InputCompany NameProject:TRUSS TR23Engineer:YPProject #AddressDate:28-OctTelpSubject:SAMPLE CONNECTIONChecker:Page:Date:ECCENTRIC SHEAR CONNECTION ANALYSIS OF BOLT GROUPLast updated: 25-Apr-02The following calculations comply with LRFD Steel Design Manual 2nd EditionTheoryUnits:USDetailsProblem Description:Testing Spreadsheet vs LRFD tablesINPUTOUTPUTBoltGroupCopyright 2005Vertical ForceHorizontal ForceUltimate Shear =64.58kipDwi HermawanPy, kip-2Px, kip2Angle to horizon, b =-45.00degmain_line94@yahoo.comex, in-5ey, in0True eccentricity, e =8.357inSolved !Bolt description:A325 SH, 3/4" Dia.Single bolt shear capacity fRn =15.9kip174.9Bolt LocationBoltXY##inin10020330640953063373683996010631166000000000000000000000000000000000000000

Xb

Input05.68023568134.090909090903-2.4789071271690003609-2.4789071271036

BoltsInstantenious Center of rotationGroup CenterApplied ForceICCGPPxPy

DetailsProblem Description:Testing Spreadsheet vs LRFD tablesSolverCharting Applied ForceSRsin(q)/sin(d)-4.062SRcos(q)/cos(d)-4.062SR*d/(e+lo)4.062-50Maximum Force at connectionfP =64.58kipGoto:InputPsin(b)-2.872Pcos(b)2.87P*lo42.48-5-2.48Connection is concentrically loaded0TheorySRsin(q)2.87SRcos(q)-2.87SR*d42.48Must be ZeroDifference0.0000Difference-0.0000Difference-0.00000.0000-50Vertical Force Py =-2.00kip Eccentricity ex =-5.00inAngle b =-0.785radBestGuess-2.520Horizontal Force Px =2.00kip Eccentricity ey =0.00inEccen. e =8.357inSingle bolt capacity multiplierPo4.0624.062Adjusted b =-0.785radDist. From 0,0 to Inst. Centerlo2.10in2.102-50Bolt ParmetersAdjusted e =8.357inSideways dist. to Inst. Centermo0.146in-2.52-2.48fRn =15.90kipInstant. Center (IC)Bolt Group Center (CG)Rmax=15.61kipXoYoXcYcdmax =7.01in4.1105.6802.7274.091S(Rsinq)S(Rcosq)S(R*d)Bolt LocationBolt forceangle tohorizonBolt to ICDistanceBoltShear ForceBoltDispl.45.67-45.67675.44Bolt ##XYqdRD, inRsinqRcosqR*dDefinition of Rangesininradinkipinkipkipkip-inBestGuess_lo=Details!$R$8100-3.777.0115.610.349.15-12.64109.42BestGuess_po=Details!$R$7203-4.134.9115.070.2412.63-8.2373.96betta=Details!$I$83061.494.1214.680.2014.631.1460.51BigX=Details!$B$19:$B$684090.895.2815.210.2611.849.5680.38ConcentricConnection=Details!$E$4530-3.335.7915.360.282.95-15.0888.92d=SQRT((X-Xo)^2+(Y-Yo)^2)633-3.532.9013.620.145.21-12.5939.52dmax=Details!$B$127361.291.169.980.069.592.7611.53e=Details!$I$98390.323.5014.230.174.5113.4949.80ex=Details!$F$6960-2.825.9915.410.29-4.87-14.6392.27ey=Details!$F$71063-2.533.2814.030.16-8.08-11.4646.00lo=Details!$P$81166-1.401.9212.060.09-11.892.0123.12mo=Details!$P$90000.000.000.000.000.000.000.00Mustbe0=Details!$R$50000.000.000.000.000.000.000.00Po=Details!$P$70000.000.000.000.000.000.000.00Print_Area=Details!$A$1:$R$460000.000.000.000.000.000.000.00Px=Details!$C$70000.000.000.000.000.000.000.00Py=Details!$C$60000.000.000.000.000.000.000.00Rn=(1-EXP(-10*d/dmax*0.34))^0.550000.000.000.000.000.000.000.00Rult=Details!$B$100000.000.000.000.000.000.000.00SumRcos=Details!$O$40000.000.000.000.000.000.000.00SumRd=Details!$Q$40000.000.000.000.000.000.000.00SumRsin=Details!$M$40000.000.000.000.000.000.000.00theta=ATAN2(X-Xo,Y-Yo)-PI()/20000.000.000.000.000.000.000.00Units=Input!$B$80000.000.000.000.000.000.000.00X=OFFSET(BigX,0,0,COUNT(BigX),1)0000.000.000.000.000.000.000.00Xc=Details!$F$120000.000.000.000.000.000.000.00Xo=Details!$D$120000.000.000.000.000.000.000.00Y=OFFSET(X,0,1)0000.000.000.000.000.000.000.00Yc=Details!$G$120000.000.000.000.000.000.000.00Yo=Details!$E$120000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.000000.000.000.000.000.000.000.00

Recalculate !eyexPxPyPeYRead DisclaimerHelpSolve !Reset SolverFor 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.InputTheory

TheoryECCENTRICALLY LOADED BOLT GROUPULTIMATE STRENGTH METHODInputDetailsInput Data:Spreadsheet Formulas:Shear force application: Px Py ex eyXc = AVERAGE(Xi)Bolt locations: Xi YiYc = AVERAGE(Yi)Single bolt shear capacity: fRnXo = -losin(b) - mocos(b) + XcYo = locos(b) - mosin(b) + YcEqulibrium equations:e = - (ey-Yc)cos(b) + (ex-Xc)sin(b)qi = atan((Yi-Yo)/(Xi-Xo)) - p/2(1)SRisin(qi) + Pusin(b) = 0di = sqrt((Yi-Yo)^2+(Xi-Xo)2)(2)SRicos(qi) + Pucos(b) = 0dmax = max(di)(3)SRidi + Pu(e+lo) = 0Dmax = 0.34 inchesDi = Dmax(di/dmax)Equations variables: Pu lo and moRi = 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.eyexPxPyPueYXloRidiCG (Xc,Yc)PrintbqiInputDetails