EGB375: DESIGN OF CONCRETE STRUCTURES | Reliable Papers

EGB375: DESIGN OF CONCRETESTRUCTURESWORKBOOKTatheer ZahraLecturer in Civil Engineering, QUT• This workbook must be accompanied (in print or electronic form) with the studentin the lecture and tutorial sessions.• This workbook is provided free of cost to the students enrolled in EGB375 atQUT; students may elect to print at their own cost.• Distribution of this workbook to other university and posting the workbook insocial media websites will be illegal as it infringes the copyright.EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 1Workbook:EGB375 Design of Concrete StructuresThis workbook is distributed free of cost to all students enrolled in EGB375 with theobjective of preparing the students prior to attending work sessions (lecture or tutorial).The students are encouraged to print and work on the hardcopy in the work sessions.Where the students are quite skilled in using e-copies to take notes including writingequations and performing calculations, they may elect to do so; however, in the examthey will have to use pen and paper – no e-tools shall be allowed. The page numbersare intentionally left blank – for you to get started working from this page!Contents: SectionWorkbook TitlePageStudy & Assessment Schedule and Grading System1Flexural Design of Singly Reinforced Beams2Flexural Design of Double Reinforced & T-Beams3Introduction to Prestressed Concrete4Flexural Design of Prestressed Concrete Beams5Shear Design of RC & PSC Beams6 & 7Serviceability of Beams8Design of RC Slabs9Design of RC Column Cross Sections10Design of Short and Long Columns11Design of RC Shallow Foundation & Retaining Walls Disclaimer:(1) This workbook is solely for the use in the QUT unit EGB375; it should neither bedistributed to outside QUT nor used in other units without the consent of thecoordinator of other units and author.(2) This workbook is not a substitute for any textbooks on concrete structures.EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 2A typical concrete structural system in buildings is shown in the above figure. Key members(or, elements) of the structural system are identified as beams, slabs, columns and walls in thefigure. A structural element resists the load that is imposed on it and provides a safe workingplace for its occupants. Civil structural engineers are responsible for choosing appropriatematerials to design structural elements. Civil construction engineers are also advised by thestructural engineers on assembling of the structural elements and conforming design details.You may work as a structural engineer for a design company or a construction company; yourrole may then be actively designing or checking the design of another engineer. The designer,design checker, the construction engineer, are all responsible for the safety of the building as ateam. If you do not have a comprehensive understanding of the design principles to defend yourdesign in the court of law to an arbitrating engineering committee, you would risk theprofessional engineering registration (Registered Professional Engineers Queensland). Thisunit covers the design principles and processes for concrete beams, slabs, columns and footing(not shown in the figure). Design of walls will be covered in an elective unit (EGB386). Anadvanced unit (EGH475) will deal with the whole structural system in detail.The workbook contains 11 sections and will be covered over a 13 week semester; a schedulefor this coverage is in P2 of this workbook – you are strongly advised to prepare for each weekas per the schedule and attend the lectures/tutorials with familiarity of the relevant section inthis workbook. Please note – it is a 12 credit point unit; you are expected to spend 12 hours/week (including the four hours of contact we offer). Please note, to enjoy learning, you muststart your learning from week 1, not when the assessment-1 is due. For timely learners, we haveplenty of time to help; never hesitate to ask any question – no question is a silly question! Wewish you an enjoyable learning experience!Dr Tatheer Zahra[email protected]Coordinator – EGB375BeamSlab/ FloorSlab/ RoofMasonryWallEGB375 Design of Concrete StructuresCivil Engineering, QUT Page 3Section 01Flexural Design ofSingly Reinforced BeamsHOMEWORK: Problem 4(a) at the end of this Section.Submit the above homework in Week 02EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 41.1 Design of Reinforced Concrete Structural Elements#1.1.1oncrete is extensively used in engineering infrastructure (bridges, paving,retaining structures and dams) projects as well as buildings of many forms andshapes. The ability of concrete to be formed at site in different shapes and sizeshas been attractive to the civil design and structural engineers over the past 100 yearsor so. During this period, concrete technology has attained significant growth; forexample, its nominal strength has grown from 10MPa levels to 100MPa and beyond.Today 200MPa strength concretes are available in the market; these concretes are veryattractive for high rise buildings as much smaller size columns (or smaller thicknesswalls) can be designed, saving useable floor space. However, the Australian ConcreteStructures Standard, AS3600 provides data for concretes of 100MPa strength or lessonly. When concretes of >100MPa strength are conceived in a design, one shouldtest the material and structural components in quality certified laboratories.#1.1.2Apart from strength, porosity of concrete is also well researched by the concretetechnologists and today high-performance concretes of low porosity and high strengthare available in the market. High performance concretes use various plasticisers asadditives for improved curing and fibres of metals, glass and/ or polymers forminimising surface shrinkage cracking. Porous concrete compromises its durability.High quality non-absorbent formwork is vital for achieving stronger and longer lifeconcrete structures.#1.1.3Concrete is weak in tension (with the tensile strength typically of only 10% of itscompressive strength) and hence require reinforcement to resist tension. Designersshould understand the exact location and amount of reinforcement required to resisttension because inappropriate positioning can result in disastrous consequences.Providing more than the required amount of reinforcement in the hope of increasing theload carrying capacity of the concrete structure will also lead to poor performance aswill be covered in this subject.CEGB375 Design of Concrete StructuresCivil Engineering, QUT Page 5#1.1.4Each project that uses reinforced concrete is unique and cannot be covered in a singlesubject over just one semester. This subject (EGB375) will, therefore, focus on thefundamental behaviour of the reinforced concrete structural elements (beams,columns & slabs) under the action effects of bending moments, shear forces and axialforces; the performance of these elements will be expressed in terms of theirdeformation, cracking and failure modes. Mathematical expressions for determiningthese performance measures will be developed. Reference will also be made to theAustralian Concrete Structures Standard, AS3600-2018.Check List: I understand……..YESAS3600 provides data for concretes up to 100MPa onlyBenefits of quality formworkBenefits of high strength concrete If not all boxes are ticked, re-read the text until you can tick the boxEGB375 Design of Concrete StructuresCivil Engineering, QUT Page 61.2 Reinforced Concrete#1.2.1einforced concrete is designed to act as a unified material consisting ofconcrete and steel. Only where the reinforcement (steel bars) work togetherwith the concrete effectively, unified behaviour can be ensured. Workingtogether requires proper bond between the steel surface and the surroundingconcrete. To enhance bond, steel surface is helically deformed – hence the bars areknown as deformed bars. The fact that the coefficient of thermal expansion of steeland concrete is same, helps maintaining the bond integrity irrespective of exposure toexternal weather.#1.2.2Steel bars used to resist tension in the major direction are known as primary (or main)reinforcements; along the perpendicular direction of the main reinforcementsecondary reinforcements (or ties/ stirrups/ ligatures) are provided; they are madefrom either deformed or smooth surfaced (round) bars.#1.2.3The area of steel reinforcing bars is one of the most important properties the designersuse often. As the bars are ‘deformed’ (i.e., not perfectly circular), it is important theirarea be read out from the manufacturers’ table/ textbooks.A 12mm diameter bar has a design area of 110mm2 – not 113mm2 as you get from the�??4??2� formula. Other diameters are 16mm, 20mm, 24mm and 32mm withcorresponding areas of 200 mm2, 310 mm2, 450 mm2 and 800 mm2. 10mm bars areused as stirrups (or, ties or links) with the design area of 78 mm2.