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Science & Math

Moment of Inertia Calculator

Calculate area moment of inertia for rectangles, circles, pipes, I-beams and tees, or mass moment of inertia for cylinders, spheres and rods.

Created by Muhammad Asif Zaheer
Reviewed by CalculatorPlanet Editorial Team

Published

Last updated

Area moment of inertia Ix
6,666.67 cm4
Area moment of inertia Ix
66,666,667 mm4
Area moment of inertia Ix
160.167 in4
Area moment of inertia Iy
1,666.67 cm4
Area moment of inertia Iy
16,666,667 mm4
Area moment of inertia Iy
40.042 in4
Polar moment of area J
8,333.33 cm4
Section area
200.00 cm2
Section modulus Zx
666.67 cm3
Radius of gyration rx
5.774 cm

How the Moment of Inertia Calculator Formula Works

Moment of inertia always measures how far something is spread from an axis, with the distance squared. That is why material far from the axis counts for much more than material near it.

The area version, the second moment of area, describes a cross-section, such as the end of a beam. It has units of length to the fourth power and depends only on the shape. The mass version describes a solid object that spins. It has units of mass times length squared and depends on the mass and where it sits.

Both follow the parallel axis theorem: to move the axis away from the centroid or centre of mass by a distance d, add area times d squared (or mass times d squared). Composite shapes are found by adding the parts, and holes are subtracted.

Area: Ix = b h3 / 12 Mass: I = m r2 / 2 Shift: I = Ic + A d2
What goes in, what comes out
  • b, hlengthWidth and height of a rectangle (depth h is cubed for Ix).
  • I, Ix, Iylength^4 or mass x length^2Moment of inertia about the stated axis.
  • Alength^2Cross-section area.
  • mmassMass of the spinning object.
  • r, LlengthRadius and length of the object.
  • + 1 more input below
Area moment of inertia Ix
6666.67
Values shown are from the worked example below
b, h
length
Width and height of a rectangle (depth h is cubed for Ix).
I, Ix, Iy
length^4 or mass x length^2
Moment of inertia about the stated axis.
A
length^2
Cross-section area.
m
mass
Mass of the spinning object.
r, L
length
Radius and length of the object.
d
length
Distance between the centroid (or centre of mass) axis and the new parallel axis.

Using the Moment of Inertia Calculator

  1. Enter shape and kind of moment of inertia

    Choose from Area: rectangle (beam section), Area: solid circle, Area: hollow circle (pipe or tube), Area: triangle, Area: I-beam (H-section), Area: T-section (tee), Area: shift to a parallel axis, Mass: solid cylinder or disk, Mass: thick-walled tube, Mass: thin ring or hoop, Mass: solid sphere, Mass: thin spherical shell, Mass: slender rod, Mass: rectangular plate, Mass: shift to a parallel axis.

  2. Enter length unit

    Choose from mm, cm, m, in, ft.

  3. Enter width b

    Type the figure you already have, then press Calculate.

  4. Enter height h

    Type the figure you already have, then press Calculate.

  5. Enter length unit

    Choose from mm, cm, m, in, ft.

  6. Enter diameter d

    Type the figure you already have, then press Calculate.

  7. Enter length unit

    Choose from mm, cm, m, in, ft.

  8. Enter outside diameter d

    Type the figure you already have, then press Calculate.

  9. Enter inside diameter d

    Type the figure you already have, then press Calculate.

  10. Enter length unit

    Choose from mm, cm, m, in, ft.

  11. Enter base b

    Type the figure you already have, then press Calculate.

  12. Enter height h

    Type the figure you already have, then press Calculate.

  13. Enter length unit

    Choose from mm, cm, m, in, ft.

  14. Enter flange width bf

    Type the figure you already have, then press Calculate.

  15. Enter flange thickness tf

    Type the figure you already have, then press Calculate.

  16. Enter web thickness tw

    Type the figure you already have, then press Calculate.

  17. Enter total depth h

    Type the figure you already have, then press Calculate.

  18. Enter length unit

    Choose from mm, cm, m, in, ft.

  19. Enter flange width bf

    Type the figure you already have, then press Calculate.

  20. Enter flange thickness tf

    Type the figure you already have, then press Calculate.

  21. Enter stem thickness tw

    Type the figure you already have, then press Calculate.

  22. Enter total height h

    Type the figure you already have, then press Calculate.

  23. Enter length unit

    Choose from mm, cm, m, in, ft.

  24. Enter centroidal i (length unit to the 4th power)

    Type the figure you already have, then press Calculate.

  25. Enter area a (length unit squared)

    Type the figure you already have, then press Calculate.

  26. Enter distance d between the axes

    Type the figure you already have, then press Calculate.

  27. Enter length unit

    Choose from mm, cm, m, in, ft.

  28. Enter mass unit

    Choose from kg, g, lb.

  29. Enter mass

    Type the figure you already have, then press Calculate.

  30. Enter radius r

    Type the figure you already have, then press Calculate.

  31. Enter length l

    Type the figure you already have, then press Calculate.

  32. Enter length unit

    Choose from mm, cm, m, in, ft.

  33. Enter mass unit

    Choose from kg, g, lb.

  34. Enter mass

    Type the figure you already have, then press Calculate.

  35. Enter outer radius

    Type the figure you already have, then press Calculate.

  36. Enter inner radius

    Type the figure you already have, then press Calculate.

  37. Enter length unit

    Choose from mm, cm, m, in, ft.

  38. Enter mass unit

    Choose from kg, g, lb.

  39. Enter mass

    Type the figure you already have, then press Calculate.

  40. Enter radius r

    Type the figure you already have, then press Calculate.

