I-beam moment of inertia
Enter top and bottom flange widths and thicknesses separately. Overall depth is hf1 + hw + hf2. Unequal flanges move the centroid; the PDF shows their individual contributions to Ix and Iy.
Free calculator & hand calculations
Moment of Inertia & Section Properties Calculator
Calculate area moment of inertia (second moment of area), centroid, and section modulus. Get a free PDF with formulas, substitutions, and hand calculations.
Free PDF hand calculationsNo sign-upmm or inches
Choose your cross-section
Build a Section Custom shapes & DXFUsing a published profile? Look up steel section properties.
Section properties
Both axes. All properties. Free hand calculations.
| Property | Value |
|---|---|
| A Area (mm²) | - |
| Cx Centroid from left (mm) | - |
| Cy Centroid from bottom (mm) | - |
| Ix Moment of inertia, x-axis (mm⁴) | - |
| Iy Moment of inertia, y-axis (mm⁴) | - |
| Sx Elastic section modulus, x (mm³) | - |
| Sy Elastic section modulus, y (mm³) | - |
| Zx Plastic section modulus, x (mm³) | - |
| Zy Plastic section modulus, y (mm³) | - |
| rx Radius of gyration, x (mm) | - |
| ry Radius of gyration, y (mm) | - |
Enter dimensions and select Calculate, or load an example.
Ixx, Iyy, Zxx and Zyy: which result do you need?
Ixx = Ix and Iyy = Iy: area moment of inertia about the centroidal axes, in mm⁴ or in⁴. On this calculator, Sx / Sy are elastic section moduli and Zx / Zy are plastic section moduli, in mm³ or in³.
Looking for a Zxx calculator? Check your reference’s definition: some software, including Strand7, labels the elastic modulus Zxx / Zyy and the plastic modulus Sxx / Syy. Match the property name and axis, rather than the letter alone. For asymmetric shapes, the elastic modulus shown here uses the farther extreme fibre: S = I / cmax.
Show your working
Free PDF hand calculations, step by step
Follow each result from the formula to the final number. Your PDF includes the section diagram, input dimensions, substituted values, intermediate steps, and results for both axes. For built-up sections, see how each part contributes to the centroid and moment of inertia.
Free to calculate. Free to download. No sign-up required. Keep the report with your design notes, compare it with your own hand calculation, or share it for review.
- Input dimensions and selected section type
- Area and centroid calculation
- Moment of inertia Ix / Iy calculation
- Parallel axis theorem breakdown where applicable
- Elastic and plastic section modulus
- Radius of gyration and final results summary
Rectangle · 200 × 400 mm
1. Formula
Ix = b × d³ / 12
2. Substitute dimensions
Ix = 200 × 400³ / 12
3. Result
Ix = 1.06667 × 10⁹ mm⁴
From dimensions to a checkable calculation
How to calculate second moment of area
The second moment of area, also called area moment of inertia, measures how a cross-section’s area is distributed about an axis. This calculator returns Ix and Iy in mm⁴ or in⁴. These are the geometric properties used in bending stiffness EI; mass moment of inertia, used for rotational motion, has different units and is a different quantity.
- Select the shape and units. Match the section’s orientation to the diagram. Enter all dimensions in mm or all in inches.
- Enter the actual dimensions. For an I-beam, channel or tee, web height is the clear height between flanges or below the flange. For a hollow section, enter both outer and inner dimensions.
- Calculate both axes. Read the centroid, Ix, Iy, elastic and plastic section moduli, and radii of gyration. Check the axis you need before using a result.
- Download the hand calculations. The free PDF shows the formulas and substituted dimensions behind the results. Copy the table for a spreadsheet, or copy a section link to reopen the same inputs.
- Check the section in a beam. Open the Beam Calculator with your calculated section, then adjust the sample beam's span, material, supports and loads to check deflection, bending stress and reactions.
