1. Mechanics of Materials II UET, Taxila Lecture No. (6)

2. Final form of hoop stress

3. Longitudinal stress final form

4. This means that: Longitudinal stress = ½ of the hoop stress Or: Hoop stress = 2 Longitudinal stress

5. From Hooke’s Law

6. From Hooke’s Law: Circumferential strain  H ) or  θ )=

7. Change in Diameter

8. (c) Change in internal volume Volumetric strain = change in volume /original volume Change in volume = volumetric strain x original volume

10. Volumetric strain = One longitudinal strain + 2 Diametral strains

11. Final form for change of volume equation: Change in volume = volumetric strain x original volume if “V” is the original volume

12. Thin rotating ring or cylinder

13. Consider a thin ring or cylinder as shown in Next Figure Rotating thin ring or thin cylinder. pcpc

14. subjected to a radial pressure p c caused by the centrifugal effect of its own mass when rotating. The centrifugal effect on a unit length of the circumference is: P c = m  2 r

15. Where: p c is the pressure, m is the mass,  is the angular velocity (rad/s) and r is the inner radius of the cylinder.

16. Thus, considering the equilibrium of half the ring shown in the figure: c c

17. where F is the hoop tension set up owing to rotation. The cylinder wall is assumed to be so thin that the centrifugal effect can be taken to be constant across the wall thickness then: c

18. Hoop stress owing to rotation This tension is transmitted through the complete circumference and therefore is restricted by the complete cross- sectional area then: where A is the cross-sectional area of the ring.

19. Now with unit length assumed, m/A is the mass of the ring material per unit volume, i.e. the Density . Then hoop stress owing to rotation:

20. Thin spherical shell under internal pressure

22. Because of the symmetry of the sphere the stresses set up owing to internal pressure will be two mutually perpendicular hoop or circumferential stresses of equal value and a radial stress.

23. As with thin cylinders having thickness to diameter ratios less than 1 : 20, the radial stress is assumed negligible in comparison with the values of hoop stress set up.

24. The stress system is therefore one of equal biaxial hoop stresses.

25. Consider, therefore, the equilibrium of the half-sphere shown in Next Figure: Half of a thin sphere subjected to internal pressure showing uniform hoop stresses acting on a surface element.

26. For Sphere

28. Change in internal volume for sphere

29. Change in internal volume As for the cylinder: change in volume = original volume x volumetric strain

30. But: volumetric strain = sum of three mutually perpendicular strains (in this case all equal)

31. Sphere volumetric strain

32. By considering V as the original volume

33. Vessels subjected to fluid pressure

34. If a fluid is used as the pressurisation medium the fluid itself will change in volume as pressure is increased.

35. This must be taken into account when calculating the amount of fluid which must be pumped into the cylinder in order to raise the pressure by a specified amount, the cylinder being initially full of fluid at atmospheric pressure.

36. Bulk Modulus The bulk modulus (K) of a substance measures the substance's resistance to uniform compression.

37. Definition of bulk modulus It is defined as the pressure increase needed to effect a given relative decrease in volume.

38. Now the bulk modulus of a fluid is as follows: bulk modulus “K” =volumetric stress/ volumetric strain

39. where, in this case, volumetric stress = pressure p And Volumetric strain = Change in volume / original volume = V/V

40. Then:

41. Change of volume under pressure = pV / k

42. The extra fluid required to raise the pressure must, therefore, take up this volume together with the increase in internal volume of the cylinder itself.

43. Then: extra fluid required to raise cylinder pressure by p =

44. For Sphere Similarly, for spheres, the extra fluid required is