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

[Baluncore]

Ultimately, we can kick or punch as hard as we want as long as the compressive breaking load of our leg or arm bones are not exceeded. Is this correct?

The momentum of the hand or foot, can apply an impulse greater than the compressive strength of the leg or arm bones. It is maintaining a steady force that tests the bone.
It will also depend on whether an injury is acceptable.

Bones tend to be long, and thinner in the middle. That permits some bending near the middle, but not at the ends.

 

[A.T.]

Ultimately, we can kick or punch as hard as we want as long as the compressive breaking load of our leg or arm bones are not exceeded. Is this correct?

These are not the main tasks that our musculoskeletal system has developed for, so more likely to cause injury. Activities like walking, running, jumping, climbing are more natural to us. And despite requiring muscle forces of multiple body weights, they rarely break bones.

 

[Rev. Cheeseman]

The momentum of the hand or foot, can apply an impulse greater than the compressive strength of the leg or arm bones. It is maintaining a steady force that tests the bone.
It will also depend on whether an injury is acceptable.

Bones tend to be long, and thinner in the middle. That permits some bending near the middle, but not at the ends.

These are not the main tasks that our musculoskeletal system has developed for, so more likely to cause injury. Activities like walking, running, jumping, climbing are more natural to us. And despite requiring muscle forces of multiple body weights, they rarely break bones.

Sorry for my English comprehension, I'm not sure if I got your point. The compressive force to break a femur from a vertical position is around 1000 lbs so can we make conclusion that the strongest push kick (not roundhouse kick) ever will be around 1000 lbs?

 

[A.T.]

... can we make conclusion that the strongest push kick (not roundhouse kick) ever will be around 1000 lbs?

No you cannot make any such conclusions about a short impact force, as already explained by @Baluncore above.

 

[Rev. Cheeseman]

No you cannot make any such conclusions about a short impact force, as already explained by @Baluncore above.

If we can't use the maximum compressive breaking load of bones alone and want to make hypothetical assumptions, what is the definitive force of the strongest ever kick or punch?

 

[A.T.]

If we can't use the maximum compressive breaking load of bones alone and want to make hypothetical assumptions, what is the definitive force of the strongest ever kick or punch?

Not in a simple way. You would need a mechanical musculoskeletal model, with a very high temporal resolution to capture the high impact accelerations, which also takes the inertia of the body segments into account.

Also note that the individual strength of bones can vary a lot, because bones adapt to the loading they experience. Athletes and and people doing martial arts can have much stronger bones than the values for the general population that you might find in references.

 

[Rev. Cheeseman]

Not in a simple way. You would need a mechanical musculoskeletal model, with a very high temporal resolution to capture the high impact accelerations, which also takes the inertia of the body segments into account.

Also note that the individual strength of bones can vary a lot, because bones adapt to the loading they experience. Athletes and and people doing martial arts can have much stronger bones than the values for the general population that you might find in references.

Ok, I can't understand the point of this statement. What does "The momentum of the hand or foot, can apply an impulse greater than the compressive strength of the leg or arm bones. It is maintaining a steady force that tests the bone. It will also depend on whether an injury is acceptable" means? Sorry, my English comprehension is bad.

 

[A.T.]

Ok, I can't understand the point of this statement. What does "The momentum of the hand or foot, can apply an impulse greater than the compressive strength of the leg or arm bones. It is maintaining a steady force that tests the bone. It will also depend on whether an injury is acceptable" means?

The steady force application to static bones in the tests is different from briefly acting impact force peaks at the foot. The brief impact force must also account for the acceleration of shank and foot, rather than just for compressing the femur.

 

[Rev. Cheeseman]

Baluncore said


No.
Buckling occurs with longer columns, with lengths over 30 times the diameter. Bones are not long enough to fit that model, especially when your test sample examples have diameter to length less than about 1 : 2.

Your problem is more like sampling. If you have a sample of strength 100±10, then how strong will two be in series 100±?

You have a live thread in the open forum. If you ask there, I will be notified, and others will also see it, and answer, and the forum will get hits from Google etc. That way, others benefit from each answer, which justifies the time needed for good answers.

No one will see this conversation, so it costs me full price, then only you benefit.

[Mentor Note: This quote above appears to be from a Personal Message conversation, not from this thread originally]

 

[Rev. Cheeseman]

According to https://books.google.com.my/books?i...of the femur is defined in FMVSS 208"&f=false, 10000 Newton force for several milliseconds is enough to break a femur from the vertical compression position. Humerus is weaker than femur in terms of vertical compression. Therefore, hypothetically, the strongest possible punch a human can generate is around 2 tonnes in my opinion before the humerus started to break.

 

[Rev. Cheeseman]

According to https://journals.sagepub.com/doi/full/10.1177/2280800018793816, a midshaft part of the humerus is sanctioned with a height-to-diameter ratio of 2 and its max compressive force is around 2.5 to 3 kiloNewton. Is it possible to imagine the max compressive force of the whole humerus? If possible, what are the formulas? Thank you.

