randerson75

11-28-2001, 10:04 PM

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/

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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/

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To unsubscribe send SIGNOFF BIOMCH-L to LISTSERV@nic.surfnet.nl

For information and archives: http://isb.ri.ccf.org/biomch-l

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