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Doubly reinforced beams...PRC-I

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madam rafia firdous is a lecturer and instructor in University of South Asia LAHORE,PAKISTAN and She gave lecture us via in this slide

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Doubly reinforced beams...PRC-I

  1. 1. Plain and Reinforced Concrete- 1 Doubly Reinforced Beams ByEngr.RafiaFirdous
  2. 2. Plain & Reinforced Concrete-1 Doubly Reinforced Beams “Beams having both tension and compression reinforcement to allow the depth of beam to be lesser than minimum depth for singly reinforced beam” • By using lesser depth the lever arm reduces and to develop the same force more area of steel is required, so solution is costly. • Ductility will be increased by providing compression steel. • Hanger bars can also be used as compression steel reducing the cost up to certain cost. • For high rise buildings the extra cost of the shallow deep beams is offset by saving due to less story height.
  3. 3. Plain & Reinforced Concrete-1 Doubly Reinforced Beams (contd…) • Compression steel may reduce creep and shrinkage of concrete and thus reducing long term deflection. • Use of doubly reinforced section has been reduced due to the Ultimate Strength Design Method, which fully utilizes concrete compressive strength. Doubly Reinforced Beam
  4. 4. Behavior Doubly Reinforced Beams Tension steel always yields in D.R.B. There are two possible cases: 1. Case-I Compression steel is yielding at ultimate condition. 2. Case-II Compression steel is NOT yielding at ultimate condition.
  5. 5. Behavior Doubly Reinforced Beams Cc T = Asfs N.A. εcu= 0.003 Strain Diagram Internal Force Diagram εs h c d b 0.85fc a Whitney’s Stress Diagram (d-d’) fs d εs’ fs’ Cs d – a/2 T = Asfs Cs=As’fs’ Cc=0.85fc’ba fs=Esεs fs’=Esεs’
  6. 6. Behavior Doubly Reinforced Beams (contd…) Case-I Both Tension & Compression steel are yielding at ultimate condition fs = fy and fs’=fy Location of N.A. Consider equilibrium of forces in longitudinal direction sc CCT  yscys f'Aba'f85.0fA    b'f85.0 f'AA a c yss   1β a c and
  7. 7. Case-I Both Tension & Compression steel are yielding at ultimate condition (contd…) c d'c 0.003 'εs   εcu= 0.003 Strain Diagram εs c εs’ d’ B D E C A Δ ABC & ADE         c d'c 0.003'εs 1 1 s β β c d'c 0.003'ε                 a d'βa 0.003'ε 1 s If εs’ ≥ εy compression steel is yielding. If εs’ < εy compression steel is NOT yielding. (1)
  8. 8. Case-I Both Tension & Compression steel are yielding at ultimate condition (contd…) Cc T = Asfy N.A. Internal Force Diagram (d-d’) Cs d – a/2 T = total tensile force in the steel 21 TTT  T1 is balanced by Cs T2 is balanced by Cc s1 CT  c2 CT 
  9. 9. Case-I Both Tension & Compression steel are yielding at ultimate condition (contd…) Cc T = Asfy N.A. Internal Force Diagram (d-d’) Cs d – a/2 Moment Capacity by Compression Steel    'dd'f'A'ddCM yssn1   'ddT1  Moment Capacity by Concrete              2 a dT 2 a dCM 2cn2          2 a dTTM 1n 2          2 a d'f'AfAM ysysn 2
  10. 10. Case-I Both Tension & Compression steel are yielding at ultimate condition (contd…) Total Moment Capacity 21 nnn MMM             2 a d'f'AfA'dd'f'AM ysysysn
  11. 11. Case-II Compression steel is not yielding at ultimate condition fs = fy and fs’< fy 'εE'f ss  b'f85.0 'f'AfA a c ssys   1β a c and a d'βa 600'f 1 s   Location of N.A.
  12. 12. Concluded
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madam rafia firdous is a lecturer and instructor in University of South Asia LAHORE,PAKISTAN and She gave lecture us via in this slide

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