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A 5x5 K matrix, K, the associated external force vector, f and the reaction force vector, Rare given as below 2 -1 0 0 0 3 R

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Given Data: - 2 -1 0 0 0 - 1 3 Ri ks -2 OO 4 0 -1 0 -2 fa R= R2 2 0 4 -1 2 4 2 R3 RA R5 Boundary Conditions: 24, +Uz -0.5, Ug² -212 +4Uz - U5 = 2 +R3 W) 444 -us = 4 +RA -Uz - u4 +245 = 2+25 Now, from Guation @ : → 444 = 4 + R4 (Us=0 ..UA 4+RA 4 fromLet RA Uq=1 Ug = 1 Ug=0 how Displacement matrix given as: u= -3.5 Therefore, R=0, R4 =0 from equation 2 -4, +GU2 - 243 = H+ RHence, the Reaction matrix matrix is Ri O 10 R₂ .: 2 = R3 -18 R5 -6.5

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