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How to find the perfect withholding amount

February 22, 2021

withholding size of the withholding tube.

The hydraulic pipe selected today is German standard, 1SN, inner hole is 12.8mm, steel wire outer diameter is 20mm, sleeve model is 01100, inner diameter is 21mm, outer circle is 28mm, core thickness is 13mm. During the crimping process, my actual seizure amount was 3.2mm, 28mm-24.8mm.


The core produced by manufacturer A becomes 12mm

The core produced by manufacturer B becomes 12.3mm

The actual crimping 3.2mm contains a gap of 1mm, and the net crimping is 2.2mm, but the deformation is large, with the smallest deformation reaching 0.7mm.

Therefore, in order to achieve the most perfect pressure tube state (ie the maximum value of elastic deformation, the minimum value of plastic deformation). 24.8mm+0.7mm=25.5mm (the most perfect withholding state)

28mm-25.5mm-1mm (actual gap) = 1.5mm (actual net withholding amount)

This is the core 13mm, two-layer tube, theoretically the most perfect crimping amount. In actual operation, it can crimp 1.7-1.8mm and deform the core 0.2-0.3mm.

You can use it as a reference.

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