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3. AXIALLY LOADED MEMBERS - University of …

PageCIVL 4135 AXIALLY LOADED Members363. AXIALLY LOADED Reading Assignment:Section and Sections and of AXIALLY LOADED structural MEMBERS carry some moment in addition to axial load-- for this discussion, restrict consideration to axial load Reinforcement of Compression Plain concrete columns prohibited: possibility of bending is always present:ACI As/Ag :whereAs= Area of longitudinal reinforcement;Ag= Total area of column cross section; Possible column configurationa. Tied-- Deformed bars or wires placed normal to column axisLongitudinalRodsand lateral tiesDifferent tied arrangementsACI 4135 AXIALLY LOADED Spacing of Ties to Prevent Longitudinal Bar BucklingA.

page CIVL 4135 38 Axially Loaded Members b. Spiral: Circulararrangementoflongitudinalbarsconfinedbyacontinuouswirewhichspi-rals around the bars for the entire length of the member;

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1 PageCIVL 4135 AXIALLY LOADED Members363. AXIALLY LOADED Reading Assignment:Section and Sections and of AXIALLY LOADED structural MEMBERS carry some moment in addition to axial load-- for this discussion, restrict consideration to axial load Reinforcement of Compression Plain concrete columns prohibited: possibility of bending is always present:ACI As/Ag :whereAs= Area of longitudinal reinforcement;Ag= Total area of column cross section; Possible column configurationa. Tied-- Deformed bars or wires placed normal to column axisLongitudinalRodsand lateral tiesDifferent tied arrangementsACI 4135 AXIALLY LOADED Spacing of Ties to Prevent Longitudinal Bar BucklingA.

2 Tied column may fail prior to steel yield if shell spalls and longitudinal bars buckle;B. Insure that bar buckling load is greater than yield load. ( cr>fy)Assume that bar buckling load is greater than yield load -- Assume a pin--pin bar between ties:Pcr= 2 EIL2 The moment of inertia of a circular bar is:I= D464and for cr=Pcr/A = Pcr/( D2/4) cr= 2E16 (L D)2 Example:Forfy=40ksi = cr cr= 2E16 (L D)2 40= 2E16 (L D)2 Solving for critical buckling conditionL=21 DSo, to prevent buckling, space ties more closely than Code requires (ACI-02, Sect )that spacing not to be greater than16 D (D = Diameter of longitudinal bar);48 tie bar diameter;Least member Code requirements are given inACI-02 Sections 4135 AXIALLY LOADED Members38b.

3 Spiral:Circular arrangement of longitudinal bars confined by a continuous wire which spi-rals around the bars for the entire length of the member;Longitudinal Rodsand spiral hoopingc. Composite and combination:Structural steel member encased in concrete or steel pipe filled with concreteACI :at least 4 bars in tied columnsat least 6 bars in spiral columnsat least 3 bars in triangular Tied or spiral columns are used in order: to prevent buckling of longitudinal bars to prevent movement of longitudinal bars during of steel bars are sometimes used to prevent congestion. It is shown that they act as aunit with area as the same as all of the bundle buildings columns generally have proportions with the ratio of length to cross sectionwidth (L/h) in the range from about 8 to 12.

4 (use of high strength, more slender column becomingmore popular.)pageCIVL 4135 AXIALLY LOADED Design compatibility between steel and concrete (to prefect bonding; no slip; mechanical in-terlocking). LOADED Uniform strain over cross section;Axial load plus vary linearly;Moment Strains vary Concrete tensile strength usually The internal forces must be in equilibrium with applied external Plane cross section remains plane after application of Theory is based on real strain--stress 4135 AXIALLY LOADED Elastic behavior of column -- ExampleSee ACI section y= (fy=40ksi)Strains the same, stresses much differentStrain (in/in) Up tofc 12f cconcrete stress--strain approximately linear.

5 This is known as the working or service load range:2. For strain equality in this range: = c= s=fcEc=fsEsorfs=EsEcfclettingfs=nfcNote: n is generally rounded to the nearest whole 4135 AXIALLY LOADED Members41 Adopting the following notation:Ag= gross section area, (in2)As=areaofsteel(in2)Ac= net concrete area =Ag-- AsP= Axial force on columnThenP=Acfc+Asfs=Acfc+nfcAs=fc(Ac+n As)Transformed areaActual SizeAs2nAs2 Transformed Sec--tionAt=Ac+nAs(n 1)As2 Transformed Sec--tionAt=Ag+(n--1)AsTransformed section in axial compression(a)(b)(c)The three bars along each of the two faces arethought of as being removed and replaced, atthe same distance from the axis of the section, withadded areas of fictitious concrete of total amountofnAs.

