EXELTECH, INC. Project: TRUSS TR23 Engineer: YP615 2nd Ave. Suite 660 Date: 19-AprSeattle, WA 98024 Subject: SAMPLE CONNECTION Checker:

tel. (206)623-9646 Date:

ECCENTRIC SHEAR CONNECTION ANALYSIS OF BOLT GROUP Last updated: 25-Apr-02

The following calculations comply with LRFD Steel Design Manual 2nd Edition

Units: SI

Problem Description: Testing Spreadsheet vs LRFD tables

INPUT OUTPUT BoltGroup

Copyright © 2001

Vertical Force Horizontal Force Ultimate Shear = 85.57 kN Yakov Polyakov, PE

Py, kN -2 Px, kN 10 -11.31 deg

ex, mm -5 ey, mm 0 True eccentricity, e = 5.527 mm

Solved !

Bolt description: A325 SH, 3/4" Dia.15.9 kN 174.9

Bolt LocationBolt X Y## mm mm

1 0 02 0 33 0 64 0 95 3 06 3 37 3 68 3 99 6 0

10 6 311 6 6

TheoryDetails

Angle to horizon, b = yakov@yakpol.net

http://yakpol.net

Single bolt shear capacity fRn =

-6.000 -4.000 -2.000 0.000 2.000 4.000 6.000 8.000

-2.000

0.000

2.000

4.000

6.000

8.000

10.000

12.000

20-0.77

4.091

7.169

Instantenious Center of rota-tionGroup CenterApplied Force

ex

Py

e

Y

Project #

Page:

Last updated: 25-Apr-02

X

b

ey

Px

P

Problem Description: Testing Spreadsheet vs LRFD tables

Maximum Force at connection 85.57 kN Goto:

Connection is concentrically loaded 0

Vertical Force Py = -2.00 kN ex = -5.00 mm -0.197 radHorizontal Force Px = 10.00 kN ey = 0.00 mm Eccen. e = 5.527 mm

-0.197 radBolt Parmeters Adjusted e = 5.527 mm

15.90 kN Instant. Center (IC) Bolt Group Center (CG)Rmax= 15.61 kN Xo Yo Xc Ycdmax = 7.83 mm 3.153 7.169 2.727 4.091

Bolt Location 16.78 -83.91 738.32

Bolt ## X Y q d R R*dmm mm rad mm kN mm kN kN kN-mm

1 0 0 -3.56 7.83 15.61 0.34 6.28 -14.29 122.222 0 3 -3.79 5.23 14.97 0.23 9.03 -11.94 78.263 0 6 -4.36 3.36 13.75 0.15 12.89 -4.78 46.234 0 9 1.04 3.65 14.01 0.16 12.12 7.04 51.085 3 0 -3.16 7.17 15.51 0.31 0.33 -15.50 111.196 3 3 -3.18 4.17 14.41 0.18 0.53 -14.40 60.137 3 6 -3.27 1.18 9.61 0.05 1.25 -9.53 11.338 3 9 0.08 1.84 11.44 0.08 0.95 11.40 21.029 6 0 -2.76 7.71 15.59 0.33 -5.75 -14.49 120.26

10 6 3 -2.54 5.05 14.90 0.22 -8.40 -12.30 75.2111 6 6 -1.96 3.08 13.44 0.13 -12.44 -5.11 41.38

fP = InputTheory

Angle b =

Adjusted b =

fRn =

S(Rsinq) S(Rcosq) S(R*d)

Bolt forceangle tohorizon

Bolt to ICDistance

BoltShear Force

BoltDispl.

