WeldGroup Eccentric

7/28/2019 WeldGroup Eccentric

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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



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



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



0.000

0.000

0.000

0.000

0.000





Yc =Details!$G$12

Yo =Details!$E$12



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)





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

Documents

WeldGroup Eccentric