Piping system design constitutes a major part of the design and engineering effort in any facility. Stress analysis is a critical component of piping design through which important parameters such as piping safety, safety of related components and connected equipment and piping deflection can be addressed. The objective of pipe stress analysis is to prevent premature failure of piping and piping components and ensuring that piping stresses are kept within allowable limits.
Download Chapter List
Fundamentals of Pipe Stress Analysis with Introduction to CAESAR II  Introduction to Pipe Stress Analysis
This chapter provides a brief introduction to pipe stress analysis and explains the need for stress analysis in piping systems. The phenomenon of overstressing in piping systems and its consequences are touched upon in detail. An attempt is also made to obtain a clear understanding of the fundamental physical parameters used in stress analysis such as force, stress and strain and modulus of elasticity. Details of the various physical quantities and units used in pipe stress analysis are also discussed along with a description of the concepts of tensile testing and yield strength of materials. The various aspects related to the thermal expansion / contraction and flexibility of piping systems are also adequately covered during the course of the discussion.
Learning objectives
Piping systems need to be analyzed for stresses, to ensure that the components of the system are not overstressed. Piping systems typically consist of straight pipes, fittings (elbow, tees, reducers), flanges, valves and accessories such as actuators. The stresses on equipment nozzles where the pipe connects to the equipment also need to be analyzed. The pipe wall resists both the internal and external forces experienced by the piping system. The force per unit metal area of the pipe wall is the resulting pipe stress. The objective of pipe stress analysis is to ensure that the stresses do not exceed allowable values specified by the design codes. Pipe stress analysis provides the necessary techniques and methods for designing piping systems without overstressing the piping components and the connected equipment.
Piping stress analysis applies to calculations that address the static and dynamic loading arising on account of the effects of temperature changes, gravity, external and internal pressures and changes in fluid flow rate.
Some of the important reasons why piping stress analysis is needed include:
Overstressing can lead to premature failure of the piping system, causing leaks and safety hazards. Overstressing can lead to cracks, breakages and other secondary failures and failures such as bowing and opening of flanges. In some cases, the failure can be catastrophic, causing the collapse of the system and with the potential for loss of life and property. Some situations may even require the entire plant to be shut down. Thus, the objective of pipe stress analysis is to ensure the safe operation of piping systems within a plant, while simultaneously meeting the performance requirements of the plant.
Overstressing in piping can result in the following:
In general, overstressing can result from many different sources. Common examples include inadequate input such as insufficient pipe thickness, overconstraint, excessive thermal expansion or presence of other loads. The remedy for overstressing can be both, to add or in certain instances remove constraints such as releasing degrees of freedom of pipe supports or hangers. Although this process is often carried out on a trial and error basis, major piping layout related problems can usually be anticipated by experienced piping engineers during the design stage itself.
1.3.1 Force
Force is a vector quantity that has both magnitude and direction. It can be defined as a push or pull on an object resulting from its interaction with another object. Force is no longer experienced when this interaction ceases. Piping systems experience both tensile and compressive forces. Forces experienced by piping systems are also known as “piping loads”. Commonly used units for force are: Newton (N), kilogram force (kg_{f}) and pound force (lb_{f}). The units of force are explained here.
Newton is the force required to accelerate a one kilogrammass at 1 m/s^{2}. Thus,


