Torsional Strength Equivalent Bar vs Tube

[tomcannon]

I want to replace a piece of 15mm round bar with CDS tube of a larger diameter to save weight, I am looking to use 3/4" CDS steel tube, but unsure what wall thickness is required to provide the same strength as the original 15mm diameter round steel bar. Length is 600mm, ends will be TIG welded.

I understand that there are factors which will influence the calculation, mainly grade of steel, the 15mm bar will be EN8 at best, more likely EN3B or similar.

If someone could provide me with an approximate answer it would be most appreciated.

Thanks

 

[rb1957]

for a tube shear stress = T/(2*A) where A is the enclosed area, in this case the area of the mid-thickness circle.

but is 3/4" really that different to 15mm ? i suspect you'll need to use 1" dia to get appreciable weight savings.

 

[ione]

You are looking for a tube having the same torsional shear constant as that of your round bar, that is a section which has the same max shear stress of your bar when the same torque is applied.

For a round bar the torsional shear constant is: PI*D^3/16

For a thin walled circular tube the torsional shear constant is: PI*Dm*t/2

Where:

D = bar diameter
Dm = average tube diameter (Do + Di)/2 = Do – t
t = tube thickness

 

[tomcannon]

I could use 7/8" tube instead of 3/4" but definitely no bigger, sorry I'm struggling with the formula above, I've lost myself not having done anything like this for a while...

 

[ione]

Well there’s indeed an error:

For a thin walled circular tube the torsional shear constant is: PI*Dm^2*t/2

So:

PI* D^3/16 = PI^Dm^2*t/2

Replace “Dm” with “(Do – t)”, where Do is the outer diameter of your tube and solve for “t”

 

[rb1957]

and in mine stress = T/(2*A*t) ! ... which is the same as ione's

so you have (with your 15mm bar) pi*(15/25.4)^3/16
and for your 3/4" tube ... pi*(0.75-t/2)^2*t/2

i'd simplify to start to (15/25.4)^3/16 = (0.75)^2*t/2 ...
t = (15/25.4)^3*2/16/(0.75)^2 = (15/25.4)^3*2/9 = 0.045" ... i'd use 0.05" (or a standard size) and check ...
(15/25.4)^3/16 < (0.75-0.05/2)^2*0.05/2

 

[tomcannon]

Guys firstly thanks very much for your time and help, but it hasn't quite clicked yet!

I think I'll sit down tonight when I'm home from work and try and get my head around it. I did a few years of this and struggled through it, and not used it at all for about 8 years now as done very little that requires proper calculations, so my appologies on that!

Regards,
TC

 

[rb1957]

use it or lose it !