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:

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