lecture5混凝土结构设计原理-英文课件
结构设计原理课程设计-混凝土梁(英文版)
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。
《混凝土结构设计原理》双语 (5)
ρ
ρ
A～B:
ρρmin
ρ (少筋梁) Mu=Mcr=const.
Brittle failure
C～D:
ρ min
ρρmax
(适筋梁) Steel→fy Concrete→fc
When ρρmax Balanced section
given by
y
h0 xb xb
u
xb
• The depth of the neutral axis is then h0
εu=0.0033
xb
h0
1 y
u
a εy=fy/Es
• The ultimate strain of concrete cu = 0.0033; and for the HPB400 steel (fy = 400N/mm2), the yield strain y =
• Over-reinforced section (超筋梁)
Concrete strain reaches εcu before steel strain reaches εy
The failure of an over-reinforced beam is initiated by the crushing of the concrete, while the steel strain is still relatively low. The failure is therefore characterised by a small deflection and by the absence of extensive cracking in the tension zone. The failure, often explosive (爆发性的), occurs with little warning.
混凝土结构设计原理-受弯构件76页PPT
第II阶段——带裂缝工作阶段
砼开裂后，受压区高度明显减小，中和轴明显上移。开裂砼退出工作,
拉应力转卸给钢筋承担。开裂后，刚度降低，挠度 f 增加的速度较快。
在第II阶段，随M增大，裂缝宽度也增大，开裂截面钢筋拉应力增大，
s 增大至屈服。Mf 曲线出现第二个转折点，记为II a状态。
2、钢筋骨架的构造
• 绑扎骨架——用细铁丝绑扎钢筋，整体现浇 • 焊接骨架——先将钢筋焊成平面骨架，再用箍筋联接
为主体骨架。主要用于预制构件 焊接骨架构造见图3-13，P52
双面焊接时为图中尺寸 单面焊时图中尺寸加倍
受力主钢筋：直径14~25mm, 特殊(如构件特别大)，可用28、32mm 采用不同直径时，两种钢筋的直径差应≥2mm，便于识别。
≥
30
d
mm（三层及三层以下）
≥14.02m5dm （三层以上）
≥20mm C
水平纵向钢筋
≥40mm ＞1.25d 主钢筋
3-2 受弯构件正截面受力过程和破坏形态
一、试验研究
受拉钢筋合力作用点（实际上就是受拉钢筋截面形心）
至受压边缘的距离 h 0 ，称为有效高度。
定义 配筋率 As
b h0
试验梁的配筋率 0 .000 9 .97 % 7
• 板内钢筋保护层厚度 2cm • 单向板应设分布钢筋。
分布钢筋的作用： ①将板面上的荷载更均匀地传递给受力钢筋； ②固定受力钢筋的位置； ③抵抗温度应力和砼收缩应力； ④承受沿长边传递的荷载。
分布钢筋的要求：
直径≥6mm 间距≤250mm
位置：与受力钢筋垂直，布置在受力钢筋内侧
主钢筋弯折处，均应布置分布钢筋。（板中一般按30°弯起 ）
(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）
混凝土结构设计原理PPT详解110页PPT
6、最大的骄傲于最大的自卑都表示心灵的最软弱无力。——斯宾诺莎 7、自知之明是最难得的知识。——西班牙 8、勇气通往天堂，怯懦通往地狱。——塞内加 9、有时候读书是一种巧妙地避开思考的方法。——赫尔普斯 10、阅读一切好书如同和过去最杰出的人谈话。——笛卡儿
Thank you
混凝土结构设计原理PPT详解
36、“不可能”这个字(法语是一个字 )，只 在愚人 的字典 中找得 到。--拿 破仑。 37、不要生气要争气，不要看破要突 破，不 要嫉妒 要欣赏 ，不要 托延要 积极， 不要心 动要行 动。 38、勤奋，机会，乐观是成功的三要 素。(注 意：传 统观念 认为勤 奋和机 会是成 功的要 素，但 是经过 统计学 和成功 人士的 分析得 出，乐 观是成 功的第 三要素 。
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。
混凝土结构设计原理-受弯构件(课件)
2、钢筋骨架的构造
