Mechanics of Materials Thick Cylinders PDF

Mechanics of Materials THICK CYLINDERS Dr. S Venkateswara Rao Civil Engineering Department 9/17/2022 1 At any point in a cylindrical pressure vessel, there are three stresses in three perpendicular directi...

Mechanics of Materials THICK CYLINDERS Dr. S Venkateswara Rao Civil Engineering Department 9/17/2022 1 At any point in a cylindrical pressure vessel, there are three stresses in three perpendicular directions Stress element Longitudinal stress l (closed ends) Tangential stress θ Hoop stress r Radial stress 1. Longitudinal Stress σL 2. Hoop Stress σh 3. Radial Stress σr 9/17/2022 2 Assumptions in thin cylinders: 1. σr is negligible compared to σL and σh 2. σh is uniform along the thickness 3. The longitudinal strain is uniform so that plane sections (transverse) remain plane before and after the application of internal pressure. In case of thick cylinders, 1 and 2 are not valid, but 3 is true. 9/17/2022 3 Thin Cylinder geometry: d = inner diameter t = thickness of cylinder wall If thin cylinder analysis Thin cylinder is subjected to internal fluid pressure ‘p’ hoop stress (tensile) longitudinal stress (tensile) (if ends are closed) radial stress (compressive) neglected 9/17/2022 4  Thin cylinders Thick cylinders 1. low internal fluid pressure medium 1 internal fluid pressure 2. radial stress is neglected when radial stress is not neglected compared to Hoop stress and and it is varying over the wall Longitudinal stress thickness of cylinder 3. 2-dimensional stress system 3-dimensional stress system Hoop stress and Longitudinal Hoop stress, Longitudinal stress stress and Radial stress 4. hoop stress is assumed to be hoop stress is varying over the constant over the wall wall thickness of cylinder thickness of cylinder 9/17/2022 5 Assumptions in Lame’s theory: The material of thick cylinder is homogeneous & isotropic The material of thick cylinder obeys Hooke’s law. Longitudinal stress is same at any point in the thick cylinder Plane sections which are normal to the axis of thick cylinder remain plane even under the internal fluid pressure i.e., longitudinal strain is same at all points i.e,. independent of the radius longitudinal strain εl = constant at all points i.e. it is not function 9/17/2022 of radius ‘x’ 6  = Constant E and μ are material constants Longitudinal stress σl is also constant from Lame’s assumption Note: at any point in the thick cylinder, the algebraic sum of the hoop stress and radial stress is equal to 2A Note: at any point in the thick cylinder, the radial stress is always compressive and its magnitude is equal to radial pressure i.e., px. 9/17/2022 7 Where A is called Lame’s constant Consider a thick cylinder of r1 = outer radius & r2 = inner radius p = internal fluid pressure acts at r2 Outer surface of thick cylinder is under atmospheric pressure i.e. datum (i.e. zero) Therefore, the pressure varies from p (at r2) to zero (at r1) 9/17/2022 8 Consider an elemental ring at radius ‘x’ and thickness ‘dx’ in the cross section of thick cylinder. px = radial pressure at radius ‘x’ px+dpx = radial pressure at radius ‘x+dx’ where dpx = change in radial pressure over ‘dx’ thickness Consider failure of thin ring having radius ‘x’ and thickness ‘dx’ Bursting force = pressure x projected area of curved surface onto horizontal plane = = = by neglecting higher order terms 9/17/2022 9 Resisting force = for equilibrium, Bursting force = Resisting force From and By integrating on both sides and 9/17/2022 10 Lame’s equations for Radial pressure and Hoop stress Thick Cylinder The ends of thick cylinder are closed with plates rigidly r1 = outer radius & r2 = inner radius Lame’s equations: hoop stress and radial pressure Thick Cylinder subjected to Thick Cylinder subjected to internal fluid pressure only external fluid pressure only ends are closed pi = internal fluid pressure acts at r2 po = external fluid pressure acts at r1 Boundary conditions: Boundary conditions: (i) at radius x = r2, radial pressure px = pi (i) at radius x = r2, radial pressure