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1、第八章受压构件,8.4Designequationsforrectangularsection,Designandcheckingofrectangularsectionwithnon-symmetricallyplacedbars,1、Largeeccentricitycompressionfailure(tensionfailure)2、Smalleccentricitycompressionfailure(compressivefailure),DesignandCheckingofRectangularsectionwithsymmetricallyplacedbars,第八章受压构件
2、,一、Designofrectangularsectionwithnon-symmetricallyplacedbars1、Largeeccentricitycompression,Given:sizes(bh)、thestrengthofthematerial(fc、fy,fy)、slendernessofthemember(l0/h)andthedesignvalueofaxialloadandmoment,ifheieib.min=0.32h0,Wecancalculateaslargeeccentricitycompressionfirst,第八章受压构件,8.4矩形截面偏心受压正截面
3、承载力计算,一、不对称配筋截面设计1、大偏心受压(受拉破坏),已知:截面尺寸(bh)、材料强度(fc、fy,fy)、构件长细比(l0/h)以及轴力N和弯矩M设计值,若heieib.min=0.32h0,一般可先按大偏心受压情况计算,BothAsandAsaretobedetermined.,Inthiscase,threeunknownsAs、Asandxaretobedeterminedfromtwoavailableformulae,sothattherearenouniquesolution.Similartodoublyreinforcementrectangularbeams,tom
4、akethearea(As+As)minimum?ie.x=xbh0,第八章受压构件,IfAs0.002bh?TakeAs=0.002bh,andconsiderAsisgiven,IfAsrminbh?TakeAs=rminbh.,WhenAsisgiven,WhenAsisgiven,twounknown(Asandx)canbesolvedfromthetwoavailableequationsuniquely。xcanbesolvedfromthesecondformulafirstly,ifx2a,then,Ifxxbh0?,IfAsrminbh?TakeAs=rminbh.,第八章
5、受压构件,Asmustbecalculatedasthefirstcaseagain,Thatwecanassumex=2a,decidedAsasbellow,Ifx2a?,As为已知时,当As已知时,两个基本方程有二个未知数As和x,有唯一解。先由第二式求解x,若x2a,则可将代入第一式得,若xxbh0?,若As若小于rminbh?应取As=rminbh。,第八章受压构件,则应按As为未知情况重新计算确定As,则可偏于安全的近似取x=2a,按下式确定As,若x2a?,As为已知时,当As已知时,两个基本方程有二个未知数As和x,有唯一解。先由第二式求解x,若x2a,则可将代入第一式得,若x
6、xbh0?,若As若小于rminbh?应取As=rminbh.,第八章受压构件,则应按As为未知情况重新计算确定As,则可偏于安全的近似取x=2a,按下式确定As,若x2a?,As为已知时,当As已知时,两个基本方程有二个未知数As和x,有唯一解。先由第二式求解x,若x2a,则可将代入第一式得,若xxbh0?,若As若小于rminbh?应取As=rminbh。,若As若小于rminbh?应取As=rminbh。,第八章受压构件,则应按As为未知情况重新计算确定As,则可偏于安全的近似取x=2a,按下式确定As,若x2a?,2、Smalleccentricitycompressioncolumn
7、sectionheieib.min=0.32h0,ThreeunknownsAs、Asandx,aretodeterminedfromtwoavailableEqs.,第八章受压构件,Smalleccentricitycompressioncolumnsection,thatxxb,ss-fy,andAsnotreachyieldstrength.So,whenxbx(2b-xb),Aswillnotreachyieldstrengthinanyway,TomakethereinforcementisminimumwetakeAs=max(0.45ft/fy,0.002bh)。,第八章受压构件
8、,2、小偏心受压(受压破坏)heieib.min=0.32h0,两个基本方程中有三个未知数,As、As和x,故无唯一解。,小偏心受压,即xxb,ss-fy,则As未达到受压屈服因此,当xbx(2b-xb),As无论怎样配筋,都不能达到屈服,为使用钢量最小,故可取As=max(0.45ft/fy,0.002bh)。,第八章受压构件,Besides,iftheeccentricityisverysmallandtheadditionaleccentricityisintheoppositedirectionofloadeccentricity.theconcreteinthesideofAswil
