受压构件8CompressiveElementorColum.ppt

第八章 受压构件,8.1 轴心受压构件的承载力计算,Chapter 8 Compressive Element or Column,8.1 轴心受压构件的承载力计算,Questions: 1 Whats about the influence of axial force on the strength and behavior of RC column (section)? 2 What else should be considered in the strength calculation of column? 3 How to determine the strength of column under the combined action of axial force and moment? 4 Whats about the shear strength when there exist axial force? 5 Reinforcement detailing?,8.1 轴心受压构件的承载力计算,A compression member is very important ,the whole structure will be collapsed when the member is failure.,受压构件（柱）往往在结构中具有重要作用，一旦产生破坏，往往导致整个结构的损坏，甚至倒塌。,8.1 轴心受压构件的承载力计算,The bearing load of an axial member is the upper limit of compressive strength. 轴心受压承载力是正截面受压承载力 的上限。 We discuss the design of the axially member first，then the uniaxial compression member. 先讨论轴心受压构件的承载力计算,然后重点讨论单向偏心受压正截面。,8.1 轴心受压构件的承载力计算,8.1 The compressive capacity under axial load, In actual structures , the member with completely axial load does not exist. Eccentricity in loading is Possibly present. However, most of members may be designed under axial compressive load.,The function of the stirrup？ The function of the longitudinal reinforcement?,Column with spiral ：,Column with ties,A spiral column has closely spaced with spirals. The function？,8.1 轴心受压构件的承载力计算,8.1 轴心受压构件的承载力计算, 在实际结构中，理想的轴心受压构件几乎是不存在的。 往往存在一定的初始偏心距。 但有些构件，如以恒载为主的等跨多层房屋的内柱、桁架中的受压腹杆等，主要承受轴向压力，可近似按轴心受压构件计算。,普通钢箍柱：箍筋的作用？ 纵筋的作用？,螺旋钢箍柱：箍筋的形状为圆形，且间距较密， 其作用？,Function of longitudinal ： Assisting bearing axial loads of concrete Minimum rebar ratio under compressive ：0.6% Resisting bending moment Reducing the efforts of creep and shrinkage of concrete under compressive load.,8.1 轴心受压构件的承载力计算,纵筋的作用： 协助混凝土受压 受压钢筋最小配筋率：0.6% (单侧0.2%) 承担弯矩作用 减小持续压应力下混凝土收缩和徐变的影响。 实验表明，收缩和徐变能把柱截面中的压力由混凝土向钢筋转移，从而使钢筋压应力不断增长。 压应力的增长幅度随配筋率的减小而增大。 如果不给配筋率规定一个下限，钢筋中的压应力就可能在持续使用荷载下增长到屈服应力水准。,8.1 轴心受压构件的承载力计算,8.1 轴心受压构件的承载力计算,一、Column with ties,A concrete column with lateral ties fails abruptly when concrete crushes, causing spalling and bukling of the longitudinal bars between ties.,Short columns with axial load,8.1 轴心受压构件的承载力计算,Short columns with axial load,Stability coefficient,Coefficient j (P.136) is relation to slender column l0/b,Reduced factor 0.9 is regarded as considering of accidental eccentricity the effects and the reliability of column with axial load.,Long columns with axial load,1、Column with ties,8.1 轴心受压构件的承载力计算,1、普通钢箍柱,轴心受压短柱,轴心受压长柱,稳定系数,稳定系数j (P.136)主要与柱的长细比l0/b有关,折减系数 0.9是考虑初始偏心(accidental eccentricity)的影响，以及主要承受恒载作用的轴压受压柱的可靠性。