Compressive force of a shorter cylindrical bone vs a longer one

[Rev. Cheeseman]

Hello,

Why is it more difficult to break a shorter bone that is cut into a cylinder than a longer cylindrical bone by pressing the bone vertically from its top despite sharing the same cross-sectional area?

 

[Frabjous]

Fracture is largest flaw dominated. So longer samples are “weaker”.
You might also be engaging with buckling instability, where longer samples are also weaker.

 

[Baluncore]

A bone is a cylindrical wall that contains a fluid. If you cut a cylinder from that, you have a short tube. A longer bone is more likely to fail by buckling, like a column. The joints at the ends of a longer bone are attached to the cylindrical fusilla of the bone by a narrower neck that forms a weak point.

Do you have any numbers on the force involved?

 

[Rev. Cheeseman]

A bone is a cylindrical wall that contains a fluid. If you cut a cylinder from that, you have a short tube. A longer bone is more likely to fail by buckling, like a column. The joints at the ends of a longer bone are attached to the cylindrical fusilla of the bone by a narrower neck that forms a weak point.

Do you have any numbers on the force involved?

Let's say the force is around 1000 lbs, which is I guess the compressive force in order to break a femur from the vertical orientation?

 

[Baluncore]

Let's say the force is around 1000 lbs, which is I guess the compressive force in order to break a femur from the vertical orientation?

How do you apply the force to the femur?
Through the joints or somehow else?

 

[Rev. Cheeseman]

How do you apply the force to the femur?
Through the joints or somehow else?

For example, we cut the joints and make them look like cylinders with flat top and bottom for both the longer and shorter femur.

 

[Rev. Cheeseman]

How do you apply the force to the femur?
Through the joints or somehow else?

Sorry @Baluncore , any thoughts?

 

[Baluncore]

... any thoughts?

You have not described your setup for preparing specimens for the test. You have not identified the diameter or length of the specimens. We don't even know what species of animal they are from.

 

[Rev. Cheeseman]

You have not described your setup for preparing specimens for the test. You have not identified the diameter or length of the specimens. We don't even know what species of animal they are from.

Let assume we are trying to find out how much compressive force to break two bulls' femurs from vertical loading by using a hydraulic press. Both bull femurs are cut into cylinders and the diameters are the same for both. But one is longer than the other.

 

[Baluncore]

What are the diameters, and what are the lengths?
Have you plotted compressive strength against length?

 

[Rev. Cheeseman]

What are the diameters, and what are the lengths?
Have you plotted compressive strength against length?

Diameter is about 6-7 inches for both, but the short one is about 3 inches in height and the taller one is about 6 inches or more. How to plot compressive strength against length? Sorry.

 

[Baluncore]

Why is it more difficult to break a shorter bone that is cut into a cylinder than a longer cylindrical bone by pressing the bone vertically from its top despite sharing the same cross-sectional area?

Is it more difficult?
What data do you have that demonstrates that?

 

[Rev. Cheeseman]

Is it more difficult?
What data do you have that demonstrates that?

This is just a hypothetical scenario, I'm not doing any physical experiments. Hence, this question whether a shorter cylindrical bone is stronger or not than a longer cylindrical bone with the same circumference as the shorter one from vertical compression along its length.

 

[Lnewqban]

Diameter is about 6-7 inches for both, but the short one is about 3 inches in height and the taller one is about 6 inches or more. How to plot compressive strength against length? Sorry.

IMHO, both samples should be equally strong to pure compression, as the longest one is not slender enough to consider buckling.

Please, see:
https://www.engineeringtoolbox.com/euler-column-formula-d_1813.html

Any difference in failure compressing forces could be influenced by the homogeneousness, cross-section shape and thickness of the calcium shelf forming the bone.
And that would be assuming that both cuts are perfectly parallel to each other and perpendicular to the main axis of the column for each sample.

 

[A.T.]

Why is it more difficult to break a shorter bone ...

Hence, this question whether a shorter cylindrical bone is stronger or not ...

