1 Spherical Pressure Vessels Pressure vessels are closed structures containing liquids or gases under essure. Examples include tanks, pipes, essurized...

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r > 10 t

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 body. Acting on this free body are the tensile stress σ in the wall of the vessel and the fluid pressure p .

The pressure that acts horizontally against the plane circular area is uniform and gives a resultant pressure force of : Where p is the gage or internal pressure (above the pressure acting in the outside of the vessel). The stress is uniform around the circumference and it is uniformly distributed across the thickness t (because the wall is thin). The resultant horizontal force is : Equilibrium of forces in the horizontal direction:

P = pπr

2

Horizontal Force = σ (2πrm )t t rm = r + 2

σ (2π rm )t = p π r 2

pr σ = 2t rm ~ r for thin walls. Therefore the formula to calculate the stress in a thin walled spherical vessels is

As is evident from the symmetry 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 pressurized spherical vessel is subjected to uniform tensile stresses σ in all directions. Stresses that act tangentially to the curved surface of a shell are known as membrane 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 vessel except near points of stress concentration.

Stresses at the Outer Surfaces. The element below has the x and y axes tangential to the surface of the sphere and the z axis is perpendicular to the surface. Thus, the normal stresses σx and σy are equal to the membrane stress σ and the normal stress σz is zero. pr The principal stresses are σ1 = σ 2 = σ =

2t

and σ3 = 0. Any 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 y axis. Elements oriented at 45o of the x or y axis have maximum shear stresses equal to σ/2 or

τ Max

σ

pr = = 2 4t

Stresses at the Inner Surface

At the inner wall the stresses in the x and y direction are equal to the membrane 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 the outer surface. pr The element is in triaxial stress σ1 = σ 2 = σ =

2t

σ3 = − p The in-plane shear stress are zero, but the maximum out-of-plane shear stress (obtained at 45o rotation about either the x or y axis) is

(σ

τ

Max

τ

Max

pr p + p) = = + 2 4t 2 p ⎛ r ⎞ = ⎜1 + ⎟ 2 ⎝ 2t ⎠

or

When the vessels is thin walled and the ratio r/t is large, we can disregard the number 1 and

τ Max

pr = 4t

Consequently, we can consider the stress state at the inner surface to be the same as the outer surface. Summary for Spherical pressure vessel with r/t large:

pr σ1 = σ 2 = 2t

As the two stresses are equal, Mohr’s circle for in-plane transformations reduces to a point

σ = σ 1 = σ 2 = constant τ max(in -plane) = 0 Maximum out-of-plane shearing stress

τ max =

1σ 2 1

pr = 4t

A compressed air tank having an inner diameter of 18 inches and a wall thickness of ¼ inch is formed by welding two steel hemispheres (see figure). (a) If the allowable tensile stress in the steel is 14000psi, what is the maximum permissible air pressure pa in the tank?. (b) If the allowable shear stress in the steel is 6000psi, what is the maximum permissible pressure pb?. (c) If the normal strain in the outer surface of the tank is not to exceed 0.0003, what is the maximum permissible pressure pc? (Assume Hooke’s law is obeyed E = 29x106psi and Poisson’s ratio is ν = 0.28) (d) Tests on the welded seam show that failure occurs when the tensile load on the welds exceeds 8.1kips per inch of weld. If the required factor of safety against failure of the weld is 2.5, what is the maximum permissible pressure pd? (e) Considering the four preceding factors, what is the allowable pressure pallow in the tank?

Solution (a)Allowable pressure based on the tensile stress in the steel. .We will use the equation pa r

σ allowed =

2t

then

2tσ allowed 2(0.25inch )(14000 psi ) = = 777.8 psi pa = r 9.0inch (b) Allowable pressure based upon the shear stress of the steel. We will use

τ allowed = pb =

σ 2

=

pb r 4t

then

(0.25inch )(6000 psi ) = 666.7 psi 4tτ allowed =4 r 9.0inch

(c) Allowable pressure based upon ε X = the normal strain in the steel. For biaxial stress then

(σ X − υσ Y )

this equation can be solved for pressure pc

(

E

εX =

substituti ng

(1 − υ )σ E

)

=

σX = σY = σ =

(1 − υ ) pr 2tE

2 tE ε allowed 2 (0 . 25 inch ) 29 x10 6 psi (0 . 0003 ) pc = = = 671 . 3 psi (9 .0 inch )(1 − 0 .28 ) r (1 − υ )

pr 2t

(d) Allowable pressure based upon the tension in the welded seam.

