CAESAR Il article 3 prepared by Eng Antar Mustafa (1).pdf

CAESAR Il article 3 prepared by Eng Antar Mustafa

1. CAESAR II
Training
Article 3
This prepared
by Engineer Antar Mustafa
[email protected]
[phone]

2. ➢Contents of article 2
1.Output Processor
2.Axial
3.Theory and Development of Pipe Stress Requirements
4.Basic Stress Concepts
5.Example
➢Contents of article 3…………………………..………page
1.Example ……………………..…..…………………….…… 3
2.Piping loads and stresses …..….……………………4
3.3D State of Stress in the Pipe Wall………..………6
4.Failure Theories. ………………………..…………….… 9
5.Maximum Stress Intensity Criterion ……………13
6. Code Stress Equations……………….……………Article 4

3. Most piping codes require stresses to be calculated using some form
of the following equations:
➢Longitudinal Stress:
➢Shear Stress:
➢Hoop Stress:
➢Example
This example calculation illustrates for a 6” nominal diameter,
standard schedule pipe (assuming the piping loads are known):
•Cross sectional properties
•Outside diameter d₀ = 168.3 mm
•Mean thickness tm = 7.112 mm
•Inside diameter dᵢ = d₀ - (2 x tm ) = 154.076mm
•Cross sectional Area

4. ➢Piping loads
Torsional Moment
Bending Moment
Axial Force
Internal Pressure
➢Stresses1.Longitudinal Stress
•Moment of Inertia
•Section Modulus

5. 4. Bending Component of Longitudinal stress
is the radius where the stress is being considered This will be at a maximum
value at the outer surface where
5. Torsional Stress The maximum torsional stress occurs at the outer radius where
at the outer surface
3. Hoop Stress
2. Shear Stress

6. 3. 3D State of Stress in the Pipe Wall
During operation, pipes are subject to all these types of stresses.
Examining a small cube of metal form the most highly stressed point of the pipe wall,
the stresses are distributed as so:
There are an infinite number of orientations
in which this cube could have been selected, each with a different combination of
normal and shear stresses on the faces. For example, there is one orientation of the
orthogonal stress axes for which one normal stress is maximised and another for which
one normal stress is minimised – in both cases; all shear stress components are zero. In
orientation in which the shear stress is zero, the resulting normal components of the
stress are termed the principal stresses. For 3-dimensional analyses, there are three of
them and they are designated S1 (the maximum), S2 and S3 (the minimum). Note that
regardless of the orientation of the stress axes, the sum of the orthogonal stress
components is always equal, i.e.:

7. The converse of these orientations is that in which the shear stress component is maximised
(there is also an orientation in which the shear stress is minimised, but this is ignored since
the magnitudes of the minimum and maximum shear stresses are the same); this is
appropriately called the orientation of maximum shear stress. The maximum shear stress in
a three dimensional state of stress is equal to ½ the difference between the largest and
smallest of the principal stresses (S1 and S3). The values of the principal and maximum shear
stress can be determined through the use of Mohr’s circle. The Mohr’s circle analysis can be
simplified by neglecting the radial stress component, therefore considering a less complex
(i.e. 2D) state of stress. A Mohr’s circle can be developed by plotting the normal vs. shear
stresses for the two known orientations (i.e. longitudinal stress vs. shear and hoop stress vs.
shear), and constructing a circle through the two points. The infinite combinations of normal
and shear stresses around the circle represent the combinations
present in the infinite number of possible orientations of the local
stress axes. A differential element at the outer radius of the pipe
(where bending and torsional stresses are maximised and
the radial normal and force-induced shear stresses are usually
zero) is subject to 2D plane stress and thus the principal stress
terms can be computed from the following Mohr’s circle:

8. The center of the circle is at and the radius is equal to
Therefore the principal stresses S1 and S2 are equal to the center of the circle, plus
or minus the radius respectively.
The principal stresses are calculated as: and
As noted above, the maximum shear
stress present in any orientation is equal to
or : Continuing our example: Mohr’s Circle of Stress
Centre of circle: Centre = = 78.07
Radius of Circle: =50.592MPa
Ƭ
Maximum Principal Stress S1

