结构设计原理课程设计-混凝土梁(英文版)

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《reinforce_concrete》课程教学大纲

《reinforce_concrete》课程教学大纲

《Structural Design:reinforced concrete》课程教学大纲一、教师或教学团队信息二、课程基本信息课程名称(中文):钢筋混凝土结构基本理论课程名称(英文):Structural Design:Reinforced concrete课程类别:□通识必修课□通识选修课■专业必修课□专业方向课□专业拓展课□实践性环节课程性质*:■学术知识性□方法技能性□研究探索性□实践体验性课程代码:2410515周学时:8 总学时:64+2周学分: 4先修课程:建筑材料、材料力学、Structural Mechanics I授课对象:土木工程(中英合作项目)本科学生三、课程简介该课程是第5学期开设的专业主干课程。

本课程主要介绍混凝土基本构件的设计原理,具体学习以概率论为基础的极限状态设计方法,学习拉、压、弯、剪、扭基本构件的极限承载力的设计计算方法,学习裂缝宽度、变形的计算方法,学习梁板结构的设计计算方法。

与后续课程《结构设计》有密切的联系,《结构设计》包括钢筋混凝土结构设计和砌体结构设计,属专业课。

通过本课程的学习,学生要了解中国、欧洲相关混凝土结构设计规范,熟悉基本设计方法,结合后续课程《结构设计》的学习,经过工程师基本训练,能够从事钢筋混凝土建筑结构设计,进一步提高外语水平,提高国际视野。

该课程包括中方授课和英方授课两部分内容,其中中方授课内容包括:钢筋和混凝土的力学性能,共同工作的原理和条件;中国现行结构设计方法;钢筋混凝土构件拉、压、弯、剪、扭极限承载力的计算方法;钢筋混凝土的变形和裂缝宽度计算方法;梁板结构设计。

英方授课内容包括:设计原理及欧洲规范简介;设计荷载(Eurocode 1);混凝土结构设计规范(Eurocode 2);钢筋混凝土板、梁、平板、柱的设计;结构整体稳定。

四、课程目标本课程旨在让学生理解和掌握钢筋混凝土设计的基础知识和基本原理,掌握以概率论为基础的极限状态设计方法,掌握钢筋混凝土拉、压、弯、剪、扭基本构件的极限承载力的设计计算方法,掌握钢筋混凝土构件裂缝宽度、变形的计算方法,掌握钢筋混凝土梁板结构的设计计算方法。

混凝土结构基本原理课程设计

混凝土结构基本原理课程设计

《混凝土梁课程设计》——计算书姓名:***学号:********班级:土木0911目录1 计算依据与基础资料 (4)1.1 标准及规范 (4)1.1.1 标准 (4)1.1.2 规范 (4)1.1.3 参考资料 (4)1.2 主要材料 (4)2 横断面布置 (5)2.1 横断面布置图 (5)2.2 预制板截面尺寸 (5)3 荷载横向分布系数、汽车冲击系数 (6)3.1 荷载横向分布系数 (6)3.2车道折减系数 (6)3.3汽车荷载冲击系数 (6)4 作用效应组合 (6)4.1 作用的标准值 (6)4.1.1 永久作用标准值 (6)4.1.2 汽车荷载效应标准值 (8)4.2 作用效应组合 (10)4.2.1 基本组合(结构承载能力极限状态设计) (10)4.2.2 作用短期效应组合(正常使用极限状态设计). 114.3 截面几何特性计算 (13)5 持久状态承载能力极限状态计算 (14)5.1 正截面抗弯承载能力 (14)5.2 斜截面抗剪承载力验算 (16)6 持久状态正常使用极限状态计算 (19)6.1弯曲裂缝计算 (19)6.2 挠度计算 (20)钢筋混凝土空心板梁设计1 计算依据与基础资料1.1 标准及规范1.1.1 标准跨径:桥梁计算跨径7.4m设计荷载:公路-Ⅰ级桥面宽度:1m(人行道))+8m(双车道)+1m(人行道)=10m结构重要性系数: 1.1环境条件:Ⅱ类1.1.2 规范《公路工程技术标准》JTG B01-2003《公路桥梁设计通用规范》JTG D60-2004(简称《通规》)《公路钢筋混凝土及预应力混凝土桥涵设计规范》JTG D62-2004(简称《预规》)1.1.3 参考资料《公路桥涵设计手册》桥梁上册(人民交通出版社2004.3)1.2 主要材料1)混凝土:C402)普通钢筋:主筋采用HRB335钢筋,其他用R235钢筋2 横断面布置2.1 横断面布置图(单位:mm)2.2 预制板截面尺寸(单位:cm)3 荷载横向分布系数、汽车荷载冲击系数(以空心板中间板为设计对象)3.1荷载横向分布系数汽车荷载:梁端剪力0.5,1/4跨、跨中弯矩0.28 人群荷载引起的梁端剪力0,1/4跨、跨中弯矩0.21 人行道板及栏杆引起梁端剪力0,1/4跨、跨中弯矩0.21。

