Machine Design Lec 6 & 7 PDF

CHAPTER 5 Pressure vessels 3.1 Introduction The pressure vessels (i.e. cylinders or tanks) are used to store fluids under pressure. The fluid being stored may undergo a change of state inside the pressure vessel as in case of steam boilers or it may combine with other reagents as in a che...

CHAPTER 5 Pressure vessels 3.1 Introduction The pressure vessels (i.e. cylinders or tanks) are used to store fluids under pressure. The fluid being stored may undergo a change of state inside the pressure vessel as in case of steam boilers or it may combine with other reagents as in a chemical plant. The pressure vessels are designed with great care because rupture of a pressure vessel means an explosion which may cause loss of life and property. The material of pressure vessels may be brittle such as cast iron, or ductile such as mild steel. 3.2 Classification of Pressure Vessels The pressure vessels may be classified as follows: 1. According to the dimensions. Thickness of the shell (𝑡) < 𝟏/𝟏𝟎 of the diameter of the shell (d), → thin shell. Thickness of the shell (𝑡) > 𝟏/𝟏𝟎 of the diameter of the shell, → thick shell. Thin shells are used in boilers, tanks and pipes. Thick shells are used in high pressure cylinders, tanks etc. 2. According to the end construction. A simple cylinder with a piston such as cylinder of a press is an example of an open end vessel. A tank is an example of a closed end vessel. Open ends, the circumferential or hoop stresses are induced by the fluid pressure. Closed ends, longitudinal stresses in addition to circumferential stresses are induced. 3.3 Stresses in a Thin Cylindrical Shell due to an Internal Pressure When a thin cylindrical shell is subjected to an internal pressure, it is likely to fail in the following two ways: 1. along the longitudinal section (i.e. circumferentially) splitting the cylinder into two troughs, as shown in Fig. 3.1 (a). 2. across the transverse section (i.e. longitudinally) splitting the cylinder into two cylindrical shells, as shown in Fig. 3.1 (b). The wall of a cylindrical shell subjected to an internal pressure has to withstand tensile stresses of the following two types: (a) Circumferential or hoop stress (σc). (b) Longitudinal stress (σt ). 3.4 Circumferential or Hoop Stress Consider a thin cylindrical shell subjected to an internal pressure as shown in (a) and (b). A tensile stress acting in a direction tangential to the circumference is called circumferential or hoop stress. In other words, it is a tensile stress on longitudinal section (or on the cylindrical walls). Let 𝑝 = Intensity of internal pressure, 𝑑 = Internal diameter of the cylindrical shell, 𝑙 = Length of the cylindrical shell, 𝑡 = Thickness of the cylindrical shell, and 𝜎𝑡1 = 𝜎𝑐 = Circumferential or hoop stress for the material of the cylindrical shell. We know that the total force acting on a longitudinal section (i.e. along the diameter X-X) of the shell = Intensity of pressure × Projected area = 𝑝 × 𝑑 × 𝑙 (i) and the total resisting force acting on the cylinder walls = 𝜎𝑡1 × 2𝑡 × 𝑙 (∵ of two sections) (ii) From equations (i) and (ii), we have 𝜎𝑡1 × 2𝑡 × 𝑙 = 𝑝 × 𝑑 × 𝑙 𝑝×𝑑 𝑝×𝑑 or 𝜎𝑡1 = or 𝑡= (iii) 2𝑡 2𝜎𝑡1 The following points may be noted: 1. In the design of engine cylinders, a value of 6 mm to 12 mm is added in equation (iii) to permit reboring after wear has taken place. Therefore 𝑝×𝑑 𝑡= + 6 𝑡𝑜 12 𝑚𝑚 2 𝜎𝑡1 2. In constructing large pressure vessels like steam boilers, riveted joints or welded joints are used in joining together the ends of steel plates. In case of riveted joints, the wall thickness of the cylinder, 𝑝×𝑑 𝑡= 2 𝜎𝑡1 × 𝜂𝑙 where 𝜂𝑙 = Efficiency of the longitudinal riveted joint. 3. In case of cylinders of ductile material, the value of circumferential stress (𝜎𝑡1) may be taken 0.8 times the yield point stress (𝜎𝑦) and for brittle materials, σt1 may be taken as 0.125 times the ultimate tensile stress (𝜎𝑢). 