Kilogram force is the force required to accelerate a one kilogrammass at 9.81 m/s^{2}. Thus,
Pound force is the force required to accelerate a one poundmass at 32.2 ft/s^{2}. Thus,
The definition of pound force creates a need for using the conversion constant g_{c} while performing calculations in the US Customary System (USCS).
The conversion factors for force units are:
1 kg_{f} = 9.81 N
1 kg_{f} = 2.205 lb_{f}
1 lb_{f} = 4.4462 N
The concept of force can be better understood with the help of the following exercise.
Sample Exercise
Problem
A 5 kg .object is moving horizontally at a speed of 10m/sec. Determine the Net force required to keep the object moving at this speed and in the same direction.
Solution
Zero N. This is because, an object in motion will maintain its state of motion and the presence of an unbalanced force results in a change in its velocity.
1.3.2 Engineering stress
Engineering stress S is the force per unit area of the metal cross section. A stress may be normal, shear or torsion, leading to corresponding deformations. While stress cannot be measured directly, deformations can be measured.
Units for engineering stress:
N/m^{2} (Pascal, Pa)
lb_{f}/in^{2 }(psi)
kg_{f}/cm^{2}
Commonly used units for stress:
Kilo pounds per square inch (ksi) = 10^{3 }psi
Megapascals (MPa) = 10^{6} Pa
Commonly used conversion factors for stress:
1 lb_{f}/in^{2} (psi) = 0.0703 kg_{f}/cm^{2} = 6.896 kPa
1 lb_{f}/in^{2} (psi) = 6.896 kPa
1 MPa = 145 psi
1 ksi = 6.88 MPa
1.3.3 Deformation of materials and engineering strain
When the elements of materials are subjected to tensile or compressive loads, they undergo small deformations. These deformations can be “elastic” or “plastic”. Within the elastic limit, the deformation is “elastic”, i.e. the material springs back to its original shape when the load is removed. Thus, elastic deformation is temporary in nature and exists only when the load is present. After the material begins to yield, the deformation is permanent and remains even after the load is removed. This is called “plastic” deformation. Figure illustrates the manner in which deformation of a material occurs, when it is subjected to tensile and compressive loads.
Tension Compression
Figure 1.1
Deformation of a material when subjected to tensile and compressive loads
Engineering strain e is the change in length divided by the original length, i.e.
Where
DL is the change in length
L_{o} is the original length
Units of strain: in/in or mm/mm.
While an object in tension has resulting tensile strain, an object in compression has resulting compressive strain. The above equation for strain is only valid if the deformation of the object is uniform throughout its volume.
1.3.4 Modulus of elasticity (E)
Modulus of Elasticity E is a material property that is indicative of the strength of the material. The modulus of elasticity values for steel and aluminum are given here. The values indicate that steel is about three times stronger than aluminum.
E_{steel} = 30 x 10^{6} psi = 2.07 x 10^{5} MPa
E_{aluminum} = 10 x 10^{6} psi = 0.70 x 10^{5} MPa
The modulus of elasticity of materials decreases with increase in temperature. This is due to the thermal expansion of materials. At higher temperatures, thermal expansion results in a lesser force being required to cause a given amount of strain, resulting in a lower modulus of elasticity. The modulus of elasticity for different materials and at various temperatures is listed in Table 1.1. Modulus of elasticity is also referred to as “Young’s Modulus”.
Table 1.1
Modulus of Elasticity of Different Materials at Various Temperatures
(Modulus of Elasticity is given in 10^{5} MPa. The values in parenthesis are in 10^{6} psi)
Material 
130°C (203°F) 
20°C (68°F) 
260°C (500°F) 
540°C (1004°F) 
810°C (1490°F) 

Carbon steels (<3% C) 
2.03 (29.5)

1.92 (27.9)

1.82 (26.4)

1.06 (15.4)

 


Low, Intermediate alloy steels

1.96 (28.5)

1.88 (27.4)

1.79 (26.0)

1.57 (22.8)

 


Austenitic stainless steels

2.06 (29.9)

1.95 (28.3)

1.80 (26.1)

1.56 (22.7)

1.23 (17.9)


Monel (67Ni, 30Cu)

1.83 (26.6)

1.79 (26.0)

1.75 (25.4)

1.10 (16.0)

 


CuproNickel (70Cu, 30Ni) 
 

1.49 (21.6)

1.40 (20.3)

 

 


Aluminum Alloys
Copper
Brass (66Cu, 34Zn)
Bronze (88Cu, 6Sn, 4.5Zn, 1.5Pb) 
0.750 (10.9)
1.15 (16.7)
1.01 (14.7)
0.945 (13.8)

0.695 (10.1)
1.10 (16.0)
0.963 (14.0)
0.894 (13.0)

0.530 (7.7)
1.01 (14.7)
0.874 (12.7)
0.805 (11.7)

 
 
 
 

 
 
 
 









