Solution manual ch02 pressure distribution in a fluid

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Solution manual ch02 pressure distribution in a fluid

Chapter 2 • Pressure Distribution in a FluidP2.1 For (he two-dimensional stress field in Fig. P2.1. let<7XX - 3000 psf ơyy = 2000 psf<7X> = 500 psfFin

Solution manual ch02 pressure distribution in a fluidnd the shear and normal stresses on plane AA cutting through at 30°.Solution: Make cut "A A" so that it just hits the bottom right comer of the elemen

t. Tins gives the freebody shown at right. Now sum forces normal and tangential to side AA. Denote side length A A as "L.”SF„.aa=0-ơaaL-(3000 sin 30 + Solution manual ch02 pressure distribution in a fluid

500 cos 30)L sin 30-(2000cos 30 + 500sin 30)L cos 30Solve for <7AA ~ 2683 Ibf/ft2 Ans. (a)SF.aa-0-taa L - (3000 cos 30 - 500 sin 30 )L sin 30 - (500

Solution manual ch02 pressure distribution in a fluid

cos 30 - 2000 sin 30 )L cos 30Solve for rAA •> 683 Ihf/ft2 Ans. (b)P2.2 For the stress field of Fig. P2.1. change the known data to cTux ■ 2000 psf. O

Chapter 2 • Pressure Distribution in a FluidP2.1 For (he two-dimensional stress field in Fig. P2.1. let<7XX - 3000 psf ơyy = 2000 psf<7X> = 500 psfFin

Solution manual ch02 pressure distribution in a fluidfreebody above, with ơh(AA) known and Ojy unknown:s *kAA = 2500L - (ưxy cos 30° + 2000 sin 30°)L sin 30°- (ơxy sin 30° + 3000 cos 30°)Lcos 30° = 0Solv

e for ƠKV - (2500 - 500 - 2250)/0.866 - -289 Ibf/ft2 Ans. (a)A J<44959Solutions Manual • Fluid Mechanics, Eighth EditionIn like manner, solve for the Solution manual ch02 pressure distribution in a fluid

shear stress on plane AA. using our result for ơiỵ:S1ỸAA = TAAL-(2000cos30° + 289sin30<>)Lsin30’> +(289cos30° + 3000SÚÌ 30°)Lcos30° = 0Solve for rAA =

Solution manual ch02 pressure distribution in a fluid

938-1515- -577 Ibf/ft2 Ans. (b)This problem and Prob. P2.1 can also be solved using Mohr’s circle.P2.3 A vertical clean glass piezometer tube has an i

Chapter 2 • Pressure Distribution in a FluidP2.1 For (he two-dimensional stress field in Fig. P2.1. let<7XX - 3000 psf ơyy = 2000 psf<7X> = 500 psfFin

Solution manual ch02 pressure distribution in a fluidimate the applied pressure in PaSolution: For water, let }■ = 0.073 N/m. contact angle 6 = 0°. and •/ = 9790 N/m' The capillary rise in the tube, from

Example 1.9 of the text, is, 2YCOSỚ 2(0.073 Ww)cos(0°)/irữ„ = ——- = 0.030 mp ỵR WMNIm'WMHSm)Then the rise due to applied pressure is less by that amo Solution manual ch02 pressure distribution in a fluid

unt: //press ■ 0.25 Ill - 0.03 111 ■ 0.22 111. The applied pressure is estimated to be p = -///press = (9790 N/n?)(0.22 111) = 2160 Pa Ans.P2.4 Pressu

Solution manual ch02 pressure distribution in a fluid

re gages, such as the Bourdon gage111 Fig. P2.4. are calibrated with a deadweight piston.If the Bourdon gage is designed to rotate the pointer10 degre

Chapter 2 • Pressure Distribution in a FluidP2.1 For (he two-dimensional stress field in Fig. P2.1. let<7XX - 3000 psf ơyy = 2000 psf<7X> = 500 psfFin