 

[Baluncore]

Is it possible to imagine the max compressive force of the whole humerus? If possible, what are the formulas?

It is probable that the maximum compressive force will be the same for all parts of the humerus. There is no advantage in having one part fail at a higher or lower compressive stress than another part.

 

[Rev. Cheeseman]

It is probable that the maximum compressive force will be the same for all parts of the humerus. There is no advantage in having one part fail at a higher or lower compressive stress than another part.

But the previous examples, especially those experiments that involved whole femurs show different results where the whole femurs need lesser force than the sanctioned femur parts due to structural differences. Thanks for all the help. Much appreciated. I'll study it by myself from now on although I'll like to hear some opinions from you all in the upcoming days if I'm curious about something.

 

[A.T.]

It is probable that the maximum compressive force will be the same for all parts of the humerus. There is no advantage in having one part fail at a higher or lower compressive stress than another part.

This argument assumes that the compression is the same everywhere along the bone, which is not the case. Bones aren't just compressed throughout by fores acting at their ends, but are also compressed partially by muscles attaching somewhere along the bone.

See the muscle attachments for the humerus:
https://en.wikipedia.org/wiki/Humerus#Shaft

 

[Tom.G]

For some people, early teenagers and younger, there is another bone breaking way called a "corn flake fracture (or break)".

An early teenager was roller skating on the sidewalk. At a street crossing, skating in a tight circle was done to slow down. The sidewalk had some raised cracks that caused the skater to fall.

Trying to control the fall, arms and hands were extended to soften the fall. The result was a "corn flake fracture" of one wrist. The Ulna and Radius, two lower arm bones, had flakes (chips) of bone broken off from the high compression force.

The cast was from the hand to above the elbow.

 

[Rev. Cheeseman]

Hello, does this article https://www.sciencedirect.com/science/article/pii/S188844152300089 tells us the axial compressive strength of human humeri? Thank you.

 

[berkeman]

Hello, does this article https://www.sciencedirect.com/science/article/pii/S188844152300089 tells us the axial compressive strength of human humeri? Thank you.

What part of that article?

 

[Rev. Cheeseman]

What part of that article?

This part, to be exact.

"Static study​

When subjected to rupture (Fig. 6), the median failure load was 1.844 N, and the mean failure time was 72 s (Table 1). When comparing the two cemented screw configurations, specimens with screws cemented in rows B and D presented a higher ultimate load (2.218 N vs. 2.105, p = 0.901) and higher stiffness (125 N/mm vs. 106 N/mm, p = 0.672). Specimens with cemented screws in rows A and E tolerated 6 mm more deformity before failure (p = .447) and the time to failure was longer (84 s vs. 56 s, p = .342); however, neither of these differences was statistically significant (Table 2).


Table 1. Median of values obtained in the study at breakage.

Variable	Median (ranges)
Breaking load (N)	1844 (804–4135.5)
Elastic load (N)	1236 (611.9–3917.6)
Maximum deformity (mm)	25 (8.7–50.6)
Stiffness (N/mm)	93 (32.2–233.4)
Energy (J)	39 (7.1–92.7)
Breakage time (s)	72 (22.5–166.9)

All specimens failed in the diaphysis by a fracture line that included at least one of the holes used for plate fixation.

A comparative analysis was performed between those specimens with an age older than 76 years (n = 6) and those younger than 76 years (n = 4) and it was observed that those specimens younger than 76 years tolerated greater deformity before failure (30.7 mm vs. 22.4 mm, p = .61), higher ultimate load (2411 N vs. 1995 N, p = .476) and absorbed greater energy (59 J vs. 30 J, p = .171), without these differences being statistically significant. Significant differences were only observed in stiffness and breakage time"

*For this purpose, an increasing compressive load was applied at a rate of 20 mm/min until failure, defined as dissociation between the bone and the implant, the appearance of a fracture line or implant breakage.

 

[A.T.]

Hello, does this article https://www.sciencedirect.com/science/article/pii/S188844152300089 tells us the axial compressive strength of human humeri? Thank you.

For a healthy intact bone? Obviously not.

Have you even looked the pictures? It's a bone with an introduced large gap, that is bridged with a plate, which requires drilling a bunch of holes into the bone. And that is exactly where the failures occur:

"All specimens failed in the diaphysis by a fracture line that included at least one of the holes used for plate fixation."

 

[Rev. Cheeseman]

For a healthy intact bone? Obviously not.

Have you even looked the pictures? It's a bone with an introduced large gap, that is bridged with a plate, which requires drilling a bunch of holes into the bone. And that is exactly where the failures occur:

I just saw the black and white photos of the humeri and yes they have holes. Hmm, sorry. So, are they actually measured the compressive strength of the stainless steel locking plates? Given that the compressive force is still around 1000 lbs/4000 Newtons?

I wonder what's the difference in compressive force needed to break healthy intact humeri, I'm trying to find similar articles on healthy intact humeri but still didn't find them

 

[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"