Parallel-Axis Theorem - Statics

1. A thin rectangular plate (8 cm × 12 cm) has a moment of inertia of $128 \text{ cm}^4$ about an axis passing through its centroid parallel to the 8 cm side. What is the moment of inertia about a parallel axis located 5 cm away from the centroidal axis?
   Answer: $2528 \text{ cm}^4$ When you encounter moment of inertia problems involving parallel axes, you're dealing with the parallel axis theorem, one of the most important tools in structural analysis. This theorem allows you to find the moment of inertia about any axis parallel to a centroidal axis. The parallel axis theorem states: $I = I_c + Ad^2$, where $I_c$ is the centroidal moment of inertia, $A$ is the cross-sectional area, and $d$ is the distance between axes. First, calculate the plate's area: $A = 8 \times 12 = 96 \text{ cm}^2$. Now apply the theorem: $I = 128 + 96 \times 5^2 = 128 + 96 \times 25 = 128 + 2400 = 2528 \text{ cm}^4$. This confirms answer A is correct. Looking at the wrong answers: Answer B ($2400 \text{ cm}^4$) represents just the $Ad^2$ term—you forgot to add the original centroidal moment of inertia. Answer C ($2656 \text{ cm}^4$) likely comes from incorrectly using the distance as 6 cm instead of 5 cm, giving $128 + 96 \times 36 = 2656$. Answer D ($628 \text{ cm}^4$) appears to use an incorrect area calculation or distance. The key study tip: Always remember that the parallel axis theorem adds to the centroidal moment of inertia—moving away from the centroid always increases the moment of inertia. Double-check that your final answer is larger than the given centroidal value, and ensure you're using the correct area and distance measurements.
2. A structural engineer needs to find the moment of inertia of an L-shaped section about its centroidal axis. The section consists of two rectangles: Rectangle A ($150\text{ mm} \times 25\text{ mm}$) positioned horizontally at the top, and Rectangle B ($25\text{ mm} \times 125\text{ mm}$) positioned vertically at the right end of Rectangle A. If the centroid of the L-section is located $33.3\text{ mm}$ from the left edge, what is the contribution of Rectangle B alone to the total moment of inertia about the centroidal y-axis?
   Answer: $2.19 \times 10^6\text{ mm}^4$ Rectangle B has dimensions $25\text{ mm} \times 125\text{ mm}$ with its centroid at $x = 150 - 12.5 = 137.5\text{ mm}$ from the left edge. Its own moment of inertia about its centroidal y-axis is $I_{yy} = \frac{h \times b^3}{12} = \frac{125 \times 25^3}{12} = 1.628 \times 10^5\text{ mm}^4$. The distance from Rectangle B's centroid to the L-section's centroidal y-axis is $d = 137.5 - 33.3 = 104.2\text{ mm}$. Area of Rectangle B = $25 \times 125 = 3125\text{ mm}^2$. Using parallel-axis theorem: $I_B = 1.628 \times 10^5 + 3125 \times 104.2^2 = 2.19 \times 10^6\text{ mm}^4$. Choice A uses wrong centroidal distance. Choice B uses incorrect moment formula. Choice C omits the parallel-axis transfer term.
3. A T-section is formed by welding two rectangular plates: a horizontal flange ($200\text{ mm} \times 20\text{ mm}$) and a vertical web ($20\text{ mm} \times 180\text{ mm}$). If the moment of inertia about the centroidal axis of the T-section is $I_c = 2.89 \times 10^7\text{ mm}^4$, what is the moment of inertia about an axis parallel to the centroidal axis but located at the bottom of the web?
   Answer: $5.67 \times 10^7\text{ mm}^4$ First find the centroid of the T-section. Flange area = $200 \times 20 = 4000\text{ mm}^2$ with centroid at 190 mm from bottom. Web area = $20 \times 180 = 3600\text{ mm}^2$ with centroid at 90 mm from bottom. Total area = 7600 mm². Centroid location: $\bar{y} = \frac{4000 \times 190 + 3600 \times 90}{7600} = 142.1\text{ mm}$ from bottom. Using parallel-axis theorem: $I_{bottom} = I_c + A_{total} \times \bar{y}^2 = 2.89 \times 10^7 + 7600 \times 142.1^2 = 5.67 \times 10^7\text{ mm}^4$. Choice A uses incorrect centroid calculation. Choice C adds the distance incorrectly. Choice D uses wrong total area.
4. An engineer is designing a reinforced concrete beam where steel reinforcement bars must be analyzed separately. Four steel bars, each with diameter $d = 20\text{ mm}$, are positioned at the corners of a $150\text{ mm} \times 150\text{ mm}$ square pattern. The square is oriented with its centroid at the origin and sides parallel to the coordinate axes. What is the polar moment of inertia of this reinforcement system about the origin?
