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: $B_1=B_3=B_4=B_6=650 \mathrm{~mm}^2 \quad B_2=B_5=1300 \mathrm{~mm}^2$

Question

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: $B_1=B_3=B_4=B_6=650 \mathrm{~mm}^2 \quad B_2=B_5=1300 \mathrm{~mm}^2$

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

Accepted Answer

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)

[latex]\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}} ight) \mathrm{d} z[/latex] (20.17)

[latex]\Delta_S=\int_L\left(\int_{ ext {sect }} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z[/latex] (20.19)

[latex]\Delta=\int_0^{2000} \frac{M_{x, 0} M_{x, 1}}{E I_{x x}} \mathrm{~d} z+\int_0^{2000}\left(\int_{ ext {section }} \frac{q_0 q_1}{G t} \mathrm{~d} s ight) \mathrm{d} z[/latex] (i)

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

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

[latex]\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} ext { at all sections of the beam }\end{aligned}[/latex]

The [latex]q_1 [/latex] shear flows in this case are given by [latex]q_0 / 44.5 imes 10^3[/latex]. Thus

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

Hence, from Eq. (i)

[latex]\Delta=\int_0^{2000} \frac{44.5 imes 10^3(2000-z)^2}{69000 imes 81.3 imes 10^6} \mathrm{~d} z+\int_0^{2000} 1.22 imes 10^{-3} \mathrm{~d} z[/latex]

giving
Δ = 23.5 mm

Question 23.5: Calculate the deflection at the free end of the two-cell bea...

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