Reinforced Concrete Structures

Transcription

1 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

2 RC Beam Design Upper reinforcement Upper reinforcement stirrup lower reinforcement 2

3 RC Beam Design 3

4 Ultimate Strength Theory compression Moment couple tension 4

5 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 5

6 Ultimate Strength Theory Necessity for reinforcement M V Deformation; cracking at tension region Cruching at compression region 6

7 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 7

8 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

9 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

10 Ultimate Strength Theory Deformation & strain Moment (Md) increases crushing deformation İnternal forces deformation İnternal forces deformation İnternal forces deformation İnternal forces cracking Starin increases cracking cracking 10

11 Ultimate Strength Theory compressive stress block Theoritical Stress distribution Virtual Stress distribution Fs Fs Neutral axis Theoric distribution (TS 500) 11

12 Ultimate Strength Theory compressive stress block beam beam Theoritical CEB model ACI TS5000 Stress distribution at the crushing/failure of concrete 12

13 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 ) 13

14 Ultimate Strength Theory Assumptions (TS500) 5- at ultimate strength; ε c = ε cu = at ultimate strength; theoric σ c -ε c 7- shape of tensional region is not important Equivalent sections 14

15 Ultimate Strength Theory Assumptions (TS500) Failure/fracture types; ε c = ε cu = 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

16 Ultimate Strength Theory Assumptions (TS500) 1. Ductile failure (tensional) Before ε c = ε cu = ; 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 = ; brittle failure occurs. Firstly concrete cruches, bar does not yield. 16

17 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 = Concrete cruching and yielding of bar are in the same time No more elongation is available in Reinf. Bar. Brittle failure 17

18 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

19 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

20 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

21 Pure bending Stress-strain / cases C : Moment increases D : beam with heavy reinforcement Brittle failure concrete concrete steel steel 21

22 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

23 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

24 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

25 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

26 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 =

27 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

28 Pure bending Tables for solution Mr = b. d2 K As = ks. Mr d 28

29 Pure bending Examples - 1 Ex. 1: 29

30 Pure bending Examples -1 Required reinf. Selected reinf. r min =0.002 A smin = =4.2cm 2 A smax = /70=12.25cm 2 30

31 Pure bending Examples -2 Ex. 2: 31

32 Pure bending Examples -2 Selected reinf. 5 f 20 ( 15.7 cm 2 ) 32

33 Pure bending Solution by dimensionless parameters 33