,8.1 轴心受压构件的承载力计算,8.1 轴心受压构件的承载力计算,P.136,8.1 轴心受压构件的承载力计算,二、 Column with spiral,A concrete column with adequate spiral reinforcement fails at a higher load and in a ductile manner, since the spiral confines the concrete core and prevent buckling of the longitudinal bars.,8.1 轴心受压构件的承载力计算,Longitudinal strength of the concrete under tri-axial loads (混凝土圆柱体三向受压状态的纵向抗压强度),8.1 轴心受压构件的承载力计算,8.1 轴心受压构件的承载力计算,At limit states:,8.1 轴心受压构件的承载力计算,达到极限状态时（保护层已剥落，不考虑）,8.1 轴心受压构件的承载力计算,The factors of the spiral column on ultimate load，when fcu,k50N/mm2 , = 2.0 ; when fcu,k=80N/mm2; a =1.7 the other value may be obtained by linear interpolation.,8.1 轴心受压构件的承载力计算,A spiral stirrup can be used to enhance the axial bearing load of column。 In order that the cover does not shell off before it reached to ultimate load, the provision is given in the code: the calculated ultimate load of a column section with spiral is not exceeded 0.5 times of a tied column.,采用螺旋箍筋可有效提高柱的轴心受压承载力。 如螺旋箍筋配置过多，极限承载力提高过大，则会在远未达到极限承载力之前保护层产生剥落，从而影响正常使用。 规范规定 按螺旋箍筋计算的承载力不应大于按普通箍筋柱受压承载力的50%。,8.1 轴心受压构件的承载力计算,If the slenderness of column is too large that it can not prevent concrete failure and bulking of bars, the provision is below in the code: Any columns with a slenderness of l0/d 12 do not consider the confined effect of the spiral stirrup., 对长细比过大柱，由于纵向弯曲变形较大，截面不是全部受压，螺旋箍筋的约束作用得不到有效发挥。规范规定 对长细比l0/d大于12的柱不考虑螺旋箍筋的约束作用。,8.1 轴心受压构件的承载力计算, The effort of spiral stirrup is relation to the area of the section Ass1and the spacing S, the provision is below in the code: Transformed area Ass0 of the spiral is no less than one quarter of longitudinal area As The spacing of the spirals should not be less than 40mm, or greater than 80mm or dcor/5., 螺旋箍筋的约束效果与其截面面积Ass1和间距S有关，为保证由一定约束效果，规范规定： 螺旋箍筋的换算面积Ass0不得小于全部纵筋As 面积的25% 螺旋箍筋的间距s不应大于dcor/5，且不大于80mm，同时为方便施工，s也不应小于40mm。,作业：P.177 8-2,第七章 受压构件,8.1 轴心受压构件的承载力计算,8.2 Behaviors under flexure and axial load 压力和弯矩共同作用下的截面受力性能,压弯构件 偏心受压构件,8.2 偏心受压构件的承载力计算,压弯构件 偏心受压构件,e0=0？ When e0，N=0，？ Failure mode of an eccentrically member with compression load is between axial loaded and flexure member.,8.2 偏心受压构件的承载力计算,8.2 Behaviors under flexure and axial load (压力和弯矩共同作用下的截面受力性能),压弯构件 偏心受压构件,偏心距e0=0时？ 当e0时，即N=0，？ 偏心受压构件的受力性能和破坏形态界于轴心受压构件和受弯构件。