(双语版)混凝土结构设计原理

(双语版)混凝土结构设计原理

(3)The dowel action V d of the longitudinal reinforcing bars crossing the diagonal crack(纵向钢筋的销栓作用传递的剪 力).
The total ultimate shear resistance of the diagonal section
(1)The shear strength on the residual concrete compression zone. V c (斜裂缝上端剪压区混凝土截面承担的剪力)
(2)The vertical component V i of the interlocking force
between the aggregates on the two sides of the diagonal crack. (斜裂缝交界面骨料的咬合与磨擦作用传递的剪力)
cross section E-E of the beam can be expressed as
My 0 I0
VS 0 bI 0
●The principal tension stress 1 and the principal compression
stress 2 can be expressed as
(作用在纵筋的销栓力可能产生沿纵筋的劈裂裂缝,使销栓作 用大大降低)
◆Consequently, the only reliable contribution to the shear resistance of a member without web reinforcement is the shear strength V c (无腹筋构件剪力的三个组成部分中,只有斜裂缝上端截面 承担的剪力最为稳定)

混凝土结构设计课程设计

混凝土结构设计课程设计

混凝土结构设计课程设计一、课程目标知识目标:1. 学生能理解混凝土结构设计的基本原理和概念,掌握混凝土结构的材料性质及力学性能。

2. 学生能够掌握混凝土结构设计的相关规范和标准,了解不同结构类型的设计要求。

3. 学生能够运用所学知识,分析混凝土结构在实际工程中的应用和问题。

技能目标:1. 学生能够运用计算机软件进行混凝土结构的初步设计和计算,具备实际操作能力。

2. 学生能够运用力学原理,解决混凝土结构设计中的简单问题,具备一定的结构分析能力。

3. 学生通过课程学习,能够进行团队合作,沟通协调,共同完成混凝土结构设计项目。

情感态度价值观目标:1. 培养学生对混凝土结构设计学科的兴趣和热情,激发学生主动学习的动力。

2. 培养学生严谨的科学态度和良好的工程意识,提高学生的职业道德素养。

3. 培养学生关注社会发展,认识到混凝土结构设计在国民经济建设中的重要性,增强学生的社会责任感。

课程性质:本课程为专业核心课程,旨在培养学生具备混凝土结构设计的基本理论和实践能力。

学生特点:学生为土木工程专业大三学生,已具备一定的力学和材料科学基础,具有较强的学习能力和实践欲望。

教学要求:结合课程性质和学生特点,注重理论与实践相结合,提高学生的实际操作能力和解决问题的能力。

在教学过程中,注重启发式教学,引导学生主动思考,培养学生的创新意识和团队合作精神。

通过课程学习,使学生能够达到上述课程目标,为将来从事混凝土结构设计工作打下坚实基础。

二、教学内容1. 混凝土结构设计基本原理:包括混凝土材料的力学性能、混凝土结构耐久性、结构设计的基本原则和方法。

教材章节:第一章 混凝土结构设计基本原理2. 混凝土结构设计规范与标准:介绍国内外混凝土结构设计规范、标准及其应用。

教材章节:第二章 混凝土结构设计规范与标准3. 混凝土结构类型及设计方法:包括梁、板、柱、墙、基础等结构的设计计算方法。

教材章节:第三章 混凝土梁设计;第四章 混凝土板设计;第五章 混凝土柱设计;第六章 混凝土剪力墙设计;第七章 基础设计4. 混凝土结构设计实例分析:分析典型混凝土结构工程案例,使学生了解实际工程中的应用和问题。

