Question 21.3: Calculate the shear flow distribution and the stringer and f......
Calculate the shear flow distribution and the stringer and flange loads in the beam shown in Fig. P.21.3 at a section 1.5 m from the built-in end. Assume that the skin and web panels are effective in resisting shear stress only; the beam tapers symmetrically in a vertical direction about its longitudinal axis.
The beam section at a distance of 1.5 m from the built-in end is shown in Fig. S.21.3. The bending moment, M, at this section is given by
M=-40\times1.5=-60\,\mathrm{kNm}Since the x axis is an axis of symmetry \cdot I_{x y}=0;\mathrm{~also~}M_{y}=0. The direct stress distribution is then, 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}={\frac{M_{x}}{I_{xx}}}y (i)
in which I_{x x}{=}2\times1000\times1\,12.5^{2}+4\times500\times112.5^{2}=50.63\times10^{6}\,\mathrm{mm}^{4}. hen, from Eq. (i), the direct stresses in the flanges and stringers are
\sigma_{z}=\pm{\frac{60\times10^{6}\times112.5}{50.63\times10^{6}}}=\pm133.3\,\mathrm{N/mm^{2}\ ,}Therefore,
P_{z,1}=-P_{z,2}=-133.3\times1000=-133300\,\mathrm{N}and
P_{z,3}=P_{z,5}=-P_{z,4}=-P_{z,6}=-133.3\times500=-66650\,\mathrm{N}From Eq. (21.9),
P_{y,r}=P_{z,r}{\frac{\delta y_{r}}{\delta z}} (21.9)
P_{y,1}=P_{y,2}=133300\times{\frac{75}{3\times10^{3}}}=3332.5\mathrm{N}and
P_{y,3}=P_{y,4}=P_{y,5}=P_{y,6}=66650\times{\frac{75}{3\times10^{3}}}=1666.3\,\mathrm{N}Thus the total vertical load in the flanges and stringers is
2\times3332.5+4\times1666.3=13\,330.2\,\mathrm{N}Hence the total shear force carried by the panels is
40\times10^{3}-13\,330.2=26669.8\,\mathrm{N}The shear flow distribution is given by Eq. (20.11), which, since I_{x y}=0,S_{x}=0,{\mathrm{~and~}}t_{\mathrm{D}}=0, reduces to
q_{s}=-\left({\frac{S_{x}I_{x x}-S_{y}I_{x y}}{I_{x x}I_{y y}-I_{x y}^{2}}}\right) ]\left(\int_{0}^{s}t_{\mathrm{D}}x\,\mathrm{d}s+\sum\limits_{r=1}^{n}B_{r}x_{r}\right) -\left(\frac{S_{y}I_{y y}-S_{x}I_{x y}}{I_{x x}I_{y y}-I_{x y}^{2}}\right)\left(\int_{0}^{S}t_{\mathrm{D}}y\operatorname{d}\!s+\sum\limits_{r=1}^{n}B_{r}y_{r}\right)+q_{S,0} (20.11)
q_{s}=-{\frac{S_{y}}{I_{x x}}}\sum\limits_{r=1}^{n}B_{r}y_{r}+q_{s,0}i.e.,
q_{s}=-{\frac{26669.8}{50.63\times10^{6}}}\sum\limits_{r=1}^{n}B_{r}y_{r}+q_{s,0}or
q_{s}=-5.27\times10^{-4}\sum\limits_{r=1}^{n}B_{r}y_{r}+q_{s,0} (ii)
From Eq. (ii),
The remaining distribution follows from symmetry. Now taking moments about the point 2 (see Eq. (17.17)),
S_{x}\eta_{0}-S_{y}\xi_{0}=\oint\!p q_{\mathrm{b}}~\mathrm{d}s+2A q_{s,0} (17.17)
26\,669.8\times100=59.2\times225\times500+29.6\times250\times225+2\times500\times225q_{s,0}from which
q_{s,0}=-36.9\,\mathrm{N/mm}Then
Finally,
P_{1}=-\sqrt{P_{z,1}^{2}+P_{y,1}^{2}}=-\left(\sqrt{1333300^{2}+3332.5^{2}}\right)\times10^{-3}=-133.3\mathrm{kN}=-P_{2}P_{3}=-\sqrt{P_{z,3}^{2}+P_{y,3}^{2}}=-\left(\sqrt{66650^{2}+1666.3^{2}}\right)\times10^{-3} =-66.7\mathrm{kN}=P_{5}=-P_{4}=-P_{6}
