Uniform built-up compression members (6.4)

Many times it is advantageous to built-up (to make) a member (column) by joining two or more members. The joining the members together is provided at discrete points at distances of $a$ (it is not realized continuously along the length).

The common built-up members used for columns are built-ups with lacings and built-ups with battenings:

1—For the whole column an initial imperfection can be considered as $e_0 = L / 500$;
2—Built-up column with lacings (left) and built-up column with battenings (right)

Buckling of all types of members involved has to be taken into account.

Chords

Against axis $\mathbf{y}$

The asssessment is made as if the profile is a uniform beam. That means $I_y = 2 I_{y,1}$

Against axis $\mathbf{z}$

  1. Web is absent, so shear stiffness is lowered.
  2. It might happen that buckling appears
    • in the scope of the whole column,
    • in the scope of particular chord between stiffening members (located at distances $a$).

For both types of built-up columns

the assessment is made in the middle of the length. The verification is

$$ \begin{equation} \frac{N_{ch,Ed}}{N_{b,Rd}} \geq 1.0, \end{equation} $$
For both built-up types the assessment is made at the middle of the length

where $N_{ch,Ed}$ is design compression force in the chord at mid-length, $N_{b,Rd}$ is design value of buckling resistance.

$$ \begin{equation} N_{ch,Ed} \simeq \frac{N_{Ed}}{2} + \frac{M_{Ed}}{h_0} \end{equation} $$

In the code is more accurate

$$ \begin{equation} N_{ch,Ed} = \frac{1}{2} N_{Ed} + \frac{M_{Ed}}{W_{eff}}\cdot A_{ch}. \end{equation} $$

Term $N_{Ed}/2$ is external load divided into two chords and $M_{Ed}$ is taken as

$$ \begin{equation} M_{Ed} = \frac{N_{Ed}\cdot e_0 + M^1_{Ed}}{1-\frac{N_{Ed}}{N_{cr}} - \frac{N_{Ed}}{S_{V}}} \end{equation} $$

Moment $M^1_{Ed}$ only if an external moment is applied, term $N_{Ed}/N_{cr}$ involves deflections (2nd order effects) and the fraction with $S_V$ represents deflections from shear; $S_V$ is shear stifness of the lacings (for details see table in code), $N_{cr} = {\pi^2 EI_{eff}}/{L^2_{cr}}$.

For built-up columns with battenings

also assessment at the support has to be made. At the support the shear force produces moment on chord $(V_{Ed}/2)\cdot(a/2)$ and on battening the moment equals to double that moment (at a node sum of moments has to be zero).

For built-up type with battening the assessment is made also at the support. The picture depicts forces close to the top support on chords and on battening

That means assessment of the chord is made for combination of normal force $N_{ch,Ed}$ together with moment $M_{ch,Ed} = V_{Ed}\cdot a / 4$. Assessment for battening member for moment $M_{bat,Ed} = V_{Ed}\cdot a/2$.

Diagonals

Also diagonals are considered for buckling. Axial forces in diagonals are derived from the shear force $V_{Ed}$ at the support.

The shear force of the built-up columns at the support

To determine the shear force at the support, we start with realizing that the deflection shape $w(x)$ (according to Euler) is a sine. Deflected shape differentiated two times brings $M(x)$, further differentiation gives $V(x)$

In order to determine forces within diagonals we have to evaluate the shear force at the support firstly. The shear force at the support is then

$$ \begin{eqnarray} w(x) &=& f(x) = e_0 \sin \frac{\pi x }{L} \\ M(x) &=& \frac{d^2 w}{dx^2} = M_{Ed} \sin \frac{\pi x }{L} \\ V(x) &=& \frac{d M}{dx} = \frac{\pi M_{Ed}}{L} \cos\frac{\pi x}{L}\\ V_{Ed} &=& \frac{\pi M_{Ed}}{L} \end{eqnarray} $$

$M_{Ed}$ is known and $V_{Ed}$ is the largest possible from $V(x)$.


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