CHAPTER OBJECTIVES Analyze the stress developed in thin-walled pressure vessels Review the stress analysis developed in previous chapters regarding axial load, torsion, bending and shear Discuss the solution of problems where several of these internal loads occur simultaneously on a member’s cross-section

CHAPTER OUTLINE Thin-Walled Pressure Vessels State of Stress Caused by Combined Loadings

8.1 THIN-WALLED PRESSURE VESSELS Cylindrical or spherical vessels are commonly used in industry to serve as boilers or tanks When under pressure, the material which they are made of, is subjected to a loading from all directions However, we can simplify the analysis provided it has a thin wall “Thin wall” refers to a vessel having an inner-radius-thickness ratio of 10 or more (r/t ≥ 10) When r/t = 10, results of a thin-wall analysis will predict a stress approximately 4% less than actual maximum stress in the vessel Assumption taken before analysis is that the thickness of the pressure vessel is uniform or constant throughout The pressure in the vessel is understood to be the gauge pressure, since it measures the pressure above atmospheric pressure, which is assumed to exist both inside and outside the vessel’s wall

8.1 THIN-WALLED PRESSURE VESSELS Cylindrical vessels A gauge pressure p is developed within the vessel by a contained gas or fluid, and assumed to have negligible weight Due to uniformity of loading, an element of the vessel is subjected to normal stresses 1 in the circumferential or hoop direction and 2 in the longitudinal or axial direction

8.1 THIN-WALLED PRESSURE VESSELS We use the method of sections and apply the equations of force equilibrium to get the magnitudes of the stress components For equilibrium in the x direction, we require pr t 1 = Equation 8-1 As shown, 2 acts uniformly throughout the wall, and p acts on the section of gas or fluid. Thus for equilibrium in the y direction, we require pr 2t 2 = Equation 8-2

8.1 THIN-WALLED PRESSURE VESSELS For Eqns 8-1 and 8-2, 1, 2 = normal stress in the hoop and longitudinal directions, respectively. Each is assumed to be constant throughout the wall of the cylinder, and each subjects the material to tension p = internal gauge pressure developed by the contained gas or fluid r = inner radius of the cylinder t = thickness of the wall (r/t ≥ 10) Comparing Eqns 8-1 and 8-2, note that the hoop or circumferential stress is twice as large as the longitudinal or axial stress When engineers fabricate cylindrical pressure vessels from rolled-form plates, the longitudinal joints must be designed to carry twice as much stress as the circumferential joints

8.1 THIN-WALLED PRESSURE VESSELS Spherical vessels The analysis for a spherical pressure vessel can be done in a similar manner Like the cylinder, equilibrium in the y direction requires pr 2t 2 = Equation 8-3 Note that Eqn 8-3 is similar to Eqn 8-2. Thus, this stress is the same regardless of the orientation of the hemispheric free-body diagram An element of material taken from either a cylindrical or spherical pressure vessel is subjected to biaxial stress; normal stress existing in two directions

8.1 THIN-WALLED PRESSURE VESSELS Spherical vessels The material is also subjected to radial stress, 3. It has a value equal to pressure p at the interior wall and decreases to zero at exterior surface of the vessel However, we ignore the radial stress component for thin-walled vessels, since the limiting assumption of r/t = 10, results in 3 and 2 being 5 and 10 times higher than maximum radial stress (3)max = p

EXAMPLE 8.1 Cylindrical pressure vessel has an inner diameter of 1.2 m and thickness of 12 mm. Determine the maximum internal pressure it can sustain so that neither its circumferential nor its longitudinal stress component exceeds 140 MPa. Under the same conditions, what is the maximum internal pressure that a similar-size spherical vessel

EXAMPLE 8.1 (SOLN) Maximum stress occurs in the circumferential direction. From Eqn 8-1, we have 1 = ; pr t p(600 mm) 12 mm 140 N/mm2 = p = 2.8 N/mm2 Note that when pressure is reached, from Eqn 8-2, stress in the longitudinal direction will be 2 = 0.51 = 0.5(140 MPa) = 70 MPa. Furthermore, maximum stress in the radial direction occurs on the material at the inner wall of the vessel and is (3)max = p = 2.8 MPa. This value is 50 times smaller than the circumferential stress (140 MPa), and as stated earlier, its effects will be neglected.

EXAMPLE 8.1 (SOLN) Here, the maximum stress occurs in any two perpendicular directions on an element of the vessel. From Eqn 8-3, we have 2 = ; pr 2t p(600 mm) 2(12 mm) 140 N/mm2 = p = 5.6 N/mm2 Although it is more difficult to fabricate, the spherical pressure vessel will carry twice as much internal pressure as a cylindrical vessel.

8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS In previous chapters, we developed methods for determining the stress distributions in a member subjected to internal axial forces, shear forces, bending moments, or torsional moments. Often, the cross-section of a member is subjected to several of these type of loadings simultaneously We can use the method of superposition to determine the resultant stress distribution caused by the loads Application of the method of superposition The stress distribution due to each loading is determined These distributions are superimposed to determine the resultant stress distribution Conditions to satisfy A linear relationship exists between the stress and the loads Geometry of the member should not undergo significant change when the loads are applied

8.2 STATE OF STRESS CAUSED BY COMBINED LOADINGS This is necessary to ensure that the stress produced by one load is not related to the stress produced by any other load

EXAMPLE 8-2 A force of 15,000 N is applied to the edge of the member shown. Neglect the weight of the member and determine the state of stress at points B and C.

