I – TOPIC AND RANGE OF application
This instruction describes the procedure and equipment for the bending test of valve 470 to our suppliers and external service providers.
II – HISTORY OF MODIFICATIONS
|DATE||INDICE||DETAILS OF MODIFICATIONS||PAGES|
III – Plan
- Equipement for the implementation
- Test process
IV – ANNEXES
|DATE CREATION /MODIF.||23/02/2015|
|CREATE OR MODIF. BY||R. Dubroca||RD|
|CHECKED BY||M. Fillioux||MF|
|APPROVED BY||M. Vevaud||MV|
- Equipment for the implementation
Machine for bending test:
Ball valve support:
- Test process
- Assemble the valve support
2) Set the valve support on the tensile machine.
3) Screw the suitable insert (according to the dimension) on the valve support.
4) Screw the valve on the insert. Use the appropriate torque as indicated in the table below. The valve’s bonnet must be on the support side (see below).
|Screwing Torque (N.m)||21||30||51||75||96|
5) Screw the suitable flexion tube to the valve.
- Install the sensor of the tensile machine, then place it in contact with the flexion tube.
7) Apply a force on the flexion tube. Increase until the valve brake and record the value.
*For the 1″ and 1″1/4 valves, don’t go until the valve brake, but stop the test when a torque of 420 N.m is reached.
The force has to be increased with a speed of S = 50 mm/min.
To be congruent, the valve must pass the followings acceptations points:
|Acceptation point (N.m)||80||110||230||340||–|
|Surveillance zone (N.m)||Between 80 – 95||Between 110 – 130||Between 230 – 250||Between 340 et 380||–|
A surveillance zone has been defined (see the table above) in order to prevent any drift of the threads resistance.
If the valves are in this zone, the supplier will be notified in order to put in place all necessary corrective actions to leave this zone for the next tests.
See illustration below:
S = 50 mm/min
10) Calculation of the torque
To find the value of breakage in N.m: T (Torque) = F (Force in Newton) x L (Length in meter between the application point, meaning the axis of the tools, and the breakage point)
If the force at break was 1240 N, and the distance between the application point and the breakage point is 0,13 meter, then the Torque: (T) N.m = F x L :
= 1240 X 0,13 = 161,2 N.m