How to calculate the moment of inertia of symmetrical and unsymmetrical I beams, with the strong and weak axis formulas and two checked worked examples.
The moment of inertia of an I beam is the property that decides how stiff the beam is in bending and, through the section modulus, how much moment it can carry. I sections put most of their steel in two flanges a long way from the neutral axis, which is exactly where area does the most good. This page works through the strong and weak axis formulas for a symmetrical I section, then the longer procedure for an unsymmetrical I section where the centroid has to be found first.
If you need the general background, including the formulas for rectangles, circles and hollow shapes, start with our guide to the moment of inertia of common shapes. This page is the I beam worked example only.
What the moment of inertia tells you about an I beam
The second moment of area, usually written \( I \), measures how far the material sits from the bending axis, squared and summed over the whole section:
$$ I_x = \int y^2 \, dA $$
It appears in both of the checks that govern a steel beam. Deflection is proportional to \( 1/I \), so doubling \( I \) halves the midspan sag for the same load. Bending stress is \( \sigma = M y / I \), and the elastic section modulus \( Z = I / y_{max} \) folds the two together. Because the distance is squared, a flange at 145 mm from the axis contributes about 21 times more than the same area would at 32 mm. That is the whole logic of the I shape.
Units are length to the fourth power. Work in millimetres and report in \( \text{mm}^4 \), typically as a multiple of \( 10^6 \text{ mm}^4 \). Section tables quote the same quantity in \( \text{cm}^4 \), and \( 1 \text{ cm}^4 = 10^4 \text{ mm}^4 \).
Moment of inertia of a symmetrical I section
The section properties of common shapes calculator runs the symmetric case live: pick the I or H section, enter the four plate dimensions and read \( I_x \), \( I_y \) and the section moduli.
For a doubly symmetric I or H section the centroid is at mid-depth, so no centroid calculation is needed. There are two equivalent ways to get \( I_x \).
Method 1: subtract the two side rectangles
Treat the section as a solid rectangle \( b \times d \) and remove the two rectangles either side of the web. All three rectangles share the same centroidal axis, so their moments of inertia subtract directly:
$$ I_x = \frac{b d^3 - (b - t_w)(d - 2 t_f)^3}{12} $$
where
\( b \) is the flange width
\( d \) is the overall depth
\( t_f \) is the flange thickness
\( t_w \) is the web thickness
Method 2: three rectangles and the parallel axis theorem
Split the section into the two flanges (A and C) and the web (B). The web's centroid is already on the x axis. Each flange's centroid sits a distance \( y_f = (d - t_f)/2 \) from the axis, so the parallel axis theorem adds an \( A y^2 \) term:
$$ I_x = 2 \left( \frac{b t_f^3}{12} + b t_f \, y_f^2 \right) + \frac{t_w (d - 2 t_f)^3}{12} $$
Both methods give the same number. Method 2 is the one that generalises to any built-up section, so it is worth being fluent in it.
Weak axis
About the y axis the three rectangles all have their centroids on the axis, so there is no parallel axis term at all:
$$ I_y = \frac{2 t_f b^3 + (d - 2 t_f) t_w^3}{12} $$
The web term is usually negligible. For the example below it is 0.2 percent of the total.
Worked example: 300 mm deep symmetrical I section
Take a welded I section with \( d = 300 \) mm, \( b = 150 \) mm, \( t_f = 10 \) mm and \( t_w = 8 \) mm.
Method 1:
$$ I_x = \frac{150 \times 300^3 - 142 \times 280^3}{12} = \frac{4.050 \times 10^9 - 3.117 \times 10^9}{12} = 77.7 \times 10^6 \text{ mm}^4 $$
Method 2, for the check. Each flange has area \( 1500 \text{ mm}^2 \) at \( y_f = 145 \) mm:
- Flange own term: \( 150 \times 10^3 / 12 = 12\,500 \text{ mm}^4 \)
- Flange transfer term: \( 1500 \times 145^2 = 31\,537\,500 \text{ mm}^4 \)
- Two flanges: \( 2 \times 31\,550\,000 = 63\,100\,000 \text{ mm}^4 \)
- Web: \( 8 \times 280^3 / 12 = 14\,634\,667 \text{ mm}^4 \)
- Total: \( I_x = 77\,734\,667 \text{ mm}^4 = 77.7 \times 10^6 \text{ mm}^4 \)
Notice that the flanges' own terms are 0.03 percent of the total. The transfer terms carry 81 percent, the web the rest.
Weak axis:
$$ I_y = \frac{2 \times 10 \times 150^3 + 280 \times 8^3}{12} = \frac{67\,500\,000 + 143\,360}{12} = 5.64 \times 10^6 \text{ mm}^4 $$
So the section is about 14 times stiffer about its strong axis than its weak axis, which is why lateral restraint matters so much for I beams. Two derived numbers you will often want:
- Elastic section modulus \( Z_x = I_x / (d/2) = 77.73 \times 10^6 / 150 = 518 \times 10^3 \text{ mm}^3 \)
- Radius of gyration \( r_x = \sqrt{I_x / A} = \sqrt{77.73 \times 10^6 / 5240} = 121.8 \) mm, with \( A = 5240 \text{ mm}^2 \)
Moment of inertia of an unsymmetrical I section
The composite section properties calculator does the parallel-axis bookkeeping below for up to three rectangles, centroid first.
