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Shear flow and shear centre in thin-wall beams

Short answer

Shear flow q = τ t is shear force per unit length, derived as q = VQ/I. The shear centre is the point where a transverse load causes bending without twist; loads away from it induce torsion. For an angle, it lies at the leg junction; for a T-section, at the flange-web intersection; for a channel, outside the web away from flanges.

Define shear flow and shear centre. Derive the expression for shear flow in a thin-walled open beam section, and explain the importance of locating the shear centre. State where the shear centre lies for an angle, a T-section and a channel section.

Shear flow

For a thin-walled section, the shear stress is assumed constant across the thickness and parallel to the wall. Shear flow qq is the shear force per unit length of the wall:

q=τ t(N/mm)q = \tau\,t\quad (\text{N/mm})

Derivation

Consider a beam element of length dzdz under shear force VV. Cut the section at a point and consider the part of the area A∗A^* beyond the cut, with first moment Q=∫A∗y dAQ = \int_{A^*} y\,dA about the neutral axis.

The bending moments on the two faces are MM and M+dMM + dM, so the normal forces on the cut part of the faces differ by

dF=dMI QdF = \frac{dM}{I}\,Q

This is balanced by the shear force on the longitudinal cut, q dzq\,dz:

q dz=dMI Q⇒q=V QI(since V=dMdz)q\,dz = \frac{dM}{I}\,Q \quad\Rightarrow\quad \boxed{q = \frac{V\,Q}{I}}\quad\left(\text{since } V = \frac{dM}{dz}\right)

and τ=VQIt\tau = \dfrac{VQ}{It}. The shear flow is zero at a free edge and changes linearly in a flange of constant thickness (flange shear flow q=V t s yf/Iq = V\,t\,s\,y_f/I), and parabolically in a web.

Shear centre

The shear centre (flexural centre) SS is the point on the cross-section through which a transverse load must act so that the beam bends without twisting. Its position is found by equating the moment of the shear flow resultants about any point to the moment of the applied shear force VV about the same point.

Importance

If the load does not pass through SS, the section twists as well as bends. Thin open sections have very low torsional stiffness, so the twist can be large and the stresses higher than for pure bending. Designers therefore place the load line through SS or add torsion allowance.

Location

SectionShear centre
Doubly symmetric (I, box, rectangle)At the centroid
One axis of symmetry (channel, T)On the axis of symmetry; channel: outside the web, on the side away from the flanges
T-sectionAt the intersection of the flange mid-line and the web mid-line (the shear flows in the two parts meet there)
Angle section (equal or unequal legs)At the junction of the two legs (corner), as the shear flow resultants of both legs pass through that point
Channel (open towards right)At distance ee to the left of the web centre-line, e=3b2tfh tw+6btfe = \dfrac{3b^2t_f}{h\,t_w + 6bt_f}

✓ Answer checked by Claude Sonnet 5.5

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