受弯构件正截面承载力计算.ppt

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1、第四章 受弯构件,5.1 概述,Chapter 5 Flexural Strength of RC Members 受弯构件正截面承载力计算,第五章 受弯构件,5.1 概述,Questions 1 Why there are different types of RC members? 2 What about the functions of reinforcement detailing? 3 How to simplify the calculation of flexure strength? 4 How to determine the maximum and minimum rei

2、nforcement ratio for RC beams? 5 How to design RC beam when the maximum or minimum reinforcement ratio are not satisfied? 6 How to design T-section RC beam?,第五章 受弯构件,5.1 概述,5.1 概 述 Introduction, Beams and slabs are all Bending Members. Generally, types of beam section include rectangular, T-section,

3、 I-section, box section, -section and section. Section of cast-in-site one-way slab is rectangular, whose depth h is slab thickness and width is unit width （b=1000mm）.,5.1 概述,第五章 受弯构件, Precast slab include hollow slab and section slab. Considering of requirement of the construction and the entirety

4、of the structure, it is sometimes selected to combine the precast method with the cast-in-site method in the application, and this is called combined beams or slabs.,5.1 概述,第五章 受弯构件, 结构中常用的梁、板是典型的受弯构件(Bending Member) 梁的截面形式常见的有矩形、T形、工形、箱形、形、形 现浇单向板为矩形截面，高度h取板厚，宽度b取单位宽度（b=1000mm） 预制板(Precast Slab)常见的

5、有空心板、槽型板等 考虑到施工方便和结构整体性要求，工程中也有采用预制和现浇结合的方法，形成叠合梁和叠合板,5.1 概述,第五章 受弯构件,第五章 受弯构件,5.1 概述,Constructional Details of Beams： 1 To ensure the durability and fire resistance of RC structures, and the bond performance between the rebar and concrete, the thickness of the concrete cover should not less than 25

6、mm; 2 To ensure the the consolidation of the casting, the clear spacing between the rebars at the bottom of the beam should not less than 25mm or the diameter d of the rebar. And for the rebars at the top, it should no less than 30mm or 1.5 times of the rebar diameter d;,Single Row,Double Rows,第五章 受

7、弯构件,5.1 概述,Constructional Details of Beams： 3 Generally, the number of the longitudinal rod at the beam bottom is not less than 2, and the ordinary diameter is 1032mm。When there are many rods, they can be placed in different rows;,Single Row,Double Rows,第五章 受弯构件,5.1 概述,梁的构造要求： 1为保证RC结构的耐久性、防火性以及钢筋与混

8、凝土的粘结性能(Bond Behavior)，钢筋的混凝土保护层(Cover)厚度一般不小于 25mm； 2为保证混凝土浇注的密实性(Consolidation)，梁底部钢筋的净距(Clear Spacing)不小于25mm及钢筋直径d，梁上部钢筋的净距不小于 30mm及1.5 d； 3 梁底部纵向受力钢筋一般不少于2根，直径常用1032mm。钢筋数量较多时，可多排配置，也可以采用并筋配置方式；,第五章 受弯构件,5.1 概述,Constructional Details of Beams ：,第五章 受弯构件,5.1 概述, When no compressive reinforcement

9、 at the top of the beam, it should be placed two hanger bars that no less than 10mm diameter; When h500mm, skin reinforcements should be arranged throughout the depth in either side of the beam. The diameter should be no less than 10mm and the spacing is 250mm;,Single Rows,Double Rows,第五章 受弯构件,5.1 概

10、述, For rectangular section: h/b=2.03.5 For T section: h/b=2.54.0。 To ensure lateral stability,Single Rows,Double Rows,第五章 受弯构件,5.1 概述, 梁上部无受压钢筋时，需配置2根架立筋(Hanger Bars)，以便与箍筋和梁底部纵筋形成钢筋骨架，直径一般不小于10mm； 梁高度h500mm时，要求在梁两侧沿高度每隔250设置一根纵向构造钢筋(Skin Reinforcement)，以减小梁腹部的裂缝宽度，直径10mm；, 矩形截面梁高宽比h/b=2.03.5 T形截面梁高

