
#1
Sep2204, 04:51 PM

P: 2

hello all, i am trying to solve for outside diameter on a hollow shaft where the inside diameter is 1.25"
the hollow shaft is the same material as a solid shaft which is 1" in diameter. Since both are the same materials and all i'm trying to do is duplicate the amount of torque and stresses that the solid shaft is able to handle what formulas and/or thumbrules would you consider best for using, thanks! p.s. an example would be a car drive shaft, the output shaft from the gear box is solid and 1" but the drive shaft is hollow and 4" with a wall thickness of about 0.1". 



#2
Sep2304, 08:43 AM

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P: 1,481

It would be a great discussion here talking about dynamic resistance of various shaped shafts. But I only can advice you online to consider the stresses present on your shaft. Supposing there is no weight bending the shaft, it is only suffering of a torsion stress. (I'm not sure if the word "torsion" exist in english). Anyway, it is a shear stress:
[tex]\tau=\frac{Tr}{I_o}[/tex] (Pa) where T=torque exerted (N.m), r=radial coordinate, Io=inertia moment (m^4). Inside Io is implicit the shape of the shaft section. So that, you must calculate the figures in the external radius (where the stress is maximum) and compare it with your safety coefficient. 



#3
Sep2304, 09:43 AM

P: 2

"r=radial coordinate" what is this and how can i get this value?
"Io=inertia moment (m^4)." is this mass to the 4th power? "calculate the figures in the external radius (where the stress is maximum) and compare it with your safety coefficient" could you explain more I'm not sure how to do what you mean (I have no other source of information) torsion is an English word, thanks! 



#4
Sep2304, 12:31 PM

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P: 1,341

hollow vs. solid shafts
Greetings !
Assuming you're dealing with cicular shafts: for a hollow shaft where the difference between inner and outer radiuses is not neglagible compared to the outer radius you substract inner Io from the outer, for a hollow shaft where the above difference is neglagible you get: 2 * pi * Radius^3 * t , where t is the above difference. In general it means that a solid shaft is better if you care about size, and the hollow shaft is better if you care about weight. Live long and prosper. 



#5
Sep2504, 12:00 PM

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P: 1,481

Drag has explained it to you much better than that I would have done.




#6
Sep1110, 02:12 AM

P: 1

tell me about that the strength of solid shaft is more than hollow shaft




#7
Sep1110, 06:00 AM

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P: 1,422

typically you want to go hollow and a larger diameter at the same time. there are several benefits:
you will have less weight (or at least the same) for an item that can handle more torsional stress. the reason is that when you have a hollow shaft you have a lower 'stress gradient' through the material. the way that works is that you have draw a line from the center to the outer edge and when you twist (torque) the shaft you have the center not moving and the outer edge moving (displacing) a lot (relatively) and the difference between them is a factor in the stress in the shaft. if you remove the center you have removed a bunch of that difference. I'm almost certain that given the same outside diameter, the solid shaft will have more torsional resistance than the hollow section. The main reason for hollow sections is to reduce mass. If memory serves, in order for a hollow section to carry similar torsional loads the diameter has to be increased, therefore if you compare a solid and a hollow section that weigh exactly the same then the hollow should be stronger because of the increased OD. If you're comparing two axles with the same OD, the solid one will deform less given both torsional and cantilevered (what you described as supporting one end and pushing down on the other) loading. For two axles of a given mass, the hollow one will deform less as mentioned above, because it will have a larger OD to compensate for mass removed from the inside. the strongest lightest section for a cylindrical object will be the biggest radius you can achieve with the thinnest wall you can achieve. Solid sections ARE stronger, but the strength increase is not proportional to the added weight vs a hollow section. IE you might have a solid axle that can carry 100 ft lbs while weighing 20 lbs. A hollow axle of the same OD might be able to carry 90 ft lbs while weighing in at only 5 lbs. I made these numbers up they're just an example. The angular deflection of a torsion hollow shaft can be expressed as θ = 584*L*T / (G (D^4 d^4)) where θ = angular shaft deflection L = length of shaft G = modulus of rigidity T = Torque D = OD d = ID So you get diminishing returns on increasing wall thickness (relationship is cubic, not linear) the torsional stiffness property of a shaft varies with the 4th power of it's outside diameter. A solid shaft will always be stiffer than a hollow shaft with all other material properties being identical. but remember, the stiffness property varies with the 4th power, so hollowing out the middle won't make it very much different. The easiest way to prove this to throw away all the other variables out of the equations and just compare the diameter variable. with that in mind, take d^4 of a solid shaft, and compare it to (Dout^4Din^4) of the hollow shaft. you'll see that they're usually very close. A solid bar is stiffer/harder to flex than a hollow bar of equal length and outside diameter. 



#8
Oct1210, 01:37 PM

P: 1

My research into this, a very short term, has provided the following
information. See the .pdf attached for formulas about solid vs. hollow. Hollow shafting is stronger as a "cantilevered" application than a simply supported each hollow shaft. Solid is stronger in the simply supported on both ends application. 



#9
Apr1911, 02:43 PM

P: 2

What is the torque of a solid shaft (plastic tube) compared to a hollow shaft (three separate internal openings) filled with fluid? (water and/or blood). What if the internal diameters are filled with sand?



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