B Compressive force of a shorter cylindrical bone vs a longer one

[A.T.]

So, are they actually measured the compressive strength of the stainless steel locking plates?

Obviously not, since it's not the plates that failed first.

 

[Rev. Cheeseman]

Obviously not, since it's not the plates that failed first.

How do we hypothesized the compressive force to fracture whole humeri? Can we use the height to diameter ratio? For instance, if the compressive force to break a humerus with height to diameter ratio of, say, around 2 is around 3 tonnes, what will be the compressive force for a whole humerus? Can we use stainless steel or concrete cylinders as references but without treating the humerus literally as steel or concrete?

 

[Baluncore]

Can we use the height to diameter ratio?

There are three simple modes of humerus failure.
1. A side-force that bends the humerus to the point it fractures.
2. An axial-force that crushes, and so reduces, the length of the bone.
3. An axial-force that first causes buckling, followed by a failure similar to a side-force.

The height to diameter ratio, coupled with the initial humerus asymmetry, decides the force needed for a buckling failure. I would expect that, under axial compression, the end joints would dislocate or fracture, before the humerus buckled.

 

[Rev. Cheeseman]

There are three simple modes of humerus failure.
1. A side-force that bends the humerus to the point it fractures.
2. An axial-force that crushes, and so reduces, the length of the bone.
3. An axial-force that first causes buckling, followed by a failure similar to a side-force.

The height to diameter ratio, coupled with the initial humerus asymmetry, decides the force needed for a buckling failure. I would expect that, under axial compression, the end joints would dislocate or fracture, before the humerus buckled.

Lets see if this makes sense. The length of men's humeri is around 13 inches or 0.3 meter, while if we divide the humerus in five parts starting from the top uppermost to the lowest part and we just remove the middle fifth (the midshaft of the humerus) the length of the sanctioned part will be around 3 inches or 0.07 meter. The radius and the thickness that we assumed for the humerus is 15 mm radius and 10mm thick. Assuming the compressive force to break the 3 inches midshaft humerus is around 30 kilonewton/3 tonnes or 3000 kg, according to this buckling calculator https://efficientengineer.com/buckling-calculator/ the compressive force to break a whole humerus which is 13 inches or 0.3 meter is 1.6 kilonewton/163 kg/359 lbs. The modulus of elasticity I put there in the buckling calculator is 0.0948 GPa.

 

[Baluncore]

Lets see if this makes sense. The length of men's humeri is around 13 inches or 0.3 meter, while if we divide the humerus in five parts starting from the top uppermost to the lowest part and we just remove the middle fifth (the midshaft of the humerus) the length of the sanctioned part will be around 3 inches or 0.07 meter. The radius and the thickness that we assumed for the humerus is 15 mm radius and 10mm thick. Assuming the compressive force to break the 3 inches midshaft humerus is around 30 kilonewton/3 tonnes or 3000 kg, according to this buckling calculator https://efficientengineer.com/buckling-calculator/ the compressive force to break a whole humerus which is 13 inches or 0.3 meter is 1.6 kilonewton/163 kg/359 lbs. The modulus of elasticity I put there in the buckling calculator is 0.0948 GPa.

I do not understand your argument.

 

[Rev. Cheeseman]

I do not understand your argument.

I'm not arguing, the compressive force to break 3 inches midshaft segment from a 13 inches whole humerus is around 3000 kg. The diameter of the humerus is 1.5 inches, so the height to diameter ratio for the 3 inches humerus is about 2. I think height to diameter ratio that's more than 2 will result to having lower compressive force.

 

[Baluncore]

The diameter of the humerus is 1.5 inches, so the height to diameter ratio for the 3 inches humerus is about 2.

The resistance to buckling must be calculated over the full length of the column. The degree of freedom must be specified for both ends.

An engineering calculator will assume a constant section, but that is not the case with a bone. The humerus is more flexible near the middle, with large-area ball joints at the ends.

I think height to diameter ratio that's more than 2 will result to having lower compressive force.

The compressive force will be the same at all points along a strut. If that was not the case, the strut would move to balance the forces.

 

[Rev. Cheeseman]

The compressive force will be the same at all points along a strut. If that was not the case, the strut would move to balance the forces.

According to https://ro.uow.edu.au/cgi/viewcontent.cgi?referer=&httpsredir=1&article=1290&context=eispapers1, the more slender (higher height-to-diameter ratio) the less the compressive force. So, I guess the same happen to humeri.

 

[Baluncore]

Bone is not a ductile metal encased in concrete.

the more slender (higher height-to-diameter ratio) the less the compressive force.

Exactly where in that dissertation, (line number), does it make that statement ?

 

[Rev. Cheeseman]

Bone is not a ductile metal encased in concrete.

Exactly where in that dissertation, (line number), does it make that statement ?

I can't find the compressive force when the fracture or failure begin, but only compressive strength.

I'm trying to find articles or studies that said the compressive force to break a whole femur is the same as breaking a midshaft only segment but still can't find it.

Saw some studies on concrete cylinders, like this https://www.researchgate.net/public...in_uniaxial_compression_of_concrete_cylinders. The conclusion below said...

"For lower ratios, the strength rapidly increases with decreasing slenderness ratio"

 

[Baluncore]

In order for global buckling to occur, the length-to-diameter ratio needs to be less than 6.

No. You have that backwards.
"In general, to avoid global buckling of galvanized and cold-formed
steel tubes in the structural applications, the L/D ratio needs to be < 6."

I'm trying to find articles or studies that said the compressive force to break a whole femur is the same as breaking a midshaft only segment but still can't find it.

Probably because it is false.

 

[Rev. Cheeseman]

No. You have that backwards.
"In general, to avoid global buckling of galvanized and cold-formed
steel tubes in the structural applications, the L/D ratio needs to be < 6."

Sorry for the mistake. Thank you for noticing. I was rushing when typing and sleepy

Probably because it is false.

Sorry, why is it false.

 

[Baluncore]

Sorry, why is it false.

You are cherry-picking with a search engine.
If it were true, you would expect to find a cherry.

 

[Rev. Cheeseman]

You are cherry-picking with a search engine.
If it were true, you would expect to find a cherry.

Is there a study where they demonstrated the compressive force of whole humeri?

 

[Baluncore]

Is there a study where they demonstrated the compressive force of whole humeri?

Probably yes.
But finding that research is your problem.