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1、Chaper 7 Torsional strength of members7.1 Introduction7.2 Experiment about pure torsion member7.3 Torsional capacity of members under pure torsion7.4 Torsional capacity of member under combined flexure, shear and torsion7.7 Detailing requirements7.1 IntroductionTwo types of torsionEquilibrium torsio
2、n (平衡扭矩)：statically determinate torsion; primary torsionIt is determined based on static equilibrium, and not related to the torsional stiffness.Compatibility torsion (协调扭矩)：statically indeterminate torsion; secondary torsionIt cannot be determined based on static equilibrium alone, and related to t
3、he torsional stiffness and deformation of members .7.2 Experiment of pure torsion1. Mechanical analysis of pure torsion before crackWhen 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. Fin
4、ally, cracks are spiral when concrete in one side crushes.Spiral cracks2. Torsional reinforcementsStirrupsLongitudinal bars2. Failure modes of torsional members with torsional reinformentsFailure modes are close related to the ratio of reinforcements.scarce-reinforced member Tension failureUnder-rei
5、nforced member Bending failureUnder-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 strengthWhen torsional reinforcements are inadequate , m
6、embers fail as soon as cracks form.Minimum amount of torsional reinforcementsPartially over-reinforced member Over-reinforced member Compression failureWhen two types of torsional reinforcements are excessive , the reinforcements do not yield before concrete crushes.When one type of torsional reinfo
7、rcements is excessive , the excessive reinforcements do not yield before concrete crushes.Check the minimum dimension of sectionControl the ratio of longitudinal bars to stirrups （）7.3 Torsional resistance of member under pure torsion 1、Cracking torque Tcr 45F1F2F3F4Principle stressIf concrete is id
8、eal 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 analogyLongitudinal bars- tension chordsStirrups-tension web membersconcrete-compression diagonals
9、(2) Torsional strength of rectangular member under pure torsionT member and -shaped member under pure torsionnThe section is divided into several rectangular section.1、Experiment and failure modesFlexure-typed failureTorsion-typed failureTorsion and shear-typed failure7.4 Torsional capacity of membe
10、r under combined flexure, shear and torsionFlexure 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
11、 extending to the neighboring side. Finally, concrete at the bottom crushes.Torsion and shear are dominantConcrete 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 to
12、rsionTVBecause 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 strengthBecause of shear, the torsion strength of member will lower than that of pure torsion.Calculation formula
13、s of torsion and shearFormulas of pure torsion and shear3、Design methods of member under combined flexure, shear and torsionMVTTorsional longitudinal bars AstLTorsional stirrupsAst1/stLongitudinal bars AsShear stirrups Asv1/svTotal stirrups7.7 Detailing requirements1、Minimum ratio of torsional longi
14、tudinal reinforcements2、Minimum ratio of shear and torsional stirrups 3、Minimum dimension of section4、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
15、 are satisfied.Shear cannot be considered (V0)Torsion cannot be considered (T0)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 fl
16、exural 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 reinforcements8、Sketch the secti
17、on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：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V0可不考虑扭矩，即T01、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n雨篷剖面见图7-16。雨篷
18、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。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2402401300
