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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]8/10/2019 Grupo de Pernos en una platina
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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.
8/10/2019 Grupo de Pernos en una platina
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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