Reinforced Concrete Structures MIM 232E Dr. Haluk Sesigür I.T.U. Faculty of Architecture Structural and Earthquake Engineering WG Ultimate Strength Theory Design of Singly Reinforced Rectangular Beams RCSD-3
RC Beam Design Upper reinforcement Upper reinforcement stirrup lower reinforcement 2
RC Beam Design 3
Ultimate Strength Theory compression Moment couple tension 4
Ultimate Strength Theory Pure bending load... Moment : couple (tension and comp. force) compression tension compression compression force Concrete is good in compression; Tension Tension force compression crushing tension cracking Reinforcment to resist tension at the bottom http://mmf2.ogu.edu.tr/atopcu/index.htm 5
Ultimate Strength Theory Necessity for reinforcement M V Deformation; cracking at tension region Cruching at compression region http://mmf2.ogu.edu.tr/atopcu/index.htm 6
Ultimate Strength Theory Deformation Moment couple As: area of tension reinf. Fs: tension force of steel Fc: compression force of concrete z: moment arm ε c = strain of concrete (shortening) ε s = strain of steel (elongation) Horizontal equilibrium ; Fc=Fs Comp. N.A. ε c = strain of concrete (shortening) ε s = strain of steel (elongation) Cross-setion tension İnternal forces crack deformation http://mmf2.ogu.edu.tr/atopcu/index.htm 7
Ultimate Strength Theory concrete Idealized Stress-Strain (σ-ε) Relationships steel fcd fyd =0.003 crushing e yd = fyd E s =0.01 rupture E s =210000MPa (modulus of elasticity) fcd = fck 1,5 fyd = fyk 1,15 8
Ultimate Strength Theory general Md moment copule (Fc;Fs) Fc=Fs Compression at top; tension at bottom Fc can be carried (f cd is good); Fs cannot be carried (f ctd is poor) cracking occur Fs should be complately carried by As (no concrete contribution is considered) ε c shortening at top; ε s elongation at bottom; by rotation of cross section 9
Ultimate Strength Theory Deformation & strain Moment (Md) increases 1 2 3 4 crushing deformation İnternal forces deformation İnternal forces deformation İnternal forces deformation İnternal forces cracking Starin increases cracking cracking http://mmf2.ogu.edu.tr/atopcu/index.htm 10
Ultimate Strength Theory compressive stress block Theoritical Stress distribution Virtual Stress distribution Fs Fs Neutral axis Theoric distribution (TS 500) http://mmf2.ogu.edu.tr/atopcu/index.htm 11
Ultimate Strength Theory compressive stress block beam beam Theoritical CEB model ACI TS5000 Stress distribution at the crushing/failure of concrete http://mmf2.ogu.edu.tr/atopcu/index.htm 12
Ultimate Strength Theory Assumptions (TS500) 1- Strain distribution is lineer (plane sections remain plane-bernoulli/navier) 2- Tensile strength of concrete is neglected ( ~0) 3- full adherence 4- Elasto-plastic curve Design strength (yielding) Strain at yielding Strain at rupture (ε su ) http://mmf2.ogu.edu.tr/atopcu/index.htm 13
Ultimate Strength Theory Assumptions (TS500) 5- at ultimate strength; ε c = ε cu = 0.003 6- at ultimate strength; theoric σ c -ε c 7- shape of tensional region is not important Equivalent sections http://mmf2.ogu.edu.tr/atopcu/index.htm 14
Ultimate Strength Theory Assumptions (TS500) Failure/fracture types; ε c = ε cu = 0.003 concrete crushing Mr is reached failure condition limit; At this point; there are 3 failure types due to yielding of reinforcement; Ductile failure (tensional) Brittle failure (compression) Failure at balance (brittle) 15
Ultimate Strength Theory Assumptions (TS500) 1. Ductile failure (tensional) Before ε c = ε cu = 0.003 ; reinforcing bar yields (ε s ε sd ) Firstly bar yields; then ε c = ε cu and concrete crushes. Ductile Failure Section deformation 2. Brittle failure (compression) Design strength (yielding) Strain at yielding Strain at rupture (ε su ) Section deformation If ε s < ε sd (before yielding) ; ε c becomes ε cu = 0.003 ; brittle failure occurs. Firstly concrete cruches, bar does not yield. http://mmf2.ogu.edu.tr/atopcu/index.htm 16
