- 2074 Chaitra · 4+8 marks
Parallel axis theorem; MOI of trapezoid with cutout
State and prove the parallel axis theorem for moment of inertia. Determine the moment of inertia about centroidal axes of the plane figure shown below. [Figure: trapezoidal plane figure with 120 mm top width, 150 mm left height, 50 mm bottom offset, bottom widths 90 mm and 130 mm, with a semicircular cutout of radius 60 mm at the bottom.]
Parallel axis theorem
Statement: the moment of inertia of an area about any axis equals its moment of inertia about a parallel axis through its centroid plus the product of the area and the square of the distance between the two axes.
I = I_G + A d²
Proof: let G be the centroid of the area A, and let X′ be the centroidal axis. Take a parallel axis X at distance d from X′. For a small element dA at distance y′ from X′, its distance from X is y = y′ + d.
I_X = ∫ y² dA = ∫ (y′ + d)² dA
= ∫ y′² dA + 2d ∫ y′ dA + d² ∫ dA
- ∫ y′² dA = I_G, the moment of inertia about the centroidal axis.
- ∫ y′ dA = A·ȳ′ = 0, because the distance of the centroid from the centroidal axis is zero.
- ∫ dA = A.
So I_X = I_G + A d². In the same way I_Y = I_G(y) + A d_x². The moment of inertia about an axis through the centroid is the smallest of all parallel axes.
Moment of inertia of the figure
Reading of the figure: trapezium with top width 120 mm, left vertical side 150 + 50 = 200 mm, base 90 + 130 = 220 mm, so the right side slopes from (120, 200) to (220, 0). A semicircular hole of radius 60 mm with its diameter horizontal at 50 mm above the base, centre at x = 90 mm (the printed values are used as given; the arc then dips 10 mm below the base line, which the figure draws only approximately). Origin at the bottom-left corner.
Divide into: rectangle 120 × 200, right triangle 100 × 200 (vertical side at x = 120), minus the semicircle (centroid 4r/3π = 25.46 mm below its diameter, so at y = 24.54 mm).
Centroid
| Part | A (mm²) | x (mm) | y (mm) | A·x | A·y |
|---|---|---|---|---|---|
| Rectangle 120 × 200 | 24,000.00 | 60.00 | 100.00 | 1,440,000.0 | 2,400,000.0 |
| Triangle 100 × 200 | 10,000.00 | 153.33 | 66.67 | 1,533,333.3 | 666,666.7 |
| Semicircle r = 60 (hole, −) | -5,654.87 | 90.00 | 24.54 | -508,938.0 | -138,743.3 |
ΣA = 28,345.13 mm² x̄ = 86.94 mm, ȳ = 103.30 mm
Moments of inertia about the centroidal axes
Own values: rectangle bh³/12, hb³/12; triangle bh³/36, hb³/36; semicircle (π/8 − 8/9π) r⁴ about its centroidal horizontal axis and πr⁴/8 about its vertical diameter.
| Part | I_x own (mm⁴) | A·dy² | I_y own | A·dx² |
|---|---|---|---|---|
| Rectangle 120 × 200 | 80,000,000.0 | 260,640.0 | 28,800,000.0 | 17,421,494.5 |
| Triangle 100 × 200 | 22,222,222.2 | 13,416,678.6 | 5,555,555.6 | 44,077,494.1 |
| Semicircle r = 60 | -1,422,450.2 | -35,078,132.4 | -5,089,380.1 | -52,865.2 |
I_xx = 79.40 × 10⁶ mm⁴, I_yy = 90.71 × 10⁶ mm⁴ (radii of gyration k_x = 52.9 mm, k_y = 56.6 mm)
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