11、宽比h/b=2.54.0。 To ensure lateral stability,第五章 受弯构件,5.1 概述, Usually, to unify the formwork size and easy to construct: Beam Width b=120、150、180、200、220、250、300、350、(mm) Beam Height h=250、300、750、800、900、(mm)。,Single Rows,Double Rows,第五章 受弯构件,5.1 概述, 为统一模板尺寸、便于施工，通常采用： 梁宽度(Width) b=120、150、180、200、220

12、、250、300、350、(mm) 梁高度(Height) h=250、300、750、800、900、(mm)。,第五章 受弯构件,5.1 概述,Constructional Details of Slabs： The thickness of the concrete cover should not less than 15mm or the diameter d of the rebar; The reinforcement diameter is usually 612mm, grade ; The reinforcement diameter may be 1418mm with

13、grade when the slab thickness is bigger.,第五章 受弯构件,5.1 概述,Constructional Details of Slabs： The spacing of the longitudinal reinforcement is 70200mm; It should have distribution bar in the direction perpendicular to the longitudinal reinforcement, so as to transfer the load uniformly to the longitudin

14、al reinforcement, fixing the longitudinal reinforcement in constructing and resist the stress due to the temperature and shrinkage.,第五章 受弯构件,5.1 概述,板的构造要求： 混凝土保护层厚度一般不小于15mm和钢筋直径d； 钢筋直径(Diameter)通常为612mm，级钢筋； 板厚度较大时，钢筋直径可用1418mm，级钢筋； 受力钢筋间距一般在70200mm之间； 垂直于受力钢筋的方向应布置分布钢筋(Distribution Bar)，以便将荷载均匀地传递

15、给受力钢筋，并便于在施工中固定受力钢筋的位置，同时也可抵抗温度和收缩等产生的应力。,第五章 受弯构件,5.1 概述,第五章 受弯构件,5.1 概述,Contents for the Design of RC Flexural Members： Flexure StrengthTo determine the section size and the longitudinal reinforcement according to the given M; (2) Shear StrengthTo determine the stirrup and the number of bent-up ba

16、rs according to the given V at the object section;,第五章 受弯构件,5.1 概述,Contents for the Design of RC Flexural Members： (3) Reinforcement DetailingTo ensure the bond performance between the reinforcement and the concrete and to make the most of the reinforcement, the detailing of the reinforcement is det

17、ermined through bending moment diagram and shear diagram; (4) Checking calculation； (5) Drawing the construction documents.,第五章 受弯构件,5.1 概述,钢筋混凝土受弯构件的设计内容： (1) 正截面受弯承载力计算 (Flexure Strength)按已知截面弯矩设计值M，计算确定截面尺寸和纵向受力钢筋； (2) 斜截面受剪承载力计算(Shear Strength)按受剪计算截面的剪力设计值V，计算确定箍筋和弯起钢筋的数量； (3) 钢筋布置(Reinforcemen

18、t Detailing)为保证钢筋与混凝土的粘结，并使钢筋充分发挥作用，根据荷载产生的弯矩图(Bending Moment Diagram)和剪力图(Shear Diagram)确定钢筋的布置； (4) 正常使用阶段的裂缝宽度和挠度变形验算(Checking Calculation); (5) 绘制施工图(Drawing the Construction Documents).,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,5.2 Basis of Flexural Strength Design,1、Basic Assumptions,A cross section that w

19、as plane before loading remains plane under load; (2) Take no account of the tensile strength of Concrete; (3) The stress-strain relationship of concrete;,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,5.2 Basis of Flexural Strength Design,(4) The stress-strain relationship of reinforcement. And the ultimate tensile

20、strain of the reinforcement is 0.01.,Based on the basic assumptions as above, it is easy to design the flexural strength theoretically. However it is quite inconvenient in practice due to the complexity of the stress-strain relationship of concrete.,一、Basic Assumptions,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,5

