612 Flexuraltorsional buckling of thinwalled columns

It is recommended that the reading of this section be delayed until after Section 11.5 has been studied.

In some instances thin-walled columns of open cross-section do not buckle in bending as predicted by the Euler theory but twist without bending, or bend and twist simultaneously, producing flexural-torsional buckling. The solution of this type of problem relies on the theory presented in Section 11.5 for the torsion of open section beams subjected to warping (axial) restraint. Initially, however, we shall establish a useful analogy between the bending of a beam and the behaviour of a pin-ended column.

The bending equation for a simply supported beam carrying a uniformly distributed load of intensity it ,, and having C.v and Cv as principal centroidal axes is

Also, the equation for the buckling of a pin-ended column about the C.v axis is (see Eq. (6.1))

Fig. 6.21 Flexural-torsional buckling of a thin-walled column.

total movement of C, mc, in the x direction is given by uc = u + C'D = u + C'C" sin a (S'C'C" ~ 90°)

uc = u + 0CS sin a = u + ys9 Also the total movement of C in the y direction is vc = v- DC" = v- C'C" cos a = v- (9CS cos a so that vc = v- xs9

Since at this particular cross-section of the column the centroidal axis has been displaced, the axial load P produces bending moments about the displaced x and y axes given, respectively, by and

From simple beam theory (Section 9.1)

where Ixx and Iyy are the second moments of area of the cross-section of the column about the principal centroidal axes, E is Young's modulus for the material of the column and z is measured along the centroidal longitudinal axis.

The axial load P on the column will, at any cross-section, be distributed as a uniform direct stress a. Thus, the direct load on any element of length 6s at a point B(jcb,jb) is f'ds acting in a direction parallel to the longitudinal axis of the column. In a similar manner to the movement of C to C" the point B will be displaced to B". The horizontal movement of B in the jc direction is then

or mb = u + {ys - yB)9 Similarly the movement of B in the y direction is vb = « - (*s - *b)0

Therefore, from Eqs (6.76) and (6.77) and referring to Eqs (6.68) and (6.69), we see that the compressive load on the element 6s at B, atds, is equivalent to lateral loads

-aiSs j-j [w + (j>s - >>b)0] in the x direction and

—o~t6s——« \u — (xs — xB)#] in the y direction dz

The lines of action of these equivalent lateral loads do not pass through the displaced position S' of the shear centre and therefore produce a torque about S' leading to the rotation 9. Suppose that the element 6s at B is of unit length in the longitudinal z direction. The torque per unit length of the column 6T(z) acting on the element at B is then given by

Integrating Eq. (6.78) over the complete cross-section of the column gives the torque per unit length acting on the column, i.e.

T(z) = - f crt^(ys- yB) ds - [ at(ys - _yB)2 ds J Sect dz*2 J Sect dz^

+ f <rti-i(*s-*b)di- [ a/(xs-xB)2^|di (6.79) J Sect dZz J Sect dz"

Expanding Eq. (6.79) and noting that a is constant over the cross-section, we obtain d2u '

T{z)= -<j—1ys\ tds + cr—jl tyB ds - a—¿ids dz1 J Sect dz J Sect dzl J Sect d2e d2v f d20 f tyB ds - cr Í

+1 0

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