Singly reinforced beam analysis

Singly reinforced beam analysis

• 1.
  CE8501 Design OfReinforced Cement Concrete Elements Unit 1-Introduction Analysis of singly reinforced beam [As per IS456:2000] Presentation by, P.Selvakumar.,B.E.,M.E. Assistant Professor, Department Of Civil Engineering, Knowledge Institute Of Technology, Salem. 1
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  Singly Reinforced Rectangularbeam - Analysis 1. Stress strain profile of rectangular beam 2. Expression for compression and tension 3. Depth of Neutral axis 4. Moment of resistance for compression and tension 5. IS 456 recommendations 6. Procedure for finding Moment of resistance 2
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  Analysis – Stressand strain Profile 3 = Z
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  Symbols 4 [Refer IS456 Pg.96]
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  Expression for compressiveforce and tension force • Compressive force • Tensile force 5 [Refer IS456 Pg.69] [Refer IS456 Pg.70]
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  Depth of Neutralaxis • Actual Neutral axis depth • Limiting Neutral Axis depth 6 [Refer IS456 Pg.96] [Refer IS456 Pg.70]
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  Depth of Neutralaxis • If Xu = Xu,max then it is balanced section • If Xu < Xu,max then it is under reinforced section • If Xu > Xu,max then it is over reinforced section 7
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  Moment of resistancefrom compression • When xu/d < xu,max/d [Under reinforced section] 8 [Refer IS456 Pg.96]
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  Moment of resistancefrom compression • When xu/d = xu,max/d [Balanced section] 9 [Refer IS456 Pg.96]
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  Moment of resistancefrom compression • When xu/d > xu,max/d [Over reinforced section] 10 [Refer IS456 Pg.96] [Refer IS456 Pg.96] [Use this formula]
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  Example 1 • Arectangular reinforced concrete section having a breadth of 350 mm is reinforced with 2 bars of 28 mm and 2 bars of 25 mm diameter at an effective depth of 700 mm. Adopting M-20 grade concrete and Fe-415 HYSD bars determine the ultimate moment of resistance of the section. Given data: b = 350 mm d = 700 mm M-20 – fck – 20 N/mm2 Fe415 – fy – 415 N/mm2 Ast = 2 [ 𝜋 4 (282) + 𝜋 4 (252)] = 2214 mm2 11
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  Step 1 :Depth of neutral axis 12 • 𝑥 𝑢 𝑑 = 0.87 𝑓𝑦 𝐴𝑠𝑡 0.36 𝑓𝑐𝑘 𝑏 𝑑 • 𝑥 𝑢 𝑑 = 0.87 ∗415 ∗2214 0.36 ∗20 ∗350 ∗700 • 𝑥 𝑢 𝑑 = 0.453 < 0.48 ( Under reinforced)
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  Step 2 :Moment of resistance • Mu = 0.87 fy Ast d [1 - 𝐴 𝑠𝑡 𝑓𝑦 𝑏 𝑑 𝑓𝑐𝑘 ] • Mu = 0.87 * 415 * 2214 * 700 [1- ( 2214∗415 350∗700∗20 )] • Mu = 456 x 106 N.mm 13
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  Example 2 • Asingly reinforced concrete beam having a width of 250 mm is rein forced with steel bars of area 3600 mm2 at an effective depth of 400 mm. If M-20 Grade Concrete and Fe-415 HYSD bars are used, compute the ultimate flexural strength of the section. Given : b = 250mm d = 400mm M-20 – fck – 20N/mm2 Fe415 – fy – 415N/mm2 Ast = 3600mm2 14
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  Step 1 :Depth of neutral axis 15 • 𝑥 𝑢 𝑑 = 0.87 𝑓𝑦 𝐴𝑠𝑡 0.36 𝑓𝑐𝑘 𝑏 𝑑 • 𝑥 𝑢 𝑑 = 0.87 ∗415∗3600 0.36 ∗20∗250∗400 • 𝑥 𝑢 𝑑 = 1.80 > 0.48 [Over reinforced section]
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  Step 2 :Moment of resistance • Mu,lim = 0.36 𝑥𝑢,𝑚𝑎𝑥 𝑑 [1 – 0.42 𝑥𝑢,𝑚𝑎𝑥 𝑑 ] b d2 fck • Mu,lim = 0.36 ∗ 0.48 [1 – 0.42 ∗ 0.48 ] 250 * 4002 *20 • Mu,lim = 110.3 x106 N.mm 16
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  Assignment Problem • Arectangular RC beam of width 250mm and overall depth of 550mm. It is reinforced with 3nos of 20mm diameter mild steel bar(Fe250). If M-20 grade concrete is used, estimate the ultimate moment of resistance of the section. 17
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  Thank You 18