21、.2 正截面受弯承载力计算的基本规定,一、基本假定 Basic Assumptions,截面应变保持平面 (2) 不考虑混凝土的抗拉强度 (3) 混凝土的受压应力-应变关系； (4) 钢筋的应力-应变关系，受拉钢筋的极限拉应变取0.01。,根据以上四个基本假定，从理论上来说钢筋混凝土构件的正截面承载力（单向和双向受弯、受压弯、受拉弯）的计算已不存在问题,但由于混凝土应力-应变关系的复杂性，在实用上还很不方便。,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,二、Equivalent Rectangular Stress Block,It is enough to design the

22、 ultimate bending moment when the value of C and its action point yc are known.,在极限弯矩的计算中，仅需知道 C 的大小和作用位置yc就足够了。,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,二、等效矩形应力图 Equivalent Rectangular Stress Block,在极限弯矩的计算中，仅需知道 C 的大小和作用位置yc就足够了。,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,二、Equivalent Rectangular Stress Block,The stress b

23、lock of concrete in compressive region can be replaced by the equivalent rectangular stress block, in which the C value and action point yc are all same to before.,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,二、等效矩形应力图 Equivalent Rectangular Stress Block,a equivalent rectangular compressive stress factor b equivale

24、nt rectangular compressive zone factor,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,二、等效矩形应力图 Equivalent Rectangular Stress Block,可取等效矩形应力图形来代换受压区混凝土应力图,等效矩形应力图的合力大小等于C，形心位置与yc一致,在极限弯矩的计算中，仅需知道 C 的大小和作用位置yc就足够了。,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,二、等效矩形应力图 Equivalent Rectangular Stress Block,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,

25、Fundamental Equations,基本方程 Fundamental Equations,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定, 相对受压区高度 Relative Compression Depth,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,Relative Compression Depth,For adequately reinforced beams, the stress of the longitudinal reinforcement ss=fy。,x reflect not only the area ratio of reinforc

26、ement to concrete(r), but the strength ratio of reinforcement to concrete. Tension Reinforcement Index,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,相对受压区高度,对于适筋梁，受拉钢筋应力ss=fy。,相对受压区高度x 不仅反映了钢筋与混凝土的面积比（配筋率r），也反映了钢筋与混凝土的材料强度比，是反映构件中两种材料配比本质的参数。 Tension Reinforcement Index,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,3、Relative Compre

27、ssion Depth,Relative compression depth is only related to the material behaviors.,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,三、相对界限受压区高度,相对界限受压区高度仅与材料性能有关，而与截面尺寸无关,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,对配置无明显屈服点钢筋的截面，其界限相对受压区高度 xb=？,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,When the reinforcement in the section has no obvious yield po

28、int, what about balanced relative compression depth? xb=？,Relative Compression Depth,Table,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,达到界限破坏时的受弯承载力为适筋梁Mu的上限,适筋梁的判别条件,这几个判别条件是等价的,本质是,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,The upper limit of the flexural capacity Mu of the proper reinforced beams,Criterion of proper reinforc

29、ed beams,These criterions are equivalent,The nature is,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,4、Minimum Reinforcement Ratio,Mcr=Mu,approximately 1-0.5x =0.98 h=1.1h0,第五章 受弯构件,5.2 正截面受弯承载力计算的基本规定,四、最小配筋率 Minimum Reinforcement Ratio,Mcr=Mu,近似取 1-0.5x =0.98 h=1.1h0,5.2 正截面受弯承载力计算的基本规定,第五章 受弯构件,ftk /fyk=1.4ft/1.1

30、fy=1.273ft/fy, 同时不应小于0.2% 对于现浇板和基础底板沿每个方向受拉钢筋的最小配筋率不应小于0.15%。,5.2 正截面受弯承载力计算的基本规定,第五章 受弯构件,ftk /fyk=1.4ft/1.1fy=1.273ft/fy, In addition,it is no less than 0.2% For the cast-in-site slabs and foundation slabs, the minimum tensile reinforcement of either direction should not less than 0.15%.,第五章 受弯构件,