6 Alternatively, as shown in figure c, we can think of the area of the steel bars as replaced withconcrete in which case one has to add to the gross concrete areaAgso obtained only (n --1)Asin orderto obtain the same total transformed , knowingAc=Ag-- AsP= fc(Ag+ (n--1)As)pageCIVL 4135 AXIALLY LOADED Example 112 12 Given4#8barsAssume:f c= 4000 psify=40ksiarea of steel = 4(area of # 8 bars)4( ) = (seeACI 318-- bar dimension table, or page iii of class notes)AsAg= (ACI-02 Sect. < < )What axial load will cause concrete to be at its maximum working stress?Solutionfc= (4000)/2 = 2000 psifs=(Es/Ec)fc=nfcEs= 29x106psi (always)Therefore n= (29,000,000/3,600,000) = 8P= fc(Ag+ (n--1)As)= 2000[144 + (8--1) ] = 332,000 lb = 332 kipsboth steel and concrete behaved , 000 4000 = 106psipageCIVL 4135 AXIALLY LOADED Example 2 For the previous example find the axial load P which produces c= s= ,016+91,640=429,655lbs=430kipsSteelP=fcA c+fsAs=2,400 (144 )+29,000 the slow rate curve of text on page 26, read stress in concrete for given strain of s< fs=Es s=29,000,000(psi) ,000psiSolutionthereforefc=2, Nominal axial load of columnPn.

7 (Pu= Pn) -- Greatest calculated loadA. Should occur when concrete stress peaks, steel reaches yield -- assume this Concrete stress will notbe f c: f cbased on test of standard cylinder; ends confined.; f cdepends on the rate of loading; Strength of actual column varies over length -- water migrates to top, causing top to beslightly weaker. use fc= cat nominal load conditionthenPN=Acfc+Asfs=Ac( c)+Asfyfor column of previous example:PN= (144 -- )( )(4000) + (40,000) = 605 kipspageCIVL 4135 AXIALLY LOADED Example 3 Consider a rectangular column subjected to axial compression. The material stress--strain relation-ships have been idealized as shown 24 PP5000 psiCONCRETE 0Ec=57,000f c psi yEs=29,000 (ksi)60 ksiSTEEL12--#14 the stress in the concrete and stress in the steel if the applied load is equal to the stress in the concrete and stress in the steel if the applied load is equal to the solution on the next 4135 AXIALLY LOADED Members45 Solution: f c=5ksi Ec= 57,000= 4030 ksi ( Sect of ACI) 12#14 bars As=27in2(from table ) Es= 29,000 ksi1.

8 Assume < 0or < elastic behavior for < yor < steel and concreteP=fcAc+fsAswhere Ac=Ag-- As=24x24 -- 27 = 549 in2P = 3100 = Ec cAc+Es sAsNote: = c= s(perfect bonding)3100 = (4030) (549) + (29000) (27) = 3100/(2,212,470 + 783,000) = < was we have:fs= Es= ,000 = ksifc= Ec= ,030 = ksi2. Assume > 0or > elastic behavior for < yor < and inelastic behavior for concreteP=fcAc+fsAs4050 = ( )x(549) + (29,000) (27) = > Assumption was correct (Concretebehaves inelastically). = < Assumption was correct ( Steel behaveselastically).fs= Es= ,000 = ksifc= ksi5000 pageCIVL 4135 AXIALLY LOADED Behavior of Spirally Reinforced ColumnsBehavior of Spirally Reinforced and Tied spiral reinforcement ratio based on tests by Richart, Brandtzeg and Brown -- 1928;(Univ.)

9 Of Illinois experimental bulletin no. 185).Using 6 x 12 cylinders, they related lateral confining pressure to axial capacity;Lateral Confining PressureAxialcapacityPsif cf*= c+ *c=Compressive strength of spirally confined core c= compressive strength ofconcrete if unconfinedf 2= lateral confinement stress in core concrete produced by spiralpageCIVL 4135 AXIALLY LOADED Members47 Spiral ColumnpageCIVL 4135 AXIALLY LOADED Members48B. What sort of lateral confinement can a given spiral provide?Consider a length of a spiral--wrapped circular section:for a length S :volume of spiral =Asp D(approximately)volume of concrete = ( D2/4)SLet s=volume of spiralvolume of concrete=4 AspDSCalculate equivalent confinement:f 2DS=2 Aspfysorf 2=( sfys)/2from previous research:f*c= + = + sfys2f*c= + is to insure that PB> , make sure spiral increases capacity of core enough to make up for loss of shell spalls: PN=Asfy+ c(Ag-- As)After shell spalls: PB=Asfy+(Acore-- As)( c+2 sfys)sAspDpageCIVL 4135 AXIALLY LOADED Members49 Set PB=PN, calculate like terms, expand.

10 Asfy+ Ag As=Asfy+Acore( +2 sfys) As( ) 2As sfysSmallIgnore the last term -- very smallthen; (Ag Acore)=Acore(2 sfys)solve for spiral reinforcement ratio we have: s= (Ag Acore)Acore(2fys)or s= fys AgAcore 1 conservatively, change to to get Eq. 10--6 ofACI-02: s= fys AgAcore 1 Eq. 10--6which says that the ratio of spiral reinforcement shall not be less than the value given by the equationabove; where fyis the specified yield strength of spiral reinforcement but not more than 60,000 Maximum Loads for Spiral ColumnPrior to spalling of shell: same as tied columnPA=PN=Asfy+ c(Ag-- As)after spalling of shell:PB=Asfy+(Acore-- As)( c+2 sfys)orPB=Asfy+(Acore-- As) c+(Acore-- As)(2 sfys)-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --pageCIVL 4135 AXIALLY LOADED Members50 The underlined term is the added capacity of the core resulting from the presence of the will sbe critical?


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