Idealize the box section shown in Fig. P.20.1 into an arrangement of direct stress carrying booms positioned at the four corners and panels which are assumed to carry only shear stresses. Hence determine the distance of the shear centre from the left-hand web.

Question

Accepted Answer

From either Eq. (20.1) or (20.2)

[latex]B_1=\frac{t_{\mathrm{D}} b}{6}\left(2+\frac{\sigma_2}{\sigma_1} ight)[/latex] (20.1)

[latex]B_2=\frac{t_{\mathrm{D}} b}{6}\left(2+\frac{\sigma_1}{\sigma_2} ight)[/latex] (20.2)

[latex]B_1=60 imes 10+40 imes 10+\frac{500 imes 10}{6}(2+1)+\frac{300 imes 10}{6}(2-1)[/latex]

i.e.

[latex]\begin{aligned}&B_1=4000 \mathrm{~mm}^2=B_4 \&B_2=50 imes 8+30 imes 8+\frac{500 imes 10}{6}(2+1)+\frac{300 imes 8}{6}(2-1)\end{aligned}[/latex]

i.e.

[latex]B_2=3540 \mathrm{~mm}^2=B_3[/latex]

Since the section is now idealized, the shear flow distribution due to an arbitrary shear load [latex]S_y[/latex] applied through the shear centre is, from Eq. (20.11), given by

[latex]\begin{aligned}q_s=&-\left(\frac{S_x I_{x x}-S_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} x \mathrm{~d} s+\sum_{r=1}^n B_r x_r ight) \&-\left(\frac{S_y I_{y y}-S_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2} ight)\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+\sum_{r=1}^n B_r y_r ight)+q_{s, 0}\end{aligned}[/latex] (20.11)

[latex]q_s=-\frac{S_y}{I_{x x}} \sum_{r=1}^n B_r y_r+q_{s, 0}[/latex] (i)

in which
[latex]I_{x x}=2 imes 4000 imes 150^2+2 imes 3540 imes 150^2=339 imes 10^6 \mathrm{~mm}^4 .[/latex]
‘Cut’ the section in the wall 12. Then

[latex]\begin{aligned}q_{\mathrm{b}, 12} &=q_{\mathrm{b}, 43}=0 \q_{\mathrm{b}, 41} &=-\frac{S_y}{I_{x x}} imes 4000 imes(-150)=1.77 imes 10^{-3} S_y \q_{\mathrm{b}, 32} &=-\frac{S_y}{I_{x x}} imes 3540 imes(-150)=1.57 imes 10^{-3} S_y\end{aligned}[/latex]

Since the shear load is applied through the shear centre the rate of twist is zero and [latex]q_{s,0}[/latex] is given by Eq. (17.28) in which

[latex]q_{s, 0}=-\frac{\oint q_{\mathrm{b}} \mathrm{d} s}{\oint \mathrm{d} s}[/latex] (17.28)

[latex]\oint \frac{\mathrm{d} s}{t}=2 imes \frac{500}{10}+\frac{300}{10}+\frac{300}{8}=167.5[/latex]

Then

[latex]q_{s, 0}=-\frac{1}{167.5} S_y\left(1.57 imes 10^{-3} imes \frac{300}{8}-1.77 imes 10^{-3} imes \frac{300}{10} ight)[/latex]

which gives
[latex]q_{s, 0}=-0.034 imes 10^{-3} S_y[/latex]

The complete shear flow distribution is then as shown in Fig. S.20.1(b).

Taking moments about the intersection of the horizontal axis of symmetry and the left-hand web
[latex]S_y x_{\mathrm{S}}=1.536 imes 10^{-3} S_y imes 300 imes 500-2 imes 0.034 imes 10^{-3} S_y imes 500 imes 150[/latex]
from which
[latex]x_S[/latex] = 225 mm

Question p.20.1: Idealize the box section shown in Fig. P.20.1 into an arrang...

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