#1.2.4Unlike concrete, steel possess same strength and stiffness (Young’s modulus)properties under tension and compression (isotropic). Steel reinforcing bars fail at500MPa stress level; but depending on their ability to deform at this maximum (yield)R Bar Dia.61012162024283236Area28781102003104506208001010 EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 7stress level, they are classified as low (L) or normal (N) ductility bars; hence aredesignated with the prefix of L or N – for example, N12/ N20/ L32 and the likes. Lowductile bars do not ‘neck’ at failure and fracture at around 0.015 strain(15,000microstrain); normal ductile bars do ‘neck’ at failure and sustain strains of upto 0.05 (50,000microstrain). Irrespective of the ultimate strain, the yield strain of bothL and N type steels occur at 0.0025 (or, 2,500microstrain). All suspended structuralelements (e.g. beams) should normally be designed using the N category steels.#1.2.5For effective functioning of steel, concrete must crack; therefore, steel reinforcementdoes not prevent formation of cracks in concrete (as may sometimes people thinkinadvertently); however, the bars help reducing crack width and maintain the integrityof the cracked reinforced concrete. Steel is also used to resist compression; compressionsteel improves ductility of beams and reduces the required size of columns.Check List: I understand……..YESFunctions of main reinforcement in RC structuresFunctions of secondary reinforcement in RC structuresEffect of bond between reinforcement and concrete in RCstructuresSteel reinforcement is also used to resist compression If not all boxes are ticked, re-read the text until you can tick the box• Why do we reinforce the concrete?– Because concrete is weak in tension………• http://au.youtube.com/watch?v=JduiSVsNYrU• You should observe:– Plain Concrete under bending is Brittle (once cracked,simply collapses)– Concrete with steel reinforcement can be Ductile (sustainsload even after cracking)– http://www.youtube.com/watch?v=PQlIipMokq4&feature=related– Bottom fibre cracks in tension– Top fibre crushes under compressionEGB375 Design of Concrete StructuresCivil Engineering, QUT Page 81.3 Properties of Concrete#1.3.1ompressive strength of concrete is easier to determine compared to otherproperties (e.g. Tensile strength). 100mm diameter × 200mm long concretecylinders are used to determine the compressive strength of concrete. Aminimum of three specimens are normally tested and their average strength representedas????????. Strain gauges may be attached on the surface of these cylinders or non-contactdigital image correlation method may be used to measure the progressive increase instrain due to the applied compressive deformation (in testing machines). The reactionforce of the concrete cylinders to the applied deformation is divided by the area (stress)and plotted against the measure strains.#1.3.2From the stress-strain curves of the concretes, make the following observations:(1) The initial slope ?? = �???????? ???????????? ????� of the graphs vary with the strength; lowerstrength concretes exhibit lower Young’s modulus (????).(2) Concretes attain peak stress (strength) at a strain value of approximately 0.002.(3) Lower strength concretes sustain their peak stress for quite large post-peakstrains; higher strength concretes shed their peak stress immediately after thepeak stress is attained. In other words, lower strength concretes are moreductile; higher strength concretes are relatively brittle.(4) Except for very low strength concrete, most lose their strength significantly at astrain level of 0.003, which is designated as ultimate strain for concrete (??????).Knowing ???? = (195 ???? 205)??????, ?? = �????????�is determined; [ ?? is ‘modular ratio’].C20MPa;25MPa;32MPa;40MPa;50MPa;65MPa;80MPa&100MPa‘ infc MPa24,000;26,700;30,100;32,800;34,800;37,400;39,600&42,200inE MPa c8.37.56.66.15.75.35.14.7nn is “Modular Ratio”EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 9Check List: I understand……..YESStress-strain curves are determined from concrete cylinders tested under axialcompressionElastic modulus of concrete is a function of its mean strengthDenser the concrete, higher will be its elastic modulusHigher the mean strength of concrete, larger its elastic modulus If not all boxes are ticked, re-read the text until you can tick the box#1.3.3The compressive strength of concrete for design purpose is taken as CharacteristicStrength (????′) – at site concretes are expected to exhibit mean strengths larger than????′.Direct (????′??) and flexural tensile strengths �????′??,??