  41. Enter length unit

    Choose from mm, cm, m, in, ft.

  42. Enter mass unit

    Choose from kg, g, lb.

  43. Enter mass

    Type the figure you already have, then press Calculate.

  44. Enter radius r

    Type the figure you already have, then press Calculate.

  45. Enter length unit

    Choose from mm, cm, m, in, ft.

  46. Enter mass unit

    Choose from kg, g, lb.

  47. Enter mass

    Type the figure you already have, then press Calculate.

  48. Enter radius r

    Type the figure you already have, then press Calculate.

  49. Enter length unit

    Choose from mm, cm, m, in, ft.

  50. Enter mass unit

    Choose from kg, g, lb.

  51. Enter mass

    Type the figure you already have, then press Calculate.

  52. Enter length l

    Type the figure you already have, then press Calculate.

  53. Enter length unit

    Choose from mm, cm, m, in, ft.

  54. Enter mass unit

    Choose from kg, g, lb.

  55. Enter mass

    Type the figure you already have, then press Calculate.

  56. Enter width a

    Type the figure you already have, then press Calculate.

  57. Enter height b

    Type the figure you already have, then press Calculate.

  58. Enter length unit

    Choose from mm, cm, m, in, ft.

  59. Enter mass unit

    Choose from kg, g, lb.

  60. Enter i about the centre of mass (mass unit x length unit squared)

    Type the figure you already have, then press Calculate.

  61. Enter mass

    Type the figure you already have, then press Calculate.

  62. Enter distance d between the axes

    Type the figure you already have, then press Calculate.

  63. Read the result

    Press Calculate to update the result. Until you do, the last result is dimmed and marked as out of date, so you always know which numbers belong to the values in the boxes.

A Real Worked Example

A rectangular timber or steel bar section is 100 mm wide and 200 mm deep. About the horizontal axis through the centroid, Ix = b h3 / 12 = 100 × 2003 / 12 = 66,666,667 mm4, which is 6,666.67 cm4 or 160.167 in4.

About the vertical axis, Iy = 200 × 1003 / 12 = 1,666.67 cm4. The area is 200 cm2, the section modulus is Zx = Ix / 10 cm = 666.67 cm3 and the radius of gyration is 5.774 cm.

1
mode
a_rect
2
lengthUnit_a_rect
mm
3
lengthUnit_a_circle
mm
4
lengthUnit_a_hollow
mm
5
lengthUnit_a_tri
mm
6
lengthUnit_a_ibeam
mm
7
lengthUnit_a_tee
mm
8
b_a_rect
100
9
h_a_rect
200
10
d_a_circle
50
11
D_a_hollow
60
12
d_a_hollow
50
13
b_a_tri
100
14
h_a_tri
150
15
bf_a_ibeam
100
16
tf_a_ibeam
8
17
tw_a_ibeam
5
18
h_a_ibeam
200
19
bf_a_tee
100
20
tf_a_tee
10
21
tw_a_tee
10
22
h_a_tee
100
23
lengthUnit_a_shift
cm
24
ic_a_shift
6666.667
25
area_a_shift
200
26
dist_a_shift
10
27
lengthUnit_m_cyl
mm
28
massUnit_m_cyl
kg
29
lengthUnit_m_hollowcyl
m
30
massUnit_m_hollowcyl
kg
31
lengthUnit_m_ring
m
32
massUnit_m_ring
kg
33
lengthUnit_m_sphere
m
34
massUnit_m_sphere
kg
35
lengthUnit_m_shell
m
36
massUnit_m_shell
kg
37
lengthUnit_m_rod
m
38
massUnit_m_rod
kg
39
lengthUnit_m_plate
m
40
massUnit_m_plate
kg
41
lengthUnit_m_shift
m
42
massUnit_m_shift
kg
43
mass_m_cyl
20
44
r_m_cyl
150
45
len_m_cyl
400
46
mass_m_hollowcyl
8
47
ro_m_hollowcyl
0.1
48
ri_m_hollowcyl
0.06
49
mass_m_ring
3
50
r_m_ring
0.25
51
mass_m_sphere
5
52
r_m_sphere
0.1
53
mass_m_shell
5
54
r_m_shell
0.1
55
mass_m_rod
2
56
len_m_rod
0.5
57
mass_m_plate
6
58
a_m_plate
0.4
59
b_m_plate
0.3
60
icm_m_shift
0.041666667
61
mass_m_shift
2
62
dist_m_shift
0.25
This calculator returns6666.67ixCm466666667ixMm4160.167ixIn41666.67iyCm416666667iyMm440.042iyIn48333.33jCm4200areaCm2666.67zxCm35.774rxCm

I-beam section

An I-beam is 200 mm deep with 100 mm wide, 8 mm thick flanges and a 5 mm web. The web height between flanges is 200 − 16 = 184 mm. Ix = [100 × 2003 − (100 − 5) × 1843] / 12 = 17,349,760 mm4, which is 1,734.98 cm4.

