PDA

View Full Version : Bone Deformation (theoretical) - SUMMARY

randerson75
11-29-2001, 01:44 AM
Firstly thanks you to all who replied to my message (see original message
below)

Firstly the major thing was that the formula below is slightly wrong (if a
formula can ever be 'slightly' wrong!) and should read as follows -

X=(M.L^2)/(2.E.I)

This is assuming that the force applied to the beam is at the very end of
the beam and causes a moment (M) of F x L.

The other thing is that to calculate I for a bone (which we all know is a
hollow cylinder) you need to use a slightly different equation to calculate
I, this is as follows (I had this one covered but wasn't going to 'reveal'
it to my students mathematically just the theory, but now I think I'll fully
cover the difference).

I=(pi.(ro^4 - ri^4))/4 where ro is the outer bone radius and ri is the inner

Hope this clears te answer up for everyone out there.

Regards Ross

-----Original Message-----
From: Ross.Anderson [mailto:ross.anderson@ul.ie]
Sent: 29 November 2001 12:04
To: BIOMCH-L@NIC.SURFNET.NL
Subject: [BIOMCH-L] Bone Deformation (theoretical)

I am about to discuss the ability of us to look at bone (from a simple point
of view) as a cantilever with a circular cross sectional area with my 3rd yr
sport and exercise students. From my old physics notes (circa 1991!) I have
the following equation for estimating the linear deviation that occurs at
the unsupported end of the beam.

X=(4.M.L^2)/(3.pi.E.r^4)

Where M is the bending moment, L is the length of the beam, E is the
stiffness of the material (Young's Modulus), and r is the radius of the
beam.

In another, more general format, this can be expressed as

X=(M.L^2)/(3.E.I)

Where I is the second moment of area calculated around the neutral axis, and
for a beam with a circular cross section is

I=(pi.r^4)/4

I have been getting results, theoretical, from these equations which do not
make sense to me. I have been using realistic values for all unknowns in
trying to calculate x (the linear deviation) but when the calculation is
done using SI units the deviation is massive (longer than the beam!). I
have tried converting units to mm and this gives very small values, I can
see mathematically what is happening but the values are highly unrealistic.

Can anyone tell me whether

1 - the formula I am using is correct, if not what is recommended.
2 - what would be a typical linear deviation for a bone?

3 - Any other comments on the methodology etc I am using here.

Regards Ross

Ross Anderson__________________________________________ _______

Dept of Sport and Exercise Sciences and
Centre for Biomedical Electronics
University of Limerick
IRELAND
Tel - +353 (0) 86 6090866 or +353 (0) 61 202810
Fax - +353 (0) 61 330431
e-Mail - ross.anderson@ul.ie
WWW - www.ul.ie/~pess/staff/ross/

---------------------------------------------------------------
To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl
For information and archives: http://isb.ri.ccf.org/biomch-l
---------------------------------------------------------------

---------------------------------------------------------------
To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl
For information and archives: http://isb.ri.ccf.org/biomch-l
---------------------------------------------------------------