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    Bolt Group 01

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    Livian Teddy
    BOLT

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    Attribution Non-Commercial (BY-NC)
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    Bolt Group 01

    Uploaded by

    Livian Teddy
    140 views4 pages
    BOLT

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    © Attribution Non-Commercial (BY-NC)

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    BOLT

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as XLS, PDF, TXT or read online from Scribd
    Download as xls, pdf, or txt
    140 views4 pages

    Bolt Group 01

    Uploaded by

    Livian Teddy
    BOLT

    Copyright:

    Attribution Non-Commercial (BY-NC)
    Download as XLS, PDF, TXT or read online from Scribd
    Download as xls, pdf, or txt
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    EXELTECH, INC. 615 2nd Ave. Suite 660 Seattle, WA 98024 tel.

    (206)623-9646

    Project: Subject:

    TRUSS TR23 SAMPLE CONNECTION

    Engineer: Date: Checker: Date:

    YP 17-Sep

    Project # Page:

    ECCENTRIC SHEAR CONNECTION ANALYSIS OF BOLT GROUP


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

    Last updated: 25-Apr-02 Theory Details

    Units:

    US

    Problem Description:

    Testing Spreadsheet vs LRFD tables


    OUTPUT
    BoltGroup Copyright 2001

    INPUT Vertical Force Horizontal Force Py, kip -2 Px, kip 10 ex, in -5 ey, in 0

    Ultimate Shear =
    Angle to horizon, b = True eccentricity, e = Solved !

    85.57 kip
    -11.31 deg 5.527 in

    Yakov Polyakov, PE

    yakov@yakpol.net
    http://yakpol.net

    Bolt description: A325 SH, 3/4" Dia. Single bolt shear capacity fRn = 15.9 kip Bolt Location X Y in in 1 0 0 2 0 3 3 0 6 4 0 9 5 3 0 6 3 3 7 3 6 8 3 9 9 6 0 10 6 3 11 6 6

    174.9 Y

    Bolt ##

    12

    10

    IC

    Px e b

    CG

    Py P
    5 10

    ey

    Px
    0 -10 -5

    Py
    -2

    ex

    X
    Bolts Instantenious Center of rotation Group Center Applied Force diagonal scale

    Problem Description: Maximum Force at connection

    Testing Spreadsheet vs LRFD tables


    fP = 85.57 kip FALSE ex = ey = -5.00 in 0.00 in Goto: Input

    Connection is concentrically loaded Vertical Force Py = Horizontal Force Px = Bolt Parmeters fRn = 15.90 kip Rmax= 15.61 kip dmax = 7.83 in -2.00 kip 10.00 kip

    Theory
    -0.197 5.527 -0.197 5.527 rad in rad in

    Angle b = Eccen. e = Adjusted b = Adjusted e = Instant. Center (IC) Bolt Group Center (CG) Xo Yo Xc Yc 3.153 7.169 2.727 4.091

    S(Rsinq) S(Rcosq) S(R*d) Bolt Location Bolt ## 1 2 3 4 5 6 7 8 9 10 11 X in 0 0 0 0 3 3 3 3 6 6 6 Y in 0 3 6 9 0 3 6 9 0 3 6 Bolt force angle to horizon q rad -3.56 -3.79 -4.36 1.04 -3.16 -3.18 -3.27 0.08 -2.76 -2.54 -1.96 Bolt to IC Distance d in 7.83 5.23 3.36 3.65 7.17 4.17 1.18 1.84 7.71 5.05 3.08 Bolt Shear Force R kip 15.61 14.97 13.75 14.01 15.51 14.41 9.61 11.44 15.59 14.90 13.44 Bolt Displ. D, in in 0.34 0.23 0.15 0.16 0.31 0.18 0.05 0.08 0.33 0.22 0.13 16.78 Rsinq kip 6.28 9.03 12.89 12.12 0.33 0.53 1.25 0.95 -5.75 -8.40 -12.44 -83.91 Rcosq kip -14.29 -11.94 -4.78 7.04 -15.50 -14.40 -9.53 11.40 -14.49 -12.30 -5.11 738.32 R*d kip-in 122.22 78.26 46.23 51.08 111.19 60.13 11.33 21.02 120.26 75.21 41.38

    Solver
    SRsin( q)/sin(d)
    -5.382

    SRcos(q)/cos(d)

    -5.382 SR*d/(e+lo)

    5.382

    Psin(b ) SRsin(q) Difference

    -1.055 1.06

    Pcos(b ) SRcos(q)

    5.28 -5.28

    P*lo SR*d

    46.44 46.44 Must be Zelo 0.0003 0.0000 BestGuess 5.382 3.102

    0.0000 Difference Po lo mo

    0.0000 Difference 5.382 3.10 in 0.186 in

    Single bolt capacity multiplier Dist. From 0,0 to Inst. Center Sideways dist. to Inst. Center

    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.

    Definition of Ranges
    BestGuess_lo =Details!$R$8 BestGuess_po =Details!$R$7 betta =Details!$I$8 BigX =Details!$B$19:$B$68 ConcentricConnection =Details!$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$7 Print_Area =Details!$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$4 theta =IF(Y-Yo=0,0,ATAN2(X-Xo,Y-Yo))-PI()/2 UnitForce =IF(Units="US","kip","kN") UnitLength =IF(Units="US","in","mm") Units =Input!$B$8 X =OFFSET(BigX,0,0,COUNT(BigX),1) Xc =Details!$F$12 Xo =Details!$D$12

    ECCENTRICALLY LOADED BOLT GROUP ULTIMATE STRENGTH METHOD 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), where Rult = 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.

    Pu

    Py Input Details b

    (Xi,Yi) CG (Xc,Yc) e Px lo mo IC (Xo,Yo) di ex qi ey Ri

    Input Data: Shear force application: Px Py ex ey Bolt locations: Xi Yi Single bolt shear capacity: fRn Equlibrium equations: (1) (2) (3) SRisin(qi) + Pusin(b) = 0 SRicos(qi) + Pucos(b) = 0 SRidi + Pu(e+lo) = 0

    Spreadsheet Formulas: Xc = AVERAGE(Xi) Yc = AVERAGE(Yi) 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 di = sqrt((Yi-Yo)^2+(Xi-Xo)2) dmax = max(di) Dmax = 0.34 inches Di = Dmax(di/dmax) Ri = fRn(1-exp(-10Di))0.55

    Equations variables: Pu lo and mo

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