Question 21.1: A wing spar has the dimensions shown in Fig. P.21.1 and carr......
A wing spar has the dimensions shown in Fig. P.21.1 and carries a uniformly distributed load of 15 kN/m along its complete length. Each flange has a cross-sectional area of 500 mm² with the top flange being horizontal. If the flanges are assumed to resist all direct loads while the spar web is effective only in shear, determine the flange loads and the shear flows in the web at sections 1 and 2 m from the free end.
Referring to Fig. P.21.1, the bending moment at section 1 is given by
M_{1}={\frac{15\times1^{2}}{2}}=7.5\,\mathrm{kNm}Thus,
P_{z,U}=-P_{z,\mathrm{L}}={\frac{7.5}{300\times10^{-3}}}=25\,\mathrm{kN}Also,
P_{y,U}=0\mathrm{~and} P_{y,{\bf L}}=-25\times{\frac{100}{1\times10^{3}}}=-2.5\,\mathrm{kN~}\left(\mathrm{see\,Eqs}\left({21.1}\right)\right)\begin{array}{c}{{P_{y,1}=P_{z,1}\tan\alpha_{1}=P_{z,1}\displaystyle\frac{\delta y_{1}}{\delta z}}}\\ {{{}}}\\ {{P_{y,2}=-P_{z,2}\tan\alpha_{2}=-P_{z,2}\displaystyle\frac{\delta y_{2}}{\delta z}}}\end{array} (21.1)
Then
P_{U}\!=\!{\sqrt{P_{z,U}^{2}+P_{y,U}^{2}}}=25\,\mathrm{kN}\,(\operatorname{tension})P_{\mathrm{{L}}}=-{\sqrt{25^{2}+2.5^{2}}}=-25.1\,\mathrm{kN}({\mathrm{compression}})
The shear force at section 1 is 15\times1=15\mathrm{~kN}. This is resisted by P_{\mathrm{y,L}}, the shear force in the web. Thus,
shear in web ={15-2.5}={12.5\,\mathrm{kN}}
Hence,
q={\frac{12.5\times10^{3}}{300}}=41.7\mathrm{kN/mm}At section 2 the bending moment is
M_{2}={\frac{15\times2^{2}}{2}}=30\,\,{\mathrm{kNm}}Hence,
P_{z,\mathrm{U}}=-P_{z,\mathrm{L}}={\frac{30}{400\times10^{-3}}}=75\,\mathrm{kN}Also,
P_{y,U}=0{\mathrm{~and~}}P_{y,\mathrm{L}}=-75\times{\frac{200}{2\times10^{3}}}=-7.5\,\mathrm{kN},Then
P_{\mathrm{U}}=75\,{\mathrm{kN}}\,({\mathrm{tension}})and
P_{L}=-\sqrt{75^{2}+7.5^{2}}=-75.4\,\mathrm{kN}\ \ \ \ \mathrm{(compression)}The shear force at section 2 is 15\times2=30\,\mathrm{kN}. Hence the shear force in the web =30 – 7.5=22.5 kN, which gives
q={\frac{22.5\times10^{3}}{400}}=56.3\,\mathrm{N/mm}