Looking for an online calculator that gives generic comparisons on the
change of tubing thickness to tubing wall. Example - how thick does a 1.125
tube have to be to equal the strength of a 1.250 x .083 or .095 tube. A link
would be very appreciated.
Reply to
Darkwing Duck
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Would the strength you be interested in be: tensile, torsion, buckling. In tension, the load bearing strength would be proportional to the cross sectional area, given the tensile strength. Calculating the cross sectional area of tubing is, of course, elementary and no calculator is needed. In torsion, the shear bearing strength is likewise elementary. In buckling, vertical beam considerations involving a special modulus involving the square of some critical dimension come into play and simple methods vanish. Have you consulted any college level strength of materials textbook. You would find a modulus formula that you could reproduce in any credible spread sheet application.
Reply to
Jack Ferman
I need to know for a racing chassis, I need to know if I can just go up on wall instead of going to a bigger OD bar to get the strength and flex I need. It's a lot easier to change wall thickness then reworking the jig and reprogramming the tube bender. I'm mainly needing a moment of inertia and modulus of steel calculator. I have found some formulas and some papers on the subject online but I'm not much of a math whiz so I was hoping to find something that was javascript or something.
Reply to
Darkwing Duck
of tubing thickness
strength of a 1.250
Calculations of Moment of Inertia, Section Modulus and cross sectional area (weight of tube) can be accomplished with a pocket calculator that can do squares and square roots. Although the formulas have variables to the fourth power or fourth root, you can square a variable twice to get the fourth power. Taking the square root twice will get the fourth root. Example: 3² = 9 and 9² = 81 which is 3 to the fourth power. Obtain the fourth root similarly using the square root feature twice in succession. (In the examples below I used my pocket calculator and show 3 or 4 decimal places even though it stores ten significant figures. So you may have a slight difference in the square and square root figures. Note: .625^4 = .625 x .625 x .625 x .625 and .0338^.25 is the fourth root of .0338)
For the Moment of Inertia the formula is I = Pi(Ro^4 - Ri^4) / 4 where Ro is the external radius, Ri is the internal radius and Pi is ?. (? = 3.14159+)
For the 1.25 OD tube with 0.083 wall, R0 = 0.625 and Ri = 0.542.
I = ?(.625^4 - .542^4) /
4 = ?(.1526 - .0863) / 4 = 0.0521 in^4
For Section Modulus the formula is Z = I /
Ro = 0.0521 / .625 = 0.0833 in^3
The cross sectional area of the tube is A = ?(Ro^2 - Ri^2). This is required to determine the weight difference in the various tube sizes.
A = ?(.625^2 - .542^2) = 0.3043 sq. in.
Now that you know what tube you are using and want to replace, we must substitute the replacement information in the above formulas. We have a 1.125 OD tube that must have a Moment of Inertia equal to or greater than 0.521 in^4 and a Section Modulus equal to or greater than 0.0833 in^3. Re-arranging the formula for Moment of Inertia, we have: Ri = (Ro^4 - (4I/?))^.25. (That is 4 times Mom. Inertia divide by Pi)
Ri = (.5625^4 - (4x.0521/?))^.25 = (.1001 - .0663)^.25 = (.0338)^.25 = 0.4288 in.
Wall thickness = (.5625 - .4288) = 0.134 inch (minimum wall thickness)
Assuming that a tube with 1.125 inch outside diameter and 0.134 inch wall exists, we check them in the formulae using .5625 - .134 = 0.4285 for Ri.
I = ?(.5625^4 - .4285^4) / 4 = 0.0522 in^4
Z = .0522 /
.5625 = 0.0927 in^3
A = ?(.5625^2 - .4285^2) = 0.4172 sq. in.
The Moment of Inertia is good. The Section Modulus has improved. The weight has increased by a factor of .4172 / .3043 = 1.371 or more than one third. This could be a problem with deflection of the tube requiring a higher Moment of Inertia. From my experience, using a smaller outside diameter hollow tube tends to create more problems than it solves.
I hope that helps,
Jim Y
Reply to
Jim Y
of tubing thickness
strength of a 1.250
(weight of tube) can
Although the
variable twice to get
Example: 3² = 9 and 9² =
square root feature
show 3 or 4 decimal
difference in the
.0338^.25 is the
the external radius,
to determine the
substitute the replacement
Moment of Inertia
than 0.0833 in^3.
(4I/?))^.25. (That is 4
exists, we check them in
has increased by a
problem with deflection of
smaller outside
================ I see my symbol for PI has been changed to a question mark. Be aware! Jim Y
Reply to
Jim Y

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