,8.2 偏心受压构件的承载力计算,8.2 Behaviors under flexure and axial load 压力和弯矩共同作用下的截面受力性能),一、Failure character,The mode of failure is related to the eccentricity e0 and the longitudinal bar ratio 1、Tensile failure,8.2 偏心受压构件的承载力计算,M is big and N is small,eccentricity e0is big,As is suitable,一、破坏特征,偏心受压构件的破坏形态与偏心距e0和纵向钢筋配筋率有关 1、受拉破坏 tensile failure,8.2 偏心受压构件的承载力计算,M较大，N较小,偏心距e0较大,As配筋合适,1、 Tensile failure,8.2 偏心受压构件的承载力计算, Cracks will first appear in the tension zone ,the stress of As will improve quickly with increase of load and will yield first. Then, the crack will be wider and the neutral axis depth will be reduced. Finally the bar As in the compression zone will yield and the concrete is crushed., There be visible warning of failure and have large deflection ,it likes the beam with suitable reinforcement. Its bearing capacity depends on the steel in the tension zone. Condition is that the eccentricity e0 is large and the longitudinal bar ratio is suitable.,1、受拉破坏,8.2 偏心受压构件的承载力计算, 截面受拉侧混凝土较早出现裂缝，As的应力随荷载增加发展较快，首先达到屈服。 此后，裂缝迅速开展，受压区高度减小 最后受压侧钢筋As 受压屈服，压区混凝土压碎而达到破坏。 这种破坏具有明显预兆，变形能力较大，破坏特征与配有受压钢筋的适筋梁相似，承载力主要取决于受拉侧钢筋。 形成这种破坏的条件是：偏心距e0较大，且受拉侧纵向钢筋配筋率合适，通常称为大偏心受压。,2、Compressive failure There are two conditions for a compression failure ： Relative eccentricity e0/h0 is small.,8.2 偏心受压构件的承载力计算, Although relative eccentricity e0/h0 is large, the amount of tension reinforcing bar is big.,As太多,2、受压破坏 产生受压破坏的条件有两种情况： 当相对偏心距e0/h0较小,8.2 偏心受压构件的承载力计算,或虽然相对偏心距e0/h0较大，但受拉侧纵向钢筋配置较多时,As太多,8.2 偏心受压构件的承载力计算,The stress of the concrete and steel in the compression zone is very large. The stress of steel in the tension zone is small. When the relative eccentricity e0/h0 is small，the steel in the tension zone will under compressive.,2、Compressive failure When relative eccentricity e0/h0 is small,Although relative eccentricity e0/h0 is large that the amount of tension reinforcing bar is big.,8.2 偏心受压构件的承载力计算, The section is destroyed due to the concrete is crushed. Load carrying capacity is decided by the concrete and bars in compression zone, and the steel does not reach the tensile yield strength，the section fails in brittle. This case is called small eccentricity compression column due to the eccentricity is usually small. We should avoid the second case in our design.,8.2 偏心受压构件的承载力计算, 截面受压侧混凝土和钢筋的受力较大， 而受拉侧钢筋应力较小， 当相对偏心距e0/h0很小时，受拉侧还可能出现受压情况。,2、受压破坏 当相对偏心距e0/h0较小,或虽然相对偏心距e0/h0较大，但受拉侧纵向钢筋配置较多时,8.2 偏心受压构件的承载力计算, 截面最后是由于受压区混凝土首先压碎而达到破坏， 承载力主要取决于压区混凝土和受压侧钢筋，破坏时受压区高度较大，受拉侧钢筋未达到受拉屈服，破坏具有脆性性质。 第二种情况在设计应予避免，因此受压破坏一般为偏心距较小的情况，故常称为小偏心受压。,受拉破坏,受压破坏,第八章 受压构件,8.2 偏心受压构件的承载力计算,受拉破坏,受压破坏,第八章 受压构件,小偏心受压： - fy ss fy， As未达到受拉屈服或受压屈服； As达到受压屈服fy ； 混凝土达到cu,大偏心受压： As达到受拉屈服fy； As一般达到受压屈服fy，当x2a时， ss fy ； 混凝土达到cu,8.2 偏心受压构件的承载力计算,受拉破坏,受压破坏,第七章 受压构件,受拉和受压破坏的共同点： 靠近轴向力一侧的混凝土均能达到极限压应变； 远离轴向力一侧的钢筋能达到抗拉极限强度的为受拉破坏，不能达到抗拉极限强度（钢筋受拉或受压，但一般达不到设计强度）的为受压破坏。 