trapezoidal channel 2m wide at the bottom and 1m deep has side slopes 1:1. Determine: i) Total pressure ii) Centre of pressure, when it is full of water (July 2014, Jan 2015) Ans: Given B = 2m Area of flow A = (B+sy)y = 3m2 The combined centroid will be located based on two triangular areas and one rectangle (shown as G1, G2, G2) y  A1  h1  A2  h2  A3  h3   A1 A2 A3 The total area A = 3m2 Area of rectangle = 2 ×1 = 2m2 1 ×1×1=0.5m2 Area of Triangle = 2       1 1 (21)0.5   (11)0.333  (11)0.333 2  2  y   4    3 0.444m 9 The total pressure P = w×A× = 9810×3×0.444 = 13080N y Centre of pressure The centroidal moment of Inertia of Rectangle and Triangle is 2 0.5m  at  IG1 from water surface 13  0.1667m 4 12 113 IG1   0.028m  from water surface 36 4 0.333m at I   h y Eq.1 A y The moment of Inertia about combined centroid can be obtained by using parallel axis theorem
trapezoidal channel 2m wide at the bottom and 1m deep has side slopes 1:1. Determine: i) Total pressure ii) Centre of pressure, when it is full of water (July 2014, Jan 2015) Ans: Given B = 2m Area of flow A = (B+sy)y = 3m2 The combined centroid will be located based on two triangular areas and one rectangle (shown as G1, G2, G2) y  A1  h1  A2  h2  A3  h3 A1  A2  A3 The total area A = 3m2 Area of rectangle = 2 ×1 = 2m2 Area of Triangle = 1 ×1×1=0.5m2 2       1 1 (21)0.5   (11)0.333  (11)0.333 y   2  2  3   4  0.444m 9 The total pressure P = w×A× y = 9810×3×0.444 = 13080N Centre of pressure The centroidal moment of Inertia of Rectangle and Triangle is IG1  2 13 12  0.1667m 4 at 0.5m from water  surface IG1  113 36  0.028m 4 at 0.333m from water  surface h  y  I Eq.1 A y The moment of Inertia about combined centroid can be obtained by using parallel axis theorem
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