Figure P.20.3 shows the cross-section of a single cell, thin-walled beam with a horizontal axis of symmetry. The direct stresses are carried by the booms B1 to B4, while the walls are effective only in carrying shear stresses. Assuming that the basic theory of bending is applicable,

Question

Figure P.20.3 shows the cross-section of a single cell, thin-walled beam with a horizontal axis of symmetry. The direct stresses are carried by the booms [latex]B_1[/latex] to [latex]B_4[/latex], while the walls are effective only in carrying shear stresses. Assuming that the basic theory of bending is applicable, calculate the position of the shear centre S. The shear modulus G is the same for all walls.

Cell area = 135 000 mm².Boom areas: [latex]B_1=B_4=450 \mathrm{~mm}^2, B_2=B_3=550 \mathrm{~mm}^2[/latex]

[latex]\begin{array}{lll}\hline ext { Wall } & ext { Length }(\mathrm{mm}) & ext { Thickness }(\mathrm{mm}) \\hline 12,34 & 500 & 0.8 \23 & 580 & 1.0 \41 & 200 & 1.2 \\hline\end{array}[/latex]

Accepted Answer

The shear centre, S, lies on the horizontal axis of symmetry, the x axis. Therefore apply an arbitrary shear load, [latex]S_y[/latex], through S (Fig. S.20.3(a)). The internal shear flow distribution is given by Eq. (20.11) which, since [latex]I_{x y}=0, S_x=0 ext { and } t_{\mathrm{D}}=0[/latex], simplifies to

[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 450 imes 100^2+2 imes 550 imes 100^2=20 imes 10^6 \mathrm{~mm}^4[/latex]
Equation (i) then becomes

[latex]q_s=-0.5 imes 10^{-7} S_y \sum_{r=1}^n B_r y_r+q_{s, 0}[/latex] (ii)

The first term on the right-hand side of Eq. (ii) is the [latex]q_b[/latex] distribution (see Eq. (17.16)). To determine [latex]q_b[/latex] ‘cut’ the section in the wall 23. Then

[latex]q_s=q_{\mathrm{b}}+q_{s, 0}[/latex] (17.16)

[latex]\begin{aligned}&q_{\mathrm{b}, 23}=0 \&q_{\mathrm{b}, 34}=-0.5 imes 10^{-7} S_y imes 550 imes(-100)=2.75 imes 10^{-3} S_y=q_{\mathrm{b}, 12} \&q_{\mathrm{b}, 41}=2.75 imes 10^{-3} S_y-0.5 imes 10^{-7} S_y imes 450 imes(-100)=5.0 imes 10^{-3} S_y\end{aligned}[/latex]

The value of shear flow at the ‘cut’ is obtained using Eq. (17.28) which, since G = constant becomes

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

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

In Eq. (iii)

[latex]\oint \frac{\mathrm{d} s}{t}=\frac{580}{1.0}+2 imes \frac{500}{0.8}+\frac{200}{1.2}=19996.7[/latex]

Then, from Eq. (iii) and the above [latex]q_b[/latex] distribution

[latex]q_{s, 0}=-\frac{S_y}{1996.7}\left(2 imes \frac{2.75 imes 10^{-3} imes 500}{0.8}+\frac{5.0 imes 10^{-3} imes 200}{1.2} ight)[/latex]

i.e.
[latex]q_{s, 0}=-2.14 imes 10^{-3} S_y[/latex]
The complete shear flow distribution is shown in Fig. S.20.3(b).

Now taking moments about O in Fig. S.20.3(b) and using the result of Eq. (20.10)

[latex]M_q=2 A q_{12}[/latex] (20.10)

[latex]\begin{aligned}S_y \xi_{\mathrm{S}}=& 2 imes 0.61 imes 10^{-3} S_y imes 500 imes 100+2.86 imes 10^{-3} S_y imes 200 imes 500 \&-2.14 imes 10^{-3} S_y imes 2(135000-500 imes 200)\end{aligned}[/latex]

which gives
[latex]\xi_{\mathrm{S}}[/latex] = 197.2 mm

Question p.20.3: Figure P.20.3 shows the cross-section of a single cell, thin...

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