� of concretes are related to theircompressive strength.Check List: I understand……..YES0.002 strain is ‘special’ to the compression behaviour of concreteMean strength is always larger than the characteristic strengthFlexural tensile strength is larger than the direct tensile strengthLower strength concretes are relatively more ductile If not all boxes are ticked, re-read the text until you can tick the box ‘ ‘.0.60fct f = fc ‘infc MPainE MPa c‘‘‘ 0.36 ( 3600)0.50cct cct f ASf ff ==Flexural Test Splitting Tension Test Compression TestEGB375 Design of Concrete StructuresCivil Engineering, QUT Page 101.4 Flexural Capacity of RC Beams#1.4.1ach structural member (for example, RC beam) possess a specific capacity toresist the imposed loading. The capacity is assessed for loadings that is NOTnormally experienced by the structural element – in other words, once-in-lifeloading will be considered for the ‘ultimate’ limit state strength (or capacity) of thestructural elements. Under the ultimate loads, the stresses in steel and concrete will beexpected to attain their yield strength and maximum stress respectively. Ultimate limitstate design ensures safety and economy (conservation of materials). The current limitstates method allows the stresses be close to yield (of steel) and ultimate strength (ofconcrete) in design. The former (in the USA still current) ‘working stress’ methodlimited the stress levels in concretes and steel well below their maximum stress levels.EAssumptions (AS3600)8.1.2 (Snipped from AS3600-2018)(a) Plane sections normal to the axis remain plane after bending.(b) The concrete has no tensile strength.(c) The distribution of compressive stress is determined from a stress-strainrelationship for the concrete in accordance with Clause 3.1.4 (see Note below).(d) The strain in compressive reinforcement CONCRETE does not exceed 0.003.3.1.4 Stress-strain curvesThe stress-strain curves for concrete shall be either –(a) assumed to be curvilinear form defined recognized simplified equations; or(b) determined from test data.For design purposes, the shape of the in situ uniaxial compressive stress-straincurve shall be modified so that the maximum stress is 0.9 fc’.Bond between steel & concrete remains PERFECT.EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 11#1.4.2Consider a RC beam that is reinforced for tension zone only.Under positive bending (sagging), compression develops at the top fibre with tensionat the bottom fibre of the cross section. The distribution of strain and stress across thedepth of the cross section is shown in the figure above. The strain diagram is just linearand the stress diagram defines a discontinuous line. The zero strain – zero stress line(neutral axis, NA) defines two triangles – one for stress in concrete above the NA andthe other for tensile stress in steel (with the concrete tension ignored).#1.4.3When the loading is increased, strain and hence stresses will increase. Let the loadingis increased until concrete attains ultimate stage (0.003 strain). The strain and stressdiagrams are now modified as follows:Assumed Stress-Strain Curves 5002000.0025sysyf MPaE GPaε=≈∴ =~0.002 to0.0025‘0.9 20 100cf MPa MPa = – Steel –Tension TestConcrete –Cylinder Compression TestLoading when Steel & Concreteremain elasticb dAst Dσ cuε stε cuσ stεσεσStrain Diagram Stress DiagramConcrete Steel????< ?????? < ?????? ????EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 12It should be noted that whilst the strain diagram is defined by a single sloped straightline, the stress diagram is not. First the concrete stress changes from triangle (prior toincreasing the loading) to a curved parabola. Second, the stress in the top fibre issomewhat smaller than the stress in a fibre below the top – in other words, max stressoccurs in a fibre below the top fibre although maximum strain occurs in the top fibre!Why? First let us know how to draw these stress and strain distribution diagrams.(1) Draw the strain distribution as it is just a sloped straight line with an angle.(2) Then for each ‘strain’ at ‘a fibre on RC beam cross section’, you determine acorresponding stress – using the stress-strain curves of the concrete cylinder.(3) Start this plotting from the neutral axis where you have zero strain – so zero stress!(4) Go upwards. As the strain increases, the stress will increase up to a point (pink linein the diagram) and after that the stress will decrease with further increase in strain).