The different physical quantities of force, stress, strain, modulus of elasticity and their respective units have already been discussed. Let us go ahead and discuss some of the other physical quantities used in pipe stress analysis.
The density of a substance is its mass per unit volume. It is represented by the symbol “r”. Density for a given substance can be calculated from the following equation,
Density (r) = Mass of the substance (m) / Volume of the substance (V)
Density has the units, lbm/ft^{3} or kg/m^{3}.
If equal masses of cotton and lead are taken (say 1 kg each), we will find that the volume of cotton is much larger than the volume of lead. This is because lead is heavier (denser) than cotton. The particles of lead are closely packed while those of cotton are more diffused.
Density tends to change with change in temperature.
The specific gravity of a substance is the ratio of the density of a substance to the density of some standard substance. The standard substance is usually water (at 4°C) for liquids and solids, while for gases it is usually air. Specific gravity is also known as Relative Density.
Relative density for liquids and solids (s) = Density of substance
Density of water at 4°C
Relative density for gases (s) = Density of substance
Density of air
Density of substance = Density of water at 4°C ´ Relative density of liquid or solid
i.e.: r (for liquids and solids) = 1000 ´ s and
r (for gases) = 1.29 ´ s
Specific gravity is a dimensionless number.
The specific weight of a substance is the weight per unit volume. It has units of kN/m^{3} or kg_{f}/m^{3} or lb_{f}/ft^{3}. The specific weight of water at standard conditions is 9.81 kN/m^{3} or 1000 kg_{f}/m^{3} or 62.4 lb_{f}/ft^{3}. The specific weight of any substance is the product of the specific gravity of the substance and the specific weight of water at standard conditions.
Table 1.2
Specific Gravity, Density and Specific Weights of Materials
Material 
Specific Gravity 
Density kg/m^{3} 
Density lb_{m}/ft^{3} 
Specific Weight kN/m^{3} 
Specific Weight kg_{f}/m^{3} 
Specific Weight lb_{f}/m^{3} 
CS (<0.3% C) 
7.84

7840 
489 
76.91 
7840 
489 
Intermediate Alloy Steels (5% Cr, Mo to 9% Cr, Mo)

7.84 
7840 
489 
76.91 
7840 
489 
Austenetic Stainless Steel

7.98

7980 
498 
78.28 
7980 
498 
Brass (66% Cu, 34% Zn)

8.75 
8750 
546 
85.84 
8750 
546 
Aluminum Alloys 
2.77 
2770 
173 
27.17 
2770 
173 
1.4.4 Poisson’s Ratio
When a material is subjected to a tensile load, it elongates. Since the volume of the material is constant, the elongation in the longitudinal direction results in compression in the lateral direction. Similarly, compression along the longitudinal direction is accompanied by elongation along the lateral direction. Poisson’s ratio is the ratio of lateral strain to the longitudinal strain and is mathematically represented as
n =  e_{lateral} / e_{longitudinal}
In the case of a perfectly incompressible material that is deformed elastically at small strains, the Poisson's ratio would be exactly 0.5. Most practical engineering materials have values between 0 and 0.5. While cork has a value close to 0, most steels have values around 0.3. Rubber has a value of almost 0.5. Some materials, mostly polymer foams, have a negative Poisson's ratio. A value of 0.3 is used for most materials. Typical Poisson’s Ratio values for some common materials are given in table 1.3.
Table 1.3
Typical Poisson’s Ratio Values for Different Materials
Material 
Poisson’s Ratio 
Rubber Lead Phosphor Bronze Copper Magnesium Molybdenum Magnesium alloy Beryllium Copper Wrought Iron Nickel Silver Aluminum 
0.48 – 0.50 0.431 0.359 0.355 0.350 0.307 0.281 0.285 0.278 0.322 0.334 
Clay 
0.3  0.45 
Zinc Brass (7030) Titanium Stainless steel 188 
0.331 0.331 0.320 0.305 
Mild steel 
0.303 
High carbon steel Nickel steel 
0.295 0.291 
Cast steel Glass Ceramic Glass Cast iron  grey Concrete Bronze Cork 
0.265 0.290 0.240 0.211 0.200 0.140 0.000 
1.4.5 Linear coefficient of thermal expansion (a)
Thermal expansion and contraction of piping systems is an important aspect of pipe stress analysis. The Linear Coefficient of Thermal Expansion a is useful in determining the thermal displacements of piping systems and connected equipment.
Coefficient of Thermal Expansion is defined as the thermal strain per unit degree change in temperature. Thermal strain is the change in length (DL) divided by the original length(L_{o}).
Table 1.4 gives the Thermal Expansion Coefficients for different materials.
Table 1.4
Thermal Expansion Coefficients for Selected Materials at 21°C (70°F)
Material 
a (mm/mm)/°C 
a (in./in.)/°F 
Carbon and Low Alloy Steel Through 3 CrMo

