-ka₁ +2k(a₂-a)+1=0 -2k(a₂-a) +k(a₂-a₂)=0 which are solved simultaneously, yielding k(a-a₂)=0 The second column of the fl

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Ka 2k A A 1 0 2k A A K A A 0 Which Are Solved Simultaneously Yielding K A A 0 The Second Column Of The Fl 1 (19.28 KiB) Viewed 54 times

-ka₁ +2k(a₂-a)+1=0 -2k(a₂-a) +k(a₂-a₂)=0 which are solved simultaneously, yielding k(a-a₂)=0 The second column of the flexibility matrix is obtained by applying a unit load to the middle block and applying the equations of equilibrium to the free body diagrams of Fig. 5-15b leading to 3 3 2k 2k aym The third column of the flexibility matrix is obtained by applying a unit load to the block whose displacement is described by x, and applying the equations of equilibrium to the free body diagrams of Fig. 5-15c, leading to