I need to reference this in a specification I'm creating, but I don't know how to without recreating it and incorporating my recreation. Ideally I'd like to reference the specific standard and specify the scope and level required.
p.s. English or Chinese (not Japanese).
I’m trying to identify the standard that this “Defect Comparator” is associated with. I want to refence it in a speciation. Any advice would be much appreciated.
Hi all,
I have a UL certified product that contains plastic parts (PA6). To attain citification all the relevant tests were done by UL including verifying the plastic material supplier was one from their list of certified suppliers.
We wish to change the plastic supplier. UL are insisting that...
Thanks all,
The plan was always to use belleville washers and a high friction washer of unknown material. Perhaps my error is proposing a high friction material.
A similar mechanism has a thumb screw and steel washer that never works well. Unable to set the pressure accurately and needs a...
The material would be clamped between a steel washer and the sliding face with a nut. More tension on the nut would result in more friction. I don’t want the material to crush or disintegrate over a short time. It will only slide 20 -50mm once a week so long term wear isn’t an issue. Positional...
I’m looking for a material to give me high coefficient of friction (against steel) and good structural integrity in a 20mm o’d ish washer. The washers will be clamped against the steel offering functionality like car brake pads on a disk. But cheap, simple and small.
Any advice the best...
I couldn’t imagine how I could loose 8mm out of tolerance in a final bend. The part is just too small. Your idea with a screws wouldn’t work because there’s no free space bellow the 3.15 flanges. I have done some tests and I think I can accept a tol as high as +0/-15mm on the 60.54dim. and it...
@drawoh
This has to sit over a curved face, like a ring clip on a hose. The rad could be made with multiple flat faces (I guess 10). But I imagine that wouldn’t help (lots of bends with a tolerance on each bend to deal with).
@mfgenggear
I'm confused. I thought you had to 1, Form the shape, 2, Anneal it to release stresses, 3, "harden" heat treat the part to make it springy.
If 65Mn spring steel is already hardened, how do they form it without it springing back to its original shape?
Why is the factory telling me...
Thanks for your help,
These are being cold formed and then ht (as above) to make them "springy".
Surely they must form them before ht otherwise they would not take form???
They can vary in either direction but normally the gap is opening but without any consistency.
The supplier tells me they...
Hi, I hope someone can help me. I’m having problems with our manufacturing partners not being able to control the final dimensions of a spring clip.
The clip is as shown in the image below. The problem is they can’t maintain the 60.54 dimension with in a tolerance of +0/-5mm. After forming the...
Medium carbon steel with a tensile strength of 800n/mm^2
Different diameters of the steel would support different torques.
What torque could I expect a 5mm dia Bar to support before any permanent deformation.
Then I’d like to be able to calculate with different tensile strengths and bar...
I wish I knew. I need help.
If I was to measure the yield torque and yield tensile load on a range are bars with differing tensile yield strengths I expect to find a correlation between tensile load and torque measured. How do I calculate this correlation?
Yes.
The problem is I’m comparing apples to pears.
I’m trying to calculate the torque that can be expected based on a specific tensile load.
Or if I have a Bar with a yield strength of “S”n/mm^2 what torque would I expect to be able to apply before yield.
So can you help me?
Racookpe1978,
pi*r^3/4 = (pi*(r^3))/4 and not ((pi*r)^3)/4. If I was to write (pi*(r^3))/4 on paper I would not use the brackets or the times. If r=2.5 then pi*r^3/4 =12.27.
I’m trying to calculate the max torque a bar can be expected to support based on its tensile strength.
Is the following correct:-
Radius of Bar (mm)= r
Moment of inertia =I=pi*r^4/4
Section modulus =z= I/r= (pi*r^4/4)/r= pi*r^3/4
Torque (Nmm)= T
Tensile Strength (N/mm^2)=S
T=S*z
I have searched the web looking for the UNS or ASTM equivilent of BS1490 LM24 & LM6 and can't seam to find an answer.
I think LM24 is the equivalent of UNS A03800 (A380.0-F) but the mechanical properties I find on LM24 do not seam to match A380.0-F on matweb.com.
And perhaps LM6 is equivalent...