For a rectangle, Ix = bd³/12 and Iy = db³/12. Rotating a 200 × 400 mm rectangle by 90° swaps the two values: its Ix becomes one quarter of the original. Load the 200 × 400 mm example to check the numbers.
For sections made from several parts, first locate the composite centroid. Then add each part’s centroidal inertia and its offset contribution, I = Σ(Ic + AΔ²). See the parallel axis method and the independent explanation in Engineering Statics, §10.4.
Choose the right geometry
Moment of inertia for eight common cross-sections
Choose from I-beams, channels, angles, tees, circles, pipes, rectangles and hollow rectangles. Each example below loads the shape and dimensions into the calculator, ready to review the results and download the hand calculations.
Channel moment of inertia
Flange widths include the web thickness. The centroid moves toward the web, so the y-axis inertia needs each part’s horizontal offset. Top and bottom flanges can have different dimensions.
Angle moment of inertia
Enter both outside leg widths and their thicknesses. The shared corner area is counted once. The reported x- and y-axes follow the legs; they generally differ from an angle’s principal axes.
T-section moment of inertia
Web height excludes the flange thickness. The centroid shifts toward the flange, so distances to the top and bottom fibres differ. The reported elastic modulus uses the larger distance.
Circle moment of inertia
Enter the full diameter, rather than the radius. The centroid lies at the centre, and symmetry gives equal values about both centroidal axes: Ix = Iy = πd⁴/64. Use this shape for a solid circular cross-section.
Pipe / CHS moment of inertia
Enter the actual outer and inner diameters. If you know the wall thickness t, the inner diameter is di = do − 2t. The concentric opening is subtracted from the outer circle, giving equal Ix and Iy.
Rectangle moment of inertia
Enter the horizontal width b and vertical depth d shown in the diagram. The centroid is at mid-width and mid-depth. Ix = bd³/12 and Iy = db³/12, so rotating the rectangle by 90° swaps the two values.
Hollow rectangle / RHS moment of inertia
Enter the outer and inner widths and depths. For uniform wall thickness t, bi = bo − 2t and di = do − 2t. The calculator subtracts a centred rectangular opening and assumes square corners.
Modelling assumptions
The section is homogeneous; flange and web components are solid rectangles. Hollow rectangles and pipes have concentric openings. Fillets, corner radii and flange tapers are excluded, so idealized dimensions may differ from published rolled-section properties. Use the steel section database when you need catalogue values. For other geometry, Build a Section lets you combine shapes, add cut-outs or import a DXF.
This page reports centroidal x/y properties, rather than principal axes, shear centre or torsion constants. Use the beam deflection formula guide to apply the appropriate inertia in a supported beam case.
Engineering reference
What does each section property mean?
The section properties calculator provides key geometric properties used in beam design, including moment of inertia, centroid, elastic section modulus, plastic section modulus, and radius of gyration. These values help determine how a cross-section behaves under bending, deflection, yielding, and buckling. Engineers use these properties to compare beam shapes, calculate bending stress, check deflection, and evaluate column slenderness. The following section types are included: I-Beam, T-Beam, Channel, Angle, Circular, Hollow Circular, Rectangular, and Hollow Rectangular.
Moment of inertia
Also called: Second moment of area
- Notation
- Ix, Iy
- Units
- length4 (in4, mm4)
Moment of inertia describes how the area of a cross-section is distributed around an axis. In beam design, it is one of the most important properties because it directly affects bending stiffness and deflection.
A higher moment of inertia means more of the material is located farther away from the neutral axis, making the section stiffer in bending. For the same material, span, and load, a beam with a larger I will deflect less than a beam with a smaller I.
This is why deeper sections are usually much more efficient in bending. Moving material away from the neutral axis increases moment of inertia significantly.
E·I
Where
- E
- = modulus of elasticity of the material
- I
- = moment of inertia of the cross-section
Bending stiffness of a beam.
