Chapter 8. Chapter Spherical Pressure Vessels

Transcription

1 Chapter 8 Chapter Spherical Pressure Vessels Here are examples of spherical pressure vessels. Let s determine how to analyze them. Consider Figures 8-2 and 8-3 below. 1

2 Figure 8-2 illustrates the pressure inside the pressure vessel. The pressure pushes uniformly against the inside surface. The only net force is the result of the pressure times the projected area of the cross-section because the radial pressure is the same everywhere, so the net radial force is zero. Figure 8-3 (a) and (b) illustrate the principle. Figure 8-3 (a) shows a cutting plane through the e pressure vessel equator. Figure 8-3 (b) shows the resulting free-body diagram. The force resulting from the pressure acting on the project area is balanced by the force resulting from the normal stress acting on the annular area. Figure 8-3 (c) shows the resulting stress element where the stresses are only normal stresses. Note: The stresses are, therefore, principal stresses. We note the pressure is gauge pressure, not absolute pressure, because the atmospheric pressures on the inside and outside cancel each other. The force acting on the projected area is: P A = pπr 2 and the force acting on the annular area is the average circumference times the thickness times the normal stress: P C = σπr m t = σπ(r + t/2)t = σπrt(r + t/(2r)) For thin-walled vessels, t/(2r) 0 and since P A = P c for equilibrium, we get the normal stress in a thin-walled spherical pressure vessel, σ: σ = pr/(2t) The state of stress on the outer surface is bi-axial, assuming the vessel is in air at normal atmospheric pressure because the atmospheric pressure effect may be neglected. Therefore, σ x = σ 1 = σ y = σ 2 = σ and σ z = σ 3 = 0 where σ 1, σ 2, and σ 3 are principal stresses (stresses on planes with no shear stress). Because the principal stresses are equal, there is no in-plane shear stress on the surface. 2

3 But, there is an out-of-plane shear stress on a 45 o degree plane obtained by rotating the stress element about the x and y axes. Referring back to Ch 7-3, the maximum shear stress on the surface then, using Eq. 7-28,a,b,c, is: τ max =σ/2 = pr/(4t) However, the state of stress on the inner surface is tri-axial in general, assuming the vessel is in air at normal atmospheric pressure, because the internal pressure effect may not be neglected. Therefore, σ x = σ 1 = σ y = σ 2 =σ and σ z = σ 3 = -p where σ 1, σ 2, and σ 3 are principal stresses (stresses on planes with no shear stress). The in-plane shear stresses are zero, but there is an out-of-plane shear stress. Referring back to Ch 7-6, the maximum shear stresses can be found by 45 o rotations about the x and y axes. So using Eq 7-58a,b: τ max = (σ + p)/2 = ((pr/2t) + p)/2 = (p/2)[r/(2t) +1] For thin-walled vessels, r/(t) >10 >> 1, so we find the state of stress on the inner surface to be the same as the state-of-stress on the outer surface: τ max =σ/2 = pr/(4t) Cylindrical Pressure Vessels Examples of cylindrical pressure vessels are shown to the left. The analysis is similar to that for spherical pressure vessels. 3

4 Figure 8-7 illustrates the pressure inside the pressure vessel. The pressure pushes uniformly against the inside surface. In this case, there are two applied forces: (1) the force resulting from the pressure times the projected area of the longitudinal cross-section, P 1, (Figure 8-7 (b) and (2) the force resulting from the pressure times the projected area of the circumferential cross-section, P 2, (Figure 8-7 (c). The forces resulting from the pressure acting on the projected areas are balanced by the forces resulting from the normal stresses acting on the respective areas. Figure 8-7 (a) shows the resulting stress element where the stresses are only normal stresses. Note: The stresses are, therefore, principal stresses. We note the pressure is gauge pressure, not absolute pressure, because the atmospheric pressures on the inside and outside cancel each other. The pressure acting on the longitudinal projected area, rb, results in a circumferential stress: P 1 = 2prb and the resulting circumferential stress acting on the longitudinal area, bt, results in a circumferential force: P C1 =2σ 1 bt Since P A = P C1 for equilibrium, we get the circumferential stress (the length dimension cancels) in a thin-walled cylindrical pressure vessel, σ 1 : σ 1 = pr/t The pressure acting on the circumferential projected area, πr 2, results in a longitudinal force: P 2 = pπr 2 and the longitudinal force acting on the circumferential annular area, 2πrt, results in a longitudinal stress,σ 2. (Note: we use the thin-walled assumption, so we use r instead of r m to calculate the areas).: P C2 = σ 2 π2rt 4

5 Since P 2 = P C2 for equilibrium, we get the longitudinal stress in a thin-walled cylindrical pressure vessel, σ 2 : σ 2 = pr/(2t) The state of stress on the outer surface is bi-axial, assuming the vessel is in air at normal atmospheric pressure, because the atmospheric pressure effect may be neglected. But, in the case of cylindrical pressure vessels, the principal stresses are not equal. Therefore, σ y = σ 1 and σ x = σ 2 and σ z = σ 3 = 0 where σ 1, σ 2, and σ 3 are principal stresses (stresses on planes with no shear stress). Note: The ratio of circumferential stress to longitudinal stress is σ 1/ σ 2 = 2. We note the radius to thickness ratio is the controlling factor. Referring to Figure 8-7 (a), the state of stress on the outer surface is biaxial, assuming the vessel is in air at normal atmospheric pressure because the atmospheric pressure effect may be neglected. Therefore, σ 1, σ 2, and σ 3 = 0 are principal stresses (stresses on planes with no shear stress). Referring to Figure 8-8 (a), because σ 1 σ 2 (as opposed to spherical vessels where σ 1 = σ 2 ), there are in-plane shear stresses that occur on a plane rotated 45 o about the z-axis (See section 7-3, eqs and 7-28 (c) so: (τ max ) z = (σ1 - σ2)/2 = pr/(4t) The maximum out-of plane shear stresses are: (See section 7-3, eq a,b) and we have (τ max ) x = σ 2 /2 = pr/(2t) (Note: This is the maximum surface shear stress) (τ max ) y = σ1/2 = pr/(4t) However, the state of stress on the inner surface is tri-axial in general, because the internal pressure effect may not be neglected. Therefore σ y = σ 1, σ x = σ 2 and σ z = σ 3 = -p where σ 1, σ 2, and σ 3 are principal stresses (stresses on planes with no shear stress. Referring back to Ch 7-6, the maximum shear stresses can be found by 45 o rotations about the x, y, and z axes. So,using Eqs and 7-58a,b: 5

6 (τ max ) x = (σ 1 (- p))/2 = ((pr/t) + p)/2 = (p/2)[r/t +1] (τ max ) y = (σ 2 (- p))/2 = ((pr/2t) + p)/2 = (p/2)[r/(2t) +1] (τ max ) z = (σ 1 σ 2 )/2 = pr/(4t) For thin-walled vessels, r/(t) >10 >> 1, so we find the state of stress on the inner surface to be the same as the state-of-stress on the outer surface: (τ max ) x = pr/(2t) (τ max ) y = pr/(4t) (τ max ) z = pr/(4t) See Examples 8-1 and 8-2 in separate files to illustrate the principles. 6