- 2073 Chaitra · 3+9 marks
Parallel axis theorem, radius of gyration; arch figure MOI
Short answer
Parallel axis theorem: I=I_G+Ad². Radius of gyration k=√(I/A). For shaded area (trapezium 60/30 cm×26 cm minus semicircle r=20 cm), centroid is at x̄=30 cm, ȳ=15.11 cm; I_xx=3.31×10⁴ cm⁴, I_yy=1.565×10⁵ cm⁴.
State parallel axis theorem and radius of gyration. Calculate the moment of inertia of the given shaded area about its centroidal axes. [Figure: shaded trapezoid-like area with 15 cm + 15 cm top, height 26 cm, 30 cm + 30 cm base, and a semicircular/arched cutout with 20 cm dimension.]
Parallel axis theorem
Statement: the moment of inertia of an area about any axis is equal to its moment of inertia about a parallel centroidal axis plus A d², where A is the area and d is the distance between the two axes: I = I_G + A d².
Radius of gyration
The radius of gyration k of an area about an axis is the distance from that axis at which the whole area can be assumed concentrated to give the same moment of inertia: I = A k², so k = √(I/A).
Moment of inertia of the shaded area
Reading of the figure: a symmetrical trapezium, base 30 + 30 = 60 cm, top 15 + 15 = 30 cm, height 26 cm, with a semicircular opening (arch) of radius 20 cm whose diameter lies on the base, centred on the axis of symmetry. y is measured from the base and x from the left corner, so the axis of symmetry is at x = 30 cm.
Trapezium: area = (60 + 30)/2 × 26 = 1170 cm², ȳ = 26(60 + 2×30)/(3 × 90) = 11.56 cm, I_x(own) = 63 469 cm⁴, I_y(own) = 219 375 cm⁴ (from the formulae for a trapezium). Semicircle (hole): area = π × 20²/2 = 628.3 cm², ȳ = 4r/3π = 8.49 cm.
Centroid
| Part | A (cm²) | x (cm) | y (cm) | A·x | A·y |
|---|---|---|---|---|---|
| Trapezium (60/30 × 26) | 1,170.00 | 30.00 | 11.56 | 35,100.0 | 13,520.0 |
| Semicircle r = 20 (hole, −) | -628.32 | 30.00 | 8.49 | -18,849.6 | -5,333.3 |
ΣA = 541.68 cm² x̄ = 30 cm (axis of symmetry), ȳ = 15.11 cm above the base.
Moments of inertia about centroidal axes
| Part | I_x own (cm⁴) | A·dy² | I_y own | A·dx² |
|---|---|---|---|---|
| Trapezium (60/30 × 26) | 63,468.9 | 14,810.4 | 219,375.0 | 0.0 |
| Semicircle r = 20 | -17,561.1 | -27,578.7 | -62,831.9 | -0.0 |
I_xx = 33,140 cm⁴ (≈ 3.31 × 10⁴ cm⁴) I_yy = 156,543 cm⁴ (≈ 1.565 × 10⁵ cm⁴) (k_x = 7.82 cm, k_y = 17.00 cm)
✓ Answer checked by Claude Opus 5.5 (high effort)
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