px = 0 (ii) at radius x = r1, radial pressure px = 0 (ii) at radius x = r1, radial pressure px = po Lame’s constants Lame’s constants Stresses: Tensile +ve 9/17/2022 Compressive –ve 11  maximum hoop stress maximum hoop stress (occurs at inner surface) (occurs at inner surface) minimum hoop stress minimum hoop stress (occurs at outer surface) (occurs at outer surface) maximum radial stress maximum radial stress (occurs at inner surface) (occurs at outer surface) minimum radial stress minimum radial stress (occurs at outer surface) (occurs at inner surface) 9/17/2022 12 Longitudinal stress: Longitudinal stress: pi = internal fluid pressure po = external fluid pressure i.e., radial pressure on the i.e., radial pressure on the outer inner surface of thick surface of thick cylinder as well cylinder as well as on lids as on lids The force exerted by fluid pressure on the lids causes longitudinal stress in the thick cylinder. Bursting force = Bursting force = This bursting force is resisted by resistance generated on the cross section of thick cylinder. This resistance is parallel to longitudinal axis of thick cylinder. Resisting force = Resisting force = for equilibrium, Bursting force = Resisting force 9/17/2022 13  for equilibrium, Bursting force = Resisting force tensile compressive Note: If the ends are opened then 9/17/2022 14 Stresses in thick cylinders: Stresses are depends on the nature of loading which may be i. Only internal pressure (Pi) ii. Only an external pressure (Pe) iii. Both internal and external pressure (Pi and Pe) Only internal pressure: a. The hoop stress will be in tensile b. The longitudinal stress will be in tensile c. The radial stress will be compressive Only external pressure: a. The hoop stress will be compressive b. The longitudinal stress will be compressive c. The radial stress will be compressive 9/17/2022 15 Both internal and external stresses: a. The hoop stress will be tensile (if Pi > Pe) b. The longitudinal stress will be tensile (if Pi > Pe) c. The radial stress will be compressive Variation of Hoop stress and Radial stress over the wall thickness of thick cylinder To join the maximum hoop stress point and minimum hoop stress point To join the maximum radial stress point and minimum radial stress point Find the value of stress at the mid-point of wall thickness of thick cylinder and judge the curve 9/17/2022 16 9/17/2022 17 Thick cylinder subjected to internal fluid pressure only Thick cylinder subjected to external fluid pressure only 9/17/2022 18 Thick cylinder subjected to both internal and external fluid pressure only (internal fluid pressure is more than the external fluid pressure) 9/17/2022 19 Orthogonal Strains Hoop strain in thick cylinder Longitudinal strain in thick cylinder Radial strain in thick cylinder Volumetric strain (fluid stored in thick cylinder) is Volume (storage capacity) is Change is storage capacity of thick cylinder is 9/17/2022 20 Note: 1) Lame’s constants A and B are positive when the thick cylinder is subjected to internal fluid pressure only. 2) Lame’s constants A and B are negative when thick cylinder is subjected to external fluid pressure only. 3) Lame’s constant A is equal to the longitudinal stress in a thick cylinder, when ends are closed. 4) The numerical difference between maximum and minimum hoop stresses in a thick cylinder is equal to the numerical difference of internal and external fluid pressure. 5) Hoop stress and Longitudinal stress are Tensile and Radial stress is Compressive when the thick cylinder is subjected to internal fluid pressure only. 6) Hoop stress, Radial stress and Longitudinal stress are Compressive when the thick cylinder is subjected to external fluid pressure only. 9/17/2022 21  h,r2 h,r1 h(tensile) O r2 r1 r(comp.) r,r2 9/17/2022 22 Q1. A thick cylinder of 320 mm inner diameter and 480 mm outer diameter, is subjected to an internal fluid pressure of 8 MPa. The Young’s modulus of elasticity and Poisson’s ratio of cylinder material are 200 GPa and 0.3 respectively. 1. Draw the variation of stresses over the wall thickness of cylinder. 