Question p.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 \times 1^2}{2}=7.5 \mathrm{kN} \mathrm{m}Thus
P_{z, \mathrm{U}}=-P_{z, \mathrm{~L}}=\frac{7.5}{300 \times 10^{-3}}=25 \mathrm{kN}Also
P_{y, \mathrm{U}}=0 \quad \text { and } \quad P_{y, \mathrm{~L}}=-25 \times \frac{100}{1 \times 10^3}=-2.5 \mathrm{kN} (see Eqs (21.1))
P_{y, 1}=P_{z, 1} \frac{\delta y_1}{\delta z} \quad P_{y, 2}=-P_{z, 2} \frac{\delta y_2}{\delta z} (21.1)
Then
P_{\mathrm{U}}=\sqrt{P_{z, \mathrm{U}}^2+P_{y, \mathrm{U}}^2}=25 \mathrm{kN} (tension)
P_{\mathrm{L}}=-\sqrt{25^2+2.5^2}=-25.1 \mathrm{kN} (compression)
The shear force at section 1 is 15 × 1 = 15 kN. This is resisted by P_{y, \mathrm{~L}}, the shear force in the web. Thus
shear in web = 15 − 2.5 = 12.5 KN
Hence
q=\frac{12.5 \times 10^3}{300}=41.7 \mathrm{kN} / \mathrm{mm}At section 2 the bending moment is
M_2=\frac{15 \times 2^2}{2}=30 \mathrm{kN} \mathrm{m}Hence
P_{z, \mathrm{U}}=-P_{z, \mathrm{~L}}=\frac{30}{400 \times 10^{-3}}=75 \mathrm{kN}Also
P_{y, \mathrm{U}}=0 \quad \text { and } \quad P_{y, \mathrm{~L}}=-75 \times \frac{200}{2 \times 10^3}=-7.5 \mathrm{kN}Then
P_U = 75 kN (tension)
and
P_{\mathrm{L}}=-\sqrt{75^2+7.5^2}=-75.4 \mathrm{kN} (compression)
The shear force at section 2 is 15 × 2 = 30 kN. Hence the shear force in the web = 30 − 7.5 = 22.5 kN which gives
q=\frac{22.5 \times 10^3}{400}=56.3 \mathrm{~N} / \mathrm{mm}