Τallowed

The allowable tensile load on the welded seam is equal to the failure load divided by the factor of safety

Τallowed

Τfailure

8.1kips / inch = = n 2.5 = 3.24kips / inch = 3240lb / inch

The corresponding allowable tensile stress is equal to the allowable load on 1inch length 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 allowed 2 )(0 . 25 inch )(12960 psi ) pd = = = 720 . 0 psi r 9 . 0 inch (e) Allowable pressure Comparing the preceding results for pa, pb, pc and pd, we see that the shear stress in pallow = 666psi. the wall governs and the allowable pressure in the tank is

Cylindrical Pressure Vessel Examples: Compressed air tanks, rocket motors, fire extinguishers, spray cans, propane tanks, grain silos, pressurized pipes, etc. We will consider the normal stresses in a thin walled circular tank AB subjected to internal pressure p.

σ1 and σ2 are the membrane stresses in the wall. No

shear stresses act on these elements because of the symmetry of the vessel and its loading, therefore σ1 and σ2 are the principal stresses.

Because of their directions, the stress σ1 is called circumferential stress or the hoop stress, and the stress σ2 is called the longitudinal stress or the axial stress.

Equilibrium of forces to find the circumferential stress:

σ 1 (2 b )t = p (2 b )r

pr σ1 = t

σ 2 (2πr )t = p (πr 2 )

Equilibrium of forces to find the longitudinal stress: σ

2

pr = 2t

This longitudinal stress is equal to the membrane stress in a spherical vessel. Then:

σ 1 = 2σ 2 We note that the longitudinal welded seam in a pressure tank must be twice as strong as the circumferential seam.

Stresses at the Outer Surface The principal stresses σ1 and σ2 at the outer surface of a cylindrical vessel are shown below. Since σ3 is zero, the element is in biaxial stress. The maximum in plane shear stress occurs on planes that are rotate 45o about the z-axis

( σ1 − σ 2 ) (τ Max )z = = 2

pr 4t

The maximum out of plane shear stresses are obtain by 45o rotations about the x and y axes respectively. σ pr

τ Max , x =

τ Max , y

1

=

2t σ pr = 2 = 2 4t 2

Then, the absolute maximum shear stress is τmax = pr / 2t , which occurs on a plane that has been rotated 45o about the x-axis.

Stresses at the Inner Surface pr pr The principal stresses are σ1 = σ2 = and σ 3= − p t 2t The three maximum shear stresses, obtained by 45o rotations about the x, y and z axes are

( σ1 − σ 3 ) (τ MAX )x = = (τ Max )y (τ Max )z

pr p + 2 2t 2 ( σ 2 − σ 3 ) pr p = = + 2 4t 2 ( σ 1 − σ 2 ) pr = = 2 4t

The first of these three stresses is the largest. However, when r/t is very large (thin walled), the term p/2 can be disregarded, and the equations are the same as the stresses at the outer surface.

Summary for Cylindrical vessels with r/t large Cylindrical vessel with principal stresses σ1 = hoop stress σ2 = longitudinal stress

Hoop stress: ∑ Fz = 0 = σ 1(2t Δx ) − p (2r Δx ) pr σ1 = t

Longitudinal stress:

( )

2 ∑ Fx = 0 = σ 2 (2π rt ) − p π r pr σ2 = 2t σ 1 = 2σ 2

Points A and B correspond to hoop stress, σ1, and longitudinal stress, σ2 Maximum in-plane shearing stress:

1 pr τ max(in − plane) = σ 2 = 2 4t Maximum out-of-plane shearing stress corresponds to a 45o rotation of the plane stress element around a longitudinal axis

pr τ max = σ 2 = 2t

A cylindrical pressure vessel is constructed from a long, narrow steel plate by 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 = 55o with the longitudinal axis. The vessel has an inner radius r=1.8m and a wall thickness t = 20mm. The material is steel with a modulus E=200GPa and a Poisson’s ratio ν=0.30. The internal pressure p is 800kPa Calculate the following quantities for the cylindrical part of the vessel: The circumferential and longitudinal stresses σ1 and σ2 respectively; The maximum in-plane and out-of-plane shear stresses The circumferential and longitudinal strains ε1 and ε2 respectively, and The normal stress σw and shear stress τw acting perpendicular and parallel, respectively, 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:

pr (800kPa )(1.8m ) σ1 = = = 72 MPa t 0.02m pr (800kPa )(1.8m ) σ2 = = = 36 MPa 2t 2(0.02m )

(τ Max )in − plane

( σ1 − σ 2 ) = =

(τ Max )out − of − plane

pr = 18MPa 2 4t ( σ1 ) pr = = = 36 MPa 2 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 εx = ε2 and εy = ε1 and also σx = σ2 and σy = σ1. Therefore the above equations can be written in the following forms:

εx =

σx E

ε y = −υ

−υ

σx E

σy E +

σy E

ε2 =

(1 − 2υ ) pr

=

(1 − 2υ )σ 2

=

(1 − 2 (0 .30 ))(36 MPa ) = 72 x10 − 6

2 tE 200 GPa E (2 − υ ) pr = (2 − υ )σ 1 = (2 − 0 .30 )(72 MPa ) = 306 x10 − 6 ε1 = (2 )(200 GPa ) 2 tE 2E

(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 cylinder with sides parallel and perpendicular to the weld is θ = 90o – α = 35o We will use the stress transformation equations:

σ x1 = σ x cos 2 θ + σ y sin 2 θ + 2τ xy sin θ cosθ τ x1 y1 = −

σ x −σ y 2

sin 2θ + τ xy cos 2θ

σ x1 + σ y1 = σ x + σ y

σ x1 = 36(cos(35))2 + 72(sin (35))2 = 47.8MPa

τ x1 y1 = −36 ⋅ sin (35) ⋅ cos(35) + 72 ⋅ sin (35) ⋅ cos(35) = 16.9 MPa 47.8 + σ y1 = 36 + 72 ⇒ σ y1 = 60.2MPa Mohr’s circle Coordinates of point A (for θ = 0) σ2 = 36MPa and shear stress = 0 Coordinates of point B (for θ = 90) σ1 = 72MPa and shear stress = 0 Center (point C) = (σ1 + σ2) / 2 = 54MPa Radius = (σ1- σ2) / 2 =18MPa A 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) of the element. The coordinates of point D σx1 = 54 – R cos 70o = 54MPa – (18MPa)(cos 70o) = 47.8MPa τx1y1 = R sin 70o = (18MPa)(sin 70o) = 16.9MPa

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 cylinder 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 cylinder shape is w = π d sin θ.

For practical reasons, the angle θ is usually in the range from 20o to 35o.

Quiz A close‐end pressure vessel has an inside diameter of 1600 mm an a wall thickness of 40mm. It is pressurized to an internal pressure of 6.5MPa an it has a centric compressive force of 750kN applied as shown. The tank is welded together along the helix making an angle of 25o to the horizontal. Determine the normal and shear stresses along the helix.

750kN

The material is steel with a modulus E = 200GPa and a Poisson’s ratio ν = 0.30 Calculate the following quantities for the cylindrical part of the vessel: (a) The circumferential and longitudinal stresses σ1 and σ2 respectively;

25O

(b) The maximum in‐plane and out‐of‐plane shear stresses (c) The circumferential and longitudinal strains ε1 and ε2 respectively, and (d) The normal stress σw and shear stress τw acting perpendicular and parallel, respectively, to the welded seam.

General State of Stresses From equilibrium principles: τxy = τyx , τxz = τzx , τzy = τyz

The most general state of stress at a point may be represented by 6 components

Normal Stresses

σx

σy

σz

Shear Stresses

τ xy

τ yz

τ xz

These are the principal axes and principal planes and the 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 and shearing stresses for rotation around each principal axis. Points A, B, and C represent the principal stresses on the principal planes (shearing stress is zero) Radius of the largest circle yields the maximum shearing stress.

