CALCS PROC073
Winch Foundation - 4 Point Fixing Forces 160404

CONTENTS

INTRODUCTION
NOTATION
INPUT TO PROCEDURE
OUTPUT FROM PROCEDURE
PROJECT EXAMPLE INPUT
PROJECT EXAMPLE OUTPUT

INTRODUCTION


This procedure is configured for a rope under tension, at specified angle leading off & instantaneous position across the drum. The rope is wound on the drum. The drum may rotate on two bearings, one at each end of the drum. The rotation, pulling the rope in or allowing it to pay-out under the control of motors & brakes. The drum, motors & brakes are then held by the base frame. The base frame is secured to supporting structure by four fixings (on rectangular disposition). The object of the proc edure is to obtain the forces acting in these four fixings.

SEE ALSO: BAS010 PROC071

THEORETICAL BASIS: Statics analysis from 1st principles with reasonable simplifying assumptions to render the problem statically determinate.

NOTATION


AzrRope angle with y-axis (about z-axis)
drRope Diameter
Ox Oy OzCo-ord System Origin: Centre of Drum
PCDrInstantaneous Rope PCD
PrRope Force
Pxw PywWinch self-weight forces
x1 y1 z11st Node +x +z
x2 y2 z22nd Node -x +z
x3 y3 z33rd Node +x -z
x4 y4 z44th Node -x -z
zd11/2 width of drum
ztPosition across drum (effective) of driving or braking torque
zrRope Transverse Co-ordinate

INPUT TO PROCEDURE


As the origin of the co-ordinate system is at the centre of the drum, the position of the base frame fixings (1 to 4) will normally be below the drum axis and will be entered with -ve values of y1.

Where the base frame sits horizontally, y co-ordintes are vertical. For the case where the base frame is fixed to a vertical surface, x co-ordintes are then vertical.

Note that zt needs to be specified according to the relative effectivity of motors/brakes at each end of the drum. For example, if motors contribute equally at each side of the drum, then zd1=0. Ifa motor acts at one side only, then zd1 will equal th e transverse offset of the motor effective driving point.

OUTPUT FROM PROCEDURE


The formulae of the calculations are self-explanatory. Note that results are provided for three instantaneous positions of the rope across the drum: -zd1, 0 +zd1. This provides a range of loadings on the fixings.

PROJECT EXAMPLE INPUT


PROCtitle("D80-363 MH Winch Frame"," ",L06_D$)
LET Pxw=0:Pyw=-3.3:Pr=22.12:dr=3.2:PCDr=75.2:Azr=RAD(10):zd1=65:zr=zd1-dr/2:zt=0:x1=45:y1=-53:z1=80.7:x2=-45:z3=-80.7
L06_D$="MH Rope Dyn Force Li0041"
PROC073



PROJECT EXAMPLE OUTPUT


D80-363 - PROC073 Winch Foundation - 4 Point Fixing Forces 150918
x_tgt= PCDr/2*COS(Azr)= 37.03
y_tgt= PCDr/2*SIN(Azr)= 6.529
rope_pos_z= zr= -63.4
Py12t= (Pr*PCDr/2)*((zt-z3)/(z1-z3))/(x1-x2)= 4.621
Py12y= (Pr*COS(Azr)*(zr-z3)+Pyw*(0-z3))/(z1-z3)= 0.685
Py12x= (Pr*SIN(Azr)*(zr-z3)-Pxw*(0-z3))/(z1-z3)*((0-y1)/(x1-x2))= 0.2425
Py34t= (Pr*PCDr/2)*((z1-zt)/(z1-z3))/(x1-x2)= 4.621
Py34y= (Pr*COS(Azr)*(z1-zr)+Pyw*(z1-0))/(z1-z3)= 17.8
Py34x= (Pr*SIN(Azr)*(z1-zr)-Pxw*(z1-0))/(z1-z3)*((0-y1)/(x1-x2))= 2.02
Py1= Py12y*(x_tgt-x2)/(x1-x2)+Py12x+Py12t= 5.487
Py2= Py12y*(x1-x_tgt)/(x1-x2)-Py12x-Py12t= -4.802
Py3= Py34y*(x_tgt-x2)/(x1-x2)+Py34x+Py34t= 22.86
Py4= Py34y*(x1-x_tgt)/(x1-x2)-Py34x-Py34t= -5.064
rope_pos_z= zr= 0
Py12t= (Pr*PCDr/2)*((zt-z3)/(z1-z3))/(x1-x2)= 4.621
Py12y= (Pr*COS(Azr)*(zr-z3)+Pyw*(0-z3))/(z1-z3)= 9.242
Py12x= (Pr*SIN(Azr)*(zr-z3)-Pxw*(0-z3))/(z1-z3)*((0-y1)/(x1-x2))= 1.131
Py34t= (Pr*PCDr/2)*((z1-zt)/(z1-z3))/(x1-x2)= 4.621
Py34y= (Pr*COS(Azr)*(z1-zr)+Pyw*(z1-0))/(z1-z3)= 9.242
Py34x= (Pr*SIN(Azr)*(z1-zr)-Pxw*(z1-0))/(z1-z3)*((0-y1)/(x1-x2))= 1.131
Py1= Py12y*(x_tgt-x2)/(x1-x2)+Py12x+Py12t= 14.18
Py2= Py12y*(x1-x_tgt)/(x1-x2)-Py12x-Py12t= -4.933
Py3= Py34y*(x_tgt-x2)/(x1-x2)+Py34x+Py34t= 14.18
Py4= Py34y*(x1-x_tgt)/(x1-x2)-Py34x-Py34t= -4.933
rope_pos_z= zr= 63.4
Py12t= (Pr*PCDr/2)*((zt-z3)/(z1-z3))/(x1-x2)= 4.621
Py12y= (Pr*COS(Azr)*(zr-z3)+Pyw*(0-z3))/(z1-z3)= 17.8
Py12x= (Pr*SIN(Azr)*(zr-z3)-Pxw*(0-z3))/(z1-z3)*((0-y1)/(x1-x2))= 2.02
Py34t= (Pr*PCDr/2)*((z1-zt)/(z1-z3))/(x1-x2)= 4.621
Py34y= (Pr*COS(Azr)*(z1-zr)+Pyw*(z1-0))/(z1-z3)= 0.685
Py34x= (Pr*SIN(Azr)*(z1-zr)-Pxw*(z1-0))/(z1-z3)*((0-y1)/(x1-x2))= 0.2425
Py1= Py12y*(x_tgt-x2)/(x1-x2)+Py12x+Py12t= 22.86
Py2= Py12y*(x1-x_tgt)/(x1-x2)-Py12x-Py12t= -5.064
Py3= Py34y*(x_tgt-x2)/(x1-x2)+Py34x+Py34t= 5.487
Py4= Py34y*(x1-x_tgt)/(x1-x2)-Py34x-Py34t= -4.802
END073

PROJECT END