P.138,8.2 偏心受压构件的承载力计算,第八章 受压构件,大偏心,小偏心,小偏心,8.2 偏心受压构件的承载力计算,对大偏压破坏，弯矩不变，轴力越大越安全； 对小偏压破坏，弯矩不变，轴力越大越不安全； 对大偏压和小偏压破坏，轴力不变，弯矩越大越不安全。,第八章 受压构件,8.2 偏心受压构件的承载力计算,2、The flexure strength calculation of bearing load The calculation of member with an eccentricity load is the same with the one under moment, we also take the assumption of plane section. According to the relation of strain and stress between concrete and steel ,we can analyze the process under compression and flexure. We can also take the equivalent rectangular stress block in the calculation of bearing load. The strength of equivalent rectangular block is a fc，the ratio of equivalent depth to neutral axis depth is b 。,8.2 偏心受压构件的承载力计算,2、正截面承载力计算 偏心受压正截面受力分析方法与受弯情况是相同的，即仍采用以平截面假定为基础的计算理论， 根据混凝土和钢筋的应力-应变关系，即可分析截面在压力和弯矩共同作用下受力全过程。 对于正截面承载力的计算，同样可按受弯情况，对受压区混凝土采用等效矩形应力图， 等效矩形应力图的强度为a fc，等效矩形应力图的高度与中和轴高度的比值为b 。,8.2 偏心受压构件的承载力计算,The limit of between tension failure and compression failure The tensile steel must simultaneously yield while the concrete strain reaches ecu. Its similarity with the limit between over-reinforced and balanced beam. So, the relative compressive depth at balanced state is:,8.2 偏心受压构件的承载力计算,受拉破坏和受压破坏的界限 即受拉钢筋屈服与受压区混凝土边缘极限压应变ecu同时达到 与适筋梁和超筋梁的界限情况类似。 因此，相对界限受压区高度仍为，,8.2 偏心受压构件的承载力计算,When x xb,When x xb,8.2 偏心受压构件的承载力计算,Tension failure 受拉破坏(大偏心受压), Compression failure 受压破坏(小偏心受压),rebar stress of tensile side 受拉侧钢筋应力ss According to plan section, it can be,8.2 偏心受压构件的承载力计算,为避免采用上式出现 x 的三次方程,考虑：当 x =xb，ss=fy；,8.2 偏心受压构件的承载力计算,当 x =b，ss=0,8.2 偏心受压构件的承载力计算,3、Relative eccentricitye0b/h0,In the member design under eccentricity load , we must decide to which failure the section belongs and use corresponding equations .,x =xb is the limit. In tensile failure, take x=b h0 and a=a，we can obtain the axial force Nb and bending moment Mb，,3、相对界限偏心距e0b/h0,偏心受压构件的设计计算中，需要判别大小偏压情况，以便采用相应的计算公式。,8.2 偏心受压构件的承载力计算,x =xb时为界限情况，取x=xbh0代入大偏心受压的计算公式，并取a=a，可得界限破坏时的轴力Nb和弯矩Mb，,8.2 偏心受压构件的承载力计算,e0b/h0 will be determined if the size of the section, materials, As and As are given。对于给定截面尺寸、材料强度以及截面配筋As和As ，界限相对偏心距e0b/h0为定值。 When e0e0b，its large eccentricity compression failure;当偏心距e0e0b时，为大偏心受压情况； When e0e0b，its small eccentricity compression failure.当偏心距e0e0b时，为小偏心受压情况。, When the size of the section and materials are given, relative eccentricity e0b/h0 will decreased with As and As When take minimum As and As，e0b/h0 will be the smallest; The minimum As is 0.45ft /fy according to the whole section area; The minimum As is 0.002 according to the whole section area. .We can approximately take h=1.05h0，a=0.05h0，, 进一步分析，当截面尺寸和材料强度给定时，界限相对偏心距e0b/h0随As和As的减小而减小， 故当As和As分别取最小配筋率时，可得e0b/h0的最小值。 