• 绑扎骨架——用细铁丝绑扎钢筋，整体现浇 • 焊接骨架——先将钢筋焊成平面骨架，再用箍筋联接
为主体骨架。主要用于预制构件 焊接骨架构造见图3-13，P52
双面焊接时为图中尺寸 单面焊时图中尺寸加倍
受力主钢筋：直径14~25mm, 特殊(如构件特别大)，可用28、32mm 采用不同直径时，两种钢筋的直径差应≥2mm，便于识别。
第III阶段——破坏阶段
钢筋屈服后，进入第III阶段。裂缝迅速开展，中和轴迅速上移，刚度
急剧下降，挠度 f 明显增大，最后发展至受压区边缘砼达到极限压应变，
梁破坏。破坏时记为IIIa 状态。
y y
应变图
s s Es s y
0
[2(
0
)
(
0
)
2
]
cu
应力图 M
tu
Mcr
M
y
My
M
xc D
≥
30
d
mm （三层及三层以下）
≥
40mm 1.25d
（三层以上）
≥20mm C
水平纵向钢筋
≥40mm ＞1.25d 主钢筋
3-2 受弯构件正截面受力过程和破坏形态
一、试验研究
受拉钢筋合力作用点（实际上就是受拉钢筋截面形心）
至受压边缘的距离 h0 ，称为有效高度。
定义 配筋率 As
bh0
试验梁的配筋率 0.0097 0.97%
第三章 受弯构件正截面承载力计算
Cross-Section Carrying Capacity of Flexural Members
注：通过学习前两章的基础知识，进而开始讨论 钢筋混凝土构件的设计。受弯构件是所有构件中 最基本的一种，掌握了它的计算，以后各章的内 容就是举一反三。所以我们要详细讨论，并且熟 练掌握本章内容。
(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。
chapter 1(混凝土结构设计原理英文课件)
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.
• 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.
混凝土结构设计原理第五章ppt课件
Nu
fc Acor 2
f y Ass0
f yAs
令2 / 2
图5-11 混凝土径向压力示意图 Nu 0.9( fc Acor 2 f y Ass0 f yAs)
α称为间接钢筋对混凝土约束的折减系数，当混凝土强度等级不超过C50时， 取α＝1.0；当混凝土强度等级为C80时，取α＝0.85；当混凝土强度等级在 C50与C80之间时，按直线内插法确定。
图5-16 不同长细比柱从加荷到破坏的N-M关系
在图5 -16中，示出了截面尺寸、配 筋和材料强度等完全相同，仅长细比不 相同的3根柱，从加载到破坏的示意图。
5.4 偏心受压构件的二阶效应
轴向压力对偏心受压构件的侧移和挠 曲产生附加弯矩和附加曲率的荷载效应称 为偏心受压构件的二阶荷载效应，简称二 阶效应。其中，由侧移产生的二阶效应， 习称P-Δ效应；由挠曲产生的二阶效应， 习称P-δ效应。
①M1/M2＞0.9或 ②轴压比N/fcA＞0.9或
③lci＞34-12(M1/M2)
3)考虑二阶效应后控制截面的弯矩设计值
《混凝土结构设计规范》规定，除排架结构柱外，
其他偏心受压构件考虑轴向压力在挠曲杆件中产生的
二阶效应后控制截面的弯矩设计值，应按下列公式计
算：
M CmnsM 2
Cm
0.7 0.3
5.3.2 偏心受压长柱的破坏类型
图5-15 长柱实测N-f曲线 偏心受压长柱在纵向弯曲影响下，可能发生失稳破坏和材料破坏两种破坏类 型。长细比很大时，构件的破坏不是由材料引起的，而是由于构件纵向弯曲失去 平衡引起的，称为“失稳破坏”。当柱长细比在一定范围内时，虽然在承受偏心 受压荷载后，偏心距由ei增加到 ei+f，使柱的承载能力比同样截面的短柱减小， 但就其破坏特征来讲与短柱一样都属于“材料破坏”，即因截面材料强度耗尽而 产生破坏。
chapter 1(混凝土结构设计原理英文课件)
1-3 cement and water
• Structural concrete uses, almost exclusively, hydraulic cement. With this cement, water is necessary for the chemical reaction of hydration. In the process of hydration, the cement sets and bonds the fresh concrete into one mass. Portland cement, which originated in England, is undoubtedly the most common form of cement. • Portland cement consists chiefly of calcium and aluminum silicates.
