10.01.03.041

Presentation On
Rectangular Beam Design

USD Method of Designing Singly & Doubly Reinforced Beam

The Following Presentation Is
Conducted for to Meet the Purpose of
Course CE-416 : Pre-Stressed Concrete
Sessional

Presented ByEzaz Ahmed
10.01.03.041
th Year 2nd Semester
4
Section-A

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 Torsion.

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.

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.

Flexural Design Equations
Singly Reinforced Beam

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

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.

10.01.03.041

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