2. Find the orthogonal stresses at any point on the inner surface of thick cylinder when the ends are opened. 3. Find the orthogonal stresses at any point on the outer surface of thick cylinder when the ends are opened. 4. Find the orthogonal stresses at any point on the inner surface of thick cylinder when the ends are closed. 5. Find the orthogonal stresses at any point on the outer surface of thick cylinder when the ends are closed. 6. Calculate the change in wall thickness of cylinder when the ends of thick cylinder are closed. 7. Calculate the change in wall thickness of cylinder when the ends of cylinder are opened. 9/17/2022 23 Solution: Outer radius = r1 = 240 mm Inner radius = r2 = 160 mm Internal fluid pressure = pi = 8 MPa= 8 N/mm2(acts at r2)Young’s modulus of elasticity of cylinder material E = 200 GPa = 2x105 MPa = 2X105N/mm2 Poisson’s ratio of cylinder material ν = 0.3 Lame’s equations: Hoop stress Radial pressure Boundary conditions (i)at x = r2 =160 mm, px = 8 Mpa (ii) at x = r2 = 240 mm, px = 0 Lame’s constants A and B A = 6.4 B = 368640 9/17/2022 24  Hoop stresses Radial stresses 20.8 MPa (tensile) 8 MPa (compressive) 12.8 MPa (tensile) 0 Longitudinal stress (when ends are closed) = 6.4 MPa (tensile) (numerically equal to Lame’s constant A) Orthogonal stresses when ends are closed when ends are opened At inner surface20.8 MPa (T), 6.4 MPa (T), 20.8 MPa (T), 0MPa, 8 MPa (C) 8 MPa (C) At outer surface12.8 MPa (T), 6.4 MPa (T), 12.8 MPa (T), 0MPa, 0 MPa 0 MPa 9/17/2022 25 Hoop strain in thick cylinder Hoop strain at inner surface x = r2 tensile Hoop strain at outer surface x = r1 9/17/2022 tensile 26 Change in inner diameter = inner diameter x hoop strain at inner surface (increase) Change in outer diameter = outer diameter x hoop strain at outer surface (increase) (f) change in wall thickness (when ends are closed) = 0.003968 mm (decrease) (g) change in wall thickness (when ends are opened) 0.0032 mm (decrease) 9/17/2022 27 H.W.A thick cylinder of 120 mm and 200 mm as inner and outer diameters respectively, is subjected to an external fluid pressure of 40 MPa. Draw the variation of stresses over the wall thickness of cylinder. (Ans: A = -62.5, B = -225000, hoop stresses = 125 MPa & 85 MPa both compressive, radial stresses 0& 40 MPa compressive) H.W.A thick cylinder of 100 mm and 180 mm as inner and outer diameters respectively, is subjected to an internal fluid pressure of 45 MPa and external radial pressure of 10 MPa. Draw the variation of stresses over the wall thickness of cylinder. (Ans: A = 5.625, B = 126562.5, hoop stresses = 56.25 MPa &21.25 MPa both tensile, radial stresses 45 MPa &10 MPa both compressive) 9/17/2022 28 Q.2. Find the thickness of metal necessary for a cylindrical shell of internal diameter 160 mm to withstand an internal preassure of 8 MPa. The maximum hoop stress in the section is not to exceed 35 MPa. Solution: r2 = 80 mm Internal Pressure Px = 8 MPa At r2 = 80 mm, P r2 = 8 MPa and σ r2 = 35 MPa Maximum hoop stress occurs at the inner radius. R1 = External radius Hoop stress = Radial stress = Apply the boundary conditions, At r2 = 80 mm, P r2 = 8 MPa and σ r2 = 35 MPa 9/17/2022 29 8 = B/802 – A 35 = B/802 + A Subtracting the eq. A = 13.5 and B = 21.5 x 6400 Px = {(21.5 x 6400) / x2} -13.5 At x1 = r2, Px = 0 0 = {(21.5 x 6400) / r2 2} -13.5 r2 = 100.96 mm Thickness of shell = 100.96 – 80 = 20.96 mm 9/17/2022 30 Q.3.A thick cylinder having 160 mm and 200 mm inner and outer diameters respectively has been designed to withstand a certain internal fluid pressure but re-boring became necessary. Determine the limit to the new inside diameter if the maximum hoop stress is not to exceed the previous value by more than 5% while the internal fluid pressure is same as before. (Ans: r = 80.8813 mm) Solution: Before re boring After re boring, the new value of inner radius = r The maximum hoop stress is not to exceed the previous value by more than 5% 9/17/2022 31 Q.4.A steel cylinder of 200 mm external diameter and 150 mm internal diameter is used for a working internal pressure of 12 MPa. Owing to external corrosion, the external diameter of the cylinder has to be machined down to 195 mm, find how much the working pressure must be reduced for the maximum hoop stress to be the same as before.