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Chapter 24

Q. p.24.1

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 and the stiffener BE and the shear flows in the panels.

Screenshot 2022-10-11 131940

Step-by-Step

Verified Solution

From the overall equilibrium of the beam in Fig. S.24.1(a)
R_{\mathrm{F}}=4 \mathrm{kN} \quad R_{\mathrm{D}}=2 \mathrm{kN}
The shear load in the panel ABEF is therefore 4 kN and the shear flow q is given by

q_1=\frac{4 \times 10^3}{1000}=4 \mathrm{~N} / \mathrm{mm}

Similarly

q_2=\frac{2 \times 10^3}{1000}=2 \mathrm{~N} / \mathrm{mm}

Considering the vertical equilibrium of the length h of the stiffener BE in Fig. S.24.1(b)

P_{\mathrm{EB}}+\left(q_1+q_2\right) h=6 \times 10^3

where P_{\mathrm{EB}} is the tensile load in the stiffener at the height h, i.e.
P_{\mathrm{EB}}=6 \times 10^3-6 h (i)
Then from Eq. (i), when h = 0, P_{\mathrm{EB}} = 6000 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

P_{\mathrm{AB}} \times 1000+R_{\mathrm{F}} z=0
whence
P_{\mathrm{AB}} = −4z 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.

Screenshot 2022-10-11 132341
Screenshot 2022-10-11 132404