Ultimate Strength Theory Assumptions (TS500) 3. Failure at balance (brittle) It is a special case. Section deformation When ε s = ε sd ; ε c = ε cu = 0.003 Concrete cruching and yielding of bar are in the same time No more elongation is available in Reinf. Bar. Brittle failure 17
Pure bending General d h concrete reinforcement b f ck ε co ε cu ε c T Pure bending region f yk steel M 18
Pure bending Stress-strain / cases ε c < ε c0 ε c < ε c0 ε c = ε c0 ε c = ε cu ε c < ε c0 ε s < ε y ε s = ε y ε s > ε y ε s < ε y ε s = ε u σ c < f ck σ c < f ck σ c = f ck f ck f ck F s = σ s A s σ s < f yk F s = σ s A s σ s = f yk F s = σ s A s σ s = f yk F s = σ s A s σ s < f yk F s = σ s A s σ s = f yk A B C D E 19
Pure bending Stress-strain / cases A : Moment small Bernoulli/Novier (elastic behaviour) B : Moment increases ε c < ε c0 concrete concrete ε s < ε y linear ε c ε cu σ c < f ck steel steel F s = σ s A s σ s < f yk ε su ε su 20
Pure bending Stress-strain / cases C : Moment increases D : beam with heavy reinforcement Brittle failure concrete concrete steel steel 21
Pure bending Stress-strain / cases E: reinforcement ratio is small, beam with weak reinforcement ratio concrete ε c ε co ε cu steel ε s = ε su 22
Pure bending Moment D (heavy reinf.) σ or f / Stress steel çelik Ductile/brittle E (weak reinf.) deflection beton concrete ε cu ε / Strain Dimensioning: TS500; concrete ε cu = %o 3 ε su = %o 10 f cd σ c beton concrete f yd çelik steel ε co (2) ε cu (3) ε c (%o ) (10) %o 23
Pure bending Moment capacity of rectangular section Brittle Failure: beam with heavy reinforcement f yd steel f cd concrete σ s < f yd ε s ε su ε cu = %o 3 Ductile Failure: σ s = f yd beam with weak reinforcement f yd f cd ε su > ε s > ε y ε s Failure at Balance / Brittle Failure: ε su ε cu = %o 3 f yd σ s = f yd f cd ε s = ε y ε s ε su Note that all the failures govern by concrete crashing ε cu = %o 3 24
Pure bending Moment capacity of rectangular section Fc: compression stress resultant in concrete Fs: tension force in reinforcement 0,85fcd ; (specimen size/experimental cond.) Fc =0,661. b. x. fcd = 0,661. b. d. kx. fcd kx=x/d ; Fs = fyd. As Equilibrium; Outer moment acts on cross-section; Fc = Fs Mr = b. d 2. fcd. kx. 0,661 0,268kx Mr = b.d2 K ; K = 1 fcd.kx. 0,661 0,268kx cm 2 /t 25
Pure bending Moment capacity of rectangular section Moment for any point; Mr = Fs. z z = Mr Fs j = z d (dimensionless) j = Mr d. Fs = b. d2. fcd. kx. 0,661 0,268kx d. As. fyd Mr = Fs. z = As. fyd. j. d As = Mr fyd.j.d As = Mr fyd.j.d = ks. Mr d ks depends on stress and strain r= As b. d reinforcement ratio r min = As b. d = 0. 8 f ctd f yd, r max = 0.85r b Ex. C20, S420 fctk=1.6mpa, fyk=420mpa fctd=fctk/1.5=1.07mpa, fyd=fyk/1.15=365mpa r min =0.002 26
Pure bending Tables for solution Parameters ks, K are dimensional. For ε c, ε s ; kx and ks coefficients are calculated and For each (fcd) K or kd; For each steel quality ( fyd ) ks are calculated and tabulated. In tables; the values for section at balance; last 3 row below the line. 23,24,25. rows are K* values for S500,S420 ve S220 respectively. If K<K* shows brittle failure; increase of section or compressive reinforcement is required. 0,85% of «reinf. at balance» should not be exceeded ( underlined ks values) 27
Pure bending Tables for solution Mr = b. d2 K As = ks. Mr d 28
Pure bending Examples - 1 Ex. 1: 29
Pure bending Examples -1 Required reinf. Selected reinf. r min =0.002 A smin =0.002.30.70=4.2cm 2 A smax =0.343.2500/70=12.25cm 2 30
Pure bending Examples -2 Ex. 2: 31
Pure bending Examples -2 Selected reinf. 5 f 20 ( 15.7 cm 2 ) 32
Pure bending Solution by dimensionless parameters 33