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I Section Moment of Inertia

The document provides a detailed calculation of the moment of inertia for an I-section girder with specified dimensions. It includes the calculations for the top flange, bottom flange, and web, leading to a total moment of inertia of 186.76 × 10^6 mm^4. The centroidal axis is located at 150 mm from the base of the girder.

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53 views1 page

I Section Moment of Inertia

The document provides a detailed calculation of the moment of inertia for an I-section girder with specified dimensions. It includes the calculations for the top flange, bottom flange, and web, leading to a total moment of inertia of 186.76 × 10^6 mm^4. The centroidal axis is located at 150 mm from the base of the girder.

Uploaded by

gatiyochu
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Problem:

An I-section girder has:


- Flange width = 200 mm
- Overall depth = 300 mm
- Flange thickness = 20 mm
- Web thickness = 20 mm

Determine the moment of inertia of the I-section about the horizontal centroidal axis (I ).

Solution:

Given:
Width of flange, b_f = 200 mm
Thickness of flange, t_f = 20 mm
Thickness of web, t_w = 20 mm
Total depth of section, H = 300 mm
Height of web = h_w = H - 2t_f = 300 - 40 = 260 mm

Centroid (neutral axis) location:


y = H / 2 = 300 / 2 = 150 mm

1. Moment of Inertia of Top Flange:


Area A = b_f × t_f = 200 × 20 = 4000 mm²
Distance from centroid = 150 - 10 = 140 mm
I_top = (b_f × t_f³) / 12 + A × d²
= (200 × 20³)/12 + 4000 × 140²
= 133333.33 + 78400000 = 78533333.33 mm

2. Moment of Inertia of Bottom Flange:


I_bottom = I_top = 78533333.33 mm

3. Moment of Inertia of Web:


Area = 20 × 260 = 5200 mm²
I_web = (t_w × h_w³) / 12 = (20 × 260³) / 12
= 29293333.33 mm

Total Moment of Inertia:


I_xx = I_top + I_bottom + I_web
= 78533333.33 + 78533333.33 + 29293333.33
= 186760000 mm

Final Answer:
I_xx = 186.76 × 10 mm

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