CALCS PROC048
BRACKET FIXED ON ARC

CONTENTS

Guidance: Application, Input, Output & Theory
Notation
Input Data Format
Project Example
PROC048 Theory

GUIDANCE

APPLICATION

The procedure has two parts: PROC048a & PROC048b use respectively radial & shear line force reaction around the supporting arc.

PROC048a is suitable where there are reasonable radial stress paths both inside & outside of the arc. For example, where a substantial fixing bracket for a hydraulic cylinder is fixed to an in-plane plate of the supporting member.

PROC048b is suitable where there is not a reasonable radial stress path inside the arc. For example, where a substantial fixing bracket is fixed (around an arc) to an unstiffened cylinder. The reactive stress distribution, around the arc, is then by shear.

INPUT TO PROCEDURE

Specification of plate thickness, arc angle, radius & applied forces

OUTPUT FROM PROCEDURE

Stress Distribution

THEORETICAL BASIS

1st Principles

NOTATION

fsshear, fixing bracket to cylinder
Nstepnumber of integration steps (20 max)
NpnfBAS026_PNF steps, where Nstep/Npnf is integer
Pforces, radial applied (PROC048b)
Px PyApplied forces parallel & normal to Arc Axes (PROC048a)
prx pryRadial Line Forces due to Px & Py around arc (PROC048a)
R rradius, arc
Tw tplate thickness, around arc
thetahalf included angle of sector (SemiArc)

INPUT DATA FORMAT

LET Ngr%=:Nk%=:R=:Tw=:SemiArc=RAD():Px=:Py=:Nstep=16:Npnf=4:PROC048a
LET Ngr%=:Nk%=:r=:t=:theta=RAD():P=:PROC048b

PROJECT EXAMPLE: Pin Plate for Hydraulic Cylinder

LET Ngr%=S355:Nk%=AISC:R=23:Tw=5:SemiArc=RAD((180+58)/2):Px=50:Py=L07/2:Nstep=16:Npnf=4:PROC048a
LET Ngr%=S355:Nk%=LR:r=23:t=5:theta=RAD(119/2):P=488.6/4:PROC048b

SlFr -1020101 PT3&48 - PROC048a STIFF RING - RADIAL REACTION ON ARC 150822 - L07
LET Px_=0:Py_=0:pr_=1/Tw:prx_=0:pry_=0:pr_max=0:pr_min=0
LET Px_=0:Py_=0:pr_=0.2:prx_=0:pry_=0:pr_max=0:pr_min=0
Ang= -SemiArc+2*SemiArc*stepi/Nstep= -2.077
Ang= -2.077+2*2.077*0/16= -2.077
Pxi= pr_*(SIN(Ang))^2*R*(SemiArc/Nstep)= 0.4568
Pxi= 0.2*(SIN(-2.077))^2*23*(2.077/16)= 0.4568
Pyi= pr_*(COS(Ang))^2*R*(SemiArc/Nstep)= 0.1403
Pyi= 0.2*(COS(-2.077))^2*23*(2.077/16)= 0.1404
LET Px_=Px_+Pxi:Py_=Py_+Pyi
LET Px_=0.9135:Py_=0.2807
prx_= -(Px/Tw)*pr_/Px_= -0.3233
prx_= -(50/5)*0.2/6.187= -0.3233
pry_= (Py/Tw)*pr_/Py_= 2.465
pry_= (244.3/5)*0.2/3.964= 2.465
pr= prx_*SIN(Ang)+pry_*COS(Ang)= -0.9124
pr= -0.3233*SIN(-2.077)+2.465*COS(-2.077)= -0.9124
StepAngPxiPyipr
0-2.0770.45680.1403-0.9124
1-1.8170.56160.03556-0.2881
2-1.5580.5971.023E-40.3555
3-1.2980.55380.043320.9753
4-1.0380.44330.15381.53
5-0.77890.29470.30251.982
6-0.51920.1470.45012.301
7-0.25960.039350.55782.466
8000.59712.465
90.25960.039350.55782.3
100.51920.1470.45011.98
110.77890.29470.30251.527
121.0380.44330.15380.9726
131.2980.55380.043320.3526
141.5580.5971.023E-4-0.291
151.8170.56160.03556-0.9151
162.0770.45680.1403-1.478
f_max_= FNmax(pr_max,pr_min,0,0,0)= 2.466
f_max_= 2.466= 2.466
UF_001= f_max_/(dac(Nk%,Kd%)*mgr(Ngr%,Fy%))= 1.0511.051
UF_001= 2.466/(0.67*3.5)= 1.0521.051
END048a
P - PROC048b BRACKET ON CYLINDER, SHEAR ON ARC - L07
fs= P/(r*t*(theta-.5*SIN(2*theta)))= 1.767
fs= 122.2/(23*5*(1.038-.5*SIN(2*1.038)))= 1.77
UF_002= fs/(dac(Nk%,Ks%)*mgr(Ngr%,Fy%))= 1.2991.299
UF_002= 1.767/(0.3886*3.5)= 1.2991.299
END048b