第7章受扭構(gòu)件的扭曲截面承載力

1、Chaper 7 Torsional strength of members,7.1 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 tor

2、sion7.7 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 torsio

3、nal stiffness.,Compatibility torsion (協(xié)調(diào)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

4、 .,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 45

5、°line to the neighboring faces. Finally, cracks are spiral when concrete in one side crushes.,,,,,,,Spiral cracks,2. Torsional reinforcements,2. Failure modes of torsional members with torsional reinforments,Failure

6、 modes are close related to the ratio of reinforcements.,Under-reinforced member , scarce-reinforced member , Over-reinforced member , Over-reinforced member,Calculation of torsional strength,Minimum amount of torsional

7、reinforcements,Partially over-reinforced member,Check the minimum dimension of section,7.3 Torsional resistance of member under pure torsion,1、Cracking torque Tcr,Principle stress,If concrete is ideal plastic material, m

8、ember 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 chord

9、sStirrups-tension web membersconcrete-compression diagonals,(2) Torsional strength of rectangular member under pure torsion,T member and Ⅰ-shaped member under pure torsion,The section is divided into several rectangula

10、r section.,,1、Experiment and failure modes,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,

11、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 b

12、ottom 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

13、flexure, shear and torsion,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,B

14、ecause 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

15、and torsion,,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 reinforce

16、ments 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)

17、,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 tor

18、sion 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 det

19、ermine the shear stirrups. 7、Add the amount of the same type of reinforcements and select reinforcements8、Sketch the section,彎型破壞,2、滿足下列條件可按構(gòu)造配置剪扭鋼筋：,3、出現(xiàn)下列情況時，可不考慮剪力或扭矩的作用,(1)當(dāng),(2) 當(dāng),可不考慮剪力，即V＝0,可不考慮扭矩，即T＝0,1、截面尺寸驗(yàn)算,二

20、者中的較大值,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,例題,雨篷剖面見圖7-16。雨篷板上承受均布荷載（已包括板的自身

21、重力）q=3.6kN/m2（設(shè)計值），在雨篷自由端沿板寬方向每米承受活荷載p=1.4kN/m（設(shè)計值）。雨篷梁截面240mmX240mm，計算跨度2.5m，采用C30混凝土，箍筋采用HPB300，縱向鋼筋采用HRB400，環(huán)境類別為二類a。經(jīng)計算知，雨篷梁玩具設(shè)計值M=14kN.m，剪力設(shè)計值V=30kN。確定雨篷梁的配筋數(shù)量。,例題圖,1、計算簡圖及內(nèi)力,2、幾何參數(shù),例題圖,3、驗(yàn)算截面尺寸,4、是否可按構(gòu)造配置剪扭鋼筋,5、驗(yàn)算是