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2003

ZA = 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