Plain & Reinforced Concrete-1 CE3601

1. Plain & Reinforced Concrete-1CE3601 Lecture # 11 &121st to 6th March 2012 Flexural Analysis and Design of Beams(Ultimate Strength Design of Beams)

2. Plain & Reinforced Concrete-1 Under-Reinforced Failure Stage-I, Un-cracked Section N.A. position is fixed, means “la” remains constant. Only “T” and “Cc” increase with the increase of load a/2 Cc la T Stage-II, Cracked Section When section cracks, N.A. moves towards compression face means “la” increases. “T” and “Cc” also increase. Internal Force Diagram

3. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) Stage-III, Yielding in Steel Occur T = Asfy remains constant and Cc also remains constant.“la” increases as the N.A. moves towards compression face because cracking continues. Failure initiates by the yielding of steel but final failure is still by crushing of concrete a/2 Cc la T Internal Force Diagram

4. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) Derivation for ρ Design Moment Capacity For tension controlled section Φ = 0.9 (1) And (2)

5. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) Put value of “a” from (1) to (2) For economical design

6. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) Let And Hence

7. Plain & Reinforced Concrete-1 Under-Reinforced Failure (contd…) By simplification We have to use –ve sign for under reinforced sections. So < 1.0 Reason For under reinforced section ρ <ρb If we use positive signρ will become greater than ρb, leading to brittle failure. ω

8. ρ R Plain & Reinforced Concrete-1 Plotting of R -ρ (3) (4) R ρ

9. Plain & Reinforced Concrete-1 Trial Method for the determination of “As” Trial # 1, Assume some value of “a” e.g. d/3 or d/4 or any other reasonable value, and put in (C) to get “As” (A) (B) Trial # 2, Put the calculated value of “As” in (A) to get “a”. Put this “a” value in (C) to get “As” (C) Keep on doing the trials unless “As” from a specific trial becomes equal to the “As” calculated from previous trial. THIS VALUE OF AS WILL BE THE FINAL ANSWER.

10. Plain & Reinforced Concrete-1 Is The Section is Under-Reinforced or NOT ? • Calculate ρ and if it is less than ρmax, section is under reinforced • Using “a” and “d” calculate εt if it is ≥ 0.005, section is under-reinforced (tension controlled) • If section is over-reinforced than in the following equation –ve term will appear in the under-root.

11. Plain & Reinforced Concrete-1 Is The Section is Under-Reinforced or NOT ? (contd…) • For tension controlled section, εt = 0.005, Using formula of Mn from concrete side If we keep d > dmin the resulting section will be under-reinforced. d > dmin means that section is stronger in compression.

12. Plain & Reinforced Concrete-1 Over-Reinforced Failure Stage-I, Un-cracked Section Stage-II, Cracked Section These two stages are same as in under-reinforced section. a/2 Cc la T Stage-III, Concrete reaches strain of 0.003 but steel not yielding We never prefer to design a beam as over-reinforced (compression controlled) as it will show sudden failure.Φ = 0.65 εs < εy fs<fy Internal Force Diagram

13. Plain & Reinforced Concrete-1 Over-Reinforced Failure Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…) (i) “a” is unknown as “fs” is not known (ii)

14. Plain & Reinforced Concrete-1 Over-Reinforced Failure Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…) εcu=0.003 Comparing ΔABC & ΔADE B C c A d- c E D εs Strain Diagram Eq # (iv) is applicable when εs < εy (iv) (iii)

15. Plain & Reinforced Concrete-1 Over-Reinforced Failure Stage-III, Concrete reaches strain of 0.003 but steel not yielding (contd…) Putting value of “fs” from (iv) to (ii) (v) Eq. # (v) is quadratic equation in term of “a”. Flexural Capacity Calculate “a” from (v) and “fs” from (iv) to calculate flexural capacity from these equations