31、5.3 正截面受弯承载力计算,5.3 Flexural Analysis and Design of Beams,1、Singly Reinforced Section,Basic Formulae,第五章 受弯构件,5.3 正截面受弯承载力计算,5.3 受弯构件正截面承载力计算 Flexural Analysis and Design of Beams,一、单筋矩形截面 Singly Reinforced Section,基本公式 Basic Formulae,第五章 受弯构件,5.3 正截面受弯承载力计算,Application Conditions,To avoid brittle fr

32、acture due to over reinforcement.,To avoid brittle fracture due to under reinforcement.,第五章 受弯构件,5.3 正截面受弯承载力计算,适用条件(Application Conditions ),防止超筋脆性破坏 To avoid brittle fracture due to over reinforcement.,防止少筋脆性破坏 To avoid brittle fracture due to under reinforcement.,第五章 受弯构件,5.3 正截面受弯承载力计算,Checking

33、Calculation,Given that: section size b，h(h0), reinforcement As and the material strength fy, fc To determine：flexural capacity MuM Indeterminate：compression depth x and flexural capacity Mu Basic formulae：,xxbh0时， Mu=？,Asrminbh，？,It is usually happened when the constructing quality is not good enoug

34、h and the strength of concrete does not reach the predicted value.,第五章 受弯构件,5.3 正截面受弯承载力计算,截面复核(Checking Calculation),已知：截面尺寸b，h(h0)、截面配筋As，以及材料强度fy、fc 求：截面的受弯承载力 MuM 未知数：受压区高度x和受弯承载力Mu 基本公式：,xxbh0时， Mu=？,Asrminbh，？,这种情况在施工质量出现问题，混凝土没有达到设计强度时会产生。,第五章 受弯构件,5.3 正截面受弯承载力计算,Design of Beam Section,Given

35、that: bending moment M To determine：section size b，h(h0), reinforcement As and the material strength fy、fc Indeterminate： compression depth x, b, h(h0), As, fy and fc Basic formulae：A total of two,No unique solution Designer should make the economic and appropriate design according to various factor

36、s, such as loading state, material supply, constructing condition.,第五章 受弯构件,5.3 正截面受弯承载力计算,截面设计(Design of Beam Section),已知：弯矩设计值M 求：截面尺寸b，h(h0)、截面配筋As，以及材料强度fy、fc 未知数：受压区高度x、 b，h(h0)、As、fy、fc 基本公式：两个,没有唯一解 (No unique solution) 设计人员应根据受力性能、材料供应、施工条件、使用要求等因素综合分析，确定较为经济合理的设计。,第五章 受弯构件,5.3 正截面受弯承载力计算,Ma

37、terial Selection： Mu of proper reinforced beams is determined by fyAs. So fc of flexural RC members should not too high. Cast-in-site beams and slabs：grade C15C25 pre-cast beams and slabs： grade C20C30, On the other hand, flexural RC members service with cracks. And due to the limit of the crack wid

38、th and the deflection, it is difficult to make the most of the strength of the high-strength reinforcement. For RC beams: grade is usually used; For RC slabs: grade is usually used.,第五章 受弯构件,5.3 正截面受弯承载力计算,材料选用(Material Selection)： 适筋梁的Mu主要取决于fyAs， 因此RC受弯构件的 fc 不宜较高。 现浇梁板：常用C15C25级混凝土 预制梁板：常用C20C30级

39、混凝土, 另一方面，RC受弯构件是带裂缝工作的， 由于裂缝宽度和挠度变形的限制，高强钢筋的强度也不能得到充分利用。 梁常用级钢筋，板常用级钢筋。,第五章 受弯构件,5.3 正截面受弯承载力计算,Determine the sizes of section The section should has suitable stiffness so as to satisfy the deflection requirement. According to engineering experience，the section depth is usually predicted according

40、to the depth to span ratio h/L. For simply supported beams, they are h=(1/10 1/16)L，b=(1/21/3)h. For simply supported slabs, it is h = (1/30 1/35)L However the range to select the section is still too large, so it is needed to make a further study to select the section from the point of view of econ

