
| (1c) | Let’s consider a different case now forapply the force ofpoints listed above (in the same format). Obtain the amplitude and frequency of vibrations of each point.at a corner L = 1m,W = 1m. Let’s clamp two faces,. Obtain the plots for all thex = 0 and y = W and
100N 0.98L ≤ x ≤ L,0 ≤ y ≤ 0.02W,z = H |
| (1d) | Let’s clamp all the side facesforce ofplacement vs time for following points in a single plot:, but this time in the middleand x = 0, .x = L, y = 0 and y = W (for L, = 1m,W = 1m),. Apply the same. Plot the dis-,
100N 0.49L ≤ x ≤ 0.51L,0.49W ≤ y ≤ 0.51W,z = H z (0.5L,0.5W,H) (0.25L,0.5W,H) (0.75L,0.5W,H) (0.5Repeat the plots fordifference in amplitudes and frequencies of all these points? If so, comment on the differences.L,0.25W,H) (0.5L,0.75W,H) , , , . Is there a (0.25L,0.25W,H) (0.25L,0.75W,H) (0.75L,0.25W,H) (0.75L,0.75W,H) Now change the dimensions again topoint change? L = 2m,W = 2m. How do the frequency and amplitudes of each |
(1 peat this forplacement of following points of the plate as a function of time on a single plot: . Once the plate is bent, we let go of the force and see the plate vibrate. Plot the,on a small patch on the top towards the right side, . Obtain the amplitude and frequency of vibration of each point. Re-, , and subject to a downward (, (L,0,H),and plot these(L,0.25W,dis-H),
0.51W,z = H z
(L,0.5W,H)(L,0.75W,H) (L,W,H)
