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Is that true, and why?

Thanks.

- Thread starter Nicholas
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- #1

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Is that true, and why?

Thanks.

- #2

jamesrc

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First thing: Both pipes will bend since neither pipe will have infinite stiffness; under a given load, both pipes will bend some finite amount.

Second thing: The character of the bending depends on the materials used and the geomoetry of the pipes. So it is possible to design a hollow pipe that is stiffer than some solid pipe. However, since you didn't state specifics, I have to assume that both pipes are made of the same material and both pipes have the same outer diameter, d. (I'm also assuming they're round.)

Now we can look more closely at the problem:

The bending stress in the beam (pipe) is given by the following:

[tex] \sigma(x) = \frac{M(x)c}{I} [/tex]

where M(x) is the external moment as a function of x (same for both since loading conditions are the same), c is the distance from the neutral axis (will equal d/2 in this problem when finding the maximum bending stress) and I is the area moment of inertia (aka second moment of area) which is a function of the cross section geometry.

(If you look at the Euler beam bending formula, you will find the same dependence of deflection on moment of inertia, which is the key to this problem.)

The moment of inertia of the solid pipe:

[tex] I_s = \frac{\pi d^4} {64}[/tex]

The moment of inertia of the hollow pipe (same outer diameter, d, along with inner diameter of d

[tex] I_h = \frac{\pi \left(d^4 - d_i^4\right)}{64} [/tex]

Since the amount of bending is inversely proportional to the moment of inertia, the pipe with the lower moment of inertia will bend more. Looking at the equations you should see that the hollow pipe will bend more.

- #3

jamesrc

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I just realized you may have meant that the loading was solely due to the weight of the pipe. If so, consider the expression for the maximum deflection in a beam under a uniform load (you can get these too if you work out the Euler beam bending equation):

if it's cantilevered:

[tex] \delta_{max} = \frac{qL^4}{8EI} [/tex]

if it's simply supported at both ends (max deflection occurs at center of beam here):

[tex] \delta_{max} = \frac{5qL^4}{384EI} [/tex]

(If you are unfamiliar with the definitions of any of these terms, please ask.) In either case (and assuming both beams are made of the same material (E)), the maximum deflection is proportional to the term q/I.

For the solid beam:

[tex] \frac q I = \frac{\frac{\rho L \pi d^2 g}{4L} }{\frac{\pi d^4}{64}} = \frac{16\rho g}{d^2} [/tex]

For the hollow beam:

[tex] \frac q I = \frac{\frac{\rho L \pi \left(d^2-d^2_i\right) g}{4L} }{\frac{\pi \left(d^4-d^4_i\right)}{64}} = \frac{16\rho g}{d^2+d_i^2} [/tex]

Assuming I haven't messed up my math here, the bending for the hollow beam is less than the bending for the solid beam. The solid beam is still more stiff, but it also undergoes more loading (because the solid beam weighs more). I hope that helps.

- #4

enigma

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Between a solid pipe and a hollow pipe

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From the equations already presented above, the maximum bending moment is proportional to:

Between a solid pipe and a hollow pipeof the same weight, the hollow pipe will be more resistant to bending. It will also have a much larger radius.

[tex]d^4 - d_{i}^4[/tex]

The mass of the pipe is proportional to the cross-sectional area which is proportional to difference in the squares of the outer and inner radius (or diameter since our constant [tex]K[/tex] is arbitrary), so we can set that equal to constant and eliminate either the outer or inner diameter from our equation:

[tex]d^2 - d_{i}^2 = K[/tex]

[tex]d_{i}^2 = d^2 - K[/tex]

[tex]d_{i}^4 = d^4 - 2Kd^2 + K^2[/tex]

[tex]d^4 - d_{i}^4 = 2Kd^2 - K^2[/tex]

Thus for a constant mass, we see the bending moment of a pipe is quadratically proportional to the outer diameter:

[tex]2Kd^2 - K^2[/tex]

Not accounting for other factors such as localized shear stress, compression load if employed as a vertical structural beam, etc, the larger the diameter hollow pipe, will have a quadratically higher maximum bending moment than a small diameter hollow pipe of the same mass.

This moment property is conceptually related to the physics of the ratio of lengths of fulcrum or ratio of length to height of a truss, visualize the diameter of the pipe as the short side of the fulcrum, where forces are more concentrated the shorter the short side relative to the long side.

A solid pipe of equivalent constant mass, would be limited to an outer diameter of [tex]\sqrt{K}[/tex], because [tex]d_{i} = 0[/tex].

If I have misstated, please free to post a correction. Hope my input is helpful.

===============

Example: compare a Schedule 40 2.5" pipe, which has an outside diameter of 2.875", inside diameter of 2.469", and wall thickness of 0.203", to a Schedule 10 3.5" pipe, which has an outside diameter of 4.0", inside diameter of 3.760", and wall thickness of 0.120". The 4" pipe has 14% less mass, but comparing [tex]d^4 - d_{i}^4[/tex] has 80% higher maximum bending moment before elasticity stress failure.

Another example with maximum bending moment calculation for GI pipe mast (30 Ksi yield strength):

http://en.wikipedia.org/wiki/Second_moment_of_area#Circular_cross_section

20' x 4" Sched 5 (1/12" wall), 15T load, 32 kg, 28T@100" bending

20' x 4" Sched 5 (1/12" wall), 28T load, 67 kg, 28T@100" bending

(3 Ksi concrete filled with 3" hollow or pvc pipe center)

20' x 2.875" Sched 10 (1/8" wall), 15T load, 32 kg, 14T@100" bending

20' x 2.875" Sched 5 (1/12" wall), 11T load, 23 kg, 10T@100" bending

T is tons

Last edited:

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- #7

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high bending stresses are generally a bad occurance. what you should relate a high bending stress to is the safety factor, always work about 2 - 4 safety factor in machinery design. safety factor is the (yield stress of material/bending stress)

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