View Full Version : Strength of hollow shaft vs. solid

arismount

4th Nov 2006, 14:17

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.

JetMech

4th Nov 2006, 16:56

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.

barit1

4th Nov 2006, 21:13

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

JetMech

5th Nov 2006, 05:13

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.

arismount

6th Nov 2006, 00:15

many thanks, gents.

lomapaseo

6th Nov 2006, 01:24

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.