41、omy.,第五章 受弯构件,5.3 正截面受弯承载力计算,截面尺寸确定 (Determine the sizes of section) 截面应具有一定刚度，满足正常使用阶段的验算能满足挠度变形的要求。 根据工程经验(Experience)，一般常按高跨比h/L来估计截面高度 简支梁(Simply Supported Beam)可取h=(1/10 1/16)L，b=(1/21/3)h 估计 简支板Simply Supported Slab)可取h = (1/30 1/35)L 但截面尺寸的选择范围仍较大，为此需从经济角度进一步分析。,When M is given: When she sect

42、ion size b、h(h0) are bigger, the needed As will be less and the reinforcement r will be smaller. But it will use more concrete and the cost of the formwork will increase. Furthermore, it will affects the headroom requirement;,第五章 受弯构件,5.3 正截面受弯承载力计算,给定M时 When M is given: 截面尺寸b、h(h0)越大，所需的As就越少，r 越小，

43、但混凝土用量和模板费用增加，并影响使用净空高度； 反之，b、h(h0)越小，所需的As就越大，r 增大。,第五章 受弯构件,5.3 正截面受弯承载力计算,第五章 受弯构件,5.3 正截面受弯承载力计算,Economical Reinforcement For beams：r =（0.51.6）% For slabs：r =（0.40.8）%,第五章 受弯构件,5.3 正截面受弯承载力计算,经济配筋率 Economical Reinforcement 梁：r =（0.51.6）% 板：r =（0.40.8）%,第五章 受弯构件,5.3 正截面受弯承载力计算,When fy , fc, b, an

44、d h(h0) are given, The indeterminate are x and As, and then it is to resolve the problems.,Problems？,To increase the section size or fc,Homework：5-1 5-4，5-5,？,第五章 受弯构件,5.3 正截面受弯承载力计算,选定材料强度 fy、fc，截面尺寸b、h(h0)后， 未知数就只有x，As，基本公式可解,Problems？,增加截面尺寸或 fc,作业：5-1 5-4，要写体会小结 5-5,？,第五章 受弯构件,5.3 正截面受弯承载力计算,2、D

45、oubly Reinforced Section,It refers to one beam section in which the tensile reinforcement and compressive reinforcement are all arranged.,compressive reinforcement,第五章 受弯构件,5.3 正截面受弯承载力计算,二、双筋矩形截面 Doubly Reinforced Section,双筋截面是指同时配置受拉和受压钢筋的情况。,第五章 受弯构件,5.3 正截面受弯承载力计算,In general, doubly reinforced s

46、ection is not economic. And it is usually used in the following situations: When the section size and the material strength are restricted and at the same time the design can not satisfy the proper reinforcement requirement; When the section will subjected to positive moment in one state and negativ

47、e moment in other state; Compressive reinforcement can improve the section ductility .,第五章 受弯构件,5.3 正截面受弯承载力计算,一般来说采用双筋是不经济的，工程中通常仅在以下情况下采用： 当截面尺寸和材料强度受建筑使用和施工条件（或整个工程）限制而不能增加，而计算又不满足适筋截面条件时，可采用双筋截面，即在受压区配置钢筋以补充混凝土受压能力的不足。 另一方面，由于荷载有多种组合情况，在某一组合情况下截面承受正弯矩，另一种组合情况下承受负弯矩，这时也出现双筋截面。 此外，由于受压钢筋可以提高截面的延性，

48、因此，在抗震结构中要求框架梁必须必须配置一定比例的受压钢筋。,第五章 受弯构件,5.3 正截面受弯承载力计算, Utilization of the strength of compressive reinforcement,When compressive reinforcement is arranged in the section, the enclosed stirrup must be placed to protect the concrete cover from premature peeling due to the buckling of the compressive,

49、 which will affect the flexural capacity of the beam.,Compressive Reinforcement,Enclosed Stirrup,第五章 受弯构件,5.3 正截面受弯承载力计算, 受压钢筋强度的利用,配置受压钢筋后，为防止受压钢筋压曲(Buckling)而导致受压区混凝土保护层过早崩落影响承载力，必须配置封闭箍筋。,当受压钢筋多于3根时，应设复合箍筋。,第五章 受弯构件,5.3 正截面受弯承载力计算, When the constructional details are satisfied in the doubly reinf

