Thick cylinders

hoop stress or circumferential stress and radial stress Lame's equation, t > d/20

1. Thick Cylinders by Venkata Sushma chinta

2. Thick cylinders The thickness of the cylinder is large compared to that of thin cylinder. i . e., in case of thick cylinders, the metal thickness ‘t’ is more than ‘d/ 20 ’, where ‘d’ is the internal diameter of the cylinder. Magnitude of radial stress ( p r ) is large and hence it cannot be neglected. The circumferential stress is also not uniform across the cylinder wall. The radial stress is compressive in nature and circumferential and longitudinal stresses are tensile in nature. Radial stress and circumferential stresses are computed by using ‘Lame’s equations’ . DJ996

3. LAME’S EQUATION Assumptions : Plane sections of the cylinder normal to its axis remain plane and normal even under pressure. Longitudinal stress ( σ L ) and longitudinal strain ( ε L ) remain constant throughout the thickness of the wall. Since longitudinal stress ( σ L ) and longitudinal strain ( ε L ) are constant, it follows that the difference in the magnitude of hoop stress and radial stress (p r ) at any point on the cylinder wall is a constant. The material is homogeneous, isotropic and obeys Hooke’s law. (The stresses are within proportionality limit).

4. Consider a thick cylinder of external radius r 1 and internal radius r 2 , containing a fluid under pressure ‘p’ as shown in the fig. Let ‘L’ be the length of the cylinder.

5. Consider an elemental ring of radius ‘r’ and thickness ‘ δ r ’ as shown in the figures. Let p r and (p r + δ p r ) be the intensities of radial pressures at inner and outer faces of the ring. p r + δ p r p r r p r + δ p r p r r σ c σ c r δ r P r p r + δ p r External pressure DJ996

6. Consider the longitudinal section XX of the ring as shown in the fig. The bursting force is evaluated by considering the projected area: ‘2rL’ for the inner face and ‘2(r+δr)L’ for the outer face . The net bursting force, P = p r . 2 r L - (p r + δ p r )2 . ( r + ) . L =( -p r . – r . δ p r - δ p r . ) 2L Bursting force is resisted by the hoop tensile force developing at the level of the strip i.e., F r = σ c 2 . L  

7. Thus, for equilibrium, P = F r (-p r . – r . δ p r - δ p r . ) 2L = σ c . 2 . . L -p r . δ r – r . δ p r - δ p r . = σ c . δ r Neglecting products of small quantities, (i.e., δ p r . δ r) σ c = - p r – ( r . δ p r )/ δ r ...…………….(1) Longitudinal strain is constant. Hence we have,   = 2a; σ c = p r + 2a …………(2)  

8. p r + 2a = - p r – ( r . δ p r )/ 2( p r + a)= – ( r . δ p r )/ = ...........(3)   Integrating (3), (-2 ×log e r ) + c = log e (p r + a) log e ( p r +a ) = -2 ×log e r + log e b => log e ( p r +a ) = log e (   i.e., p r  a    ............(5) Substituting it in equation 2, we get radial stress, p r  -a   Hoop stress, σ c  a   ...........(4) σ c = p r + 2a σ c = -a + 2a  

9. Points to remember: Variations of Hoop stress and Radial stress are parabolic across the cylinder wall. At the inner edge, the stresses are maximum. The value of ‘Permissible or Maximum Hoop Stress’ is to be considered on the inner edge.

10. Q1.A steel cylinder of 160mm internal diameter and 320mm outer diameter. If it is subjected to an internal pressure of 150MPa, find the maximum radial and tangential stress and the maximum shear stress. Assume the ends are closed. r i = 80mm r o =160mm P i =150MPa radial stress, p r  -a The boundary conditions to solve constants are : At radius r= r i , internal pressure is applied so p r =150 At radius r= r o , no external pressure is applied acting , so p r =0  

11. p r  -a 150=-a+ ….(1) 0 =-a+ …..(2) Solving equation (1) and (2) a= 50 ,b= 1280000 p r  -50   σ c = a σ c =50 Hoop stress at inner radius is r i = 80mm σ c (max) = 50 Hoop stress at outer radius is r o =160mm σ c (min) = 50  

12. Maximum shear stresses τ max = =  

13. Q2. A cylinder is 150 mm ID and 450 mm OD. The internal pressure is 160 MPa and the external pressure is 80 MPa. Find the maximum radial and tangential stresses and the maximum shear stress. The ends are closed r i = 75mm r o =225mm P i =160MPa P o =80MPa radial stress, p r  -a The boundary conditions to solve constants are : At radius r= r i =75mm , internal pressure is applied so p r =160 MPa At radius r= r o =225mm ,external pressure is applied , so p r =80 MPa   160=-a+ ….(1) 80 =-a+ …..(2)   Solving equation (1) and (2) a= -70 ,b= 506250 p r  7  

14. σ c = a σ c =-70 Hoop stress at inner radius is r i = 75mm σ c (max) = -70 Hoop stress at outer radius is r o =225mm σ c (min) = -70   a= -70 ,b= 506250 Maximum shear stresses τ max =  

15. Q3. A cylinder has an ID of 100 mm and an internal pressure of 50 MPa. Find the needed wall thickness if the factor of safety n is 2.0 and the yield stress is 250 MPa. Use the maximum shear stress theory, i.e. maximum shear stress = yield strength/2n. r i = 50mm r o =? r o - r i = t P i =50MPa P o =0MPa Maximum shear stresses τ max = = yield strength/2n = 125 = 125 = 75  

16. radial stress, p r  -a ; hoop stress is σ c = a The boundary conditions to solve constants are : At radius r= r i =50mm , internal pressure is applied so p r =50 MPa At radius r= r i =50mm , hoop stress is σ c =62.5 MPa   50= - a+ ….(1) 75 =a+ …..(2)   Solving equation (1) and (2) a= 12.5,b= 156250 p r  -12.5   At radius r= r o , No external pressure is applied , so p r =0 MPa  -12.5   On solving r o = 111.8 mm => t= 111.8 -50=61.8mm