Steel Buildings in Europe

Title Appendix B Worked Example: Design of a truss node with gusset 15 of 44 5 - 94 Buckling resistance The gusset is made similar to an embedded column of characteristics: Area 2 3 , 4297 ,5mm  g A Height h c = 112 mm (see Figure B.13) Second moment of area I c,zz = 80578 mm 2 We should satisfy: M1 3,g y 3,g,b,Rd 3,g,Ed   A f N N   Where  is the reduction factor for the relevant buckling curve EN 1993-1-1 6.3.1.1 With a buckling length of 2 h c , the slenderness is given by: c 2 c y 2 c 4 EI h A f    = 0,677 The buckling curve to use is curve c and the imperfection is:  = 0,49     2 0,2) 0,51         = 0,846 2 2 1        = 0,739 Table 6.1 EN 1993-1-1 6.3.1.2 Then: 1127 kN 609 4 3,g,b,Rd 3,g,Ed    N N , 3.3.4. Connection N3 – Checking of bolts with regard to the gusset component Design shear force F V,Ed for each bolt Due to the orientation of the axial force N 3,Ed , the load on each bolt is not parallel to the edge of gusset. Also, the components of the design shear load will be performed in a suitable basis. EN 1993-1-8 Table 3.4 3) In first the components are calculated in the basis   h v   , located at the centre of gravity of the joint and oriented in agreement with the principal directions of the fasteners which are also the principal directions of the angles (See Figure B.14). Then a change of basis is performed from the initial   h v   , to the basis   h v , (see Figure B.15). In the basis   h v   , the normal force N 3,g,Ed causes a horizontal shear load for each bolt b i : 5 3,g,Ed N,bi,h N F   = 101,57 kN