--- title: "Principal MI of 15 cm right triangle" subject: "Strength of Materials" chapter: 2 key: "semester_3/strength_of_materials:principal-mi-right-triangle-15x15" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-right-triangle-15x15 site: BCT Engineering --- # Principal MI of 15 cm right triangle Strength of Materials · Chapter 2: Geometrical Properties of Sections - Asked in: 2066 Jestha (old course), 8 marks - Video: [CE 270 8.3 Principal Axes: Application](https://www.youtube.com/watch?v=9PK6DWAkj2A) — Aaron Hanai - 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-right-triangle-15x15 - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2 ## Question Calculate principal moment of inertia about the X-Y axes and locate the principal axes of the right angled triangle section with 15cm x 15cm sides. [Figure: right-angled triangle with the right angle at the origin, legs of 15 cm along the X and Y axes.] ## Short answer I_max = 6328 cm^4 about axis through the right angle at 135° to X (parallel to hypotenuse); I_min = 2109 cm^4 about axis at 45° to X (line to hypotenuse midpoint). ## Answer Right angle at the origin O, legs of 15 cm along the X and Y axes (so $b = h = 15$ cm). The principal axes are required for the X–Y axes at O. ```text Y 15|\ | \ | \ O---- 15 X ``` #### Moments and product of inertia about X–Y $$ \begin{aligned} I_x &= \frac{bh^3}{12} = \frac{15\times15^3}{12} = 4218.750\ \text{cm}^4\\ I_y &= \frac{hb^3}{12} = 4218.750\ \text{cm}^4\\ I_{xy} &= \frac{b^2h^2}{24} = \frac{15^2\times15^2}{24} = 2109.375\ \text{cm}^4 \end{aligned} $$ #### Principal moments of inertia $$ \begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} = 4218.750\pm\sqrt{0 + 2109.375^2}\\ I_{max} &= 6328.125\ \text{cm}^4,\qquad I_{min} = 2109.375\ \text{cm}^4 \end{aligned} $$ #### Principal axes $$ \tan2\theta_p = \frac{-2I_{xy}}{I_x-I_y} = \frac{-2(2109.375)}{0} = -\infty\;\Rightarrow\;2\theta_p = -90^\circ\ \text{or}\ 90^\circ $$ so $\theta_p = 45^\circ$ and $135^\circ$. Checking with $I(\theta) = \frac{I_x+I_y}{2} + \frac{I_x-I_y}{2}\cos2\theta - I_{xy}\sin2\theta$: at $\theta = 135^\circ$ it is $I = 4218.750 - 2109.375\sin270^\circ = 6328.125$ (maximum), and at $\theta = 45^\circ$ it is $2109.375$ (minimum). **Answer: $I_{max} = 6328.12\ \text{cm}^4$ about the axis through O at $135^\circ$ to X (parallel to the hypotenuse), and $I_{min} = 2109.38\ \text{cm}^4$ about the axis through O at $45^\circ$ to X (the line from O to the mid-point of the hypotenuse).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For the 15x15 right triangle with its legs on the axes, how do Ix and Iy compare, and what is Ixy? A. Ix = bh^3/36; Ixy = 0 B. Ix is larger than Iy; Ixy = 0 C. Ix = Iy = bh^3/12 = 4218.75 cm^4; Ixy = b^2h^2/24 = 2109.375 cm^4 D. Ix = Iy = 2109 cm^4 Answer: C. The triangle is symmetric in b and h so Ix = Iy, and Ixy is positive since it is in the first quadrant. (Key point: Inertias) 2. Why is Imax = 6328.125 and Imin = 2109.375 cm^4? A. Imax = Ix + Iy B. Centre 4218.75 ± radius |Ixy| since Ix = Iy C. Imax = Ix, Imin = Iy D. Centre ± Ix Answer: B. With Ix = Iy the radius is sqrt(0 + Ixy^2) = 2109.375. (Key point: Principal moments) 3. Why are the principal axes at 45° and 135°? A. Ix - Iy = 0, so tan 2θ is infinite, giving 2θ = ±90° B. Ixy = 0 C. The hypotenuse is at 45° so it must be principal D. The triangle is isosceles only Answer: A. The denominator Ix - Iy is zero, so 2θ = 90° and θ = 45° or 135°. (Key point: Orientation) 4. Which axis carries the maximum moment? A. The axis at 45° through the hypotenuse mid-point B. The x axis C. The y axis D. The axis at 135° (parallel to the hypotenuse) Answer: D. I(135°) = 4218.75 - 2109.375 sin270° = 6328.125, the maximum; the 45° axis gives the minimum. 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