Thick Walled cylinders and Spheres
Thick Walled CylindersUnder the action of radial Pressures at the surfaces, the three Principal Stresses will be : and . These Stresses may be expected to vary over any cross-section and equations will be found which give their variation with the radius r. It is assumed that the longitudinal Strain e is constant. This implies that the cross-section remains plain after straining and that this will be true for sections remote from any end fixing. Let u be the radial shift at a radius r. i.e. After Straining the radius r becomes (r + u). and it should be noted that u is small compared to r. The increase in Hoop Strain ( Circumferential) is given by:-
Internal Pressure OnlyPressure Vessels are found in all sorts of engineering applications. If it assumed that the Internal Pressure is at a diameter of and that the external pressure is zero ( Atmospheric) at a diameter then using equation (22)
The Error In The "thin Cylinder" FormulaIf the thickness of the cylinder walls is t then and this can be substituted into equation (43)
- The cylinder of a Hydraulic Ram has a 6 in. internal diameter. Find the thickness required to withstand an internal pressure of 4 tons/sq.in. The maximum Tensile Stress is limited to 6 tons/sq.in. and the maximum Shear Stress to 5 tons/sq.in. If D is the external diameter, then the maximum tensile Stress is the hoop Stress at the inside. Using equation (43)
- Find the ratio of thickness to internal diameter for a tube subjected to internal pressure when the ratio of Pressure to Maximum circumferential Stress is 0.5. Find also the alteration of thickness of metal in such a tube 8 in internal diameter when the pressure is 5 tons/sq.in.
- The maximum permitted Stress permitted in a thick Cylinder of radii 8 in. and 12 in. is 2000 lb./sq.in. If the external pressure is 600 lbs/sq.in. what Internal pressure can be applied? Plot curves showing the variation of hoop and radial stresses throughout the material. Using Equations (22) and (23)
- Two thick steel cylinders A and B , closed at the ends have the same dimensions, the inside being 1.6 times the inside. A is subjected to internal pressure only and B to external pressure only. Find the ratio of these two pressures:-
- When the greatest Circumferential Stress has the same numerical value.
- When the greatest circumferential Strain has the same numerical value.
Cylinder BUsing equation (22) the External Pressure Likewise as the Internal Pressure is zero. Solving these equations for a and b:-
The Plastic Yielding Of Thick Tubes.If the internal pressure of a thick cylinder is increased sufficiently, yielding will occur starting at the Internal surface and spreading outwards until the whole cross-section becomes plastic. The Strains generated will not initially be excessive since there will be a ring of elastic material around the plastic zone. Collapse will only happen once the final stage is reached. If the Pressure for complete plasticity can be estimated and used as the "collapse" pressure then a design pressure can be derived by dividing by a suitable "Load Factor". This system was used in the pages on "The Plastic Theory of Bending". A useful application of partial plasticity can be used in the manufacture of gun barrels and pressure vessels. The process involves deliberately overstraining the component with a high internal pressure during manufacture. The result is to impart compressive stresses into the inner layers and this has the effect of reducing the Hoop Stress which occur under normal working conditions. Note the same effect can be achieved by shrinking one tube over another. Assumptions made in the Theory of Plastic Yielding.
- Yield takes place when the maximum Stress difference ( or Shear Stress) reaches the value corresponding the yield in simple tension. This assumption has been found to be in good agreement with experimental results for ductile material.
- The Material exhibits a constant yield Stress in tension and there is NO Strain hardening. i.e. it is an ideal elastic material.
- The longitudinal Stress in the tube is either zero or lies algebraically between the Hoop and Radial Stresses. From this it follows that the maximum Stress difference is determined by the Hoop and Radial Stresses only.
Partially Plastic WallConsider a thick tube of internal radius and external radius subjected to an internal pressure only of such a magnitude, that the material below a radius of is in the plastic state. (i.e. is the boundary between the inner plastic region and the outer elastic region) If is the radial Stress at , it is stated by the elastic theory for internal pressure only (Equation (45))that the maximum Stress difference is (i.e.) just reaching the yield conditions at
- A gun barrel of 4 in. bore and with a wall thickness of 3 in, is subjected to an internal pressure sufficient to cause yielding in two thirds of the metal. Calculate this pressure and show the variation of Stresses across the wall. What are the pressures required for initial ield and complete yield? Assume that yield occurs due to maximum shear stress and neglect Strain Hardening. In simple Tension equals 25 tons/sq.in. Using Equation (108) which gives the pressure required to produce a given depth of yielding where
Compound Tubes.It can be seen from the figure that the Hoop falls off appreciably as the material near the outside of the tube is not being stressed to the limit. In order to even out the stresses the tube may be made in two parts.one part being shrunk on to the other on to the other (after heating). By this means the inner tube is put into compression and the pouter tube in tension. When an internal pressure is then applied it causes a tensile Hoop Stress to be superimposed on the "Shrinkage" Stresses and the resultant Stress is the algebraic sum of the two sets. In general, the procedure is to use the knowledge of the radial pressure at the common surface, to calculate the stresses due to shrinkage in each component. The Stresses due the application of internal pressure are calculated in the usual way. Provided that the tubes are made of the same material, the two tubes may be treated as one. The radial pressure at the common surface due to shrinkage is related to the diametral "interference" before the tubes are fitted together. If is the Compressive Hoop stress at the outside of the inner tube and is the tensile Hoop Stress at the inside of the outer tube then due to shrinkage the inner tube diameter is decreased by:-
- A tube that is 4 in. inside and 4 in. outside diameter is strengthened by shrinking on a second tube of 8 in. outside diameter. The compound tube is to withstand an internal pressure of 5000 lb./sq.in. and the shrinkage allowance is to be such that the final maximum stress in each tube is to be the same. Calculate this stress and show in a diagram the variations of Hoop Stress in the two tubes. What is the initial difference of diameters prior to assembly ?
A Hub Shrunk Onto A Solid ShaftThe Shaft will be subjected to an external pressure and if are the Hoop and Radial Stresses at a radius r, the equilibrium equation (14) will be obtained as for a "Thick Cylinder".
- A Steel shaft 2 in. in diameter is to be pressed into a cast iron hub of 6 in. diameter and 4 in. lomng, so that no slip occurs under a torque of 20 ton-in. Find the necessary force fit allowance and the maximum circumferential Stress in the hub. for both. The coefficient of friction between th the surfaces is 0.3 If after assembly the shaft is subjected to an axial compressive Stress of 5 tons /sq.in., what is the resulting increase in the maximum circumferential Hub Stress
If is the increase in the maximum Hoop Stress in the hub whne a axial stress of 5 tons/sq.in. is applied to the shaft, then the corresponding increase in Radail Pressure at the inside surface is determined by the dimensions of the hub.