受弯构件的斜截面承载力.ppt

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,4 Flexural Strength of Reinforced Concrete Members,with Flexure and Shear,受弯构件正截面承载力计算,Wan Shengwu,Department of Civil Engineering ,WUST,2011.09,CONCRETE STRUCTURAL FUNDAMENTALS,混凝土结构设计原理,4,Flexural Strength of RC,Members with Flexure and Shear,Contents,4.1,Preface,概述,4.2 Basic Detailing of RC Members with Flexure and Shear,受弯构件基本构造要求,4.3 Analysis for Normal Section of Flexural Members under Loading,受弯构件正截面受力过程分析,4.4 Calculation for Normal Section of Flexural Members,受弯构件正截面承载力计算,4.5 Calculation for Rectangular Section Beam with Tension Reinforcement only,单筋矩形截面受弯构件正截面承载力计算,4.6Rectangular Beam with Compression Reinforcement,双筋矩形截面受弯构件正截面受弯承载力计算,4.7 T-Section Beam with Tension Reinforcement,only,T,形截面受弯构件正截面受弯承载力计算,E,mphases,Analyze of Flexural Strength,Study on Test,Rectangular beam with tension reinforcement only,Rectangular beam with compression,reinforcement,Flanged section beam with tension reinforcement only,4.1,Preface,F,lexural,M,embers受弯构件：承受荷载作用下产生的弯矩,（,M,）,和剪力,（,V,）,的构件。,常见,Beam and slab,正截面Normal Section：垂直于轴线的截面。,受力特点：受弯构件在,M,作用下容易产生,正截面破坏,，在,M,和,V,共同作用下容易产生斜截面破坏。,受弯构件,设计问题,：,保证构件在,M,作用下不发生,正截面破坏,（本章）；,保证构件在,M,和,V,共同作用下不发生,斜截面破坏第5章）；,保证构件在荷载作用下能,满足正常使用,的要求（第9章）,本章要解决的问题,通过结构计算确定构件的,截面尺寸、纵向受力钢筋的数量以及满足相应的构造要求,和绘制施工图,。,F,lexural members,Beam,and,slab,4.1,Preface,Beam,4.1,Preface,Slab,4.1,Preface,4.2 Basic Detailing of,RC Members,with Flexure and Shear,受弯构件的基本构造要求,4.2.1 Detailing of Beam,（,1,）,Typs of Section,Rectangle section,4.2,Basic Detailing of,RC Members,with Flexure and Shear,T-section,、,I,-section,、,Coss,-section,O,thers,L-section、,-section、-section、Basket-section,（,2,）,Size of section,满足承载力和刚度要求,Depth of beams section h,：按梁的连接方式不同，依据跨度确定。,Example，简支梁,h=L/12,，悬挑梁,h=L/6,，框架梁和连续梁另外有规定。为施工方便，截面高度应符合模数，h=200，250，300，等，以50mm模数递增；当h800mm时，以100mm模数递增。,Width of beams section b,：h确定后，b按h/b确定。,Rectangle section：b=（1/21/4）h；T-section：b=（1/2.51/4）h,梁宽b同样应符合模数要求，b=150，200，250，等，以50mm模数递增，特殊情况，如圈梁b可以取为墙体宽度,。,4.2,Basic Detailing of,RC Members,with Flexure and Shear,（,3,）,Steel bar in Beam,Longitudinal main bar,Longitudinal Reinforcement with,Tension,：布置在梁的受拉区，抵抗,W,在正截面产生的拉应力；,Longitudinal Reinforcement with,Compression,：布置在梁的受压区，协助砼抵抗,W,在正截面产生的压应力。,设计要求,：普通钢筋砼梁中一般采用直径,1025mm,之间的钢筋，级别,HRB335,（,II,）、,HRB400,（,III,）。但对于承力构件，受拉钢筋,d,宜大于,12mm,，受压钢筋宜大于,14mm,。同一跨度的梁中间可以采用直径不同的钢筋，但直径种类不宜超过,2,种。对于受拉纵筋宜采用直径较小的钢筋，以利于抗裂度要求。,4.2,Basic Detailing of,RC Members,with Flexure and Shear,Stirrup and Bent up of main bar,抵抗剪力作用，下章介绍。