Spherical Pressure Vessels Pressure vessels are closed structures containing liquids or gases under essure. Examples include tanks, pipes, essurized cabins, etc. Shell structures : When essure vessels have walls that are thin in comparison to their radii and length. In the case of thin walled essure vessels of spherical shape the ratio of radius r to wall thickness t is greater than 0. A sphere is the theoretical ideal shape for a vessel that resists internal essure. To determine the stresses in an spherical vessel let us cut through the sphere on a vertical diameter plane and isolate half of the shell and its fluid contents as a single free bod. Acting on this free bod are the tensile stress in the wall of the vessel and the fluid essure p. r t > 0
The essure that acts horizontall against the plane circular area is uniform and gives a resultant essure force of : Where p is the gage or internal essure (above the essure acting in the outside of the vessel). The stress is uniform around the circumference and it is uniforml distributed across the thickness t (because the wall is thin). The resultant horizontal force is : P pπr ( r ) Horizontal Force π t rm r + ( ) Equilibrium of forces in the horizontal direction: πr m t t m pπr r m ~ r for thin walls. Therefore the formula to calculate the stress in a thin walled spherical vessels is t
As is evident from the smmetr of a spherical shell that we will obtain the same equation regardless of the direction of the cut through the center. The wall of a essurized spherical vessel is subjected to uniform tensile stresses in all directions. Stresses that act tangentiall to the curved surface of a shell are known as membrane stresses. Limitations of the thin-shell theor:. The wall thickness must be small (r/t > 0). The internal essure must exceed the external essure. 3. The analsis is based onl on the effects of internal essure. 4. The formulas derived are valid throughout the wall of the vessel except near points of stress concentration.
The element below has the x and axes tangential to the surface of the sphere and the z axis is perpendicular to the surface. Thus, the normal stresses x and are equal to the membrane stress and the normal stress z is zero. The incipal stresses are Stresses at the Outer Surfaces. and 3 0. An rotation element about the z axis will have a shear stress equals to zero. To obtain the maximum shear stresses, we must consider out of plane rotations, that is, rotations about the x and axis. Elements oriented at 45 o of the x or axis have maximum shear stresses equal to / or t Max 4t
Stresses at the Inner Surface At the inner wall the stresses in the x and direction are equal to the membrane stress x, but the stress in the z direction is not zero, and it is equal to the essure p in comession. This comessive stress decreases from p at the inner surface to zero at the outer surface. The element is in triaxial stress 3 p t The in-plane shear stress are zero, but the maximum out-of-plane shear stress (obtained at 45 o rotation about either the x or axis) is Max Max ( + p ) p + r t 4 t + p or
When the vessels is thin walled and the ratio r/t is large, we can disregard the number and Max 4t Consequentl, we can consider the stress state at the inner surface to be the same as the outer surface. Summar for Spherical essure vessel with r/t large: t
As the two stresses are equal, Mohr s circle for in-plane transformations reduces to a point max(in-plane) constant Maximum out-of-plane shearing stress 0 max 4t
A comessed air tank having an inner diameter of 8 inches and a wall thickness of ¼ inch is formed b welding two steel hemispheres (see figure). (a) If the allowable tensile stress in the steel is 4000psi, what is the maximum permissible air essure p a in the tank?. (b) If the allowable shear stress in the steel is 6000psi, what is the maximum permissible essure p b?. (c) If the normal strain in the outer surface of the tank is not to exceed 0.0003, what is the maximum permissible essure p c? (Assume Hooke s law is obeed E 9x0 6 psi and Poisson s ratio is ν 0.8) (d) Tests on the welded seam show that failure occurs when the tensile load on the welds exceeds 8.kips per inch of weld. If the required factor of safet against failure of the weld is.5, what is the maximum permissible essure p d? (e) Considering the four eceding factors, what is the allowable essure p allow in the tank?
Solution (a)allowable essure based on the tensile stress in the steel..we will use the equation p allowed a par t t r allowed then ( 0.5inch)( 4000 psi) 9.0inch 777.8 psi (b) Allowable essure based upon the shear stress of the steel. We will use p allowed b 4t r allowed pbr 4t (c) Allowable essure based upon the normal strain in the steel. For biaxial stress 4 then ( 0.5inch)( 6000 psi) ε X then 9.0inch υ ( ) X ε E X Y 666.7 psi ( υ ) ( υ ) E substituti ng te X Y t this equation can be solved for essure p c p te ε r ( υ ) 6 ( 0.5inch )( 9 x0 psi )( 0.0003 ) ( 9.0inch )( 0.8 ) allowed c 67. 3 psi
(d) Allowable essure based upon the tension in the welded seam. The allowable tensile load on the welded seam is equal to the failure load divided b the factor of safet Τ Τ allowed allowed Τfailure 8.kips / inch n.5 3.4kips / inch 340lb / inch The corresponding allowable tensile stress is equal to the allowable load on inch length of weld divided b the cross-sectional area of a inch length of weld: p Τ (.0inch ) (.0inch )( t ) ( 340 lb / inch )(.0inch ) (.0inch )( 0.5 inch ) allowed allowed 960 t r ( )( 0.5 inch )( 960 psi ) allowed d 70. 0 9.0inch (e) Allowable essure Comparing the eceding results for p a, p b, p c and p d, we see that the shear stress in the wall governs and the allowable essure in the tank is p allow 666psi. psi psi
Clindrical Pressure Vessel Examples: Comessed air tanks, rocket motors, fire extinguishers, sa cans, opane tanks, grain silos, essurized pipes, etc. We will consider the normal stresses in a thin walled circular tank AB subjected to internal essure p. and are the membrane stresses in the wall. No shear stresses act on these elements because of the smmetr of the vessel and its loading, therefore and are the incipal stresses. Because of their directions, the stress is called circumferential stress or the hoop stress, and the stress is called the longitudinal stress or the axial stress.
Equilibrium of forces to find the circumferential stress: ( b ) t p ( b ) This longitudinal stress is equal to the membrane stress in a spherical vessel. Then: We note that the longitudinal welded seam in a essure tank must be twice as strong as the circumferential seam. t Equilibrium of forces to find the longitudinal stress: r ( ) ( πr t p πr ) t
Stresses at the Outer Surface The incipal stresses and at the outer surface of a clindrical vessel are shown below. Since 3 is zero, the element is in biaxial stress. The maximum in plane shear stress occurs on planes that are rotate 45 o about the z-axis ( ) Max z ( ) 4t The maximum out of plane shear stresses are obtain b 45 o rotations about the x and axes respectivel. Max, x Max, t 4t Then, the absolute maximum shear stress is max / t, which occurs on a plane that has been rotated 45 o about the x-axis.
The incipal stresses are Stresses at the Inner Surface t t The three maximum shear stresses, obtained b 45 o rotations about the x, and z axes are and 3 ( ) MAX ( ) Max ( ) Max z x ( ) ( ) ( ) 3 3 + t + 4t 4t p p p The first of these three stresses is the largest. However, when r/t is ver large (thin walled), the term p/ can be disregarded, and the equations are the same as the stresses at the outer surface.
Summar for Clindrical vessels with r/t large Clindrical vessel with incipal stresses hoop stress longitudinal stress Hoop stress: F z 0 t ( t Δx) p( r Δx) Longitudinal stress: F x 0 t ( ) ( ) π rt p π r
Points A and B correspond to hoop stress,, and longitudinal stress, Maximum in-plane shearing stress: max( in plane) 4t Maximum out-of-plane shearing stress corresponds to a 45 o rotation of the plane stress element around a longitudinal axis max t
A clindrical essure vessel is constructed from a long, narrow steel plate b wrapping the plate around a mandrel and then welding along the edges of the plate to make an helical joint (see figure below). The helical weld makes an angle α 55 o with the longitudinal axis. The vessel has an inner radius r.8m and a wall thickness t 0mm. The material is steel with a modulus E00GPa and a Poisson s ratio ν0.30. The internal essure p is 800kPa Calculate the following quantities for the clindrical part of the vessel: The circumferential and longitudinal stresses and respectivel; The maximum in-plane and out-of-plane shear stresses The circumferential and longitudinal strains ε and ε respectivel, and The normal stress w and shear stress w acting perpendicular and parallel, respectivel, to the welded seam.
Solution Circumferential and longitudinal stresses: (b) Maximum Shear Stress The largest in-plane shear stress is obtained from the equation The largest out-of-plane shear stress is obtained from the equation: ( ) Max ( ) Max t t in plane out of plane ( 800kPa)(.8m) 0.0m ( 800kPa)(.8m) ( 0.0m) ( ) ( ) 7MPa 36MPa 8MPa 4t 36MPa t This 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 the vessel. Using the equations We note that ε x ε and ε ε and also x and. Therefore the above equations can be written in the following forms: x ε x υ E E x ε υ + E E
ε ε ( υ ) ( υ ) ( ( 0.30 ))( 36 MPa ) te E ( υ ) ( υ ) ( 0.30 )( 7 MPa ) te E ( )( 00 GPa ) 00 GPa 7 x0 306 x0 6 6 (d) Normal and shear stresses acting on the weld seam The angle θ for the stress element at point B in the wall of the clinder with sides parallel and perpendicular to the weld is θ 90 o α 35 o We will use the stress transformation equations: x x cos θ + sin θ + + x + x x x x sin θ + x cos θ sinθ cosθ
( cos( 35) ) + 7( sin( 35) ) 47. MPa 36 sin( 35) cos( 35) + 7 sin( 35) cos( 35) 6. MPa x 36 8 x 9 47.8 + Mohr s circle 36 + 7 60. MPa Coordinates of point A (for θ 0) 36MPa and shear stress 0 Coordinates of point B (for θ 90) 7MPa and shear stress 0 Center (point C) ( + ) / 54MPa Radius ( - ) / 8MPa A counterclockwise angle θ 70 o (measured on the circle from point A) locates point D, which corresponds to the stresses on the x face (θ 35 o ) of the element. The coordinates of point D x 54 R cos 70 o 54MPa (8MPa)(cos 70 o ) 47.8MPa x R sin 70 o (8MPa)(sin 70 o ) 6.9MPa
Note: When seen in a side view, a helix follows the shape of a sine curve (see Figure below). The pitch of the helix is p π d tan θ, where d is the diameter of the circular clinder and θ is the angle between a normal to the helix and a longitudinal line. The width of the flat plate that wraps into the clinder shape is w π d sin θ. For actical reasons, the angle θ is usuall in the range from 0 o to 35 o.
