--- title: "Principal axes and MI with Mohr's circle" subject: "Strength of Materials" chapter: 2 key: "semester_3/strength_of_materials:principal-axes-mi-mohr-6m-section" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-axes-mi-mohr-6m-section site: BCT Engineering --- # Principal axes and MI with Mohr's circle Strength of Materials · Chapter 2: Geometrical Properties of Sections - Asked in: 2070 Chaitra, 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-axes-mi-mohr-6m-section - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2 ## Question Find the principal axes and principal moments of inertia about axes through centroid of the given figure. Verify your results using Mohr's circle. [Figure: section 6 m wide (3 m + 3 m at the top), left edge vertical 6 m (2 m + 2 m + 2 m), with a sloping cut from the top-middle to the lower right; right side 2 m, 2 m, 2 m.] ## Short answer Centroid at (2.600, 2.667) m. I_x=82.67, I_y=76.20, I_xy=-22.00 m^4. Principal moments: I_max=101.7, I_min=57.20 m^4. Principal axes through centroid at 40.82° (I_max) and -49.18° (I_min) from +x axis. Verified by Mohr's circle. ## Answer **Dimensions and assumptions (read from the figure):** - Dimensions in m. Polygon (0,0), (6,0), (6,2), (3,6), (0,6): width 6, left side 6, top width 3, sloping edge from the top-middle to the right side at 2 m height. #### Properties of the parts (x, y measured from the reference origin; $I_G$ about the part's own centroid; area in $m^2$, I in $m^4$) | Part | Sign | A | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ | |---|---|---|---|---|---|---|---| | Section | + | $30.00$ | 2.600 | 2.667 | $82.67$ | $76.20$ | $-22.00$ | #### Centroid $$ \begin{aligned} A &= \sum A_i = 30.00\ \text{m}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{78.00}{30.00} = 2.600\ \text{m} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{80.00}{30.00} = 2.667\ \text{m} \end{aligned} $$ #### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ | |---|---|---|---| | Section (+) | $82.67$ | $76.20$ | $-22.00$ | | **Total about centroid** | **$82.67$** | **$76.20$** | **$-22.00$** | #### Principal moments of inertia and principal axes $$ \begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 79.43\pm22.24 \\ I_{max} &= 101.7\ \text{m}^4,\qquad I_{min} = 57.20\ \text{m}^4 \end{aligned} $$ Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{44.00}{6.467}$, so $2\theta_p = 81.64^\circ$ (and $-98.36^\circ$). - The axis of $I_{max}$ makes 40.82° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -49.18° (perpendicular), both passing through the centroid (2.600, 2.667) measured from the reference origin. #### Mohr's circle check - Centre $C = \left(\dfrac{I_x+I_y}{2},\,0\right) = (79.43,\ 0)$; radius $R = 22.24$. - Plot $X(I_x,\ I_{xy}) = (82.67,\ -22.00)$ and $Y(I_y,\ -I_{xy}) = (76.20,\ 22.00)$; the line XY is a diameter. - The circle cuts the horizontal axis at $I_{max} = 101.7$ and $I_{min} = 57.20$, equal to the values found analytically. The angle from CX to the $I_{max}$ end of the horizontal diameter is $2\theta_p = 81.64^\circ$ (turned from the x axis towards the principal axis in the same sense as the physical rotation). **Answer: $I_{max} = 101.7\ \text{m}^4$, $I_{min} = 57.20\ \text{m}^4$; principal axes through the centroid at 40.82° (for $I_{max}$) and -49.18° (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 must the centroid (X, Y) of the whole section be found first? A. Because the principal moments asked for are about axes through the centroid, and parallel-axis shifts need the centroid location B. Because the principal axes always coincide with the reference axes through the origin C. Because the product of inertia about the origin is always zero D. Because the centroid is the point where Imax occurs Answer: A. I_x, I_y, I_xy are required about centroidal axes, so each part's distance d_x, d_y is measured from the overall centroid (2.600 m, 2.667 m). (Key point: Centroid first) 2. Which expression gives the principal moments of inertia from I_x, I_y and I_xy? A. (I_x+I_y)/2 ± sqrt((I_x-I_y)^2 + I_xy) B. (I_x-I_y)/2 ± sqrt(((I_x+I_y)/2)^2 + I_xy^2) C. (I_x+I_y)/2 ± sqrt(((I_x-I_y)/2)^2 + I_xy^2) D. sqrt(I_x^2 + I_y^2) ± I_xy Answer: C. The principal values are the centre of Mohr's circle plus and minus its radius. (Key point: Principal moment formula) 3. With I_x = 82.67, I_y = 76.20 and I_xy = -22.00 m^4, which statement matches the answer? A. I_max = 82.67 and I_min = 76.20 m^4; the I_max axis is along the x axis B. I_max = 57.20 m^4 and I_min = 101.7 m^4 with both axes at 45 degrees C. I_max and I_min are both equal to (I_x+I_y)/2, and the axes are not defined D. I_max = 101.7 and I_min = 57.20 m^4; the I_max axis is at 40.82° from +x and the I_min axis is perpendicular to it Answer: D. Imax and Imin come from the formula, and the principal axes are perpendicular, with the I_max axis at 40.82° (I_min axis at -49.18°). (Key point: Result and axis angle) 4. In the Mohr's circle verification, how are the centre, radius and principal values obtained? A. Centre = I_xy on the vertical axis, radius = (I_x+I_y)/2 B. Centre = I_x on the I axis, radius = I_y; I_max = I_x and I_min = I_y C. Centre = 0, radius = sqrt(I_x*I_y); I_max = R and I_min = -R D. Centre = (I_x+I_y)/2 on the I axis, radius = distance to point X(I_x, I_xy); I_max = C + R and I_min = C - R Answer: D. Points X(I_x, I_xy) and Y(I_y, -I_xy) form a diameter; the circle meets the horizontal axis at C ± R, matching the analytical values. Here C = 79.43, R = 22.24, so I_max = 101.7 and I_min = 57.20. 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