Question

Problem 1. A 1 meter inner diameter steel pressure tank with 8 mm wall thickness is subject to a internal pressure of 1.5 MPa and due to the piping weight a torsion of 10 N.m is acting as shown in the figure. The length of the cylinder is 3 meters. Determine: The state of stress in the cylinder wall The state of stress if the normal Cartesian system is rotate 25° The principal stress and the principal angle. The maximum in plane shear stress Draw the Mohr Circle for the state of stress and identify the principal state of stress angle. T = 10 N.m

Answer 1

given datu- Torsion, Tolon-m Inner Diameter, Deim wall thickness, t = 8 mm Internal pressure, p=1.5mpa Cylinder length, L= 3m Assumption: * Thin cylinder CD > 200 =j 1000 mm 160mm) i Subjeded to internal pressure Material Homogenous, Isotropic moterial Solution wall State at cylinder of Stress Ty T Try g o → Jx Tu T txy Jy Ci) Normal Stress longitutional driection (Ov) Thin when the to Cylinder ü subjected internal pressure the longitutional stress is given by, PD AL Longitutional stress, Ox=DL 1-5 x 1000 4x8 Ох = 46.815 mpa Cui Cum fe renuai Dviectui, Cry) cii) Normal Stress in Рp cui cum ferenuri stress, Th: Jy = 2t

1.5 X1000 Jy = 2 x 8 Jy 93.75 mpa Clil shear due to torsion: stress IR Shear stress, Txy = For hollow shaht J= plolar MOI Joc de-d," T Polar MOI, JE I (de4-4,4) R. Di 32 pis im 4-19) (1.0164 = 32 D2 Di + 2t 9 4 mm D2 = 1.016m J= 6. 435 X10 3 10 XID txy x (0.5X1000) 61435x109 - 3 мро Txy = 0. 777 X10

State of stress in Cylinder wall A Jy Try: 0.777 xlo'mpa Ja=46.875 MPa Jie Jy = 93.75 mpa 0 = 25° obliqua plane b State stress in of LOS 20 + txy Sin 20 -3 To = ( ostety) + (ose they ( 46.815 793-75) J+1 (46816 46875 - 93.75 5) cs(2825)+ 0-777810 Sin (2x85 - 3 0 595 XD 70.3125+ (-23.43) x0.6427 + To = 55.25 MPa Jx - Jy COS20 To -( Sin 20 + txy 2 - 3 46-875-93.75 Sin (24&5) + 0.777x10 cos (2251 to: 17.95 mpa to Txy >To 1250 TE x se Txg Jy te

© Principle Principle Angle stress and + Try vilo = (5x409) ( 02109) = V(52764) + (40.87593.75 V (40 875 2 46.815+93.75 + +10.177x103) 2 70.31 + 23.437 Ti 93.747 MPa ( Tensile) To 46.873 mpa (Tensile) 2txy Principle plane Angle, 20p = Jx-Ty 2x0-77781003 46-875- 93.75 OP 179,99 89.999 2 P2 = 179.999 of Inplane maximum Shear Stress (tmax) Ti-J2 Tmans 2 93.747–46.873 2 Tmar = 23 437 Mpa

Mohr cicle Os you ,o) - (46.815 4 93-95, o centne C = 70.31, 0) Radius, R = voz-oyetemis - v 915-09-19 5)*601710) R = 23.437 Mpa Jy=93.75 mpa BC Ty, Try) Try it dx x = 46.875 MPa "A Cox, - Try) Ty t cmpa) B (93-75, 0.778103) E 270.31,0) F A (46.87,0. Txil 2011 Timpa Ji: OF: 93.747 MPa Principle Stress, T2 - OE = 46.873 MPa. Principle Angle, OPI = 89.99° OP2 = 179.999