,Longitudinal Tie rod,作用：,固定箍筋的正确位置并形成钢筋骨架，保证施工时必要的刚度；抵抗可能有的比较小的弯矩和温度变化产生的裂缝。,设置：,按构造要求设置，一般设置两根，置于截面受压区域角部。直径按照跨度确定，一般不小于10mm。如果截面配置有受压钢筋，架立筋不再单独另设，受压钢筋可以兼做架立筋。,4.2,Basic Detailing of,RC Members,with Flexure and Shear,Longitudinal Tie rod at sides of Beam,设置：,当矩形截面梁高h或T形截面梁肋高,h,w,450mm,，在梁两侧应设置直径不小于10mm的构造纵筋，钢筋截面面积为,bh或bh,w,的0.1%,，间距不超过200mm，两侧对称布置。,作用：,抵抗温度变化和砼收缩产生拉应力，阻止在梁腹产生过大的竖向裂缝。,4.2,Basic Detailing of,RC Members,with Flexure and Shear,截面有效高度：,h,0,=h-a,s,（,4,）,Distance of Reinforcement in Beam,4.2,Basic Detailing of,RC Members,with Flexure and Shear,Typs of Section,矩形、空心板、折板、槽型板、双,T,形板。,（,1,）,Depth of Slab h,：,满足承载力和刚度要求，按跨度确定。如，单跨简直板，,h=L/30,，悬挑板,h=L/12,。板厚,h,同样应符合模数要求，,b=60,，,70,，,80,，,等，以,10mm,模数递增。,4.2.,2,Detailing of,Slab,4.2,Basic Detailing of,RC Members,with Flexure and Shear,（2,）,Steel bar in Slab,：,Main Reinforcement,：数量按计算确定，常采用,HPB300,级，也可采用,HRB335,级钢筋、冷轧钢丝或冷拔低碳钢丝。当采用,HPB300,级钢筋时，直径不应小于,6mm,，常用,6,、,8,、,10,、,12mm,。,Distribution Reinforcement,：置于受力钢筋内侧。,作用：,a.,固定受力钢筋的正确位置，将板中荷载更加均匀的传递给受力钢筋；,b.,抵抗温度变化或砼收缩产生的、垂直于受力筋方向的拉应力，提高板的抗裂度。,设置：,单位长度上分布钢筋的截面面积不宜小于单位宽度上受力钢筋截面积的,15%,且不宜小于该方向板截面面积的,0.15%,。,4.2,Basic Detailing of,RC Members,with Flexure and Shear,（,3,）,Distance of Main Reinforcement in Slab,受力钢筋间距：不宜过小（不方便施工），至少,70mm,，也不宜太大（受力不均匀）。,一般，当板厚,h150mm,时，间距,200mm,；,当板厚,h150mm,时，间距,1.5hmm,；且不宜大于,250mm,。,4.2,Basic Detailing of,RC Members,with Flexure and Shear,4.2.3 Thickness of Protective Covering,含义：钢筋外边缘至砼表面的距离。,（,1,）,作用,：保证钢筋和砼的粘结；保护钢筋不发生锈蚀和有害介质的侵蚀；遇到火灾时使构件中钢筋升温缓慢。,（,2,）,确定原则,：防火要求和使用环境。,（,3,）,厚度,C,：纵向受力的普通钢筋和预应力钢筋，其砼保护层厚度不应小于钢筋的公称直径，且应符合,混凝土结构设计规范,（,GB50010-2010,）的规定要求。,4.2,Basic Detailing of,RC Members,with Flexure and Shear,4.2.4 Ratio of Tension Reinforcement,定义,：纵向受拉钢筋总截面积（,As,）与梁截面积,bh,0,（,T,形和,I,形截面应扣除受压翼缘面积）的比值，称为纵向受拉钢筋的配筋百分率，简称配筋率，用,表示。,反映截面钢筋的含量，也是影响梁受力性能的一个重要指标。,4.2,Basic Detailing of,RC Members,with Flexure and Shear,4.3,Analysis for Normal Section of Flexural Members under Loading,受弯构件正截面受力过程分析,回顾材料力学知识，对于完全弹性材料的简支梁，受到荷载作用，截面性能：,应变,应力,挠度,砼不是完全弹性材料，也不是理想的塑性材料，而是具有弹性和塑性性能的,弹塑性体,，因此不能完全依靠材料力学知识，必须结合砼梁受力试验，通过对实验分析来找出截面的,应力应变,关系，,M和f,的关系。,4.3,Analysis for Normal Section of Flexural Members under Loading,Moment,diagram,Shear,diagram,Test beam,Failure of normal section:vertical crack in mid-span of beam,M,Failure of inclined section:inclined crack near the support of beam;M,、,V,Deflection and crack width,Other,4.3.1,Study on Test,dial indicator,Survey points,of,strain,dial indicator,displacement meter,Model of Test,Moment