Where it is used
- Beam deflection calculations
- Bending stiffness, EI
- Comparing how efficiently different shapes resist bending
- Major-axis and minor-axis beam behavior
Centroid
Also called: Geometric center
- Notation
- Cx, Cy, x̄, ȳ
- Units
- length (in, mm)
The centroid is the geometric center of the cross-section's area. It is the point where the area can be considered balanced.
For symmetric shapes, such as rectangles, circles, and standard I-shapes, the centroid is usually easy to locate because it lies on the axes of symmetry. For asymmetric shapes, such as channels, angles, tees, or custom built-up sections, the centroid may be offset from the obvious center of the shape.
The centroid is important because many section properties are calculated about centroidal axes. In elastic bending, the neutral axis passes through the centroid for a homogeneous section.
Where it is used
- Locating the neutral axis for elastic bending
- Calculating moment of inertia for built-up or composite shapes
- Determining extreme-fiber distances for section modulus
- Understanding bending behavior of asymmetric sections
Plastic neutral axis
The equal-area axis used for plastic bending
- Notation
- yp
- Units
- length (in, mm)
The plastic neutral axis is the line where the yielded compression force and yielded tension force balance. For a homogeneous material with the same yield strength in tension and compression, this means the plastic neutral axis divides the cross-section into two equal areas.
That is different from the centroid. The centroid is an area-weighted balance point used for elastic bending. The plastic neutral axis is an equal-area divider used for plastic bending.
For symmetric sections, the centroidal neutral axis and plastic neutral axis usually land on the same line. For asymmetric sections, such as unequal-flange I-beams, tees, channels, and angles, they can be noticeably different.
Elastic centroid
ȳ = ΣAiyi / ΣAi
Finds the area-weighted location of the section. This is the axis used for elastic neutral axis, moment of inertia, and elastic section modulus.
Plastic neutral axis
Aabove = Abelow = A / 2
Finds the line where the yielded area above and below the axis are equal. This is the axis used for plastic section modulus.
Where it is used
- Locating the equal-area axis for plastic bending
- Calculating plastic section modulus Zx and Zy
- Checking plastic moment capacity, Mp = ZFy
- Understanding why asymmetric sections need a separate plastic axis
Elastic section modulus
- Notation
- Sx, Sy
- Units
- length3 (in3, mm3)
Elastic section modulus measures how efficiently a cross-section resists bending stress while the material remains in the elastic range.
It is calculated by dividing the moment of inertia by the distance from the neutral axis to the extreme fiber:
S = I / c
Where
- I
- = moment of inertia
- c
- = distance from the neutral axis to the farthest edge of the section
Elastic section modulus is directly used in the bending stress equation:
σ = M / S
Where
- σ
- = bending stress
- M
- = applied bending moment
- S
- = elastic section modulus
A larger section modulus means the section can resist a larger bending moment before reaching the same bending stress.
Where it is used
- Elastic bending stress checks
- Beam design under service or allowable stress conditions
- Comparing bending strength of different cross-sections
- Determining stress at the extreme top or bottom fiber
Plastic section modulus
- Notation
- Zx, Zy
- Units
- length3 (in3, mm3)
Plastic section modulus is used to calculate the plastic bending capacity of a cross-section after the full section has yielded.
Unlike elastic section modulus, which is based on the centroidal neutral axis and elastic stress distribution, plastic section modulus is based on the plastic neutral axis. The plastic neutral axis divides the cross-section so that the compressive and tensile forces are balanced.
For materials with similar tension and compression yield strength, this usually means the plastic neutral axis divides the cross-section into two equal areas.
Plastic moment capacity is calculated as:
Mp = Z × Fy
Where
- Mp
- = plastic moment capacity
- Z
- = plastic section modulus
- Fy
- = yield strength of the material
Plastic section modulus is typically greater than elastic section modulus because it represents the section's capacity after yielding has spread through the full cross-section.