(Ans: p = 10.9931 mm) Solution: Before external corrosion After external corrosion p = new value of int. fluid pressure D = 195 mm (outer diameter) 9/17/2022 32 Q.5.If a cylinder of internal diameter d, wall thickness t and subjected to internal pressure only, is assumed to be a thin cylinder, what is the greatest value for the thickness and internal diameter ratio, if the error in the estimated maximum hoop stress is not to exceed 5%. Solution: internal fluid pressure = and wall thickness and inner diameter ratio for thin cylinder for thick cylinder 9/17/2022 33 Maximum hoop stress in thick cylinder error in the estimated maximum hoop stress is not to exceed 5% 9/17/2022 34  or A closed cylinder has an internal diameter 420 mm and an external diameter 520 mm. It is 1.5 m long and is subjected to an internal pressure of 8 MPa. Draw the distribution of hoop and radial stresses across the thickness. Determine the change in internal volume and thickness. Take E = 200 GPa and ν = 0.3 9/17/2022 35 Compound cylinders  To enhance the strength of existing thick cylinder. One thick cylinder over another thick cylinder. Initially the inner radius of outer cylinder is less than the outer radius of inner cylinder. Outer cylinder is subjected to rise of temperature so that the outer cylinder slides freely on the inner cylinder. After sliding the outer cylinder over the inner cylinder, the temperature is brought back to room temperature. This process is called shrink-fitting. After shrink-fit, both surfaces will settle at radius of r3 and both cylinders 9/17/2022 act together as single cylinder. 36 Due to shrink-fit, outer cylinder exerts radial pressure on the inner cylinder. This radial pressure is called shrinkage pressure (ps), which is at common surface i.e., radius of r3. Therefore, the inner cylinder is under hoop compression and outer cylinder is under hoop tension. 9/17/2022 37 Inner Cylinder Outer Cylinder Inner radius of inner cylinder = Outer radius of outer cylinder = Radius of both cylinders at common surface (junction) = Before Fluid is admitted Radial pressure (shrinkage pressure) at common surface = Lami’s constants: and Lami’s constants: and Boundary conditions: Boundary conditions: 1) at x = r2, px = 0 1) at x = r1, px = 0 2) at x = r3, px =ps  2) at x = r3, px =  find and (both are negative) find and (both are positive) 9/17/2022 38 Hoop stresses: Hoop stresses:.. (Compressive).. (Tensile).. (Compressive).. (Tensile) After Fluid is admitted Both cylinders act together as a single cylinder i.e., compound cylinder Radial pressure (fluid pressure) at inner surface of compound cylinder = Lami’s constants: A and B Boundary condition: at x = r2 , px = Boundary condition: at x = r1 , px = 0 9/17/2022 find A and B 39 Hoop Stresses: (Tensile) (Tensile) at common surface (Tensile) due to internal fluid pressure in the compound cylinder, the radial pressure at common surface is 9/17/2022 40 Final Stresses: Tensile +ve Compressive -ve Inner Cylinder Outer Cylinder Hoop stress at x = r2 x = r 3 x = r3 at x = r1 Due to shrinkage pressure ps (C) (C) (T) (T) Due to Internal fluid pressure pf (T) (T) (T) (T) Total Inner Cylinder Outer Cylinder Radial stress at x = r2 x = r3 x = r3 at x = r1 Due to shrinkage pressure ps 0 - ps - ps 0 Due to Internal fluid pressure pf - pf (.. ) (.. ) 0 Total (All are compressive) 9/17/2022 41  Shrink-fit allowance (Initial difference of radii at junction) Inner Cylinder Outer Cylinder Inner radius of inner cylinder = Outer radius of outer cylinder = r2 r1 Radius of both cylinders at common surface (junction) = r3 9/17/2022 42 Inner Cylinder Outer Cylinder Inner radius of inner cylinder = Outer radius of outer cylinder = Radius of both cylinders at common surface (junction) = Before Fluid is admitted Radial pressure (shrinkage