diagram,Shear diagram,Test beam,Behavior under load:,Load Deflection,Stress Strain,Influence factors,4.3,Analysis for Normal Section of Flexural Members under Loading,4.3,Analysis for Normal Section of Flexural Members under Loading,Actually measure diagram of,D,eflection,（,M-f,关系曲线）,特点：,转折点,a,：,弹性阶段消失，裂缝出现,转折点,b,：,钢筋屈服,顶点：,最大弯矩（极限弯矩）M,u,Data processing of Test,Actually measure diagram of,Actually measure diagram of,strain distribution,实测钢筋应力,（,M-,s,关系曲线）,4.3,Analysis for Normal Section of Flexural Members under Loading,实测截面应变分布,4.3,Analysis for Normal Section of Flexural Members under Loading,根据这三个特征点，,为了分析方便,将曲线划分成三个阶段：,第I阶段：,弹性阶段,荷载小，M和f都小，M和f变化快慢相同，M、f成比例。,第,II,阶段：,正常使用阶段,裂缝出现，并随着荷载的增加而加宽加长，挠度也随荷,载的加大而加大但是增加量不同，挠度变化快，M变化慢。由于荷载的增加量主要在这一阶段，故称之为梁带裂缝的工作阶段。,第,III,阶段：,破坏阶段,荷载增加不是很多，但是挠度急剧增加，裂缝宽度也不断加大，裂缝和挠度的加大已不再适合继续加载不适合使用。实际上已开始破坏。,4.2.1 Three Working Stages of under,-,reinforced Beam,4.3,Analysis for Normal Section of Flexural Members under Loading,4.3,Analysis for Normal Section of Flexural Members under Loading,（,1,）,Stress Distribution of Normal Section,截面应力分布,试验可测得应变，不能测应力。利用材料力学知识，借助砼受压应力-应变曲线，仿射截面应力分布，推断应力分布图,Stage,I,:,弹性阶段,，三角形分布,Stage,I,e,:裂缝出现，砼退出工作，弹性消失，进入第II阶段。转折点,a,对应的弯矩称为开裂弯矩（,M,cr,）。此阶段应力图形作为,抗裂度,计算,依据,。,4.3,Analysis for Normal Section of Flexural Members under Loading,Stage,II,:裂缝已经出现，钢筋承担拉力，随着荷载的加大，钢筋的拉应力也逐渐增加，裂缝会加宽加长，并向受压区延伸。随着弯矩的增加，,受压区砼表现出塑性性能,，应力图形为曲线。,第,II,阶段是带裂缝工作的，而且随着弯矩增加，挠度,也,在增加。此阶段为荷载增幅较大阶段，代表梁的正常工作应力状态，使用阶段裂缝宽度和挠度的控制应该处在这一阶段，因此，该阶段的应力图形作为,裂缝宽度和挠度验算的依据,。随着荷载的继续增加，钢筋的拉应力达到屈服应力，钢筋屈服，到达,II,e,阶段（即转折点,b,，对应弯矩称为屈服弯矩,M,y,），随后进入第,III,阶段。,4.3,Analysis for Normal Section of Flexural Members under Loading,Stage,III,:荷载继续增加，弯矩加大。钢筋屈服后，抵抗拉应力能力明显减弱，裂缝迅速加宽加长，挠度急剧增加，受压区高度减小。由于内力臂加大，产生抵抗弯矩的受压区面积减小，致使受压区砼压应力增加，当砼的压应力超过了极限压应力时，砼被压碎，梁宣告破坏。,砼被压碎时，,受压边缘,实测应变结果（0.0030.004）比轴心受压极限压应变（,0,=0.002）大，主要原因是,边缘砼未受约束，靠近中和,轴,砼受约束，,压,区,砼受压处于,非,均匀受力状态,。规范对压区砼极限压应变取为,cu,=0.0033,，,据,此，可以推断,压区,砼的,极限,压应力,cu,超过了轴心受压时的,极限压应力,（,f,c,）,。对应峰值,（转折点,c,）,的弯矩称为极限弯矩,Mu,。,第,III,e,阶段的应力图形是,承载力计算依据,。,4.3,Analysis for Normal Section of Flexural Members under Loading,（,3,）,Strain Characteristic of Normal Section,截面应变特征,弯矩很小时呈三角形分布，裂缝出现后，近似三角形，应证了平截面假设。因裂缝出现的原因，截面中和轴随弯矩的增加而上升,Actually measure diagram of,strain distribution,4.3,Analysis for Normal Section of Flexural Members under Loading,4.,3.3,Three Failure Modes of Normal section of RC Beam,（,1,）,Under-reinforced Beam,4.3,Analysis for Normal Section of Flexural Members under Loading,The tensile steel reaches,fy,before the concrete reaches its maximum capacity.,钢筋配置合适的梁，上述试验就是适筋梁。