randerson75

11-29-2001, 06:42 PM

In my previous summary (slightly premature I guess) I stated that - 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 (see below). In fact the following

statement holds true.

If the deflection is due to a point load applied to the end of a beam, x=(F

L^3)/(3 E I). If it is due to a point

moment: x=(M L^2)/(2 E I). The equations are different because the point

force results in a linearly decreasing moment in the beam from end to end,

while the point moment is a constant moment along the entire length.

Regards Ross

Original Summary

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

bone radius.

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/

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

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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 (see below). In fact the following

statement holds true.

If the deflection is due to a point load applied to the end of a beam, x=(F

L^3)/(3 E I). If it is due to a point

moment: x=(M L^2)/(2 E I). The equations are different because the point

force results in a linearly decreasing moment in the beam from end to end,

while the point moment is a constant moment along the entire length.

Regards Ross

Original Summary

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

bone radius.

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/

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

---------------------------------------------------------------