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Re: Solid Rollbar (long Poindexter answer :-)

To: autox@autox.team.net
Subject: Re: Solid Rollbar (long Poindexter answer :-)
From: TeamZ3@aol.com
Date: Tue, 5 Oct 1999 21:33:01 EDT
Stress comparison of both solid and tubular circular sections of constant 
radius is a function of the section modulus of each member, and a direct 
relationship can be obtained by equaling the diameter relationship  functions 
between the solid and tubular sections (there is a constant that is equal in 
both equations and it cancels out):

tubular relationship = {(OD)4th power - (ID)4th power}/OD
solid relationship = (OD)3rd power

setting the tubular and solid relationship equal to each other, and knowing 
that the tubular diameters are outer dia = 1.5 in and inner dia = 1.25 in 
(1/8" wall thk) we can solve the equivalent diameter for a solid member:

solid OD = [{(tubular OD)4th power - (tubular ID)4th power}/(OD)]1/3rd power
solid OD = [{(1.5)4th power - (1.25)4th power}/(1.5)]0.333 power
solid OD = 1.20 inches

and the weight relationship between the two can be based on equivalent area, 
which is a function of (diameter)2nd power you will quickly see that the 
solid member is 209% heavier given the same density material.

However, if you want to evaluate the solid and tubular members based on 
deflection, the corresponding comparison is a function of the moment of 
inertia equation, and setting them equal creates a similar situation with a 
slightly different formula:

tubular relationship = (OD)4th power - (ID)4th power
solid relationship = (OD)4th power

solid OD = {(1.5)4th power - (1.25)4th power} 0.25 power
solid OD = 1.27 inch (234% heavier)

It should now be clear why tubular members are used; they're considerably 
lighter for a given design load.  The difference in the two size variations 
demonstrates that the tubular member handles deflection slightly better than 
stress compared to an equivalent solid member.

Mark Sipe

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