,破坏特点：,破坏始于受拉区钢筋的屈服，然后经历一个较长的过程，直到受压边缘砼压应变达到极限值，砼被压碎，才宣告梁的破坏。裂缝有一个充分发展、挠度有一个充分增大的过程，破坏前有征兆，属于,延性破坏,或者,塑性破坏,。适筋梁中钢筋和砼的强度都得到充分的利用,。,tension failure or plastic failure ductile factor=(,f,u,-,f,y,)/,f,y,4.3,Analysis for Normal Section of Flexural Members under Loading,The compression concrete reaches its maximum capacity,before the tensile steel reaches,f,y,.,Beam fails suddenly without warning.,compression failure,or,brittle failure,4.3,Analysis for Normal Section of Flexural Members under Loading,（,2,）,Over-reinforced Beam,Not allowed in design,The compression concrete does not reach its maximum capacity,while,the tensile steel gets its harden stage or fracture.,large crack,brittle failure,4.3,Analysis for Normal Section of Flexural Members under Loading,（,3,）,Scarce-reinforced Beam,Not allowed in design,KEYS,Three Working Stages of under reinforced Beam,Three Failure Modes of Normal section of RC Beam,1.Under-reinforced,2.Over-reinforced,3.Scarce-reinforced,4.2 Study on Test,4.,4,Flexure Strength Calculation of RC Beam,4.,4,.1,Basic Assumptions,（,1,）,A plane section before bending remains a plane section after bending.,（,2,）,In calculating the ultimate moment capacity of a beam,the tensile strength of concrete is neglected.,（,3,）,The steel is assumed to be uniformly strained to the strain that exists at the level of the centroid of the steel.,s,=E,s,s,f,y,su,=0.01,s,s,0,0.01,y,f,y,4.4,Flexure Strength Calculation of RC Beam,（,4,）,The compressive stressstrain relation of concrete:,4.4,Flexure Strength Calculation of RC Beam,4.,4,.2Equation for Normal Section of Flexural Members of Under-reinforced Bea,m,Actual Compressive Stress Block,Rectangular Equivalent Compressive Stress Block,4.4,Flexure Strength Calculation of RC Beam,等效条件,Compressive force C equality,Force C action position y,c,unchangeable,4.4,Flexure Strength Calculation of RC Beam,按等效矩形应力图形计算,按实际应力图形计算,钢筋拉力：,T,=,f,y,A,s,砼压力：,实际计算,按等效应力矩形,由,X=0,T,-C=0,f,y,A,s,=,1,f,c,bx,由,M,AS,=0,M,u,=,1,f,c,bx,（,h,0,-0.5x,）,或者 由,M,c,=0,Mu=,f,y,A,s,（,h,0,-0.5x,）,x,受压区计算高度,equivalent rectangular compressive stress factor,equivalent rectangular compressive zone factor,混凝土,强度等级,C50,C55,C60,C65,C70,C75,C80,0.8,0.79,0.78,0.77,0.76,0.75,0.74,1.0,0.99,0.98,0.97,0.96,0.95,0.94,4.4,Flexure Strength Calculation of RC Beam,4.4,Flexure Strength Calculation of RC Beam,4.4.3BalancedDepthofCompression,ZoneofSection,x,b,andMaximumReinforcementRatio,ma,x,界限受压区高度,x,b,与最大配筋率,（,1,）,Balanced Failure,Referring again to Figure,if there were just enough steel to put the neutral axis at location where the yield strain in the steel and the maximum concrete strain,of,0.0033,existed at same time,the cross section would be said to be,balanced,Failure,.,（,2,）,Balanced depth of compression zone of section,x,b,相对界限受压区高度仅与材料性能有关，与截面尺寸无关。