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Q2/ Find the equivalent stiffness for the system shown in figure. K, = 235 N/m K = 155 N/m kg = 350 N/m Ke = 140 N/m kg = 325
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235 Ki 155 K₂ mu 350 K3 100 325 KG 140 w K5 Ky K, and K2 K₃ are patated series 1 Keg = t - - wat Keg, K2 K3 + kik3 t Kik2 KIK(155x350) + (835*350) + (2354155 325 +140 140X325 (835) (155) (350) = 0.01356 + 0.010219 0.0238 Keg=0.0238 K6 = 100 these two

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