10.93 
6.07 
Intermediate Alloy Steel 5 CrMo Through 9 CrMo

10.30 
5.73 
Austenitic Stainless Steel 18 Cr8 Ni

16.40 
9.11 
Copper

16.68 
9.27 
Aluminum 
22.85 
12.69 
Sample Exercise
Problem
A steel rod of 10 mm diameter is subjected to a tensile load of 5000 N. Calculate the following:
Solution
A.
B.
Increase in length,
The physical quantities used in pipe stress analysis and their units are listed in Table 1.5.
Table 1.5
Physical Quantities and Units Used in Pipe Stress Analysis
Physical Quantity 
Symbol 
SI System 
USCS 
Length 
L 
Meter (m) 
Feet (ft) 
Diameter 
D 
Millimeter (mm) 
Inch (in) 
Thickness 
Dx 
Millimeter (mm) 
Inch (in) 
Mass 
M 
Kilogram (kg) 
Pound mass (lb_{m}) 
Weight 
W 
Newtons (N) 
Pound force (lbf) 
Time 
t 
Seconds (s) 
Seconds (sec) 
Temperature 
T 
Degree Celcius (°C) 
Degree Farenheit (°F) 
Area 
A 
Square meter (m^{2}) 
Square feet (ft^{2}) 
Volume 
V 
Cubic meter (m^{3}) 
Cubic feet (ft^{3}) 
Density 
r 
kg / m^{3} 
lbm / ft^{3} 
Acceleration 
a 
Meters/sec^{2} (m/s^{2}) 
Feet/sec^{2} (ft/sec^{2}) 
Force 
F 
Newton (N) 
Pound force (lbf) 
Pressure 
P 
Pascal (Pa) 
Pounds/in^{2} (psi) 
Stress 
s 
Megapascal (Mpa) 
Pounds/in^{2} (psi) 
Strain 
e 
mm/mm 
in/in 
Work 
W 
Newtonmeter (N.m) 
Foot pound force (ftlbf) 
Energy 
E 
Joule (J) 
British thermal unit (Btu) 
Modulus of Elasticity 
E 
MPa 
Kilopounds / in^{2} (ksi) 
Moment 
M 
N.m 
ftlbf 
Moment of Inertia 
I 
mm^{4} 
in^{4} 
Section Modulus 
Z 
mm^{3} 
in^{3} 
Unit Prefixes:
Kilo (k) = 10^{3} Micro (m) = 10^{6}
Mega (M) = 10^{6} Nano (n) = 10^{9}
Giga (G) = 10^{9 }Milli (m) = 10^{3}
Tensile tests are conducted on material specimens to determine material properties such as modulus of elasticity and yield strength. The yield strength of a material is frequently used in determining allowable stresses for piping systems. Tensile tests are conducted using procedures and guidelines established by the “American Society for Testing of Materials (ASTM)”. Tensile tests are carried out using “Universal Testing Machine” or UTM.
The result of tensile testing is a “Stress – Strain Curve”. The stress – strain curve for a typical ductile material such as mild steel is shown in Figure 1.2.
Figure 1.2
Stress – Strain Curve for a Ductile Material (Source: “Introduction to Pipe Stress Analysis”, Sam Kannappan, John Wiley & Sons, 1986)
The following points can be observed from the Stress – Strain Curve for the ductile material shown in the figure.
Figure 1.3 illustrates the stressstrain curve for a nonductile material such as cast iron. In the case of a ductile material, there is significant plastic deformation after yielding and before failure. In contrast, failure occurs without significant plastic deformation in the case of a nonductile material. The area under the stressstrain curve is a measure of the energy required to cause failure. It is clear that this area is much larger for ductile materials as compared to nonductile materials.
Figure 1.3
Stress – Strain Curve for a NonDuctile Material (Source: “Introduction to Pipe Stress Analysis”, Sam Kannappan, John Wiley & Sons, 1986)
1.5.1 Yield strength based on 0.2% offset
Sometimes, the results from tensile testing of materials do not exhibit a sharp, welldefined yield point. In such cases, the “0.2% Offset Method” is used in determining the yield point. This is based on the observation that most materials can have a plastic strain of 0.2% without failing. 0.2% strain is equivalent to a strain of 0.002. The technique involving the 0.2% offset method is illustrated in Figure 1.4. A strain value of 0.002 is used as the starting point and a line parallel to the linear portion of the stress – strain curve is drawn. The intersection of this line with the stress – strain curve gives the yield point.
Figure 1.4
Yield Strength Based on the 0.2% Offset Method
The Yield Strength and Tensile Strength of selected piping materials are given in Table 1.6.
Table 1.6
Yield Strength and Tensile Strength of Selected Piping Materials
Material 
Specification 
TS (ksi) 
TS (MPa) 
YS (Ksi) 
YS (MPa) 
Carbon Steel 
A106 Gr.B

