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**Physics, Astronomy & Cosmology / Re: Is a hollow tube stronger than a solid bar?**

« **on:**23/02/2018 18:58:02 »

Because this is the second or third result that comes up on Google, I feel a duty to set the record straight with aggressive equations despite the fact that this thread is many years old.

In all respects, a solid bar will be stronger (and stiffer) than a hollow bar of equal diameter according standard beam theory (which assumes linear elastic behavior). This is not arguable. I'll work out the math for examples of tensile strength/stiffness, compressive strength/stiffness, and bending strength/stiffness.

***Tension***

Again, larger area corresponds to greater stiffness.

***Compression***

Compressive stiffness is the same as tensile stiffness.

For compressive strength, there are 2 primary failure modes: Euler buckling and Squash (yield) buckling. Johnson's rule determines where and how to transition between these two modes, but as I will show, a solid bar is stronger for both and there is no need to apply this (it's somewhat intuitive to me and the math will be too difficult to type without a proper typesetter so trust that this is the case)

For Euler,

For Squash buckling, the load corresponding to the yield stress is determined the same way as in tension.

I'll make a note here that Euler buckling is catastrophic and is more likely to occur for beams with smaller area moment of inertia (a result of Johnson's rule which predicts departure from Euler's at half of Euler's yield stress) and therefore pipes are more likely to experience this failure mode than squash buckling. I think someone in the linked thread was trying to get at this but didn't back it up with evidence or equations but rather by intuition.

***Bending***

For critical load, we use the equation

For stiffness, there are various different loading cases/configurations for a beam, but I'll use the example of a cantilever because the equations are simpler.

Now for the reason that I believe this rumor is so widespread:

People are correct in stating that the centers of rods have generally less effect on strength/stiffness. If you look at the area moment of inertia equations, for a circle I is proportional to r^4. That means doubling the radius increases strength/stiffness 16 fold for the equations above which contain I. Since area moment of inertia can be analyzed with superposition, cutting out the center circle makes the new area moment of inertia proportional to (R^4-r^4) where R and r are the outer and inner diameters respectively. Of course, r has a much smaller effect than R.

What does this mean? This means that in any equation above which contains I, the strength to weight ratio of a pipe is much higher than for a rod. For any equation which uses A, cross-sectional area actually increases proportionally to weight so the strength to weight ratios are pretty much the same. Why is strength to weight ratio important? Of course, designing a building, for example, means the weight of the structures at the top of the building are going to be weighing down on the structures at the bottom of the building. In animals/plants, lighter means fewer resources are required to achieve the same strength, as well as allowing for better mobility (recall F=ma ... same muscle force garners greater acceleration if mass is decreased). The Eiffel Tower took inspiration from bone structures and showcased the ability of a mostly empty structure to support so much weight.

In conclusion, saying a pipe of equal MASS is stronger than a rod is generally true (depending on which metric of strength is being used, they may be the same). Saying a pipe of equal diameter is stronger than a rod is flat out false.

I hope I was able to clear up any misconceptions.

For reference, I'm finishing up an undergrad mechanical engineering degree and I've been taking and TAing multiple structural analysis courses every semester for multiple years. For those of you who don't trust theory/equations, yes, I've also seen these tests done in real testing machines for materials like steel, aluminum, titanium, plastic, carbon fiber, etc. and you'd be shocked how closely the results match the equations. The biggest departures from theory generally appear in heat treatment (which generally just changes the stiffness and/or yield point but does not change the fact that the material still follows linear elastic behavior), nonlinear behavior (i.e. loading past yield and some strange materials), and non-isotropic materials (i.e. wood, carbon fiber, etc). For ordinary steel, copper, or aluminum pipes/rods, this theory 100% applies.

In all respects, a solid bar will be stronger (and stiffer) than a hollow bar of equal diameter according standard beam theory (which assumes linear elastic behavior). This is not arguable. I'll work out the math for examples of tensile strength/stiffness, compressive strength/stiffness, and bending strength/stiffness.

***Tension***

(stress) = F/A

The cross-sectional area of a solid bar of equal diameter is larger, thus the critical force F required to reach the yield stress of a material is higher for a solid bar. This means more tensile strength.(strain) = (stress)/(young's modulus)

Rearranging to get stiffness,(deflection) = F / (stiffness)

where (stiffness) = (young's modulus)*(area) / (length)Again, larger area corresponds to greater stiffness.

