Max Bending Moment of a Pipe

Using torsion isn't going to work, as you are bending, not twisting the pipe. Torque would be applicable to the head of the frozen bolt.

I think you should try a Bending of a Cantilever Beam equation.

Can you post a sketch of the set-up? It could save a lot of false leads by people misunderstanding the problem.

Ok, so I dont know if this image is clear enough however for you. However all measurements are in mm.

The length is 180 mm

Outer diameter = 15 mm / Raduis 7.5 mm

Inner Diameter = 5 mm / Radius 2.5 mm

and the applied force = 41.8 N/m

Like I said I am finding it difficult to discover whether or not the thickness of this pipe is strong enough to withstand the force 41.8 N/m or to put it another way what is the pipes max bending moment or shear stress.

I hope this helps you guy, and thanks for responding.

Samurai 7

" ... and the applied force = 41.8 N/m "

Sorry to be picky, but N/m isn't a unit of force (it's actually the unit of a spring constant). Do you mean 41.8N ?

Yup. Definitely a cantilevered beam. Try this calculator. If the answer comes out greater than the yield strength of the stainless steel alloy you are using, the tube will fail.

Thanks for your help Thermopylae, but how do I know from the results if the answer is lesser or greater than the yield strength? The result I have for stress at specific point (right at the fixed end) is 22.57 N/m^2. So Im guessing that as long as this particular thickness of stainless steel can withstand this level of force then the pipe wont bend or deform??? The yield strength of this particular stainless steel (304L) is 290 MPa or 40 ksi. So how do I use the answers to compare this? Thanks once again!

Bending is OK. It's when the yield is exceeded that the problem arises. Deformation to a point is fine, but if you exceed the yield, the deformation becomes permanent. Not a good thing. You need to select a pipe that can handle no more than 60% of yield.

Thanks ronseto, I was thinking of duplex or AISI 302 stainless steel as they both have a yield strength of 520MPa! However I still do not know which formulas to use to find these results and given the calculator the Thermopylae provided, I do not know how to decipher the results of the figures I inputed! Do you have any sugggestions? Regards, Samurai7.

τUse these formulas:

Max bending stress (at support) due to bending σ = W·l/Z

Where W = FORCE (in your case N - NOT N/m)

l = length from support to load

Z = section modulus of pipe = (pi/32)(D⁴ - d⁴)/D

where D = outside diameter of pipe

d = inside diameter of pipe.

Max Shear stress τ = 2·V/A

where V = shear force (in N - same as above W)

A = cross-sectional area of pipe

ENSURE CONSISTENT UNITS

Make sure above stresses (multiplied by safety factor) are below the yield strength and shear strength of your material.

Thanks again for all your help, Especially the last Guest input...except that is for one thing: If the applied force on the pipe is stated as 41.8 N/m, how then do i put this figure in to the equation if I have to use Newtons as a unit of force? Is there a conversion or is it just 41.8 N? Sorry to drag this out but believe me I have searched the entire web before continuing this discussion on and on! Samurai7

I've said it before, and I'll say it again: N/m isn't a unit of force !

"If the applied force on the pipe is stated as 41.8 N/m"

Something is amiss - who "stated" this? Where does this value come from? Because with those above equations you need a specific point load force of N.

The only thing I can think of, other than a typo or straight out error of units, is that this "force" of 41.8 N/m is specified as a uniform load and not a point load. But this doesn't seem appropriate for a wrench calculation.

To reiterate - Who exactly "stated" this 41.8 N/m "force"?

I think that the wrench has to apply a TORQUE of 41.8 Nm. So that the force will be 41.8/0.18= 232 N.

In fact the above mentioned torque is the bending load on the critical section.

The bending stress will be 128 N/mm² which has to be compared to the yield strength of material.

It is possible to increase the central bore and reduce the weight.

Thank you very much nickname, that was just what I was searching for! After going back to the aircraft publications at work I did indeed find that the applied force was 41.8 Nm and as such I have now worked my equations and I will be using AISI 302 cold rolled or Duplex stainless steel, as they both around 520 MPa yield and an ultimate strength of 860 MPa which is way above the safety standard. My results showed that the yield necessary need to be more than 128 MPa (128 N mm^2) and a shear force resilience of more than 66.8 N/mm^2 to withstand the applied force 41.8 Nm. I apologize to all for my error with regards to Nm and N/m and I would like to thank you all for attempting to help me regardless. Samurai7.

Careful now, please use the correct nomenclature / terms / definitions (or it will cause difficulties - as you have seen) -

Without doing the calculations to make sure your numbers are correct:

Make sure that with the above equations you use a force of 232 N

DO NOT USE THOSE EQUATIONS WITH A TORQUE OF 41.8 Nm

TORQUE (Nm) AND FORCE (N) ARE COMPLETELY DIFFERENT QUANTITIES

Now that I showed you the errors it would be good for you to list your computations in order to get all indications for the correct version and see where you made errors and which kind.

Thanks again, Yes I did use 232 N in both the Yield and shear computations as you stated, which worked out fine as the pipe has been measured in mm. Thanks again for all your help, Samurai7