2526x
004398
2020-04-01

Question

How does the calculation of the moments of inertia differ when the cross-section consists of several unconnected or connected partial cross-sections?


Answer:

If the cross-section consists of several unconnected partial sections, the sum of the moments of inertia is calculated without the parallel axis theorem components. The cross-section shown in Image 01 consists of two angle sections that are not connected to each other.

The individual angle sections have the following moments of inertia:

Iy,1,2 = 180.39 cm4 (referred to the centroidal axes y, z)

Iz,1,2 = 65.05 cm4 (referred to the centroidal axes y, z)

The moments of inertia of the entire cross-section result in:

Iy,1+2 = 2 ⋅ Iy,1,2 = 2 ⋅ 180.39 = 360.78 cm4 (referred to the centroidal axes y, z)

Iz,1+2 = 2 ⋅ Iz,1,2 = 2 ⋅ 65.05 = 130.11 cm4 (referred to the centroidal axes y, z)

If the cross-section consists of several connected partial sections, the sum of the moments of inertia is calculated with the parallel axis theorem components. The cross-section shown in Image 02 consists of two connected angle sections.

The individual angle sections have the following cross-section properties:

A1,2 = 16.25 cm²

yS,0,1,2 = ±2.30 cm (referred to the zero point)

zS,0,1,2 = 3.07 cm (referred to the zero point)

Iy,1,2 = 180.39 cm4 (referred to the centroidal axes y, z)

Iz,1,2 = 65.05 cm4 (referred to the centroidal axes y, z)

The cross-section properties of the entire cross-section result in:

yS,0,1+2 = 0.00 cm (referred to the zero point)

zS,0,1+2 = 3.07 cm (referred to the zero point)

Iy,1+2 = 2 ⋅ Iy,1,2 + 2 ⋅ A1,2 ⋅ (zS,0,1,2 − zS,0,1+2)2

Iy,1+2 = 2 ⋅ 180.39 + 2 ⋅ 16.25 ⋅ (3.07 − 3.07)2 = 360.78 cm4 (referred to the centroidal axes y, z)

Iz,1+2 = 2 ⋅ Iz,1,2 + 2 ⋅ A1,2 ⋅ (yS,0,1,2 − yS,0,1+2)2

Iz,1+2 = 2 ⋅ 65.05 + 2 ⋅ 16.25 ⋅ (2.30 − 0.00)2 = 301.46 cm4 (referred to the centroidal axes y, z)