Math Problem Statement

Determine the moment of inertia Ix for the given symmetrical I-beam cross-section.

Solution

To determine the moment of inertia IxI_x for the given cross-section (symmetrical I-beam), we can break the section into simpler geometric shapes, usually rectangles, and then apply the parallel axis theorem to calculate the total moment of inertia about the centroidal axis.

Steps:

  1. Divide the section into three parts:

    • Flange 1 (top): A rectangle of width 60 mm and height 20 mm.
    • Flange 2 (bottom): Another rectangle of width 120 mm and height 20 mm.
    • Web (middle): A vertical rectangle of width 20 mm and height 60 mm.
  2. Calculate the moment of inertia for each part about their own centroidal axes:

    • For a rectangle, the moment of inertia about the centroidal axis (parallel to its base) is: Irect=112×base×height3I_{\text{rect}} = \frac{1}{12} \times \text{base} \times \text{height}^3
    • Apply this formula to each part.
  3. Use the parallel axis theorem to shift the moment of inertia of the flanges to the centroidal axis of the entire section: I=Icentroid+Ad2I = I_{\text{centroid}} + A \cdot d^2 where AA is the area of the part, and dd is the distance between the centroid of the part and the centroid of the entire section.

  4. Sum the contributions from the web and the two flanges to get the total moment of inertia IxI_x.

Let me calculate the values step by step.

Geometry dataTo determine the moment of inertia IxI_x for the given cross-section (symmetrical I-beam), we can break the section into simpler geometric shapes, usually rectangles, and then apply the parallel axis theorem to calculate the total moment of inertia about the centroidal axis.

Steps:

  1. Divide the section into three parts:

    • Flange 1 (top): A rectangle of width 60 mm and height 20 mm.
    • Flange 2 (bottom): Another rectangle of width 120 mm and height 20 mm.
    • Web (middle): A vertical rectangle of width 20 mm and height 60 mm.
  2. Calculate the moment of inertia for each part about their own centroidal axes:

    • For a rectangle, the moment of inertia about the centroidal axis (parallel to its base) is: Irect=112×base×height3I_{\text{rect}} = \frac{1}{12} \times \text{base} \times \text{height}^3
    • Apply this formula to each part.
  3. Use the parallel axis theorem to shift the moment of inertia of the flanges to the centroidal axis of the entire section: I=Icentroid+Ad2I = I_{\text{centroid}} + A \cdot d^2 where AA is the area of the part, and dd is the distance between the centroid of the part and the centroid of the entire section.

  4. Sum the contributions from the web and the two flanges to get the total moment of inertia IxI_x.

Let me calculate the values step by step.

Geometry

  • Flange 1: 60 mm × 20 mm
  • Flange 2: 120 mm × 20 mm
  • Web: 20 mm × 60 mm
  • Total height = 100 mm (20 mm + 60 mm + 20 mm)

Let me compute the moment of inertia IxI_x.The total moment of inertia (

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Math Problem Analysis

Mathematical Concepts

Geometry
Structural Analysis
Moment of Inertia
Centroid Calculation
Parallel Axis Theorem

Formulas

Moment of Inertia for a rectangle about its centroidal axis: I_rect = (1/12) * base * height^3
Parallel Axis Theorem: I = I_centroid + A * d^2

Theorems

Parallel Axis Theorem

Suitable Grade Level

Undergraduate (Engineering)