混凝土结构 中英文 PPT 演示文稿

混凝土结构 中英文 PPT 演示文稿

几个小伙伴为您讲课啦!!!165•Plain concrete is formed from a hardened mixture of cement ,water ,fine aggregate, coarse aggregate (crushed stone or gravel),air, and often other admixtures. The plastic mix is placed and consolidated in the formwork, then cured to facilitate the acceleration of the chemical hydration reaction lf thecement/water mix, resulting in hardened concrete.•素混凝土是由水泥、水、细骨料、粗骨料(碎石或卵石)、空气,通常还有其他外加剂等经过凝固硬化而成。

将可塑的混凝土拌合物注入到模板内,并将其捣实,然后进行养护,以加速水泥与水的水化反应,最后获得硬化的混凝土。

•The finished product has high compressive strength, and low resistance to tension, such that its tensile strength is approximately one tenth lf its compressive strength. Consequently, tensile and shear reinforcement in the tensile regions of sections has to be provided to compensate for the weak tension regions in the reinforced concrete element.•其最终制成品具有较高的抗压强度和较低的抗拉强度。

chapter 3 reinforced concrete beams(混凝土结构设计原理英文课件)

chapter 3 reinforced concrete beams(混凝土结构设计原理英文课件)
ysis
• For purposes of analysis and design, the ACI Code, section 8.10, has established limits on the effective flange width as follows: • 1. the effective flange width must not exceed one-fourth of the span length of the beam, and the effective overhanging flange width on each side of the web must not exceed eight times the thickness of the slab nor onehalf of the clear distance to the next beam. In other words, the effective flange width must not exceed: • A. one – fourth of the span length • B. b 16h
As ,min 3 f c' fy bw d 200 bw d fy
• Note that for T-beams, bw represents the width of the web. Also note that the first expression controls only is • f c' 4440psi . For negative moment (flange in tension) ' 6 f c • As ,min the smaller of bw d or 3 f bd

(双语版)混凝土结构设计原理资料

(双语版)混凝土结构设计原理资料

1

Байду номын сангаас
2

2 2
4
2


2

2 2
4
主拉应力的作用方向与梁纵轴的夹角
tan 2 ( 2 )
分析图5.2有: 单元2位于中和轴处,该单元上只作用有剪应力(shear
stress),而无正应力(direct stress),主拉应力与主压应力 相互垂直与梁轴线成450角,主拉应力与主压应力的大小相等,
◆In order that the member not to be failed by shear, the following condition must be fulfilled
V Vu Vc
●Take moment about the compression force of the concrete Cc (对混凝土压力作用点取矩), the moment acting on the diagonal section M C is resisted by
(1)The shear strength on the residual concrete compression zone. Vc(斜裂缝上端剪压区混凝土截面承担的剪力)
(2)The vertical component Vi of the interlocking force
between the aggregates on the two sides of the diagonal crack. (斜裂缝交界面骨料的咬合与磨擦作用传递的剪力)
(4)斜裂缝出现后,纵筋间的混凝土可能产生沿纵筋的撕裂 裂缝 。
随着荷载的继续增加,剪压区混凝土承受的剪应力和压应力也 继续增加,混凝土处于剪压复合受力状态,当达到此种状态的极限 强度时,剪压区混凝土发生破坏,亦即发生斜截面破坏。

(最新整理)chapter1(混凝土结构设计原理英文课件)