1-4 aggregates
• Aggregates are classified as fine or coarse. Fine aggregate is generally sand and may be categorized as consisting of particles that would be retained on a No. 4 sieve(four openings per linear inch). Coarse aggregate consists of particles that would be retained on a No. 4 sieve. The maximum size of course aggregate in reinforced concrete is governed by various ACI Code requirements. These requirements are established primarily to ensure that the concrete can be placed with ease into the forms without any danger of jam-up between adjacent bars or between bars and the sides of the forms.
《混凝土结构设计原理》双语 (13)
• Variance （方差）
Variance=
E[( X
E(X )2 ]
E[ X
2 ] [E(X )]2
2 X
• Standard deviation（标准差） X
• Coefficient of variation （变异系数）
X X E(x) X
s
s
s
s
Pf P(Z 0) [ f R (r)dr] fS (S)dS FR (S) fS (S)dS
00
0
First-order second-moment method
Z=R-S
Z R S
Z
2 R
2 S
βis called reliability index (可靠度指标)
GB50100-2002
第4.1.3条 混凝土轴心抗压，轴心抗拉强度标准值fck,ftk应 按表4.1.3采用。
强度种 类
混凝土强度标准值(N/mm2) 混凝土强度等级
表4.1.3
C15 C20 C25 C30 C35 C40 C45 C50 C55 C60 C65 C70 C75 C80
fck
10.0 13.4 16.7 20.1 23.4 26.8 29.6 32.4 35.5 38.5 41.5 44.5 47.4 50.2
3.4 Classification of design methods
• 水准Ⅰ • 水准Ⅱ • 水准Ⅲ
半概率法 近似概率法 全概率法
3.5 Limit State Design
• Two principal types of limit state: • o Ultimate limit state: The whole structure or its
lecture5混凝土结构设计原理-英文课件
June 11, 2003 CVEN 444
Lecture Goals
Rectangular Beams Safety factors Loading and Resistance Balanced Beams
Flexural Stress
The compressive zone is modeled with a equivalent stress block.
b1 0.85 for fc 4000 psi
Flexural Stress – Rectangular Example
From equilibrium (assume the steel has yielded)
C T 0.85 f cba f y As a f y As 0.85 f cb 0.85 4 ksi 12 in
Flexural Stress – Non-Rectangular Example
Compute the b1 value
f c 4000 psi b1 0.85 0.05* 1000 psi 6000 psi 4000 psi 0.85 0.05* 1000 psi 0.75
Flexural Stress – Non-Rectangular Example
Find the neutral axis
c
a
b1
5.0 in. 6.67 in. 0.75
Safety Provisions
Structures and structural members must always be designed to carry some reserve load above what is expected under normal use.
2019精品混凝土结构设计原理 第5章 受压构件的截面承载力英语
f
' y
As'
f y As
Nub
Nub
1 fcbbh0
f
' y
As'
f y As
N Nub ,
N Nub ,
1x
N
Nu
1 fcbx
f
' y
As'
f y As
2)x
1 1 1400 ei
l0 h
2
1
2e0
h0
Ne
1 fcbx h0
x 2
f
' y
As'
h0
as'
3)M u , M u Ne0
b
�5.4
5
2
N Nu 1 fcbx
x
3
1
x N
1 fcb
x bh0 x bh0
M N l0 As As
�5.4
5
x
x xb bh0
x N
1 fcb
As 1 x 2as
2 x 2as
As
As
Ne 1 fcbx(h0
f yh0 as
N Nu 1 fcbx
Ne
Nue
1 fcbx(h0
x) 2
fyAsh0
as
3
x bh0
x 2as
5.4
5
l0 b Nu
�5.6
�5.4
1
1
1 2
As As
N Nu 1 fcbx
Ne
Nue
1 fcbx(h0
x) 2
fyAsh0
as
3
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
Flexural Stress – Rectangular Example
Determine the area of steel, #7 bar has 0.6 in2.
As 4 0.6 in 2 2.4 in 2
The b value is b1 = 0.85 because the concrete has a fc =4000 psi.
