12Ei L 3. Question In the diagram below the stiffness coefficients 12EI/L^3 and 6EI/L^2? 67074 In the diagram below are the stiffness coefficients 12EI/L^3 and 6EI/L^2? Or are they reactions? This is a clarification question no calculation required.

Calculation Of Stiffness In Structural Elements Skill Lync 12ei l 3
Calculation Of Stiffness In Structural Elements Skill Lync from Skill-Lync

PDF file12EI/L3 12EI/L3 + x ∆i 1 + 4EI/L 2EI/L 6EI/L2 6EI/L2 x θi x δj AE/L 1 AE/L + 6EI/L2 6EI/L2 1 12EI/L3 12EI/L3 x ∆j 1 + 2EI/L 6EI/L2 6EI/L2 4EI/L x θj Fxi Fyi Mi Fxj Fyj Mj Ł Member Equilibrium Equations i j = E I A L x δi AE/L AE/L 1.

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As an added note take stiffness of column that is pinfix k=3EI/L^3 fixfix k=12EI/L^3 Fixing the column bases makes it 4 times stiffer Tradeoff is you now have a foundation and anchor bolts that need to be designed for a moment I would use 1 above and it isn’t bad to assume if your assumption is correct2011051820100506.

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PDF fileL4IL( L[3(I ) M)L2I ( 2 )ML I ] 12EI (L I L I ) + Ω + + Ω − Ω −Ω − θ= + [17] Substituting θAθBand θCin the slopedeflection equations we get the memberend moments as follows MMAB A=− [18] BC AB AB BA AB BC BC CB BC AB A AB BC C BA BC AB AB BC IL( 2 )IL(2 )ILM ILM M 2(I L I L ) Ω + Ω + Ω +Ω + + =− + [19].

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PDF filethen C1 = q L 3 / 24 the equation of slope is q v’ = CCC (L3 6 L x2 + 4 x3) 24 EI integrating again it is obtained q v = CCC (L3 x 2 L x3 + x4) + C 2 24 EI boundary condition v = 0 at x = 0 thus we have C2 = 0 then the equation of deflection is q v = CCC (L3 x 2 L x3 + x4) 24 EI.