Question 24.1: The beam shown in Fig. P.24.1 is simply supported at each en......
The beam shown in Fig. P.24.1 is simply supported at each end and carries a load of 6000 N. If all direct stresses are resisted by the flanges and stiffeners and the web panels are effective only in shear, calculate the distribution of axial load in the flange ABC, the stiffener BE, and the shear flows in the panels.
From the overall equilibrium of the beam in Fig. S.24.1(a),
R_{ F }=4 kN R_{ D }=2 kNThe shear load in the panel ABEF is therefore 4 kN and the shear flow q is given by
q_{1}={\frac{4\times10^{3}}{1000}}=4\mathrm{N/mm}Similarly,
q_{2}={\frac{2\times10^{3}}{1000}}=2\mathrm{N/mm}Considering the vertical equilibrium of the length h of the stiffener BE in Fig. S.24.1(b),
P_{\mathrm{{EB}}}+(q_{1}+q_{2})h=6\times10^{3}where P_{\mathrm{EB}} is the tensile load in the stiffener at the height h, i.e.,
P_{\mathrm{{EB}}}=6\times10^{3}-6h (i)
Then from Eq. (i), when h=0, P_{\mathrm{EB}}=6000\mathrm{\scriptsize~N}, and when h=1000 mm, P_{\mathrm{EB}}=0. Therefore the stiffener load varies linearly from zero at B to 6000 N at E.
Consider now the length z of the beam in Fig. S.24.1(c). Taking moments about the bottom flange at the section z,
whence,
P_{\mathrm{AB}}=-4z\mathrm{N}Thus P_{\mathrm{AB}} varies linearly from zero at A to 4000 N (compression) at B. Similarly, P_{\mathrm{{CB}}} varies linearly from zero at C to 4000 N (compression) at B.
