<Table> <Tr> <Th> Description </Th> <Th> Figure </Th> <Th> Area moment of inertia </Th> <Th> </Th> </Tr> <Tr> <Td> A filled circular area of radius r </Td> <Td> </Td> <Td> I x = π 4 r 4 (\ displaystyle I_ (x) = (\ frac (\ pi) (4)) r ^ (4)) I y = π 4 r 4 (\ displaystyle I_ (y) = (\ frac (\ pi) (4)) r ^ (4)) I z = π 2 r 4 (\ displaystyle I_ (z) = (\ frac (\ pi) (2)) r ^ (4)) </Td> <Td> I z (\ displaystyle I_ (z)) is the Polar moment of inertia . </Td> </Tr> <Tr> <Td> An annulus of inner radius r and outer radius r </Td> <Td> </Td> <Td> I x = π 4 (r 2 4 − r 1 4) (\ displaystyle I_ (x) = (\ frac (\ pi) (4)) \ left ((r_ (2)) ^ (4) - (r_ (1)) ^ (4) \ right)) I y = π 4 (r 2 4 − r 1 4) (\ displaystyle I_ (y) = (\ frac (\ pi) (4)) \ left ((r_ (2)) ^ (4) - (r_ (1)) ^ (4) \ right)) I z = π 2 (r 2 4 − r 1 4) (\ displaystyle I_ (z) = (\ frac (\ pi) (2)) \ left ((r_ (2)) ^ (4) - (r_ (1)) ^ (4) \ right)) </Td> <Td> For thin tubes, r ≡ r 1 ≈ r 2 (\ displaystyle r \ equiv r_ (1) \ approx r_ (2)) and r 2 ≡ r 1 + t (\ displaystyle r_ (2) \ equiv r_ (1) + t). So, for a thin tube, I x = I y ≈ π r 3 t (\ displaystyle I_ (x) = I_ (y) \ approx \ pi (r) ^ (3) (t)). I z (\ displaystyle I_ (z)) is the Polar moment of inertia . </Td> </Tr> <Tr> <Td> A filled circular sector of angle θ in radians and radius r with respect to an axis through the centroid of the sector and the center of the circle </Td> <Td> </Td> <Td> I x = (θ − sin ⁡ θ) r 4 8 (\ displaystyle I_ (x) = \ left (\ theta - \ sin \ theta \ right) (\ frac (r ^ (4)) (8))) </Td> <Td> This formula is valid only for 0 ≤ θ (\ displaystyle \ theta) ≤ π (\ displaystyle \ pi) </Td> </Tr> <Tr> <Td> A filled semicircle with radius r with respect to a horizontal line passing through the centroid of the area </Td> <Td> </Td> <Td> I x = (π 8 − 8 9 π) r 4 ≈ 0.1098 r 4 (\ displaystyle I_ (x) = \ left ((\ frac (\ pi) (8)) - (\ frac (8) (9 \ pi)) \ right) r ^ (4) \ approx 0.1098 r ^ (4)) I y = π r 4 8 (\ displaystyle I_ (y) = (\ frac (\ pi r ^ (4)) (8))) </Td> <Td> </Td> </Tr> <Tr> <Td> A filled semicircle as above but with respect to an axis collinear with the base </Td> <Td> </Td> <Td> I x = π r 4 8 (\ displaystyle I_ (x) = (\ frac (\ pi r ^ (4)) (8))) I y = π r 4 8 (\ displaystyle I_ (y) = (\ frac (\ pi r ^ (4)) (8))) </Td> <Td> I x (\ displaystyle I_ (x)): This is a consequence of the parallel axis theorem and the fact that the distance between the x axes of the previous one and this one is 4 r 3 π (\ displaystyle (\ frac (4r) (3 \ pi))) </Td> </Tr> <Tr> <Td> A filled quarter circle with radius r with the axes passing through the bases </Td> <Td> </Td> <Td> I x = π r 4 16 (\ displaystyle I_ (x) = (\ frac (\ pi r ^ (4)) (16))) I y = π r 4 16 (\ displaystyle I_ (y) = (\ frac (\ pi r ^ (4)) (16))) </Td> <Td> </Td> </Tr> <Tr> <Td> A filled quarter circle with radius r with the axes passing through the centroid </Td> <Td> </Td> <Td> I x = (π 16 − 4 9 π) r 4 ≈ 0.0549 r 4 (\ displaystyle I_ (x) = \ left ((\ frac (\ pi) (16)) - (\ frac (4) (9 \ pi)) \ right) r ^ (4) \ approx 0.0549 r ^ (4)) I y = (π 16 − 4 9 π) r 4 ≈ 0.0549 r 4 (\ displaystyle I_ (y) = \ left ((\ frac (\ pi) (16)) - (\ frac (4) (9 \ pi)) \ right) r ^ (4) \ approx 0.0549 r ^ (4)) </Td> <Td> This is a consequence of the parallel axis theorem and the fact that the distance between these two axes is 4 r 3 π (\ displaystyle (\ frac (4r) (3 \ pi))) </Td> </Tr> <Tr> <Td> A filled ellipse whose radius along the x-axis is a and whose radius along the y - axis is b </Td> <Td> </Td> <Td> I x = π 4 a b 3 (\ displaystyle I_ (x) = (\ frac (\ pi) (4)) ab ^ (3)) I y = π 4 a 3 b (\ displaystyle I_ (y) = (\ frac (\ pi) (4)) a ^ (3) b) </Td> <Td> </Td> </Tr> <Tr> <Td> A filled rectangular area with a base width of b and height h </Td> <Td> </Td> <Td> I x = b h 3 12 (\ displaystyle I_ (x) = (\ frac (bh ^ (3)) (12))) I y = b 3 h 12 (\ displaystyle I_ (y) = (\ frac (b ^ (3) h) (12))) </Td> <Td> </Td> </Tr> <Tr> <Td> A filled rectangular area as above but with respect to an axis collinear with the base </Td> <Td> </Td> <Td> I x = b h 3 3 (\ displaystyle I_ (x) = (\ frac (bh ^ (3)) (3))) I y = b 3 h 3 (\ displaystyle I_ (y) = (\ frac (b ^ (3) h) (3))) </Td> <Td> This is a result from the parallel axis theorem </Td> </Tr> <Tr> <Td> A filled triangular area with a base width of b and height h with respect to an axis through the centroid </Td> <Td> </Td> <Td> I x = b h 3 36 (\ displaystyle I_ (x) = (\ frac (bh ^ (3)) (36))) I y = b 3 h 36 (\ displaystyle I_ (y) = (\ frac (b ^ (3) h) (36))) </Td> <Td> </Td> </Tr> <Tr> <Td> A filled triangular area as above but with respect to an axis collinear with the base </Td> <Td> </Td> <Td> I x = b h 3 12 (\ displaystyle I_ (x) = (\ frac (bh ^ (3)) (12))) I y = b 3 h 12 (\ displaystyle I_ (y) = (\ frac (b ^ (3) h) (12))) </Td> <Td> This is a consequence of the parallel axis theorem </Td> </Tr> <Tr> <Td> An equal legged angle, commonly found in engineering applications </Td> <Td> </Td> <Td> I x = I y = t (5 L 2 − 5 L t + t 2) (L 2 − L t + t 2) 12 (2 L − t) (\ displaystyle I_ (x) = I_ (y) = (\ frac (t (5L ^ (2) - 5Lt + t ^ (2)) (L ^ (2) - Lt + t ^ (2))) (12 (2L - t)))) I (x y) = L 2 t (L − t) 2 4 (t − 2 L) (\ displaystyle I_ ((xy)) = (\ frac (L ^ (2) t (L-t) ^ (2)) (4 (t - 2L)))) I a = t (2 L − t) (2 L 2 − 2 L t + t 2) 12 (\ displaystyle I_ (a) = (\ frac (t (2L - t) (2L ^ (2) - 2Lt + t ^ (2))) (12))) I b = t (2 L 4 − 4 L 3 t + 8 L 2 t 2 − 6 L t 3 + t 4) 12 (2 L − t) (\ displaystyle I_ (b) = (\ frac (t (2L ^ (4) - 4L ^ (3) t + 8L ^ (2) t ^ (2) - 6Lt ^ (3) + t ^ (4))) (12 (2L - t)))) </Td> <Td> I (x y) (\ displaystyle I_ ((xy))) is the often unused product of inertia, used to define inertia with a rotated axis </Td> <Td> </Td> </Tr> <Tr> <Td> A filled regular hexagon with a side length of a </Td> <Td> </Td> <Td> I x = 5 3 16 a 4 (\ displaystyle I_ (x) = (\ frac (5 (\ sqrt (3))) (16)) a ^ (4)) I y = 5 3 16 a 4 (\ displaystyle I_ (y) = (\ frac (5 (\ sqrt (3))) (16)) a ^ (4)) </Td> <Td> The result is valid for both a horizontal and a vertical axis through the centroid, and therefore is also valid for an axis with arbitrary direction that passes through the origin . </Td> </Tr> </Table> <Tr> <Th> Description </Th> <Th> Figure </Th> <Th> Area moment of inertia </Th> <Th> </Th> </Tr> <Tr> <Td> A filled circular area of radius r </Td> <Td> </Td> <Td> I x = π 4 r 4 (\ displaystyle I_ (x) = (\ frac (\ pi) (4)) r ^ (4)) I y = π 4 r 4 (\ displaystyle I_ (y) = (\ frac (\ pi) (4)) r ^ (4)) I z = π 2 r 4 (\ displaystyle I_ (z) = (\ frac (\ pi) (2)) r ^ (4)) </Td> <Td> I z (\ displaystyle I_ (z)) is the Polar moment of inertia . </Td> </Tr> <Tr> <Td> An annulus of inner radius r and outer radius r </Td> <Td> </Td> <Td> I x = π 4 (r 2 4 − r 1 4) (\ displaystyle I_ (x) = (\ frac (\ pi) (4)) \ left ((r_ (2)) ^ (4) - (r_ (1)) ^ (4) \ right)) I y = π 4 (r 2 4 − r 1 4) (\ displaystyle I_ (y) = (\ frac (\ pi) (4)) \ left ((r_ (2)) ^ (4) - (r_ (1)) ^ (4) \ right)) I z = π 2 (r 2 4 − r 1 4) (\ displaystyle I_ (z) = (\ frac (\ pi) (2)) \ left ((r_ (2)) ^ (4) - (r_ (1)) ^ (4) \ right)) </Td> <Td> For thin tubes, r ≡ r 1 ≈ r 2 (\ displaystyle r \ equiv r_ (1) \ approx r_ (2)) and r 2 ≡ r 1 + t (\ displaystyle r_ (2) \ equiv r_ (1) + t). So, for a thin tube, I x = I y ≈ π r 3 t (\ displaystyle I_ (x) = I_ (y) \ approx \ pi (r) ^ (3) (t)). I z (\ displaystyle I_ (z)) is the Polar moment of inertia . </Td> </Tr>

How to calculate the moment of inertia of a circle
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