受拉钢筋As按构件全截面面积计算的最小配筋率为0.45ft /fy， 受压钢筋按构件全截面面积计算的最小配筋率为0.002。 近似取h=1.05h0，a=0.05h0，代入上式可得，,A minimum amount of e0b/h0 is e0b,min/h0=0.4100.313, We take the smaller eccentricity e0b,min/h0=0.32 in calculation; (计算中取偏小值) When e0e0b，it calculate as large eccentricity compression;（按小偏心受压计算） When e0e0b, it calculate as small eccentricity compression. （先按大偏心受压计算）,8.2 偏心受压构件的承载力计算,4、Interaction relation of Nu -Mu Nu-Mu相关曲线,对于给定的截面、材料强度和配筋，达到正截面承载力极限状态时，其压力和弯矩是相互关联的，可用一条Nu-Mu相关曲线表示。根据正截面承载力的计算假定，可以直接采用以下方法求得Nu-Mu相关曲线：,取受压边缘混凝土压应变等于ecu； 取受拉侧边缘应变； 根据截面应变分布，以及混凝土和钢筋的应力-应变关系，确定混凝土的应力分布以及受拉钢筋和受压钢筋的应力； 由平衡条件计算截面的压力Nu和弯矩Mu； 调整受拉侧边缘应变，重复和,8.2 偏心受压构件的承载力计算,理论计算结果 等效矩形计算结果,8.2 偏心受压构件的承载力计算,The relationship of Nu-Mu illustrated the rule of bearing load of the positive section:,Any point in the curve stand for one combined internal force of the positive section. If the point is inside the curve that the section is safe. If its outside the curve, the ability of section is not enough (unsafe).,8.2 偏心受压构件的承载力计算, When the moment equals zero that the axial load carrying capacity gets to largest, that is N0（point A）； When the axial load is 0 that is M0（point C）；,Nu-Mu相关曲线反映了在压力和弯矩共同作用下正截面承载力的规律，具有以下一些特点：, 相关曲线上的任一点代表截面处于正截面承载力极限状态时的一种内力组合。 如一组内力（N，M）在曲线内侧说明截面未达到极限状态，是安全的； 如（N，M）在曲线外侧，则表明截面承载力不足；,8.2 偏心受压构件的承载力计算, 当弯矩为零时，轴向承载力达到最大，即为轴心受压承载力N0（A点）； 当轴力为零时，为受纯弯承载力M0（C点）；, Bearing load Mu is related to the axial load N; When N is small，Mu will increasing with the improve of N（CB段）； When N is small，Mu is decreasing with the increase of N （AB段）；,8.2 偏心受压构件的承载力计算, Bearing load Mu reach to maximum at point B (Nb，Mb) ，its a balanced failure point； In CB（NNb）is tension failure， In AB（N Nb）is compression failure；,截面受弯承载力Mu与作用的轴压力N大小有关； 当轴压力较小时，Mu随N的增加而增加（CB段）； 当轴压力较大时，Mu随N的增加而减小（AB段）；,8.2 偏心受压构件的承载力计算,截面受弯承载力在B点达(Nb，Mb)到最大，该点近似为界限破坏； CB段（NNb）为受拉破坏， AB段（N Nb）为受压破坏；, For symmetry reinforcement section Nb is consistent when reach to the limit failure. 对于对称配筋截面，达到界限破坏时的轴力Nb是一致的。,8.2 偏心受压构件的承载力计算, If the section dimension and material strength are not changed, the Nu-Mu will change to outwards as the increase of the reinforcement.如截面尺寸和材料强度保持不变，Nu-Mu相关曲线随配筋率的增加而向外侧增大；,小结：,判断大、小偏心破坏形式的方法：,8.2 偏心受压构件的承载力计算,8.3 附加偏心距和偏心距增大系数,8.3 Additional eccentricity and eccentricity magnifying coefficient 附加偏心距和偏心距增大系数,Accidental eccentricity ea （附加偏心距） e0+ea is called initial eccentricity ei（初始偏心距） are considered in the calculation.,1、 Accidental eccentricity附加偏心距,参考以往工程经验和国外规范，附加偏心距取ea=20mm or h/30 两者中的较大值，此处h是指偏心方向的截面尺寸。