Where it is used
- Plastic design of steel beams
- Plastic moment capacity checks
- Limit-state design
- Compact steel sections capable of developing full plastic capacity
Radius of gyration
- Notation
- rx, ry
- Units
- length (in, mm)
Radius of gyration describes how far the area of a cross-section is distributed from an axis. It is calculated from the moment of inertia and area:
r = √(I / A)
Where
- I
- = moment of inertia about the axis
- A
- = cross-sectional area
You can think of radius of gyration as the distance from the axis where the entire area could be concentrated and still produce the same moment of inertia.
In structural design, radius of gyration is especially important for compression members and column buckling. It is used in the slenderness ratio:
KL / r
Where
- K
- = effective length factor
- L
- = unsupported length
- r
- = radius of gyration
A larger radius of gyration means the section is more efficient at resisting buckling about that axis.
Where it is used
- Column slenderness checks
- Buckling calculations
- Comparing weak-axis and strong-axis stability
- Compression member design
Quick summary
The main section properties and axes at a glance.
| Property | Meaning | Main use | Units |
|---|---|---|---|
| I | How area is distributed about an axis | Deflection and bending stiffness | length4 |
| C | Geometric center of the section | Neutral axis and section property calculations | length |
| Plastic neutral axis | Equal-area axis for plastic bending | Plastic section modulus and plastic moment capacity | length |
| S | Elastic bending strength measure | Bending stress, σ = M / S | length3 |
| Z | Plastic bending strength measure | Plastic moment capacity, Mp = Z·Fy | length3 |
| r | Area distribution measure for buckling | Slenderness ratio, KL / r | length |
Formula reference
Section formula reference
Simple symmetric shapes use direct closed-form formulas. Built-up sections, such as I-beams, channels, angles, and tees, are split into rectangles and calculated with the parallel axis theorem.
Simple closed-form shapes
Use these formulas directly when the section matches the labelled diagram. For circular sections, symmetry means the x- and y-axis values are the same.
Rectangle
- A
- = b · d
- Ix
- = b · d³ / 12
- Iy
- = d · b³ / 12
- Sx
- = b · d² / 6
- Sy
- = d · b² / 6
- Plastic section modulus
- Zx = b · d² / 4; Zy = d · b² / 4
- rx
- = d / √12
- ry
- = b / √12
Hollow rectangle / RHS
- A
- = bo·do − bi·di
- Ix
- = (bo·do³ − bi·di³) / 12
- Iy
- = (do·bo³ − di·bi³) / 12
- Sx
- = Ix / (do/2)
- Sy
- = Iy / (bo/2)
- Plastic section modulus
- Zx = (bodo² − bidi²) / 4
Zy = (dobo² − dibi²) / 4 - rx
- = √(Ix / A)
- ry
- = √(Iy / A)
Circle
- A
- = π · d² / 4
- Ix = Iy
- = π · d⁴ / 64
- Sx = Sy
- = π · d³ / 32
- Plastic section modulus
- Zx = Zy = d³ / 6
- rx = ry
- = d / 4
Pipe / CHS
- A
- = π · (do² − di²) / 4
- Ix = Iy
- = π · (do⁴ − di⁴) / 64
- Sx = Sy
- = I / (do/2)
- Plastic section modulus
- Zx = Zy = (do³ − di³) / 6
- rx = ry
- = √(do² + di²) / 4
Composite method
Built-up shapes are easiest to follow as a repeatable composite-section method: decompose the section, find the composite centroid, then shift each part to the final axes.
I-beam example for composite sections
This I-beam shows the general composite-section workflow used for I-beams, channels, angles, tees, and other shapes made from simple rectangles.
- Split into top flange, web, and bottom flange.
- Find each part's area Ai and centroid coordinates x̄i, ȳi.
- Composite centroid: x̄ = ΣAix̄i / ΣAi and ȳ = ΣAiȳi / ΣAi.
