Shear flow in a closed box beam

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

$$I_x = 2(b\,t)\left(\frac h2\right)^2 + 2\,\frac{t\,h^3}{12} = 1.200\times10^{7} + 6.667\times10^{6} = 1.8667\times10^{7}\ \text{mm}^4$$

(b) Shear flow

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

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

$$q_{corner} = \frac{30000}{1.8667\times10^{7}}\times(5\times60\times100) = 48.21\ \text{N/mm}$$

Web, $y$ measured down from the corner: $Q = (5\times60\times100) + 5\,y\,(100 - y/2)$. At the neutral axis ($y = 100$) this gives $30000 + 25000 = 55000$ mm$^3$:

$$q_{NA} = \frac{30000\times55000}{1.8667\times10^{7}} = 88.39\ \text{N/mm}$$

        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:

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

The two webs carry $30$ kN, which is 100 percent of $V$. The flanges carry equal and opposite horizontal forces that cancel, so they give no net vertical force.

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