Check with the parallel axis theorem: the two flanges give 2 × (100 × 83 / 12 + 800 × 962) and the web adds 5 × 1843 / 12. The sum is the same 17,349,760 mm4. Iy is 133.525 cm4 and the area is 25.2 cm2.

mode: a_ibeamlengthUnit_a_rect: mmlengthUnit_a_circle: mmlengthUnit_a_hollow: mmlengthUnit_a_tri: mmlengthUnit_a_ibeam: mmlengthUnit_a_tee: mmb_a_rect: 100h_a_rect: 200d_a_circle: 50D_a_hollow: 60d_a_hollow: 50b_a_tri: 100h_a_tri: 150bf_a_ibeam: 100tf_a_ibeam: 8tw_a_ibeam: 5h_a_ibeam: 200bf_a_tee: 100tf_a_tee: 10tw_a_tee: 10h_a_tee: 100lengthUnit_a_shift: cmic_a_shift: 6666.667area_a_shift: 200dist_a_shift: 10lengthUnit_m_cyl: mmmassUnit_m_cyl: kglengthUnit_m_hollowcyl: mmassUnit_m_hollowcyl: kglengthUnit_m_ring: mmassUnit_m_ring: kglengthUnit_m_sphere: mmassUnit_m_sphere: kglengthUnit_m_shell: mmassUnit_m_shell: kglengthUnit_m_rod: mmassUnit_m_rod: kglengthUnit_m_plate: mmassUnit_m_plate: kglengthUnit_m_shift: mmassUnit_m_shift: kgmass_m_cyl: 20r_m_cyl: 150len_m_cyl: 400mass_m_hollowcyl: 8ro_m_hollowcyl: 0.1ri_m_hollowcyl: 0.06mass_m_ring: 3r_m_ring: 0.25mass_m_sphere: 5r_m_sphere: 0.1mass_m_shell: 5r_m_shell: 0.1mass_m_rod: 2len_m_rod: 0.5mass_m_plate: 6a_m_plate: 0.4b_m_plate: 0.3icm_m_shift: 0.041666667mass_m_shift: 2dist_m_shift: 0.251734.98ixCm417349760ixMm441.683ixIn4133.53iyCm41335250iyMm43.208iyIn41868.5jCm425.2areaCm2173.5zxCm38.297rxCm

Hollow circle (pipe or tube)

A tube has a 60 mm outside diameter and a 50 mm inside diameter, a 5 mm wall. I = pi (D4 − d4) / 64 = pi x (604 − 504) / 64 = 329,376 mm4, or 32.938 cm4 about any axis through the centre.

The area is 8.639 cm2 and the polar value is J = 65.875 cm4. A solid 60 mm bar has I = 63.62 cm4, so the tube keeps 51.8 percent of the stiffness with 30.6 percent of the material.

mode: a_hollowlengthUnit_a_rect: mmlengthUnit_a_circle: mmlengthUnit_a_hollow: mmlengthUnit_a_tri: mmlengthUnit_a_ibeam: mmlengthUnit_a_tee: mmb_a_rect: 100h_a_rect: 200d_a_circle: 50D_a_hollow: 60d_a_hollow: 50b_a_tri: 100h_a_tri: 150bf_a_ibeam: 100tf_a_ibeam: 8tw_a_ibeam: 5h_a_ibeam: 200bf_a_tee: 100tf_a_tee: 10tw_a_tee: 10h_a_tee: 100lengthUnit_a_shift: cmic_a_shift: 6666.667area_a_shift: 200dist_a_shift: 10lengthUnit_m_cyl: mmmassUnit_m_cyl: kglengthUnit_m_hollowcyl: mmassUnit_m_hollowcyl: kglengthUnit_m_ring: mmassUnit_m_ring: kglengthUnit_m_sphere: mmassUnit_m_sphere: kglengthUnit_m_shell: mmassUnit_m_shell: kglengthUnit_m_rod: mmassUnit_m_rod: kglengthUnit_m_plate: mmassUnit_m_plate: kglengthUnit_m_shift: mmassUnit_m_shift: kgmass_m_cyl: 20r_m_cyl: 150len_m_cyl: 400mass_m_hollowcyl: 8ro_m_hollowcyl: 0.1ri_m_hollowcyl: 0.06mass_m_ring: 3r_m_ring: 0.25mass_m_sphere: 5r_m_sphere: 0.1mass_m_shell: 5r_m_shell: 0.1mass_m_rod: 2len_m_rod: 0.5mass_m_plate: 6a_m_plate: 0.4b_m_plate: 0.3icm_m_shift: 0.041666667mass_m_shift: 2dist_m_shift: 0.2532.94ixCm4329376ixMm40.791ixIn432.94iyCm4329376iyMm40.791iyIn465.88jCm48.64areaCm210.98zxCm31.953rxCm

T-section (tee)

A tee has a 100 × 10 mm flange on top of a 10 mm thick stem, 100 mm total height. The stem is 90 mm tall. The centroid is 71.316 mm above the bottom of the stem, found by weighting the flange and stem areas.

Adding each part's own I and its area times the squared distance to that centroid gives Ix = 1,800,044 mm4, or 180.00 cm4. About the vertical axis Iy = 84.083 cm4. The area is 19 cm2.

