
#1
Feb913, 07:38 PM

P: 8

for a cantilever beam with a point load and a moment as shown in the fig.how do we calculate the force and moment,During loading the applied moment, M, remains proportional to P, such that M=10P. yield strength, young's modulus are given.
calculated Y and I for the the I section, and with yield strength σ found M (maximum bending moment). since the maximum BM is at the fixed end for a cantilever ,equated this to the other moments to find P and M ....it would be of great help if someone could help. 



#2
Feb913, 07:55 PM

P: 5

I'll have to grab my old mechanics of materials textbook to verify all this, but that should be a pretty good start for you. 



#3
Feb913, 07:57 PM

P: 197

You need a free body diagram to see what's going on. You should have reactions in the x and y direction and a moment. Equations of equilibrium will give you what you're looking for. That should help point you in the right direction




#4
Feb913, 08:35 PM

P: 8

Determining the force and moment
well to draw a FBD the forces and moments are not specified...




#5
Feb913, 08:37 PM

P: 8

if i do sigmaF=0 i''ll get constants i.e P and M...




#6
Feb913, 09:18 PM

P: 5

Well some design criteria is then needed. Is the beam loaded to yield or to ultimate? Otherwise, yeah you'll just get constants for M and P




#7
Feb913, 10:06 PM

P: 8

yes exactly it is loaded to yield...




#8
Feb913, 10:54 PM

P: 5

OK. If its yield strength is the criteria, then set σ=yield strength for the beam material.
Then, σ=Mc/I will yield the moment about the cantilevered support. This moment Msupport = (P x Beam Length) + M Since M is 10 times the Magnitude of P, You now have 2 equations with 2 unknowns. You can solve for M and P. 



#9
Feb913, 11:05 PM

P: 8

well actually i tried this in some other way it did not work(took p= aσ and got m.... ) now it all makes sense ... i owe you something more than thanks ....helping others is not so easy ...please continue ...



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