9、lbefailurefirstInthiscasetheentiresectionmaybeundercompression,accordingthedistributionofstressintheFig,takethemomentaboutthepointofAs.,e=0.5h-a-(e0-ea),h0=h-a,第八章受压构件,另一方面,当偏心距很小时,如果附加偏心距ea与荷载偏心距e0方向相反,则可能发生As一侧混凝土首先达到受压破坏的情况。此时通常为全截面受压,由图示截面应力分布,对As取矩,可得,,e=0.5h-a-(e0-ea),h0=h-a,第八章受压构件,WhenAsisde
10、termined,thereareonlytwounknownxandAs,thentheanswerisuniqueaccordingx,therearethreecase.,Ifx(2b-xb),ss=-fy,thebasicequationischangedtotheonebellow,,Ifxh0h,takex=h,anda=1,usingthebasicEq.solveAs,第八章受压构件,xandAsmustbedeterminedagain,确定As后,就只有x和As两个未知数,故可得唯一解。根据求得的x,可分为三种情况,若x(2b-xb),ss=-fy,基本公式转化为下式,,若
11、xh0h,应取x=h,同时应取a=1,代入基本公式直接解得As,第八章受压构件,重新求解x和As,ThewaytosolvexandAsiscomplicated.Usingtherelativecompressiondepthx,,Whenintherangex=xb1.1,,第八章受压构件,FortheconcreteC50andthereinforcement,asisbetween0.40.5,take0.45,thechangeofas=x(1-0.5x)issmall。,由基本公式求解x和As的具体运算是很麻烦的。迭代计算方法用相对受压区高度x,,在小偏压范围x=xb1.1,,第八
12、章受压构件,对于级钢筋和C50混凝土,as在0.40.5之间,近似取0.45,as=x(1-0.5x)变化很小。,As(1)的误差最大约为12%。如需进一步求较为精确的解,可将As(1)代入基本公式求得x,,第八章受压构件,取as=0.45,试分析证明上述迭代是收敛的,且收敛速度很快。,二、Checkingofrectangularsectionwithun-symmetricallyplacedbars,Whenthesizeofthesection(bh)、thereinforcementAsandAs、thematerialstrength(fc、fy,fy)andtheslendern
13、essofthemember(l0/h)areknown,accordingtotheorderoftheaxialandthemomentactonthemember,therearetwocase1、ThedesignvalueNisgiven,tofindtheultimatemomentM,第八章受压构件,二、不对称配筋截面复核,在截面尺寸(bh)、截面配筋As和As、材料强度(fc、fy,fy)、以及构件长细比(l0/h)均为已知时,根据构件轴力和弯矩作用次序,截面承载力复核分为两种情况:1、给定轴力设计值N,求弯矩作用平面的弯矩设计值M,第八章受压构件,二、Checkingofre
14、ctangularsectionwithun-symmetricallyplacedbars,Whenthesizeofthesection(bh)、thereinforcementAsandAs、thematerialstrength(fc、fy,fy)andtheslendernessofthemember(l0/h)areknown,accordingtotheorderoftheaxialandthemomentactonthemember,therearetwocase1、ThedesignvalueNisgiven,tofindtheultimatemomentM,第八章受压构件,
15、二、Checkingofrectangularsectionwithun-symmetricallyplacedbars,Whenthesizeofthesection(bh)、thereinforcementAsandAs、thematerialstrength(fc、fy,fy)andtheslendernessofthemember(l0/h)areknown,accordingtotheorderoftheaxialandthemomentactonthemember,therearetwocase1、ThedesignvalueNisgiven,tofindtheultimatemo
16、mentM,第八章受压构件,2、TofindtheultimateaxialforceNforagiveneccentricitye0,,二、不对称配筋截面复核,在截面尺寸(bh)、截面配筋As和As、材料强度(fc、fy,fy)、以及构件长细比(l0/h)均为已知时,根据构件轴力和弯矩作用次序,截面承载力复核分为两种情况:1、给定轴力设计值N,求弯矩作用平面的弯矩设计值M,第八章受压构件,2、给定轴力作用的偏心距e0,求轴力设计值N,二、不对称配筋截面复核,在截面尺寸(bh)、截面配筋As和As、材料强度(fc、fy,fy)、以及构件长细比(l0/h)均为已知时,根据构件轴力和弯矩作用次序,截面承载力复核分为两种情况:1、给定轴力设计值N,求弯矩作用平面的弯矩设计值M,第八章受压构件,2、给定轴力作用的偏心距e0,求轴力设计值N,1、TofindtheultimatemomentMforagivenaxialforceNForsizeofthesection、the