60 
414 
30 
207 
Carbon Steel 
API 5L Gr.B

60 
414 
35 
241 
Carbon Steel 
API 5LX Gr.X52

66 
455 
52 
359 
Low and Intermediate Alloy Steel 
A333 Gr.3 
65 
448 
35 
241 
Low and Intermediate Alloy Steel 
A334 Gr.8 
100 
689 
75 
517 
Low and Intermediate Alloy Steel 
A369 Gr.FP1 
55 
379 
30 
207 
Stainless Steel

A312 Gr.TP304 
75 
517 
30 
207 
Stainless Steel

A312 Gr.TP310 
75 
517 
30 
207 
Stainless Steel 
A312 Gr.TP316L 
70 
483 
25 
172 
1.5.2 Hooke’s Law
This law states that
“Within the elastic limit, the strain of a material is proportional to the applied stress. This can be represented as
e a S
Using the reciprocal of Modulus of Elasticity as the constant of proportionality, Hooke’s Law can be mathematically written as
From Hooke’s Law, it can be concluded that for a given applied stress, the engineering strain will be lesser for a material having higher Modulus of Elasticity.
Along with this, various pipe properties such as DN (or Nominal Diameter), wall thickness and pipe schedule play a very significant role in Stress Analysis.
Sample Exercise
Problem
A steel rod of 25 mm diameter indicates a strain of 0.001 when subjected to a tensile load. Find the applied load. E_{steel} is 2.03 x 10^{5} MPa.
Solution
1.6 Thermal effects and flexibility of piping systems
Piping systems should have the flexibility to expand or contract as required, due to differences between the operating and installation temperatures. This flexibility is achieved by providing loops in the pipe routing as shown in Figure 1.5 or by providing expansion bellows as shown in Figure 1.6. The stiff piping system illustrated in Figure 1.7 lacks flexibility. This will result in overstressing of the system due to thermal expansion.
Figure 1.5
Providing Flexibility for Piping Systems by Using Expansion Loops (Source: “Introduction to Pipe Stress Analysis”, Sam Kannappan, John Wiley & Sons, 1986)
Figure 1.6
Providing Flexibility for Piping Systems by Using Expansion Bellows (Source: “Introduction to Pipe Stress Analysis”, Sam Kannappan, John Wiley & Sons, 1986)
Figure 1.7
Piping System that Lacks Flexibility (Stiff Piping) (Source: “Introduction to Pipe Stress Analysis”, Sam Kannappan, John Wiley & Sons, 1986)
1.6.1 Calculating thermal growth
Tables in piping codes provide thermal data in the form of thermal expansion/ contraction, in mm/m and in/100 ft, between 21°C (70°F) and indicated temperatures. This data is used for determining the displacement in piping systems on account of thermal expansion/contraction. The thermal data for common piping materials is presented in Table 1.7.
Table 1.7
Total Thermal Expansion between 21°C (70°F) and Indicated Temperatures for Common Piping Materials
Temperature °C °F 
Carbon Steel mm/m in./100ft 
Inter. Alloy Steel mm/m in./100ft 
Austenitic SS mm/m in./100ft 
184 300 
1.90 2.24 
1.70 2.10 
3.00 3.63 
129 200 
1.40 1.71 
1.30 1.62 
2.30 2.73 
93 200 
0.80 0.99 
0.80 0.94 
1.20 1.46 
204 400 
2.20 2.70 
2.10 2.50 
3.20 3.80 
316 600 
3.80 4.60 
3.50 4.24 
5.20 6.24 
427 800 
5.60 6.70 
5.10 6.10 
7.30 8.80 
538 1000 
7.40 8.89 
6.70 8.06 
9.60 11.48 
649 1200 
9.20 11.10 
8.30 10.00 
11.80 14.20 
760 1400 
11.10 13.34 
10.00 12.05 
14.10 16.92 
Fundamentals of Pipe Stress Analysis with Introduction to CAESAR II  Introduction to Pipe Stress Analysis
This chapter provides a brief introduction to pipe stress analysis and explains the need for stress analysis in piping systems. The phenomenon of overstressing in piping systems and its consequences are touched upon in detail. An attempt is also made to obtain a clear understanding of the fundamental physical parameters used in stress analysis such as force, stress and strain and modulus of elasticity. Details of the various physical quantities and units used in pipe stress analysis are also discussed along with a description of the concepts of tensile testing and yield strength of materials. The various aspects related to the thermal expansion / contraction and flexibility of piping systems are also adequately covered during the course of the discussion.