EXAMPLE 8-2 (SOLN) Internal loadings P = 15 000 N M = 750 00 N.mm F = 15 000 N M = 750 00 N.mm P = 15 000 N Internal loadings Member is sectioned through B and C. For equilibrium at section, +↑  Fy = 0; F = 15 000 N +  MBC = 0; M = (15000)(50)= 750 000 N·mm

EXAMPLE 8-2 (SOLN) Stress components 1. Axial stress Uniform normal-stress distribution due to normal force is shown.  = P/A = (15000)/(100x40) = 3.75 N/mm2 = 3.75 MPa 2. Maximum bending stress Maximum normal stress distribution due to bending moment is shown. I = bh3 12 = (40)(100)3 = 3.33x106 mm4 M ymax I max= (750 000)(±50) 3.33x106 = 11.25 N/mm2 = 11.25 MPa

EXAMPLE 8.2 (SOLN) Superposition B = 7.5 MPa Superposition If above normal-stress distributions are added algabraically, resultant stress distribution is shown. Although not needed here, the location of the line of zero stress can be determined by proportional triangles, i.e., C = 15 MPa 7.5 MPa x = 15 MPa (100 mm – x) x = 33.3 mm

EXAMPLE 8-3 500 mm x y 800 mm P = 1500 N z y 50 mm 100 mm The solid rectangular beam shown has a cross-sectional area 100x50 mm. If it is subjected to the loading shown, determine the maximum stress in segment BC. Cross-sectional area: A = (50)(100) = 5000 mm2 Moment of inertia: I = = bh3 12 (50)(100)3 = 4.167x106 mm4

= + EXAMPLE 8-3 y x y x Stress components in segment BC: Axial Load (negative) Bending (negative) M = – P(500) N.mm = – 750 kN.mm x 500 mm P = 1500 N 800 mm Stress components in segment BC: 1. Compressive stress due to axial load: s = – P/A = –1500/5000 = –0.3 N/mm2 s = –0.3MPa 2. Bending stress: s =  =  =  1.8 N/mm2 Mc I (–750x103)(50) 4.167x106 s = 1.8 MPa s = –1.8 MPa

EXAMPLE 8-3 500 mm x y 800 mm P = 1500 N = x y + P M Axial Load (negative) Bending (negative) M = – P(500) N.mm = – 750 kN.mm s = –0.3MPa s = 1.5 MPa s = –2.1 MPa = s = 1.8 MPa s = –1.8 MPa + Maximum bending stress at the upper part of the beam: s = – 0.3 + 1.8 = 1.5 Mpa (tensile) Maximum bending stress at the lower part of the beam: s = –0.3 – 1.8 = –2.1 MPa (compressive)

EXAMPLE 8-4 The solid rod shown has a radius of 0.75 cm. If it is subjected to the loading shown, determine the stress at point A.

EXAMPLE 8-4 Cross-sectional area: A = r2 = (7.5)2 = 176.7 mm2 Internal loadings Rod is sectioned through point A. Using free-body diagram of segment AB, the resultant internal loadings can be determined from the six equations of equilibrium.

EXAMPLE 8-4 Internal loadings The normal force (500 N) and shear force (800 N) must act through the centroid of the cross-section and the bending-moment components (8000 N·cm) and 7000 N·cm) are applied about centroidal (principal) axes. In order to better “visualize” the stress distributions due to each of these loadings, we will consider the equal but opposite resultants acting on AC.

EXAMPLE 8-3 Stress components 1. Normal stress due to axial load Normal stress distribution is shown. For point A, we have A = P/A = 500/176.7 = 2.83 MPa (tensile)

EXAMPLE 8.5 (SOLN) 2. Shear stress due to shear force Maximum shear stress occurs at the neutral axis and it is determined from the shaded semi-circular area. Hence, the height of the centroid is r = (7.5) = 3.183 mm 4 3 First moment about neutral axis Q = y’A’ = y’(A/2) =(3.183)(176.7/2) = 281.2 mm3 Moment of inertia about the neutral axis I = r4 = (7.54) = 2485 mm4  4 Shear stress maximum: A = VQ/It = ..... = 6.04 MPa

EXAMPLE 8.5 (SOLN) 3. Bending moments For the 8000 N·cm component, point A lies on the neutral axis, so the normal stress is A = 0 For the 7000 N·cm component, c = 0.75 cm, so normal stress at point A, is A = Mc/I = … = 211.26 MPa

EXAMPLE 8-4 4. Torsional moment At point A, A = c = 0.75 cm, thus A = Tc/J = … = 169.01 MPa

EXAMPLE 8-4 Superposition When the above results are superimposed, it is seen that an element of material at point A is subjected to both normal and shear stress components sA = 214.09 MPa tA = 175.05 MPa