|May29-09, 12:04 PM||#1|
Cantilever Beam Deflection
1. The problem statement, all variables and given/known data
As in the attachment; cantilever beam with two UDLs and a point load of 1000N/m, 1000N, and 500N/m respectively. First UDL acts over the first 1m, point load at 1m, and the next udl over the next meter - so the total beam length is 2m. Max tip deflection is 2mm, what is the minimum bending stiffness (EI) required.
2. Relevant equations
Bernoulli-Euler Bending Theory
3. The attempt at a solution
EIv'' = (1000*1*0.5) + 1000*1 + (500*1*1.5)
EIv'' = 2250
EIv' = 2250x + A
EIv = 1125x^2 + Ax + B
Boundary Conditions v(0)=v'(0)=0
EI = 22.5 E5
Answer is 98.95E4
I'm guessing my mistake is in either the boundary conditions or the actual initial creating of the EIv'' equation.
Any help would be hugely appreciated!
|May29-09, 02:47 PM||#2|
We haven't seen your attachmnet yet, but from your description, your boundary conditions are OK, but your EIv" equation is not. You cannot just take the maximum moment (2250 N-m) at the fixed end and apply it in the equation. The moment must be written as a function of x. I haven't done this calc in awhile, but it would appear best to me that you look at each load case separately: A udl of 500 N/m across the entire beam, then the point load of 1000 N at x = 1m, then a udl of 500 N/m over the first 1 m. Determine the deflection at the free end for each case, and add them up via superposition. You're probably going to need to find the slope of the deflection curve at 1 m out to solve for the free end deflection for the last 2 loadings.
BTW, welcome to PF!
|May29-09, 03:11 PM||#3|
I'll have another go at sorting the equations out - all the previous examples I've done have been without any physical numbers and have only had single loads which is where my confusion has come from I guess. Don't think you fully understand (no doubt my rubbish explanation) where the loads are acting, however when the attachment is approved it will all be clear.
|May29-09, 06:11 PM||#4|
Cantilever Beam Deflection
I see your attachment and you explained it well. You have
(1.) a uniform distributed load of 500 N/m along the entire beam length, L;
(2.) a concentrated 1000 N load at its center (at L/2), and then
(3.) an additional 500 N/m uniformly distributed load over the first meter (I've taken the 1000 N/m shown and subtracted out the 500 N/m already included in (1.) above. I believe it is easier to solve the problem this way).
So now calculate the deflection (as a function of EI) at the tip, for each load case, add them up , and set them equal to 0.002 m to solve for EI. In calculating the deflection, you must write the moment as a function of x for each load case, before integrating. For example , for the 1000N load, P, the moment, M , is P(L/2) - Px, between x = 0 and x = L/2, and M = 0 between x = L/2 and x = L. Double integrate over the first half of the beam to get the deflection, v, at L/2, single integrate to get the slope of the deflection curve, v', at L/2, and then calculate the tip deflection using trig (the 2nd part of the beam has no moment, but it goes along for the ride and deflects accordingly). Continue for the other loads. Requires patience...and time.
|Sep12-11, 01:25 PM||#5|
How do I start a post, I have an engineering question about cantilever beams?
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