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Testing spreadsheet vs LRFD manual
Maximum Force at connection fP = 180.95 kip
Connection is concentrically loaded FALSE
Vertical Force Py = 30.00 kip Eccentricity ex = 12.00 in Angle d = 1.162 rad
Horizontal Force Px = 13.00 kip Eccentricity ey = 14.00 in Eccen. e = 7.231 inWeld size 0.25 in Adjusted d = 1.162 rad
Unit Weld Strength Adjusted e = 7.231 in
fRn = 5.568 kip/in Instant. Center Weld Group Centroid
Xo Yo Xc Yc
-5.520 13.963 2.158 9.474
Number of elements in the longest weld = 20
Total number of elements = 38 Element properties
Weld Properties
Centroid Length
Weld Node 1 Node 2 Length angle X Y L
## X1 Y1 X2 Y2 LW alphaW in in in1 0 0 0 20 20.000 1.5708 0.000 0.500 1.000
2 0 20 8 20 8.000 0.0000 0.000 1.500 1.000
3 0 0 10 0 10.000 0.0000 0.000 2.500 1.000
0.000 3.500 1.000
0.000 4.500 1.000
0.000 5.500 1.000
0.000 6.500 1.000
0.000 7.500 1.000
0.000 8.500 1.000
0.000 9.500 1.000
0.000 10.500 1.000
0.000 11.500 1.0000.000 12.500 1.000
0.000 13.500 1.000
0.000 14.500 1.000
0.000 15.500 1.000
0.000 16.500 1.000
0.000 17.500 1.000
0.000 18.500 1.000
0.000 19.500 1.000
0.500 20.000 1.000
1.500 20.000 1.000
2.500 20.000 1.000
3.500 20.000 1.000
4.500 20.000 1.000
5.500 20.000 1.000
6.500 20.000 1.000
7.500 20.000 1.000
0.500 0.000 1.000
1.500 0.000 1.000
2.500 0.000 1.000
3.500 0.000 1.000
4.500 0.000 1.000
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5.500 0.000 1.000
6.500 0.000 1.000
7.500 0.000 1.000
8.500 0.000 1.000
9.500 0.000 1.000
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Solver Charting Applied Force
SRsin(q)/sin(d) -180.949 SRcos(q)/cos(d) -180.948 SR*lr/(e+lo) 180.947 12 14
Psin(d) 166.029 Pcos(d) 71.95 P(e+lo) 2906.17 12 23.41
SRsin(q) -166.03 SRcos(q) -71.95 SR*lr 2906.16 Must be Zero
Difference -0.0018 Difference -0.0001 Difference 0.0013 0.0000 12 14
Best Guess 16.08 14
Ultimate shear force Po 180.947 kip 180.95
Dist. From 0,0 to Inst. Center lo 8.83 in 8.8298 12 14
Sideways dist. to Inst. Center mo -1.066 in 16.08 23.41
Min Dmax/r
0.0010023
SR SRsin(b)
Angle to
horizon
Angle of
resultant to
horizon
Angle to
resultant
force
Dist. To IC Dmax Dmax/lr D P=D /Dmax 7.983 -3.028
a b q lr D R Rsin(b)
rad rad rad in in in kip kip
1.571 -2.752 1.182 14.55 0.017 0.001 0.015 0.878 7.983 -3.028
1.571
1.571
1.571
1.571
1.571
1.571
1.571
1.571
1.571
1.5711.571
1.571
1.571
1.571
1.571
1.571
1.571
1.571
1.571
0.000
0.000
0.0000.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
0.000
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0.000
0.000
0.000
0.000
0.000
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Yc =Details!$G$12
Yo =Details!$E$12
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R(DEGREES(theta)+6,-0.65)*w,0.17*w)
N(theta),1.5))*POWER(p_ratio*(1.9-0.9*p_ratio),0.3)
)
),1)
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InputDetails
Input Data: Spreadsheet Formulas:
Shear force application: Px Py ex ey Continuous weld formulas
X1,X2,Y 1,Y 2 - Cont. weld coordinates LW = SQRT(SUMSQ(X2-X1,Y 2-Y 1))
Single bolt shear capacity: fR n aW=atan2(X2-X1,Y 2-Y 1)
Xc=SUM(LW*(X1+X2)/2)/SUM(LW)
Equlibrium equations: Y c=SUM(LW*(Y 1+Y 2)/2)/SUM(LW)
Xo = -losin(b) - mocos(b) + Xc
(1) SRsin(b) + Posin(d) = 0 Y o = locos(b) - mosin(b) + Y c
(2) SRcos(b) + Pocos(d) = 0(3) SRlr + P0(e+lo) = 0 Unit-weld elements formulas
b = ATAN2(X-Xo,Y-Yo)-p /2
Equations variables: Po lo and mo q = aW - b
e = -(ey-Y c)cos(b) + (ex-Xc)sin(b)
Excel Solver is used to find the roots qi = atan((Y i-Y o)/(Xi-Xo)) - p /2
of the equations. lr = SQRT((Y-Y o)2+(X-Xo)
2)
VBA routine 'ReadWelds' divides Dmax=min(1.087*(DEGREES(q)+6)-0.65w,0.17w)
continuous welds into elements. p=(lr/Dmax)(Dmax /lr)min
Default number of elements is 20 R=fRn*L(1+0.5*SIN1.5(q))[p(1.9-0.9p)]0.3
per longest weld, provides sufficient fRn = 0.75*0.6*FEXX*0.707*w*f1 [force/length]
accuracy. f1 - electrode strength adjustment coefficient (see sheet 'Input' for table)
ECCENTRICALLY LOADED WELD GROUP
ULTIMATE STRENGTH METHOD, LRFD 2nd EDITION
This spreadsheet is using the InstanteniousCenter of Rotation Method to determineshear capacity of weld group. This methodis described in LRFD Code (2nd Edition,
Volume II, p. 8-154). In theory, the weldgroup rotates around IC, and thedisplacements of each unit-weld element isproportional to the distance to IC.To determine the critical element, the ratio
Dmax /lr is computed for each element,where
Dmax = 1.087w(q+6)-0.65<=0.17wlr = distance from IC to unit-weld element.w = leg size of the weld.q = the angle between resisting force inunit-weld segment and its axis.
The element with the smallest ratio
reaches the ultimate capacity first. Thedeformation of the other elements canthen be computed as D=lr(Dmax /lr)min
The resisting force for each element canbe found from :R=FEXX(1+0.5sin1.5q)[p(1.9-0.9p)]0.3, wherep=D /Dmax
D = total deformation of the unit-weld
element.
Solving the system of three equilibriumequations we can find location of IC andultimate shear force of weld group.
q(Xi,Yi)
IC (Xo,Y o)
b
d
ey
ex
Px
Py
Pu
e
Y
X
lo
mo
R i
lri
CG (Xc,Y c)
Unit-Weld element