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Steiner's Theorem — Definition, Formula & Examples

Steiner's Theorem, also called the Parallel Axis Theorem, states that the moment of inertia of a shape about any axis equals its moment of inertia about a parallel axis through the centroid, plus the area (or mass) times the square of the distance between the two axes.

Let IGI_G be the second moment of area (or moment of inertia) of a planar region about an axis through its centroid, AA its total area, and dd the perpendicular distance from the centroid to a parallel axis. Then the second moment of area about the parallel axis is I=IG+Ad2I = I_G + A d^2. For mass moments of inertia, AA is replaced by total mass mm.

Key Formula

I=IG+A d2I = I_G + A\,d^2
Where:
  • II = Moment of inertia about the target parallel axis
  • IGI_G = Moment of inertia about the centroidal axis (parallel to the target)
  • AA = Total area of the cross-section (or mass m for mass moment of inertia)
  • dd = Perpendicular distance between the centroidal axis and the target axis

How It Works

You use Steiner's Theorem whenever you need the moment of inertia about an axis that does not pass through the centroid. First, find (or look up) the moment of inertia about the centroidal axis parallel to your target axis. Then measure the perpendicular distance dd between the two axes. Finally, add the product Ad2Ad^2 (or md2md^2) to the centroidal value. This approach is especially powerful for composite shapes: compute each sub-shape's centroidal moment, shift it to the common axis using Steiner's Theorem, and sum the results.

Worked Example

Problem: A rectangle has width 6 cm and height 4 cm. Its centroidal moment of inertia about a horizontal axis through its center is known. Find the moment of inertia about the base (bottom edge).
Step 1: Compute the centroidal moment of inertia about a horizontal axis through the center of the rectangle.
IG=bh312=6×4312=38412=32 cm4I_G = \frac{bh^3}{12} = \frac{6 \times 4^3}{12} = \frac{384}{12} = 32 \text{ cm}^4
Step 2: Find the distance from the centroid to the base. The centroid is at half the height.
d=h2=42=2 cmd = \frac{h}{2} = \frac{4}{2} = 2 \text{ cm}
Step 3: Apply Steiner's Theorem with area A = 6 × 4 = 24 cm².
I=IG+A d2=32+24×22=32+96=128 cm4I = I_G + A\,d^2 = 32 + 24 \times 2^2 = 32 + 96 = 128 \text{ cm}^4
Answer: The moment of inertia about the base is 128 cm4128 \text{ cm}^4. (This matches the known formula bh33\frac{bh^3}{3} as a check: 6×643=128\frac{6 \times 64}{3} = 128.)

Why It Matters

Steiner's Theorem is essential in structural and mechanical engineering when analyzing beams, columns, and machine parts whose reference axis rarely coincides with the centroid. In courses like Statics, Strength of Materials, and Dynamics, it appears whenever you calculate bending stress, deflection, or rotational kinetic energy of composite cross-sections.

Common Mistakes

Mistake: Applying the theorem between two non-centroidal axes directly, e.g., computing I2=I1+Ad2I_2 = I_1 + Ad^2 where neither axis passes through the centroid.
Correction: One of the two axes must be the centroidal axis. To shift between two non-centroidal axes, first shift back to the centroid (IG=I1−Ad12I_G = I_1 - Ad_1^2), then shift out to the new axis (I2=IG+Ad22I_2 = I_G + Ad_2^2).