. The Civil engineer and architect's journal, scientific and railway gazette. Architecture; Civil engineering; Science. 1840.] THE CIVIL ENGINEER AND ARCHITECTS JOURNAL. 79 Example 1.âGiven the dejith of water on the outside of a dam equal 20 feet, that inside equal 6 feet, and tile giitli GO feet, what is tlie cirecti\'e pressure against the dam ? We have c+(i:=2G, câd-=.\i, and ^=G0 therefore l-25x(e+^)X(c-i)Xg^l25x2Gxl4xGO^^^^^^^^^ 14x30 =45,500x30=T,3G5,00011)s. for tlie pfFective pressure. When the inside and outsiiU' girths dilter, by putting g fur tlie out- side girth, and ^'for that in
. The Civil engineer and architect's journal, scientific and railway gazette. Architecture; Civil engineering; Science. 1840.] THE CIVIL ENGINEER AND ARCHITECTS JOURNAL. 79 Example 1.âGiven the dejith of water on the outside of a dam equal 20 feet, that inside equal 6 feet, and tile giitli GO feet, what is tlie cirecti\'e pressure against the dam ? We have c+(i:=2G, câd-=.\i, and ^=G0 therefore l-25x(e+^)X(c-i)Xg^l25x2Gxl4xGO^^^^^^^^^ 14x30 =45,500x30=T,3G5,00011)s. for tlie pfFective pressure. When the inside and outsiiU' girths dilter, by putting g fur tlie out- side girth, and ^'for that inside, we get in this case 125(c-gâi'g') for the effective pressure. Example 2.âGiven the height of water on tlie sill to the upper gates of a lock above, 10 feet and girth 21 feet; below 4 feet and girth 25 feetâwhat is the effective pressure on the gates? â, . , ]25fc=gârf=g') 125(100x24â10x25) The pressure is equal ⢠^â-S~'z= i =rl25(100x24â8x25)= 125x1000= 125,0001bs. the pressure re- quired. Example 3.âFind the effective pressure against a coffer dam, the exterior depth and girth respectively being 2/ and 120 feet; and the interior depth 5 feet, and girth 100 feet. â , ,, , , 125(27x27x120â5x5x100) Here by the formula; ^ = ' 2 125(729x00â25x50)=:125x424,90=5,311,250lbs. the pressure re- quired. Problem III. To Jind the centre of pressure in a given depth of water: or iliat point where the force of the whole pressure is equal to the sum of the forces arising from the pressures at different depths from the surface. The whole pressure (problem 1) is represented by a right angled triangle having its base and perpendicular each equal to the depth of water, and as the pressure at each point along tlie depth is propor- tional to the depth of sucli jioint from the surface, or which is the same thing to a line parallel to the base at that point meeting the hypothenuse ; the centre of pressure is evidently on the same liori- zontal line with the centre of gravity of
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