Parallel-Axis Theorem

I = I ˉ + A d^{2}

Moment of inertia about an axis parallel to a centroidal axis is the centroidal I plus area times the square of the distance between axes. Never subtract Ad² unless transferring toward the centroid.

$I$ (m⁴): Inertia about the parallel axis
$I ˉ$ (m⁴): Centroidal inertia
$A$ (m²): Area
$d$ (m): Distance between axes

Step-by-step solved example

Rectangle 100 mm × 200 mm. I about the base (axis parallel to the 100 mm side)?

Shift Ī to a parallel axis with +Ad² (never skip the centroidal Ī).

1. Centroidal
b=0.10, h=0.20, Ī = bh³/12 = 6.667×10⁻⁵ m⁴.
2. Transfer
A = 0.020 m², d = 0.10 m, Ad² = 2.00×10⁻⁴.
3. Sum
I = 6.667e-5 + 2.00e-4 = 2.667×10⁻⁴ m⁴ = bh³/3.
$I = 2.667 t im es 1 0^{- 4} t e x t m^{4}$

Answer: 2.667×10⁻⁴ m⁴

10 practice questions

0/10 correct

1.Parallel-axis theorem requires the reference axis at the centroid of

2.Ad² is always

3.To move I toward the centroid you

4.If d = 0, I equals

5.A = 10, d = 2, Ī = 5. I =

6.Composite I is the

7.A hole is handled by

8.Wrongly using a non-centroidal Ī in the theorem

9.d is measured

10.Mass moment of inertia uses