***Compression***

Compressive stiffness is the same as tensile stiffness.

For compressive strength, there are 2 primary failure modes: Euler buckling and Squash (yield) buckling. Johnson's rule determines where and how to transition between these two modes, but as I will show, a solid bar is stronger for both and there is no need to apply this (it's somewhat intuitive to me and the math will be too difficult to type without a proper typesetter so trust that this is the case)

For Euler,

P

where P_{cr}=_{cr}is the failure load, E is young's modulus, I is the area moment of inertia, K is a constant depending on the loading conditions, and L is the length of the rod/pipe. E, K, and L are the same for both because those depend on the conditions in which the pipe is set up. I is larger for a solid pipe (google "2nd moment of area" for the equations) .For Squash buckling, the load corresponding to the yield stress is determined the same way as in tension.

I'll make a note here that Euler buckling is catastrophic and is more likely to occur for beams with smaller area moment of inertia (a result of Johnson's rule which predicts departure from Euler's at half of Euler's yield stress) and therefore pipes are more likely to experience this failure mode than squash buckling. I think someone in the linked thread was trying to get at this but didn't back it up with evidence or equations but rather by intuition.

***Bending***

For critical load, we use the equation

[tex]\sigma_{max} = \frac{Mc}{I}[\tex]

where [tex]\sigma_{max}[\tex] is the maximum stress in the cross-section of the rod, M is the moment (aka torque) at a location in the pipe/rod, c is the radius of the rod, and I is the area moment of inertia. Yielding occurs when [tex]\sigma_{max}[\tex] exceeds yield stress. For the same loading conditions and equal diameter, a solid rod has larger I thus it will have less stress in the beam. Thus, the critical load which causes yielding (which influences M) is higher for a solid rod.For stiffness, there are various different loading cases/configurations for a beam, but I'll use the example of a cantilever because the equations are simpler.

(stiffness) = [tex]\frac{3EI}{L^3}[\tex]

and, again, I is larger for a solid beam resulting in greater stiffness.Now for the reason that I believe this rumor is so widespread:

People are correct in stating that the centers of rods have generally less effect on strength/stiffness. If you look at the area moment of inertia equations, for a circle I is proportional to r^4. That means doubling the radius increases strength/stiffness 16 fold for the equations above which contain I. Since area moment of inertia can be analyzed with superposition, cutting out the center circle makes the new area moment of inertia proportional to (R^4-r^4) where R and r are the outer and inner diameters respectively. Of course, r has a much smaller effect than R.

What does this mean? This means that in any equation above which contains I, the strength to weight ratio of a pipe is much higher than for a rod. For any equation which uses A, cross-sectional area actually increases proportionally to weight so the strength to weight ratios are pretty much the same. Why is strength to weight ratio important? Of course, designing a building, for example, means the weight of the structures at the top of the building are going to be weighing down on the structures at the bottom of the building. In animals/plants, lighter means fewer resources are required to achieve the same strength, as well as allowing for better mobility (recall F=ma ... same muscle force garners greater acceleration if mass is decreased). The Eiffel Tower took inspiration from bone structures and showcased the ability of a mostly empty structure to support so much weight.

In conclusion, saying a pipe of equal MASS is stronger than a rod is generally true (depending on which metric of strength is being used, they may be the same). Saying a pipe of equal diameter is stronger than a rod is flat out false.

I hope I was able to clear up any misconceptions.

For reference, I'm finishing up an undergrad mechanical engineering degree and I've been taking and TAing multiple structural analysis courses every semester for multiple years. For those of you who don't trust theory/equations, yes, I've also seen these tests done in real testing machines for materials like steel, aluminum, titanium, plastic, carbon fiber, etc. and you'd be shocked how closely the results match the equations. The biggest departures from theory generally appear in heat treatment (which generally just changes the stiffness and/or yield point but does not change the fact that the material still follows linear elastic behavior), nonlinear behavior (i.e. loading past yield and some strange materials), and non-isotropic materials (i.e. wood, carbon fiber, etc). For ordinary steel, copper, or aluminum pipes/rods, this theory 100% applies.

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