Question 21.2: If the web in the wing spar of Problem 21.1 has a thickness ......
If the web in the wing spar of Problem 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\times1^{2}}{2}}=7.5\,\mathrm{kNm}The second moment of area of the beam cross-section at section 1 is
I_{x x}=2\times500\times150^{2}+{\frac{2\times300^{3}}{12}}=2.7\times10^{7}\,{\mathrm{mm}}^{4}The direct stresses in the flanges in the z direction are, from Eq. (16.17),
\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 (16.17)
\sigma_{z,U}=-\sigma_{z,\mathrm{L}}={\frac{7.5\times10^{6}\times150}{2.7\times10^{7}}}=41.7\,\mathrm{N/mm^{2}}Then
P_{z,U}=41.7\times500=20850\,\mathrm{N}=P_{U} (tension)
Also,
P_{\mathrm{z,L}}=-20850\,\mathrm{N} (compression)
Hence,
P_{y, L}=-20850\times\frac{100}{1\times10^{3}}=-2085\,\mathrm{N} (compression)
Therefore, the shear force in the web at section 1 is given by
S_{y}=-15\times1\times10^{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\times10^{7}}\left[\int_{0}^{s}2(150-s)\mathrm{d}s+500\times150\right]Hence,
q=4.8\times10^{-4}\bigl(300s-s^{2}+75000\bigr)The maximum value of q occurs when s=150 mm, i.e.,
q_{\mathrm{max}}=46.8\,\mathrm{N/mm}