16. Plain & Reinforced Concrete-1

17. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method Data • Dimensions, b, h, d and L (span) • fc’, fy, Ec, Es • As Required • ΦbMn • Load Carrying Capacity

18. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…) Solution Step # 1Calculte the depth of N.A assuming the section as under-reinforced and and

19. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…) Solution Step # 2 Calculate εs and check the assumption of step# 1 For extreme point If εs ≥ εy, the assumption is correct If εs ≤ εy, the section is under-reinforced. So “a” is to be calculated again by the formula of over reinforced section

20. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…) Solution Step # 3 Decide Φ factor For εs ≥ 0.005, Φ = 0.9 (Tension controlled section) For εs ≤ εy, Φ = 0.65 (Compression controlled section) For εy ≤ εs ≤0.005, Interpolate value of Φ (Transition Section) Step # 4 Calculate ΦbMn For under-reinforced Section For over-reinforced Section

21. Plain & Reinforced Concrete-1 Capacity Analysis of Singly Reinforced Rectangular Beam by Strength Design method (contd…) Alternate Method Step # 1 to step # 3 are for deciding whether the section is tension over reinforced or under-reinforced. Alternatively it can be done in the following manner. • Calculate ρ and ρmaxif ρ < ρmax section is under-reinforced. • Calculate dmin, if d ≥ dmin, section is tension controlled

22. Plain & Reinforced Concrete-1 Selection of Steel Bars for Beams • When different diameters are selected the maximum difference can be a gap of one size. • Minimum number of bars must be at least two, one in each corner. • Always Place the steel symmetrically. • Preferably steel may be placed in a single layer but it is allowed to use 2 to 3 layers. • Selected sizes should be easily available in market • Small diameter (as far as possible) bars are easy to cut and bend and place.

23. Plain & Reinforced Concrete-1 Selection of Steel Bars for Beams (contd…) • ACI Code Requirements There must be a minimum clearance between bars (only exception is bundled bars). • Concrete must be able to flow through the reinforcement. • Bond strength between concrete and steel must be fully developed. Minimum spacing must be lesser of the following • Nominal diameter of bars • 25mm in beams & 40mm in columns • 1.33 times the maximum size of aggregate used. We can also give an additional margin of 5 mm.

24. Plain & Reinforced Concrete-1 Selection of Steel Bars for Beams (contd…) • A minimum clear gap of 25 mm is to be provided between different layers of steel • The spacing between bars must not exceed a maximum value for crack control, usually applicable for slabs What is Detailing? • Deciding diameter of bars • Deciding no. of bars • Deciding location of bent-up and curtailment of bars • making sketches of reinforcements. Not O.K. 25mm

25. Plain & Reinforced Concrete-1 Concrete Cover to Reinforcement Measured as clear thickness outside the outer most steel bar. Purpose • To prevent corrosion of steel • To improve the bond strength • To improve the fire rating of a building • It reduces the wear of steel and attack of chemicals specially in factories.

26. Plain & Reinforced Concrete-1 Concrete Cover to Reinforcement (contd…) ACI Code Minimum Clear Cover Requirements • Concrete permanently exposed to earth, 75 mm • Concrete occasionally exposed to earth, • # 19 to # 57 bars 50 mm • # 16 and smaller bars 40 mm • Sheltered Concrete • Slabs and Walls 20 mm • Beams and Columns 40 mm

27. Plain & Reinforced Concrete-1 Number of Bars in a Single Layer (for beams) Rounded to lower whole number bw = width of web of beam For I-Shape beam width of bottom flange should be used in place of bw .

28. Plain & Reinforced Concrete-1 Example A singly reinforced rectangular beam has a width of 228 mm and effective depth of 450 mm. fc’ = 17.25 MPa, fy = 420 MPa. Calculate flexural capacity for the following three cases. • 2 # 25 bars (SI size) • 3 # 25 + 2 # 15 (SI) • Capacity for balanced steel

29. Concluded