Answers (a) er=15.30-250.2/r^2 , etheta=15.30+250.2/r^2(b) delta c = 8.12x10^-4 in 6 in2 in0 The thin pressurized ring shown is an axisymmetric problem. The normalstresses as functions of r are0,= 625 -562512σe = 625 +56252%₂=0If the material constants are E = 29 X 10° lb/in² and v =r≤3 in, determine (a) the normal strains as functions of r and (b) the circum-ference at r = 3 in before and after the application of the pressure p.= 0.29, for 1 in SO431-431L.STAT41015/0

The normal stresses are The material properties are

The thin pressurized ring shown is an axisymmetric problem. The normal stresses as functions of rare σ, = 625 - 5625 1² 00=625 + 5625 2² σ₂ = 0 If the material constants are E = 29 x 10° lb/in² and v = 0.29, for 1 in ≤ r≤3 in, determine (a) the normal strains as functions of r and (b) the circum- ference at r = 3 in before and after the application of the pressure p.

The thin pressurized ring shown is an axisymmetric problem. The normal stresses as functions of rare σ, = 625 - 5625 1² 00=625 + 5625 2² σ₂ = 0 If the material constants are E = 29 x 10° lb/in² and v = 0.29, for 1 in ≤ r≤3 in, determine (a) the normal strains as functions of r and (b) the circum- ference at r = 3 in before and after the application of the pressure p.

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter7: Analysis Of Stress And Strain

Section: Chapter Questions

Problem 7.7.6P

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