Longitudinal Reinforcement Sample Calculation
In the sample project, the reinforced concrete calculation of the 1st floor S11 column will be made by using the method explained in Adnan Çakıroğlu - Erkan Özer's book " Formularies of bearing strength in rectangular reinforced concrete sections under the effect of oblique bending and axial force ".
Preliminary information
fck = 30 N/mm2 = 0.306 tf/cm2
fyk = 420 N / mm2 = 4,283 tf / cm2
Md2= 7.933 tfm = 793.3 tfcm
Md3 = -25.458 tfm=- 2545.8 tfcm
Nd= -5.925 tf
concrete cover (p) = 4.5 cm
Column dimensions = 35 cm / 70 cm
n = Nd / (b h fck), mx= Mxd / (bh2fck) , my = Myd /(b2hfck)
n= 5.925 / (35 * 70 * 0.306) = 0.0079
mx = 2545.8 / (35*70*70*0.306) = 0.0485
my = 793.3 / (35*35*70*0.306) = 0.0302
Since n <0.2;
k = (2 - 6.65n) my + 2.65n + 0.20
k = (2- 6.65 * 0.0079) * 0.0302 + 2.65 * 0.0079 + 0.20 = 0.28
m= mx + k * my (my<=mx)
m = 0.0485 + 0.28*0.0302 = 0.056
u = 2.86m+2.92 n2-1.48n
y = 2.86 * 0.056 + 2.92 * 0.0079 * 0.0079 -1.48 * 0.0079 = 0.151
alfa = ( h- 2*p ) / h
alfa = 61 / 70 = 0.87
k2 = [ (m/n) (0.8/alfa) +0.2] / (m/n+0.2)
k2 = ((0.056 / 0.0079) (0.8 / 0.87) +0.2 / (0.056 / 0.079 + 0.2) = 0.92
→ k1 = 1.15 is read from the sheet given in the booklet according to the placement of reinforcement equal on 4 sides.
As = k1 k2 ubh fck / fyk
As= 1.15 * 0.92 * 0.151 * 35 * 70 * 0.306 / 4.823
As= 24.83 cm2 > 0.01 Ac = 0.01 * 35 * 70 = 24.50 cm2
16 reinforcement will be used:
selected reinforcement = 24.83 / 2.01 → 12.35 -> 14fi16 (equal distribution of reinforcement on 4 sides)
ideCAD report:
