Spherical Pressure Vessels
Spherical Pressure VesselsPressure vessels are closed structures containing liquids or gases underpressure. Examples include tanks, pipes, pressurized cabins, etc.Shell structures : When pressure vessels have walls that are thin incomparison to their radii and length. In the case of thin walledpressure vessels of spherical shape the ratio of radius r to wallthickness t is greater than 10. A sphere is the theoretical ideal shapefor a vessel that resists internal pressure.r 10tTo determine the stresses in an spherical vessel let us cut through the sphereon a vertical diameter plane and isolate half of the shell and its fluid contentsas a single free body. Acting on this free body are the tensile stress σ in thewall of the vessel and the fluid pressure p .
The pressure that acts horizontally against the plane circulararea is uniform and gives a resultant pressure force of :Where p is the gage or internal pressure (above the pressureacting in the outside of the vessel).The stress is uniform around thecircumference and it is uniformly distributedacross the thickness t (because the wall isthin). The resultant horizontal force is :Equilibrium of forces in the horizontal direction:P pπr2Horizontal Force σ (2πrm )ttrm r 2σ (2π rm )t p π r 2prσ 2trm r for thin walls.Therefore the formula tocalculate the stress in a thinwalled spherical vessels is
As is evident from the symmetry of a spherical shell that we will obtain thesame equation regardless of the direction of the cut through the center.The wall of a pressurized spherical vessel is subjected to uniform tensilestresses σ in all directions.Stresses that act tangentially to the curved surface of a shell are known asmembrane stresses.Limitations of the thin-shell theory:1. The wall thickness must be small (r/t 10)2. The internal pressure must exceed the external pressure.3. The analysis is based only on the effects of internal pressure.4. The formulas derived are valid throughout the wall of the vesselexcept near points of stress concentration.
Stresses at the Outer Surfaces.The element below has the x and y axes tangential to the surface of thesphere and the z axis is perpendicular to the surface.Thus, the normal stresses σx and σy are equal to the membrane stress σ andthe normal stress σz is zero.prThe principal stresses areσ1 σ 2 σ 2tand σ3 0. Any rotation element about the z axis will have a shear stressequals to zero.To obtain the maximum shear stresses, we mustconsider out of plane rotations, that is, rotationsabout the x and y axis.Elements oriented at 45o of the x or y axis havemaximum shear stresses equal to σ/2 orτ Maxσpr 2 4t
Stresses at the Inner SurfaceAt the inner wall the stresses in the x and y direction are equal to themembrane stress σx σy σ, but the stress in the z direction is not zero,and it is equal to the pressure p in compression.This compressive stress decreases from p at the inner surface to zero at theouter surface.prThe element is in triaxial stressσ1 σ 2 σ 2tσ3 pThe in-plane shear stress are zero, but themaximum out-of-plane shear stress (obtainedat 45o rotation about either the x or y axis) is(στMaxτMaxprp p) 24t2p r 1 2 2t or
When the vessels is thin walled and the ratior/t is large, we can disregard the number 1 andτ Maxpr 4tConsequently, we can consider the stressstate at the inner surface to be the same asthe outer surface.Summary for Spherical pressure vessel with r/t large:prσ1 σ 2 2t
As the two stresses are equal, Mohr’s circle for in-plane transformationsreduces to a pointσ σ 1 σ 2 constantτ max(in -plane) 0Maximum out-of-plane shearing stressτ max 1σ2 1pr 4t
A compressed air tank having an inner diameter of 18 inches and a wall thickness of ¼inch is formed by welding two steel hemispheres (see figure).(a) If the allowable tensile stress in the steel is 14000psi, what is the maximumpermissible air pressure pa in the tank?.(b) If the allowable shear stress in the steel is 6000psi, what is the maximumpermissible pressure pb?.(c) If the normal strain in the outer surface of the tank is not to exceed 0.0003, what isthe maximum permissible pressure pc? (Assume Hooke’s law is obeyed E 29x106psiand Poisson’s ratio is ν 0.28)(d) Tests on the welded seam show that failureoccurs when the tensile load on the welds exceeds8.1kips per inch of weld. If the required factor ofsafety against failure of the weld is 2.5, what is themaximum permissible pressure pd?(e) Considering the four preceding factors, what isthe allowable pressure pallow in the tank?
Solution(a)Allowable pressure based on the tensile stress in the steel. .We will use theequationpa rσ allowed 2tthen2tσ allowed 2(0.25inch )(14000 psi ) 777.8 psipa r9.0inch(b) Allowable pressure based upon the shear stress of the steel. We will useτ allowed pb σ2 pb r4tthen(0.25inch )(6000 psi ) 666.7 psi4tτ allowed 4r9.0inch(c) Allowable pressure based upon ε X the normal strain in the steel. Forbiaxial stressthen(σ X υσ Y )this equation can be solved for pressure pc(EεX substituti ng(1 υ )σE) σX σY σ (1 υ ) pr2tE2 tE ε allowed2 (0 . 25 inch ) 29 x10 6 psi (0 . 0003 )pc 671 . 3 psi(9 .0 inch )(1 0 .28 )r (1 υ )pr2t
(d) Allowable pressure based upon thetension in the welded seam.ΤallowedThe allowable tensile load on the weldedseam is equal to the failure load divided bythe factor of safetyΤallowedΤfailure8.1kips / inch n2.5 3.24kips / inch 3240lb / inchThe corresponding allowable tensile stress is equal to the allowable load on 1inchlength of weld divided by the cross-sectional area of a 1inch length of weld:σ allowed Τallowed (1 . 0 inch(1 . 0 inch )(t )) (3240 lb / inch )(1 . 0 inch ) 12960(1 . 0 inch )(0 . 25 inch )psi(2 t σ allowed2 )(0 . 25 inch )(12960 psi )pd 720 . 0 psir9 . 0 inch(e) Allowable pressureComparing the preceding results for pa, pb, pc and pd, we see that the shear stress inpallow 666psi.the wall governs and the allowable pressure in the tank is
Cylindrical Pressure VesselExamples: Compressed air tanks, rocket motors, fireextinguishers, spray cans, propane tanks, grain silos,pressurized pipes, etc.We will consider the normal stresses in a thin walledcircular tank AB subjected to internal pressure p.σ1 and σ2 are the membrane stresses in the wall. Noshear stresses act on these elements because of thesymmetry of the vessel and its loading, therefore σ1and σ2 are the principal stresses.Because of their directions, the stressσ1 is called circumferential stressor the hoop stress, and the stressσ2 is called the longitudinal stressor the axial stress.
Equilibrium of forces to find the circumferential stress:σ 1 (2 b )t p (2 b )rprσ1 tσ 2 (2πr )t p (πr 2 )Equilibrium of forcesto find the longitudinalstress:σ2pr 2tThis longitudinal stress is equal to themembrane stress in a sphericalvessel. Then:σ 1 2σ 2We note that the longitudinal welded seam in a pressure tank must betwice as strong as the circumferential seam.
Stresses at the Outer SurfaceThe principal stresses σ1 and σ2 at the outer surface of a cylindrical vesselare shown below. Since σ3 is zero, the element is in biaxial stress.The maximum in plane shear stress occurs on planes that are rotate 45oabout the z-axis(σ1 σ 2 )(τ Max )z 2pr4tThe maximum out of plane shear stressesare obtain by 45o rotations about the x and yaxes respectively.σprτ Max , x τ Max , y1 2tσpr 2 24t2Then, the absolute maximum shear stress isτmax pr / 2t , which occurs on a plane thathas been rotated 45o about the x-axis.
Stresses at the Inner SurfaceprprThe principal stresses areσ1 σ2 and σ 3 pt2tThe three maximum shear stresses,obtained by 45o rotations about thex, y and z axes are(σ1 σ 3 )(τ MAX )x (τ Max )y(τ Max )zpr p 22t 2(σ 2 σ 3 ) pr p 24t 2(σ 1 σ 2 ) pr 24tThe first of these three stresses is the largest.However, when r/t is very large (thin walled), theterm p/2 can be disregarded, and the equationsare the same as the stresses at the outersurface.
Summary for Cylindrical vessels with r/t largeCylindrical vessel with principal stressesσ1 hoop stressσ2 longitudinal stressHoop stress: Fz 0 σ 1(2t Δx ) p (2r Δx )prσ1 tLongitudinal stress:( )2 Fx 0 σ 2 (2π rt ) p π rprσ2 2tσ 1 2σ 2
Points A and B correspond to hoop stress, σ1,and longitudinal stress, σ2Maximum in-plane shearing stress:1prτ max(in plane) σ 2 24tMaximum out-of-plane shearing stresscorresponds to a 45o rotation of the plane stresselement around a longitudinal axisprτ max σ 2 2t
A cylindrical pressure vessel is constructed from a long, narrow steel plateby wrapping the plate around a mandrel and then welding along the edges ofthe plate to make an helical joint (see figure below). The helical weld makesan angle α 55o with the longitudinal axis. The vessel has an inner radiusr 1.8m and a wall thickness t 20mm. The material is steel with a modulusE 200GPa and a Poisson’s ratio ν 0.30. The internal pressure p is 800kPaCalculate the following quantities for thecylindrical part of the vessel:The circumferential and longitudinal stressesσ1 and σ2 respectively;The maximum in-plane and out-of-planeshear stressesThe circumferential and longitudinal strains ε1 and ε2 respectively, andThe normal stress σw and shear stress τw acting perpendicular and parallel,respectively, to the welded seam.