(最新整理)chapter1(混凝土结构设计原理英文课件)
• Portland cement consists chiefly of calcium and aluminum silicates.
2021/7/26
7
1-4 aggregates
• In ordinary structural concretes the aggregates occupy approximately 70% to 75% of the volume of the hardened mass. Gradation of aggregate size to produce close packing is desirable because, in general, the more densely the aggregate can be packed, the better are the strength an durability.
2021/7/26
3
• The compressive strength of concrete is relatively high. Yet it is a relatively brittle material, the tensile strength of which is small compared with its compressive strength. Hence steel reinforcing rods (which have high tensile and compressive strength) are used in combination with the concrete; the steel will resist the tension and the concrete the compression. Reinforced concrete is the result of this combination of steel and concrete. In many instances, steel and concrete are positioned in members so that they both resist compression.
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Project of Principles of Structure Design -------Reinforced Concrete BeamName:Wang PengzhiClass :2011210701NO. :08SCHOOL OF HIGHWAY COLLEGECHANG’AN UNIVERSITYDECEMBER 7, 2013ProjectKnown: the member is a concrete T beam (simple support beam), the standard span is, calculating span. The reinforcement distribution placed and section dimensions of normal section flexural capacity are shown in the diagram. Main steel bar is HRB335 (),, erection steel bar is HRB335, an eight-layer steel skeleton is welded. Concrete is C30, grade Ⅰenvironment,. For support section, shearing calculated value V0 =γ0Vd,0 = 342kN,bending moment calculated value M0 =γ0Md,0 = 0;For mid-span section, shearing calculated value Vl/2 =γ0Vd, l/2 =71.0kN, bending moment calculated value Ml/2 =γ0Md, l/2 = 1000kN.m.Solution:(1)Check of the sectional dimensionsTo meet the construction requirements, the bottommost layer of steel bar must be through the bearing section.And the effective depth of bearing section is (set the thickness of cover)So, the sectional dimensions meet the construction requirements.(2)Check that if necessary to equip the web reinforcementMid-span section:Bearing section:So,The stirrup must be used with some zone of mid-span. As for other zones, the construction requirements are suit.(3) Distribution of the envelope diagram of shear forceFrom the envelope diagram of shear force and , and .The distance between section and mid-span section can be get asIt ’s justifiable to place stirrup reinforcement in the zone of 1023mm length.And calculated shearing force of the section with a distance to the support center can be obtained proportionally from the envelope diagram of shear force.Hence, is responsible for the 60% load and is responsible for the rest load. .Besides, the session length of bent-up reinforcement is 2980 mm.(4) Stirrup reinforcement designThe stirrup is taken for reinforcement design. Its section area is .The longitude reinforcement ratio p and effective depth of diagonal section can be the average of the support ’s section value and mid-span section value.Support section: ,Mid-span section:So, the average value isHence, the spacing of stirrup is 342k N71k N114.76k N50029801017128.2k N192.3k N Figure 1 Allocation diagram of shearing forceAnd the design value is eventually determined as 300mm with concerning the requirements of construction.Web reinforcement ratio (take) isBesides thatObviously, the design meets the requirements.In conclusion, the stirrup spacing is 100mm of the zone from the support center to the mid-span direction at the distance 1000mm. as for the rest zone; the stirrup spacing is 300mm.(5)Bent-up reinforcement designAs the erection steel bar (HRB335) of weld reinforcement skeleton is and the distance between C.G. of compression rebar and outer fiber of concrete is.Now, it’s planned to bent up the steel bar N1~N3. And a sheet of the calculating values is listed here.The detailed calculation process is in the following:The perpendicular distance between upper bent-up point and lower bent-up point As the bent-up angle is,the distance to the center of support for first rowFor the second row of steel bar,The distance to the center of support for first rowAs for the second row’s calculated shearing value of distributionNotation: the designed length of bent-up session is 2980mm.Required area of bent-up barsThe distance from intersection point of bent-up bar and axis to the center of supportThe calculation mode of other arrow is the same with the second arrow. And the results are placed in the sheet.Since the main steel bars have been set, the corresponding calculation of normal section carrying capacity can be done and the results are in the following sheet.fulcrumsectionAnd the preliminary design of bent-up steel bar is shown in the diagram as well as the diagram of resistance bending moment and the envelope of bending moment. Plot parallel lines by representing carrying capacity of flexural member and substitute each into the intersection of parallels and envelope of bending moment which is named as i ,j,…,q(theatrical points ). Substitute each intoand got distance from i ,j,…,q mid-span section.Check that if initial position of bent-up steel bar is qualified for the requirements by each point.(a)For first arrow steel bar (2N3)The abscissa of fully utilized point k is, and the abscissa of bent-up point 1 is, so the point 1 is left to the point l andThe abscissa of non-utilized point l is, while the abscissa of the intersection of 2N3 bar and axis isHence, the position of bent-up point 1 is qualified.