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FORCE SYSTEM FORCE SYSTEM RESULTANTS RESULTANTS Reduction of a Simply Distributed Load Reduction of a Simply Distributed Load

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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
PRE-REQUISITE KNOWLEDGE PRE-REQUISITE KNOWLEDGE Units of measurement Units of measurement Integration of functions over an area Integration of functions over an area

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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 . 0 2 1 2 2 6 0 50 0 = = = = = = = = γ γ Water Z p = (γZ) Floor

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NONLINEAR LOAD DISTRIBUTION NONLINEAR LOAD DISTRIBUTION In general, a distributed load can be replaced by a resultant force as follows: The given distributed load Step 1 Reduce the y-dimension w = w(x) = (A)[p(x)]

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