If you don't even know whether it is true, why did you ask for the reason?

 

[Rev. Cheeseman]

If you don't even know whether it is true, why did you ask for the reason?

I don't know whether its true or not because it looks like some of you all are not sure about that (For instance, why is it more difficult to break a shorter bone), I thought some of you guys have those experiences regarding this that's why I ask you all the question.

 

[Rev. Cheeseman]

IMHO, both samples should be equally strong to pure compression, as the longest one is not slender enough to consider buckling.

Please, see:
https://www.engineeringtoolbox.com/euler-column-formula-d_1813.html

Any difference in failure compressing forces could be influenced by the homogeneousness, cross-section shape and thickness of the calcium shelf forming the bone.
And that would be assuming that both cuts are perfectly parallel to each other and perpendicular to the main axis of the column for each sample.

Thank you, appreciate it.

 

[Rev. Cheeseman]

IMHO, both samples should be equally strong to pure compression, as the longest one is not slender enough to consider buckling.

Please, see:
https://www.engineeringtoolbox.com/euler-column-formula-d_1813.html

Any difference in failure compressing forces could be influenced by the homogeneousness, cross-section shape and thickness of the calcium shelf forming the bone.
And that would be assuming that both cuts are perfectly parallel to each other and perpendicular to the main axis of the column for each sample.

Found this online. Is this ok? https://efficientengineer.com/buckling-calculator/

 

[Baluncore]

And that would be assuming that both cuts are perfectly parallel to each other and perpendicular to the main axis of the column for each sample.

The sample will also need to be placed in the testing machine with the axis of the machine, coaxial with the sample. Failure to do that will result in asymmetric force applied to the sample.

The ratio of length to thickness, does not satisfy the description of a column. The bone has a wall thickness that is probably less than one third of the diameter. The bone wall is more likely to splinter or split than to buckle.

 

[Lnewqban]

Perhaps the age of the animal will be another factor to consider in the type of fracture to be observed, as the density of the tissue should be lower, and the fragility of the bone's wall should be greater for a naturally aged bone.

The second part of the following video compares how more and less plastic materials fail, compared to a bone.

 

[Rev. Cheeseman]

Perhaps the age of the animal will be another factor to consider in the type of fracture to be observed, as the density of the tissue should be lower, and the fragility of the bone's wall should be greater for a naturally aged bone.

The second part of the following video compares how more and less plastic materials fail, compared to a bone.

According to this link https://books.google.com.my/books?id=on_SBwAAQBAJ&pg=PA207&dq=Yamada+femur+compressive+"5050"+kg&hl=en&newbks=1&newbks_redir=0&source=gb_mobile_search&sa=X&ved=2ahUKEwiygLmAn4-HAxXqR2wGHcH8ALEQ6AF6BAgLEAM#v=onepage&q=Yamada femur compressive "5050" kg&f=false, the max human femur compressive force in kg is 5050 kg. But according to https://www.sciencedirect.com/science/article/pii/S001048252300714X, the compressive load is much lower which is at 4546 N or just around 400 kg. In the former link, they cut the femurs into cylinders with flat top and bottom by cutting the top and bottom of the femurs leaving the midshaft part of the femurs only and use the midshaft parts for the experiment. I believe that's the reason why they got the 5050 kg figure.

 

[Rev. Cheeseman]

Perhaps the age of the animal will be another factor to consider in the type of fracture to be observed, as the density of the tissue should be lower, and the fragility of the bone's wall should be greater for a naturally aged bone.

The second part of the following video compares how more and less plastic materials fail, compared to a bone.

Regarding the moose femoral bone in the video, perhaps if they use a whole moose femur the compressive force will be lower?

 

[A.T.]

