MachineCalcs

Square Tube Deflection Calculator

Bending deflection, stress and second moment of area of a square or rectangular hollow section (HSS tube) used as a beam — simply-supported central load or cantilever end load. Metric and imperial. Free, no signup.

Calculator

Support condition and load: simply supported with a central point load, or a cantilever with the load at the free end.

Outside width of the tube — the dimension along the bending axis.

mm

Outside height (depth) of the tube — the dimension that resists bending. Orient the larger dimension vertically.

mm

Wall thickness. Must be less than half of both b and h.

mm

Span between supports (simply supported) or the cantilever length from the fixed end.

mm

Applied point load — at mid-span (simply supported) or at the free end (cantilever).

N

Elastic modulus of the material — steel ≈ 200 GPa, aluminum ≈ 69 GPa.

GPa

Results

Default result
Edit inputs
Max deflection(δ)
0.9992mm

δ = F·L³ / (48·E·I)

Maximum elastic deflection at the load point.

Also computed

Max bending stress(σ)59.95MPa

M = F·L/4

σ = M·c / I at the extreme fibre (c = h/2).

Second moment of area(I)20.85cm⁴

I = (b·h³ − bᵢ·hᵢ³) / 12 about the bending axis.

Method notes 4 notes
  • Simply supported, central point load. Second moment of area I = (b·h³ − bᵢ·hᵢ³)/12 with bᵢ = b − 2t and hᵢ = h − 2t.
  • Simply supported with a central point load: δ = F·L³/(48·E·I), maximum moment M = F·L/4 at mid-span.
  • Max bending stress σ = M·c / I with c = h/2 (distance to the extreme fibre). Bending is about the axis parallel to width b — the depth h resists bending, so orient the larger dimension vertically.
  • Euler–Bernoulli elastic beam theory — small deflections, material below yield, self-weight ignored. Add the beam self-weight as a distributed load for long, lightly loaded spans.

A square or rectangular hollow section used as a beam has area moment I = (b·h³ − bᵢ·hᵢ³)/12 for outside b×h and wall t (bᵢ = b−2t, hᵢ = h−2t). It deflects δ = F·L³/(48·E·I) under a simply-supported central load (or F·L³/(3·E·I) as a cantilever), with peak bending stress σ = M·c/I at c = h/2. This calculator returns deflection, stress and I.

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All Structural

How to use this calculator

  1. Pick the load case. Choose simply supported with a central point load, or a cantilever with the load at the free end.
  2. Enter the section size. Enter the outside width b, outside height h and wall thickness t of the tube.
  3. Enter the span and load. Enter the span (or cantilever length) L, the point load F, and the material’s Young’s modulus E.
  4. Read the results. Read the maximum deflection δ, maximum bending stress σ and the second moment of area I.

How it works

A square or rectangular hollow section (HSS tube) used as a beam resists bending through its second moment of area. For a hollow rectangle it is the outer rectangle minus the inner void: I = (b·h³ − bᵢ·hᵢ³) / 12, with bᵢ = b − 2t and hᵢ = h − 2t. The depth h enters as a cube, so a deeper tube is dramatically stiffer for the same wall.

The maximum elastic deflection then depends on how the beam is supported. Simply supported with a central point load: δ = F·L³ / (48·E·I). A cantilever with the load at the free end: δ = F·L³ / (3·E·I). The peak bending stress is σ = M·c / I, with c = h/2 and the maximum moment M = F·L/4 (simply supported) or M = F·L (cantilever). Because N·mm / mm⁴ × mm = MPa, the stress comes straight out in MPa.

Worked example

Verified against the live calculator

A 50×50×3 mm steel tube (E = 200 GPa), simply supported over L = 1,000 mm with a 2,000 N central load. The inner dimensions are bᵢ = hᵢ = 50 − 2×3 = 44 mm, so I = (50·50³ − 44·44³)/12 = (6,250,000 − 3,748,096)/12 ≈ 208,492 mm⁴. The deflection is δ = 2000 × 1000³ / (48 × 200,000 × 208,492) ≈ 0.999 mm. The moment is M = F·L/4 = 500,000 N·mm, so the bending stress is σ = 500,000 × 25 / 208,492 ≈ 59.95 MPa. The calculator returns exactly these numbers.

Frequently asked questions

How do I calculate the deflection of a square tube?

First find the second moment of area of the hollow section, I = (b·h³ − bᵢ·hᵢ³)/12, where bᵢ = b − 2t and hᵢ = h − 2t. Then for a simply-supported beam with a central point load, δ = F·L³/(48·E·I); for a cantilever with an end load, δ = F·L³/(3·E·I). For a 50×50×3 mm steel tube spanning 1,000 mm under a 2,000 N central load, I ≈ 208,492 mm⁴ and δ ≈ 1.0 mm.

What is the second moment of area of a rectangular tube?

A rectangular hollow section is the outer rectangle minus the inner void, so I = (b·h³ − bᵢ·hᵢ³)/12 about the horizontal axis, with bᵢ = b − 2t and hᵢ = h − 2t. The depth h enters as a cube, so making the tube taller (greater h) stiffens it far more than making it wider for the same wall thickness.

Simply supported or cantilever — which load case do I use?

Use simply supported (central point load) when the tube rests on a support at each end and the load sits at mid-span — δ = F·L³/(48·E·I). Use cantilever when one end is fixed (welded or clamped) and the load hangs off the free end — δ = F·L³/(3·E·I). For the same span and load the cantilever deflects 16× as much, because it is far less restrained.

How do I find the bending stress in the tube?

Bending stress is σ = M·c/I, where M is the maximum bending moment and c = h/2 is the distance to the extreme fibre. The moment is M = F·L/4 for a simply-supported central load, or M = F·L at the fixed end of a cantilever. Compare σ against the material yield strength with a safety factor.

Which way should I orient a rectangular tube?

Orient the larger dimension (greater depth) vertically, in line with the load. Because depth enters the inertia as h³, a 40×80 tube stood on its 40 mm face is far stiffer and stronger than the same tube laid on its 80 mm face — same material, very different deflection. This calculator bends about the axis parallel to the width b.

Does the calculator include the tube's own weight?

No — it models a single point load and ignores self-weight, which keeps the formulas exact and transparent. For long, lightly-loaded spans the beam's own mass matters; add it as a separate distributed-load deflection. The result is also elastic (Euler–Bernoulli) and valid only while the material stays below yield.

Method & assumptions

  • Euler–Bernoulli elastic beam theory — small deflections, plane sections remain plane, and the material stays below yield.
  • Single concentrated point load only: at mid-span (simply supported) or at the free end (cantilever). Self-weight is ignored — add it as a distributed load for long, lightly-loaded spans.
  • Bending is about the axis parallel to the width b; the depth h resists bending. Orient the larger (deeper) dimension vertically to use the section efficiently.
  • I = (b·h³ − bᵢ·hᵢ³)/12 assumes a true hollow rectangle with uniform wall t and sharp corners; real HSS has rounded corners, so the catalogue I is slightly different.
  • No shear deflection, local buckling, web crippling or stress concentration at the load or supports — slender beams only.
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