Page 466 - Aircraft Stuctures for Engineering Student

11.2 Built-in end of a closed section beam 447

I.Ornm 2
22 000 N

vL

- U
1.6rnrn
4
1.0 rnrn
100 rnrn
(a) (b)
Fig. 11.2 (a) Beam cross-section at built-in end; (b) notation and sign convention.

It is helpful at the start of the problem to sketch the notation and sign convention
as shown in Fig. 11.2(b). The walls of the beam are flat and therefore p and $J are
constant along each wall. Also the thickness of each wall is constant so that the
shear flow q is independent of s in each wall. Let point 1 be the origin of the axes,
then, writing 13‘ = dO/dz, u’ = du/& and v‘ = dv/dz, we obtain from Eq. (1 1.1)
q12 = 1.6Gv’ (i)
q23 = i.o~(m 0.8868’ - 0.886~’ - 0.5~’) (E)
x
q34 = i.2~poo 0.866e’ - .~r) (iii)
x
q41 = 1.OGu’ (4
For horizontal equilibrium
500 x 0.886q41 - 500 x O.866q23 = 0
giving

q41 = q23
For vertical equilibrium
375q12 - 125q34 - 25oq23 = 22 000
For moment equilibrium about point 1
500 x 375 x 0.886q23 + 125 x 500 x 0.886q34 = 22000 x 100
or

3q23 + q34 = 40.6 (vii)
Substituting for qlz etc. from Eqs (i), (ii), (iii) and (iv) into Eqs (v), (vi) and (vii), and
solving for 8‘, u‘ alid v’, gives 8‘ = 0.122/G, u‘ = 9.71/G, v‘ = 42.9/G. The values of#,
u’ and v’ are now inserted in Eqs (i), (ii), (iii) and (iv), giving qI2 = 68.5N/mm,
q23 = 9.8 N/mm, q34 = 11.9 N/mm, q41 = 9.8 N/mm from which
q2 = 42.8 N/mm2, ~23 = 7-41 = 9.8 N/mm2: r34 = 9.9 N/mm2