50、orced section, the mark of Mu is still while ecu appears at the outmost concrete fiber in compression region. When the reinforcement yields before the compression concrete reaches ecu, the failure modes will be same to that of proper reinforced RC beams and the ductility is obvious. In the capacity

51、design, the equivalent rectangular stress block is still valid.,第五章 受弯构件,5.3 正截面受弯承载力计算, 双筋截面在满足构造要求的条件下，截面达到Mu的标志仍然是受压边缘混凝土达到ecu。 在受压边缘混凝土应变达到ecu前，如受拉钢筋先屈服，则其破坏形态与适筋梁类似，具有较大延性。 在截面受弯承载力计算时，受压区混凝土的应力仍可按等效矩形应力图方法考虑。,第五章 受弯构件,5.3 正截面受弯承载力计算,When x xb, the equilibrium equation for the section force is:

52、,第五章 受弯构件,5.3 正截面受弯承载力计算,当相对受压区高度x xb时，截面受力的平衡方程为，,第五章 受弯构件,5.3 正截面受弯承载力计算,As described in the axial compressive members, the compressive strength of reinforcement fy 400 MPa. To make the best of the strength of compressive reinforcement, the strain should not less than 0.002. According to plane sec

53、tion assumption，,ecu=0.0033,第五章 受弯构件,5.3 正截面受弯承载力计算,如轴心受压构件所述，钢筋的受压强度fy 400 MPa。 为使受压钢筋的强度能充分发挥，其应变不应小于0.002。 由平截面假定可得，,ecu=0.0033,第五章 受弯构件,5.3 正截面受弯承载力计算,Fundamental Formulae,fyAs,第五章 受弯构件,5.3 正截面受弯承载力计算,基本公式 (Fundamental Formulae),fyAs,第五章 受弯构件,5.3 正截面受弯承载力计算,Fundamental Formulae,Singly Reinforced

54、 Part,Pure Reinforcement Part,第五章 受弯构件,5.3 正截面受弯承载力计算,基本公式(Fundamental Formulae),第五章 受弯构件,5.3 正截面受弯承载力计算,Resolution of Doubly Reinforced Section,第五章 受弯构件,5.3 正截面受弯承载力计算,Resolution of Doubly Reinforced Section,第五章 受弯构件,5.3 正截面受弯承载力计算,Resolution of Doubly Reinforced Section,第五章 受弯构件,5.3 正截面受弯承载力计算,Fle

55、xural capacity of pure reinforcement part is independent to that of concrete. So the failure modes is not related to As2.,Fundamental Formulae,Singly Reinforced Part,Pure Reinforcement Part,第五章 受弯构件,5.3 正截面受弯承载力计算,单筋部分,受压钢筋与其余部分受拉钢筋As2组成的“纯钢筋截面”的受弯承载力与混凝土无关 因此截面破坏形态不受As2配筋量的影响，理论上这部分配筋可以很大，如形成钢骨混凝土构

56、件。,基本公式 (Fundamental Formulae),纯钢筋部分,第五章 受弯构件,5.3 正截面受弯承载力计算,Application Conditions, To avoid brittle fracture due to over reinforcement., To make the most of the compressive reinforcement.,It is not needed for doubly reinforced section to check the minimum reinforcement.,第五章 受弯构件,5.3 正截面受弯承载力计算,适用条

57、件 (Application Conditions), 防止超筋脆性破坏 To avoid brittle fracture due to over reinforcement., 保证受压钢筋强度充分利用 To make the most of the compressive reinforcement.,双筋截面一般不会出现少筋破坏情况，故可不必验算最小配筋率。,第五章 受弯构件,5.3 正截面受弯承载力计算, Checking Calculation Given that：b, h, a, a, As, As , fy, fy, fc To resolve：MuM Indetermina

58、te：x , Mu Problem：When x xb, Mu=?,当 x 2a 时，Mu =? It is a safe design according to the following formula:,第五章 受弯构件,5.3 正截面受弯承载力计算, 截面复核 已知：b、h、a、a、As、As 、fy、 fy、fc 求：MuM 未知数：受压区高度 x 和受弯承载力Mu两个未知数，有唯一解。 问题：当x xb时，Mu=?,当 x 2a 时，Mu =? 可偏于安全的按下式计算,第五章 受弯构件,5.3 正截面受弯承载力计算, Design of Beam Section,Given tha

