DESIGN OF SINGLY REINFORCED CONCRETE BEAM SECTION (beta version)

RHC ENGINEERING

DESIGN OF SINGLY REINFORCED CONCRETE BEAM SECTION

Tension-Controlled Section

Sections are tension-controlled when the net tensile strain in the extreme tension steel is equal to or greater than 0.005 just as the concrete in compression reaches its assumed strain limit of 0.003.

Material Properties:

fc = MPa

fy = MPa

es = Reinf. modulus of elasticity

f =

Section Properties:

B = mm

H = mm

dc = mm

Applied Moment:

Mua = kNm

Select Bar Diameter:

Db = 8 10 12 14 16 20 25 28 32 mm, Diameter of bars

Singly Reinforced Section properties:

ec = 0.003 Concrete compressive strain

es = 0.005 Reinf. tensile strain

dt = 342.00 mm Effective depth

dt = h - d_c

ct = 128.25 mm Neutral axis depth

ct = (ec/(ec+es_)) * dt = (0.003/0.003+0.005) * 342.00 mm

beta = 0.850 Equivalent depth factor

beta = 0.85 - 0.05 * (fc - 28) / 7

Ymax = 109.01 mm Equivalent compressive block depth

Ymax = beta * ct = 0.850 * 128.25 mm

As = 1103.10 mm^2 Maximum area of steel for singly reinforced section

As = (0.85 * fc * b * Ymax) / fy

Mur = 119.88 kNm Maximum resistance moment for singly reinforced section

Mur = f * (0.85 * fc * b * Ymax) * [dt - (Ymax/2)] / 1000000

Section works as a Singly Reinforced Section because of:

The applied moment Mua = 100.00 kNm is less than the maximum resistance moment for the section Mur = 119.88 kNm.

Calculate Required Area of Steel:

Y = 87.68 mm Equivalent compressive block depth

Y = dt - sqrt(dt^2 - [(2 * Mua * 1000000)/(phi * 0.85 * fc * b)])

As(req) = 887.28 mm^2 Required area of steel

As(req) = 0.85 * fc * b * Y / fy = 887.28 mm^2

Asmax = 3420.00 mm^2 Maximum area of steel for the beam section

Asmax = 0.04 * b * dt = 3420.00 mm^2

Asmin1 = 227.60 mm^2 Minimum area of steel for the beam section

Asmin1 = 0.25 * [sqrt(fc)/fy] * b * dt = 227.60 mm^2

Asmin2 = 285.00 mm^2 Another minimum area of steel

Asmin2 = (1.4/fy) * b * dt = 285.00 mm^2

Asmin3 = 1183.04 mm^2 Another minimum area of steel

Asmin3 = (4/3) * As(req) = 1183.04 mm^2

As(used) = 887.28 mm^2 Used area of steel

As(used) = max(Asmin, As(req)) = 887.28 mm^2

Calculate Quantity of Bars:

Nb = 5 Number of bars

Nb = roundup(As(used) / (0.25 * p * (Db)^2), 0) = 5