hwsoln16 - 85[8.4 Consider the thick-walled cylinder of Hg...

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Unformatted text preview: 85. [8.4] Consider the thick-walled cylinder of Hg. P.8.4 constrained by immovable walls at the ends and subject to an . ‘ internal pressure so as to undergo plane strain. It is best here Etc use cylindrical coordinates. The strains are considered for fan element shown on a slice in Fig. P.8.4 as follows: a" 5 normal strain in radial direction see 5 normal strain in transverse direction 8’9 E shear strain between line segments dr and r :19 of the element showu Figure 18.4. If the following data are known for a point in the pipe at 9 = 30", en, = —.002 a” = .00 0:"9 = .001 what are the strains on and a”, at the point? Er fit Redefine axes as shoWn to right in order to use formulations of text. 5‘: i. ' = — “a .- Exx '002 7' 1-: = .003 yy 10‘ Exy — .001 _ -.002 + .003 + -.002 - .003 F')(')(' " 2 2 cos(-60°) + .001 sin(-60°) .0005 - (.0025)(.500) - (.001)(.866) = .0005 - .00125 - .000866 = -.001616 ey.y. = .0005 + .00125 + .000866 = .002516 ‘ The desired strains are: -.001616 ——.—_ .002616 "8.6. [8.4] In Problem 85, what are the principal strains in the xy plane at the point of interest? Let err = ex.x. = -.002 see = ey.y. = .003 are = exw. = .001 ta" 29 = —.302;?303 = ’ i%%% = "400 29 = '21.8" 6 = -10.9° - 002 + .003 -.002 - .003 .52 """§' * 2' - cos(-21.8) + (.001)s1'n(-21.8) .0005 - (.0025)(.928) + (.001)(—.371) .0005 - .00232 - .000371 =' = .0005 + .00232 + .000371 = 5.10. [8.4] Given : ex,-= —-1xmos .i a” = .0002 ex), = -.0001 ‘Using Mohr’s circle as an aid, find the principal strains and the strains {Cir taxes I ’y ’ rotated 30“ counterclockwise from axes xy.‘; _ S x ‘ 5 2 2 1/2 r — [(—X—T—u) +(exy) 1 = [(2'.5)2+12]1/2 x 10'4 = 2.593x10'4 -4 tan 29) = ,'.¢ = 21.8cl 2.5x10 .‘ly Ea . 6E - r cos(60°-21.8°) = - % x 10'4-2.693x1o'4 cos(38.2) = -2.616 x 10'4 ____.~..—.— . 55 + r coé 38.2 = - % x 10‘4 + 2.593x10'4 cos(38.2) = 1.616 x 10" I = -r sin(38.2) = -2.693x10"sin(3 .2) = -1.665x10'4 4 -2.616x1o' 1.616xm'4 yy ———- ('I II +1.665x1o'4 = 65 + r = - % x 10’4 + 2.693x10'4 = 2,133x10'4 = 66 - r = - i x 10'4 - 2.693x10‘4 2 - 3.193x10‘4 EEiflEiEfll_&1§§ @‘k ...
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This note was uploaded on 07/15/2008 for the course ME 311 taught by Professor Pitteressi during the Spring '08 term at Binghamton.

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hwsoln16 - 85[8.4 Consider the thick-walled cylinder of Hg...

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