4. In designing steam boilers, the wall thickness calculated by the above equation may be compared with the minimum plate thickness as provided in boiler code as given in the following table. Note: If the calculated value of 𝑡 is less than the code requirement, then the latter should be taken, otherwise the calculated value may be used. 3.5 Longitudinal Stress Consider a closed thin cylindrical shell subjected to an internal pressure as shown in Fig. (a) and (b). A tensile stress acting in the direction of the axis is called longitudinal stress. In other words, it is a tensile stress acting on the transverse or circumferential section Y-Y (or on the ends of the vessel). Let 𝜎𝑡2 = Longitudinal stress. In this case, the total force acting on the transverse section (i.e. along Y-Y) = Intensity of pressure × Cross-sectional area (i) and the total resisting force (ii) From equations (𝑖) and (𝑖𝑖), we have ∴ If ηc is the efficiency of the circumferential joint, then From above we see that the longitudinal stress is half of the circumferential or hoop stress. Therefore, the design of a pressure vessel must be based on the maximum stress i.e. hoop stress. Example. 1 A thin cylindrical pressure vessel of 1.2 m diameter generates steam at a pressure of 1.75 N/mm2. Find the minimum wall thickness, if (a) the longitudinal stress does not exceed 28 MPa; and (b) the circumferential stress does not exceed 42 MPa. Solution. Given: 𝑑 = 1.2 𝑚 = 1200 𝑚𝑚; 𝑝 = 1.75 𝑁/𝑚𝑚2; 𝜎𝑡2 = 28 𝑀𝑃𝑎 = 28 𝑁/𝑚𝑚2; 𝜎𝑡1 = 42 𝑀𝑃𝑎 = 42 𝑁/𝑚𝑚2 (a) When longitudinal stress (𝜎𝒕𝟐) does not exceed 28 Mpa We know that minimum wall thickness, (b) When circumferential stress (𝜎𝒕𝟏) does not exceed 42 MPa We know that minimum wall thickness, Example. 2 A thin cylindrical pressure vessel of 500 mm diameter is subjected to an internal pressure of 2 N/mm2. If the thickness of the vessel is 20 mm, find the hoop stress, longitudinal stress and the maximum shear stress. Solution. Given: 𝑑 = 500 𝑚𝑚; 𝑝 = 2 𝑁/𝑚𝑚2; 𝑡 = 20 𝑚𝑚 Hoop stress We know that hoop stress, Longitudinal stress We know that longitudinal stress, Maximum shear stress We know that according to maximum shear stress theory, the maximum shear stress is one-half the algebraic difference of the maximum and minimum principal stress. Since the maximum principal stress is the hoop stress (𝜎𝑡1) and minimum principal stress is the longitudinal stress (𝜎𝑡2), therefore maximum shear stress, 3.6 Change in Dimensions of a Thin Cylindrical Shell due to an Internal Pressure When a thin cylindrical shell is subjected to an internal pressure, there will be an increase in the diameter as well as the length of the shell. Let 𝑙 = Length of the cylindrical shell, 𝑑 = Diameter of the cylindrical shell, 𝑡 = Thickness of the cylindrical shell, 𝑝 = Intensity of internal pressure, 𝐸 = Young’s modulus for the material of the cylindrical shell, and μ = Poisson’s ratio. The increase in diameter of the shell due to an internal pressure is given by, The increase in length of the shell due to an internal pressure is given by, It may be noted that the increase in diameter and length of the shell will also increase its volume. The increase in volume of the shell due to an internal pressure is given by Example. 3 Find the thickness for a tube of internal diameter 100 mm subjected to an internal pressure which is 5/8 of the value of the maximum permissible circumferential stress. Also find the increase in internal diameter of such a tube when the internal pressure is 90 N/mm2. Take 𝐸 = 205 𝑘𝑁/𝑚𝑚2 and 𝜇 = 0.29. Neglect longitudinal strain. Solution. Given: 5 𝑝 = × 𝜎𝑡1 = 0.625 𝜎𝑡1; 8 𝑑 = 100 𝑚𝑚; 𝑝1 = 90 𝑁/𝑚𝑚2; 𝐸 = 205 𝑘𝑁/𝑚𝑚2 = 205 × 103 𝑁/𝑚𝑚2; 𝜇 = 0.29 Thickness of a tube We know that thickness of a tube, Increase in diameter of a tube We know that increase in diameter of a tube, 3.7 Thin Spherical Shells Subjected to an Internal Pressure Consider a thin spherical shell subjected to an internal pressure as shown. Let 𝑉 = Storage capacity of the shell, 𝑝 = Intensity of internal pressure, 𝑑 = Diameter of the shell, 𝑡 = Thickness of the shell, 𝜎𝑡 = Permissible tensile stress for the shell material. In designing thin spherical shells, we have to determine 1. Diameter of the shell 2. Thickness of the shell. 