pressure) at common surface = Lami’s constants: and Lami’s constants: and Boundary conditions: Boundary conditions: 1) at x = r2, px = 0 1) at x = r1, px = 0 2) at x = r3, px =ps  2) at x = r3, px =  find and (both are negative) find and (both are positive) 9/17/2022 43 Hoop stresses: Hoop stresses:.. (Compressive).. (Tensile).. (Compressive).. (Tensile) Hoop strain at junction: (from Hoop strain at junction: (from stresses) stresses (Compressive) (tensile) 9/17/2022 44 Hoop strain at junction: (from Hoop strain at junction: (from deformations) deformations) difference between the difference between the initial outer radius of inner initial inner radius of outer cylinder and r3 i.e., the cylinder and r3 i.e., the negative deformation due to positive deformation due the shrink-fit of outer to the shrink-fit of outer cylinder over the inner cylinder over the inner cylinder. cylinder. 9/17/2022 45 Shrink-fit allowance 9/17/2022 46 The initial difference in diameters at junction of both cylinders = Temperature of the outer cylinder to be raised = t Coefficient of thermal expansion of outer cylinder material =  9/17/2022 47 Q. Find the ratio of thickness to internal diameter for a tube subjected to internal pressure when the ratio of internal pressure to maximum circumferential stress is 0.5. Find the change in thickness of metal in such a tube of 200 mm internal diameter when the internal pressure is 75 MPa. E = 200 GPa, ν = 0.3. and   Then 9/17/2022 48  increase 150 – 75 = 75 MPa increase Decrease in thickness = (decrease) 9/17/2022 49 Q. A compound cylinder is made by shrinking a cylinder of external diameter 300 mm and internal diameter of 250 mm over another cylinder of external diameter 250 mm and internal diameter 200 mm. The radial pressure at the junction after shrinking is 8 MPa. Find the final stresses set up across the section, when the compound cylinder is subjected to an internal fluid pressure of 84.5 Mpa. Ans: For outer cylinder : External radius r1 = 150 mm Radius at the junction r3 = 125 mm For inner cylinder: Internal radius r2 = 100 mm Radial pressure due to shrinkage at the junction ps = 8MPa Fluid pressure in the compound cylinder pf = 84.5 MPa 9/17/2022 50 i) Stresses due to shrinking in the outer and inner cylinders before the fluid pressure is admitted. a) Lame’s equations for outer cylinder: Boundary conditions: 1) at x = r1, px = 0 2) at x = r3, px =  0 = (b1/1502) – a1 ---- (1) 8 = (b1/1252) – a1 ---- (2) Subtracting (2) from (1) b1 = 409090.9 a1 = 18.18 Hoop stress σx = (409090.9/x2) + a1 Hoop stress in the outer cylinder due to shrinking 9/17/2022 51 σ150 = (409090.9/1502) + 18.18 = 36.36 MPa (Tensile) σ125 = (409090.9/1252) + 18.18 = 44.36 MPa (Tensile) b) Lame’s equations for the inner cylinder: 1) at x = r2, px = 0 2) at x = r3, px =ps  0 = (b2/1002) – a2 ----(3) 8 = (b2/1252) – a2 ----(4) Subtracting (3) from (4) b2 = -222222.22 a2 = -22.22 9/17/2022 52 Hoop stress σx = -(222222.2/x2) – a2 σ125 = (-222222.2/1252) – 22.22 = -36.44 MPa (Compressive) σ100 = (-222222.2/1002) – 22.22 = -44.44 MPa (Compressive) ii) Stresses due to fluid pressure alone When the fluid pressure is admitted inside the compound cylinder, the two cylinders together will be considered as one single unit. Boundary condition: at x = r2 , px = pf Boundary condition: at x = r1 , px = 0 9/17/2022 53 84.5 = B/1002 – A ------ (5) 0 = B/ 1502 – A ------ (6) Subtracting (6) from (5) B = 1521000 A = 67.6 Hoop stress σx = (1521000/x2) + A σ100 = (1521000/1002) + 67.6 = 219.7 MPa (Tensile) σ125 = (1521000/1252) + 67.6 = 164.94 MPa (Tensile) σ150 = (1521000/1502) + 67.6 = 135.2 MPa (Tensile) 9/17/2022 54 The resultant stresses will be the algebraic sum of the initial stresses due to shrinking and those due to internal fluid pressure Inner Cylinder: F100 = σ100 due to shrinkage + σ100 due to internal fluid pressure = -44.44 + 219.7 = 175.26 MPa (Tensile) F125 = σ125 due to shrinkage + σ125 due to internal fluid pressure = -36.44 + 164.94 = 128.5 MPa (Tensile) Outer Cylinder: F125 = σ125 due to shrinkage + σ125 due to