(5) From stress-strain curves you know that the max stress in concretes occurred forstrains in the range of 0.0020. So, for an ultimate strain of 0.003, the stress will belower than that for a strain (0.002 for example). This is the reason for the parabolicstress distribution in the compression zone. Steel stress will increase up to yield, butthen will remain constant until failure of the RC beam (i.e., it will not fall after peakstress as happens in concrete)#1.4.4The stress diagram at ultimate stage of RC beam is too complex for design calculation(mainly as it involves a parabola). Therefore, design standards simplify the parabolainto an equivalent rectangle as below:Loading increased such that steel& concrete are at post-peakb dAst Dσ cuε stε cuσ stεσεσEGB375 Design of Concrete StructuresCivil Engineering, QUT Page 13The strain diagram remains linear until collapse; it is assumed collapse occurs whenconcrete strain at top fibre is reached 0.003.#1.4.5The strain diagram shows that the concrete strength ????′ is attained at a fibre below thetop fibre. The parabola (green) is converted into an equivalent rectangle (dashed-blue)of width ??2????′and depth ??????, Compressive force (C) will act at the centroid of thisrectangle. Tension force (T) will act at the centre line of the reinforcement with themaximum value capped at ????????????.C = Compression in concreteT = Tension in SteelC = Tε stε cuσ st γ dnT= (2= ‘α2 fcC 2 γ dnC f d b α γ c n ‘ )( )T A f st syMax 0.003EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 14#1.4.6The coefficients ??2 and ?? are defined as functions of ????′as:These empirical formulae are found in Cl.8.1.3,AS3600 (2018). To fully define the rectangular stress block, werequire four variables:????′, ??2, ?? and ????. The first three variables aredefined already & ???? is determined by solving equation C = T. This is simply“Horizontal Equilibrium of Forces” covered under “Engineering Statics” unit –ENB110 in your first year!!#1.4.7The moment capacity of the section (or beam) is determined by taking moment of allforces about the compression force line (C). This again is from ENB110 Statics!!????= ???? = �????????????� × �?? – ????2??�, Where “z” is “lever-arm” (See figure below)‘2 20.85 0.00150.67α fcα= –≥0.97 0.0025 ‘0.67γ fcγ= –≥Calculation of MuCTDetermine fromLever12-a ::rm2nuuu undndkdkM T dd C Tz dM T zγγ=   = × –     =×–==γ dn( )For Under-reinforced Beams0.85M T d u ≈AS3600 limitsthis factor toMax of 0.36‘ Zα2 fc???? = Ductility factor???? = Depth ofoutermost steel layer????????EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 15#1.4.8In the calculation of ????, it is inherently assumed that the tension steel is at yield (i.e.,?????? = ??????. It is therefore important that we check if the assumption iscorrect. This is done by determining the strain in tension steel from the strain diagram.Knowing ???? and letting ?????? = 0.003 (representing ultimatestage), the strain in steel ??????can be determined from similar triangles.d??????0.003?? – ????=????The ?????? determined must be ≥ 0.0025 to validate theassumption of the tension steel at yield. If not, it implies that theε st tension steel is not yet at yield and the compression concrete will fail by reaching ?????? =0.003prior to steel yielding. Such a failure, controlled by concrete crushing, is brittleand is discouraged in design. AS3600 has provisions to ensure tension steel will be atyield prior to concrete compression failure. The provision is in Cl. 8.1.5. It can beinferred that AS3600 discourages design of sections whose ????0 > 0.36 in which ????0 =�??????0� – the ratio of the depth of neutral axis (at ultimate stage) from the top compressionfibre (????) to the distance between the top compression fibre and the centroid of theoutermost layer of the tensile steel (??0). Note the term ‘outermost’ – yes, there canbe several ‘layers’ of tension steel – see below:Tension SteelNeutral Axis Dd0dn bCentroid of bottom most row of tensile steelMax0.003 dn EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 16EXAMPLE 1.1:Determine the ultimate moment capacity of the section shown below:Assume tension steel is at yield.∴ ?? = ???????????? = [(8 ×. . . . . . . ) × … . . ./1000]= 800?????? = (??2????′) × (??????) × ????2 = 0.85 – (0.0015????′)= 0.85 – (0.0015 × 32) =. . . . . . .How does this compares with 0.67???2 =. . . . . . . . ..?? = 0.97 – 0.0025????′ =. . . . . . . . . .Is this > 0.67? ? OK?? (Y/N)?? =. . . . . . . . . ..Equate C = T(. . . . . .×. . . . . . . ) × (. . . . . . . .× ????) ×. . . . . . =. . . . . . . . . . . . . .???? = 100???? ?? = ?? –= 690 –= 645.5??????????. . . .×. . . . . . .22 ????= ?? × ?? = (800,000 × 645.5)/106 = 516.4??????Check if tension steel is at yield:?????? =0.003????