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# Presentation on rectangular beam design by USD method

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• 1. Presentation On Rectangular Beam Design USD Method of Designing Singly & Doubly Reinforced Beam
• 2. The Following Presentation Is Conducted for to Meet the Purpose of Course CE-416 : Pre-Stressed Concrete Sessional
• 3. Presented ByEjaj Ahmed 10.01.03.041 th Year 2nd Semester 4 Section-A
• 4. Factors Affecting Design • Compressive Strength of Concrete, f‘c (3000-5000psi) • Yield Strength of Steel, fy(40-60 grade) • Spacing of Reinforcement------ (≥ maximum diameter of the bars or 1-1.5in or 1.50 times maximum size of aggregate) • Concrete Cover------ ( for main bars 1in and stirrups 0.75 in) • Beam Width • Beam Depth, d Physical Factors • Steel Area Strength Factors Important Considerations In Design •Factored Loads (Generally 1.2 Dead Load + 1.6 Live Load which implies factored moment Mu) •Capacity Reduction Factors,  ɸ = 0.9 for Bending in Reinforced Beam; = 0.75 for Shear and Tension.
• 5. Important Definitions • Balanced Steel Ratio  When due to same load tensile steel theoretically reach its yield and concrete attains its ultimate strain 0.003.  ρb = 0.85 * (f’c/fy) * [{0.003/0.003+(fy/Ey)}*d]. • Under Reinforced Beam  When steel begin to yield even though concrete compressive strain < 0.003.  Steel ratio <<<< ρb.  Failure with warning • Over Reinforced Beam  When steel begin to yield with concrete compressive failure.  Steel ratio >>>> ρb.  Sudden Failure.
• 6. Design Types of RCC Beams Singly Reinforced Beam Doubly Reinforced Beam Steel Only at Tension Zone i.e. below neutral zone. Steel in Compression as well as Tension Zone. Design procedure is General Flexural Design. General Flexural Design both for top and bottom fibre. Generally Rectangular Beam. Both Rectangular and T-beam. Design as when ɸMn ≥ Mu Design as when ɸMn ≤ Mu.
• 7. Flexural Design Equations Singly Reinforced Beam
• 8. Doubly Reinforced Beam
• 9. Now, to design doubly reinforced beam steel for both concrete-steel couple and steel-steel couple is determined. As it is the case where ɸMn ≤ Mu. So, Mu should be combination of moment by concrete-steel and steel-steel couple. Which is--Mu = Mn1 + Mn2 Where, Mn1 = Asfy(d-0.5*a); As1 = Required Steel Area = ρbd Mn2 = Mu – Mn1 =As'fy(d-d’); generally d’ = 3in
• 10. Shear Design Equations • Using FEM formula find Vu. • Calculate Va = {ɸ*2√(f’c)bd}/1000 kip. • For #3 or ϕ10 bar Spacing S= (ɸAvfyd)/(Vu-Va) • Maximum Spacing is the smallest of Smax = (Avfy)/0.74√(f’c)b ≤ (Avfy)/50b; Smax = d/2; Smax = 24 in.  This spacing will be used when Vu < Va.