lecture5混凝土结构设计原理英文PPT课件.ppt

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1、Lecture 5-Flexure,June 11,2003CVEN 444,Lecture Goals,Rectangular Beams Safety factorsLoading and ResistanceBalanced Beams,Flexural Stress,The compressive zone is modeled with a equivalent stress block.,Flexural Stress,The equivalent rectangular concrete stress distribution has what is known as a b1

2、coefficient is proportion of average stress distribution covers.,Flexural Stress,Requirements for analysis of reinforced concrete beams,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

3、 Stress,Example of rectangular reinforced concrete beam.,(1)Setup equilibrium.,Flexural Stress,Example of rectangular reinforced concrete beam.,(2)Find flexural capacity.,Flexural Stress,Example of rectangular reinforced concrete beam.,(2)Find flexural capacity.,Flexural Stress,Example of rectangula

4、r reinforced concrete beam.,(3)Need to confirm es ey,Flexural Stress Rectangular Example,Example of rectangular reinforced concrete beam.,Given a rectangular beamfc=4000 psify=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.,Flexural Stress Rectan

5、gular Example,Determine the area of steel,#7 bar has 0.6 in2.The b value is b1=0.85 because the concrete has a fc=4000 psi.,Flexural Stress Rectangular Example,From equilibrium(assume the steel has yielded),The neutral axis is,Flexural Stress Rectangular Example,Check to see whether or not the steel

6、 has yielded.,Check the strain in the steel,Steel yielded!,Flexural Stress Rectangular Example,Compute moment capacity of the beam.,Flexural Stress Non-Rectangular Example,For the given beam with concrete rated at fc=6 ksi and the steel is rated at fs=60 ksi.d=12.5 in.,For a non-rectangular beam,(a)

7、Determine the area of the steel for a balanced system for shown area of concrete.(b)Determine the moment capacity of the beam.Mn(c)Determine the NA.,Flexural Stress Non-Rectangular Example,The area of the concrete section is,For a non-rectangular beam,The force due to concrete forces.,Flexural Stres

8、s Non-Rectangular Example,Using equilibrium,the area of the steel can be found,Flexural Stress Non-Rectangular Example,Find the center of the area of concrete area,Flexural Stress Non-Rectangular Example,The moment capacity of the beam is,Flexural Stress Non-Rectangular Example,Compute the b1 value,

9、Flexural Stress Non-Rectangular Example,Find the neutral axis,Safety Provisions,Structures and structural members must always be designed to carry some reserve load above what is expected under normal use.,Safety Provisions,There are three main reasons why some sort of safety factor are necessary in

10、 structural design.1 Consequences of failure.2 Variability in loading.3 Variability in resistance.,Consequences of Failure,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

11、load paths.,A number of subjective factors must be considered in determining an acceptable level of safety.,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

12、between the as-built dimensions and those found in structural drawings.Effects of simplification made in the derivation of the members resistance.,Variability in Resistance,Comparison of measured and computed failure moments based on all data for reinforced concrete beams with fc 2000 psi.,Margin of

13、 Safety,The distributions of the resistance and the loading are used to get a probability of failure of the structure.,Margin of Safety,The term Y=R-Sis called the safety margin.The probability of failure is defined as:and the safety index is,Loading,SPECIFICATIONSCities in the U.S.generally base th

14、eir building code on one of the three model codes:Uniform Building CodeBasic Building Code(BOCA)Standard Building Code,Loading,These codes have been consolidated in the 2000 International Building Code.Loadings in these codes are mainly based on ASCE Minimum Design Loads for Buildings and Other Stru

15、ctures has been updated to ASCE 7-02.,Loading,The loading variations are taken into consideration by using a series of“load factors”to determine the ultimate load,U.,Loading,The equations come from ACI code 9.2 on loading(4.6 in your book),D Dead LoadW Wind LoadL Live LoadLr Roof LoadF Fluid Pressur

16、e R Rain LoadE Earthquake LoadT Temperature LoadS Snow LoadH Soil Load,Loading,The most general equation for the ultimate load,U(Mu)that you will see is going to be:,Resistance,The load factors will generate the ultimate load,which is used in the design and analysis of the structural member.Mu Ultim

17、ate MomentMn Nominal Momentf Strength Reduction Factor,Resistance,The strength reduction factor,f,varies from member to member depending whether it is in tension or compression or the type of member.The code has been setup to determine the reduction.,Three possibilities in Inelastic Behavior,Compres

18、sion Failure-(over-reinforced beam)Tension Failure-(under-reinforced beam)Balanced Failure-(balanced reinforcement),Inelastic Behavior,Compression Failure,The concrete will crush before the steel yields.This is a sudden failure.The beam is known as an over-reinforced beam.,Inelastic Behavior,Tension

19、 Failure,The reinforcement yields before the concrete crushes.The concrete crushes is a secondary compression failure.The beam is known as an under-reinforced beam.,Inelastic Behavior,Balanced Failure,The concrete crushes and the steel yields simultaneously.The beam is known as an balanced-reinforce

20、d beam.,Inelastic Behavior,Which type of failure is the most desirable?,The under-reinforced beam is the most desirable.fs=fyes eyYou want ductility,system deflects and still carries load.,Balanced Reinforcement Ratio,rbal,rbal=unique r value to get simultaneous ec=0.003&es=ey,Use similar triangles:

21、,Balanced Reinforcement Ratio,rbal,The equation can be rewritten to find cb,Nominal Moment Equation,The equation can be rewritten in the form:,Nominal Moment Equation,The equation can be rewritten in the form:,Use the ratio r=b/d and r,Nominal Moment Equation,Use w=r fy/fc and,Use the ratio r=b/d an

22、d R,Strain Limits Method for Analysis,The strength reduction factor,f,will come into the calculation of the strength of the beam.,Limitations on Reinforcement Ratio,r,The selection of the steel will be determined by the Lower Limit on r ACI 10.5.1 ACI Eqn.(10-3)fc&fy are in psi,Limitations on Reinfo

23、rcement Ratio,r,Lower Limit on r ACI 10.5.1Lower limit used to avoid“Piano Wire”beams.Very small As(Mn Mcr)beam fails right away because Mn Mcr,Additional Requirements for Lower Limit on r,If As(provided)4/3 As(required)based on analysis,then As(min)is not required.for As(provided)See ACI 10.5.3,Add

24、itional Requirements for Lower Limit on r,Temperature and Shrinkage reinforcement in structural slabs and footings(ACI 7.12)place perpendicular to direction of flexural reinforcement.GR 40 or GR 50 Bars:As(T&S)=0.0020 AgGR 60 or Welded Wire Fabric(WWF):As(T&S)=0.0018 Ag Ag-Gross area of the concrete

25、,Example,Given:fc=3 ksi&fy=40 ksi and As=4 in2Determine:(1)(2),Determine if the beam will satisfy ACI code.If fc=6 ksi?,Example,Given:fc=3 ksi&fy=40 ksi and As=4 in2,The minimum steel ratio is,Example,Given:fc=3 ksi&fy=40 ksi and As=4 in2,The neutral axis is,Example,The strain in the steel is,There

26、for the beam is in the compression zone and f would be 0.65,however c/d ratio is greater than 0.375 so the beam will need to be redesigned.,Example,c/d=0.615,Example,Given:fc=6 ksi&fy=40 ksi and As=4 in2,The minimum steel ratio is,Example,Given:fc=6 ksi&fy=40 ksi and As=4 in2,The neutral axis is at,Example,The strain in the steel will be,There for the beam is in the tension zone and f will be 0.9.,Example,c/d=0.349,