   Answer: $1.89 \times 10^6\text{ mm}^4$ Each bar is located at distance $r = \frac{150\sqrt{2}}{2} = 106.1\text{ mm}$ from the origin (half the diagonal of the square). For each circular bar about its own centroid: $J_0 = \frac{\pi d^4}{32} = \frac{\pi \times 20^4}{32} = 1.571 \times 10^4\text{ mm}^4$. Cross-sectional area of each bar = $\frac{\pi \times 20^2}{4} = 314\text{ mm}^2$. Using parallel-axis theorem for polar moments: $J = J_0 + A \times r^2$. For each bar: $J = 1.571 \times 10^4 + 314 \times 106.1^2 = 4.69 \times 10^5\text{ mm}^4$. Total for four bars: $J_{total} = 4 \times 4.69 \times 10^5 = 1.89 \times 10^6\text{ mm}^4$. Choice A omits the bars' own polar moments. Choice B uses incorrect distance calculation. Choice D uses rectangular moment formula instead of polar.
5. A cantilever beam cross-section is modified by removing a triangular piece from its rectangular base. The original rectangle is $80\text{ mm} \times 120\text{ mm}$. The removed triangle has a base of $40\text{ mm}$ (along the bottom edge of the rectangle) and height of $60\text{ mm}$, with its apex $60\text{ mm}$ above the bottom edge. What is the moment of inertia of the modified section about the original centroidal axis of the rectangle?
   Answer: $1.09 \times 10^7\text{ mm}^4$ Original rectangle $I = \frac{80 \times 120^3}{12} = 1.152 \times 10^7\text{ mm}^4$. For the triangle, its centroidal axis is at $\frac{h}{3} = 20\text{ mm}$ from its base, so the triangle's centroid is at $20\text{ mm}$ from the bottom of the rectangle. Distance from triangle's centroid to rectangle's centroidal axis = $60 - 20 = 40\text{ mm}$. Triangle's moment about its own centroidal axis: $I = \frac{bh^3}{36} = \frac{40 \times 60^3}{36} = 2.4 \times 10^5\text{ mm}^4$. Triangle area = $\frac{1}{2} \times 40 \times 60 = 1200\text{ mm}^2$. Triangle's moment about rectangle's centroidal axis: $I_{triangle} = 2.4 \times 10^5 + 1200 \times 40^2 = 2.16 \times 10^6\text{ mm}^4$. Modified section: $I = 1.152 \times 10^7 - 2.16 \times 10^6 = 1.09 \times 10^7\text{ mm}^4$. Choice A uses incorrect triangle centroid. Choice C adds instead of subtracting. Choice D uses wrong triangle moment formula.
6. A circular area with radius 4 cm is cut from a larger rectangular area (16 cm × 12 cm). The circle's center is located at (4 cm, 3 cm) from the rectangle's centroidal axes. If the rectangle's centroidal moment of inertia about x-axis is $I_{x,rect} = 2048 \text{ cm}^4$ and the circle's centroidal moment of inertia is $I_{x,circle} = 201 \text{ cm}^4$, what is the moment of inertia of the remaining area about the x-axis?
   Answer: $1394 \text{ cm}^4$ When you encounter composite area problems in statics, you're dealing with the principle of superposition: the moment of inertia of a shape with material removed equals the original shape's moment of inertia minus the removed portion's moment of inertia about the same axis. For the removed circle, you need its moment of inertia about the rectangle's x-axis, not its own centroidal axis. Using the parallel axis theorem: $I_x = I_{centroidal} + Ad^2$, where $d$ is the distance between axes. The circle's area is $\pi r^2 = \pi(4)^2 = 16\pi \text{ cm}^2$, and its center is 3 cm from the rectangle's x-axis. Therefore: $I_{x,circle} = 201 + 16\pi(3)^2 = 201 + 144\pi = 653.3 \text{ cm}^4$. The remaining area's moment of inertia is: $I_x = 2048 - 653.3 = 1394.7 \text{ cm}^4$. Answer A ($1847 \text{ cm}^4$) results from incorrectly subtracting only the circle's centroidal moment of inertia without applying the parallel axis theorem. Answer C ($1621 \text{ cm}^4$) likely comes from calculation errors in the parallel axis theorem application. Answer D ($2702 \text{ cm}^4$) incorrectly adds the circle's contribution instead of subtracting it. Remember: when material is removed from a composite area, always use the parallel axis theorem to transfer the removed portion's moment of inertia to the reference axis before subtracting. The distance between centroidal axes is crucial for accurate calculations.

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