(?? – ????) =0.003100(. . . . . . . . . -. . . . . . . . ) = 0.0177> 0.0025 Assumption OK8N16,Ast=1600mm2 d‘ 32f MPa c = = 730-40= 690mm40mmb = 350mm EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 17Homework & Tutorial(1) Why ‘non-absorbent’ formwork is essential for quality for ‘design compliant’concrete construction? (PRACTICE PROBLEM)(2) Why is concrete reinforced? Why is steel commonly used as reinforcement?(PRACTICE PROBLEM)(3) What is “mean strength” of concrete and how it differs from the “characteristicstrength”? What strength do you use in the structural design and why? Whichone is larger and why? (PRACTICE PROBLEM)(4) Calculate the moment capacity (????)of a rectangular, singly reinforced beamswith the following design data:(a) ????′ = 32??????; ?? = 250????; ?? = 350????; 4??20 in a single layer.(HOMEWORK)(b) ????′ = 32??????; ?? = 250????; ?? = 350????; ??0 = 375????; 4??20 in twolayers of 2??20 in each layer. (TUTORIAL)(c) ????′ = 60??????; ?? = 250????; ?? = 350????; 4??20 in a single layer.(PRACTICE)Member Actions & CapacityM – Bending Moment (kNm)V – Shear Force (kN)N – Axial Force (kN)Superscript * – “action”Subscript u – “ultimate”Section PropertiesB, D, d – Dimensions (mm)Z – Section Modulus (mm3)I – Moment of Inertia (mm4)Beam Geometry & LoadingL – Span (m)w – Distributed Load (kN/m)P – Concentrated Load (kN)Material Propertiesf – “strength” (MPa)σ – “Stress” (MPa)E – Young’s Modulus (GPa)ε – strain (constant – micron)EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 18Section 02Flexural Design ofDoubly Reinforced & T BeamsHOMEWORK: Problem 2 at the end of this SectionSubmit the above homework in Week 03EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 192.1 Under-Reinforced, Balanced and Over-Reinforced Beams#2.1.1f tension steel yields first prior to concrete attaining its ultimate strain of 0.003, theRC beam cross section is said to be under reinforced. If the tension steel justyields when the concrete reaches its ultimate strain of 0.003, the section is said tobe balanced. If the tension steel remains elastic (strain lower than 0.0025), when theconcrete strain reaches 0.003, the section is said to be over-reinforced.In singly reinforced concrete beam design, under reinforced beams are alwaysencouraged as it will provide the best possible ductile failure mode. Balanced and overreinforced sections will exhibit brittle failure mode and hence will be unsafe; Further,over-reinforced sections use tension steel ineffectively (as the stress level will be belowyield of 500MPa) and hence will be uneconomical. It should therefore be rememberedthat reinforcing beams more than essential will neither be safe nor be economical.If you do not understand how ?????? = 0.545 was obtained, use the below space andcheck.“Under-Reinforced”, “Balanced” and “Over-Reinforced” Beams• In balanced beams Steel yieldsjust when the compressionfailure in concrete occurs(epsiloncu=0.003)0.003>0.0025 =0.0025 0.545EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 20#2.1.2Design of RC beams for bending can therefore be challenging when you face increasedbending moment demand (e.g., owners’ wish for increasing seating in an auditorium).Increasing tensile steel area is the best option (???? = ?? × ??) as it will increase “T”. Theother option of increasing “z” will require increase of depth of section which will meanincrease of self-weight and loss of head room (below the beam) in buildings and ‘flyover’ bridges crossing highways/ railway lines – that limits the height of the passingvehicles under those bridges. Other options include increase the concrete grade – it willNOT be very effective at all (You have solved a problem in last week tutorial). Steelshould not be increased arbitrarily as RC beams must be under-reinforced[???? ≤(?????? = 0.545) ≤ 0.36]. What can we do in such situations??2.2 Doubly Reinforced Beams#2.2.1o calculate ????we take moment of all forces about “T” – this is because wehave two “C” forces but just one “T” force. This means that we need todetermine????&????.Two points should be observed in the above equations:(1) Compression force in steel (final of the set of above equations) is calculated eitherusing elastic modulus and strain or from yield strength; this is because compressionsteel may or may not be at yield – depending on ??????; if it is ≥ 0.0025, we use ??????-else we use ??????????as the stress.( 2 ‘ )( )( ) ( )0.5(or)u c c s scc c nc ns sc s sc sc syM C d d C d dC f d bd dC A E A fα γγε= – + –= = =T AstbdDε stTAscdsc0.003ε scdn CcCs EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 21(2) We should determine ???? ; this is done from ???? + ???? = ?? with the inherentassumption of ?? = ????????????