Find the center of the area of concrete area
yA y A
i i
i
6 in. 3 in.1.5 in. 10 in. 2 in. 4 in. 6 in. 3 in. 10 in. 2 in. 2.8158 in.
Consequences of Failure
A number of subjective factors must be considered in determining an acceptable level of safety. Potential loss of life. Cost of clearing the debris and replacement of the structure and its contents. Cost to society. Type of failure warning of failure, existence of alternative load paths.
Example of rectangular reinforced concrete beam. (2) Find flexural capacity.
M n T moment arm a As fs d 2
Flexural Stress
Example of rectangular reinforced concrete beam. (3) Need to confirm es > ey
Variability in Resistance
Comparison of measured and computed failure moments based on all data for reinforced concrete beams with fc > 2000 psi.
Margin of Safety
60 ksi ey 0.00207 Es in in the steel
fy
d c es 0.003 Steel yielded! c 15.5 in. 4.152 in. 0.003 0.0082 0.000207 4.152 in.
Safety Provisions
There are three main reasons why some sort of safety factor are necessary in structural design.
[1] Consequences of failure. [2] Variability in loading. [3] Variability in resistance.
Flexural Stress – Non-Rectangular Example
Find the neutral axis
c
a
b1
5.0 in. 6.67 in. 0.75
Safety Provisions
Structures and structural members must always be designed to carry some reserve load above what is expected under normal use.
Flexural Stress
The equivalent rectangular concrete stress distribution has what is known as a b1 coefficient is proportion of average stress distribution covers.
Variability in Loading
Frequency distribution of sustained component of live loads in offices.
Variability in Resistance
Variability of the strengths of concrete and reinforcement. Differences between the as-built dimensions and those found in structural drawings. Effects of simplification made in the derivation of the members resistance.
ey
c
y
Es a
es
d c e
c
b1
c
ey
Flexural Stress – Rectangular Example
Example of rectangular reinforced concrete beam. Given a rectangular beam fc = 4000 psi fy = 60 ksi (4 #7 bars) b = 12 in. d = 15.5 in. h= 18 in. Find the neutral axis. Find the moment capacity of the beam.
2 60 ksi 2.4 in
3.53 in.
The neutral axis is a 3.53 in. c 4.152 in. b1 0.85
Flexural Stress – Rectangular Example
Check to see whether or not the steel has yielded.
Flexural Stress
Example of rectangular reinforced concrete beam. (2) Find flexural capacity.
T As fs C 0.85 f c ab As fs a 0.85 f cb
Flexural Stress
[1] Stress-Strain Compatibility – Stress at a point in member must correspond to strain at a point.
[2] Equilibrium – Internal forces balances with external forces
Flexural Stress – Non-Rectangular Example
The moment capacity of the beam is
Mn T d y 193.8 kips 12.5 in. 2.8158 in. 1869 k-in. 155.75 k-ft.
Using equilibrium, the area of the steel can be found
T C 0.85 f c Ac fs As 0.85 f c Ac As fs 193.8 kips 2 As 3.23 in 60 ksi
Flexural Stress – Non-Rectangular Example
Lecture 5 - Flexure
June 11, 2003 CVEN 444
Lecture Goals
Rectangular Beams Safety factors Loading and Resistance Balanced Beams
Flexural Stress
The compressive zone is modeled with a equivalent stress block.
b1 0.85 for f c 4000psi
f c 4000 b1 0.85 0.05* 0.65 1000
Flexural Stress
Requirements for analysis of reinforced concrete beams
2
Flexural Stress – Non-Rectangular Example
For a non-rectangular beam For the given beam with concrete rated at fc = 6 ksi and the steel is rated at fs = 60 ksi. d = 12.5 in. (a) Determine the area of the steel for a balanced system for shown area of concrete.
Ac 6 in. 3 in. 10 in. 2 in. 38 in 2
The force due to concrete forces.
C 0.85 f c Ac
0.85 6 ksi 38 in 2 193.8 kips.
Flexural Stress – Non-Rectangular Example
Flexural Stress – Non-Rectangular Example
Compute the b1 value
f c 4000 psi b1 0.85 0.05* 1000 psi 6000 psi 4000 psi 0.85 0.05* 1000 psi 0.75