Constant stress beam design 2

[borjame]

I'm trying to get an equation that describes the shape of a beam for constant stress. The beam (cross section doesn't matter) is really a boom centrifugally loaded. I want to determine how much the cross sectional area varies when the mass on the end of the boom is varied, the RPM varies, and the radius of the end mass's CG varies. It seems since the mass of the boom comes into play when calcualting stress at the base of the radially loaded beam, that this might be a differential eqn but I'm not sure. Any help would be appreciated, thanks.

 

[GregLocock]

I can't understand your mechanism - are you trying to work out an optimum CSA profile for the arm of the rotor (ie tension), or for the shaft on which the rotor is carried (ie bending)?

Anyway, even for slightly simpler cases once you introduce self-loading then the differential equations become unsolvable analytically (eg the space elevator strength member calculation, I think), so you end up just simulating small slices of the structure, iteratively. Excel excels at this. You can develop solvable equations for simple cases, but not if they include much complexity in the section or the load case.
Cheers

Greg Locock

 

[vonlueke]

borjame: I assume beam is rotating in a horizontal plane about end A, mass is attached at end B, and beam cross section is rectangular. Let x axis be oriented along beam longitudinal axis with origin at point A. Let L = beam length, f = frequency (Hz) = RPM/60, m = mass of object attached at beam end B, rho = beam material density (e.g., for typical steel, rho = 7850 kg/m^3), b = beam cross-sectional width, h = beam cross-sectional height, pi = 3.14159, g = 9.80665 m/s^2. Velocity and centripetal acceleration at any point on beam as a function of x is v = 2(pi)(x)(f), and a_n(x) = (v^2)/x = 4(pi^2)(f^2)(x), respectively.

Thus F1 = beam uniform axial force due to mass attached at end B = m*a_n(L) = 4(pi^2)(f^2)(L)(m). F2 = beam uniform axial force at end A due to beam's own mass = integral[(a_n(x))(rho)(b)(h)(dx)] = 2(pi^2)(f^2)(rho)(b)(h)(L^2). Thus, P = beam end A uniform axial force = F1 + F2 = 4(pi^2)(f^2)(L)[m + 0.5(rho)(b)(h)(L)].

M = bending moment at end A due to gravity = (m)(g)(L) + 0.5(rho)(b)(h)(g)(L^2) = (g)(L)[m + 0.5(rho)(b)(h)(L)].

Normal stress at upper extreme fiber of beam at end A is sigma = (M*c/I) + P/A, where I = (b)(h^3)/12, c = h/2, and A = (b)(h). Substituting from the above, maximum stress in beam is sigma = [6(g)(L) + 4(pi^2)(f^2)(h)(L)][m + 0.5(rho)(b)(h)(L)]/[(b)(h^2)].

 

[GregLocock]

vonlueke I like your first para - but why would you get bending? is this not just pure tension? I'm imagining a centrifuge.

Cheers

Greg Locock

 

[GregLocock]

if so then dA/dx=rho*omega^2*(L-x)/sigma

where A is the cross sectional area at distance x inboard of the mass

and A(0)=m*L*omega^2/sigma

Cheers

Greg Locock

 

[vonlueke]

Greg: Here's the reason I included bending moment. If frequency f = 0, the cantilever would be deflecting (drooping) downward toward earth. There is some relatively high value of f at which the centrifugal force would be great enough to tend to remove moment M in my previous post and cause mass m to rise back up to the horizontal plane. But I conservatively assumed the frequency is not that high.

 

[borjame]

This is all good stuff but I guess I need to simplify things and make them a bit harder to get what I'm looking for. First, ignore gravity by putting the problem in space (no bending moments). Second, you need to include the weight of the beam itself in the problem and that's where it gets tricky. Solving for just the mass on the end of the boom (picturing a centrifuge is perfect) is pretty straightforward but when you solve that problem you now have a beam of varying cross section in the problem. That beam has it's own CG and associated component in the stress tensor at the base of itself. A good analogy is the rocket problem- you need so much fuel to get a satellite into space, but then you need more fuel to account for the weight of the fuel for the satellite, and then you need fuel to account for the new fuel, and so on. As you add cross sectional area to account for the mass on the beam, you need more cross sectional area to support the support beam, and so on. I hope I'm not rambling, just trying to fully explain the problem.

 

[vonlueke]

My solution above is for a prismatic beam, which, now as I understand it, is not what you're looking for. For my prismatic solution above, since gravitational acceleration g is a parameter, just set g therein to zero for the prismatic, zero-gravity solution.

Also, my above solution neglects bending stresses developed as a result of inertial loads while you are accelerating the beam up to the operating frequency.