mode: a_teelengthUnit_a_rect: mmlengthUnit_a_circle: mmlengthUnit_a_hollow: mmlengthUnit_a_tri: mmlengthUnit_a_ibeam: mmlengthUnit_a_tee: mmb_a_rect: 100h_a_rect: 200d_a_circle: 50D_a_hollow: 60d_a_hollow: 50b_a_tri: 100h_a_tri: 150bf_a_ibeam: 100tf_a_ibeam: 8tw_a_ibeam: 5h_a_ibeam: 200bf_a_tee: 100tf_a_tee: 10tw_a_tee: 10h_a_tee: 100lengthUnit_a_shift: cmic_a_shift: 6666.667area_a_shift: 200dist_a_shift: 10lengthUnit_m_cyl: mmmassUnit_m_cyl: kglengthUnit_m_hollowcyl: mmassUnit_m_hollowcyl: kglengthUnit_m_ring: mmassUnit_m_ring: kglengthUnit_m_sphere: mmassUnit_m_sphere: kglengthUnit_m_shell: mmassUnit_m_shell: kglengthUnit_m_rod: mmassUnit_m_rod: kglengthUnit_m_plate: mmassUnit_m_plate: kglengthUnit_m_shift: mmassUnit_m_shift: kgmass_m_cyl: 20r_m_cyl: 150len_m_cyl: 400mass_m_hollowcyl: 8ro_m_hollowcyl: 0.1ri_m_hollowcyl: 0.06mass_m_ring: 3r_m_ring: 0.25mass_m_sphere: 5r_m_sphere: 0.1mass_m_shell: 5r_m_shell: 0.1mass_m_rod: 2len_m_rod: 0.5mass_m_plate: 6a_m_plate: 0.4b_m_plate: 0.3icm_m_shift: 0.041666667mass_m_shift: 2dist_m_shift: 0.25180ixCm41800044ixMm44.325ixIn484.08iyCm4840833iyMm42.02iyIn4264.09jCm419areaCm225.24zxCm33.078rxCm7.132ybarCm

Mass: solid cylinder

A solid steel cylinder has a mass of 20 kg, a radius of 150 mm and a length of 400 mm. About its own axis I = m r2 / 2 = 20 × 0.152 / 2 = 0.2250 kg m2, which is 5.339 lb ft2.

About a transverse axis through its centre, I = m (3 r2 + L2) / 12 = 20 × (3 × 0.0225 + 0.16) / 12 = 0.3792 kg m2. The lengthwise axis is easier to spin because the mass is closer to it.

mode: m_cyllengthUnit_a_rect: mmlengthUnit_a_circle: mmlengthUnit_a_hollow: mmlengthUnit_a_tri: mmlengthUnit_a_ibeam: mmlengthUnit_a_tee: mmb_a_rect: 100h_a_rect: 200d_a_circle: 50D_a_hollow: 60d_a_hollow: 50b_a_tri: 100h_a_tri: 150bf_a_ibeam: 100tf_a_ibeam: 8tw_a_ibeam: 5h_a_ibeam: 200bf_a_tee: 100tf_a_tee: 10tw_a_tee: 10h_a_tee: 100lengthUnit_a_shift: cmic_a_shift: 6666.667area_a_shift: 200dist_a_shift: 10lengthUnit_m_cyl: mmmassUnit_m_cyl: kglengthUnit_m_hollowcyl: mmassUnit_m_hollowcyl: kglengthUnit_m_ring: mmassUnit_m_ring: kglengthUnit_m_sphere: mmassUnit_m_sphere: kglengthUnit_m_shell: mmassUnit_m_shell: kglengthUnit_m_rod: mmassUnit_m_rod: kglengthUnit_m_plate: mmassUnit_m_plate: kglengthUnit_m_shift: mmassUnit_m_shift: kgmass_m_cyl: 20r_m_cyl: 150len_m_cyl: 400mass_m_hollowcyl: 8ro_m_hollowcyl: 0.1ri_m_hollowcyl: 0.06mass_m_ring: 3r_m_ring: 0.25mass_m_sphere: 5r_m_sphere: 0.1mass_m_shell: 5r_m_shell: 0.1mass_m_rod: 2len_m_rod: 0.5mass_m_plate: 6a_m_plate: 0.4b_m_plate: 0.3icm_m_shift: 0.041666667mass_m_shift: 2dist_m_shift: 0.250.225iKgm25.3393iLbft22250000iGcm2768.86iLbin20.3792i2Kgm28.9978i2Lbft2

Mass: slender rod

A slender rod has a mass of 2 kg and a length of 0.5 m. About its centre, perpendicular to the rod, I = m L2 / 12 = 2 × 0.25 / 12 = 0.0417 kg m2.

About one end, I = m L2 / 3 = 0.1667 kg m2. That is four times larger, and it matches the parallel axis theorem: 0.0417 + 2 × 0.252 = 0.1667 kg m2. In lb ft2 the two values are 0.9888 and 3.9551.

mode: m_rodlengthUnit_a_rect: mmlengthUnit_a_circle: mmlengthUnit_a_hollow: mmlengthUnit_a_tri: mmlengthUnit_a_ibeam: mmlengthUnit_a_tee: mmb_a_rect: 100h_a_rect: 200d_a_circle: 50D_a_hollow: 60d_a_hollow: 50b_a_tri: 100h_a_tri: 150bf_a_ibeam: 100tf_a_ibeam: 8tw_a_ibeam: 5h_a_ibeam: 200bf_a_tee: 100tf_a_tee: 10tw_a_tee: 10h_a_tee: 100lengthUnit_a_shift: cmic_a_shift: 6666.667area_a_shift: 200dist_a_shift: 10lengthUnit_m_cyl: mmmassUnit_m_cyl: kglengthUnit_m_hollowcyl: mmassUnit_m_hollowcyl: kglengthUnit_m_ring: mmassUnit_m_ring: kglengthUnit_m_sphere: mmassUnit_m_sphere: kglengthUnit_m_shell: mmassUnit_m_shell: kglengthUnit_m_rod: mmassUnit_m_rod: kglengthUnit_m_plate: mmassUnit_m_plate: kglengthUnit_m_shift: mmassUnit_m_shift: kgmass_m_cyl: 20r_m_cyl: 150len_m_cyl: 400mass_m_hollowcyl: 8ro_m_hollowcyl: 0.1ri_m_hollowcyl: 0.06mass_m_ring: 3r_m_ring: 0.25mass_m_sphere: 5r_m_sphere: 0.1mass_m_shell: 5r_m_shell: 0.1mass_m_rod: 2len_m_rod: 0.5mass_m_plate: 6a_m_plate: 0.4b_m_plate: 0.3icm_m_shift: 0.041666667mass_m_shift: 2dist_m_shift: 0.250.0417iKgm20.9888iLbft2416667iGcm2142.38iLbin20.1667i2Kgm23.9551i2Lbft2