(b)For second arrow steel bar (2N2)The abscissa of fully utilized point j is, and the abscissa of bent-up point 2 is, so the point 2 is left to the point j andThe abscissa of non-utilized point k is, while the abscissa of the intersection of 2N2 bar and axis isHence, the position of bent-up point 2 is qualified.(c)For third arrow steel bar (2N1)The abscissa of fully utilized point I is x=0mm, and the abscissa of bent-up point 3is, so the point 3 is left to the point i andThe abscissa of non-utilized point j is, while the abscissa of the intersection of 2N3 bar and axis is1885..Hence, the position of bent-up point 3 is qualified.In conclusion, the check above shows that the preliminary design position of bent-up point is verified.As the resisting envelope diagram is quite far away from the bending moment one, it is necessary to add diagonal steel bars to enhance the capacity. And the layout is shown in the diagram below.Figure 3 Reinforcement layout diagram(6)Check of carrying capacity for diagonal section(1)Check one – the point(a)Find the top position of diagonal section.According to the figure, the abscissa of normal section with a distance 500mm2 to the center of support is. And the effective height ofnormal section is. Let the inclined section projected length.So, the top position of diagonal section A is found and its abscissa is.Check the shear capacity of diagonal sectionThe shear force and bending moment of A section can be calculated as followings.The effective height of normal section A is. (Main steel bars are) So, the real generalized shear spam ratio m and projected length c areThe percentage of difference value isPlot the to-be-checked diagonal section and get the bevelThere are (2N6+2N5) main steel bar with the diagonal section and the corresponding reinforcement ration p isAnd web reinforcement ration (take) isAnd the bend-up steel bar intersecting with diagonal section A are 2N3and 2N4.According to the unit requirements, the inclined section carrying capacity can be got by subtitling results listed above.Therefore, the shearing capacity of diagonal section with a distance 500mm (i.e. h/2) to the center of support is verified.(2)Check one – the bent-up point(a)Find the top position of diagonal section.According to the figure, the abscissa of normal section with a distance 1630mm to the center of support is. And the effective height of normal section is. Let the inclined section projected length. So, the top position of diagonal section A is found and its abscissa is.(b)Check the shear capacity of diagonal sectionThe shear force and bending moment of A section can be calculated as followings.The effective height of normal section A is. (Main steel bars are) So, the real generalized shear spam ratio m and projected length c arePlot the to-be-checked diagonal section and get the bevelThere are (2N6+2N5) main steel bar with the diagonal section and the corresponding reinforcement ration p isAnd web reinforcement ration (take) isAnd the bend-up steel bar intersecting with diagonal section A are 2N3and 2N4and 2N2.According to the unit requirements, the inclined section carrying capacity can be got by subtitling results listed above.Therefore, the shearing capacity of diagonal section with a distance 500mm (i.e. h/2) to the center of support is verified.(3)Check one – the bent-up point(a)Find the top position of diagonal section.According to the figure, the abscissa of normal section with a distance 3230mm to the center of support is. And the effective height of normal section is. Let the inclined section projected length. So, the top position of diagonal section A is found and its abscissa is.(b)Check the shear capacity of diagonal sectionThe shear force and bending moment of A section can be calculated as followings.The effective height of normal section A is. (Main steel bars are) So, the real generalized shear spam ratio m and projected length c arePlot the to-be-checked diagonal section and get the bevelThere are (2N6+2N5) main steel bar with the diagonal section and the corresponding reinforcement ration p isAnd web reinforcement ration (take) isAnd the bend-up steel bar intersecting with diagonal section A are 2N4. According to the unit requirements, the inclined section carrying capacity can be got by subtitling results listed above.Therefore, the shearing capacity of diagonal section with a distance 500mm (i.e. h/2) to the center of support is verified.(4)Check one – the variable point of stirrup spacing(a)Find the top position of diagonal section.According to the figure, the abscissa of normal section with a distance 1000mm to the center of support is. And the effective height of normal section is. Let the inclined section projected length. So, the top position of diagonal section A is found and its abscissa is.(b)Check the shear capacity of diagonal sectionThe shear force and bending moment of A section can be calculated as followings.Project of principles of structure design 201121070108 Wang PengzhiThe effective height of normal section A is. (Main steel bars are) So, the real generalized shear spam ratio m and projected length c arePlot the to-be-checked diagonal section and get the bevelThere are (2N6+2N5) main steel bar with the diagonal section and the corresponding reinforcement ration p isAnd web reinforcement ration (take) isAnd the bend-up steel bar intersecting with diagonal section A are 2N4and 2N3.According to the unit requirements, the inclined section carrying capacity can be got by subtitling results listed above.Therefore, the shearing capacity of diagonal section with a distance 500mm (i.e. h/2) to the center of support is verified.10。

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