SolutionCircumferential and longitudinalstresses:(b) Maximum Shear StressThe largest in-plane shear stressis obtained from the equationThe largest out-of-plane shearstress is obtained from theequation:pr (800kPa )(1.8m )σ1 72 MPat0.02mpr (800kPa )(1.8m )σ2 36 MPa2t2(0.02m )(τ Max )in plane(σ1 σ 2 ) (τ Max )out of planepr 18MPa24t(σ1 ) pr 36 MPa2tThis last stress is the absolute maximum shear stress in the wall of the vessel.(c) Circumferential and longitudinal strains.Assume the Hook’s law applies to the wall of thevessel. Using the equationsWe note that εx ε2 and εy ε1 and also σx σ2 andσy σ1. Therefore the above equations can be writtenin the following forms:εx σxEε y υ υσxEσyE σyE
ε2 (1 2υ ) pr (1 2υ )σ 2 (1 2 (0 .30 ))(36 MPa ) 72 x10 62 tE200 GPaE(2 υ ) pr (2 υ )σ 1 (2 0 .30 )(72 MPa ) 306 x10 6ε1 (2 )(200 GPa )2 tE2E(d) Normal and shear stresses acting onthe weld seamThe angle θ for the stress element atpoint B in the wall of the cylinder withsides parallel and perpendicular to theweld isθ 90o – α 35oWe will use the stress transformation equations:σ x1 σ x cos 2 θ σ y sin 2 θ 2τ xy sin θ cosθτ x1 y1 σ x σ y2sin 2θ τ xy cos 2θσ x1 σ y1 σ x σ y
σ x1 36(cos(35))2 72(sin (35))2 47.8MPaτ x1 y1 36 sin (35) cos(35) 72 sin (35) cos(35) 16.9 MPa47.8 σ y1 36 72 σ y1 60.2MPaMohr’s circleCoordinates of point A (for θ 0) σ2 36MPaand shear stress 0Coordinates of point B (for θ 90) σ1 72MPaand shear stress 0Center (point C) (σ1 σ2) / 2 54MPaRadius (σ1- σ2) / 2 18MPaA counterclockwise angle 2θ 70o (measured on the circle from point A)locates point D, which corresponds to the stresses on the x1 face (θ 35o) ofthe element.The coordinates of point Dσx1 54 – R cos 70o 54MPa – (18MPa)(cos 70o) 47.8MPaτx1y1 R sin 70o (18MPa)(sin 70o) 16.9MPa
Note:When seen in a side view, a helix followsthe shape of a sine curve (see Figurebelow). The pitch of the helix is p π dtan θ, where d is the diameter of thecircular cylinder and θ is the anglebetween a normal to the helix and alongitudinal line. The width of the flatplate that wraps into the cylinder shapeis w π d sin θ.For practical reasons, the angle θ isusually in the range from 20o to 35o.
QuizA close‐end pressure vessel has an inside diameter of 1600 mm an a wall thickness of40mm. It is pressurized to an internal pressure of 6.5MPa an it has a centric compressiveforce of 750kN applied as shown. The tank is welded together along the helix making anangle of 25o to the horizontal. Determine the normal and shear stresses along the helix.750kNThe material is steel with a modulus E 200GPa and aPoisson’s ratio ν 0.30Calculate the following quantities for the cylindrical part ofthe vessel:(a) The circumferential and longitudinal stresses σ1 and σ2respectively;25O(b) The maximum in‐plane and out‐of‐plane shear stresses(c) The circumferential and longitudinal strains ε1 and ε2respectively, and(d) The normal stress σw and shear stress τw actingperpendicular and parallel, respectively, to the weldedseam.
General State of StressesFrom equilibrium principles:τxy τyx , τxz τzx , τzy τyzThe most general state of stress at a point may berepresented by 6 componentsNormal StressesσxσyσzShear Stressesτ xyτ yzτ xzThese are the principal axes and principal planes andthe normal stresses are the principal stresses.
σ x τ yx τ zx σ σ ij τ xy σ y τ zy τ xz τ yz σ z σ 1 0 0 σ ' σ ij' 0 σ 2 0 Eigen values 0 0 σ 3 The three circles represent the normal andshearing stresses for rotation around eachprincipal axis.Points A, B, and C represent the principalstresses on the principal planes (shearing stressis zero)Radius of the largest circle yields the maximumshearing stress.1τ max σ max σ min2
In the case of plane stress, the axisperpendicular to the plane of stress is aprincipal axis (shearing stress equal zero).If the points A and B (representing theprincipal planes) are on opposite sides ofthe origin, thenthe corresponding principal stresses are themaximum and minimum normal stressesfor the elementthe maximum shearing stress for theelement is equal to the maximum “in-plane”shearing stressplanes of maximum shearing stressare at 45o to the principal planes.
If A and B are on the same side of theorigin (i.e., have the same sign), thenthe circle defining σmax, σmin, and τmax forthe element is not the circlecorresponding to transformations withinthe plane of stressmaximum shearing stress for theelement is equal to half of themaximum stressplanes of maximum shearingstress are at 45 degrees to theplane of stress
Failure TheoriesWhy do mechanical components fail? Mechanical components fail because theapplied stresses exceeds the material’s strength (Too simple).What kind of stresses cause failure? Under any load combination, there is always acombination of normal and shearing stresses in the material.Mohr’s circle for centric axialloading:σx P, σ y τ xy 0Aσ x σ y τ xy P2AMohr’s circle for torsionalloading:σ x σ y 0 τ xy σx σy TcJTcτ xy 0J
What is the definition of Failure?Obviously fracture but in somecomponents yielding can also beconsidered as failure, if yieldingdistorts the material in such a waythat it no longer functions properlyWhich stress causes the material tofail?Usually ductile materials arelimited by their shear strengths.While brittle materials (ductility 5%) are limited by their tensilestrengths.Stress at which point?
Yield Criteria for Ductile Materials under Plane Stress ConditionsMaximum Shear Stress Theory (MSST)Failure occurs when the maximum shear stress in the part (subjected to plane stress)exceeds the shear stress in a tensile test specimen (of the same material) at yield.Maximum shear stress criteria: The structural component is safe as long as themaximum shear stress is less than the maximum shear stress in a tensile test specimenat yield, i.e.,σyFor σa and σb with the same sign,τ τy maxSy2τ yield τ max 2σa2For σa and σb with opposite signs,τ max σa σb2σ Y2The maximum absolute shear stress is always theradius of the largest of the Mohr’s circle.orσb2 σy2
Distortion Energy Theory (DET)Based on the consideration of angular distortion of stressed elements. The theorystates that failure occurs when the distortion strain energy in the material exceeds thedistortion strain energy in a tensile test specimen (of the same material) at yield.Maximum distortion energy criteria: Structural component is safe as long as thedistortion energy per unit volume is less than that occurring in a tensile test specimenat yield.Von Mises effective stress : Defined as theuniaxial tensile stress that creates the samedistortion energy as any actual combinationof applied stresses.ud uY1 21 2σ a σ aσ b σ b2 σ Y σ Y 0 026G6G()()σ a2 σ aσ b σ b2 σ Y2σ VM σ 12 σ 22 σ 32 σ 1σ 2 σ 2σ 3 σ 3σ 1σ VM σ 12 σ 32 σ 3σ 12Dσ VM σ x 2 σ y 2 σ xσ y 3τ xy2
Given the material SY , σx , σv and τxy find the safety factors for all the applicablecriteria.a. Pure aluminumSY 30 MPa σ x 10 MPa σ y 10 MPa τ xy 0 MPaσ 1 10 MPaσ 3 10 MPaτ Max 10 MPaDuctile-1010Use either the Maximum Shear Stress Theory(MSST) or the Distortion Energy Theory (DET)MSST TheorySy3030n 1.5σ 1 σ 3 10 ( 10) 20DET Theoryσ VM σ x 2 σ y 2 σ xσ y 3τ xy2 300 17.32MPan Syσ VM30 MPa 1.7317.32 MPa
b. 0.2%C Carbon Steel SY 65 Ksiσ x 5 Ksi σ y 35 Ksi τ xy 10 KsiIn the plane XY the principal stresses are -1.973Ksiand -38.03Ksi with a maximum shear stress in theXY plane of 18.03KsiIn any orientationσ 1 0 Ksi σ 2 1.973Ksi σ 3 38.03Ksiτ Max 19.01KsiMSST TheoryDET TheoryDuctileSy65n 1.71σ 1 σ 3 0 ( 38.03)σ VM σ 12 σ 3 2 σ 1σ 3 38.03Ksin Syσ VM 65 Ksi 1.7138.03MPa
Spherical Pressure Vessels Shell structures: When pressure vessels have walls that are thin in comparison to their radii and length. In the case of thin walled pressure vessels of spherical shape the ratio of radius r to wall thickness t is greater than 10. A sphere is the theoretical ideal shape for a vessel that resists internal pressure.
for designing the composite pressure vessels shapes of pressure vessels [27]. During the past two decades, several authors have performed detailed analyses of composite pressure vessels by using the theory of orthotropic plates. Composite pressure vessels tend to fail in their composite pressure vessels parts; the design of these parts is the .
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