What Else to Know

Two kinds of moment of inertia

The phrase moment of inertia covers two different quantities, and most confusion comes from mixing them. The second moment of area, also called the area moment of inertia or the 2nd moment of inertia, describes a cross-section. It tells you how well a beam's cross-section resists bending. Its units are length to the fourth power, such as mm4 or in4, and it depends only on the shape of the cross-section. The mass moment of inertia, also called rotational inertia, describes how hard it is to change the spin of a solid object. Its units are mass times length squared, such as kg m2 or lb ft2, and it depends on the mass and how that mass is spread around the axis.

This calculator keeps the two apart. The Area modes at the top of the shape list return second moment of area, section modulus and related section properties. The Mass modes return rotational inertia in kg m2, lb ft2, g cm2 and lb in2. The comparison table on this page lists what each quantity is used for. If you are checking a beam or a bracket cross-section, use the area modes. If you are working on a flywheel, wheel, rod or any spinning part, use the mass modes.

Comparison of second moment of area for beams and mass moment of inertia for spinning objects

Both share one idea: material far from the axis counts more than material near it, because the distance is squared. That one fact explains why deep beams are stiff, why I-beams are shaped the way they are, and why a ring is harder to spin than a solid disk of the same mass.

Second moment of area: rectangles, circles and triangles

For a rectangle of width b and height h, the second moment of area about the horizontal axis through the centroid is Ix = b h3 / 12. The height is cubed, so depth is far more valuable than width. Doubling the depth gives eight times the Ix, while doubling the width only doubles it. The 100 × 200 mm example gives Ix = 6,666.7 cm4, but turned on its side with 200 mm as the width, it gives only 1,666.7 cm4.

Moment of inertia formulas for a rectangle, solid and hollow circles, and the parallel axis theorem

For a solid circle of diameter d, I = pi d4 / 64, the same about any axis through the centre. The fourth power makes small increases in diameter matter: a 100 mm bar has sixteen times the I of a 50 mm bar. A hollow circle, such as a pipe or tube, is the solid outer circle minus the empty inner circle: I = pi (D4 − d4) / 64. The tube in the examples keeps about 52 percent of a solid bar's I with a much smaller share of the material.

Worked example: a 100 by 200 mm rectangle has Ix of 66,666,667 mm4, which is 6,666.7 cm4

For a triangle with base b and height h, the centroid sits one third of the height above the base, and Ix = b h3 / 36 about the centroidal axis. About the base it is b h3 / 12. The calculator reports Iy = h b3 / 48, which applies to an isosceles triangle with the apex above the middle of the base. These formulas are standard results, and the Wikipedia articles on the second moment of area and on the list of moments of inertia give the derivations and more shapes.

Bar chart of Ix for a 100 mm wide rectangle at depths from 100 to 300 mm, showing the cube effect

I-beam, T-beam and irregular shapes with the parallel axis theorem

Real sections such as I-beams, channels and tees are built from rectangles. Two methods give the same answer. The subtraction method treats an I-beam as the full outer rectangle minus the two side pockets: Ix = [bf h3 − (bf - tw)(h - 2 tf)3] / 12. The composite method adds each part's own I plus its area times the squared distance to the common axis. For the I-beam example both give 1,735.0 cm4.

I-beam with 1,735 cm4 compared with a solid square bar of the same area, with the shapes and bars

That second method is the parallel axis theorem: I = Ic + A d2. It is the tool for any T-beam, L-section, box section or irregular shape you can divide into simple parts. First find the centroid of the whole shape by area-weighting the parts. Then, for each part, add its own centroidal I and its area times the squared distance from its centroid to the whole shape's centroid. Holes are subtracted. The T-section mode does this for a tee and reports the centroid height; the shift mode lets you apply the theorem to any part you have worked out yourself.

Bar chart of I for a 60 mm tube with different hole sizes as a share of the solid 60 mm bar

If you work in a spreadsheet, set up one row per part with columns for width, height, area, distance and I, then sum. The sheet will match this page. For an outline that cannot be divided into simple parts, use CAD software or the polygon method described in the Wikipedia article on the second moment of area. The table of I-beam against solid bar shows the payoff of the shape: the same cross-section area as a solid bar, with about 33 times the Ix.

Section modulus, radius of gyration and polar moment

The calculator reports a few related section properties alongside Ix and Iy. The section modulus Zx = Ix / c divides Ix by c, the distance from the axis to the farthest edge, and links a bending moment to the peak bending stress. For the 100 × 200 mm rectangle c is 100 mm and Zx = 666.7 cm3. The radius of gyration rx = √(Ix / A) is a single distance that describes how spread out the area is; for that rectangle it is 57.7 mm. The polar moment J = Ix + Iy describes a section's geometric resistance to twisting about the centroid.

These numbers are section properties. They are inputs to a structural calculation, not the calculation itself. In structural engineering the second moment of area is an input to the bending stress and deflection of a beam, and a larger value means a stiffer section for the same material, span and load. A real check also needs the load, span, support conditions, material properties, safety factors and the rules in the applicable code.

This page does not do that check, and none of its numbers are a structural design. Treat them as estimates, use published section tables for real products, follow local building codes and the manufacturer's data, and have a qualified structural engineer check any work where failure could hurt someone or cost real money.

