Formulario del corso diviso per capitoli

Pagina 1

Force method - Compendium of solutions for beam displacements Load case ii displ. at midspan (f) displ. at end (δ) ij 11 3 l EJ  2 16 Mlf EJ 21 6 l EJ   2 1,0 16 Fl EJ  3 48 Flf EJ 3 1,0 24 p l EJ   45 384 p lf EJ  1,0 24 M l EJ    11 2 l EJ  2 8 Mlf EJ 1,0 l T h      2,0 l T h     3 3 Fl EJ  2 2 Fl EJ 2 2 M l EJ  U M l M EJ    4 8 p l EJ  Ex: 1 11 1 12 2 10M M        F p p Welded Joints Butt-weld assessment ߪ௜ௗ݇∙ߪ௔ௗ௠ 2 // 2 // 2 // 3  id Fillet-weld assessment Admissible Stresses Beam profiles Axis yy Axis xx Linear mass Profile identification Axis-symmetric elastic problems (disks) General stress solution Solid disk with external pressure pe Hollow disk with external and internal pressure pe and pi 222 22 22 22 222 22 22 22 1)( 1)( rrr rrpp rr rprp rrr rrpp rr rprp ie eiie ie eeii ie eiie ie eeii r         Rotating solid disk at constant speed ࣓ cost – No pressure 2222 222 8 31 8 3 )(8 3 rr rr e er      Rotating hollow disk at constant speed ࣓ cost – No pressure )3 31(8 3 )(8 3 2 2 22 222 2 2 22 222 rr rrrr rr rrrr ie ei ie eir        TBN The maximum value of ߪ௥ is reached for ݎൌ ඥݎ௜ݎ௘ rF'rC1C2r2382r3dF'drFrrC1C2r21382r2 2