From Section Geometry to Beam Bending

On Engivault’s section tools, x is horizontal and y is vertical in the cross-section. The beam’s length runs perpendicular to this cross-section. The second moment Iₓ weights area by squared distance from x; Iᵧ uses distance from y. For a solid rectangle of width b and height h, Iₓ = bh³/12 and Iᵧ = hb³/12. Subtract the concentric inner rectangle for a hollow section. These are area moments in m⁴, not mass moments in kg·m². See Engineering Statics’ shape derivations.

For these symmetric sections, pair Iₓ with c = h/2 and Iᵧ with c = b/2. The elastic section modulus S = I/c gives maximum bending-stress magnitude |M|/S. Flexural rigidity EI controls elastic curvature. E is Young’s modulus, not a strength limit. These relations require the appropriate neutral axis and elastic bending assumptions; see MIT’s beam stress derivation.

Original calculated example: use outer width 100 mm and height 200 mm, with a centered 80 mm × 180 mm void. This idealized shape has sharp corners and 10 mm walls. Its area is 5600 mm². It is not a catalog representation of a manufactured tube with rounded corners.

Set a simply supported 3 m span, E = 200 GPa, a 1000 N downward point load at midspan and 500 N/m downward over the entire span. The line load must include any self-weight you intend to model; the calculator does not add it. Compare independent principal-axis load cases with the same load magnitudes. For y-axis bending, orient the section so the applied transverse load produces bending about y.

Both cases have 1250 N upward reaction at each end and a 1312.5 N·m maximum moment. The y-axis case deflects 3.092 times as much, while its stress is 1.546 times as large. The ratios differ because the extreme-fibre distance also changes.

Those beam links transfer only inertia and fibre distance. Review the span, support, material modulus and loads before pressing Calculate. The circular-section tool offers the same handoff for solid circles and concentric annuli.

For the simply supported case above, M maximum = PL/4 + wL²/8 and downward deflection maximum = PL³/(48EI) + 5wL⁴/(384EI), both at midspan. The cantilever option instead places P at the free tip: root moment magnitude = PL + wL²/2 and tip deflection = PL³/(3EI) + wL⁴/(8EI). These use superposition of the MIT standard elastic beam cases.

Within this model, doubling E halves deflection but leaves the statically determined moment and bending stress unchanged. Doubling both loads doubles reactions, stress and deflection. A negative load or an off-center point load is outside this tool’s implemented cases.

The model assumes a uniform homogeneous slender beam, constant EI, small deflection and principal-axis loading without torsion. A low calculated bending stress alone does not check buckling, fatigue, connections or code compliance. I and c do not determine shear-stress distribution, so this tool reports shear force only. Select material properties and applicable limits for the actual product and service conditions.