Jobst-
I have been puzzling over this failure and I still think it could (place
stress on could) be a combined load failure without wheel contact or a
damaged tube. I spent some time in the garage crumpling beer cans (I had to
empty them first). Using my drill press and a flat piece of wood on top of
the can, I slowly applied an eccentric load. I can get a crack sort of noise
and a horizontal fold similar to those in the photos. A beer can is much
thinner walled of course and the down tube may act in some other way. I have
considered a detailed analysis but I lack adequate information. My
calculated buckling stress is a high number - probably higher than the
proportional limit. Here is what I did:
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A 1st rough analysis in an effort to quantify failure parameters. I have
guessed or approximated many of the variables - maybe someone would like to
refine them. Maybe somebody out there has done FEM analysis or frame testing
and can offer an opinion on possible magnitude of frame forces and frame
bending moments.
Maybe this is junk science and should go to the wastebasket. If so, please
put it there.
Anyone who is so inclined is free to check and correct my assumptions,
algebra, math, etc.
I assume that the tire did not contact the tube and that there was no
earlier damage to the tube (no dent).
Variables (guesses at best):
d = 1.75 in = diameter of down tube at failure
r = radius = d/2 = 0.875 in (smaller if oval tube)
t = 0.04 in = thickness of down tube at failure (WAG)
E = 10,000,000 psi = Modulus of elasticity
Section properties based on round tube:
A = Cross-sectional area of tube = Pi[r^2-(r-t)^2] = 0.21 in^2
I = Tube moment of inertia = pi*[d^4-(d-2*t)^4]/64 = 0.079 in^4
Stress limits:
Scr = Elastic critical stress = 0.3*E*t/r = 137,000 psi. (see note)
Sy = 40,000 psi = yield stress (WAG)
Solve for required axial force and end moment (eccentricity of axial force)
for failure:
S = stress = P/a + P*e*r/I
Where e is the eccentricity of the applied compressive force.
Solving for e:
e = (S-P/A)*(I/P*r)
M = P*e = end moment
To get a feel for magnitude of force and eccentricity. Try an axial force on
the tube of p = 3000 lbs and a limiting stress of S = 40,000 psi (yield?)
e = (40,000 - 3000/0.21) * 0.079/(2000 * 0.875) = 1.2 in
An axial force of 3000 lbs doesn't seem unlikely to me. The bike and rider
together weigh something over 400 lbs and the fork acts like a lever during
braking applying a compressive force to the down tube and a tension force to
the top tube. The frame is an indeterminate structure and I will leave its
analysis to others. An end moment equivalent to a 1.2 inch (less than a
diameter) eccentricity doesn't seem excessive either. However, The bending
moments in the tube are more complex and there may be inflection points.
Note:
Critical buckling stress (Roark & Young 5th ed. Table 35 case 14) Scr =
0.3*E*t/r where r is the radius of the tube at the location being
investigated. Method suggested by R&Y is based on failures occurring at ~
50% of theoretical critical stress. This could easily be off by 25%. The
high result indicates that elastic buckling was not likely. As the
proportional limit is exceeded the material stiffness drops off (tangent
modulus) and the tube will probably buckle at a lower stress.
