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Principal MI and orientation of L-section

Short answer

Centroid at (18.00, 20.00) mm; I_x=4.000×10⁵, I_y=3.600×10⁵, I_xy=-2.160×10⁵ mm⁴. Principal moments: I_max=5.969×10⁵ mm⁴, I_min=1.631×10⁵ mm⁴; axes at 42.35° (I_max) and -47.65° (I_min) from +x.

Calculate the principal MoI and their orientation for the following section about X-Y axes. [Figure: L-shaped section in mm with origin O at the bottom left corner; horizontal leg 60 mm long and 10 mm thick; vertical leg 12 mm wide.]

Dimensions and assumptions (read from the figure):

  • Dimensions in mm; origin O at the bottom-left corner. Horizontal leg 60 × 10 (x 0 to 60, y 0 to 10); vertical leg 12 × 50 above it (x 0 to 12, y 10 to 60). Both legs are 60 mm long overall.

Properties of the parts (x, y measured from the reference origin; IGI_G about the part's own centroid; area in mm2mm^2, I in mm4mm^4)

PartSignAxˉ\bar xyˉ\bar yIx,GI_{x,G}Iy,GI_{y,G}Ixy,GI_{xy,G}
Horizontal leg 60×10+600.0600.030.005.000500050001.800×1051.800\times10^{5}00
Vertical leg 12×50+600.0600.06.00035.001.250×1051.250\times10^{5}7200720000

Centroid

A=∑Ai=1200 mm2Xˉ=∑AixiA=216001200=18.00 mmYˉ=∑AiyiA=240001200=20.00 mm\begin{aligned} A &= \sum A_i = 1200\ \text{mm}^2 \\ \bar X &= \frac{\sum A_i x_i}{A} = \frac{21600}{1200} = 18.00\ \text{mm} \\ \bar Y &= \frac{\sum A_i y_i}{A} = \frac{24000}{1200} = 20.00\ \text{mm} \end{aligned}

Moments and product of inertia about the centroidal axes

With dx=xi−Xˉd_x = x_i-\bar X, dy=yi−Yˉd_y = y_i-\bar Y and the parallel-axis theorems Ix=Ix,G+Ady2I_x = I_{x,G}+Ad_y^2, Iy=Iy,G+Adx2I_y = I_{y,G}+Ad_x^2, Ixy=Ixy,G+AdxdyI_{xy} = I_{xy,G}+Ad_xd_y:

PartIx,G+Ady2I_{x,G}+A d_y^2Iy,G+Adx2I_{y,G}+A d_x^2Ixy,G+AdxdyI_{xy,G}+A d_x d_y
Horizontal leg 60×10 (+)1.400×1051.400\times10^{5}2.664×1052.664\times10^{5}−1.080×105-1.080\times10^{5}
Vertical leg 12×50 (+)2.600×1052.600\times10^{5}9360093600−1.080×105-1.080\times10^{5}
Total about centroid4.000×1054.000\times10^{5}3.600×1053.600\times10^{5}−2.160×105-2.160\times10^{5}

Principal moments of inertia and principal axes

Imax,min=Ix+Iy2±(Ix−Iy2)2+Ixy2=3.800×105±2.169×105Imax=5.969×105 mm4,Imin=1.631×105 mm4\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} \\ &= 3.800\times10^{5}\pm2.169\times10^{5} \\ I_{max} &= 5.969\times10^{5}\ \text{mm}^4,\qquad I_{min} = 1.631\times10^{5}\ \text{mm}^4 \end{aligned}

Orientation: tan⁡2θp=−2IxyIx−Iy=4.320×10540000\tan 2\theta_p = \dfrac{-2I_{xy}}{I_x-I_y} = \dfrac{4.320\times10^{5}}{40000}, so 2θp=84.71∘2\theta_p = 84.71^\circ (and −95.29∘-95.29^\circ).

  • The axis of ImaxI_{max} makes 42.35° with the +x axis (anticlockwise positive); the axis of IminI_{min} is at -47.65° (perpendicular), both passing through the centroid (18.00, 20.00) measured from the reference origin.

Values about the given origin

About the axes through the origin O: Ix=8.800×105I_x = 8.800\times10^{5}, Iy=7.488×105I_y = 7.488\times10^{5}, Ixy=2.160×105 mm4I_{xy} = 2.160\times10^{5}\ \text{mm}^4 (see parts table with the parallel-axis theorem). Principal values about O: Imax=1.040×106I_{max} = 1.040\times10^{6}, Imin=5.887×105I_{min} = 5.887\times10^{5} at -36.55° to the x axis.

Answer: Imax=5.969×105 mm4I_{max} = 5.969\times10^{5}\ \text{mm}^4, Imin=1.631×105 mm4I_{min} = 1.631\times10^{5}\ \text{mm}^4; principal axes through the centroid at 42.35° (for ImaxI_{max}) and -47.65° (for IminI_{min}) from the +x axis.

✓ Answer checked by Claude Opus 5.5 (high effort)

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