On 11/16/14, 5:33 AM, Stan Stockton wrote:
Wondering if any mechanical engineers out there could calculate this:
If you had 60 feet of 2 inch OD x .125 wall 6061-T6 (held solid at
one end) how much rotational torque (foot pounds?) at the other,
free end would it take to twist the mast 3, 5 and 10 degrees
respectively?
Working on something and don't want to find out too far into it that
it won't work. This is not a matter of whether something will fall
or cause damage to anyone or anything - only loss of time and
effort.
Torsional deflection is
Theta = 32 * L * T / (G *pi * (D^4-d^4))
G is modulus of rigidity, D is OD, d is ID, T is torque, L is length
(shear modulus, not E which is compression/tension modulus)
www.amesweb.info/Torsion/TorsionalStressCalculator.aspx has a calculator
which can work with mixed units
6061-t6 modulus is 68.9 GPa (10,000 ksi)
http://asm.matweb.com/search/SpecificMaterial.asp?bassnum=MA6061t6
Shear modulus is 26 GPa (3770 ksi)
A bit of iterating with the amesweb calculator...
50 lb ft -> 10.098 deg
25 lb ft -> 5.049
15 -> 3.029
at 50 lb ft ->stress is 17.183 MPa, well below the failure stress of 207 MPa
http://www.engineeringtoolbox.com/modulus-rigidity-d_946.html gives a
shear modulus of 24 GPa, so the deflection would be slightly more.
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