MEM361W09_10.HW4-soln

# MEM361W09_10.HW4-soln - MEM 361 Engineering Reliability...

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Unformatted text preview: MEM 361: Engineering Reliability Home work #4; due: 02/04/10 From text: Problems 3.12, 3.18, 3.19, 3.21 (read text around eqn.3.47) Problem 5.2: solution 3.12: f x ( x)dx f y ( y)dy; f x ( x) 1;0 x 1 a. f y ( y ) f x ( x) | dx / dy | with y x 2 dx / dy 0.5 y 1/2 f y ( y ) 0.5 y 1/2 ;0 y 1 zero otherwise. f z ( z ) f x ( x) | dx / dz | with z 3x dx / dy 1/ 3 f z ( z ) 1/ 3;0 z 3 zero otherwise. 3.18: f ( ) fl (l ) b. dl 1 l 2 /2 ; fl (l ) e ;l 0 d 2 cl 2 d 2cldl dl 1 d 2cl 1 l 2 /2 1 f ( ) e 2cl Thus, l 2 3.19: = 10 hr., = 2 hr. /c 1 e /2 c 2c 2 / c 0.1 F ( L10 ) ( z t t ) z 1.28 , from Appendix C. Hence, 3.21: L10 10 1.28 L10 7.44hr. 2 total _ load d l w 19.8kips 2 2 2 total _ load d l2 w 2.81 t _ l 1.676kips. ...
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MEM361W09_10.HW4-soln - MEM 361 Engineering Reliability...

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