1. Diameter of the shell We know that the storage capacity of the shell, 2. Thickness of the shell As a result of the internal pressure, the shell is likely to rupture along the center of the sphere. Therefore the force tending to rupture the shell along the center of the sphere or bursting force, (i) and the resisting force of the shell = Stress × Resisting area = σt × π d.t (ii) Equating equations (𝒊) and (𝒊𝒊), we have If η is the efficiency of the circumferential joints of the spherical shell, then Example.4 A spherical vessel 3 meters diameter is subjected to an internal pressure of 1.5 N/mm2. Find the thickness of the vessel required if the maximum stress is not to exceed 90 MPa. Take efficiency of the joint as 75%. Solution. Given: 𝑑 = 3 𝑚 = 3000 𝑚𝑚; 𝑝 = 1.5 𝑁/𝑚𝑚2; 𝜎𝑡 = 90 𝑀𝑃𝑎 = 90 𝑁/𝑚𝑚2; 𝜂 = 75% = 0.75 We know that thickness of the vessel, 3.8 Change in Dimensions of a Thin Spherical Shell due to an Internal Pressure Consider a thin spherical shell subjected to an internal pressure as shown in Fig. 3.4. Let 𝑑 = Diameter of the spherical shell, 𝑡 = Thickness of the spherical shell, 𝑝 = Intensity of internal pressure, 𝐸 = Young’s modulus for the material of the spherical shell, and 𝜇 = Poisson’s ratio. Increase in diameter of the spherical shell due to an internal pressure is given by, (i) and increase in volume of the spherical shell due to an internal pressure is given by, 𝛿𝑉 = Final volume – Original volume By neglecting higher terms Substituting the value of 𝛿𝑑 from equation (i), we have Example.5 A seamless spherical shell, 900 mm in diameter and 10 mm thick is being filled with a fluid under pressure until its volume increases by 150 × 103 mm3. Calculate the pressure exerted by the fluid on the shell, taking modulus of elasticity for the material of the shell as 200 kN/mm2 and Poisson’s ratio as 0.3. Solution. Given: 𝑘𝑁 𝑁 𝑑 = 900 𝑚𝑚; 𝑡 = 10 𝑚𝑚; 𝛿𝑉 = 150 × 103 𝑚𝑚3; 𝐸 = 200 = 200 × 103 ; 𝜇 = 0.3 𝑚𝑚2 𝑚𝑚2 Let 𝑝 = Pressure exerted by the fluid on the shell. We know that the increase in volume of the spherical shell (𝛿𝑉), 3.9 Thick Cylindrical Shells Subjected to an Internal Pressure When a cylindrical shell of a pressure vessel is subjected a very high internal fluid pressure, then the walls of the cylinder must be made extremely heavy or thick. In thin cylindrical shells, we have assumed that the tensile stresses are uniformly distributed over the section of the walls. But in the case of thick wall cylinders as shown in Fig. (a), the stress over the section of the walls cannot be assumed to be uniformly distributed.  They develop both tangential and radial stresses with values which are dependent upon the radius of the element under consideration. The distribution of stress in a thick cylindrical shell is shown in Figs. (b) and (c). We see that the tangential stress is maximum at the inner surface and minimum at the outer surface of the shell. The radial stress is maximum at the inner surface and zero at the outer surface of the shell. In the design of thick cylindrical shells, the following equations are mostly used: 1. Lame’s equation 2. Birnie’s equation 3. Clavarino’s equation 4. Barlow’s equation The use of these equations depends upon the type of material used and the end construction. Let 𝑟𝑜 = Outer radius of cylindrical shell 𝑟𝑖 = Inner radius of cylindrical shell 𝑡 = Thickness of cylindrical shell = 𝑟𝑜 − 𝑟𝑖 𝑝 = Intensity of internal pressure 𝜇 = Poisson’s ratio 𝜎𝑡 = Tangential stress 𝜎𝑟 = Radial stress. 1. Lame’s equation. Assuming that the longitudinal fibers of the cylindrical shell are equally strained. Lame has shown that the tangential stress at any radius 𝑥 is, and radial stress at any radius 𝑥,  Since we are concerned with the internal pressure (𝑝𝑖 = 𝑝) only, substituting the value of external pressure, 𝑝𝑜 = 0. Then the tangential stress at any radius 𝑥, (i) and the radial stress at any radius 𝑥, (ii) It is seen that the tangential stress is always a tensile stress whereas the radial stress is a compressive stress. We know that the tangential stress is maximum at the inner surface of the shell (i.e. when 𝑥 = 𝑟𝑖) and it is minimum at the outer surface of the shell (i.e. when 𝑥 = 𝑟𝑜). Substituting the value of 𝑥 = 𝑟𝑖 and 𝑥 = 𝑟0 in equation (𝒊), (i we find that the maximum tangential stress at the inner surface of the shell, and the minimum tangential stress at the outer surface of the shell, We also know that the radial stress is maximum at the inner surface of the shell and zero at the outer