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
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