Shear centre of a channel section

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.