--- title: "Orientation of principal axes and MI of composite section" subject: "Strength of Materials" chapter: 2 key: "semester_3/strength_of_materials:principal-mi-two-triangles-composite-2066b" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-two-triangles-composite-2066b site: BCT Engineering --- # Orientation of principal axes and MI of composite section Strength of Materials · Chapter 2: Geometrical Properties of Sections - Asked in: 2066 Bhadra (old course), 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-two-triangles-composite-2066b - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2 ## Question Determine the orientation of principal axes and the principal moment of inertia about centroidal axes of the composite section shown in figure. [Figure: composite section in cm: left triangle of height 6 cm + 2 cm with base 6 cm; central rectangle 12 cm long; right triangle with horizontal extent 6 cm and vertical extent 7 cm.] ## Short answer Centroid at (11.79, 2.812) cm; I_x=366.4, I_y=3829, I_xy=-45.01 cm⁴. Principal moments: I_max=3830 cm⁴, I_min=365.8 cm⁴; axes at 89.26° (I_max) and -0.74° (I_min) from +x. ## Answer **Dimensions and assumptions (read from the figure):** - Dimensions in cm. Left triangle: vertices (0,0), (6,0), (6,8) (base 6, height 6 + 2). Central rectangle 12 × 6 (x 6 to 18, y 0 to 6). Right triangle: vertices (18,0), (24,0), (18,7) (horizontal 6, vertical 7). #### Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $cm^2$, I in $cm^4$) | Part | Sign | A | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ | |---|---|---|---|---|---|---|---| | Left triangle | + | $24.00$ | 4.000 | 2.667 | $85.33$ | $48.00$ | $32.00$ | | Rectangle 12×6 | + | $72.00$ | 12.00 | 3.000 | $216.0$ | $864.0$ | $0$ | | Right triangle | + | $21.00$ | 20.00 | 2.333 | $57.17$ | $42.00$ | $-24.50$ | #### Centroid $$ \begin{aligned} A &= \sum A_i = 117.0\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{1380}{117.0} = 11.79\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{329.0}{117.0} = 2.812\ \text{cm} \end{aligned} $$ #### Moments and product of inertia about the centroidal axes With $d_x = x_i-\bar X$, $d_y = y_i-\bar Y$ and the parallel-axis theorems $I_x = I_{x,G}+Ad_y^2$, $I_y = I_{y,G}+Ad_x^2$, $I_{xy} = I_{xy,G}+Ad_xd_y$: | Part | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ | |---|---|---|---| | Left triangle (+) | $85.84$ | $1506$ | $59.18$ | | Rectangle 12×6 (+) | $218.5$ | $867.0$ | $2.777$ | | Right triangle (+) | $61.98$ | $1456$ | $-107.0$ | | **Total about centroid** | **$366.4$** | **$3829$** | **$-45.01$** | #### Principal moments of inertia and principal axes $$ \begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 2098\pm1732 \\ I_{max} &= 3830\ \text{cm}^4,\qquad I_{min} = 365.8\ \text{cm}^4 \end{aligned} $$ Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{90.03}{-3463}$, so $2\theta_p = 178.51^\circ$ (and $-1.49^\circ$). - The axis of $I_{max}$ makes 89.26° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -0.74° (perpendicular), both passing through the centroid (11.79, 2.812) measured from the reference origin. **Answer: $I_{max} = 3830\ \text{cm}^4$, $I_{min} = 365.8\ \text{cm}^4$; principal axes through the centroid at 89.26° (for $I_{max}$) and -0.74° (for $I_{min}$) from the +x axis.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why do the triangles have non-zero own Ixy (+32 and -24.5) while the rectangle has zero? A. The triangles are subtracted B. The rectangle is bigger C. The triangles have no symmetry axes aligned with x-y, and they are mirror-oriented so signs differ D. Triangles have no Ixy Answer: C. The rectangle has symmetry axes parallel to x and y, so its Ixy is zero; right triangles do not, and their sign depends on orientation. (Key point: Triangle products) 2. How is Xbar = 11.79 cm obtained? A. Midpoint of the 24 cm length B. Sum of A*x (1380) divided by total area 117 cm^2 C. Average of 4, 12 and 20 D. Center of the rectangle only Answer: B. The areas are 24, 72 and 21 cm^2 and the weighted x-values give 1380/117. (Key point: Centroid) 3. Ixy total is only -45 cm^4 compared with Ix, Iy. What does this imply for the principal axes? A. They are nearly parallel to the x-y axes, with Imax almost equal to Iy B. They are at 45° C. They cannot be found D. Imax equals Ix Answer: A. A small Ixy means a small rotation: Imax ≈ 3830 about nearly the vertical axis (89.26°) and Imin ≈ 365.8 nearly horizontal. (Key point: Small Ixy) 4. What is added to each part's own Ixy to refer it to the section centroid? A. A*dx^2 B. A*dy^2 C. Ix - Iy D. A*dx*dy Answer: D. Ixy = Ixy,G + A*dx*dy, where dx and dy are the offsets between part and section centroids. 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