Grupo de Pernos en una platina

8/10/2019 Grupo de Pernos en una platina

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EXELTECH, INC. Project: TRUSS TR23 Engineer: YP Project #

615 2nd Ave. Suite 660 Date: 3-Nov

Seattle, WA 98024 Subject: SAMPLE CONNECTION Checker: Page:

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 Theory

Units: SI Details

Problem Description:

INPUT OUTPUT BoltGroup

Copyright 2001

Vertical Force Horizontal Force Ultimate Shear = 64.58 kN Yakov Polyakov, PE

Py, kN -2 Px, kN 2 Angle to horizon, b= -45.00 deg [email protected]

ex, mm -5 ey, mm 0 True eccentricity, e = 8.357 mm http://yakpol.net

Solved !

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

Single bolt shear capacity fRn = 15.9 kN 174.9

Bolt Location

Bolt X Y

## mm mm

1 0 0

2 0 3

3 0 6

4 0 9

5 3 0

6 3 3

7 3 6

8 3 99 6 0

10 6 3

11 6 6

Testing Spreadsheet vs LRFD tables

X

b

IC

CG

Py

Px

P

-7

-2

3

8

13

-10 -5 0 5 10

Bolts

Instantenious Center of rotation

Group Center

Applied Force

ey

ex

Px

Py

P

e

Y
mailto:[email protected]://yakpol.net/http://yakpol.net/mailto:[email protected]


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Problem Description: Testing Spreadsheet vs LRFD tables

Maximum Force at connection fP = 64.58 kN Goto: Input

Connection is concentrically loaded FALSE Theory

Vertical Force Py = -2.00 kN Eccentricity ex = -5.00 mm Angle b = -0.785 rad

Horizontal Force Px = 2.00 kN Eccentricity ey = 0.00 mm Eccen. e = 8.357 mm

Adjustedb= -0.785 rad

Bolt Parmeters Adjusted e = 8.357 in

fRn = 15.90 kN Instant. Center (IC) Bolt Group Center (CG)

Rmax= 15.61 kN Xo Yo Xc Yc

dmax = 7.01 mm 4.110 5.680 2.727 4.091

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

Bolt Location

Bolt force

angle to

horizon

Bolt to IC

Distance

Bolt

Shear

Force

Bolt

Displ.45.67 -45.67 675.44

Bolt ## X Y q d R D, in Rsinq Rcosq R*d

mm mm rad mm kN mm kN kN kN*mm1 0 0 -3.77 7.01 15.61 0.34 9.15 -12.64 109.42

2 0 3 -4.13 4.91 15.07 0.24 12.63 -8.23 73.96

3 0 6 1.49 4.12 14.68 0.20 14.63 1.14 60.51

4 0 9 0.89 5.28 15.21 0.26 11.84 9.56 80.38

5 3 0 -3.33 5.79 15.36 0.28 2.95 -15.08 88.92

6 3 3 -3.53 2.90 13.62 0.14 5.21 -12.59 39.52

7 3 6 1.29 1.16 9.98 0.06 9.59 2.76 11.53

8 3 9 0.32 3.50 14.23 0.17 4.51 13.49 49.80

9 6 0 -2.82 5.99 15.41 0.29 -4.87 -14.63 92.27

10 6 3 -2.53 3.28 14.03 0.16 -8.08 -11.46 46.00

11 6 6 -1.40 1.92 12.06 0.09 -11.89 2.01 23.12


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Solver

SRsin(q)/sin(d) -4.062 SRcos(q)/cos(d) -4.062 SR*d/(e+lo) 4.062

Psin(b) -2.872 Pcos(b) 2.87 P*lo 42.48

SRsin(q) 2.87 SRcos(q) -2.87 SR*d 42.48 Must be Zelo

Difference 0.0000 Difference 0.0000 Difference 0.0000 0.0000

BestGuess

Single bolt capacity multiplier Po 4.062 4.062Dist. From 0,0 to Inst. Center lo 2.10 mm 2.102

Sideways dist. to Inst. Center mo 0.146 mm

Definition of Ranges

BestGuess_lo =Details!$R$8BestGuess_po =Details!$R$7

betta =Details!$I$8

BigX =Details!$B$19:$B$68

ConcentricConn=Details!$E$4

d =SQRT((X-Xo)^2+(Y-Yo)^2)

dmax =Details!$B$12

e =Details!$I$9

ex =Details!$F$6

ey =Details!$F$7

lo =Details!$P$8

mo =Details!$P$9

Mustbe0 =Details!$R$5

Po =Details!$P$7Print_Area =Details!$A$1:$R$46

Px =Details!$C$7

Py =Details!$C$6

Rn =(1-EXP(-10*d/dmax*0.34))^0.55

Rult =Details!$B$10

SumRcos =Details!$O$4

SumRd =Details!$Q$4

SumRsin =Details!$M$4

theta =ATAN2(X-Xo,Y-Yo)-PI()/2

Units =Input!$B$8

X =OFFSET(BigX,0,0,COUNT(BigX),1)

Xc =Details!$F$12

Xo =Details!$D$12Y =OFFSET(X,0,1)

Yc =Details!$G$12

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


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InputDetails

Input Data: Spreadsheet Formulas:

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

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

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

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

(2) SRicos(qi) + Pucos(b) = 0 dmax= max(di)(3) SRidi+ Pu(e+lo) = 0 Dmax= 0.34 inches

Di= Dmax(di/dmax)

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

ECCENTRICALLY LOADED BOLT GROUP

ULTIMATE STRENGTH METHOD

mo

(Xi,Yi)

IC (Xo,Yo)

This spreadsheet is using theInstantenious Center of RotationMethod to determine shear capacity of

bolt group. This method is described inLRFD Code (2nd Edition, Volume II, p.8-28). In theory, the bolt group rotatesaround IC, and the displacements ofeach bolt is proportional to thedistance to IC. The load deformationrelationship of the bolt:R=Rult(1-exp(-10D)

0.55), whereRult= fRnbolt ultimate shear strength.

D= total deformation of the bolt(inches), experementally determined

Dmax= 0.34 inches.Solving the system of three equilibriumequations we can find location of IC

and ultimate shear force of bolt group.Please, be aware of possibly wrongsolution when instantenious center istoo far from bolt group center.

ey

ex

Px

PyPu

e

Y

X

lo

Ridi

CG (Xc,Yc)

b

qi

Grupo de Pernos en una platina

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