D, in Rsinq Rcosq

Solver-5.382 -5.382 5.382

-1.055 5.28 P*lo 46.44

1.06 -5.28 46.44 Must be Zelo

Difference 0.0000 Difference 0.0000 Difference 0.0003 0.0000BestGuess

Single bolt capacity multiplier Po 5.382 5.382Dist. From 0,0 to Inst. Center lo 3.10 mm 3.102Sideways dist. to Inst. Center mo 0.186 mm

Definition of RangesBestGuess_lo =Details!$R$8BestGuess_po =Details!$R$7betta =Details!$I$8BigX =Details!$B$19:$B$68ConcentricConn =Details!$E$4d =SQRT((X-Xo)^2+(Y-Yo)^2)dmax =Details!$B$12e =Details!$I$9ex =Details!$F$6ey =Details!$F$7lo =Details!$P$8mo =Details!$P$9Mustbe0 =Details!$R$5Po =Details!$P$7Print_Area =Details!$A$1:$R$46Px =Details!$C$7Py =Details!$C$6Rn =(1-EXP(-10*d/dmax*0.34))^0.55Rult =Details!$B$10SumRcos =Details!$O$4SumRd =Details!$Q$4SumRsin =Details!$M$4theta =IF(Y-Yo=0,0,ATAN2(X-Xo,Y-Yo))-PI()/2UnitForce =IF(Units="US","kip","kN")UnitLength =IF(Units="US","in","mm")Units =Input!$B$8X =OFFSET(BigX,0,0,COUNT(BigX),1)Xc =Details!$F$12Xo =Details!$D$12

SRsin(q)/sin(d) SRcos(q)/cos(d) SR*d/(e+lo)

Psin(b) Pcos(b)

SRsin(q) SRcos(q) SR*d

For calculation purposes external force (Po) is always positive and rotates boltgroup contrclockwise about it's center (Xc,Yc). Eccentricity (e) is always positive as well. The distance (lo) from boltgroup center to instantenious center (Xo,Yo) can be negative but not less than -e.

ECCENTRICALLY LOADED BOLT GROUPULTIMATE STRENGTH METHOD

Input Data: Spreadsheet Formulas:

Equlibrium equations:

(1)

(2)

(3)

InputDetails

Shear force application: Px Py ex ey Xc = AVERAGE(Xi)

Bolt locations: Xi Yi Yc = AVERAGE(Yi)

Single bolt shear capacity: fRn Xo = -losin(b) - mocos(b) + Xc

Yo = locos(b) - mosin(b) + Yc

e = - (ey-Yc)cos(b) + (ex-Xc)sin(b)

qi = atan((Yi-Yo)/(Xi-Xo)) - p/2

SRisin(qi) + Pusin(b) = 0 di = sqrt((Yi-Yo)^2+(Xi-Xo)2)

SRicos(qi) + Pucos(b) = 0 dmax = max(di)

SRidi + Pu(e+lo) = 0 Dmax = 0.34 inches

Di = Dmax(di/dmax)

Equations variables: Pu lo and mo Ri = fRn(1-exp(-10Di))0.55

mo

(Xi,Yi)

IC (Xo,Yo)

This spreadsheet is using the Instantenious Center of Rotation Method to determine shear capacity of bolt group. This method is described in LRFD Code (2nd Edition, Volume II, p. 8-28). In theory, the bolt group rotates around IC, and the displacements of each bolt is proportional to the distance to IC. The load deformation relationship of the bolt:R=Rult(1-exp(-10D)0.55), whereRult = fRn bolt ultimate shear strength.D = total deformation of the bolt (inches), experementally determined Dmax = 0.34 inches.Solving the system of three equilibrium equations we can find location of IC and ultimate shear force of bolt group. Please, be aware of possibly wrong solution when instantenious center is too far from bolt group center.

ey

ex

Px

PyPu

e

Y

X

lo

Ridi

CG (Xc,Yc)

b

qi

This spreadsheet is using the Instantenious Center of Rotation Method to determine shear capacity of bolt group. This method is described in LRFD Code (2nd Edition, Volume II, p. 8-28). In theory, the bolt group rotates around IC, and the displacements of each bolt is proportional to the distance to IC. The load deformation relationship of the bolt:R=Rult(1-exp(-10D)0.55), whereRult = fRn bolt ultimate shear strength.D = total deformation of the bolt (inches), experementally determined Dmax = 0.34 inches.Solving the system of three equilibrium equations we can find location of IC and ultimate shear force of bolt group. Please, be aware of possibly wrong solution when instantenious center is too far from bolt group center.