internal fluid pressure = 44.36 + 164.94 = 209.3 MPa (Tensile) F150 = σ150 due to shrinkage + σ150 due to internal fluid pressure = 36.36 + 135.2 = 171.56 MPa (Tensile) 9/17/2022 55 Q. A compound cylinder is made by shrinking a jacket on to a cylinder. For the compound cylinder, the outer and inner radii are 100 mm and 60 mm, and the radius at the junction is 80 mm. Before the fluid pressure of 40 MPa is applied, the radial pressure at the junction is 10 MPa. Determine the final stresses in the cylinder. Also calculate the difference in the diameters of tubes before the jacket is shrunk on to the cylinder and the temperature at which this can be done. Take E = 200 GPa and α = 12x10-6/°C 9/17/2022 56 Q. A steel cylinder of 300 mm external diameter is to be shrunk to another cylinder of 150 mm internal diameter. After shrinking, the diameter at the junction is 250 mm and radial pressure at the common junction is 28 MPa. Find the original difference in radii at the junction. Take E = 200 Gpa. Ans: for Outer cylinder: px = (b1/r32) – a1 At x = 150 mm, px = 0 At x = 125 mm, px = 28 MPa. (b1/1502) – a1 = 0 (b1/1252) – a1 = 28 Then b1 = 1432000, a1 = 63.6 9/17/2022 57 for inner cylinder: px = (b2/r32) – a2 At x = 75 mm, px = 0 At x = 125 mm, px = 28 MPa. (b2/752) – a2 = 0 (b2/1252) – a2 = 28 Then b2 = -216100, a2 = - 43.75 Difference of radii at junction = (σh)outer = (b1/r32) + a1 = (σh)inner = (b2/r32) + a2 = Difference in radii at junction = 0.13 mm 9/17/2022 58 Q. A steel tube of 200 mm external diameter is to be shrunk onto another steel tube of 60 mm internal diameter. The diameter at the junction after shrinking is 120 mm. Before shrinking ob, the difference of diameters at the junction is 0.08 mm. Calculate the radial pressure at the junction and the hoop stresses developed in the two tubes after shrinking on. Take E = 200 GPa. Ans: Original difference of radii at the junction = 2r3/E (a1 – a2) = 0.04 for Outer cylinder: px = (b1/x2) – a1 9/17/2022 59 At x = 100 mm, px = 0 (b1/1002) – a1 = 0 then b1 = 1000a1 ------ (1) At x = 60 mm, px = ps (b1/602) – a1 = ps ------ (2) for Inner cylinder: px = (b2/x2) – a2 At x = 30 mm, px = 0 (b2/302) – a2 = 0 then b2 = 900a2 ------ (3) At x = 60 mm, px = ps (b2/602) – a2 = ps ------ (4) 9/17/2022 60 Equating equations (2) and (4) (b1/602) – a1 = (b2/602) – a2 (b1 –b2)/3600 = (a1 – a2) From (1) and (3) (1000a1 – 900 a2) / 3600 = (a1 – a2) a2 = - (27/64) a1 ------ (5) a1 = - 46.88 a2 = 19.77 b1 = - 42192 b2 = 197700 ps = (b1/602) – a1 = 35.16 MPa 9/17/2022 61 Hoop stresses: For outer cylinder: (σh)outer = (b1/x2) + a1 σ100 = (197700/1002) + 19.77 = 39.54 N/mm2 σ60 = (197700/602) + 19.77 = 74.68 N/mm2 For outer cylinder: (σh)inner = (b2/x2) + a2 σ60 = - (42192/602) – 46.88 = - 58.6 N/mm2 σ30 = - (42192/302) – 46.88 = - 93.76 N/mm2 Q. A steel tube of 240 mm outer diameter is to be shrunk on to another tube of 80 mm internal diameter. The common diameter at the junction of two tubes is to be 160 mm after shrinking-on. The original difference of diameters of the two tubes at the junction is 0.08 mm before shrinking-on. Find the stresses in the two tubes after shrinking-on. The Young’s modulus of tube material is E= 200 GPa. 9/17/2022 62 Q. Two thick steel cylinders A and B, closed at the ends, have the same dimensions, the outer diameter being 1.6 times the inner diameter. The cylinder A is subjected to internal pressure only and the cylinder B is subjected to external pressure only. Find the ratio of these pressures, (a) when the greatest circumferential stress has the same numerical value, and (b) when the greatest circumferential strain has the same numerical value.ν = 0.304. (Ans: (a) 1.4382 (b) 1.16394) Solution: for both cylinders r1 = 1.6 r2 Cylinder A is subjected to internal fluid pressure pi only ----------- (1) 9/17/2022 63  --------- (2) Cylinder B is subjected to external fluid pressure po only ------- (3) 9/17/2022 64  --------------- (4) For numerically same stress, equate eq. (1) and (3) For numerically same strains, equate eq. (2) and (4) 9/17/2022 65