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@= acgular W = x SE45 1.0 69607 f = 1 = 22.36 H = 3.56 H 2x ax f we need T = 220
@= acgular W = x SE45 1.0 69607 f = 1 = 22.36 H = 3.56 H 2x ax f we need T = 220 W= wi- - 22.36 1-0.05 mnd/see = 22-33d7 Question - Im F(t) ar XXX 22.33- 3.55 2x 19041.21 = kg k suspended spring of stiffness, k=4011/ 4. beam is 2.1m x 10 N/mm by 0.15m by 6 m beam has a 20N mass: a means of Determine: [6 marks] (i) The steady-state amplitude of vibration (ii) The maximum dynamic stress in the columns, assume the girder is rigid [5 marks] M=6946.97kg Section modulus 1.7x10-mm a Figure Q4(a) L=4.572 A one-sto weightless properties which (b) Briefly discuss the three (3) categories of prescribed dynamic loadings in structures [9 marks] m 4000kg VQUESTION FIVE (20 MARKS) [3 marks] (a) Describe three (3) forms of buckling in structural instabilities. (b) Discuss the origin of earthquake [4 marks] (c) The Figure Q5(c)shows structural frame of one bay for a storey building with two suspended RC floors. The concrete columns are of heights 2m & 3m as shown. Calculate the natural frequencies a
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