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## Q. 23.5

Calculate the deflection at the free end of the two-cell beam shown in Fig. 23.15 allowing for both bending and shear effects. The booms carry all the direct stresses while the skin panels, of constant thickness throughout, are effective only in shear.

Take E = 69 000 N/mm² and G = 25 900 N/mm²
Boom areas: $B_1=B_3=B_4=B_6=650 \mathrm{~mm}^2 \quad B_2=B_5=1300 \mathrm{~mm}^2$

## Verified Solution

The beam cross-section is symmetrical about a horizontal axis and carries a vertical load at its free end through the shear centre. The deflection Δ at the free end is then, from Eqs (20.17) and (20.19)

$\Delta_M=\frac{1}{E} \int_L\left(\frac{M_{y, 1} M_{y, 0}}{I_{y y}}+\frac{M_{x, 1} M_{x, 0}}{I_{x x}}\right) \mathrm{d} z$  (20.17)

$\Delta_S=\int_L\left(\int_{\text {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s\right) \mathrm{d} z$  (20.19)

$\Delta=\int_0^{2000} \frac{M_{x, 0} M_{x, 1}}{E I_{x x}} \mathrm{~d} z+\int_0^{2000}\left(\int_{\text {section }} \frac{q_0 q_1}{G t} \mathrm{~d} s\right) \mathrm{d} z$  (i)

where
$M_{x, 0}=-44.5 \times 10^3(2000-z) \quad M_{x, 1}=-(2000-z)$
and
$I_{x x}=4 \times 650 \times 125^2+2 \times 1300 \times 125^2=81.3 \times 10^6 \mathrm{~mm}^4$
also
$S_{y, 0}=44.5 \times 10^3 \mathrm{~N} \quad S_{y, 1}=1$

The $q_0 \text { and } q_1$ shear flow distributions are obtained as previously described (note dθ/dz = 0 for a shear load through the shear centre) and are

\begin{aligned}&q_{0,12}=9.6 \mathrm{~N} / \mathrm{mm} \quad q_{0,23}=-5.8 \mathrm{~N} / \mathrm{mm} \quad q_{0,43}=50.3 \mathrm{~N} / \mathrm{mm} \\&q_{0,45}=-5.8 \mathrm{~N} / \mathrm{mm} \quad q_{0,56}=9.6 \mathrm{~N} / \mathrm{mm} \quad q_{0,61}=54.1 \mathrm{~N} / \mathrm{mm} \\&q_{0,52}=73.6 \mathrm{~N} / \mathrm{mm} \text { at all sections of the beam }\end{aligned}

The $q_1$ shear flows in this case are given by $q_0 / 44.5 \times 10^3$. Thus

\begin{aligned}\int_{\text {section }} \frac{q_0 q_1}{G t} \mathrm{~d} s=& \frac{1}{25900 \times 2 \times 44.5 \times 10^3}\left(9.6^2 \times 250 \times 2+5.8^2 \times 500 \times 2\right.\\&\left.+50.3^2 \times 250+54.1^2 \times 250+73.6^2 \times 250\right) \\=& 1.22 \times 10^{-3}\end{aligned}

Hence, from Eq. (i)

$\Delta=\int_0^{2000} \frac{44.5 \times 10^3(2000-z)^2}{69000 \times 81.3 \times 10^6} \mathrm{~d} z+\int_0^{2000} 1.22 \times 10^{-3} \mathrm{~d} z$

giving
Δ = 23.5 mm