Triangle Section Properties Calculator
Enter the base, the height and the horizontal position of the apex to describe any triangle — isosceles, right-angled or scalene — and get its area, centroid, second moments, product of inertia, principal axes and section moduli, drawn to scale.
Section dimensions
Enter the dimensions in the selected unit — the drawing and the results update as you type. The drawing uses centroidal coordinates: the bold gridlines are the x and y axes through the centroid C.
Section
- Section with its dimensions
- Centroid C, centroidal axes x and y
- Principal axes 1 and 2, angle θp
- Shear centre S
The triangle is where the product of inertia first matters. An isosceles triangle (apex above the middle of the base, c = b/2) has Ixy = 0 and its axes of symmetry are principal; move the apex to a corner to make a right triangle and Ixy = −b²h²/72 appears, rotating the principal axes away from the horizontal and vertical. The drawing shows the rotation live as you drag the apex offset.
Whatever the apex position, the centroid stays at one third of the height above the base and Ix = bh³/36 about the centroidal axis parallel to the base — only Iy and Ixy change with c. Set c = 0 or c = b for a right triangle, c < 0 or c > b for an obtuse one.
How the triangle properties are calculated
- Area
- A = b·h / 2
- Centroid
- x̄ = (b + c) / 3 from the left corner, ȳ = h / 3
- Second moment about the centroidal x axis
- Ix = b·h³ / 36
- Second moment about the centroidal y axis
- Iy = b·h·(b² − b·c + c²) / 36
- Product of inertia
- Ixy = b·h²·(2·c − b) / 72 (zero for an isosceles triangle)
- Principal moments
- I1,2 = (Ix + Iy)/2 ± √[((Ix − Iy)/2)² + Ixy²]
Principal moments and axes follow from Ix, Iy and Ixy with I1,2 = (Ix + Iy)/2 ± √[((Ix − Iy)/2)² + Ixy²] and tan 2θp = −2·Ixy/(Ix − Iy); the radii of gyration are r = √(I/A) and the elastic section moduli S = I/c for each extreme fibre. Definitions of every property are in the glossary on the section properties overview.
Assumptions. Base horizontal at the bottom. The apex offset c is measured from the left corner of the base; it may be negative or larger than b. No torsion constant is given (only the equilateral triangle has a simple closed form).
Other sections
Frequently asked questions
Why is the centroidal moment of inertia bh³/36 and not bh³/12?
bh³/12 is the second moment of a triangle about its base. Moving to the centroid, h/3 above the base, subtracts A·(h/3)² = (bh/2)(h²/9) = bh³/18, leaving bh³/36. The values about the base and about the apex (bh³/4) are useful when combining a triangle with other shapes, but bending stresses always need the centroidal value.
What does the principal angle mean for a right triangle?
For a right triangle the horizontal and vertical centroidal axes are not principal: because I_xy ≠ 0, a moment applied about the x axis produces curvature about y too (unsymmetrical bending). The principal axes are the pair of perpendicular axes through the centroid for which I_xy = 0; bending about them decouples. The calculator draws them in green with the angle θp measured counter-clockwise from x.
How do I enter an equilateral triangle?
Set h = b·√3/2 ≈ 0.866·b and c = b/2. For b = 100 that is h = 86.6 and c = 50. The results then show I_x = I_y and the principal angle 0°, since a regular polygon is isotropic in bending.
References & further reading
- List of centroids — Wikipedia — centroid positions of the elementary plane shapes.
- List of second moments of area — Wikipedia — closed-form Ix and Iy for the common shapes, used to check this calculator.
- Second moment of area — Wikipedia — definition, parallel-axis theorem and the sign convention for the product of inertia.
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