Mass moment of inertia: cylinders, spheres, rods and rings

A spinning object resists a change of rotation in proportion to its mass moment of inertia, so the same push produces less angular speed-up for a larger I. For a point mass, I = m r2. For an extended object, add up m r2 over all its parts, which gives the standard results listed in the table of mass formulas: a solid cylinder or disk about its axis is m r2 / 2, a solid sphere is 2 m r2 / 5, a thin spherical shell is 2 m r2 / 3, a thin ring is m r2, and a slender rod is m L2 / 12 about its centre or m L2 / 3 about one end. These match the OpenStax University Physics text and the Wikipedia list of moments of inertia.

Bar chart of I for a sphere, cylinder, thin shell and ring with the same 10 kg mass and 0.1 m radius

The comparison table shows how shape matters at equal mass and radius. A 10 kg object of radius 0.1 m has I = 0.040 kg m2 as a solid sphere, 0.050 as a solid cylinder, 0.067 as a thin shell and 0.100 as a thin ring. The ring is 2.5 times the sphere because all its mass sits at the full radius.

Axis choice matters too. A cylinder has one value about its own axis and another about a transverse axis through the centre, and the calculator returns both. A rod has one value at the centre and four times as much at the end. For three-dimensional bodies not on the list, split them into simple solids and combine them with the parallel axis theorem.

Parallel axis theorem and units

The parallel axis theorem lets you move from the axis through the centroid or centre of mass to any parallel axis. For area, I = Ic + A d2. For mass, I = Icm + m d2. The shift is always an increase, and it grows with the square of the distance. For the 100 × 200 mm rectangle (Ic = 6,666.7 cm4, A = 200 cm2), an axis at the base, 100 mm away, gives 26,666.7 cm4, which is b h3 / 3, four times the centroidal value. Use the shift modes when you have worked out Ic and A yourself, for example for a part of a composite section.

Curve of I against distance from the centroid for the 100 by 200 mm rectangle using the parallel axis theorem

Units need care. Second moment of area scales with the fourth power of the length unit, so 1 in4 is 41.623 cm4 and 1 cm4 is 10,000 mm4. Mass moment of inertia scales with mass times the square of the length, so 1 kg m2 is 23.730 lb ft2. Pick a length unit for each shape once, enter every dimension in it, and read the result in whichever unit you need. The unit table on this page lists the common factors.

For the shift modes, the entered I, area and distance must use the same length unit (and mass unit for the mass version). The most common error is entering a cm4 value while the length unit is set to mm.

Using the calculator and its limits

Choose a shape from the list. The Area modes ask for the dimensions of the cross-section: width and height for a rectangle, diameters for circles, flange, web and depth for an I-beam or tee. The Mass modes ask for the mass and size of the object. The default values give the 100 × 200 mm rectangle, so you see 6,666.7 cm4 straight away. Results show in several units at once, and rows that do not apply to the chosen shape are hidden.

Limits of the moment of inertia calculator: estimates only, follow code, have an engineer check

The calculator assumes uniform material and sharp corners, and ignores fillets, rounded edges, holes you have not entered and welds. Real rolled sections have root radii and tapered flanges, so published tables for a product are more accurate than a simple I-beam rectangle model. The mass shapes assume a uniform density. Treat every result as an estimate, and follow the applicable standards, local building code and the manufacturer's data for real work.

For related tools, the rafter length calculator gives the span and length of a framing member whose bending stiffness depends on its section, the board foot calculator works out the volume of timber sections, and the percentage calculator finds the percent change between two section sizes.

