混凝土设计基本原理-第7章受扭构件的扭曲截面承载力.ppt

1、第第7 7章章 受扭构件承载力计算受扭构件承载力计算Chaper 7 Torsional strength of members7.1（概述）（概述）Introduction7.2（纯扭构件试验研究）（纯扭构件试验研究）Experiment about pure torsion member7.3（纯扭构件承载力）（纯扭构件承载力）Torsional capacity of members under pure torsion7.4（弯、剪、扭构件承载力）（弯、剪、扭构件承载力）Torsional capacity of member under combined flexure,shear

2、and torsion7.7（构造要求）（构造要求）Detailing requirements7.1 Introduction（概述）两类扭矩（Two types of torsion）Equilibrium torsion(平衡扭矩)：根据静力平衡条件确定根据静力平衡条件确定的扭矩、初始扭矩的扭矩、初始扭矩statically determinate torsion;primary torsion由静力平衡确定，与刚由静力平衡确定，与刚度无关度无关It is determined based on static equilibrium,and not related to the tors

3、ional stiffness.Compatibility torsion(协调扭矩)：statically indeterminate torsion（超静定扭矩）（超静定扭矩）;secondary torsion无法根据静力平衡确定，与无法根据静力平衡确定，与刚度及变形相关刚度及变形相关It cannot be determined based on static equilibrium alone,and related to the torsional stiffness and deformation of members.7.2 Experiment of pure torsion

4、1.（开裂前纯扭构件受力分析）Mechanical analysis of pure torsion before crack主拉应力超过抗拉强度后，开裂，主拉应力超过抗拉强度后，开裂，并沿并沿45 45 斜向发展，和上下底面斜向发展，和上下底面相交后进行发展，形成螺旋形斜相交后进行发展，形成螺旋形斜裂缝，构件破坏。裂缝，构件破坏。When the principle stresses exceed the tensile strength,cracks form at the middle point of longer side and extend diagonally along 45

5、line to the neighboring faces.Finally,cracks are spiral when the member reaches its ultimate capacity.Spiral cracks2.Torsional reinforcements（受扭钢筋）（受扭钢筋）StirrupsLongitudinal bars2.受扭破坏模式scarce-reinforced failure（少筋破坏）（少筋破坏）Tension failureUnder-reinforced failure（适筋破坏）（适筋破坏）Bending failureUnder-reinf

6、orced member,scarce-reinforced member,Partially over-reinforced member,Over-reinforced member When member is adequately reinforced,the reinforcements yield before concrete crushes.Calculation of torsional strength（承载力计算）（承载力计算）When torsional reinforcements are inadequate,members fail as soon as crac

7、ks form.Minimum amount of torsional reinforcements（最小钢筋用量）最小钢筋用量）Partially over-reinforced failure（部分（部分超筋）超筋）Over-reinforced Failure（全部超筋（全部超筋）Compression failureWhen two types of torsional reinforcements are excessive,the reinforcements do not yield before concrete crushes.When one type of torsion

8、al reinforcements is excessive,the excessive reinforcements do not yield before concrete crushes.Check the minimum dimension of section（最小截面尺寸验算）（最小截面尺寸验算）Control the ratio of longitudinal bars to stirrups（）7.3 Torsional capacity of member under pure torsion（纯扭构件抗扭承载力）1、开裂扭矩（Cracking torque)Tcr 45F1

9、F2F3F4tttcrWfbhbfT)3(6.2ttcrWfT7.0tPrinciple stressIf concrete is ideal plastic material,member cracks when principle stresses at all points of section exceed tensile strength(若混凝土时理想塑性，则截若混凝土时理想塑性，则截面上所有点的主拉应力达到抗拉强度构件才开裂）面上所有点的主拉应力达到抗拉强度构件才开裂）2、Torsional capacity of member with torsional reinforcem

10、ents（配筋受扭构件承（配筋受扭构件承载力）载力）(1)力学模型-空间桁架（space truss analogy）Longitudinal bars-tension chords（受拉弦杆）Stirrups-tension web members（受拉腹杆）concrete-compression diagonals（受压斜杆）cosDDsinD2dcorTqtA cosFFD 13sincorDqb coscoscotsin13corcorqbFFDqb cotcorFFqh 12cotcot2corcorcorTuFquA1-3片片水平力和竖向力平衡水平力和竖向力平衡3F1F箍筋cor

