第7章受扭构件的扭曲截面承载力.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,*,Chaper 7 Torsional strength of members,7.1,Introduction,7.2,Experiment about pure torsion member,7.3,Torsional capacity of members under pure torsion,7.4,Torsional capacity of member under combined flexure,shear and torsion,7.7,Detailing requirements,7.1 Introduction,Two types of torsion,Equilibrium torsion(,平衡扭矩,),：,statically determinate torsion;primary torsion,It is determined based on static equilibrium,and not related to the torsional stiffness.,Compatibility torsion(,协调扭矩,),：,statically indeterminate torsion;secondary torsion,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,1.Mechanical analysis of pure torsion before crack,When the principle stresses exceed the tensile strength,cracks form at the middle point of longer side and extend diagonally along 45line to the neighboring faces.Finally,cracks are spiral when concrete in one side crushes.,Spiral cracks,2.Torsional reinforcements,Stirrups,Longitudinal bars,2.Failure modes of torsional members with torsional reinforments,Failure modes are close related to the ratio of reinforcements.,scarce-reinforced member,Tension failure,Under-reinforced member,Bending failure,Under-reinforced member,scarce-reinforced member,Over-reinforced member,Over-reinforced member,When member is adequately reinforced,the reinforcements yield before concrete crushes.,Calculation of torsional strength,When torsional reinforcements are inadequate,members fail as soon as cracks form.,Minimum amount of torsional reinforcements,Partially over-reinforced member,Over-reinforced member,Compression failure,When two types of torsional reinforcements are excessive,the reinforcements do not yield before concrete crushes.,When one type of torsional reinforcements is excessive,the excessive reinforcements do not yield before concrete crushes.,Check the minimum dimension of section,Control the ratio of longitudinal bars to stirrups,（,）,7.3 Torsional resistance of member under pure torsion,1,、,Cracking torque T,cr,45,F1,F2,F3,F4,Principle stress,If concrete is ideal plastic material,member cracks when principle stresses at all points of section exceed tensile strength,2,、,Torsional capacity of member with torsional reinforcements,(1)Mechanical model-space truss analogy,Longitudinal bars-,tension chords,Stirrups,-tension web members,concrete-,compression diagonals,(2)Torsional strength of rectangular member under pure torsion,T member and-shaped member under pure torsion,The section is divided into several rectangular section.,1,、,Experiment and failure modes,Flexure-typed failure,Torsion-typed failure,Torsion and shear-typed failure,7.4 Torsional capacity of member under combined flexure,shear and torsion,Flexure is dominant.,Concrete at the bottom cracks,then extending to the neighboring sides.Finally,concrete at the top crushes.,Torsion is dominant,at the same time the amount of reinforcements at the top is little.,Concrete in the longer side cracks,then extending to the neighboring side.Finally,concrete at the bottom crushes.,Torsion and shear are dominant,Concrete in the longer side cracks,then extending to the neighboring side top.Finally,concrete in the opposite side crushes.,2,、,Torsional capacity of member under combined flexure,shear and torsion,T,V,Because of torsion,the shear strength of member will lower than that of pure shear,.(1.5-,t,),(1)Correlative relationship of shear and torsion,t,-,Reduction coefficient of torsional strength,Because of shear,the torsion strength of member will lower than that of pure torsion.,Calculation formulas of torsion and shear,Formulas of pure torsion and shear,3,、,Design methods of member under combined flexure,shear and torsion,M,V,T,Torsional longitudinal bars A,stL,Torsional stirrups,A,st1,/s,t,Longitudinal bars As,Shear stirrups A,sv1,/s,v,Total stirrups,7.7 Detailing requirements,1,、,Minimum ratio of torsional longitudinal reinforcements,2,、,Minimum ratio of shear and,torsional stirrups,3,、,Minimum dimension of section,4,、,Shear and torsion reinforcements can be arranged according to detailing requirements when the following conditions are satisfied.,5,、,Shear or torsion cannot be considered when,the following conditions are satisfied.,Shear cannot be considered(V,0),Torsion cannot be considered(T,0),Design Procedure,1,、,Check the,minimum dimension of section,；,2,、,Check if,shear and torsion reinforcements can be arranged according to detailing requirements,；,3,、,Check,whether shear or torsion can be neglected,.,4,、,Compute the flexural strength to determine the flexural reinforcements.,5,、,Compute torsion strength to,determine the,torsional stirrups and longitudinal bars,.,6,、,Compute the shear strength to,determine the shear stirrups,.,7,、,Add the amount of the same type of reinforcements and select reinforcements,8,、,Sketch the section,扭剪型破坏,扭型破坏,弯型破坏,2,、满足下列条件可按,构造配置剪扭钢筋,：,3,、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,V,0,可不考虑扭矩，即,T,0,1,、,截面尺寸验算,二者中的较大值,4,、,最小配筋率,When,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 stirrups,例题,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/m,2,（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,。经计算知，雨篷梁玩具设计值,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,2,、几何参数,例题图,3,、验算截面尺寸,4,、是否可按构造配置剪扭钢筋,5,、验算是否可忽略剪力或者扭矩,构造要求,6,、受弯纵筋计算,7,、受剪计算,计算公式,8,、受扭计算,计算公式,9,、受剪扭钢筋最小配筋率验算,构造要求,10,、配筋,240,240,1300,