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| DSH 525
; v5 H1 D. F4 v; ?; C
4 B6 f; L/ w6 h | Applied force for internal parts
( [- U/ x+ D& I$ u2 T) v. E | 13.2# Y, [( j4 v5 T0 w
| 60065(ed.6)
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Standard:
# m, p4 L9 t# v Z' J" A, @IEC60065 6ed./ O% T& D4 F3 A4 r- p4 m t0 `
Sub clause:
, l# q1 T# a9 } T) a- N13.2
8 t/ m' ]- h4 W8 ]Sheet No.
( s+ S' Y: ~# R* e/ X7 G525& ]0 v, v- N6 L; h, O5 N
Subject; E7 U9 ?6 F- @& `4 a
Applied force for internal parts
; t' X+ o5 m, a8 u/ \Key words:& M, F+ _" d; ]3 a" d' b* {
force for internal parts
( z4 c, a" @4 kDecision taken at the
8 R' F! u; f: e* B+ n$ H" `40th meeting 2003
: G# X4 v$ _" g# h9 n9 S7 d) |Question:
( Q8 ]- X2 q1 t* D8 mIn clause 13.2, it is specified that 2N for internal parts and 30N for the outside are applied
8 e9 g9 v* {; Y* L( Z2 ysimultaneously while taking measurement of clearance. However, it is not specified that forces are8 Z" ?* G: r1 V7 @7 G6 c
applied to internal parts while taking measurements between two internal parts, ex. primary circuit+ I0 F- Y: J5 E: s4 c
component and secondary circuit component.
& D3 j; |. L, i3 EWhich is the appropriate method to measure clearance between two internal parts ?
9 O& E/ X# e6 K2 z! e$ U4 Qa) 2N forces are applied to the both parts simultaneously. s" K1 D/ E. ^0 J# K" ]5 |
b) First, a 2N force is applied to one part and removed. And then a 2N is applied to the other one.% O. ~/ f; G) g
c) Other opinion
- i% F8 X" `# F5 G; O+ s1 D( [0 i+ iDecision:
2 ?! }' r% J( {: ]# y2 q* rFirst, a 2N force is applied to one part and removed. Then a 2N force is applied to the other
# S5 n3 x/ C( f# Hpart.
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