Circular Plate Bending Moment and Stress
Dear CEians,
I had a small project going onto my final semester subject. Its a about a Circular Plate attached to the shaft spinning at 1280rpm 1Hp Motor. After that, load being applied uniformly onto the Circular plate. The Diameter of the plate is 125mm and the load applied is at diameter of 100mm. Load of 50kN being applied uniformly.
As i know, the plate intertia is to be 0 or infinit (correct me if i'm wrong). and i got to found out this formula Mr=P(1+ν)log(R/r)/4π. and i dunno how to applied it to have the moments results.
I really have to idea how the calculation would be. Bending Moments of beam i would say yes, i can calculate. but for circular Plate, i would say NO. Any guide would be appreciate.
I had a small project going onto my final semester subject. Its a about a Circular Plate attached to the shaft spinning at 1280rpm 1Hp Motor. After that, load being applied uniformly onto the Circular plate. The Diameter of the plate is 125mm and the load applied is at diameter of 100mm. Load of 50kN being applied uniformly.
As i know, the plate intertia is to be 0 or infinit (correct me if i'm wrong). and i got to found out this formula Mr=P(1+ν)log(R/r)/4π. and i dunno how to applied it to have the moments results.
I really have to idea how the calculation would be. Bending Moments of beam i would say yes, i can calculate. but for circular Plate, i would say NO. Any guide would be appreciate.
Replies
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juma1987Circular flat plates Circular flat plates
Cross section size
A = pi R2
Edge supported around the circumference, load distributed uniformly over the surface
Maximum stress
σ=0.39F/(t^2)
Deflection
d=0.221F(R^2)/(E(t^3))
Rotation at support position
Ф=FR/(piE(t^3))
Edge fixed around circumference, load distributed uniformly over the surface
Maximum stress
σ=0.24F/(t^2)
Deflection
d=0.0543F(R^2)/(E(t^3))
Edge supported around the circumference, load concentrated at the center
Maximum stress
σ=(0.48F/t^2)(1+1.3Ln(R/0.325t)-0.0185(t^2/R^2))
Deflection
d=0.55F(R^2)/(E(t^3))
Rotation at support position
Ф=0.667FR/(E(t^3))
Edge fixed around circumference, load concentrated at the center
Maximum stress
σ=(0.62F/t^2)(Ln(R/0.325t)+0.0264(t^2/R^2))
Deflection
d=0.22F(R^2)/(E(t^3))
Strength check
smax ≤ Sy / ks
ks ≤k
where F is the load, t the plate thickness, Ks the safety factor
Source: Autodesk Inventor Handbook -
maxzDear Juma~!!!
Your guides is very useful to me. being finding this for a long time. Never know it can be found in Autodesk Inventor Handbook. Our Library is sucks. Thankz anyway.#-Link-Snipped-##-Link-Snipped-# -
yadavundertaker mohitmaxzDear Juma~!!!
Your guides is very useful to me. being finding this for a long time. Never know it can be found in Autodesk Inventor Handbook. Our Library is sucks. Thankz anyway.#-Link-Snipped-#
only crazy engineer can solve that so be regular on CE and help others also... -
juma1987ill be around ty 🎉
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maxzFor the manual calculation part i'm done, and the result is quite satisfied...
But i need to compare it with my simulation testing using Inventor 2010...
How to apply this format into Inventor and simulate it... i have no idea...
I kinda more in Manufacturing and Management for engineerings... Hands on testing is weak for me.. Please help me...
Guides step by step would be much appreciate...
I tried to simulate follow the Help section in Inventor, but it is far different from wat i expected...
The value and the stress is to big compare to my manual calculation. thx -
juma1987well you can use the design acelerator formulas or use FEA, can you make a drawing of what u want with details and post here? so maybe i can explain the process to make it (if i know, and if i dont maybe someone that know it can help).
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maxzbtw wat is design accelerator formulas??? where can i get it?
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