lecture5混凝土结构设计原理-英文课件

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。

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。