23.1 Principles of Stiffener/Web Construction 621Fig.23.3
Equilibrium of stiffener CJG in the beam of Example 23.1.
FromtheequilibriumofstiffenerJK,wehave(q 1 −q 2 )× 250 =4000sin60◦=3464.1N (i)andfromtheequilibriumofstiffenerHKD,
200 q 1 + 100 q 2 =4000cos60◦=2000N (ii)SolvingEqs.(i)and(ii),weobtain
q 1 =11.3N/mm q 2 =−2.6N/mmTheverticalshearforceinthepanelBCGFisequilibratedbytheverticalresultantoftheshearflowq 3.
Thus,
300 q 3 =4000cos60◦=2000Nfromwhich
q 3 =6.7N/mmAlternatively,q 3 maybefoundbyconsideringtheequilibriumofthestiffenerCJG.FromFig.23.3,
300 q 3 = 200 q 1 + 100 q 2or
300 q 3 = 200 ×11.3− 100 ×2.6fromwhich
q 3 =6.7N/mmTheshearflowq 4 inthepanelABFEmaybefoundusingeitheroftheabovemethods.Thus,considering
theverticalshearforceinthepanel,
300 q 4 =4000cos60◦+ 5000 =7000N