lecture3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》双语 (15)
10.1.3 Concept of prestressed Concrete
10.1.4 Classification of Prestressed concrete
• b)Reduced deflections under service condition. • c)The pre-compression in concrete tends to
reduce the diagonal tension (Shear failure). • d) Prestressed concrete members are lighter. • e) Absence or near absence of cracks.
10.4 Materials used in P.C member.
• a) High strength 增加预应力,减轻自重。 • b) Small shrinkage and creep. • c) Hard faster .
10.4.2 Prestressed reinforcement.
Two concepts: Transfer and Transmission length.
When the tendons is released, the force in the tendons is transferred to concrete by bond stress. “自 锚”。
The length required to transmit the full tendon stress to the concrete is called the transmission length. Ltr.
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。
3混凝土结构设计原理课件
轴心受压构件承载力计算
轴心受力构件 轴心受压长柱稳定系数 主要与柱的长细比 有关, 轴心受压长柱稳定系数φ 主要与柱的长细比 l0 / b 有关, 稳定系数的定义如下: 稳定系数的定义如下: l Nu ϕ= s Nu
《规范》给出的稳定系数与长细比的关系 规范》
l0/b ≤8 10 12 14 16 18 20 22 24 26 28 l0/d ≤7 8.5 10.5 12 14 15.5 17 19 21 22.5 24 l0/i 28 35 42 48 55 62 69 76 83 90 97 φ ≤1.0 0.98 0.95 0.92 0.87 0.81 0.75 0.7 0.65 0.6 0.56 l0/b 30 32 34 36 38 40 42 44 46 48 50 l0/d 26 28 29.5 31 33 34.5 36.5 38 40 41.5 43 l0/i 104 111 118 125 132 139 146 153 160 167 174 φ 0.52 0.48 0.44 0.4 0.36 0.32 0.29 0.26 0.23 0.21 0.19
轴心受压构件承载力计算
轴心受力构件
2/2
普通钢箍柱 Tied Columns
螺旋钢箍柱 Spiral Columns 轴心受压构件承载力计算
轴心受力构件
1 短柱与长柱
短柱( 短柱(Short Columns)是如何形成 的? ) 我们通常将柱的截面尺寸与柱长之比较小的柱,称为短柱。 我们通常将柱的截面尺寸与柱长之比较小的柱,称为短柱。在实 短柱 际结构中,带窗间墙的柱、高层建筑地下车库的柱子, 际结构中,带窗间墙的柱、高层建筑地下车库的柱子,以及楼梯 间处的柱都容易形成短柱。 间处的柱都容易形成短柱。
chapter 3 reinforced concrete beams(混凝土结构设计原理英文课件)
• 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
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.
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.
混凝土结构原理第3章结构设计方法课件
3.2 结构的极限状态
1. 承载力能力极限状态 结构或构件达到最大承载能力或者达到不适于继续承载的
变形状态。 ◆结构或构件达到最大承载力(包括疲劳) ◆结构整体或其中一部分作为刚体失去平衡(如倾覆、滑移) ◆结构塑性变形过大而不适于继续使用 ◆结构形成几何可变体系(超静定结构中出现足够多塑性铰) ◆结构或构件丧失稳定(如细长受压构件的压曲失稳) ◆遭受偶然作用引起的结构连续倒塌。
3.5.2 结构的失效概率与可靠度
重叠区的大小反映了抗力R
p
和荷载效应S之间的概率关
系,即结构的失效概率
S
R
μS mR-mS
μR
S,R
从结构安全的角度,提高结构构件的抗力R,减小抗力R和荷载效应S
的离散程度,可以提高结构的可靠程度,即提高μR- μS,减小σR 、 σS可使 失效概率降低。
3.5.2 结构的失效概率与可靠度
对于结构设计而言,如何设计的安全呢? 荷载取值越大,内力值就越大,构件截面尺寸也愈大,结构 愈安全; 材料强度取值越低,结构所需截面越大,结构愈安全.
Sm Sm+1.645σ
fm-1.645σ fm
荷载标准值: 材料强度标准值:
Sk Sm 1.645s Sm (1 1.645 ) fk fm 1.645s fm(11.645 )
3.5.5 正常使用极限状态设计表达式
对于正常使用极限状态,结构构件应分别按荷载效应 的标准组合、频遇组合、准永久组合或标准组合并考虑长 期作用影响,采用下列设计表达式:
S≤C 式中,S——正常使用极限状态的荷载组合效应的设计 值(如变形、裂缝宽度、应力等的效应设计值);C——结 构构件达到正常使用要求所规定的变形、裂缝宽度和应力 等的限值。
《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
混凝土结构设计原理PPT详解共110页
▪
26、要使整个人生都过得舒适、愉快,这是不可能的,因为人类必须具备一种能应付逆境的态度。——卢梭
混凝土结构设计原理PPT详解
11、用道德的示范来造就一个人,显然比用法律来约束他更有价值。—— 希腊
12、法律是无私的,对谁都一视同仁。在每件事上,她都不徇私情。—— 托马斯
13、公正的法律限制不了好的自由,因为好人不会去做法律不允许的事 情。——弗劳德
14、法律是为了保护无辜而制定的。——爱略特 15、像房子一样,法律和法律都是相互依存的。——伯克
▪
27、只有把抱怨环境的心情,化为上进的力量,才是பைடு நூலகம்功的保证。——罗曼·罗兰
▪
28、知之者不如好之者,好之者不如乐之者。——孔子
▪
29、勇猛、大胆和坚定的决心能够抵得上武器的精良。——达·芬奇
▪
30、意志是一个强壮的盲人,倚靠在明眼的跛子肩上。——叔本华
谢谢!