According to this link https://books.google.com.my/books?id=on_SBwAAQBAJ&pg=PA207&dq=Yamada+femur+compressive+"5050"+kg&hl=en&newbks=1&newbks_redir=0&source=gb_mobile_search&sa=X&ved=2ahUKEwiygLmAn4-HAxXqR2wGHcH8ALEQ6AF6BAgLEAM#v=onepage&q=Yamada femur compressive "5050" kg&f=false, the max human femur compressive force in kg is 5050 kg. But according to https://www.sciencedirect.com/science/article/pii/S001048252300714X, the compressive load is much lower which is at 4546 N or just around 400 kg. In the former link, they cut the femurs into cylinders with flat top and bottom by cutting the top and bottom of the femurs leaving the midshaft part of the femurs only and use the midshaft parts for the experiment. I believe that's the reason why they got the 5050 kg figure.

The femur bone is not a straight cylinder, but has a substantial bent. Compressing the entire femur introduces bending moments, which is very different from compressing a short, approximately cylindrical, section.

 

[Rev. Cheeseman]

The femur bone is not a straight cylinder, but has a substantial bent. Compressing the entire femur introduces bending moments, which is very different from compressing a short, approximately cylindrical, section.

Therefore, the shorter and stable it is, the less likely for it to bend?

 

[A.T.]

Therefore, the shorter and stable it is, the less likely for it to bend?

Yes, but it's not about a "longer cylindrical bone" as in your OP, about a but longer section of a curved bone.

 

[Rev. Cheeseman]

Yes, but it's not about a "longer cylindrical bone" as in your OP, about a but longer section of a curved bone.

So if we want realistic results, we should've pay attention on the experiments that involved breaking whole femurs from vertical positioning because our femurs are not perfectly parallel cylinders with flat top and bottom.

 

[A.T.]

So if we want realistic results, we should've pay attention on the experiments that involved breaking whole femurs from vertical positioning because our femurs are not perfectly parallel cylinders with flat top and bottom.

You have to use the experiment that is relevant to your question.

Note that in the body the bone is also subject to forces from muscles, which attach all around the bone. So compressing it just from both ends doesn't represent what happens in the body.

 

[Rev. Cheeseman]

You have to use the experiment that is relevant to your question.

Note that in the body the bone is also subject to forces from muscles, which attach all around the bone. So compressing it just from both ends doesn't represent what happens in the body.

Understood, thank you for those infos. Much appreciated.

 

[A.T.]

Understood, thank you for those infos. Much appreciated.

You're welcome. Just to give you an example on why the muscle forces matter:

From: https://www.sicot-j.org/articles/sicotj/full_html/2016/01/sicotj150120/F1.html

The femur has a crooked upper end (femoral neck) (B). If you apply only vertical forces at the the joint surfaces, the neck will break off rather quickly. But in the active body you usually have large muscle forces pulling at those hills around the neck (trochanter) with horizontal components, so the compression is well aligned with the neck there.

Also, when you look at the shaft of the femur from the side (A) you see it bowing forward in the middle, which would lead to buckling and early breaking if loaded only axially at the ends of the shaft. But in the active body you have many muscles pulling at the back side of the femur (linea aspera), which opposes the buckling.

In both cases the shape of the bone developed to minimize the local bending moments under the combined load from joints contact and muscles / ligaments.

 

[Rev. Cheeseman]

You're welcome. Just to give you an example on why the muscle forces matter:

View attachment 347856
From: https://www.sicot-j.org/articles/sicotj/full_html/2016/01/sicotj150120/F1.html

The femur has a crooked upper end (femoral neck) (B). If you apply only vertical forces at the the joint surfaces, the neck will break off rather quickly. But in the active body you usually have large muscle forces pulling at those hills around the neck (trochanter) with horizontal components, so the compression is well aligned with the neck there.

View attachment 347857

Also, when you look at the shaft of the femur from the side (A) you see it bowing forward in the middle, which would lead to buckling and early breaking if loaded only axially at the ends of the shaft. But in the active body you have many muscles pulling at the back side of the femur (linea aspera), which opposes the buckling.

View attachment 347858

In both cases the shape of the bone developed to minimize the local bending moments under the combined load from joints contact and muscles / ligaments.

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?

 

[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?