- Apply the parallel axis theorem about both centroidal axes: Ix = Σ(Īx,i + Ai·dy,i²) Iy = Σ(Īy,i + Ai·dx,i²)
- Compute S, Z, and r from the final A, centroid, Ix, and Iy values.
Engineering reference
Parallel axis theorem
Use the parallel axis theorem when a simple shape's own centroidal axis is not the same axis used for the full section. It is the key step for built-up sections, cut-outs, and asymmetric shapes.
The theorem shifts a known centroidal moment of inertia to a parallel target axis. In this page, it connects the simple formulas above to the composite I-beam example below.
I = Ī + A · d²
Where
- I
- = moment of inertia about the target axis
- Ī
- = moment of inertia about the shape's own centroidal axis
- A
- = area of the shape
- d
- = perpendicular distance between the two parallel axes
The A·d² term is the shift. The farther a part sits from the final neutral axis, the more it contributes to the total moment of inertia.
This is why flanges matter so much in an I-beam: they place area far from the neutral axis, which makes the A·d² contribution large.
About a horizontal x-axis
Ix = Īx + A · dy²
Use the vertical offset dy. This is the form used for most beam bending about the strong axis.
About a vertical y-axis
Iy = Īy + A · dx²
Use the horizontal offset dx. If all parts share the same vertical centerline, this shift is zero.
Composite sections
For built-up sections, calculate each simple part about the same final centroidal axis, then add the contributions.
Ix = Σ ( Īx,i + Ai · dy,i² )
Iy = Σ ( Īy,i + Ai · dx,i² )
1
Split the section into rectangles, circles, flanges, webs, or other simple parts.
2
Find each part's area, centroid, and centroidal Ī.
3
Locate the full-section centroid before measuring any shifts.
4
Shift and sum every part to the same final x-axis or y-axis.
Worked examples
From formulas to numbers
Two examples that show the formulas above being plugged in step by step. The first is a solid rectangle; the second is an asymmetric I-beam built up using the parallel axis theorem.
Rectangle, b = 200 mm, d = 400 mm
A simple rectangle 200 mm wide and 400 mm deep. We plug the inputs straight into the formulas from the rectangle card above and compute every section property about both centroidal axes.
Inputs
- b
- = 200 mm
- d
- = 400 mm
Area
A = 80,000 mm²
Formula A = b · d
Substitute = 200 · 400
About the x-axis
Moment of inertia, Ix
Ix = b · d³ / 12
= 200 · 400³ / 12
≈ 1.067 × 10⁹ mm⁴
Section modulus, Sx
Sx = b · d² / 6
= 200 · 400² / 6
≈ 5.333 × 10⁶ mm³
Plastic modulus, Zx
Zx = b · d² / 4
= 200 · 400² / 4
= 8.000 × 10⁶ mm³
Radius of gyration, rx
rx = d / √12
= 400 / √12
≈ 115.5 mm
About the y-axis
Moment of inertia, Iy
Iy = d · b³ / 12
= 400 · 200³ / 12
≈ 2.667 × 10⁸ mm⁴
Section modulus, Sy
Sy = d · b² / 6
= 400 · 200² / 6
≈ 2.667 × 10⁶ mm³
Plastic modulus, Zy
Zy = d · b² / 4
= 400 · 200² / 4
= 4.000 × 10⁶ mm³
Radius of gyration, ry
ry = b / √12
= 200 / √12
≈ 57.7 mm
I-beam moment of inertia: asymmetric section worked example
An I-section with a wide top flange (250 × 20), a narrower bottom flange (200 × 15), and a 300 × 10 web. Because the flanges are different sizes, the centroid sits above the geometric centre, and the plastic neutral axis sits in a different location again. Follow the seven steps below — using the parallel axis theorem — to compute every section property: A, ȳ, Ix, Iy, Sx, Sy, Zx, Zy, rx, ry.
