CALCS PROC046
LUG/DIAPHRAGM

NOTATION

x1 y1coords of support plate bottom left corner
x3 y3coords of support plate top right corner
Tthickness, plate, weld or supporting cyl
Kefactor, effectivity, plate side
SKeplate side no, to which Ke relates
Px Pylug forces, parallel to x & y axes
dx/dyplate aspect ratio
M/2Aconstant, batho shear
P1nshear force, nominal direct & batho plate sides (sim. 2,3&4)
Ppshear force, perturbation
P1pshear force, nominal + pert. on plate sides (sim. 2,3&4)

APPLICATION

Lug fixed to rectangular diaphragm & subject to pin shear

OUTPUT

Diaphragm perimeter shear stress

SEE ALSO

PROC023

THEORY

refs

GUIDANCE

The procedure provides shear forces, derived from direct and Batho distributions, which support a rectangular plate acted on by lug forces. The nominal force one any one plate side may be modified by an effectivity factor in the range from 0 to 1 and correcting perturbation forces applied to the other sides to maintain equalibrium. Ke may be set to 0 when only 3 sides of the plate are supported. Set 0

INPUT FORMAT

LET Nm=:Nk=:x1=:y1=:x3=:y3=:T=:Ke=:SKe=:Px=:Py=:PROC046

EXAMPLE

Call Statement

LET Nm=1:Nk=1:x1=5:y1=3:x3=21:y3=13:T=1.2:Ke=.4:SKe=2:Px=23:Py=15:PROC046

Output

PROC046 LUG/DIAPHRAGM, Rev 010296
dxDIVdy= (x3-x1)/(y3-y1)= 1.6
Mdiv2A= (-Px*(y3+y1)/2+Py*(x3+x1)/2)/(2*(x3-x1)*(y3-y1))= 0.03438
P1n= Py/2+(y3-y1)*Mdiv2A= 7.844
P2n= Px/2+(x3-x1)*Mdiv2A= 12.05
P3n= Py/2-(y3-y1)*Mdiv2A= 7.156
P4n= Px/2-(x3-x1)*Mdiv2A= 10.95
Pp= (1-Ke)*P2n= 7.23
P1p= P1n+Pp/dxDIVdy= 12.36
P2p= P2n-Pp= 4.82
P3p= P3n-Pp/dxDIVdy= 2.638
P4p= P4n+Pp= 18.18
fs1= ABS(P1p)/(T*(y3-y1))= 1.03
fs2= ABS(P2p)/(T*(x3-x1))= 0.251
fs3= ABS(P3p)/(T*(y3-y1))= 0.2198
fs4= ABS(P4p)/(T*(x3-x1))= 0.9469
UF= FNmax(fs1,fs2,fs3,fs4,0)/(Fy(Nm)*Ks(Nk))= 0.7575
END046