Reference Tables

The two quantities called moment of inertia: do not mix them up
Second moment of areaMass moment of inertia
Also calledArea moment of inertia, 2nd moment of inertiaRotational inertia
SymbolI (or Ix, Iy), J for polarI
Unitsmm4, cm4, in4, m4kg m2, lb ft2, g cm2
Depends onCross-section shape onlyMass and how it is spread about the axis
Used forBeam bending, deflection and bending stressSpinning parts, flywheels, rolling, pendulums
Calculator modesRectangle, circle, hollow circle, triangle, I-beam, T-section, shiftCylinder, tube, ring, sphere, shell, rod, plate, shift
Second moment of area about the centroidal axes for the section shapes in the calculator (b = width, h = height, d = diameter)
ShapeIx (about the horizontal centroidal axis)Iy (about the vertical centroidal axis)Area
Rectangleb h3 / 12h b3 / 12b h
Solid circlepi d4 / 64pi d4 / 64pi d2 / 4
Hollow circle (D outside, d inside)pi (D4 − d4) / 64pi (D4 − d4) / 64pi (D2 − d2) / 4
Triangle (base b, height h)b h3 / 36h b3 / 48 (isosceles)b h / 2
I-beam (flange bf x tf, web tw, depth h)[bf h3 − (bf - tw)(h - 2 tf)3] / 12[2 tf bf3 + (h - 2 tf) tw3] / 122 bf tf + (h - 2 tf) tw
T-sectionflange + stem + parallel axis terms[tf bf3 + (h - tf) tw3] / 12bf tf + tw (h - tf)
Mass moment of inertia for the solid shapes in the calculator (m = mass, r = radius, L = length)
ShapeAxisI
Solid cylinder or diskIts own axism r2 / 2
Solid cylinder or diskThrough the centre, perpendicular to the axism (3 r2 + L2) / 12
Thick-walled tube (r1 inside, r2 outside)Its own axism (r12 + r22) / 2
Thin ring or hoopIts own axism r2
Thin ring or hoopAcross a diameterm r2 / 2
Solid sphereAny axis through the centre2 m r2 / 5
Thin spherical shellAny axis through the centre2 m r2 / 3
Slender rodThrough the centre, perpendicularm L2 / 12
Slender rodThrough one end, perpendicularm L2 / 3
Rectangular plate a x bThrough the centre, perpendicular to the platem (a2 + b2) / 12
Ix of a 100 mm wide rectangle as the depth grows: the depth is cubed
Depth h (mm)Ix (cm4)Ix (in4)Area (cm2)Ix compared with h = 100 mm
100833.320.021001.00 times
1502,812.567.571503.38 times
2006,666.7160.172008.00 times
25013,020.8312.8325015.62 times
30022,500.0540.5630027.00 times
Solid circular sections: I rises with the fourth power of the diameter
Diameter (mm)I (cm4)Area (cm2)Section modulus Zx (cm3)Radius of gyration (mm)
251.924.911.536.25
5030.6819.6312.2712.50
75155.3244.1841.4218.75
100490.8778.5498.1725.00
1502,485.05176.71331.3437.50
A 60 mm outside diameter tube with different wall thicknesses: most of the stiffness comes from the outer material
Inside diameter (mm)Wall (mm)I (cm4)Area (cm2)I as a share of the solid 60 mm bar
030.063.6228.27100.0 percent
3015.059.6421.2193.8 percent
4010.051.0515.7180.2 percent
505.032.948.6451.8 percent
543.021.885.3734.4 percent
Same material, different shape: the 100 x 200 mm I-beam of the example against a solid square bar with the same area
SectionArea (cm2)Ix (cm4)Ix per cm2 of area
I-beam 100 × 200, flange 8, web 525.21,735.068.8
Solid square bar, 50.2 mm side25.252.92.1
Solid rectangle 100 × 200200.06,666.733.3
Section properties the calculator reports for the rectangle and I-beam examples
PropertyRectangle 100 x 200 mmI-beam 100 x 200 mm
Ix (cm4)6,666.71,735.0
Iy (cm4)1,666.7133.5
Polar J = Ix + Iy (cm4)8,333.31,868.5
Area (cm2)200.025.2
Section modulus Zx (cm3)666.7173.5
Radius of gyration rx (mm)57.783.0
Parallel axis theorem for the 100 x 200 mm rectangle (Ic = 6,666.7 cm4, A = 200 cm2): I about an axis d away from the centroid
Distance d (mm)A x d^2 (cm4)I = Ic + A d^2 (cm4)Share added by the shift
00.06,666.70 percent
251,250.07,916.716 percent
505,000.011,666.743 percent
10020,000.026,666.775 percent
15045,000.051,666.787 percent
20080,000.086,666.792 percent
Same mass, same radius, different shape: a 10 kg object of radius 0.1 m spinning about its own axis
ShapeFormulaI (kg m2)Compared with the solid sphere
Solid sphere2 m r2 / 50.04001.00
Solid cylinder or diskm r2 / 20.05001.25
Thin spherical shell2 m r2 / 30.06671.67
Thin ring or hoopm r20.10002.50
A 1 kg slender rod: the end axis gives four times the centre value (parallel axis theorem with d = L/2)
Length L (m)About the centre (kg m2)About one end (kg m2)About the centre (lb ft2)About one end (lb ft2)
0.250.00520.02080.12360.4944
0.500.02080.08330.49441.9775
1.000.08330.33331.97757.9101
1.500.18750.75004.449417.7978
2.000.33331.33337.910131.6405
Unit conversions for the two kinds of moment of inertia
KindFromToMultiply by
Areacm4mm410,000
Areain4cm441.6231
Areain4mm4416,231
Aream4cm4100,000,000
Masskg m2lb ft223.7304
Masskg m2lb in23,417.17
Masskg m2g cm210,000,000
Masslb ft2lb in2144

Frequently Asked Questions About the Moment of Inertia Calculator

What is a moment of inertia calculator?

It finds how strongly a shape resists bending or spinning. This calculator has two separate sets of modes. Area modes give the second moment of area of a cross-section, used for beams. Mass modes give the rotational moment of inertia of a solid object, used for spinning parts. Pick the shape, enter its size, and the results appear in several units at once.

What is the second moment of area?

The second moment of area, also called the area moment of inertia or 2nd moment of inertia, is the sum of each tiny piece of a cross-section's area multiplied by its distance from an axis squared. It depends only on shape. A 100 × 200 mm rectangle has Ix = b h3 / 12 = 6,666.7 cm4 about its horizontal centroidal axis.

What is the difference between area and mass moment of inertia?

Area moment of inertia (second moment of area) describes a cross-section and has units of length to the fourth power, such as mm4. It is used for bending. Mass moment of inertia describes how mass is spread about a rotation axis and has units of mass times length squared, such as kg m2. It is used for rotation. They share a name and a pattern, a distance squared, but they are different quantities.

What is the moment of inertia formula for a rectangle?

About the horizontal axis through the centroid, Ix = b h3 / 12, where b is the width and h is the height. About the vertical axis, Iy = h b3 / 12. For 100 mm wide and 200 mm deep, Ix = 66,666,667 mm4, which is 6,666.7 cm4. Doubling the depth multiplies Ix by 8.

How do you calculate the moment of inertia of an I-beam?

Subtract the two empty side pockets from the full rectangle: Ix = [bf h3 − (bf - tw)(h - 2 tf)3] / 12. For a 100 mm flange, 8 mm flange thickness, 5 mm web and 200 mm depth, Ix = 1,735.0 cm4. The I-beam mode above does this, and the composite method with the parallel axis theorem gives the same answer.

How do I find the moment of inertia of a T-beam?

