Constant stress beam design
Constant stress beam design
(OP)
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.





RE: Constant stress beam design
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
RE: Constant stress beam design
Thus F1 = beam uniform axial force due to mass attached at end B = m*an(L) = 4(pi^2)(f^2)(L)(m). F2 = beam uniform axial force at end A due to beam's own mass = integral[(an(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)].
RE: Constant stress beam design
Cheers
Greg Locock
RE: Constant stress beam design
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
RE: Constant stress beam design
RE: Constant stress beam design
RE: Constant stress beam design
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.
RE: Constant stress beam design
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
RE: Constant stress beam design
RE: Constant stress beam design
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
RE: Constant stress beam design
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.
RE: Constant stress beam design
borjame should be quite satisfied now.
prex
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