1. 1 Chapter 2: Application of stress EBB 334 Mechanical Metallurgy Assoc. Prof. Dr. Zuhailawati Hussain

2. 1.Introduction We will now investigate some practical examples of structures and components in state of plane stress or strain, building upon the concepts presented in Chapter 2. Stresses and strains in the walls of thin pressure vessels are examined.

3. 2. Spherical Pressure Vessels Pressure vessels are closed structures containing liquids or gases under pressure Examples 0 tanks, pipes and pressurized cabins in aircraft and space vehicles. When pressure vessels have walls that are thin in comparison to their overall dimensions, they are included within a more general category known as shell structures. Other examples of shell structures are roof domes, airplane wings and submarine hulls. In this section, we consider thin-walled pressure vessels of spherical shape, like the compressed-air tank shown Figure 1. Fig.1 Spherical pressure vessel

4. Pressure vessels are considered to be thin-walled when the ratio radius r to wall thickness (Figure 2) is greater than 10. We assume in the following discussions that the internal pressure p exceeds the pressure acting on the outside of the shell. Otherwise, the vessel may collapse inward due to buckling. A sphere is the theoretically ideal shape for a vessel that resists internal pressure. “natural” shape like soap bubble. To determine the stresses in a spherical vessel, let us cut through the sphere on a vertical diametric plane (Figure 3a) and isolate half of the shell and its fluid contents as a single body (Figure 3b). Figure 2: Cross section of spherical pressure vessel showing inner radius r, wall thickness t and internal pressure p

5. Acting on this free body are the tensile stresses σ in the wall and the fluid pressure p. This pressure acts horizontally against the plane circular area of fluid remaining inside the hemisphere. Since the pressure is uniform, the resultant pressure force P (Figure 3b) is P = p(πr 2 )(a) where r is the inner radius of the sphere. Pressure p is not absolute pressure inside the vessel but is the net internal pressure, or the gage pressure. Figure 3: Tensile stress σ in the wall of a spherical pressure vessel

6. If the internal and external pressures are the same, no stresses are developed in the wall of the vessel-only the excess of internal pressure over thermal pressure has any effect on these stresses. Because of the symmetry of the vessel and its loading (Figure 3.b), the tensile stress σ is uniform around the circumference. Furthermore, since the wall is thin, we can assume with good accuracy that the stress is uniformly distributed across the thickness t. The accuracy of this approximation increases as the shell becomes thinner and decreases as it become thicker. The resultant of the tensile stresses σ in the wall is a horizontal force equal to the stress σ times the area over which it acts, or σ(2πr m t) where t is the thickness of the wall and r m is its mean radius: r m = r + t/2 Thus, equilibrium of forces in the horizontal direction (Figure 3b) gives ∑F horiz = 0,σ(2πr m t) – p(πr 2 ) = 0(c) From which we obtain the tensile stresses in the wall of the vessels: σ = (pr2)/ (2r m t)(d)

7. Since our analysis is valid only for thin shells, we can disregard the small difference between the two radii appearing in Eq. (d) and replace r by r m or replace r m by r. The stresses are closer to the theoretically exact stresses if we use the inner radius r instead of the mean radius r m. Therefore, we will adopt the following formula for calculating the tensile stresses in the wall of a spherical shell: σ = (pr)/ (2t)(1) The wall of a pressurized spherical vessel is subjected to uniform tensile stresses σ in all directions. This stress condition is represented in Figure3c by the small stress element with stresses σ acting in mutually perpendicular directions. Stresses that act tangentially to the curved surface of a shell, such as the stresses σ shown in Figure 3c, are known as membrane stress. The name arises from the fact that these are only stresses that exist in true membrane, such as soap films.

8. Stresses at the Outer Surface The outer surface of a spherical pressure vessel is usually free of any loads. Therefore, the element shown in Figure 3c is in biaxial stress. To aid in analysing the stresses acting on this element, we show it again in Figure 4a, where a set of coordinate axes is oriented parallel to the sides of the element. The x and y axes are tangential to the surface of the sphere, and the z axis is perpendicular to the surface. Thus, the normal stresses σ x and σ y are the same as the membrane stresses σ, and the normal stress σ z is zero. No shear stresses act on the sides of this element. If we analyse the element of Figure 4a by using the transformation equations for plane stress, we find: σ x1 = σ and τ x1y1 = 0 as expected. In other words, when we consider elements obtained by rotating the axes about the z, the normal stresses remain constant and there are no shear stresses. Every plane is a principal plane and every direction. Thus the principal stresses for the element are σ1 = σ2 = (pr)/(2t)σ3 = 0(2a,b) The stresses σ1 and σ2 lie in the xy plane and the stress σ3 acts in the z direction. Figure 4: Stresses in a spherical pressure vessel (a) the outer surface and (b) the inner surface

9. To obtain the maximum shear stresses, we must consider out-of-plane rotations, that is, rotations about the x and y axes (because all in-plane shear stresses are zero). Elements oriented by making 45° rotations about the x and y axes have maximum shear stresses equal to σ/2. Therefore, τ max = σ/2 = (pr)/ (4t)(3) These stresses are the largest shear stresses in the element.