Plate girders, crane beams and composite steel sections often have a top flange wider or thicker than the bottom one. The centroid then sits off mid-depth, and you have to find it before you can apply the parallel axis theorem. The procedure is:
- Split the section into rectangles: top flange, web, bottom flange.
- Measure each rectangle's centroid height \( y_i \) from the bottom face.
- Find the section centroid \( \bar{y} = \sum A_i y_i / \sum A_i \).
- For each rectangle, add its own \( b h^3 / 12 \) to its transfer term \( A_i (y_i - \bar{y})^2 \).
- Sum the three.
$$ I_x = \sum \left( \frac{b_i h_i^3}{12} + A_i \left( y_i - \bar{y} \right)^2 \right) $$
The weak axis is unaffected by the asymmetry: \( I_y \) is still the sum of the three \( h b^3 / 12 \) terms, because every rectangle is centred on the y axis.
Worked example: 300 mm deep section with unequal flanges
Top flange 200 mm by 12 mm, web 276 mm by 8 mm, bottom flange 120 mm by 12 mm. Overall depth is 300 mm. Measuring from the bottom face:
- Top flange: \( A_1 = 2400 \text{ mm}^2 \), \( y_1 = 294 \) mm
- Web: \( A_2 = 2208 \text{ mm}^2 \), \( y_2 = 150 \) mm
- Bottom flange: \( A_3 = 1440 \text{ mm}^2 \), \( y_3 = 6 \) mm
$$ \bar{y} = \frac{2400 \times 294 + 2208 \times 150 + 1440 \times 6}{6048} = \frac{1\,045\,440}{6048} = 172.9 \text{ mm} $$
The centroid is 172.9 mm above the bottom face and 127.1 mm below the top, pulled upward by the heavier top flange. Now the five terms:
- Top flange: \( 200 \times 12^3/12 + 2400 \times 121.1^2 = 28\,800 + 35\,221\,420 \)
- Web: \( 8 \times 276^3/12 + 2208 \times 22.9^2 = 14\,016\,384 + 1\,153\,567 \)
- Bottom flange: \( 120 \times 12^3/12 + 1440 \times 166.9^2 = 17\,280 + 40\,091\,481 \)
$$ I_x = 90.5 \times 10^6 \text{ mm}^4 $$
An unsymmetrical section has two section moduli, one for each extreme fibre:
- Top: \( Z_{top} = 90.53 \times 10^6 / 127.1 = 712 \times 10^3 \text{ mm}^3 \)
- Bottom: \( Z_{bot} = 90.53 \times 10^6 / 172.9 = 524 \times 10^3 \text{ mm}^3 \)
The smaller one governs an elastic stress check, so the bottom flange reaches yield first under sagging moment even though it is the lighter flange. For the weak axis, \( I_y = (12 \times 200^3 + 276 \times 8^3 + 12 \times 120^3)/12 = 9.74 \times 10^6 \text{ mm}^4 \).
Where the number goes in design
Once you have \( I_x \), three things follow:
- Section moment capacity. AS 4100 Clause 5.2.1 gives the nominal section capacity as \( M_s = f_y Z_e \), where the effective section modulus \( Z_e \) depends on the section slenderness and is built from \( Z = I / y_{max} \). EN 1993-1-1 Clause 6.2.5 does the same job with \( W_{el} \) for Class 3 sections and \( W_{pl} \) for Class 1 and 2.
- Deflection. Serviceability limits in AS/NZS 1170.0 Appendix C and EN 1990 Annex A1 are expressed as span ratios, and every deflection formula has \( E I \) in the denominator. A 10 percent error in \( I \) is a 10 percent error in deflection.
- Buckling. The weak axis \( I_y \) feeds the radius of gyration and the elastic buckling load \( \pi^2 E I / L^2 \), which is why the 14 to 1 ratio above matters for columns and for lateral torsional buckling of beams.
Rolled universal beams come with all of this tabulated, but the formulas still matter for welded sections, strengthened members with cover plates, and any time you need to check a tabulated value. For hollow sections, the same subtraction logic is worked through in our RHS and SHS moment of inertia calculator and the circular hollow section calculator.
Run it in CalcTree
The section properties of common shapes calculator and the composite section properties calculator run both procedures live. Both procedures above are a short set of formulas with units attached, which makes them a natural live, checkable CalcTree page: enter the plate sizes, see the centroid, \( I_x \), \( I_y \) and the two section moduli update, and hand the result straight to a member check such as our steel member design to AS 4100. Browse the template library for the section property calculators that already exist.
