Recent content by LDC1972

Hi Oxon. Long time ago I did this but, yes, calculate as per the textbook for approximation of wall thickness. Then take that value and repeat with appropriate equation for the approximated thickness to get true thickness. Hope that helps?

I know that feeling! At least after this unit it becomes way more sensible. PM'd...

No I got a merit for that exam, just got Q2 wrong. Unfortunately the feedback just said "incorrect" so can't help! For the whole module I got a distinction as the other 3 TMA's were graded at distinction.

Sorry Jock, this post isn't going to help you much! I got this question wrong entirely. I never looked at it again after so can't help at all! After that module it was indeed back to zooming through modules though as they are a lower level! I highly advise choosing the hydraulic module...

I am in awe! And you've taught me a lot which is what it's all about really eh :-)

Wow, I've been trying for hours again and your help is amazing. Thank you so much!

The masses are definitely on the same plane (not 180 degree opposed) Thank so much for trying this one so hard for me! I like me original answer, because the values look reasonable, but my second answer the values look ridiculous! Wish I could contribute intelligently, but I'm beyond the...

I think I've changed me mind... Going back to Ra = mass x position length / shaft length = 8 x position length / 200 x 1000 = 5 KN Ra = 1000 / 8000 = 0.125 Ra = 8 x 0.125 / 0.2 x 1000 = 5 Kn And using new figures gives: Ra = mass x position length / shaft length = 20 kg x 0.5 m / 2 m = 5...

There's a second and final part to this problem and just wanted to run past you guys my solution to see if I've done OK? Exact same system except now to be balanced by 2 masses, 0.5m from each end and radius now 100mm. I have done: Shaft length = 2 m 157.1 rad/s Ra = 5 Kn Rb = 3 Kn...

It was "summing the moments that threw me. I've done it right t the start of the course for beams, but something didn't "click". I've managed to get a migraine now (been at ti since 4:30am and it's now 11am!) So I'll be back in an hour and follow your moments properly. The course I'm doing...

Is it not possible the position is as per 8 Kg but the actual weight due to radius is only 1.62 Kg?

I'm so confused :-( I'm doing the equation as follows: Ra = 5 Kn (we know this). So: Ra = 1.62 kg x 617 cm / 200cm x 1000 = 5 Kn. BUT the shaft is only 200cm long, so my 617cm is madness... Please help. I'm following an example in the book exactly as above, but there's works!

Thanks for the very fast reply, I'm still hammering away. I really thought I'd cracked this :-(

Homework Statement A shaft rotates at 1500rpm between 2 bearings, Ra and Rb. Ra force is 5 Kn and Rb is 3 Kn. A single mass is to be used to balance the shaft, to give reactions of zero. Mass to be placed at 90 degrees to shaft and at 200 mm radius. What is the required mass and position...

V = ∏r^2 x length simplify to r^2 = v / ∏ x length, that's your radius. σ1 = yield strength / factor of safety and as σ1 = Pr / t then t = Pr / σ1 and that's your wall thickness using thin wall. Cool? Once you get 231mm or so you've cracked it, then move onto thick walled with now...