How to choose the test force

Simple algorithm:
1000(mpa)*314(mm2)=314000(N)
314000N=314KN
314KN is equivalent to 31.4 tons of experimental force

The unit of tensile strength is kg/mm2 (kgf/mm^2)
1kgf/mm^2=9.8MPa
kgf (kilogram force)
Some customers often mention MPa
For example: My material's tensile strength is 1000MPA. What kind of test force should I choose?

At this time, you must confirm the specifications of the test workpiece sample.
The following is a 20mm diameter rod material to calculate the approximate test force required
The cross-sectional area of ​​rod material with a diameter of 20mm is 3.14*(20/2)*(20/2)=314mm2
1>100kgf/mm2 (1000mpa)*314mm2 (20mm diameter bar cross-sectional area)=31400kgf/314mm2,
2>31400kgf is equivalent to 314000N (31400*9.8N)
3>314000N=314KN
4>314KN is equivalent to 31.4 tons of experimental power
It is calculated that the rod material with a tensile strength of 1000mpa and a diameter of 20mm requires about 31.4 tons of test force to be pulled off.