surface of the shell. Substituting the value of 𝑥 = 𝑟𝑖 and 𝑥 = 𝑟0 in equation (𝒊𝒊), we find that the maximum radial stress at the inner surface of the shell, (ii 𝜎𝑟 max = −𝑝 (compressive) and the minimum radial stress at the outer surface of the shell, 𝜎𝑟 min = 0 In designing a thick cylindrical shell of brittle material (e.g. cast iron, hard steel and cast aluminium) with closed or open ends, the maximum normal stress theory failure is used. The tangential stress induced in the cylinder wall, Since 𝑟𝑜 = 𝑟𝑖 + 𝑡, therefore substituting this value of 𝑟𝑜 in the above expression. we get Then (iii) (iii) The value of σt for brittle materials may be taken as 0.125 times the ultimate tensile strength (σu).  In case of cylinders made of ductile material, Lame’s equation is modified according to maximum shear stress theory. According to this theory, the maximum shear stress at any point in a strained body is equal to one-half the algebraic difference of the maximum and minimum principal stresses at that point. We know that for a thick cylindrical shell, The maximum principal stress at the inner surface, and the minimum principal stress at the outer surface, 𝜎𝑡 min = −𝑝 Then the maximum shear stress, as 𝑟𝑜 = 𝑟𝑖 + 𝑡 then or  Then (iv) The value of the shear stress (τ) is usually taken as one-half the tensile stress (σt). Therefore the above expression may be written as (v) 2. Birnie’s equation. In case of open-end cylinders (such as pump cylinders, gun barrels etc.) made of ductile material (i.e. low carbon steel, brass, bronze, and aluminium alloys), the allowable stresses cannot be determined by means of maximum-stress theory of failure. In such cases, the maximum-strain theory is used. According to this theory, the failure occurs when the strain reaches a limiting value and Birnie’s equation for the wall thickness of a cylinder is The value of σt may be taken as 0.8 times the yield point stress (σy). 3. Clavarino’s equation. This equation is also based on the maximum-strain theory of failure, but it is applied to closed-end cylinders (or cylinders fitted with heads) made of ductile material. According to this equation, the thickness of a cylinder, In this case also, the value of σt may be taken as 0.8 σy. 4. Barlow’s equation. This equation is generally used for high pressure oil and gas pipes. According to this equation, the thickness of a cylinder, 𝑝. 𝑟𝑜 𝑡 = 𝜎𝑡 For ductile materials, σt = 0.8 σy For brittle materials σt = 0.125 σu where σu is the ultimate stress. Example.6 A cast iron cylinder of internal diameter 200 mm and thickness 50 mm is subjected to a pressure of 5 N/mm2. Calculate the tangential and radial stresses at the inner, middle (radius = 125 mm) and outer surfaces. Solution. Given: di = 200 mm or ri = 100 mm; t = 50 mm; p = 5 N/mm2 The outer radius of the cylinder, ro = ri + t = 100 + 50 = 150 mm Tangential stresses at the inner, middle and outer surfaces It is known that the tangential stress at any radius 𝑥, Then the tangential stress at the inner surface (i.e. when 𝑥 = ri = 100 mm), The tangential stress at the middle surface (i.e. when 𝑥 = 125 mm), and tangential stress at the outer surface (i.e. when 𝑥 = ro = 150 mm), Radial stresses at the inner, middle and outer surfaces We know that the radial stress at any radius 𝑥, Then the radial stress at the inner surface (i.e. when x = ri = 100 mm), r(inner) = – p = – 5 N/mm2 = 5 MPa (compressive) Ans. The radial stress at the middle surface (i.e. when 𝑥 = 125 mm) and the radial stress at the outer surface (i.e. when 𝑥 = ro = 150 mm), r(outer) = 0 Ans. Example.7 A hydraulic press has a maximum capacity of 1000 kN. The piston diameter is 250 mm. Calculate the wall thickness if the cylinder is made of material for which the permissible strength may be taken as 80 MPa. This material may be assumed as a brittle material. Solution. Given: W = 1000 kN = 1000 × 103 N; d = 250 mm; t = 80 MPa = 80 N/mm2 First of all, let us find the pressure inside the cylinder (p). We know that load on the hydraulic press (W), Then Let ri = inside radius of the cylinder = d / 2 = 125 mm As the material is assumed as a brittle material and the cylinder is open end. Then the wall thickness of the cylinder according to Lame,