,4.4,Flexure Strength Calculation of RC Beam,（,3,）,Balanced Moment,M,b,(界限弯矩),（,4,）,Balanced Reinforcement Ratio,max,(,最大,配筋率),为避免超筋，设计时必须：,max,4.4,Flexure Strength Calculation of RC Beam,4.4.4,Minimum Reinforcement Ratio,min,（,1,）,min,确定方法,（,2,）,混凝土结构设计规范,（,GB50010-2010,）规定的限值见第,8.5.1,条。,对于受弯构件,min,=0.2%，0.45ft/fy,max,基础底板可适当降低，,min,=0.15%,设计时,为,避免超筋,必须满足配筋率,min,4.4,Flexure Strength Calculation of RC Beam,4.3 Flexure Strength Calculation of RC Beam,KEYS,Basic Assumptions,Equivalent Stress Distribution,Balanced Failure,Limitations,4.,5,Calculation for Rectangular Section Beam with Tension Reinforcement only,（,1,）,Basic Formula,4.,5,.1.Basic Formula and its Limitation,4.,5,Calculation for Rectangular SectionBeam with Tension Reinforcement only,（,2,）,Limitations,Avoiding over-reinforced failure,Avoiding scarce-reinforced failure,4.,5,Calculation for Rectangular SectionBeam with Tension Reinforcement only,4.5.2,Application of Basic Formula,两类问题：,截面设计,和,截面复核,（,1,）,Design of Section,With know:,M,、,f,c,、,f,y,、,b,、,h,Calculation:A,s,Steps,：计算,h,0,h,0,=h-a,s,计算,x,判别适用条件，如果,x,x,b,=,b,h,0,，,截面处于适筋状态,如果,x,x,b,=,b,h,0,，,截面处于超筋状态，改变,bh,或提高材料强度,计算,As,As,=,1,f,c,bx/f,y,验算最小配筋率,=A,s,/(bh,0,),min,=0.2%,，,0.45f,t,/f,y,max,，,如果满足要求，选择钢筋直径和根数；如果不满足，则按下式,重求,A,s,=,min,bh,0,，然后选择钢筋直径和根数。,4.,5,Calculation for Rectangular SectionBeam with Tension Reinforcement only,也可按下述步骤进行由,x=,h,0,代入基本平衡方程并,解联立方程,令,计算步骤见下面框图,Begin,Read:,IF,T,F,Over-,reinforced,IF,END,T,F,Scare-,reinforced,4.,5,Calculation for Rectangular SectionBeam with Tension Reinforcement only,4.,5,Calculation for Rectangular SectionBeam with Tension Reinforcement only,Example 4.1,A beam has the cross section,bh,=250650mm.The steel is grade HRB335.It is to carry design moment of 152.0kN.m.The concrete strength is grade C30.The beam is to be,laied,on the first environment.Determine amount of steel required.,Solution:,Calculate h,0,：,If the steel bar is arranged in single row,then,h,0,=h-a,s,=650-35=615mm,Calculate x,is under-reinforcement beam,，,Calculate tension bars,4.,5,Calculation for Rectangular SectionBeam with Tension Reinforcement only,Check,rinforcement,ratio,from,min,=0.2%,，,0.45f,t,/f,y,max,=,0.2%,，,0.21%,max,=,0.21%,min,ratio satisfied,Choose numbers of steel bars,：,Select 3 20,（,A,S,实,=941mm,875.9mm,），,satisfied to carry.,b,min,=320+254=160mm,min,ratio satisfied,Choose numbers of steel bars,：,Select 3 25,（,A,S,实,=1471mm,1445.6mm,），,satisfied to carry.,b,min,=325+254=175mm,x,b,=,b,h,0,，,截面处于超筋状态，改变,bh,或提高材料强度,极限承载力,取,x=,x,b,，,极限承载力,如果,MM,u,，则构件安全，否则不安全。