1 τ max = σ max − σ min 2

In the case of plane stress, the axis perpendicular to the plane of stress is a principal axis (shearing stress equal zero). If the points A and B (representing the principal planes) are on opposite sides of the origin, then the corresponding principal 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 45o to the principal planes.

If A and B are on the same side of the origin (i.e., have the same sign), then the circle defining σmax, σmin, and τmax 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

Why 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 any load combination, there is always a combination of normal and shearing stresses in the material. Mohr’s circle for centric axial loading: σx =

P , σ y = τ xy = 0 A

σ x = σ y = τ xy =

P 2A

Mohr’s circle for torsional loading: σ x = σ y = 0 τ xy = σx =σy =

Tc J

Tc τ xy = 0 J

What is the definition of Failure? Obviously fracture but in some components yielding can also be considered as failure, if yielding distorts the material in such a way that it no longer functions properly Which stress causes the material to fail? Usually ductile materials are limited by their shear strengths. While brittle materials (ductility < 5%) are limited by their tensile strengths. Stress at which point?

Yield Criteria for Ductile Materials under Plane Stress Conditions Maximum 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 the maximum shear stress is less than the maximum shear stress in a tensile test specimen at yield, i.e., σy For σa and σb with the same sign, τ <τy = max Sy 2

τ yield =

τ max =

2

σa 2

For σa and σb with opposite signs,

τ max =

σa −σb 2

σ < Y 2

The maximum absolute shear stress is always the radius of the largest of the Mohr’s circle.

or

σb 2

<

σy 2

Distortion Energy Theory (DET) Based on the consideration of angular distortion of stressed elements. The theory states that failure occurs when the distortion strain energy in the material exceeds the distortion strain energy in a tensile test specimen (of the same material) at yield. Maximum distortion energy criteria: Structural component is safe as long as the distortion energy per unit volume is less than that occurring in a tensile test specimen at yield. Von Mises effective stress : Defined as the uniaxial tensile stress that creates the same distortion energy as any actual combination of applied stresses.

ud < uY 1 2 1 2 σ a − σ aσ b + σ b2 < σ Y − σ Y × 0 + 02 6G 6G

(

)

(

)

σ a2 − σ aσ b + σ b2 < σ Y2

σ VM = σ 12 + σ 22 + σ 32 − σ 1σ 2 − σ 2σ 3 − σ 3σ 1 σ VM = σ 12 + σ 32 − σ 3σ 1 2D

σ 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 applicable criteria. a. Pure aluminum

SY = 30 MPa σ x = 10 MPa σ y = −10 MPa τ xy = 0 MPa

σ 1 = 10 MPa

σ 3 = −10 MPa

τ Max = 10 MPa

Ductile -10

10

Use either the Maximum Shear Stress Theory (MSST) or the Distortion Energy Theory (DET)

MSST Theory

Sy

30 30 n= = = = 1.5 σ 1 − σ 3 10 − (−10) 20 DET Theory

σ VM = σ x 2 + σ y 2 − σ xσ y + 3τ xy2 = 300 = 17.32MPa n=

Sy

σ VM

30 MPa = = 1.73 17.32 MPa

b. 0.2%C Carbon Steel SY = 65 Ksi

σ x = −5 Ksi σ y = −35 Ksi τ xy = 10 Ksi

In the plane XY the principal stresses are -1.973Ksi and -38.03Ksi with a maximum shear stress in the XY plane of 18.03Ksi

In any orientation σ 1 = 0 Ksi σ 2 = −1.973Ksi σ 3 = −38.03Ksi

τ Max = 19.01Ksi MSST Theory

DET Theory

Ductile

Sy

65 n= = = 1.71 σ 1 − σ 3 0 − (−38.03) σ VM = σ 12 + σ 3 2 − σ 1σ 3 = 38.03Ksi n=

Sy

σ VM

=

65 Ksi = 1.71 38.03MPa

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