Inputs
- bf1
- = 250 mm
- hf1
- = 20 mm
- hw
- = 300 mm
- tw
- = 10 mm
- bf2
- = 200 mm
- hf2
- = 15 mm
All dimensions are labelled in the figure. Distances are measured from the bottom fibre upward.
1
Decompose and find each part's area and centroid
Split the I-beam into three rectangles — top flange, web, bottom flange — whose properties we already know how to compute. For each one, find its area Ai and the height ȳi of its centroid above the bottom fibre. The thumbnail on each row highlights the part being calculated.
Top flange
A1 = bf1 · hf1 = 250 · 20 = 5,000 mm²
ȳ1 = hf2 + hw + hf1/2 = 15 + 300 + 10 = 325 mm
Web
A2 = tw · hw = 10 · 300 = 3,000 mm²
ȳ2 = hf2 + hw/2 = 15 + 150 = 165 mm
Bottom flange
A3 = bf2 · hf2 = 200 · 15 = 3,000 mm²
ȳ3 = hf2 / 2 = 7.5 mm
Total area ΣA = 5,000 + 3,000 + 3,000 = 11,000 mm²
2
Locate the composite centroid ȳ
Take the area-weighted average of each part's own centroid. The result is the height of the composite x-axis above the bottom fibre — shown as the dashed red line in the figure.
ΣA·ȳi = 5,000·325 + 3,000·165 + 3,000·7.5
= 1,625,000 + 495,000 + 22,500
= 2,142,500 mm³
ȳ = ΣA·ȳi / ΣA = 2,142,500 / 11,000
≈ 194.77 mm above the bottom fibre
3
Distance from each part centroid to the composite x-axis
For the parallel-axis term we need dy,i = ȳi − ȳ. A positive value means the part centroid is above x̄; a negative value means below. The arrows show the magnitude and direction of each.
dy,1 = ȳ1 − ȳ = 325 − 194.77 = +130.23 mm
dy,2 = ȳ2 − ȳ = 165 − 194.77 = −29.77 mm
dy,3 = ȳ3 − ȳ = 7.5 − 194.77 = −187.27 mm
The web is almost on x̄ (dy,2 is small). The two flanges are far from x̄ — that's where most of the inertia comes from.
4
Centroidal Ix and Iy of each part
Each rectangle's own moment of inertia about its own centroidal axes uses the standard b·h³/12 formula. About x: width × (height)³/12. About y: height × (width)³/12.
About the x-axis (Īx,i = bi·hi³/12)
Īx,1 = bf1 · hf1³ / 12 = 250 · 20³ / 12
≈ 1.667 × 10⁵ mm⁴
Īx,2 = tw · hw³ / 12 = 10 · 300³ / 12
= 2.250 × 10⁷ mm⁴
Īx,3 = bf2 · hf2³ / 12 = 200 · 15³ / 12
≈ 5.625 × 10⁴ mm⁴
About the y-axis (Īy,i = hi·bi³/12)
Īy,1 = hf1 · bf1³ / 12 = 20 · 250³ / 12
≈ 2.604 × 10⁷ mm⁴
Īy,2 = hw · tw³ / 12 = 300 · 10³ / 12
= 2.500 × 10⁴ mm⁴
Īy,3 = hf2 · bf2³ / 12 = 15 · 200³ / 12
= 1.000 × 10⁷ mm⁴
5
Apply the parallel axis theorem and sum
For Ix, shift each part from its own centroid to the composite x-axis by adding Ai·dy,i². For Iy, no shift is needed because all three rectangles already share the same vertical y-axis (dx,i = 0 for every part).
Ix — parallel-axis terms
A1·dy,1² = 5,000 · 130.23² ≈ 8.480 × 10⁷
A2·dy,2² = 3,000 · 29.77² ≈ 2.659 × 10⁶
A3·dy,3² = 3,000 · 187.27² ≈ 1.052 × 10⁸
Ix = Σ(Īx,i + Ai·dy,i²)
≈ 2.154 × 10⁸ mm⁴
Iy — no parallel-axis shift
Every part is centred on the y-axis, so each dx,i is zero and we just sum the centroidal Īy,i values from Step 4.
