Calculation report includes formulas, steps, and hand-calculation breakdown.
Engineering report
Step-by-step PDF calculation report
Download a clean engineering report showing the selected section, input dimensions, formulas, intermediate steps, and final section property results. The report is formatted like a hand calculation package so you can review how each value was calculated.
Use the PDF report for design notes, internal review, calculation backup, or project documentation.
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
Calculation ReportPDF
PropertyValue
Ix12.4e6Sx82,500Zx95,700
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.
I
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
C
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
P
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
S
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
Z
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
r
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
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)
rx
= √(Ix / A)
ry
= √(Iy / A)
Circle
A
= π · d² / 4
Ix = Iy
= π · d⁴ / 64
Sx = Sy
= π · d³ / 32
rx = ry
= d / 4
Pipe / CHS
A
= π · (do² − di²) / 4
Ix = Iy
= π · (do⁴ − di⁴) / 64
Sx = Sy
= I / (do/2)
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.
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.
Example 1 · Solid rectangle
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.
Asymmetric I-beam composite, full section properties
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
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
All nine section properties of the asymmetric I-beam, alongside the labelled figure with the composite centroid drawn in.
Use the calculated area, moment of inertia, and neutral-axis distances in the Beam Calculator to solve reactions, shear, bending moment, deflection, and stress.
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.
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.