- or tension steel is at yield (which will often correct –but still should be checked finally).We should notice that by adding compression steel, ????reduces; more the compressionsteel, less the????is. Lesser the????, higher will be the ??????. In other words, IF WEWANT/ NEED, we can keep adding MORE tension steel! This is good because itwill increase the moment capacity of section WITHOUT increasing the ???? of beam!!Check List: I understand……..YESThe definition of lever-armWhy concrete stress block is approximated as rectangularUnder-reinforced beams behave better than the over-reinforced beamsProvision of compression steel is very helpful for strength and ductilityDoubly reinforced beams will provide higher ????and lower ???? If not all boxes are ticked, re-read the text until you can tick the boxEGB375 Design of Concrete StructuresCivil Engineering, QUT Page 222.3 T-Beams#2.3.1 ectangular RC beams are seldom used as isolated structural elements; mostlythey are designed as supports for the suspended RC slabs. Slabs and beams arepoured (concreted) together in construction practice; as such they act as anunit with ‘some’ (how much?) part of the slab acting with the rectangular beamsthem look as T shaped beams.bbwtFlange outstand??????integralmaking #2.3.2Assume a series of three parallel beams (blue) are to be constructed with a reinforcedconcrete slab (pink) poured over them. The side view of the beams and slab and theirend view are as shown below:The 0.1L (or, 10%of span)‘outstand’ (fromthe edges of theweb) is themaximum widthof slab that can actwith the beam. The “flange width” can be maximum of ?????? = ???? + 0.2?? in which ????isthe width of the web (beam). Should half of the clear spacing between the webs (????/2)is less than 0.1L, then ????2will be the maximum width of slab that can act with the beam;this will make ?? = ???? + ????.R 0.1L 0.1LSb LAstFlangeWeb EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 23#2.3.3 Structurally these T shaped beams may act simply as rectangular beam or as true Tbeam depending on whether the NA falls within the flange or into the web.The diagrams above qualitatively show the behaviour of T beams when the NA lieswithin the flange or web respectively. As the concrete below the NA (tension zone) isdisregarded, when the NA lies within the flange, the whole of the web is omitted,making it as a rectangular beam.bbwtAst?????? T Beams: Neutral Axis at SlabT beams whose NA lies either in the slab orvery close to the slab in the web are• Regarded as rectangular beams ofeffective width “b”• Generally under-reinforced (very difficultto over-reinforced these “T” beams!)bbwConcrete in tthis zone(below NA) isdisregardedany way!“T” Beam is reallya “rectangular”beam! Ast dbbwtAst??????EGB375 Design of Concrete StructuresCivil Engineering, QUT Page 24#2.3.4When the beam acts as true T beam, two compression forces, one from flange (suffix f)and the other from web (suffix w) develop. Moment capacity isworked out by taking moment of all forces about the tensionforce. This is analogous to the doubly reinforced beams wherewe have dealt with two compression forces.First, we determine???? from ???? + ???? = ?? . The moment ofresistance of the T section is determined as below:????= ????(?? – 0.5??) + ????(?? – 0.5??????).Check List: I understand……..YESWhere NA falls within the flange, the beam behaves as rectangularFor ‘true’ T beam action, NA must lie in the webGenerally true T beams are always ductile If not all boxes are ticked, re-read the text until you can tick the boxT Beams: Neutral Axis at WebT beams whose NA lies well within the web are• “true” T beams!bbw tCf CwAst dnAs you see from the force diagram, the calculation for thiscase of T beams will be as per “Doubly ReinforcedBeams” – the only difference is that instead of one steelforce and one concrete force in the compression zone, wenow have two concrete forces!gdn‘T’ or recTangular?Cf CwA B C D T or recTangular? SecTionS a/ B/ c & DEGB375 Design of Concrete StructuresCivil Engineering, QUT Page 252.4 Capacity Design Method#2.4.1 esign standards around the world use some form of limit state design,originally developed in the former USSR. British, European and most Asianstandards name the design method as “partial safety method” in which safetyre used for various materials depending on their quality of manufacturings (concrete is site manufactured/ cured, hence