Īy,1 ≈ 2.604 × 10⁷
Īy,2 = 2.500 × 10⁴
Īy,3 = 1.000 × 10⁷
Iy = Σ Īy,i
≈ 3.607 × 10⁷ mm⁴
6
Locate the plastic neutral axis
For plastic bending, do not use the elastic centroid. Instead, find the horizontal line that divides the total area into two equal halves. Here, half the area is 5,500 mm², so the plastic neutral axis falls inside the web just below the top flange.
Total area A = 11,000 mm²
Area on each side of the plastic neutral axis = A / 2 = 11,000 / 2 = 5,500 mm²
Area in top flange = 250 · 20 = 5,000 mm²
Additional area needed from web = 5,500 − 5,000 = 500 mm²
Depth into web = 500 / tw = 500 / 10 = 50 mm
Plastic neutral axis from bottom = hf2 + hw − 50 = 15 + 300 − 50
Plastic neutral axis = 265 mm above the bottom fibre
7
Section moduli (S, Z) and radii of gyration (r)
The elastic properties use the centroidal axis from Step 2. The plastic properties use the plastic neutral axis from Step 6. Note that Sx is governed by the larger fibre distance cbot = ȳ = 194.77 mm; the section is asymmetric so ctop = (335 − 194.77) = 140.23 mm.
About the x-axis
Elastic modulus, Sx (governed by cbot)
Sx = Ix / cbot
= 2.154×10⁸ / 194.77
≈ 1.106 × 10⁶ mm³
Plastic modulus, Zx (plastic neutral axis at 265 mm)
Zx = Σ Ai·|yi − yp|
= 5,000(60) + 500(25) + 2,500(125) + 3,000(257.5)
≈ 1.398 × 10⁶ mm³
Radius of gyration, rx
rx = √(Ix / A)
= √(2.154×10⁸ / 11,000)
≈ 139.93 mm
About the y-axis
Elastic modulus, Sy (extreme fibre = bf1/2)
Sy = Iy / (bf1/2)
= 3.607×10⁷ / 125
≈ 2.885 × 10⁵ mm³
Plastic modulus, Zy (plastic neutral axis on y-axis)
Zy = Σ bi²·hi / 4
= 250²·20/4 + 10²·300/4 + 200²·15/4
= 4.700 × 10⁵ mm³
Radius of gyration, ry
ry = √(Iy / A)
= √(3.607×10⁷ / 11,000)
≈ 57.26 mm
Final results
Key section properties of the asymmetric I-beam, alongside the labelled figure with the composite centroid drawn in.
- A
- 11,000 mm²
- Ix
- 2.154 × 10⁸ mm⁴
- Sx
- 1.106 × 10⁶ mm³
- Zx
- 1.398 × 10⁶ mm³
- rx
- 139.93 mm
- ȳ
- 194.77 mm
- Iy
- 3.607 × 10⁷ mm⁴
- Sy
- 2.885 × 10⁵ mm³
- Zy
- 4.700 × 10⁵ mm³
- ry
- 57.26 mm
Questions & answers
Moment of inertia calculator FAQ
What does this moment of inertia and section properties calculator calculate?
This section properties calculator returns cross-sectional area, centroid location, Ix and Iy moment of inertia, elastic section modulus Sx and Sy, plastic section modulus Zx and Zy, and radius of gyration rx and ry for common beam shapes including I-sections, channels, angles, tees, rectangles, hollow rectangles, circles, and pipes.
Does the PDF report show how the section properties were calculated?
Yes. The Download PDF Calculation Report option includes the input dimensions, formulas, intermediate calculation steps, and final results for the selected cross-section. For built-up shapes such as I-sections, channels, tees, and angles, the report shows the centroid calculation and parallel axis theorem breakdown where applicable.
