1.ВиоуanсyTable 5.1 BuoyaпсyWB = BF – WwHeight of Wood – 0.299mCross-SectionalDepthSubmerged(m)SpecificGravity ofBuoyBF = y x V,Area ofTrialWeights (N)BFSpecimen(m^2)Where:Wb =weight of the specimen3.950.003450.185Ww =weight of the plumb bob/weights23.850.003450.176buoyant forceunit weight of the liquid/watersubmerged volume of the specimenBF =34.020.003450.190Y%3DV

Given Data: Trial Weight (N) Cross-sectional Area Depth Submerged(m) 1 3.95 0.00345 0.185 2…

Table 5.1 Buoyancy WB = BF – Ww Height of Wood - 0.299m Cross-Sectional Depth Submerged (m) Specific Gravity of Buoy BF = y x V, Area of Trial Weights (N) BF Specimen (m^2) 0.00345 Where: weight of the specimen weight of the plumb bob/weights buoyant force unit weight of the liquid/water Wb = 1 3.95 0.185 Ww = 0.00345 0.00345 2 3.85 0.176 BF = 3 4.02 0.190 V submerged volume of the specimen

Table 5.1 Buoyancy WB = BF – Ww Height of Wood - 0.299m Cross-Sectional Depth Submerged (m) Specific Gravity of Buoy BF = y x V, Area of Trial Weights (N) BF Specimen (m^2) 0.00345 Where: weight of the specimen weight of the plumb bob/weights buoyant force unit weight of the liquid/water Wb = 1 3.95 0.185 Ww = 0.00345 0.00345 2 3.85 0.176 BF = 3 4.02 0.190 V submerged volume of the specimen

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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