Question

Problem 1 [45 marks] A cylinder (= 230 MPa) which is initially free from residual stress has internal radius 0 = 20 mm and external radius a = 50 mm. Determine: a) The maximum internal pressure without yielding the cylinder. Give your answer in MPa. (5 marks] b) The fully plastic internal pressure. Give your answer in MPa. [5 marks] C) The radius to which the material yields, when the pressure is 150 MPa. Give your answer in mm. [10 marks] The cylinder is loaded by an internal pressure, p:, which, when released, causes a residual stress in the hoop direction of -0.5oy at the internal radius. Determine: d) The internal pressure, p. Give your answer in MPa [25 marks]

Answer 1

Solution Given data- 2,230MPa, internal, radius b = 20mm external realius a = 50mm since < 20. so it is thick Cylinden (t= abb= 50-20 = 30mm max. internal pressure without yielding the cylinder Analysis g b fie Bursting force fs = P. 17.62 Resisting force, FR z бу. IT (926²) for equilibrium, Bursting force C. - () = ef= Resisting force Pb² (9262) from eah the mon, internal pressure without yielding the glinder sy. (d=6²) Pe 62 230x(501203) boss 202 Pi = 1207.5mpa fully plautic internal prefere if the pressure inside the cycinden becomes more then 1207.5 mpa et meie dead to bursting of finden, so prebine more than 1207.5 mpa is plastic internal pressure. -

p.bh na 50mm is given when pressure 150 mpa vodicus at which materials 4 yield if the stress inside the cylinder reaches to yield stength of material it will lead to yleiding of matewal. = 23ompa at pressure of 150 mpe material start yielding at co so it oy . 150%. 62 230 - 22621 bs? 50²- 62 50²6² 0.65217 62 1.652175 2 50² மார்க்க b = 38.9 mm So at b= 38.9mm malewcu mies start yielding. (radius 0.5oy B also Now the cylinders is loaded with in internal pressure of p. in the internal which causes a residual stress of racus, NOW we have to determine that internal pressure p From Lame's egn where, A and B ane Lomés constant X² and x is the radial distance from centre of section] Og +A X2 here ox= hoop stress or Residual stress Given cylinder is looded by prepare pat an internal radius So from eah - wat and at x=a, Px = 0 [because there is no pressure at outer radius of 111 X= b Q = P ylinder

* put in ean_oo) f (V) e lo solo ♡ ond - By soluing eah now and A B ܬܐܘ 810+B) floo - 11 P . 62 a² 6² B= p6² P 9²6² (02bY P. Ab? B * (62-62) Pb (a?63) ond A = ( a2b3 92 VI) A = 11 VIID p6² + A OX = From ean W Pa²b² (626) x (876) riven at internal rading fx = -os By =-0.5X230 =-115 MPa So at X= b, the internal pressure, from eah will P.(50). Gool Sot - 28) + (gost (50720) [ 30+504] 50209 P. (2012 115 -115 p= 83.275 MPa here ve sign indicates compresine nature of previne.