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1、Strip Guiding Theory,Fundamental Behaviour of Strip on Rolls Some Sketches for Discussion,EMG - ST-V - THK - 14. May 2000 - R1.2,1 of 15,2 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,Vc,Vc,Va,Strip speed Vs = Circumference speed Vc,Instant speed of strip displacement Va = Vc * tan ,3 of 15,EMG - ST
2、-V - THK - 14. May 2000 - R1.2,IDEAL CONDITIONS Under perfectly aligned rolls and ideal strip ( without geometric defects ) . no lateral strip displacement will happen during running strip REAL CONDITIONS Not perfectly aligned rolls and natural strip defects . cause strip displacement and need corre
3、ction of these effects APPLICATION OF STRIP GUIDING SYSTEMS,4 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,Initial Camber,Steady state ( running strip ),90,90,Strip displacement D due to a cambered strip,THE STRIP ENTERS THE NEXT ROLL ( IN THE STEADY STATE ) AT AN ANGLE OF 90 DEGREES TO ITs AXIS,5 o
4、f 15,EMG - ST-V - THK - 14. May 2000 - R1.2,THE STRIP ENTERS THE NEXT ROLL ( IN THE STEADY STATE ) AT AN ANGLE OF 90 DEGREES TO ITs AXIS,90,90,CL,Entry Span L,Strip deviation C due to a misaligned roll,6 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,THE STRIP ENTERS THE NEXT ROLL ( IN THE STEADY STAT
5、E ) AT AN ANGLE OF 90 DEGREES TO ITs AXIS,Strip correction C due to an inclined roll,90,90,Centerline Strip,CL,Entry Span L,Correction C,Strip correction C : +/- C = L * K * sin where: K = 0,66 . 0,75,7 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,CL,Entry Span L,Pre-Entry Span Lp ,90,90,Roll A,Roll
6、 B,Roll C,Due to the inclination of the guide roll ( C ) , a moment transfer is induced by the roll in the entry span ( B ) into the pre entry span, which heavily decreases the correction capactiy of roll ( C ). The curvature of the strip in L and Lp depends from the strip width and thickness as wel
7、l as from the strip tension, the material characteristics and the moment of inertia around the vertical axis.,To AVOID the PREENTRY SPAN PROBLEM : Take care, that the Entry Span L is as long or longer than the Pre-Entry Span Lp if guide structures with I - Efects are applied.,8 of 15,EMG - ST-V - TH
8、K - 14. May 2000 - R1.2,+/- C,Correction Capacity +/- C = Ue * sin ,Ue,Proportional Guide Structure,9 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,Proportional Integral Guide Structure,+/- C,System Radius,+,=,10 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,Proportional Integral Guide Structure,+/- C
9、,Long entry span,System Radius,Correction Capacity +/- C = L * K * sin where: K = 0,66 . 0,75 and L max. effective is about 15 to 20 m,11 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,90 =Proportional behaviour of the guide 90 + = Proportional plus integral behaviour of the guide where is normally be
10、tween 3 . 15 Length of entry and exit spans have to be checked,Proportional Guide Structure with additional integral effect,12 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,Calculation of the additional integral effect of an inclined P- Roll,Geometrics: - shifting angle of a Proportional Guide Roll (
11、 P-Roll ) is . - inclination angle of a P-Roll is . INTEGRAL ANGLE for the APPROACHING STRIP is . sin = sin * sin = sin-1 ( sin * sin ),90 + ,13 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,90,90 + ,Short Exit Span,Long Entry Span,Application of the additional integral effect of a P-Roll with short exit span,Example for an application of a Proportional, but inclined Guide Roll ( P-Roll ) to acheive an ( additional ) integral correction ( long entry span ) and having a short exit span,14 of 15,EMG - ST-V - THK - 14. May 2000 - R1.2,Steady state ( run
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