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1 B. N6 _* d0 B! v) v% j1 ^3 ~
Standard:1 t& S# `$ S8 k' o8 ~- H
IEC60065 6ed.& m" Y' F L) @; p' s# k; c. \$ }
Sub clause:( k% B7 Z6 h& G# T, y/ u
13.2
9 P# N( ]. j% } U8 X% E8 }3 pSheet No.- x3 K8 I+ S: ^* W' d6 {0 X
525) A7 C0 y& Q0 z& Z$ V
Subject
4 p3 p$ s9 e4 r6 @& h, W TApplied force for internal parts: c0 Q% G i( R5 D: _/ P! z U
Key words:# W; S1 j9 [& e( [4 K6 B5 R) w. N
force for internal parts" h' o8 c" h, k
Decision taken at the
- ], ~8 T5 k0 b7 d: V4 t40th meeting 20033 K1 X* D. m1 f! ^6 Q+ e, y
Question:
M5 q o& h* a1 V! ZIn clause 13.2, it is specified that 2N for internal parts and 30N for the outside are applied
6 g3 c+ Z" p6 S0 dsimultaneously while taking measurement of clearance. However, it is not specified that forces are
' \( F" H$ e' e q6 Sapplied to internal parts while taking measurements between two internal parts, ex. primary circuit
) X( p( h# Z6 ncomponent and secondary circuit component.
6 Q3 z- T! x8 V) x g, [Which is the appropriate method to measure clearance between two internal parts ?/ E2 N4 N4 q( y' @* l" J6 G
a) 2N forces are applied to the both parts simultaneously.
& t. y2 k5 S, f; p" ~b) First, a 2N force is applied to one part and removed. And then a 2N is applied to the other one.
3 N% t1 }$ i( h, S6 rc) Other opinion
* ~8 G; o) N r8 u6 N1 P, NDecision:
, p9 p. o6 C UFirst, a 2N force is applied to one part and removed. Then a 2N force is applied to the other
# d5 s' f/ i# r$ ppart.& P/ N, d# `# }, {" {5 A- [! q- x
' \ [8 u- H; f4 m9 g. l6 B
9 J$ P* S$ ?- r. T% @
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发表于 2012-11-8 18:42
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