Steel Buildings in Europe

Title Appendix B Worked Example: Design of a truss node with gusset 17 of 44 5 - 96 This shear load F M,bi is resolved in the basis   h v   , :      5 1 2 i i 1,a,Ed M,bi,h r v M F horizontal component     5 1 2 i i 1,a,Ed v' M,bi, r M h F vertical component With i h  and i v  coordinates of centre of bolt b i . And we obtain (see Table B.2): M,bi,h N,bi,h V,bi,h ,Ed      F F F Horizontal shear force, M,bi,v V,bi,v ,Ed    F F Transverse shear force, 2 V,bi,v ,Ed 2 V,bi,h ,Ed V,bi,Ed     F F F Resulting shear force Table B.2 Connection N3 – Gusset component – Design shear forces in kN in the basis   h v   , . Bolt b 1 b 2 b 3 b 4 b 5 b 6 i h  81,25 16,25 -48,75 48,75 -16,25 -81,25 i v  -30 -30 -30 30 30 30 i r  86,61 34,12 57,24 57,24 34,12 86,61 M,bi F -98,34 -38,74 -64,99 -64,99 -38,74 -98,34 M,bi,h  F 34,06 34,06 34,06 -34,06 -34,06 -34,06 M,bi,v  F 92,25 18,45 -55,35 55,35 -18,45 -92,25 N,bi F 101,57 101,57 101,57 101,57 101,57 101,57 V,bi,Ed F 164,03 136,88 146,49 87,30 69,98 114,31 V,bi,h ,Ed  F 135,63 135,63 135,63 67,50 67,50 67,50 V,bi,v ,Ed  F 92,25 18,45 -55,35 55,35 -18,45 -92,25 The change of basis is performed with: ) cos( ) sin( 3 V,bi,v ,Ed 3 V,bi,h ,Ed V,bi,h,Ed        F F F ) sin( ) cos( 3 V,bi,v ,Ed 3 V,bi,h ,Ed V,bi,v,Ed       F F F Where  3 = 41,3° (See Figure B.6) Table B.3gives the results.