,Beginning,Read:,Scare-,reinforced,Over-,reinforced,END,T,T,F,F,4.,5,Calculation for Rectangular SectionBeam with Tension Reinforcement only,计算步骤见下面框图,4.,5,Calculation for Rectangular SectionBeam with Tension Reinforcement only,Example 4.3,A simple supported beam with rectangular section has size,bh=200500mm,.,The calculation span is l=5.5m.There are,3,22(,As=1140mm,),tension steel bars in it.Use,C25,cretet,and,HRB335,steel.,The,beam is to be,laied,on the first environment.Determine design line load to carry,q=？,Solution:,Check,rinforcement,ratio,min,=0.2%，0.45f,t,/f,y,max,=0.2%，0.19%,max,=0.2%,ratio satisfied,400N/mm,）作为抗压钢筋，则抗压强度设计值只能取为400N/mm,。,4.6.3,Basic Formula and its Limitations,4.,6,Rectangular,B,eam with,C,ompression,R,einforcement,(1),Basic,Equation,4.,6,Rectangular,B,eam with,C,ompression,R,einforcement,(2),Limitations,Avoiding over-reinforced failure,To ensure yielding of compression steel before crash of concrete:,4.,6,Rectangular,B,eam with,C,ompression,R,einforcement,4.6.3 Application of Basic Formula,(1),DesignofSection,Case 1,With know:,Calculation:,IF,x,x,b,=,b,h,0,Set,x=,x,b,=,b,h,0,Why?,There are three variables As,As and x unknown in two coupled equations.To minimize the total steel area,we take,x=,x,b,=,b,h,0,or,Design of Section,(Case 1),4.,6,Rectangular,B,eam with,C,ompression,R,einforcement,Begin,Read:,END,Set=,b,There are three variables As,As and x unknown in two coupled equations.To minimize the total steel area,we take=,b,.,IF,Why?,4.,6,Rectangular,B,eam with,C,ompression,R,einforcement,Case,2,With know:,Calculation:,两种途径解决,一种先将应力图形分解，然后叠加,另一种方法直接利用基本方程求解,4.,6,Rectangular,B,eam with,C,ompression,R,einforcement,双筋截面应力图形分解,4.,6,Rectangular,B,eam with,C,ompression,R,einforcement,求,A,s1,Case 2,按分解应力图形求解步骤,求,求,M,1,求,x,判别适用条件,：,如果,截面处于适筋状态,求,As,如果,取,（说明受压钢筋过多）,或者按单筋截面计算,如果,说明超筋，此时按,Case 1,求解,选择钢筋直径和根数,4.,6,Rectangular,B,eam with,C,ompression,R,einforcement,Case 2,直接利用基本公式求解步骤,求,x,判别适用条件,：,求,As,如果,截面处于适筋状态,如果,取,（说明受压钢筋过多）,或者按单筋截面计算,如果,说明超筋，此时按,Case 1,求解,选择钢筋直径和根数,It means that is too sufficient,that it cant reach its yield strength.,Case 2,Design of Section,（直接利用公式求解步骤框图