Find the centroid height first, then add the flange and stem contributions using the parallel axis theorem: I = sum of (I of each part + area x distance squared). For a 100 × 100 mm tee with 10 mm flange and 10 mm stem, the centroid is 71.32 mm from the bottom and Ix = 180.0 cm4. Use the T-section mode above.

How do I find the moment of inertia of a pipe?

Treat a pipe as a hollow circle: I = pi (D4 − d4) / 64, with D the outside and d the inside diameter. For a 60 mm outside and 50 mm inside diameter, I = 32.94 cm4. The same formula applies to any round tube, and the hollow circle mode above also gives area and section modulus.

What is the moment of inertia of a cylinder?

For a solid cylinder spinning about its own axis, I = m r2 / 2. About a transverse axis through its centre, I = m (3 r2 + L2) / 12. A 20 kg cylinder with a 150 mm radius and 400 mm length gives 0.225 kg m2 about its axis and 0.3792 kg m2 about the transverse axis. For a cylinder cross-section in bending, use the circle modes instead.

How do you find the moment of inertia of an irregular shape?

Split the shape into simple parts such as rectangles and circles, find the centroid of the whole, then add each part's own I plus its area times the squared distance to the common axis. Subtract any holes. This composite method works for any section you can divide. For truly arbitrary outlines, use a polygon formula or CAD software.

Can I calculate moment of inertia in Excel?

Yes. The formulas are short enough for any spreadsheet cell. For a rectangle, type =B1*H13/12 with the width in B1 and the height in H1. For a hollow circle use =PI()*(D4-d4)/64. For a composite section, add one row per part with Ic + A*d2 and sum the column. This page checks your sheet in seconds, for example 6,666.7 cm4 for 100 × 200 mm.

What is the parallel axis theorem?

It moves a moment of inertia from an axis through the centroid (or centre of mass) to a parallel axis a distance d away: I = Ic + A d2 for area, I = Icm + m d2 for mass. A 1 kg rod 1 m long has 0.0833 kg m2 about its centre and 0.3333 kg m2 about an axis 0.5 m away, which is its end.

What is the section modulus?

The elastic section modulus is Zx = Ix / c, where c is the distance from the neutral axis to the farthest edge. It links bending moment to peak bending stress. The 100 × 200 mm rectangle has Zx = 666.7 cm3. This page reports Zx for the area shapes. It is a section property, not a design check.

What is the radius of gyration?

The radius of gyration is r = √(I / A), a single distance that describes how far the area is spread from the axis. For the 100 × 200 mm rectangle, rx = 57.7 mm. It is used in column buckling checks, which this page does not perform.

What units are used for moment of inertia?

Second moment of area uses length to the fourth power: mm4, cm4, m4 or in4. One in4 is 41.623 cm4. Mass moment of inertia uses mass times length squared: kg m2, g cm2, lb ft2 or lb in2. One kg m2 equals 23.730 lb ft2. The calculator converts every result for you.

How do I calculate a 3D moment of inertia?

For a solid body in three dimensions, the moment of inertia depends on which axis it spins about, so one object has several values. The mass modes give the standard axes of common solids: a cylinder about its own axis and across its centre, a sphere about any centre axis, and a rod about its centre or end. For other shapes, split the body into simple solids and add them with the parallel axis theorem.

Why is an I-beam stiffer than a solid bar of equal weight?

Because I depends on distance squared, material far from the centre counts much more. The example I-beam has 1,735 cm4 with an area of 25.2 cm2, while a solid square bar of the same area has only 52.9 cm4. The same material, placed in the flanges, gives about 33 times the bending stiffness.

What is the moment of inertia of a solid sphere?

For a solid sphere about any axis through its centre, I = 2 m r2 / 5. A 5 kg sphere of radius 0.1 m has I = 2 × 5 × 0.01 / 5 = 0.020 kg m2. A thin hollow shell of the same mass and radius has I = 2 m r2 / 3 = 0.0333 kg m2, because all its mass sits at the full radius.

What is polar moment of inertia?

The polar second moment of area J is the resistance of a section to twisting and equals Ix + Iy for any shape about the centroid. For a solid circle, J = pi d4 / 32. The 100 × 200 mm rectangle has J = 8,333.3 cm4 by this sum. Note that J = Ix + Iy is a geometric property. Torsion of non-circular sections needs a different constant.

Can this calculator size a beam or structure?

No. It gives section properties and rotational inertia as estimates. Sizing a beam needs loads, span, material, support conditions, deflection limits and code factors. Follow the applicable standards and local building code, use the manufacturer's published section tables for real products, and have a qualified structural engineer check any structural work.

Does the centroid matter for the moment of inertia?

Yes. The standard section values, Ix and Iy, are taken about axes through the centroid. For symmetric shapes the centroid is at the centre. For a T-section it is not, so the calculator finds it first and reports its height above the bottom. To get I about a different axis, use the shift mode and the parallel axis theorem.

What is the moment of inertia of a thin ring or hoop?

About its own axis, a thin ring of mass m and radius r has I = m r2, because all the mass sits at the same distance from the axis. Across a diameter it is half of that, m r2 / 2. A 3 kg ring of radius 0.25 m gives 0.1875 kg m2 about its axis and 0.09375 kg m2 across a diameter.

References

  1. [1] Wikipedia contributors. “Second moment of area.” Wikipedia. Accessed 2026-10-07.
  2. [2] Wikipedia contributors. “List of moments of inertia.” Wikipedia. Accessed 2026-10-07.
  3. [3] OpenStax. “University Physics Volume 1, 10.5 Calculating Moments of Inertia.” OpenStax, Rice University. Accessed 2026-10-07.