59、t：M, b, h, a和a, , fy, fy , fc To resolve： As, As,Indeterminate：x、 As 、 As Fundamental Formulae：2,Following the singly reinforced section design,x = xb,Then give,Give x = 0.8xb,第五章 受弯构件,5.3 正截面受弯承载力计算,截面设计,已知：弯矩设计值M，截面b、h、a和a，材料强度fy、 fy 、 fc 求：截面配筋,未知数：x、 As 、 As 基本公式：两个,按单筋计算,x = xb,即取,宜取x = 0.8xb,第

60、五章 受弯构件,5.3 正截面受弯承载力计算,Given that：M，b、h、a、a，fy、 fy 、 fc、As To determine：As,Indeterminate：x、 As,按As未知重算,若x2a,Homework： 5-6 5-7 5-9,第五章 受弯构件,5.3 正截面受弯承载力计算,已知：M，b、h、a、a，fy、 fy 、 fc、As 求：As,未知数：x、 As,按As未知重算,若x2a,作业： 5-6 5-7 5-9,第五章 受弯构件,5.3 正截面受弯承载力计算,三、T-Section, It has no effect on the flexural capa

61、city when concrete of the tension region is cut off. It can save concrete and reduce the self weight., When there has a number of tensile reinforcement, the bottom of the section may be enlarged so as to become an I-section. Its design is same to that of T-section.,第五章 受弯构件,5.3 正截面受弯承载力计算,三、T形截面(T-S

62、ection), 挖去受拉区混凝土，形成T形截面，对受弯承载力没有影响。 节省混凝土，减轻自重。, 受拉钢筋较多，可将截面底部适当增大，形成工形截面。工形截面的受弯承载力的计算与T形截面相同。,第五章 受弯构件,5.3 正截面受弯承载力计算, 受压翼缘(Compression Flange )越大，对截面受弯越有利(x减小，内力臂增大） 但试验和理论分析(Experimental and Theoretical Analysis)均表明，整个受压翼缘混凝土的压应力增长并不是同步的。, 翼缘处的压应力与腹板处受压区压应力相比，存在滞后现象， 随距腹板(Stem)距离越远，滞后程度越大，受压翼缘压

63、应力的分布是不均匀的。,第五章 受弯构件,5.3 正截面受弯承载力计算, 计算上为简化采有效翼缘宽度bf Effective Flange Width 认为在bf 范围内压应力为均匀分布， bf 范围以外部分的翼缘则不考虑。 有效翼缘宽度也称为翼缘计算宽度 它与翼缘厚度hf 、梁的宽度l0、受力情况(单独梁、整浇肋形楼盖梁)等因素有关。,第五章 受弯构件,5.3 正截面受弯承载力计算, To simplify the calculation, it is usually used effective flange width bf . It is considered that the com

64、pressive stress is uniform in the length of bf . Effective flange width is also called calculation width of flange. It is related to hf , l0 and the loading state.,第五章 受弯构件,5.3 正截面受弯承载力计算,第五章 受弯构件,5.3 正截面受弯承载力计算,Type 1 T-section,Type 2 T-section,Boundary State,第五章 受弯构件,5.3 正截面受弯承载力计算,第一类T形截面,第二类T形截面

65、,界限情况,第五章 受弯构件,5.3 正截面受弯承载力计算,Type 1 T-section,Design formula is same to that of rectangular section with beam width bf.,To protect the beam from brittle failure due to over reinforced, it is needed that x xb. To protect the beam from brittle failure due to under reinforced, it is needed that Asrminbh，where b is the width of stem.,For I-section and inverted T-section, the tensile reinforcement should satisfy the following equation: Asrminbh + (bf - b)hf,第五章 受弯构件,5.3 正截面受弯承载力计算,第一类T形截面,计算公式与宽度等于bf的矩形截面相同,为防止超筋脆