**C. The Design Resistance of the Joint.**

In the case of 50% < λ0*v*= 52.1% < 80%:

$$p\_{1,eff} = t\_w + 2r + 7t\_f \\ f\_{y0}/f\_{y1} = 6.5 + 2 \cdot 12 + 7 \cdot 11 \cdot 355/355 = 107.5 \text{ mm} \\ \therefore \, b\_1 = 50 \text{ mm}$$

Adopted *p*1,*eff* = 50 mm.

$$b\_{\varepsilon,av} = \frac{10}{b\_j/t\_j} \frac{f\_{\circ j}t\_j}{f\_{y\circ}t\_i} \\ b\_i = \frac{10}{60/4} \frac{355 \cdot 4}{355 \cdot 3} \cdot 50 = 44.4 \text{ mm} \quad < b\_1 = 50 \text{ mm}.$$

The brace failure.

$$\begin{array}{c} \text{N}\_{1,\text{Rd}} = f\_{\text{y}1}t\_1(p\_{1,eff} + b\_{\text{c,av}} + h\_1 - 2t\_1) / \gamma\_{\text{M5}} = \\ 355 \cdot 3(50 + 44.4 + 60 - 2 \cdot 3) / 1.0 = 158000 \text{ N} = 158.0 \text{ kN} \end{array}$$

$$N\_{2,Rd} = N\_{1,Rd} \frac{\sin \theta\_1}{\sin \theta\_2} = 158.0 \cdot \frac{0.7698}{0.6431} = 189.1 \text{ kN}.$$

Checking the braces resistance.

$$\frac{103.2}{158} = 0.65 < 1.0, \; \frac{136.1}{189.1} = 0.72 < 1.0.$$

The brace bending and the axial force resistance.

$$M\_0 = \pm 0.5(N\_{02.Ed} - N\_{01.Ed})e = \pm 0.5 \cdot (-159.9 + 121.6) \cdot 30 = \pm 574.5 \text{ kNmm},$$

$$M\_{0,pl} = W\_{pl.0} \cdot f\_{\text{y}} / \gamma\_{M1} = 165.2 \cdot 10^3 \cdot 355 / 1.0 = 586.5 \cdot 10^2 \text{ kNmm},$$

$$N\_{0,pl} = A\_0 \cdot f\_{\text{y}} / \gamma\_{M1} = 34.0 \cdot 10^2 \cdot 355 / 1.0 = 1207 \cdot 10^3 \text{ N} = 1207 \text{ kN},$$

$$\frac{N\_0}{N\_{0,pl}} + \frac{M\_0}{M\_{0,pl}} = \frac{159.9}{1207} + \frac{574.5}{58650} = 0.14 < 1.0.$$
