I 1 i i 1 i hence z is also follows log normal

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Unformatted text preview: excessive settlement (i.e. probability of no settlement)? 2 Solution. We get, S 0.09 0.02 0.12 0.15 0.0454 2 2 2 2 Also we have, 1 0.09 2 0.00405 2 1 2 B ln 5.5 0.02 1.70455 2 1 I ln 0.5 0.12 2 0.70035 2 1 2 M ln 30 0.15 3.38995 2 P ln 1.2 So, we obtain S 0.00405 1.70455 0.70035 3.38995 2.3898 Now mean settlement of footing, 1 2 0.092 m 2 S exp S S exp 2.3898 1 0.0454 2 2 And the Coefficient of variation, S S 0.0454 0.2131 . So, the reliability, R P S 15 cm ln 0.15 / 100 2.3898 R 2.31 0.9895 (from the cumulative standard 0.2131 normal distribution table used in Problem 1). Problem 4. A model of one-story building is shown in figure whose total mass M concentrated at roof level. When subjected to earthquake ground shaking, the building will vibrate about its original position including velocity components, U and V of the mass with a resultant velocity, W U 2 V 2 . If U and V are random variable with standard normal distribution N 0,1 , determine the probability distribution of the resultant kinetic energy of the mass during an earthquake. M V Elevation W Plan U M Fig. 1. Force Diagrams Solution. The resultant kinetic energy, K 1 MW 2 mW 2 m U 2 V 2 2 Let A mU 2 and B mV 2 , so K A B Taking a substitution, a mu 2 , we get u Thus we obtain, a m du du 1 1 or . da da 2 ma 2...
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This note was uploaded on 03/18/2014 for the course CE 5730 taught by Professor Dr.rajibmaity during the Spring '13 term at Indian Institute of Technology, Kharagpur.

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