110
《混凝土结构设计原理》双语 (12)
Failure process
v a 2t
The rapid expansion at the minimum volume level, where the tensile strain at right angles(正交) to the direction of maximum compressive stress soon exceeds the magnitude of the compressive strain is a failure feature of both uniaxial (单轴)and more general triaxial (三向)compressive behavior
➢ ACI Secant modulus when stress is 0.4fc’
Poisson’s ratio (泊松比:侧向应变与轴向应变之比)
➢Whenσ<0.5fc: 0.1~0.2, ➢Whenσ>0.5fc: larger then 0.5. ➢Slight lower for high strength concrete ➢ usually letγ=1/6 ➢规范: γ=0.2
Concrete in the “critical’ zones of compression always failed by “splitting”(劈裂)---never by “crashing” (压碎)
illustration of the theory ------Transverse confine can (侧向约束) greatly increase axial compressive strength
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Concrete Mixing and Proportioning
Economical takes into account effective use
of materials, effective operation, and ease of handling. The cost of producing good quality concrete is an important consideration in the overall cost of the construction project.
Concrete Mixing and Proportioning
Failure Mechanism of Concrete
Shrinkage Microcracks are the initial shrinkage cracks due to carbonation shrinkage, hydration shrinkage, and drying shrinkage.
Ec ( psi) = 33 w1.5
f 'c ( psi)
Ec ( psi) = 57,000
f 'c ( psi)
Concrete Properties
Concrete strain at max. compressive stress, o For typical curves in compression o varies between 0.0015-0.003 For normal strength concrete, o ~ 0.002
fc f’c 0.45f’c Ec
o
u
Concrete Properties
Maximum useable strain, u ACI Code: u = 0.003 Used for flexural and axial compression
fc Ec
f’c
0.45f’c
o
Concrete Mixing and Proportioning
Failure Mechanism of Concrete Matrix Microcracks - are microcracks that occur in the matrix. The propagate from 20% fc. Occur up to 30-45 % ultimate strength of concrete. Matrix microcracks start bridge one another at 75%. Aggregate microcracks occur just before failure (90%).
Concrete Mixing and Proportioning
Slump test - The measurement of the consistency of the mix is done with the slump-cone test. The recommend consistency for various classes of concrete structures .
Concrete Mixing and Proportioning
5. Types of Cement
Type IV: Low Heat of Hydration Gradually heats up, less distortion (massive structures). Type V: Sulfate Resisting For footings, basements, sewers, etc. exposed to soils with sulfates.
Concrete Mixing and Proportioning
The influence of ingredients on properties of concrete.
Concrete Mixing and Proportioning
3. Workability
Workability measured by slump test 12” slump
o u
c
Concrete Properties
Modulus of Elasticity, Ec Corresponds to secant modulus at 0.45 f’c ACI 318-02 (Sec. 8.5.1):
1.5
Ec ( psi) = 33 w
f 'c ( psi)
[b] the separation of the specimen into columnar pieces by what is known as splitting or columnar fracture. [c] Combination of shear and splitting failure.
1. 2. 3. 4.
1 2 3 4 Layer 1: Fill 1/3 full. 25 stokes Layer 2: Fill 2/3 full. 25 stokes Layer 3: Fill full. 25 stokes Lift cone and measure slump (typically 2-6 in.)
Concrete Properties
1. Uniaxial Stress versus Strain Behavior in Compression fc
f’c 0.45f’c Ec
12”
6”
o
u
c
Concrete Properties
The standard strength test generally uses a cylindrical sample. It is tested after 28 days to test for strength, fc. The concrete will continue to harden with time and for a normal Portland cement will increase with time as follows:
Quality Workability Economical
Concrete Mixing and Proportioning
Quality of concrete is measured by its
strength and durability. The principal factors affecting the strength of concrete , assuming a sound aggregates, W/C ratio, and the extent to which hydration has progressed. Durability of concrete is the ability of the concrete to resist disintegration due to freezing and thawing and chemical attack.
Lecture 3 - Fundamentals
June 6, 2003 CVEN 444
Lecture Goals
Concrete Properties Steel Properties
Concrete Mixing and Proportioning
In the design of concrete mixes, three principal requirements for concrete are of importance:
Concrete Mixing and Proportioning
Failure Mechanism of Concrete Bond Microcracks are extensions of shrinkage microcracks, as the compression stress field increases, the shrinkage microcracks widen but do not propagates into the matrix. Occur at 15-20 % ultimate strength of concrete.
chemically releasing water from fine aggregates.
Concrete Mixing and Proportioning
5. Types of Cement
Type I: General Purpose Type II: Lower heat of hydration than Type I Type III: High Early Strength Higher heat of hydration quicker strength (7 days vs. 28 days for Type I)
u
Concrete Properties
Typical Concrete Stress-Strain Curves in Compression
Concrete Properties
Types of compression failure
There are three modes of failure. [a] Under axial compression concrete fails in shear.
Concrete Mixing and Proportioning
Workability of concrete may be defined as a