Out of plane: bending becomes torsion

This is a static copy of the chapter for search engines. The interactive version has animated figures, check questions, and buttons that load the example into the calculator.

Everything so far kept the loads inside the plane of the frame. Now push the bar sideways, out of that plane. This is the case that makes bent beams genuinely different from straight ones — and it is the case that catches people out, because the ordinary beam formula gives an answer that is badly wrong.

Load a bent bar perpendicular to its plane and the corner turns bending into torsion. This is the situation a straight beam simply cannot produce — and it is why bent brackets so often fail in a way the beam formula does not predict.

The mechanism

Take an L-shaped bracket: a column of height b, an arm of length a, clamped at the base, with a force F_z pushing on the tip perpendicular to the plane.

In the arm, F_z is a plain transverse load. The arm bends about its own normal: M_⊥ = F_z·(distance to the tip), and there is no torsion at all.
At the corner, the moment vector delivered by the arm points along the axis of the column. A moment along the axis of a member is a torque. So the column is twisted by T = F_z·a — constant all the way down.
The column also bends, because F_z itself still has to travel down to the base: M_⊥ = F_z·(distance to the base), maximum F_z·b at the clamp.

So the base of the column carries bending and torsion simultaneously, and neither can be ignored.

The tip deflection

The tip moves out of plane by three separate contributions, which simply add up:

w = F_z·a³3EI_⊥ + F_z·a²·bGJ_t + F_z·b³3EI_⊥

bending of the arm, twist of the column seen at the end of the lever arm a, and bending of the column. The middle term is the one a beam calculation misses entirely, and for an open section it is often the largest of the three.

Why the section shape suddenly matters enormously

The torsional stiffness of an open section (an I-beam, a channel, an angle, a T) is pitifully small: J_t ≈ Σ b_it_i³/3. A closed section (a tube, a box) gets J_t from Bredt’s formula and is typically one to three orders of magnitude stiffer in torsion at the same weight.

Section	≈ J_t	Comment
Ø80 × 5 tube	≈ 260 cm⁴	closed — excellent in torsion
100 × 60 × 5 box	≈ 150 cm⁴	closed
IPE 160	≈ 3.6 cm⁴	open — roughly 70× worse than the tube

The rule of thumb is short and blunt: if a bent bar is loaded out of its plane, use a closed section.

Example — the same crank, two sections

First a Ø80 × 5 tube, then an IPE 160 of comparable weight. The internal actions are identical (the structure is statically determinate), but look at the angle of twist and the out-of-plane deflection.

⚠ FrameLab computes St Venant torsion only. In a real open section, warping is restrained at a clamp and at a rigid corner, which makes the member stiffer than computed and adds a warping normal stress σ_ω in the flange tips. The computed twist is therefore an upper bound and the computed shear stress is on the safe side, but the flange-tip normal stress is not included. For open sections in torsion, treat the result as a first estimate.

Example — a double kink

A Z-shaped hook loaded out of plane: torsion appears in the middle leg only, and disappears again after the second corner.