Structural Engineering

Draft 8.4Forces 161 Mext = Cd)C= Mext d (8.4-a) T=C = (808)k.ft (9:2)ft =  88 k (8.4-b) 8 Becausetheframeshape (andthusd(x)) is ...

Draft Chapter 9 DESIGN PHILOSOPHIES of ACI and AISC CODES 9.1 Safety Provisions 1 Structuresandstructuralmembersmustalways be de ...

Draft 9.3UltimateStrengthMethod 165  < all= yld F:S: (9.1) whereF:S:is thefactorof safety. 10 Majorlimitationsof thismetho ...

Draft 9.3UltimateStrengthMethod 167 Figure9.3:FrequencyDistributionsof LoadQandResistanceR Failurewouldoccurfornegative valuesof ...

Draft 9.3UltimateStrengthMethod 169 Type of Load/Member AISC DL+ LL;Members 3.0 DL+ LL;Connections 4.5 DL+ LL+ WL;Members 3.5 DL ...

Draft 9.4Example 171 9.4 Example Example9-1: LRFDvs ASD To illustratethedierencesbetweenthetwo designapproaches,letus considert ...

Draft Chapter 10 BRACED ROLLED STEEL BEAMS 1 Thischapterdealswiththebehavioranddesignoflaterallysupportedsteelbeamsaccord- ingto ...

Draft 10.2FailureModesandClassicationof SteelBeams 175         w w u M p M =(wL )/8p 2 M=(wL )/8 2 + + ...

Draft 10.3CompactSections 177 Figure10.6: Stress-straindiagramformoststructuralsteels 13 Whentheyieldstressis reachedat theextre ...

Draft 10.5SlenderSection 179 Draft         ),+-." 0 % !#"$%&"''(),+-/." 0 % 1 ' 2342 $ 5  5   67 89 : 89 ; ...

Draft 10.6Examples 181 ComputethefactoredloadmomentMu. Fora simplysupportedbeamcarrying uniformlydistributedload, Mu=wuL 2 =8 ...

Draft Chapter 11 REINFORCED CONCRETE BEAMS 11.1 FailureModesforR/CBeams. 1 Recallingthatconcretehasa tensilestrength(f 0 t) abou ...

Draft 11.1Introduction 185 11.1.3 Analysisvs Design 11 In R/Cwe always consideroneof thefollowingproblems: Analysis: Givena cert ...

Draft 11.2CrackedSection,UltimateStrengthDesignMethod 187 Figure11.3:CrackedSection,LimitState Figure11.4:WhitneyStressBlock ...

Draft 11.2CrackedSection,UltimateStrengthDesignMethod 189 11.2.3 Analysis GivenAs,b,d,f 0 c, andfydeterminethedesignmoment: ac ...

Draft 11.2CrackedSection,UltimateStrengthDesignMethod 191 Check equilibriumof forcesin thexdirection(Fx= 0) a= Asfs : 85 f 0 ...

Draft 11.3ContinuousBeams 193 Check equilibriumof forces: a= Asfy : 85 f 0 c b = (2:42)in 2 (40)ksi (:85)(3)ksi(11:5)in = 3: 3 ...

Draft 11.4ACICode 195 whereloadcombinationsshallincludebothfullvalueandzerovalueof L to determinethemore severecondition,and U= ...

Draft Chapter 12 PRESTRESSED CONCRETE 12.1 Introduction. 1 Beams withlonger spansarearchitecturallymoreappealingthan thosewithsh ...

Draft 12.1Introduction 199 If we considerthefollowing: Anunstressedsteelcableof lengthls A concretebeamof lengthlc Prestressthe ...