Which moment of inertia should I use, Ix or Iy?
Use the moment of inertia about the axis that matches the bending direction you are checking. Ix is about the horizontal centroidal x-axis and is often the strong-axis value for beams loaded vertically. Iy is about the vertical centroidal y-axis and is often the weak-axis value. For biaxial bending or columns, check both axes.
Is moment of inertia the same as second moment of area?
For beam cross-sections, moment of inertia usually means area moment of inertia, also called the second moment of area. It is a geometric property with units of length to the fourth power, such as mm4 or in4. It is different from mass moment of inertia, which is used in dynamics.
How do I use the calculator for I-beams, channels, tees, and angles?
Select the matching shape, then enter the flange, web, leg, and thickness dimensions shown in the diagram. For built-up shapes such as I-beams, channels, tees, and angles, the calculator finds each part's area and centroid, locates the composite centroid, and uses the parallel axis theorem to calculate Ix and Iy.
How are hollow rectangular sections, pipes, and cut-outs calculated?
Hollow sections are calculated by subtracting the inner void from the outer solid shape. For a hollow rectangular section or RHS, the inner rectangle is removed from the outer rectangle. For a pipe or CHS, the inner circle is removed from the outer circle. Inner dimensions must be smaller than the corresponding outer dimensions.
What is section modulus, and how is it different from moment of inertia?
Moment of inertia describes bending stiffness and is used in deflection calculations through E·I. Section modulus is calculated as S = I / c, where c is the distance from the neutral axis to the extreme fiber. Section modulus is used directly in bending stress checks with σ = M / S.
What is the difference between elastic section modulus S and plastic section modulus Z?
Elastic section modulus S is used for elastic bending stress checks before yielding. Plastic section modulus Z is used to estimate plastic moment capacity after yielding has spread through the full cross-section. Z is mainly useful for ductile, compact sections such as suitable steel shapes and should not be used by itself as a complete code design check.
Is the plastic neutral axis the same as the centroid?
No. The centroid is the area-weighted geometric center and is used for elastic properties such as moment of inertia and elastic section modulus. The plastic neutral axis is the equal-area line used for plastic section modulus. For symmetric sections they often coincide, but for asymmetric sections they can be different.
What units should I use for section properties?
Enter all dimensions in one consistent length unit, such as all millimeters or all inches. The calculator returns area in unit2, section modulus and plastic modulus in unit3, moment of inertia in unit4, and radius of gyration and centroid distances in the same length unit. For an existing value, use the section area converter or moment of inertia unit converter.
Can I use these results for beam deflection, bending stress, or buckling checks?
Yes. Use Ix or Iy with the material modulus E for beam deflection and bending stiffness. Use Sx or Sy for bending stress checks. Use area A for axial stress checks, and use rx or ry for column slenderness and buckling checks. The required axis depends on the direction of bending or buckling.
Can I send these section properties to the Beam Calculator?
Yes. After calculating a cross-section, choose Analyze in Beam Calculator to use the calculated area, moment of inertia, centroid, and section modulus in a full beam analysis workflow for reactions, shear, bending moment, deflection, and stress.
Why does changing beam depth affect moment of inertia so much?
Moment of inertia is very sensitive to where material is placed relative to the neutral axis. For many simple shapes, depth appears as a cubed or fourth-power term in the section property formulas. That is why deeper I-beams, tubes, and hollow sections can be much stiffer in bending without a proportional increase in area.
What are the limitations of this section properties calculator?
The calculator provides geometric section properties, not a complete member design by itself. Beam capacity and serviceability also depend on material properties, span, supports, loads, bracing, local buckling, connection details, and the design code or safety factors you are using.
Continue your beam design
From section properties to full beam analysis
Take your calculated x-axis properties into a sample beam, then adjust the span, material, supports and loads to check deflection, bending stress and reactions.