Shear Centers for Thin- Wall Beam Cross Sections

Shear flow

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

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

Derivation

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

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

$$dF = \frac{dM}{I}\,Q$$

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

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

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

Shear centre

The shear centre (flexural centre) $S$ 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 $V$ about the same point.

Importance

If the load does not pass through $S$, 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 $S$ or add torsion allowance.

Location

Data

$h = 160$ mm, $t_w = 8$ mm, $b = 80$ mm, $t_f = 10$ mm, $V = 20$ kN; the channel is open to the right and symmetric about the horizontal axis.

   +-------- b = 80 -------+   t_f = 10
   |                              ^
   |  web t_w = 8                 | h = 160
   |                              v
   +-----------------------+
   S      (shear centre at distance e to left of web)

Second moment of area (thin-wall approximation)

$$I_x = \frac{t_wh^3}{12} + 2\,b\,t_f\left(\frac h2\right)^2 = 2730667 + 10240000 = 12970667\ \text{mm}^4$$

(The own inertia of the flanges, $bt_f^3/12$, is negligible.)

(a) Shear flow

Flange (distance $s$ from the free edge): $q = \dfrac{V\,(t_f s)(h/2)}{I_x}$, linear from zero at the tip to the maximum at the junction:

$$q_{junction} = \frac{20000\times(80\times10)\times80}{12970667} = 98.68\ \text{N/mm}$$

Web at the neutral axis (adds the half web):

$$q_{NA} = \frac{V}{I_x}\left[b\,t_f\frac h2 + t_w\frac h2\cdot\frac h4\right] = \frac{20000}{12970667}\left[64000 + 25600\right] = 138.16\ \text{N/mm}$$

The distribution in the web is parabolic between $98.7$ N/mm at the junctions and $138.2$ N/mm at mid-depth.

Maximum shear stress in the web: $\tau = q_{NA}/t_w = 17.27$ MPa.

(b) Position of the shear centre

The resultant force in each flange is the area of the triangle of the shear flow:

$$F_f = \frac12\,q_{junction}\,b = \frac12\times98.68\times80 = 3947\ \text{N}$$

The web force is $\approx V = 20$ kN (vertical). Take moments about the web centre-line: the web force passes through it, and the two flange forces ($F_f$, in opposite directions) form a couple $F_f\,h$. The load $V$ at distance $e$ from the web must give the same moment:

$$V\,e = F_f\,h \ \Rightarrow\ e = \frac{3947\times160}{20000} = 31.58\ \text{mm}$$

General expression:

$$e = \frac{b^2h^2t_f}{4I_x} = \frac{3b^2t_f}{h\,t_w + 6\,b\,t_f} = \frac{3\times6400\times10}{160\times8 + 6\times80\times10} = \frac{192000}{6080} = 31.58\ \text{mm}$$

The shear centre lies 31.6 mm to the left of the web centre-line, outside the section on the side opposite to the flanges.

(c) Check

$F_fh/V = 3947.4\times160/20000 = 31.58$ mm, equal to $e$. $\checkmark$

Answer: $q_{flange,max} = 98.7$ N/mm, $q_{web,max} = 138.2$ N/mm ($\tau_{max} = 17.3$ MPa); shear centre $e = 31.6$ mm from the web centre-line, on the side away from the flanges.

Data

$b = 120$ mm, $h = 200$ mm, $t = 5$ mm, $V = 30$ kN.

(a) Shear centre

The section is symmetric about both the vertical and horizontal axes, so the shear centre coincides with the centroid (centre of the box). A load through the vertical axis of symmetry therefore causes no twist.

Second moment of area

$$I_x = 2(b\,t)\left(\frac h2\right)^2 + 2\,\frac{t\,h^3}{12} = 1.200e+07 + 6.667e+06 = 1.8667e+07\ \text{mm}^4$$

(b) Shear flow

By symmetry the shear flow is zero at the middle of the top (and bottom) flange, where we make the imaginary cut. Then $q = VQ/I_x$ with $Q$ the first moment of the area from the cut.

Top flange, $s$ from the middle: $Q = t\,s\,\dfrac h2$

$$q_{corner} = \frac{30000}{1.8667e+07}\times(5\times60\times100) = 48.21\ \text{N/mm}$$

Web, $y$ below the corner: $Q = (5\times60\times100) + 5\,y\,(100 - y/2)$, which at the neutral axis ($y = 100$) is $30000 + 25000 = 55000$ mm$^3$:

$$q_{NA} = \frac{30000\times55000}{1.8667e+07} = 88.39\ \text{N/mm}$$

        0 ----- 48.2 ->     (top flange: linear)
        |                |
   88.4 |                | 88.4   (web: parabolic, max at NA)
        |                |
        0 ----- 48.2 ->

(c) Maximum shear stress and force in webs

$$\tau_{max} = \frac{q_{NA}}{t} = \frac{88.39}{5} = 17.68\ \text{MPa}$$

Vertical force in one web:

$$F_w = \frac{V}{I_x}\int_0^h \left[30000 + 5\,y(100 - y/2)\right]dy = 1.607\times10^{-3}\times9.333\times10^6 = 15000\ \text{N}$$

Two webs carry $30$ kN, i.e. 100 percent of $V$. The flanges carry equal and opposite horizontal forces that cancel and give no net vertical force.

Answer: Shear centre at the centroid; $q = 0$ at mid-flange, $48.2$ N/mm at the corners, $88.4$ N/mm at the neutral axis; $\tau_{max} = 17.7$ MPa; the two webs carry the whole 30 kN (15 kN each).