11、Nqb假设钢筋均屈服tan12styvcorTNAfsA cot2corstlycorTuFA fA tan11yvstcorystlf A uf A s T12yvstcorufA As 1ystlyvstcorf A sf A u 1Ncorbcotcorb 3F1F箍筋/(cot/)tantan12corcorcorTNN nqbbsqssA(2)Torsional capacity of rectangular member under pure torsion（矩形截面纯扭构件受扭承载力）corstyvttscuAsAfWfTTT12.135.0yvycorststlffuAsA1T

12、 member and-shaped member under pure torsion（T形和工形截面构件纯扭）nThe section is divided into several rectangular section.（截面分割成几个矩形部分）TWWTWebttww.TWWTflangeecompressivttff.TWWTflangeTensilettff.)3(62bhbWw)(22 bbhWfftf)(22bbhWfftftftftwtWWWW1、Experiment and failure modesFlexure-typed failure（弯型弯型破坏）破坏）Torsi

13、on-typed failure（扭型扭型破坏）破坏）Torsion and shear-typed failure（扭剪型破坏）7.4 Torsional capacity of member under combined flexure,shear and torsion（弯、剪、扭构件承载力）Flexure is dominant.（弯矩占（弯矩占优势）优势）Concrete at the bottom cracks,then extending to the neighboring sides.Finally,concrete at the top crushes.（底部先开裂，向相邻

14、面（底部先开裂，向相邻面延伸，顶部混凝土压碎）延伸，顶部混凝土压碎）Torsion is dominant,at the same time the amount of reinforcements at the top is little.（扭（扭矩占优势，同时顶矩占优势，同时顶部钢筋很少）部钢筋很少）Concrete in the longer side cracks,then extending to the neighboring side.Finally,concrete at the bottom crushes.（长边先开裂，延（长边先开裂，延伸到相邻面，底部混凝土压碎）伸到相邻

15、面，底部混凝土压碎）Torsion and shear are dominant（扭、（扭、剪占优势）剪占优势）Concrete in the longer side cracks,then extending to the neighboring side top.Finally,concrete in the opposite side crushes.（长边开裂，长边开裂，延伸到相邻面，对边混凝土压碎）延伸到相邻面，对边混凝土压碎）2、Torsional capacity of member under combined flexure,shear and torsion（弯、剪、扭构件

16、承载力）TVBecause of torsion,the shear strength of member will lower than that of pure shear（扭矩的存在会降低混凝土抗剪承载力）（扭矩的存在会降低混凝土抗剪承载力）.(1.5-t)(1)Correlative relationship between shear and torsion剪、扭相关性剪、扭相关性 t-Reduction coefficient of torsional capacity（抗扭承载了降低系数）（抗扭承载了降低系数）Because of shear,the torsion streng

17、th of member will lower than that of pure torsion.（剪力的存在会降低混凝土抗扭承载力）（剪力的存在会降低混凝土抗扭承载力）5.100ccccVVTT受扭承载力降低系数0tccTT受剪承载力降低系数510tccVV.抗扭承载力抗扭承载力降低系数降低系数.0 5.0 5抗剪承载力抗剪承载力降低系数降低系数Calculation formulas of torsion and shear010)5.1(7.0hsnAfbhfVsvyvtt010)5.1(175.1hsnAfbhfVsvyvttcorstyvtttAsAfWfT12.135.005.0

18、15.1TbhVWttFormulas of pure torsion and shear0)1(2.015.1TbhVWtt3、Design methods of member under combined flexure,shear and torsion（弯矩扭构件设计方法）MVT受扭纵筋AstL受扭箍筋Ast1/st纵筋As或者As，As受剪箍筋Asv1/sv全部箍筋7.7 Detailing requirements（构（构造）造）1、受扭纵筋最小配筋率ytstltlffVbTbhA6.0min,min,2、剪扭箍筋配箍率yvtsvsvff/28.0min,cctfWTbhV25.0

19、8.003、最小截面尺寸4、构造配置剪扭箍筋的条件ttfWTbhV7.005、不考虑剪力或者扭矩的情形Shear cannot be considered(V0)035.0bhfVt01875.0 bhfVorcttWfT175.0Torsion cannot be considered(T0)例题n雨篷剖面见图7-16。雨篷板上承受均布荷载（已包括板的自身重力）q=3.6kN/m2（设计值），在雨篷自由端沿板宽方向每米承受活荷载p=1.4kN/m（设计值）。雨篷梁截面240mmX240mm，计算跨度2.5m，采用C30混凝土，箍筋采用HPB300，纵向钢筋采用HRB400，环境类别为二类a。

