. Applied calculus; principles and applications . Denoting by P the total stress on either semicircle, andtaking St = Sc = S; P = lim 5j -y 2x^y = — I Va^ — y^ydyA2^=o ^0 o, a Jo y>P=—-j Va^-y^yHya Jo = 2^r^(^2)V^J3^+^%in-i^TP^^^-2Exer-\ 2S TTO^ a 16 Swa Sird 8 64 ^ 64/6 32 * Hence the couple formed by the forces P has an arm,2^ = TQT^d^ and M = -^Sd^ • Y^Trd = ~o^j the section modulus. Example 4. — Find the total water pressure upon the endof a circular right cylinder immersed lengthwise, one elementof the cylinder just at the surface of the water. Find thecenter of pressure of the cir
. Applied calculus; principles and applications . Denoting by P the total stress on either semicircle, andtaking St = Sc = S; P = lim 5j -y 2x^y = — I Va^ — y^ydyA2^=o ^0 o, a Jo y>P=—-j Va^-y^yHya Jo = 2^r^(^2)V^J3^+^%in-i^TP^^^-2Exer-\ 2S TTO^ a 16 Swa Sird 8 64 ^ 64/6 32 * Hence the couple formed by the forces P has an arm,2^ = TQT^d^ and M = -^Sd^ • Y^Trd = ~o^j the section modulus. Example 4. — Find the total water pressure upon the endof a circular right cylinder immersed lengthwise, one elementof the cylinder just at the surface of the water. Find thecenter of pressure of the circular area. 362 INTEGRAL CALCULUS The intensity of pressure at a depth y being wy^approximate pressure on a strip iswy2x ^y. 0 the. 2wxy ^y) ny=v 0 P = 2w f {2ay-y^)^ydy = 2wa ( \2ay-y^)^dyJo Jo 7ra^ = 2wa-— = wira^; (Ex. 13, Exercise XXV.) 2w I (2ay-y^)y^dy ia2w I {2ay-y^)^ydy — _ Jo Jo ^ ~ P ~ wira^ = i^^ =^a = ^d. (Ex. 13, Exercise XXV.) WTTO 8 EXERCISE XLI. 1. (6) The pressure upon one side of the gate of a dry dock, thewetted area being a rectangle 80 ft. long and 30 ft. deep, is to be foundexactly. Take w = 62^ lb. for the weight of a cubic foot of water. (c) Find the depth of the center of pressure. Ans. (c) 20 ft. (a) Find the pressure approximately by a limited number of terms. (See Art. 154.)
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