This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: FORCE SYSTEM FORCE SYSTEM RESULTANTS RESULTANTS Reduction of a Simply Distributed Load Reduction of a Simply Distributed Load LEARNING OBJECTIVES LEARNING OBJECTIVES Be able to find an equivalent force for a Be able to find an equivalent force for a simply distributed load simply distributed load PREREQUISITE KNOWLEDGE PREREQUISITE KNOWLEDGE Units of measurement Units of measurement Integration of functions over an area Integration of functions over an area LINEAR LOAD DISTRIBUTION LINEAR LOAD DISTRIBUTION The unit weight of water (γ) is 62.4 pcf , in SI units it is 1 gm/cm 3 . The pressure (p) in the water is the same in all direction and is equal to the unit weight (γ) multiplied by the depth (Z); p = (γZ) Swimming Pool The total force due to the water pressure is equal to the integral of the pressure {p = (γZ)} over the area of the swimming pool wall. Water Z p = (γZ) Floor LINEAR LOAD DISTRIBUTION LINEAR LOAD DISTRIBUTION For a swimming pool depth of 6 feet and width of 50 feet, the total force can be calculated as follows: tons lb F yz dz Zdy pdA F A z y 08 . 28 56160 ) 6 )( 50 )( 4 . 62 ( 5 ....
View
Full
Document
This note was uploaded on 02/03/2012 for the course CE 221 taught by Professor Buch during the Summer '08 term at Michigan State University.
 Summer '08
 Buch

Click to edit the document details