Torsional stiffness of a square section

[izax1]

This should be simple enough, but I am struggling with finding the torsional stiffness (polar moment of Inertia)for a closed square channel with different thicknesses on top and bottom. A little background: We have a beam with square closed section, but need too design for higher loads (bending and twisting of beam) and to accomplish the higher stiffness required, we plan to weld on a flat beam on top of the beam section. (thus different thickness on top and bottom of the section) I am concerned about the stresses in the welding. Thankful for any hints in the right direction?

 

[Engdoitbetter]

Hi izax1,

I was writing a more complex message regarding basic theory of torsion of rectangular section, but then I realised that for closed sections you can use only the formula of Bredt for shear stress:

tau = T/(2*s*A) where T is the torsional moment, s the thickness of the section and A the area enclosed by the mean line of the section profile. s is the local thickness so tau varies according to it along the profile.

I suggest you to look for it on your own so that you better understand it.

Anyway, as first approximation I would say that the weld is loaded by the sum of the shear stresses on the added thickness.

Hope it helps.

Stefano

 

[rb1957]

note the welds (along the length of the reinforcing straps) see shear along their length and across (tension ?) as the weld is the only loadpath into the strap (draw a FBD of the strap)

 

[drawoh]

izax1,

Torsional stiffness of all sorts of weird sections, is covered somewhere in Roark's Equations for Stress and Strain.

--
JHG

 

[Engdoitbetter]

I have to amend my previous post, the weld is loaded by the shear stress calculated in the cross section of the weld with thw formula above i.e. the cross-section behaves as if it were continuous

 

[izax1]

Thanks for your responses. I have looked through Roark&Young and I can only find Torsional stiffness for regular or close to regular sections. The closst I can find is a hollow rectangle with t1 in the upper and lower ealls and t on the sides.

Engdoitbetter: Wouldnt the max stress in the section be midways between the corners in the thinnest wall? The beam is subject to both bending and torsion (load applied above the beam and at the mid length and at an angle to vertical).

The max stress for bending and torsion is not appearing in the same location. Any experience with superimposing bending (max. axial stresse in tension or compression) and torsional stresses? The reaon why I am so picky here, is that we are uitilising the beam to max, and I have to do be sure that we can use that beam. If we cannot, costs will rocket.

 

[rb1957]

shear stress due to torque will be constant on each face (changes with thickness changes only).

bending stress will be constant on top and bttm faces.

torsion shear stress probably lowest on the top and bttm faces (which are thicker), where bending stresses are highest. i think these'll be your critical locations, though you can look at the sides as well. maybe the sides close to the top/bttm faces ... where the torsion will be higher and the bending lower.

combine the bending normal stress and the torsion shear stress as a principal stress.

 

[GregLocock]

The torsional constant in thin walled, non reentrant, non circular, sections with varying wall thicknesses is not correctly represented by a simple formula. Hopefully a through reading of Roark or Timoshenko will make that clear.

Your get out of jail card is that a rectangular box of sides a and b, and thickness t1 and t2, can be considered, in torsion, to be 4 narrow beams in bending, two each of sizes a by t1 and b by t2, each being bent in its deep direction.

Alternatively you could use finite heffalump.

The torsional behaviour of weirdo sections is fundamental to modern car body design, you could probably find something relevant at the SAE.



Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[Engdoitbetter]

izax1,

if torsional stress is going ti be constant on the section, apart from thickness variation, I believe that the critical point is where the bending stress is max, as rb1957 pointed out.

Anyway, you should calculate the general expression for ideal stress (for example von mises = sqrt(sigma^2+3*tau^2) for beams) in every location and then find its maximum. Obviously,critical point changes according to the failure criteria you employ.

Furthermore, for torsional stiffness, look again for second Bredt formula, which states it for thin-walled hollow section.

 

[izax1]

Thanks for all your very useful and helpful answers.
Engdoitbetter: If you really go down to the nitty-gritty stuff, the method of combining the bending (Normal) stress and Torsion (shear)stress can be discussed. The normal stress from bending and the shear stress for torsion will not be in tha same locations (bending on the outer (top) surface and shear (probably) in the corner Stress Concentration at the inner suface) so calculating an equiv stress with max bending stress and max shear will result in conservative values.
Yes, I know I am streching it now, but that might be exactly what we need to do. And, yes, I might need to turn to FE, but that is kind of lost battle for an originally (at least that's what I thought) simple problem.

 

[Engdoitbetter]

izax1,

I'm not sure if I understood your post properly (I'm not English Native speaker )
Anyway, it always depends on how much effort you want to put into it, so you can:

- calculate ideal stress as a function on the section and then find its maximum etc etc OR
- sum simply max shear stress and max bending stress OR
etc etc.

So my point is not to criticize any of these methods or Engineers who use them (far from my thoughts!), but that we should always aware of what is left out when doing an approximation. So, if FE analysis is worth the money, then do it

Regards.

Stefano

 

[amorrison]

Re: Stresses in/from welding

Consider a larger starting original piece and then milling the side thicknesses as required.

No welding = no welding stresses.

 

[rb1957]

i'm lost ... ok, you know that the peak bending stress occurs somewhere on the section and the peak torsion occurrs somewhere else (i do think you're splitting hairs, but split away).

so, what's stopping you calculating the bending stress at the torsion peak, and the torsion stress at the bending peak, and combining these (as a principal stress) ?

if you Really want to split the hair, you could notice how each stress component (i believe they're linear over the thickness) is changing and so determine the peak principal stress.

i think this overlooks a key issue with your section, how does the load get into the reinforcing (welded on) strap / you've done a lot of analysis by now assuming the strap is fully effective, is it ? (i don't think so)

 

[izax1]

Yes, I know Im splitting hairs, and I do that to save alot of hazzle and money. What Im trying to get at here, is that I can gain some if my assumption is correct, that the max bending and the max shear stress is not at the same location in the section. (See Peterson SCF third edition Chart 5.22) And I hope Im not the only one here to know that! Or have I missed something? If that is the case, I can do as rb1957 suggests, to calculalte bending at max. torsion location and torsion at max bending location. And when I have sorted this out, I will certainly look at the weldings. But I have to be sure that I have the stresses sorted out correctly, before I look at the weldings.
I would really appreciate any opinions of my assumptions.

 

[chicopee]

Give us a sketch of the beam showing the reinforcement and its attaching welds. The sketch should show the longitudinal and transverse views as well as the external forces and supports.

 

[rb1957]

maximum bending stress is at the extreme fiber.

maxium tosrion shear stress is at mid-thickness. ok, you might be looking at peterson, showing you a torsion stress concentraion in the corner of the tube. if you're looking at an ultimate design case (as opposed to a fatigue load) then i'd expect you've gone plastic in the corners, but you should be ok away from the concentration. of course, you've got a problem if you've gone plastic through the thickness, but if most of the section is elastic then that should be ok.

funny how we're no longer talking about the subject of the thread !

 

[Engdoitbetter]

I agree that a sketch might be useful...

 

[rb1957]

looking at peterson (2nd ed,chart 5.22) he's gt Kt fr the corner ... for a constant thickness tube ... not the same section as your's ... different enough (IMHO).

looking at the chart, i'd probably go to the reference to see how they're calculating the reference shear stress (what is the torsion stress on an angle ?)
in your case i'depect the rerece stres to be the previous expression, = T/(2*A*t).

 

[Adrian016]

You could pull out the cross sectional properties of the beam using CAD software. Iterate through different sizes to get values for different heights, widths, and heights.

 

[rb1957]

"i'depect the rerece stres to be" = i'd expect the reference stress to be ... sigh