A doubly reinforced concrete beam has effective cover d’ to the ...
Epth of 50 mm on both the compression and tension sides. The beam has a width of 250 mm, an overall depth of 500 mm, and is reinforced with two sets of bars. The first set consists of 4 bars of 20 mm diameter on the tension side, and the second set consists of 4 bars of 20 mm diameter on the compression side. The concrete has a compressive strength of 25 MPa, and the yield strength of the steel is 500 MPa.
To design the beam, the following steps can be followed:
1. Determine the effective depth of the beam:
Effective depth = overall depth - effective cover depth
Effective depth = 500 mm - 50 mm - 50 mm = 400 mm
2. Calculate the moment capacity of the beam using the balanced reinforcement method:
Assuming balanced reinforcement, the tensile and compressive forces in the steel are equal at the ultimate limit state.
As = Ast = Asc
where As = area of steel required
Ast = area of steel on tension side
Asc = area of steel on compression side
Assuming a neutral axis depth of d, the moment capacity of the beam can be calculated as:
M = 0.68 fck bd^2 + Ast (fy/γs) (d - d')
where fck = concrete compressive strength
fy = steel yield strength
γs = partial safety factor for steel
Taking γs = 1.15, the moment capacity can be calculated as:
M = 0.68 x 25 x 250 x 400^2 + 4 x π/4 x 20^2 x 500 x (500 - 50 - 20) x 500/1.15
M = 138,400 kNm
3. Check deflection:
The maximum allowable deflection can be taken as L/250, where L is the span of the beam. Assuming a span of 6 meters, the maximum allowable deflection is:
δmax = 6000/250 = 24 mm
Using standard formulas for deflection, the actual deflection can be calculated as:
δ = (5wL^4)/(384EI)
where w = self-weight of the beam
E = modulus of elasticity of concrete
I = moment of inertia of the beam
Assuming a self-weight of the beam of 25 kN/m and a modulus of elasticity of 28,000 MPa, the moment of inertia can be calculated as:
I = (1/12)bh^3 + Ast (d - d')^2 + Asc (d - d')^2
I = (1/12) x 250 x 400^3 + 4 x π/4 x 20^2 x (400 - 50 - 20)^2 + 4 x π/4 x 20^2 x (50 + 20)^2
I = 72.9 x 10^6 mm^4
Substituting the values into the deflection formula, the actual deflection is:
δ = (5 x 25 x 6^4)/(384 x 28000 x 72.9 x 10^6) = 4.3 mm
Since the actual deflection is less than the maximum allowable deflection, the beam
A doubly reinforced concrete beam has effective cover d’ to the ...
Can anyone explain this?