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Today we are going to learn about Doubly Reinforced Beam Design in Ultimate Stress Design (USD) method. Many complications in design always been seen in different tutorials and today we are going to show you how simply you can design a doubly reinforced beam for a structure.
Before going to Beam Design fist have a tour to Beam characteristics.
What is a BEAM?
A beam is generally a structural member that carries load along its transverse axis. It resists the load through bending and through the compressive stress carrying capacity of concrete and through the tension carrying capacity of reinforcement.
What is CONCRETE?
A Concrete is a stone like material which is artificially prepared by using a bonding material and inert materials. Bonding material are generally Cement(Ordinary Portland Cement (OPC) and Portland Composite Cement (PCC)) and inert materials are generally crushed stone,sand etc.
What is Reinforcement?
High yield reinforcement is a building material that is prepared from mild steel. Reinforcement in beam carries tension that comes in the total life of the structure.
What is SINGLY REINFORCED AND DOUBLY REINFORCED BEAM?
Singly reinforced beam is one in which the steel buried in the concrete lies only in the tension side of the section.
Sometimes the upcoming moment from the load is such that only a singly reinforced beam can not support it. In this case, Reinforcement is also provided in the compression side of the beam. This type of beam is called doubly reinforced beam.
Assumptions in Beam Design:
1. Plane sections remain plane after Bending.
2.Strain distribution is linear for both steel and concrete and varies directly with the distance from the Neutral Axis.
3. All of the compressive force are taken up by the concrete only and all of the tensile stress are taken up by the reinforcement only.(N.B. Concrete has a capacity of taking tensile stress which is almost 10-15 percent of its compressive strength carrying capacity)
Evaluation of Design Parameters:
C = T
or, 0.85fc’ ba = As fy
or, a = As fy / (0.85fc’ b)
= pd fy / (0.85 fc’) (putting p = As / bd)
= w fc’ [ 1 – 0.59 w] bd^2 (putting w = r fy / fc’)
Mn = Kn bd2 (putting Kn = w fc’ [ 1 – 0.59 w)
Mu = f Mn
=f Kn bd^2 , where f = Strength Reduction Factor
From strain diagram, similar triangles
cb / d = 0.003 / (0.003 + fy / Es) ; Es = 29×10^6 psi
cb / d = 87,000 / (87,000+fy)
Pb = Asb / bd
= 0.85fc’ ab / (fy. d)
= β1 ( 0.85 fc’ / fy) [ 87,000 / (87,000+fy)]
P =0.75 P b
As(min) = 3Öfc’ x bwd / fy or 200 bwd / fy
rmin = 3Öfc’ / fy or 200 / fy