GSA2003标准规范下载简介
GSA2003 美国联邦总务署 新联邦办公楼和重大现代化项目的连续倒塌分析和设计指南2003ZA = 38.27 25
= 38.27 25
×××桥施工组织设计y=144Aa.WH=41,331,600
TableC.1.Calculated A values for FaceA
Forareinforced concreteframe,C,=0.03,thus
Thebase shearfor FaceBis calculatedas
Step1.Using Figure C.4, determine Ln, where
T = C,(H)0.75 0.564 =115,113 (lb)=115kip onFaceA. 2.000T
T = C,(H)0.75 0.658 元y =98,668 (lb)=99kip on FaceA 2.000T
L therangesfrompointXtopointsn(1through25asdepicted in FigureC.4)(roundedtothenearestfoot) SD 二 minimum defended standoff distance (ft) H total height of building (ft) W 三 width of considered face (ft)
Step 2.Using Table B.1, select the An values consistent with the ranges (Ln) intheStep 1
Step 2.Using TableB.1, select the An values consistent with the ranges (Ln)determined in the Stepl
Thecalculated Ln and An values are summarized in Table C.2,resulting ina tota Avalue
Calculatethe averageAvalue,Aave,using e
Step 4.Calculate, using equation B.2
区A= 845.39
ZAn = = 33.8 25
y=144AaweWH=18,252,000
Table C.2. CalculatedAvaluesforFaceB
Step5.Determine the required base shear value for resisting abnormal loads using equation B.3.
For a steel frame (Ct=0.035 and T =0.658).
V,= 元 (1b) 2,000T T = C,(H)0.75 = 0.564 (as previously calculated)
=50,834(lb)51kiponFaceB 2.000T
FigureC.3.Performbaseshearcalculationforbothfaces(A&B)ofthebuilding
FigureC.4.Generalizedgeometryparametersforbase shearcalculation
Buildingparameters
leterminepreliminarycolumnsizesasdescri lowingexample:
Theprocedurewillbe demonstratedforLoad Level2inbothmajordirectio
Step1.From Figure C.6, determine values of Ln.where,
Step2.UsingTableB.l, select theAnvalues consistent withthe ranges,Ln,determined inStep1.
culatedL,andA,valuesforFaceAaresumma
Table C.3.Calculated Ln and A,values for FaceA
Theremainingstepsshouldbeperformedindependentlyforeach storylevel(n).The examplewilldemonstratethestepsforstorylevelone StoryLevel l:
remainingstepsshouldbeperformedindependentlyforeachstorylevel(n).The ple will demonstrate the steps for story level one
culateAsn using equation B.4 (adjust for actual
Step5.Calculatetheadjustedbentstory stiffness(K2n)usingequationB.6.
WI =Unitweight assumed for equationB.5 derivation =100psf w2n =Adjustedunitweightforstory1=126psf
TA = 126 psf where,
K,=16.5595A22 =109,157(lb/in）
= 86,633 (lb/in) W2
actualunitweight=70psf Tributary plan area of bent = 30 ft x 75ft =
eC.6.Distanceparametersneededfordeterm
Step6.Oncetheadjustedbent storystiffnesshasbeencalculated,therequired column stiffness canbedeterminedbyusingequationB.7
K.al. =K,, /N =14,439
Step7.Therequiredmoment of inertiaforeach columncanthenbe calculated using equation B.8.
Forreinforcedconcrete,Ecanbedetermineda
Keol ,H (in*) col, 12E
Hn=Storyheight=168inches E=modulus ofelasticity(psi)
E=wl533/f=3,834,254ps
Kcol,H. =1488 in* col, 12E
Usingthismomentof inertiavalue,thefollowingequationcanbeusedfor sizingreinforcedconcretecolumns:
where, b=column width (in) d = column effective depth (in) p= positive (and negative) reinforcing ratio
I col n = [5.5p +0.083] (in4)
Foronepercent steel (p=0.01)and a squarecolumn,b=d=12 inches.
For Face B
Step2.Using Table B.1, select the An values consistent with the ranges, Ln,determined in Step1.
calculatedLnandAnvaluesaresummarizedin
Table C.3.Calculated LnandAvalues forFaceA.
Theremainingsteps shouldbeperformed independentlyforeachstoryley (n).The example will demonstrate the steps for story level four.
eAsn using equation B.4(adjust for actual bay
Step5.Calculatetheadjustedbentstorystiffness(K2n)usingequatio
Ki.=16.5595A2 = 11,551 (lb/in)
Ki. = 16.5595A32
W K. = 9,167 (lb/in) W2n
K. = 9,167 (lb/in) W2,
vhere, wi = Unit weight assumed for equation B.5 derivation = 100 psf
TA = 126 psf
Wan=actual unitweight=70psf (psf) TA = Tributary plan area of bent = 15 ft x 150 ft = 2250 ft?
actual unitweight = 70 psf (psf) Tributary plan area of bent = 15 ft x 150 ft =
Step6.Oncetheadjustedbentstory stiffnesshasbeencalculated,therequiredcolumn stiffness canbe determined by using equationB.7.
Kcol.=Kzn/N=1,528
N=numberofcolumnsinbent=6(75/15+1)
Step7.The required moment of inertia for each column can then be calculated using equation4.8.
Hn=Storyheight = 144inches E modulusof elasticity=3,834,254psi Therefore.
KeolnH, (in*) col, 12E
KcolnH, = 99.2 in 12E
Usingthismoment of inertiavalue,thedimensionscanbedeterminedfrom
bd [5.5p +0.083] (in*) col n
rcent steel (p=0.01)and a squarecolumn,b=
Appendix D. Structural Steel Connections
TableD.1.Steelmomentframeconnectiontypes
Proprietary
FR =Fully Rigid Moment Connection
R= Fully Rigid Moment Connec
(a) WUF FullyRigid Connection
(b) Welded Flange Plate
(d) Bolted Flange Plate
(f) Top and Bottom Haunch
igureD.1.Fullyrigidmomentconnections
(g) Reduced Beam Section FigureD.1.Fully rigid moment connections (continued)
(a) Bolted or Riveted Angle
GB／T 41979.2-2022标准下载FigureD.2.Partiallyrigid moment connectio
(a)FullyRigid Connectior Figure
Proprietary Moment Frame Connections
ProprietaryMomentFrameConnection
b)TypicalShearOnlyConnection
FigureD.3Weak axis connections!
高大模板施工方案 （106P）.docFigureD.4.SidePlateTM moment connection system
FigureD.5.SlottedWebTMmomentconnectic