I have an application that will be submerged in seawater and is made of EN24 T. Originally the part was not plated and during surface, it fell foul to corrosion, but worked.
We have made a new component (the material has had to stay the same), and have decided to have it Electroless Nickel...
Dear All,
I am looking at a project in where we will be using EN24 T (BS970 817M40) material in a seawater environment, and where it will be artic conditions. The material will be tested in accordance with the temperatures i.e -10 degrees c.
With that in mind, to protect this material against...
I am wondering if you could help.
I have a component (Female Cover) that has a parallel OD and a Taper Bore. The Taper Bore inner surface is subject to a constant pressure due to a component (Rod) that slides inside the bore, with a corresponding male taper (similar to a morse taper). The whole...
Again, thanks for your help.
The problem that I am looking at states that the Bolt must be 40 diameter and have a preload of 70% of the material yield, hence that is where I got the 420 kN pre load from. To achieve this, then I must apply a larger load/stretch to achieve this retained loading...
Cheers guys for the help.
What I have trouble getting my head round is that the joint/bolt must have a preload of 70% of it's yield. However, to create that clamp force (taking losses into consideration), the applied load takes it higher than the materials yield - therefore failure.
If I had...
Cheers guys, and the help is much appreciated.
I can fulled understand that say a 100Kg load with a mu=0.25 would take 25 Kg of force to move it, but if mu=1 then it would take 100Kg of force to move something on a more rougher surface - makes sense.
Its been a confusion that an application...
Cheers guys.
Desertfox, if the Shear Force was 500 kN and the total Clamping Load from the Bolts was 300 kN i.e 100 kN per Bolt.
Take CoF 0.5
I am confused, please bear with me. The way I saw it...
The shear force on the plates with friction from the plates would be 500 kN × 0.5 = 250 kN
As the...
Just a quick question, and grateful for any guidance.
I have an simple application that has 3 Bolts all have the same diameter, d, and has a slight clearance with the hole, D. They are responsible for clamping 2 plates together, or it could be 3 dependent on design.
They are tensioned, and a...
I will submit a sketch and calcs later. Thanks.
I calculated that the centrifugal force could be greater that the frictional force on the nut face, hence the slight possibilty of bending.
I have an application that has 12 bolt in a rotational flange that has speeds up to 3000 rpm.
As the nuts are external, and have a relative weight to them, I have calculated the total bending stress in one bolt as around 500 N/mm2 - this is subject to a centrifugal force if the nuts move due to...
RB1957
Thank you for your help.
I take it by if you really mean only one force at a time then there's no interaction then the vector doesn't apply - only if say Fx and FY were at the same time?
The Forces only act at anyone time, not in combination.
Hypothetically, what would happen if the...
sorry Fz shouldn't be a negative.
Only one force will acting at anyone time.
Rb1957,I'm sorry if I don't fully understand the way to combine the two forces, as I have said, I'm trying to educate myself along the way.
Please enlighten me where I have gone wrong with my figures as I thought I...
Cheers Fox and rb1957 for your help. I am learning very quickly thanks to your help.
Fox – the holes are spaced like that due to a cylindrical housing that will be in the middle of the plate. My task was to ignore that housing.
The working loads are a lot less than this, these loads only apply...
Hi Fox, and thanks.
The plates are 140mm thick each one.
I take it that to combine the Shear and the Tensile Stress you add the two results together.
Shear Force Through Bolt + Tensile Force = 414 N/mm2 + 588 N/mm2 = 1002 N/mm2 - Is that correct? This is over the Yield but just below the UTS...
