I wanted to calculate the shear strength of a screw in an assembly of which you will find the diagram attached. We apply a force of 250 Kg on the Ø140mm aluminium tube. The 168mm aluminium tube is fixed to the ground, so it is considered fixed. the screw for the assembly is an M10x40 Stainless Steel A70 screw that is tightened in a drawn steel with an 18x10 section. Do you have a calculation solution to check the one I made. And is it possible to simulate it under SW simulation.
It all depends on whether you are talking about the resistance of the bolt to shear or the calculation of the shear plane which depends on the number of bolts as seen in steel frame assembly
He, professor, will tell you better than me: but a screw should NEVER work in shear like a pin does (even if it is not uncommon to see screws working in shear ;-(( ... )
In your assembly, if I understand it correctly, your M10 screw will only have 6 threads engaged in the 10 thickness stretch, but for a 10 screw you need twice the diameter, i.e. 20 mm. This means that the number is 50% lower In addition, you are in the case of an aluminum-steel assembly where it is rather the aluminum that will deform a little more than the steel.
I'll spare you the calculation of the friction resulting from a tightening but in a pragmatic way it means that you need in theory 3 screws (at a vista) I hope your stretched is made of stainless steel too: otherwise the oxidation you seem to want to avoid will not be guaranteed if you have three different metals, one of which would rust
How to prove it with Solidworks simulation? By using the "bolt connector" between the two pieces and applying the force vertically to the top end of the inner tube. Don't forget a safety coeff of 2.5 for this type of assembly. You can have the CS from the simulation directly in solidworks.
Small precaution for the simulation , you must not have screws in the holes, it is the bolt connectors that will replace them.
For my part @Zozo_mpI've always heard that 2.5x the pitch is enough as a catch thread and ideally based on the height of a nut of the same section: here M10-> 8.4mm so the 10mm stretch is enough for me, unless something escapes me.
I also have a spit document with the resistance of an M10 threaded rod (Steel) to 1610daN shear for information. So even in stainless steel for me you have margin. Do the math with the table on the stainless steel but can doubt the result.
On the other hand, for the resistance of aluminum, the calculation is different.
Oops! You're right! the length of threads in gear is equal to the diameter for hard steel, 1.5 for cast iron and copper alloys and 2 x the Ø for aluminum and its alloys. The stainless steel screw is a little stronger than class 8.8 of steel.
What safety coeff do you think you have with a single screw with a fairly substantial clearance between the two male and female tubes with vibrations so outdoors.
That said, given the assembly, it is not specified if it is a single tapped hole or two. We can think of only one if the stretched serves as a bolt that is otherwise inaccessible.
For the simulation , not really necessary for a classic case of single-hole screwing , my suggestion of the bolt connector remains valid.
Kind regards
PS: we didn't pay a lot but we're having a good time ;-)
You're right, especially since I didn't notice it, but as drawn, you don't see how the stretched could fit between the two tubes, except machining by milling on the inner tube.
Something escapes me because the stretched should be inside the small tube or there is an extra rib in the small tube but the dimensions of the first drawing don't seem good to me. You should have a view of the end of the two profiles and see where the sliding stretch is placed in the inner rib of the small tube.
Thanks for your help! You will find attached a cross-sectional view as requested. If I could avoid a simulation and get a result by calculation, it would suit me cisailement_vis_m10_-_2.png
For the raw calculation, the answer of @sbadenisis quite appropriate.
On the other hand , @stefbenonote that the grip calculation would be appropriate.
Personally I would be quite inclined to either put two screws: one at the top of the stretch and one at the bottom. Because as you have a copious play between the two tubes, you will have leverage effects and vibrations that you will only be able to see by the simulation on your stretch. The risk is not great, but it all depends on the stresses to which the two tubes are subjected (absolute static or subjected to any movements and to what frequencies)
Unless there is a notion of adjustment (is it an adjustable system?), with such a setup, I would take a smaller coaxial screw with a mechanical pin (so Ø10 pin and M6 screw).
Rough calculation of the average shear stress of the screw: Tau = Faxial / Section = 45 MPa The elastic limit of A2-70 stainless steel being about 450 MPa, there is no need to worry about that.
If a sufficient pressure force is desired to prevent slippage by adhesion between the two tubes, a normal force equal to N = Faxial / f = 20.8 kN must be exerted by tightening the screw, if f = 0.12 is taken. Let be a normal stress in the screw of about Sigma = Faxial / S = 370 MPa. The safety margin is not very large...
You can also think about flipping the layout of the assembly. We thus have two oblique contact zones between the tubes, and by a quick estimate of the "wedge effect", an almost doubling of the contact force (N1 and N2 of the same order as Fs). For the same clamping force Fs, the slip safety margin is multiplied by 2. In this configuration, it would also be necessary to check by simulation the stresses and deformations in the left part of the assembly where the walls of the tubes will be more stressed than in the first arrangement. A more detailed analysis is only possible if you have the geometry of the tubes, or the manufacturer and references...
As the stretched bolt serves as a blind bolt (bolt not accessible and not positionable at height), a second screw at the bottom close to the ground would cost almost nothing either
