Question p.21.2: If the web in the wing spar of P.21.1 has a thickness of 2 m...
If the web in the wing spar of P.21.1 has a thickness of 2 mm and is fully effective in resisting direct stresses, calculate the maximum value of shear flow in the web at a section 1 m from the free end of the beam.
The bending moment at section 1 is given by
M=\frac{15 \times 1^2}{2}=7.5 \mathrm{kN} \mathrm{m}The second moment of area of the beam cross-section at section 1 is
I_{x x}=2 \times 500 \times 150^2+\frac{2 \times 300^3}{12}=2.7 \times 10^7 \mathrm{~mm}^4The direct stresses in the flanges in the z direction are, from Eq. (16.18)
\sigma_z=\left(\frac{M_y I_{x x}-M_x I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right) x+\left(\frac{M_x I_{y y}-M_y I_{x y}}{I_{x x} I_{y y}-I_{x y}^2}\right) y (16.18)
\sigma_{z, \mathrm{U}}=-\sigma_{z, \mathrm{~L}}=\frac{7.5 \times 10^6 \times 150}{2.7 \times 10^7}=41.7 \mathrm{~N} / \mathrm{mm}^2Then
P_{z, \mathrm{U}}=41.7 \times 500=20850 \mathrm{~N}=P_{\mathrm{U}} (tension)
Also
P_{z, \mathrm{~L}}=-20850 \mathrm{~N} (compression)
Hence
P_{y, \mathrm{~L}}=-20850 \times \frac{100}{1 \times 10^3}=-2085 \mathrm{~N} (compression)
Therefore, the shear force in the web at section 1 is given by
S_y=-15 \times 1 \times 10^3+2085=-12915 \mathrm{~N}
The shear flow distribution is obtained using Eq. (21.6). Thus, referring to Fig. S.21.2
q_s=-\frac{S_{y, w}}{I_{x x}}\left(\int_0^s t_{\mathrm{D}} y \mathrm{~d} s+B_1 y_1\right) (21.6)
q=\frac{12915}{2.7 \times 10^7}\left[\int_0^s 2(150-s) \mathrm{d} s+500 \times 150\right]Hence
q=4.8 \times 10^{-4}\left(300 s-s^2+75000\right)
The maximum value of q occurs when s = 150 mm, i.e.
q_{\max }=46.8 \mathrm{~N} / \mathrm{mm}
