--- title: "Principal MI and axes of section with semicircular cut-out" subject: "Strength of Materials" chapter: 2 key: "semester_3/strength_of_materials:principal-mi-section-with-semicircle-50cm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-section-with-semicircle-50cm site: BCT Engineering --- # Principal MI and axes of section with semicircular cut-out Strength of Materials · Chapter 2: Geometrical Properties of Sections - Asked in: 2073 Shrawan, 8 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-section-with-semicircle-50cm - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2 ## Question Obtain the principle moment of inertia and draw principle axes for the plane figure given below. [Figure: shaded plane figure 50 cm + 50 cm tall (100 cm total) with a semicircular cut-out at the bottom between horizontal dimensions 30 cm, 50 cm, 50 cm, 20 cm, 40 cm and a notch on the right side.] ## Short answer Centroid at (44.13, 48.55) cm. Using parallel axis theorem, I_x=5.966×10^6, I_y=5.986×10^6, I_xy=-1.640×10^6 cm^4. Principal moments: I_max=7.617×10^6, I_min=4.336×10^6 cm^4. Principal axes through centroid at 45.17° (I_max) and -44.83° (I_min) from +x axis. ## Answer **Dimensions and assumptions (read from the figure):** - Dimensions in cm. Plate (0,0), (100,0), (100,50), (70,50), (70,100), (0,100) (100 high, notch 30 × 50 at the top right). Semicircular cut-out of radius 20 on the bottom edge, centre (40, 0), removed. #### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ | |---|---|---|---|---|---|---|---| | Plate | + | $8500$ | 43.82 | 45.59 | $6.918\times10^{6}$ | $6.059\times10^{6}$ | $-1.544\times10^{6}$ | | Semicircle cut-out | − | $628.3$ | 40.00 | 8.488 | $17561$ | $62832$ | $0$ | #### Centroid $$ \begin{aligned} A &= \sum A_i = 7872\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{3.474\times10^{5}}{7872} = 44.13\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{3.822\times10^{5}}{7872} = 48.55\ \text{cm} \end{aligned} $$ #### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ | |---|---|---|---| | Plate (+) | $6.992\times10^{6}$ | $6.060\times10^{6}$ | $-1.536\times10^{6}$ | | Semicircle cut-out (−) | $-1.026\times10^{6}$ | $-73542$ | $-1.039\times10^{5}$ | | **Total about centroid** | **$5.966\times10^{6}$** | **$5.986\times10^{6}$** | **$-1.640\times10^{6}$** | #### Principal moments of inertia and principal axes $$ \begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 5.976\times10^{6}\pm1.640\times10^{6} \\ I_{max} &= 7.617\times10^{6}\ \text{cm}^4,\qquad I_{min} = 4.336\times10^{6}\ \text{cm}^4 \end{aligned} $$ Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{3.281\times10^{6}}{-19840}$, so $2\theta_p = 90.35^\circ$ (and $-89.65^\circ$). - The axis of $I_{max}$ makes 45.17° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at -44.83° (perpendicular), both passing through the centroid (44.13, 48.55) measured from the reference origin. **Answer: $I_{max} = 7.617\times10^{6}\ \text{cm}^4$, $I_{min} = 4.336\times10^{6}\ \text{cm}^4$; principal axes through the centroid at 45.17° (for $I_{max}$) and -44.83° (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 axes always coincide with the reference axes through the origin B. Because the product of inertia about the origin is always zero C. Because the centroid is the point where Imax occurs D. Because the principal moments asked for are about axes through the centroid, and parallel-axis shifts need the centroid location Answer: D. 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 (44.13 cm, 48.55 cm). (Key point: Centroid first) 2. How is a cut-out (hole) treated in the composite-area method? A. Its area and moments of inertia are taken negative (subtracted) while using its own centroid for the parallel-axis terms B. Its area is added because a cut-out is part of the outline C. It is ignored because its I_G about its own centroid is small D. Only its area is subtracted; its moment of inertia is still added Answer: A. A removed region contributes -A, -A*x and -(I_G + A d^2) etc. to the sums. (Key point: Cut-out handling) 3. With I_x = 5.966e6, I_y = 5.986e6 and I_xy = -1.640e6 cm^4, which statement matches the answer? A. I_max = 5.966e6 and I_min = 5.986e6 cm^4; the I_max axis is along the x axis B. I_max = 4.336e6 cm^4 and I_min = 7.617e6 cm^4 with both axes at 45 degrees C. I_max = 7.617e6 and I_min = 4.336e6 cm^4; the I_max axis is at 45.17° from +x and the I_min axis is perpendicular to it D. I_max and I_min are both equal to (I_x+I_y)/2, and the axes are not defined Answer: C. Imax and Imin come from the formula, and the principal axes are perpendicular, with the I_max axis at 45.17° (I_min axis at -44.83°). 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