I assume you want only a constant-velocity solution neglecting any start-up inertial loads.

 

[GregLocock]

OK, well my solution was nearly correct for what you want.


I missed an A out, which makes solving the equation more difficult, but still easily within my jaded memory's ability.

Consider the outermost slice of beam. using r for rho, w for omega, s for sigma


A*s= m*L*w*w


Consider the next slice, area A+dA, at a radius of L-x. It has to supply the same force as the previous slice, plus a bit more to accelerate the previous slice

(A+dA)*s=m*L*w*w+A*dx*r*(L-x+dx/2)*w*w

hence
dA/A=r/s*w*w*(L-x)dx

which is solvable if you integrate both sides. I get a solution A=m*L*w*w*e^(r/s*w*w*(L*x-1/2*x^2))

Cheers

Greg Locock

 

I have three questions on Greg's answer. Is the origin of the x coordinate at the outboard end of his beam? Could someone define the "s" variable, and how do we calculate "s"? Also, why does the units of the final "A" equation come out to be newtons instead of area? Or am I doing something wrong? Thanks.

 

[GregLocock]

origin of x in my terminology is at the mass, not the axis of rotation.

s is the working stress for the material. You'd probably use a factor of ignorance of at least 10 if this is man-rated.

The units come out wrong in the last line partly because I've forgotten to divide through by s. Oops. All these errors are transcription errors, easily corrected by the trained analyst!

A=m/s*L*w*w*e^(r/s*w*w*(L*x-1/2*x^2))



Cheers

Greg Locock

 

[austim]

borjame,

I don't know whether you were able to read my post to the identical thread that you started in "structural engineering - other topics". While GregLocock was developing his "mechanical" response, I was doing very nearly the same thing a few miles away, but in the structural forum.

Unfortunately, one of the likely results of "cross posting" (starting multiple threads on one topic on two or more forums) is that cross posted threads will be "red-flagged" and deleted (together with any responses to them). So there my response is - gone.

My approach was slightly different to Greg's, but we reach very similar conclusions. That is very good, since one of my concerns was that you should get at least one other solution independently to act as a check on mine. (I am only too conscious of the deterioration of my skills in calculus over the many years since I graduated).

Rather than work in terms of Greg's independent variable x (the distance from the tip of the boom) I started with the radius to the centre of rotation as my basic distance variable. my assumed variables are:

M : the mass attached to the boom tip
Rm : the radius to the CG of the attached mass
Rt : the radius to the tip of the boom
At : the cross-section area of the boom at the tip
A : the cross-section area of the boom at radius r
Ft : the tensile force applied by the attached mass at the tip
ó : the constant tensile stress (this was a genuine greek sigma when it was first entered, but the eng-tips preview changed it :-()
w : the angular velocity (would be a greek omega if I could type it)
gamma : the density of the boom material (mass/unit volume)

then Ft=Rm*w^2*M
ó = Ft/At

consider the equilibrium of a thin slice within the length of the boom, with midplane at radius r, area A, thickness dr.

the difference in area between opposite faces of the slice = dA/dr * dr = dA; the difference in tensile force = -ó*dA, which must balance the inertia load from the mass of the slice itself = r*w^2*gamma*A*dr.

ie -ó*dA = r*w^2*gamma*A*dr, giving dA/A = -(w^2*gamma/ó)*rdr

Integrating both sides between the limits r and Rt gives

ln(At/A) = -(w^2*gamma/ó/2)*(Rt^2-r^2)

At/A=e^[w^2*gamma/ó/2*(r^2-Rt^2)]

A=At/e^[w^2*gamma/ó/2*(r^2-Rt^2)]


if you replace Rt by L, and r by L-x, to match Greglocock's variables,
this becomes A=At/e^[w^2*gamma/ó/2*({L-x}^2)-L^2]
= At/e^[w^2*gamma/ó/2*(-2Lx+x^2)] = At*e^[gamma/ó*w^2(Lx-1/2*x^2)]

This is very close to Greglocock's solution, but not exactly the same. My At corresponds to Greg's m*L*w*w/s. I believe that there is a divisor of s missing from Greg's final formula.

If you take x=0 in A=m*L*w*w*e^(r/s*w*w*(L*x-1/2*x^2), you get A=m*L*w*w*e^0 = m*L*w*w, which is a force, not an area.

Apart from that relatively minor discrepancy, Greg and I reach the same result.

 

[prex]

I can confirm austim's solution, that I checked with a separate one, obtained by numerical integration in Excel.
borjame should be quite satisfied now. prex
motori@xcalcsREMOVE.com

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