Quiz A close end essure vessel has an inside diameter of 600 mm an a wall thickness of 40mm. It is essurized to an internal essure of 6.5MPa an it has a centric comessive force of 750kN applied as shown. The tank is welded together along the helix making an angle of 5 o to the horizontal. Determine the normal and shear stresses along the helix. 750kN The material is steel with a modulus E 00GPa and a Poisson s ratio ν 0.30 Calculate the following quantities for the clindrical part of the vessel: 5 O (a) The circumferential and longitudinal stresses and respectivel; (b) The maximum in plane and out of plane shear stresses (c) The circumferential and longitudinal strains ε and ε respectivel, and (d) The normal stress w and shear stress w acting perpendicular and parallel, respectivel, to the welded seam.
General State of Stresses From equilibrium inciples: x x, xz zx, z z The most general state of stress at a point ma be reesented b 6 components Normal Stresses Shear Stresses x x z z xz These are the incipal axes and incipal planes and the normal stresses are the incipal stresses.
ij x x xz x z zx z z ' Eigen values ' ij 0 0 0 0 0 0 3 The three circles reesent the normal and shearing stresses for rotation around each incipal axis. Points A, B, and C reesent the incipal stresses on the incipal planes (shearing stress is zero) Radius of the largest circle ields the maximum shearing stress. max max min
In the case of plane stress, the axis perpendicular to the plane of stress is a incipal axis (shearing stress equal zero). If the points A and B (reesenting the incipal planes) are on opposite sides of the origin, then the corresponding incipal stresses are the maximum and minimum normal stresses for the element the maximum shearing stress for the element is equal to the maximum in-plane shearing stress planes of maximum shearing stress are at 45 o to the incipal planes.
If A and B are on the same side of the origin (i.e., have the same sign), then the circle defining max, min, and max for the element is not the circle corresponding to transformations within the plane of stress maximum shearing stress for the element is equal to half of the maximum stress planes of maximum shearing stress are at 45 degrees to the plane of stress
Failure Theories Wh do mechanical components fail? Mechanical components fail because the applied stresses exceeds the material s strength (Too simple). What kind of stresses cause failure? Under an load combination, there is alwas a combination of normal and shearing stresses in the material. Mohr s circle for centric axial loading: P x, x 0 A x x P A Mohr s circle for torsional loading: Tc x 0 x J x Tc x J 0
What is the definition of Failure? Obviousl fracture but in some components ielding can also be considered as failure, if ielding distorts the material in such a wa that it no longer functions operl Which stress causes the material to fail? Usuall ductile materials are limited b their shear strengths. While brittle materials (ductilit < 5%) are limited b their tensile strengths. Stress at which point?
Yield Criteria for Ductile Materials under Plane Stress Conditions Maximum Shear Stress Theor (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 ield. Maximum shear stress criteria: The structural component is safe as long as the maximum shear stress is less than the maximum shear stress in a tensile test specimen at ield, i.e., S max < For a and b with the same sign, ield a b max or < For a and b with opposite signs, a b Y max < The maximum absolute shear stress is alwas the radius of the largest of the Mohr s circle.
Distortion Energ Theor (DET) Based on the consideration of angular distortion of stressed elements. The theor states that failure occurs when the distortion strain energ in the material exceeds the distortion strain energ in a tensile test specimen (of the same material) at ield. Von Mises effective stress : Defined as the uniaxial tensile stress that creates the same distortion energ as an actual combination of applied stresses. 3 3 3 3 3 + + + VM VM Maximum distortion energ criteria: Structural component is safe as long as the distortion energ per unit volume is less than that occurring in a tensile test specimen at ield. ( ) ( ) 0 0 6 6 Y b b a a Y Y b b a a Y d G G u u < + + < + < D 3 x x x VM + +
Given the material S Y, x, v and x find the safet factors for all the applicable criteria. a. Pure aluminum S 30 MPa 0MPa 0MPa MPa Y x x 0 0MPa 3 0MPa Max 0MPa Ductile -0 DET Theor n VM 0 S VM Use either the Maximum Shear Stress Theor (MSST) or the Distortion Energ Theor (DET) MSST Theor x n + 30MPa 7.3MPa S x 3 +.73 30 0 ( 0) 3 x 300 30 0.5 7.3MPa
b. 0.%C Carbon Steel S 65 Ksi 5Ksi 35Ksi Ksi Y x x 0 In the plane XY the incipal stresses are -.973Ksi and -38.03Ksi with a maximum shear stress in the XY plane of 8.03Ksi In an orientation Ksi.973Ksi 0 3 Max 9. 0Ksi Ductile 38.03Ksi MSST Theor n S 3 65 0 ( 38.03).7 DET Theor VM n S VM + 3 65Ksi.7 38.03MPa 3 38.03Ksi