--- title: "Parallel axis theorem for product of inertia" subject: "Strength of Materials" chapter: 2 key: "semester_3/strength_of_materials:parallel-axis-theorem-product-of-inertia" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/parallel-axis-theorem-product-of-inertia site: BCT Engineering --- # Parallel axis theorem for product of inertia Strength of Materials · Chapter 2: Geometrical Properties of Sections - Asked in: 2069 Asar, 4 marks - Video: [Parallel Axis Theorem - Moment of Inertia - Strength of Materials](https://www.youtube.com/watch?v=4M94dgM8OCk) — Ekeeda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/parallel-axis-theorem-product-of-inertia - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2 ## Question State and prove parallel axis theorem for product of inertia. ## Short answer The product of inertia about any axes X-Y equals the product of inertia about parallel centroidal axes plus A·x̄·ȳ, where x̄, ȳ are centroid coordinates in X-Y. Proof: substitute X=x+x̄, Y=y+ȳ into integral ∫XY dA, use centroidal properties ∫x dA=0, ∫y dA=0. ## Answer #### Statement The product of inertia of an area about any pair of perpendicular axes $X$–$Y$ is equal to the product of inertia about the parallel centroidal axes plus the product of the area and the coordinates of the centroid with respect to the $X$–$Y$ axes: $$ I_{XY} = I_{xy} + A\,\bar x\,\bar y $$ where $I_{xy}$ is the product of inertia about the centroidal axes $x$, $y$ (parallel to $X$, $Y$), and $(\bar x, \bar y)$ are the coordinates of the centroid G in the $X$–$Y$ system, with signs. #### Proof Let $dA$ be an element with coordinates $(x, y)$ with respect to the centroidal axes. Its coordinates with respect to the $X$–$Y$ axes are $$ X = x + \bar x,\qquad Y = y + \bar y $$ Then $$ \begin{aligned} I_{XY} &= \int XY\,dA = \int (x + \bar x)(y + \bar y)\,dA\\ &= \int xy\,dA + \bar y\int x\,dA + \bar x\int y\,dA + \bar x\bar y\int dA \end{aligned} $$ Since $x$ and $y$ are measured from the centroid, $\int x\,dA = 0$ and $\int y\,dA = 0$. Also $\int xy\,dA = I_{xy}$ and $\int dA = A$. Therefore $$ I_{XY} = I_{xy} + A\bar x\bar y\qquad\blacksquare $$ For a composite section the same formula is applied to each part and the results are added: $I_{xy} = \sum (\bar I_{xy,i} + A_i d_{x,i}d_{y,i})$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does the parallel axis theorem for product of inertia state? A. I_XY = I_xy + A x̄ ȳ, where I_xy is about the centroidal axes and (x̄, ȳ) are the signed coordinates of the centroid B. I_XY = I_xy + A x̄^2 ȳ^2 C. I_XY = I_xy + A (x̄ + ȳ)^2 D. I_XY = I_xy - A x̄ ȳ, with absolute values of x̄ and ȳ Answer: A. The product of inertia shifts by the area times the product of the centroid's signed coordinates. (Key point: Statement) 2. In the proof, how are the coordinates of an element dA in the X-Y system written? A. X = x - x̄ and Y = y + ȳ, with x measured from the origin B. X = x + x̄ and Y = y + ȳ, where x, y are measured from the centroid C. X = x·x̄ and Y = y·ȳ D. X = x + ȳ and Y = y + x̄ Answer: B. Adding the centroid coordinates to the centroidal coordinates gives the position relative to X-Y. (Key point: Proof set-up) 3. Why do the terms ȳ∫x dA and x̄∫y dA vanish in the proof? A. Because x̄ and ȳ are zero for every area B. Because ∫dA = 0 for any area C. Because x and y are measured from the centroid, so the first moments ∫x dA and ∫y dA are zero D. Because the product of inertia is always zero about centroidal axes Answer: C. The first moment of an area about its own centroidal axis is zero, which removes two of the four terms. (Key point: Why terms vanish) 4. How is the theorem used for a composite section? A. Add only the I_xy,G of each part, since the A d_x d_y terms cancel B. Multiply the product of inertia of each part by the total area C. Use the largest part's product of inertia only D. Apply I_xy,i + A_i d_x d_y to each part with signs and add the results Answer: D. The sum over parts of (own I_xy + A d_x d_y) gives the product of inertia of the whole section. 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