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Torsional deflection of an octagon shaped tube or housing 1

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John2002

Mechanical
Feb 10, 2002
20
Hello,

I would be grateful if someone could please help me calculate the amount of torsional deflection (in degrees) of an octagon shaped tube or housing for a given amount of torque. The housing is an 8 sided octagon circumscribed "around" a 0.905 inch diameter circle.

The housing has a 3/16" thick cam-plate doweled to it at one end (load end) and a perpendicular lever connected at the other end. The cam is connected to springs and rotation of the housing via the .375" diameter lever causes the springs to exert a torque force on the cam and housing in the opposite direction of the lever.

A rectangular bar with holes is notched into the bottom of the cam plate. The springs are connected to the rectangular bar.

The total length of the housing will be 1.260 inches. The distance from the center of the thickness of the cam plate at the load end of the housing to the center of the perpendicular lever at the other end will be 1.117 inches.

The housing will have a .688" diameter hole machined to a depth of 0.990 inches from the lever end, to receive bearings that the housing rotates about. If the housing were round, it would have a 0.109" thick "wall" surrounding the bearings, and for all practical purposes I guess that is exactly what I have. The distance from the octagon "points" to the ID of the housing is 0.146". The housing is made from 7075-T7 aluminum.

The springs will exert a maximum force of about 135 inch-pounds of torque on the cam and housing.

I would like to know how much torsional deflection (in degrees) that 135 inch-pounds of torque will produce on the cam and housing. Any formulas for a problem like this would also be very helpful. I would like to get results for both an aluminum and steel housing if possible.

Does the octagon shape have any significant effect on torsional deflection when compared to a tube with a .688" ID and a .905" OD?

Thank you for your help.

John
 
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Roark gives a formula for the principal second moments of area of a hollow regular polygon with n sides, in terms of the thickness (t), the semi angle subtended by one flat (alpha), and the width of one external flat (a). (note - (a) is NOT the width across flats).

These second moments are both equal, and are given by :

I1=I2=n*a^3*t/8*(1/3+1/(tan(alpha))^2)*(1-3*(t*tan(alpha))/a+4*((t*tan(alpha))/a)^2-2*((t*tan(alpha))/a)^3)

The second polar moment of area (say J) will then be the sum of these two (ie just replace the "8" in the above formula with a "4").

For an octagon, n = 8 and alpha = 22.5 degrees. I will leave you to figure out the value of a.

The torsional deflection of the tube will then be given by:

T/J = G*theta/L

where T is the torque, G the shear modulus, theta the deflection angle (radians) and L the length.

These formulae are not 100% accurate according to the theory of elasticity, but should be good enough for your purposes.


For a circular tube, J is given by

pi/32*(Do^4-Di^4)

where Do is the OD and Di is the ID.

I'll leave you (or someone else) to make the comparison !
 
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