Skip to main content

Thin - walled pressure vessels

A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections.
If the thickness of the wall of the cylindrical vessel is less than 1/15 to 1/20 of its internal diameter,the cylindrical vessel is known as Thin Cylinder. 

Tangential Stress (Circumferential Stress) or Hoop stress

Consider the tank shown being subjected to an internal pressure p. The length of the tank is L and the wall thickness is t . Considering the right half of the tank: 
The forces acting on this right half of the vessel are the total pressures F caused by the internal pressure p, and the total tension T in the walls.
The projected area subjected to internal pressure = A = DL
F = pxA = p.DL
The tangential stress in the walls = st
T =  sAwall st . tL
But F = 2T
 p.DL = 2(st . tL) 
The tangential stress st = pD/2t
If there exist an external pressure po and an internal pressure pi, the the Tangential stress may be expressed as: 
st = (po - pi ) D/2t

Longitudinal Stress

Consider the free body diagram in the transverse section of the tank


The total force acting at the rear of the tank must equal to the total longitudinal stress on the wall. Since is the thickness wall t is so small compared to diameter of vessel D 
The area of the wall  AwallpDt
the total longitudinal stress on the wall 
PL = sLpDt
The total internal force acting at the rear of the tank = F
F = p. Area of the end = p ( pD2/4)
There fore PL = F
sLpDt = p ( pD2/4)
The longitudinal stress sL = pD/4t
If there exist an external pressure and an internal pressure , the formula may be expressed as: 
s= (po - pi ) D/4t
Note:
It can be observed that the tangential stress/hoop stress is twice that of the longitudinal stress. 

SPHERICAL SHELL 

If a spherical tank of diameter and thickness contains gas under a pressure of , the stress at the wall can be expressed as
= 4tpD

Comments

  1. Nice information, many thanks to the author. It is incomprehensible to me now, but in general, the usefulness and significance is overwhelming. Thanks again and good luck ! Reaction Kettles || Process Piping || Site Fabrication

    ReplyDelete

Post a Comment

Popular posts from this blog

Relation between Modulus of Elasticity and Modulus of Rigidity

Modulus of Elasticity (E)   It is the ratio between Normal stress to Normal strain within the elastic limit. Elastic Modulus E = Normal stress/Normal strain E =  s/e Modulus of Rigidity (G)  It is the ratio between Shear stress to Shear strain within the elastic limit. Rigidity Modulus G = Shear stress/ Shear strain G =    Ƭ / ø   Relation between Modulus of Elasticity and Modulus of Rigidity: Consider a solid cube  PQRS is  subjected to a shearing force F.  Let  Ƭ     be the shear stress produced in the faces PQ and RS due to this shear force. The complementary shear stress consequently produced in the vertical faces PS and RQ is also equal to same and shown in figure as Ƭ   Due to the pure shearing force, the cube is deformed PQRS to PQR'S' . The point   S moved to S' and point R moved to R' as shown in fig.  The shear strain = The angle of distortion  ø                          ø = RR'/ RQ ---(1) Shear strain = Shear stress /Rigidity modulus                

PORTAL METHOD and CANTILEVER METHOD

The behavior of a structure subjected to horizontal forces depends on its height to width ratio. The deformation in low-rise structures, where the height is smaller than its width, is characterized predominantly by shear deformations. In high rise building, where height is several times greater than its lateral dimensions, is dominated by bending action. To analyze the structures subjected to horizontal loading we have two methods. Portal method  and Cantilever method 1. PORTAL METHOD The portal method is an approximate analysis used for analysing building frames subjected to lateral loads such as Wind loads/ seismic forces.  Since shear deformations are dominant in low rise structures, the method makes simplifying assumptions regarding horizontal shear in columns.  Each bay of a structure is treated as a portal frame, and horizontal force is distributed equally among them. Assumptions in portal method   1. The points of inflection are located at the mid-height of each column above th

Shearing Stresses Distribution in Circular Section

Show that the shearing stress developed at the neutral axis of a beam with circular cross section is  τ max = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis.  Solution : Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam Shear stress at a section will be given by following formula as mentioned here Where, F = Shear force (N) τ = Shear stress (N/mm2) A = Area of section, where shear stress is to be determined (mm2) ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m) Q = A. ȳ = Moment of the whole shaded area about the neutral axis I = Moment of inertia of the given section about the neutral axis (mm4) For circular cross-section, Moment of inertia, I = ПR4/4 b = Width of the given section where shear stress is to be determined. Let us consider on