20、经计算知，雨篷梁弯矩设计值M=14kN.m，剪力设计值V=30kN。确定雨篷梁的配筋数量。例题图1、计算简图及内力kNt5916.5)12.03.1(4.1)12.02/3.1(3.16.3mkNl tT.725.25916.52.mkNTkNVmkNM.7,30,.14内力：2、几何参数36210608.44608000)3(6mmbhbWt220/43.1/3.14 19545240 ,25mmNfmmNfmmhmmCtc，mmhbcorcor170352240例题图3、验算截面尺寸575.325.054.246080008.0100000071952401000308.00ctfWTbh

21、V4、是否可按构造配置剪扭钢筋001.17.016.24608000100000071952401000300ttfWTbhV需计算配置剪扭钢筋5、验算是否可忽略剪力或者扭矩TmkNwftt.153.110608.443.1175.0175.06不能忽略扭矩VkNbhft42.2319524043.135.035.00不能忽略剪力构造要求6、受弯纵筋计算2min,211152113602.222403.14mmAmmfbxfAsycs2102(1 0.5)14.3 240 1950.518(1 0.5 0.518)50.14.ucbbMf bhkN mMkN m因此，改梁可以设计成单筋截面（1

22、）验算改梁是否可以设计成单筋截面（2）配筋计算bfMhhxc12002mmx2.222403.140.110142195195627、受剪计算0.124.119524010608.4.10710305.015.15.015.16630bhWTVtt0.1t010)5.1(7.0hsnAfbhfVvyvttkNbhftt42.2319524043.15.07.0)5.1(7.00062.02195270100042.231000301vvsA计算公式8、受扭计算2.1取46.01702702.12.110608.443.135.01072661tstsAyvycorststlffuAsA1212

23、80360270340246.02.1mmffsuAAyyvcorststl计算公式2972.0240103010736VbTytstlstffVbTbhA.6.0min,min,率要求受扭纵筋满足最小配筋%,235.036043.1972.06.0min,st%15.028.0yvtsvff要求：受剪扭箍筋最小配筋率9、受剪扭钢筋最小配筋率验算522.046.0062.011tstvsvsAsA筋率要求满足受剪扭箍筋最小配构造要求10、配筋522.011tstvsvsAsA2280 mmAstl2211mmAsmm1500.52278.5s 10，箍筋采用223042113280933280

24、mmmm底层：中间层层纵筋采用三层布置，顶mmHRBmmHRB14,400210,4002直径根底层：直径根中间层顶层Design Procedure（设计过程）（设计过程）1、Check the minimum dimension of section；（检查截面尺寸）（检查截面尺寸）2、Check if shear and torsion reinforcements can be arranged according to detailing requirements；（是否可以构造配置剪扭钢筋）是否可以构造配置剪扭钢筋）3、Check whether shear or torsion

25、can be neglected.（是否可忽略是否可忽略V或或T）4、Compute the flexural strength to determine the flexural reinforcements.（计算抗弯钢筋）计算抗弯钢筋）5、Compute torsion strength to determine the torsional stirrups and longitudinal bars.（计算抗扭钢筋）（计算抗扭钢筋）6、Compute the shear strength to determine the shear stirrups.（计算抗剪箍筋）（计算抗剪箍筋）7

26、、Add the amount of the same type of reinforcements and select reinforcements（配筋）（配筋）8、Sketch the section（绘制截面配筋图）绘制截面配筋图）0107.0hsnAfbhfVsvyvt010175.1hsnAfbhfVsvyvtcorstyvttAsAfWfT12.135.0扭剪型破坏扭剪型破坏扭型破坏扭型破坏弯型破坏弯型破坏2、满足下列条件可按构造配置剪扭钢筋：ttfWTbhV7.003、出现下列情况时，可不考虑剪力或扭矩的作用(1)当(2)当可不考虑剪力，即V0035.0bhfVt01875.

27、0bhfVc或可不考虑扭矩，即T0ttWfT175.0cctfWTbhV25.08.001、截面尺寸验算ytstltlffVbTbhA6.0min,min,yvtsvsvff/28.0min,)%/45(%2.0minytff或二者中的较大值4、最小配筋率010)5.1(7.0hsnAfbhfVsvyvtt010)5.1(175.1hsnAfbhfVsvyvttcorstyvtttAsAfWfT12.135.005.015.1TbhVWtt0)1(2.015.1TbhVWttyvycorststffuAsA1When is from 0.333 to 3，all torsional reinforcements can yield at ultimate torsion moment.Chinese Code:=0.6-1.7-Ratio of strength of longitudinal bars to stirrups24024013002/6.3mkNq mkNp/4.1