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Plain & Reinforced Concrete-1 CE-313

Plain & Reinforced Concrete-1 CE-313. Lecture # 6 3 rd March 2006 Flexural Analysis and Design of Beams By Engr. Azhar. Plain & Reinforced Concrete-1. Cracked Transformed Section When tensile stress cross f r , crack appear on the tension face of beam.

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Plain & Reinforced Concrete-1 CE-313

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  1. Plain & Reinforced Concrete-1CE-313 Lecture # 63rd March 2006 Flexural Analysis and Design of Beams By Engr. Azhar

  2. Plain & Reinforced Concrete-1 Cracked Transformed Section When tensile stress cross fr, crack appear on the tension face of beam. • Tensile strength of concrete is neglected. • All the concrete on the tension side of N.A. is neglected. Only the Shaded area of cross-section is considered effective. Concrete N.A Steel

  3. Plain & Reinforced Concrete-1 Cracked Transformed Section (contd…) b εc fc Cc kd N.A. h d nAs la = jd fs T = Asfs εs Cracked Transformed Section Internal Force Diagram Stress Diagram Strain Diagram When the tension side is cracked the concrete becomes ineffective but the strains goes on increasing. The steel comes in to action to take the tension.

  4. Plain & Reinforced Concrete-1 b Cracked Transformed Section (contd…) Calculations for“k”, taking top face as reference kd = [b x kd x kd/2 + nAsd]/ [b x kd + nAs] b x (kd)2 + nAs kd = b x (kd)2/2 + nAsd b x (kd)2/2 = nAsd - nAs kd b x (kd)2/2 = (1-k) nAsd bd2/2 k2 = (1-k) nρbd2 k2 = 2(1-k) nρ k2 + 2nρ k - 2nρ = 0 kd d nAs Cracked Transformed Section ρ = As / bd

  5. Plain & Reinforced Concrete-1 b Cracked Transformed Section (contd…) k2 + 2nρ k - 2nρ = 0 k = [ -2nρ ± √ (4n2ρ2 + 8nρ) ] /2 k = [ -nρ ± √ (n2ρ2 + 2nρ) ] k = √ (n2ρ2 + 2nρ) -nρ Taking only positive value as distance can’t be negative kd d nAs Cracked Transformed Section

  6. Plain & Reinforced Concrete-1 fc Cracked Transformed Section (contd…) Value of “j” jd = d – kd/3 J = 1 – k/3 Cc = volume of stress diagram Cc = ½ kd x fc x b Cc = fc/2 x kdb T = Asfs = nfcAs b Cc kd d N.A jd T = Asfs

  7. Plain & Reinforced Concrete-1 Cracked Transformed Section (contd…) Resisting Moment Capacity For longitudinal equilibrium T = Cc Mr = T x jd = Cc x jd Mr = Asfs x jd = fc/2 x kdb x jd ------------- 1 FOR MAXIMUM VALUE RESISTING MOMENT Put in eq-1 fs = 0.5 fy for grade 300 steel fs = 0.4 fy for grade 420 steel fs = Maximum allowable tensile stress in steel fc= 0.45 fc’ fc = Maximum allowable concrete stress

  8. Concluded

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