10. Stresses at the Inner Surface At the inner surface of the wall of a spherical vessel, a stress element (Figure 4b) has the same membrane stresses σ x and σ y as does an element at the outer surface (Figure 4a). In addition, a compressive stress σ z equal to the pressure p acts in the z direction (Figure 4b). This compressive stress decreases from p at the inner surface of the sphere to zero at the outer surface. The element shown in Figure 4b is in triaxial stress with principal stresses σ1 = σ2 = (pr)/ (2t)σ = -p(e,f) The in-plane shear stresses are zero, but the maximum out-of- plane shear stress (obtained by a 45° rotation about either the x or y axis) is τ max = (σ + p)/2 = (pr)/4t + p/2 = p/2 (r/2t + 1)(g) When the vessel is thin-walled and the ratio r/t is large, we can disregard the number 1 in comparison with the term r/2t. In other words, the principal stress σ3 in the z direction is small when compared with the principal stress σ1 and σ2. Consequently, we can consider the stress state at the inner surface to be the same as at the outer surface (biaxial stress). This approximation is consistent with the approximate nature of thin- shell theory and therefore, we will use Eqs. (1), (2) and (3) to obtain the stresses in the wall of a spherical pressure vessel.

11. Example 1 A compressed-air tank having an inner diameter of 450 mm and a wall thickness of 7 mm is formed by welding two steel hemispheres (Figure 5). a)If the allowable shear stress in the steel is 40 MPa, what is the maximum permissible air pressure p a in the tank? b) If the allowable shear stress in the steel is 40 MPa, what is the maximum permissible pressure p b ? c)If the normal strain at the outer surface of the tank is not to exceed 0.0003, what is the maximum permissible pressure p c ? (Assume that Hooke’s law is valid and that the modulus of elasticity for the steel is 210 GPa and Poisson’s ratio is 0.28). d)Tests on the welded seam show that failure occurs when the tensile load on the welds exceed 1.5MN/m per inch of weld. If the required factor of safety against failure of the weld is 2.5, what is the maximum permissible pressure p d ? e)Considering the four preceding factors, what is the allowable pressure p allow in the tank? Figure 5: Spherical pressure vessel. (Attachments and supports are not shown)

12. Solution (a)Allowable pressure based upon the tensile stress in the steel. The maximum tensile stress in the wall of the tank is given by the formula σ = pr/ 2t (see Eq. 1). Solving this equation for the pressure in terms of the allowable stress, we get Thus, the maximum allowable pressure based upon tension in the wall of the tank is p a = 7.1 MPa. (Note that in a calculation of this kind, we round the downward, not upward.) (b) Allowable pressure based upon the shear stress in the steel. The maximum shear stress in the wall of the tank is given by Eq. 3, from which we get the following equation for the pressure: Therefore, the allowable pressure based upon shear is p b = 4.9 MPa.

13. (c) Allowable pressure based upon the normal strain in the steel. The normal strain is obtained from Hooke’s law for biaxial stress: Substituting σx = σy = σ = pr/2t (see Figure 4a), we obtain This equation can be solved for the pressure p c : Thus, the allowable pressure based upon the normal strain in the wall is p c = 5.4 MPa

14. (d) Allowable pressure based upon the tension in the welded seam. The allowable tensile load on the welded seam is equal to the failure load divided by the factor of safety: The corresponding allowable tensile stress is equal to the allowable load on a one-inch length of weld divided by the cross-sectional area of a one-inch length of weld: Finally, we solve for the internal pressure by using Eq. 1 This result gives the allowable pressure based upon tension in the welded seam.

15. (d) Allowable pressure. Comparing the preceding results for pa, pb, pc and pd, we see that shear stress in the wall governs and the allowable pressure in the tank is p allow = 4.9 MPa This example illustrates how various stresses and strains enter into the design of a spherical pressure vessel. Note: When the internal pressure is at its maximum allowable value (6.33 psi), the tensile stresses in the shell are Thus, at the inner surface of the shell (Figure 4b), the ratio of the principal stress in the z direction (4.9 MPa) to the in-plane principal stresses (78.8 MPa) is only 0.062. Therefore, our earlier assumption that we can disregard the principal stress σ3 in the z direction and consider the entire shell to be in biaxial stress is justified.

16. Cylindrical Pressure Vessels Cylindrical pressure vessels with a circular cross section (Fgure 6) are found in industrial settings (compressed air tank and rocket motors), in homes (fire extinguishers and spray cans), and in the countryside (propane tanks and grain silos). Pressurized pipes, such as water-supply pipe and penstocks are also classified as cylindrical pressure vessels. Figure 6: Cylindrical pressure vessels with circular cross sections

17. Example (Exercise)