PPRuNe Forums

Go Back   PPRuNe Forums > Flight Deck Forums > Tech Log
Forgotten your Username/Password?

Tech Log The very best in practical technical discussion on the web


Reply
 
Thread Tools
Old 4th Nov 2006, 13:17   #1 (permalink)
 
Join Date: Dec 2005
Location: U.S.
Posts: 144
Strength of hollow shaft vs. solid

A while back someone posted the rule(s) about the strength (torsional, tensile, & shear) of a hollow shaft (such as a rotor mast) with a given i.d. & o.d., vs. the strength of a solid shaft with the same o.d. Would some kind soul give us that again? Many thanks in advance.
arismount is offline   Reply
Old 4th Nov 2006, 15:56   #2 (permalink)
 
Join Date: Oct 2006
Location: Sydney, Australia
Age: 38
Posts: 14
G’day Arismount ,

I think the most important aspect of helicopter mast design is the amount of shear stress (tau) generated in the cross section due to the torque being delivered through the mast to the rotor from the engines. The formula relating these quantities is;

tau = (T*p) / Ip where tau = shear stress (Mpa)
T = torque (Nm)
p = radius of cross-
section at which
you want to know
tau
Ip = Polar moment of
inertia (mm^4)

With a round shaft subjected to a given torque, the highest shear stress (tau) will occur at the maximum radius of the cross-section, so “p” is usually taken as “r” which is the outer radius of the shaft.

Ip, or the polar moment of inertia, is a basically a mathematical quantity that describes the distribution of the cross sections area about the cross section axis. For a solid shaft;

Ip = [pi*(r^4)] / 2 r = radius
= [pi*(d^4)] / 32 d = diameter

For a hollow shaft;

Ip = [pi/2]*[ro^4 – ri^4]
= [pi/32]*[do^4 – di^4] ri = inner radius
ro = outer radius
di= inner diameter
do = outer radius

In metric units, the quantity Ip will have units of mm^4. You must design the shaft section such that the shear stress is below the shear strength of the material you are using. You can also use this equation to work out the amount of torque the shaft will withstand if you know the section details and the shear strength of the material.

The tensile strength of the shaft (due to helicopter weight?) is simply the cross sectional area of the section multiplied by the tensile strength of the material.

JetMech is offline   Reply
Old 4th Nov 2006, 20:13   #3 (permalink)
 
Join Date: Feb 2005
Location: flyover country USA
Age: 72
Posts: 4,184
For a shaft carrying primarily torsion loads, it is the outer elements (greatest radii) that carry the highest stress, so the center core elements contribute little to the function of the shaft. Might as well get rid of the core, make it tubular, and enjoy a better strength-to-weight ratio.

(same may be said for compression struts with column loads i.e. bending)

"Simplicate and add lightness" - William Stout, designer of the Ford Tri-Motor
barit1 is offline   Reply
Old 5th Nov 2006, 04:13   #4 (permalink)
 
Join Date: Oct 2006
Location: Sydney, Australia
Age: 38
Posts: 14
The formatting of my previous post went a bit haywire due to my post being written in MS Word and then posted across. For clarity;

Ip, or the polar moment of inertia, is a basically a mathematical quantity that describes the distribution of the cross sections area about the cross section axis. For a solid shaft;

Ip = [pi*(r^4)] / 2
= [pi*(d^4)] / 32

r = radius (mm)
d = diameter (mm)

For a hollow shaft;

Ip = [pi/2]*[ro^4 – ri^4]
= [pi/32]*[do^4 – di^4]

ri = inner radius (mm)
ro = outer radius (mm)
di= inner diameter(mm)
do = outer diameter(mm)

It is true that the inner elements of a solid shaft carry very little shear stress. In fact, for a solid shaft, the shear stress varies in a linear manner from zero in the centre to maximum at the outer radius.

A hollow shaft is much more structually efficient in terms of amount of torque carried for the weight of material in the shaft, but the outer diameter will be greater. Removing the core of a solid shaft "shifts" the shear stress distribution, thus, to get the same maximum shear stress on the outside, a hollow shaft will need to be of greater radius.

The shear stress distribution in a hollow shaft is still linear, but it now starts at some non-zero value on the inner radius, and increases (linearly) to the maximum value on the outer radius.

Last edited by JetMech; 5th Nov 2006 at 04:24.
JetMech is offline   Reply
Old 5th Nov 2006, 23:15   #5 (permalink)
 
Join Date: Dec 2005
Location: U.S.
Posts: 144
excellent info

many thanks, gents.
arismount is offline   Reply
Old 6th Nov 2006, 00:24   #6 (permalink)
 
Join Date: Mar 2002
Location: Florida
Posts: 3,993
For all the reasons pointed out above +

A solid shaft typically has much less fatigue life due to poor metallurgical properties at its innermost fibres (casting/forging lapses). This is also why the engine compressor and turbine disks do not have solid shapes all the way to their center point.
lomapaseo is offline   Reply
Reply
 
 
 


Thread Tools


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off



All times are GMT. The time now is 11:24.


vBulletin® v3.8.7, Copyright ©2000-2014, vBulletin Solutions, Inc.
SEO by vBSEO 3.6.1
© 1996-2012 The Professional Pilots Rumour Network