## Engineering workshop #4: Bending Stresses

20 May 2010

In his latest installment, Bob Johnson, Technical Director of consultancy and training company DAMT, moves into the realms of Finite Element Analysis (FEA)

In April’s article, we examined direct stress and shear stress – the two types of stress that exist when we consider stresses in a Cartesian coordinate system. We recognised that the notion of shear was necessary in order to allow the all-important direct stresses to act at any given angle (and not just parallel to the X, Y or Z directions). We will take this idea further when we look at principal

stresses so beware!

This month we will perform hand calculations and Finite Element Analysis (FEA) on a classic karabiner. The karabiner is used by climbers as a rope-joining mechanism – see {fig.1}. With the gate closed, the karabiner is rated at 21kN end-to-end and 10kN acrossgate. With the gate open (not to be recommended!) the end-to-end rating drops to 7kN.

The first thing to recognise with the karabiner closedgate is that the load path is not unique. In other words, when we apply an end load to this device, some of the load will travel through the forged aluminium “C” section and some of the load will pass through the gate. We could estimate that the load split is 50/50 but we don’t know – a plastic gate for instance may attract much less load than

50%. Thus, with the gate in the closed position, we have what is termed an “INDETERMINATE” structure (so-called because we can’t readily determine where the load will go).

When the gate is open we now have a unique load path through the main forged “C” section and the structure becomes “DETERMINATE” (we can determine where the load will go). In our everyday engineering we should be aware of the issue of determinacy – determinate structures can be addressed with hand calculations while indeterminate structures generally can’t. Indeterminate structures may be much more susceptible to manufacturing limits and fits but we’ll leave it there for now.

{fig.2} shows the karabiner with the gate removed (so that we can perform hand calculations on the determinate “C” section). Given that we would like to calculate the stresses along the straight portion of the “C” section then a cut has been made in order that free-body analysis can determine the internal forces at the cut.

It can be seen that the quadrant is in force and moment equilibrium when a direct force of 7,000N and a bending moment of (7,000 x 19.5) Nmm is applied to the cut section. This free-body diagram instructs us to carry out a direct stress calculation (carried out in green text) and a bending moment calculation (carried out in red text).

The bending stress (at +/- 1045MPa) is far greater than the direct stress (at +74MPa) and we should watch out for bending as a major source of stress from now on. BEWARE!

While the aforementioned direct stress calculation was very straightforward (force/area) the bending moment calculation was slightly more involved and we should now examine what was done in that regard. The bending stress induced when (say) a bar of steel is subjected to a bending moment, M, is not a straightforward function of the area of the bar.

The bending stress is proportional to the moment, M, the distance from the centre, y, and the second moment of area, I. Dealing with y first then this is the linear distance from the centroid of the section (the centre-of-gravity if you made a plywood replica of your section) to the point where you want to determine the stress (usually the top or bottom of the section).

The second moment of area is derived by considering the section as an infinite number of thin “leaves” (like the core of a transformer) and calculating the area of each “leaf” and multiplying that by the offset distance squared. In other words, the first moment of area is area x distance and the second moment of area is the area x distance x distance (area x distance-squared). For a rectangle (breadth, b x depth, d) the second moment of area is bd3/12 and that for a solid circle is ðD4/64. Check the bending stress calculation on {fig.2}.

Staying with {fig.2} we should be able to appreciate that the direct stress and the bending stress are defined in the same direction (parallel to the global X direction).

This allows us to add the two stresses together giving a resultant stress at the inside of the “C” as 1,119MPa (tension) and a resultant stress of minus 971MPa (compression) at the outside. Check that you agree.

Finally I have carried out Finite Element Analysis of the open gate karabiner (subject to 7kN loading) and the predicted stresses away from the end of the radius (1,120MPa) show excellent agreement with our hand calculations (see { fig.3}). What is happening where the straight part of the “C” joins the curved part of the “C”?

We have demonstrated the danger of bending – as soon as you see two forces incorporating an offset (see {fig.2}) then you know that you have a bending problem. Always re-check your calculations to see that you haven’t forgotten a bending problem. More next time…

R P Johnson BSc MSc NRA MIMechE CEng – Technical Director, DAMT Limited

Stress analysis of karabiner carried out using the ROSHAZ program.

### Comments on this article:

I’m working on a FEA of a carabiner for a class, so I imagine I’ve taken a closer look at this article than many. I think there’s a unit error in your calculations. If I’m not mistaken, in fig.2 you’ve got (N*mm*mm)/(mm^4) = N/mm^2, not N/m^2 = Pa. This would make your bending stress 10.45MPa instead of 1045MPa. Makes me a little curious about the FEA values, are they still accurate?

Cheers.

**Posted by** Thomas on Thursday 04 2010 at 05:14 AM

To anyone wondering about Thomas’s question:

1 N/m^2 is 1 Pa, however the units are in mm.

1000 mm in a m, 1000^2 =10^6

10^6*1 Pa = 1 MPa

N/mm^2 is 1 MPa

You can confirm the unit calculation here as well:

http://www.wolframalpha.com/input/?i=N*mm*mm/mm^4

**Posted by** Ash Smith on Thursday 09 2012 at 11:51 AM

I need design calculations for carabiner. Can you provide me Sir ?

**Posted by** Sushant on Thursday 24 2016 at 01:08 PM

A couple of questions about this example. I do buy the reasoning and explanation behind the mechanics here, specifically that the bending stress state is far more severe and influences the cross sectional design of the carabiner’s spine. (Most lightweight ones have Tee section)

However, as a climber who owns and handles these devices regularly there are a couple of problems for me. When the gate is in the closed position the structure is *not* indeterminate since there is clearance (at least 1-2mm on the majority of top name brand ones I own) between the hook and the pin on the gate. This means that the carabiner body needs to deform *significantly* before the pin would engage in the hook. Next, the gate itself is a pretty flimsy piece, and I’m sure the 2mm pin could be pulled out (in shear) pretty easily. (Now that would be a useful simulation to have also run!)

I’m pretty sure that the design of the gate is to ensure that the rope does not inadvertently shift and detach from the body. In fact the strength of the carabiner is maximized only if the loads are applied in the ideal location (as in your diagram). The geometry of the carabiner is such that the rope mainly stays nestled in the correct orientation.

All in all I think even this simple part demonstrates the difficulty we still have in using CAE tools to simulate simple, real world problems. There are many complications at work here (the large displacement, the contact with the pin, the changing directions of the loads)

Posted byGreg Brown on Tuesday 10 2010 at 06:16 PM