9. Maximum Principal Stress =128.662MPa
Or from the Mohr’s circle above, S1 = 78.07 + 50.59 = 128.66 Mpa
➢Maximum Principal Stresses S2 and S3
27.477 MPa Or from the Mohr’s circle,
➢Failure Theories
•The calculated stresses are not much use on their own, until they are compared to material
allowables. Material allowable stresses are related to strengths as determined by material
uniaxial tests, therefore calculated stresses must also be related to the uniaxial tensile test.
•This relationship can be developed by looking at available failure theories.
•This relationship can be developed by looking at available failure theories.
• There are three generally accepted failure theories which may be used to predict the
onset of yielding in a material:

10. •Octahedral Shear or Von Mises theory
•Maximum Shear or Tresca Theory
•Maximum Stress or Rankine Theory

11. •These theories relate failure in an arbitrary 3D stress state in a material to failure in the
stress state found in a uniaxial tensile test specimen, since it is that test that is most
commonly used to determine the allowable strength of commonly used materials.
•Failure of a uniaxial tensile test specimen is deemed to occur when plastic deformation
occurs, i.e. when the specimen yields; that is, release of the load does not result in the
specimen returning to its original state.
•The three failure theories state: Von Mises: “Failure occurs when the octahedral shear
stress in a body is equal to the octahedral shear stress at yield in a uniaxial tension test”
➢The octahedral shear stress is calculated as:
➢In a uniaxial tensile test specimen at the point of yield:
➢Therefore the octahedral shear stress in a uniaxial tensile test specimen at failure is
calculated as:
➢Therefore under the Von Mises theory: Plastic deformation occurs in a
➢3-Dimensional stress state whenever the octahedral shear stress exceeds

12. ➢Tresca: “Failure occurs when the maximum shear stress in a body is equal to the
maximum shear stress at yield in a uniaxial tension test.
The maximum shear stress is calculated as:
In a uniaxial tensile test specimen at the point of yield: ;

Therefore: Therefore, under Tresca theory
Plastic deformation occurs in a 3-Dimensional stress
state whenever the maximum shear stress exceeds
➢Rankine: “Failure occurs when the maximum tensile stress in a body is equal to
the maximum tensile stress at yield in a uniaxial tension test”
The maximum tensile stress is the largest, positive principal stress, S1 (by definition,
S1 is always the largest of the principal stresses.)
In a uniaxial tensile test specimen at the point of yield:
•Therefore, under Rankine theory: Plastic deformation occurs in a 3-Dimensional
stress state whenever the maximum shear stress exceeds

13. •Maximum Stress Intensity Criterion
Most of the piping codes use a slight modification of the maximum shear stress theory
for flexibility related failures. Repeating, the maximum shear stress theory predicts that
failure occurs when the maximum shear stress in a body equals
the maximum shear stress existing at failure during the uniaxial tensile test.
Recapping, the maximum shear stress in a body is given by:
For a differential element at the outer surface of the pipe,
the principal stresses were computed earlier as:
As seen previously, the maximum shear stress theory states that during the uniaxial
tensile test the maximum shear stress at failure is equal to one-half of the yield stress, so
the following requirement is necessary:

14. Multiplying both sides by 2 creates the stress intensity, which is an artificial parameter
defined simply as twice the maximum shear stress. Therefore the Maximum Stress
Intensity Criterion, as adopted by most piping codes, dictates the following requirement:
Note that when calculating only the varying stresses for
fatigue evaluation purposes, the pressure components drop out of the equation.
If an allowable stress based upon a suitable factor of safety is used, the Maximum Stress
Intensity criterion yields an expression very similar to that specified by the B31.3 code:
Where: longitudinal normal stress due to bending
=shear stress due to torsion, allowable stress for loading case
Continuing our example for the 6” diameter, standard wall pipe, in which longitudinal,
shear and hoop stresses were calculated:
, Assuming that the yield stress of the pipe material is 206 MPa
(30,000 psi) at operating temperature, and a factor of safely of 2/3 is to be used, the
following calculations must be made:
,
The 101.185 MPa is the calculated stress intensity in the pipe wall, while the 137.33
MPa is the allowable stress intensity for the material at the specified temperature.
In this case, the pipe would appear to be safely loaded under these conditions.