,8.3 附加偏心距和偏心距增大系数,8.3 附加偏心距和偏心距增大系数,由于施工误差、计算偏差及材料的不均匀等原因，实际工程中不存在理想的轴心受压构件。为考虑这些因素的不利影响，引入附加偏心距ea(accidental eccentricity)，即在正截面压弯承载力计算中，偏心距取计算偏心距e0=M/N与附加偏心距ea之和，称为初始偏心距ei (initial eccentricity)，,参考以往工程经验和国外规范，附加偏心距ea取20mm与h/30 两者中的较大值，此处h是指偏心方向的截面尺寸。,一、附加偏心距,二、eccentricity magnifying coefficient, second-order effect will appear due to lateral deflection ，then cause the additional moment If the column is slender, the additional moment can not be neglected. The lateral deflection of middle showed in Fig. is f . For the section in middle, the eccentricity of axial load is ei + f ，and the moment is M =N ( ei + f ).,8.3 附加偏心距和偏心距增大系数,二、偏心距增大系数, 由于侧向挠曲变形，轴向力将产生二阶效应(second-order effect)，引起附加弯矩 对于长细比较大的构件，二阶效应引起附加弯矩不能忽略。 图示典型偏心受压柱，跨中侧向挠度为 f . 对跨中截面，轴力N的偏心距为ei + f ，即跨中截面的弯矩为 M =N ( ei + f ).,8.3 附加偏心距和偏心距增大系数,二、, If the section and the initial eccentricity are the same, but the slenderness l0/h and the lateral deformation are different, the influence of second-order effect is different and will cause different failure.,8.3 附加偏心距和偏心距增大系数,Eccentricity magnification factor,二、Eccentricity magnification factor偏心距增大系数, 在截面和初始偏心距相同的情况下，柱的长细比l0/h (slenderness)不同，侧向挠度 f 的大小不同，二阶效应的影响程度会有很大差别，将产生不同的破坏类型。,8.3 附加偏心距和偏心距增大系数,For the slenderness of short column l0/h5 Lateral deformation f compare with initial eccentricity ei is very small. The moment in the middle M=N(ei+f ) is increasing with the increase of the N,8.3 附加偏心距和偏心距增大系数, Until the section reaches ultimate limit states and occurs failure. For a short column ,the deformation f is negligible.,对于长细比l0/h5的短柱 侧向挠度 f 与初始偏心距ei相比很小, 柱跨中弯矩M=N(ei+f ) 随轴力N的增加基本呈线性增长， 直至达到截面承载力极限状态产生破坏。 对短柱可忽略挠度f 影响。,8.3 附加偏心距和偏心距增大系数,For the slenderness l0/h of the middle slender column is 530 f can not be neglected comparing with ei . f is increasing as the increase of axial load , the speed of the moment in the middle of column span M = N( ei + f ) is bigger than that of N., M is increasing quickly as the increase of N and appears non linear Although it will reach the ultimate limit states that the carrying capacity is lower than the short column. For slender column , we must consider the additional deformation in design.,8.3 附加偏心距和偏心距增大系数,长细比l0/h =530的中长柱 f 与ei相比已不能忽略。 f 随轴力增大而增大，柱跨中弯矩M = N ( ei + f ) 的增长速度大于轴力N的增长速度， 即M随N 的增加呈明显的非线性增长, 虽然最终在M和N的共同作用下达到截面承载力极限状态，但轴向承载力明显低于同样截面和初始偏心距情况下的短柱。 因此，对于中长柱，在设计中应考虑附加挠度 f 对弯矩增大的影响。,8.3 附加偏心距和偏心距增大系数,8.3 附加偏心距和偏心距增大系数,For the slenderness of the slender column l0/h 30 The influence of lateral deformation f is big. Lateral deformation f i