assinment 1 key

# assinment 1 key - 1 a Area to be sheared 3d 2 3 0.5 2 0.589...

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Unformatted text preview: 1) a) Area to be sheared 3d 2 3 0.5 2 0.589 4 4 in 2 Shear stress ( b) = F A 0.1” 1” Area to be sheared= 3 = 40000 3d 2 3 0.5 2 c) Area to be sheared= 0.589 4 4 =45000 2) Answers may vary Possible designs i) Multiple drains ii) Using rounding cover iii) Installing the filters on the sides of the pool in 2 http://swimming.about.com/od/swimmingpoolsandspas/qt/mnpoolsafetyact.htm 3) Ft RT 35’ 90lb RB 200lb O x a) ∑ ∑ TmAO RT 90 ( ) ∑ RT =393.75+ 145 =11.62+5.9 =290 fr 8.75 33.9’ b) Tip ladder away from the wall with climber 3’ from top 3’ x’=0.75’ x=8.75-0.75=8’ TmAO RT 90 ( ) RT = 58.81 lbF 4) a) b) 5) a) surface speed= b) Using 1 All calenders, irrespective of size of the rolls or their configuration, shall be stopped within a distance, as measured in inches of surface travel, not greater than 1 3/4 percent of the peripheral no load surface speeds of the respective calender rolls as determined in feet per minute. ...
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## This note was uploaded on 02/19/2011 for the course IE 11 taught by Professor G during the Spring '11 term at Minnesota.

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