--- title: "Principal MI and orientation of Z-type section" subject: "Strength of Materials" chapter: 2 key: "semester_3/strength_of_materials:principal-mi-z-section-cm" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-mi-z-section-cm site: BCT Engineering --- # Principal MI and orientation of Z-type section Strength of Materials · Chapter 2: Geometrical Properties of Sections - Asked in: 2075 Chaitra, 10 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-z-section-cm - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2 ## Question Calculate the principal moments of inertia of the section given in figure and their orientation. Assume horizontal and vertical axes to be the given x and y axes and the bottom left corner of the section to be the origin for the purpose of your calculation. [Figure: section made of a vertical left strip 12 cm high, a top flange, and a bottom flange; widths 6 cm, 8 cm and 10 cm overall; flange thicknesses 4 cm (top) and 2 cm (bottom); inner dimensions 6 cm and 8 cm.] ## Short answer I_max = 894.4 cm^4, I_min = 105.6 cm^4 about centroid. Principal axes: -24.76° (I_max) and 65.24° (I_min) from +x axis. ## Answer **Dimensions and assumptions (read from the figure):** - Dimensions in cm; origin O at the bottom-left corner. Bottom flange 6 × 2 (x 0 to 6, y 0 to 2); web 2 wide (x 4 to 6) from y = 2 to 8 (inner height 6); top flange 6 × 4 (x 4 to 10, y 8 to 12). Overall 10 wide, 12 high. #### 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 | $\bar x$ | $\bar y$ | $I_{x,G}$ | $I_{y,G}$ | $I_{xy,G}$ | |---|---|---|---|---|---|---|---| | Bottom flange 6×2 | + | $12.00$ | 3.000 | 1.000 | $4.000$ | $36.00$ | $0$ | | Web 2×6 | + | $12.00$ | 5.000 | 5.000 | $36.00$ | $4.000$ | $0$ | | Top flange 6×4 | + | $24.00$ | 7.000 | 10.00 | $32.00$ | $72.00$ | $0$ | #### Centroid $$ \begin{aligned} A &= \sum A_i = 48.00\ \text{cm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{264.0}{48.00} = 5.500\ \text{cm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{312.0}{48.00} = 6.500\ \text{cm} \end{aligned} $$ #### 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 | $I_{x,G}+A d_y^2$ | $I_{y,G}+A d_x^2$ | $I_{xy,G}+A d_x d_y$ | |---|---|---|---| | Bottom flange 6×2 (+) | $367.0$ | $111.0$ | $165.0$ | | Web 2×6 (+) | $63.00$ | $7.000$ | $9.000$ | | Top flange 6×4 (+) | $326.0$ | $126.0$ | $126.0$ | | **Total about centroid** | **$756.0$** | **$244.0$** | **$300.0$** | #### Principal moments of inertia and principal axes $$ \begin{aligned} I_{max,min} &= \frac{I_x+I_y}{2}\pm\sqrt{\left(\frac{I_x-I_y}{2}\right)^2+I_{xy}^2} \\ &= 500.0\pm394.4 \\ I_{max} &= 894.4\ \text{cm}^4,\qquad I_{min} = 105.6\ \text{cm}^4 \end{aligned} $$ Orientation: $\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{-600.0}{512.0}$, so $2\theta_p = -49.52^\circ$ (and $130.48^\circ$). - The axis of $I_{max}$ makes -24.76° with the +x axis (anticlockwise positive); the axis of $I_{min}$ is at 65.24° (perpendicular), both passing through the centroid (5.500, 6.500) measured from the reference origin. #### Values about the given origin About the axes through the origin O: $I_x = 2784$, $I_y = 1696$, $I_{xy} = 2016\ \text{cm}^4$ (see parts table with the parallel-axis theorem). Principal values about O: $I_{max} = 4328$, $I_{min} = 151.9$ at -37.45° to the x axis. **Answer: $I_{max} = 894.4\ \text{cm}^4$, $I_{min} = 105.6\ \text{cm}^4$; principal axes through the centroid at -24.76° (for $I_{max}$) and 65.24° (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 principal moments asked for are about axes through the centroid, and parallel-axis shifts need the centroid location D. Because the centroid is the point where Imax occurs Answer: C. 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 (5.5 cm, 6.5 cm). (Key point: Centroid first) 2. How is the product of inertia of one part transferred to the section's centroidal axes? A. I_xy = I_xy,G + A d_x^2 B. I_xy = I_xy,G + A (d_x + d_y) C. I_xy = I_xy,G, because the product of inertia does not change with axis shift D. I_xy = I_xy,G + A d_x d_y, keeping the signs of d_x and d_y Answer: D. The parallel-axis theorem for product of inertia adds A times the product of the signed centroidal offsets. (Key point: Parallel-axis for Ixy) 3. 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)^2 + I_xy^2) B. (I_x+I_y)/2 ± sqrt((I_x-I_y)^2 + I_xy) 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: A. The principal values are the centre of Mohr's circle plus and minus its radius. (Key point: Principal moment formula) 4. With I_x = 756, I_y = 244 and I_xy = 300 cm^4, which statement matches the answer? A. I_max = 756 and I_min = 244 cm^4; the I_max axis is along the x axis B. I_max = 894.4 and I_min = 105.6 cm^4; the I_max axis is at -24.76° from +x and the I_min axis is perpendicular to it C. I_max = 105.6 cm^4 and I_min = 894.4 cm^4 with both axes at 45 degrees D. I_max and I_min are both equal to (I_x+I_y)/2, and the axes are not defined Answer: B. Imax and Imin come from the formula, and the principal axes are perpendicular, with the I_max axis at -24.76° (I_min axis at 65.24°). 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