# Sections

All sections used in the elements in the program are listed in the structure tree. Any section can be defined with Add New Section in the sections row of the structure tree.

The sections for reinforced concrete elements are automatically defined from the given size values when the object is defined. For example, when a 40/60 column is defined, it is seen as a rectangular section in the Section list.

Section sizes used in the analysis for this section will be calculated manually by formulas and compared with the program.

### Location of the Add New Section Command

You can access the command under the **Structure Tree.**

### Manual Calculation of 40/60 Rectangular Section Sizes

#### Values given by the program:

#### Manual control:

The second moment of inertia around the axis I

_{2 =}HW^{3}/12 = 40 * 60^{3}/12 = 720000 cm4 (7.2 A + 5 cm4)Moment of inertia around the 3 axis I

_{3}= w^{3}d / 12 = 40^{3}* 60/12 = 320000 cm4 (3.2E + 5 cm4)Torsion moment of inertia K = 30 * 20

^{3}[16/03 - 20/30 * * 3:36 (1-20^{4}/12/30^{4}] = 751249.38 cm4 (7512 cm4 A + 5)

Cross-sectional area A = w. d = 40 * 60 = 2400 cm2

Cutting area in 2 direction = 5/6. b. w = 5/6 * 40 * 60 = 2000 cm2

Cutting area in 3 direction = 5/6. w. b = 5/6 * 40 * 60 = 2000 cm2

Radius of inertia in 2 axis = √I

_{2}/ A = √720000 / 2400 = 17.32 cm2Radius of inertia in 3 axis = √I

_{3}/ A = √320000 / 2400 = 11.54 cm2The strength of the elastic torque of the second axis = bh

^{2}/6 = 40 * 60 * 60/6 = 24000 cm3Elastic strength moment in 3 axis = b

^{2}h / 6 = 40 * 40 * 60/6 = 16000 cm32 plastic structural strength axis torque = bh

^{2}/4 = 40 * 60 * 60/4 = 36000 cm3Plastic strength moment in the 3 axis = b

^{2}h / 4 = 40 * 60 * 60/4 = 24000 cm3

### Manual Calculation of Sample L Profile Section Sizes

Location of the neutral axis

x1= (50*5*2.5 + 45*5*27.5) / (50*5+45*5) = 14.34 mm

y1= (50*5*25+45*5*2.5)/(50*5+45*5)=14.34 mm

With the parallel axis theorem, the 3-axis moment of inertia

yg1 = 25-14.34 = 10.16mm (distance of the parallel axis to the center of gravity)

yg2 = 14.34-2.5 = 11.84 mm (distance of the parallel axis to the center of gravity)

I_{3} =5*50^{3}/12+5*50*10.66^{2} + 45*5^{3}/12+45*5*11.84^{2} = 112502.74 mm4 →** **(1.125E+5mm4)

With the parallel axis theorem, the 2-axis moment of inertia

xg1 = 14.34-2.5 = 11.84mm (distance of the parallel axis to the center of gravity)

xg2 = 27.5-14.34 = 13.16 mm (distance of the parallel axis to the center of gravity)

I_{2} =5^{3}*50/12+5*50*11.84^{2} + 45^{3}*5/12+45*5*13.16^{2} = 112502.74 mm4 → (1.125E+5mm4)

Cross-sectional area

A= 5*50+45*5=475 mm4

Radius of inertia in the 2 axis = √112502.74 / 475 = 15.39 cm2

Radius of inertia in 3 axis = √112502.74 / 475 = 15.39 cm2

**Next Topic**