Learning objectives
Piping systems need to be analyzed for stresses, to ensure that the components of the system are not overstressed. Piping systems typically consist of straight pipes, fittings (elbow, tees, reducers), flanges, valves and accessories such as actuators. The stresses on equipment nozzles where the pipe connects to the equipment also need to be analyzed. The pipe wall resists both the internal and external forces experienced by the piping system. The force per unit metal area of the pipe wall is the resulting pipe stress. The objective of pipe stress analysis is to ensure that the stresses do not exceed allowable values specified by the design codes. Pipe stress analysis provides the necessary techniques and methods for designing piping systems without overstressing the piping components and the connected equipment.
Piping stress analysis applies to calculations that address the static and dynamic loading arising on account of the effects of temperature changes, gravity, external and internal pressures and changes in fluid flow rate.
Some of the important reasons why piping stress analysis is needed include:
Overstressing can lead to premature failure of the piping system, causing leaks and safety hazards. Overstressing can lead to cracks, breakages and other secondary failures and failures such as bowing and opening of flanges. In some cases, the failure can be catastrophic, causing the collapse of the system and with the potential for loss of life and property. Some situations may even require the entire plant to be shut down. Thus, the objective of pipe stress analysis is to ensure the safe operation of piping systems within a plant, while simultaneously meeting the performance requirements of the plant.
Overstressing in piping can result in the following:
In general, overstressing can result from many different sources. Common examples include inadequate input such as insufficient pipe thickness, overconstraint, excessive thermal expansion or presence of other loads. The remedy for overstressing can be both, to add or in certain instances remove constraints such as releasing degrees of freedom of pipe supports or hangers. Although this process is often carried out on a trial and error basis, major piping layout related problems can usually be anticipated by experienced piping engineers during the design stage itself.
1.3.1 Force
Force is a vector quantity that has both magnitude and direction. It can be defined as a push or pull on an object resulting from its interaction with another object. Force is no longer experienced when this interaction ceases. Piping systems experience both tensile and compressive forces. Forces experienced by piping systems are also known as “piping loads”. Commonly used units for force are: Newton (N), kilogram force (kg_{f}) and pound force (lb_{f}). The units of force are explained here.
Newton is the force required to accelerate a one kilogrammass at 1 m/s^{2}. Thus,


Kilogram force is the force required to accelerate a one kilogrammass at 9.81 m/s^{2}. Thus,
Pound force is the force required to accelerate a one poundmass at 32.2 ft/s^{2}. Thus,
The definition of pound force creates a need for using the conversion constant g_{c} while performing calculations in the US Customary System (USCS).
The conversion factors for force units are:
1 kg_{f} = 9.81 N
1 kg_{f} = 2.205 lb_{f}
1 lb_{f} = 4.4462 N
The concept of force can be better understood with the help of the following exercise.
Sample Exercise
Problem
A 5 kg .object is moving horizontally at a speed of 10m/sec. Determine the Net force required to keep the object moving at this speed and in the same direction.
Solution
Zero N. This is because, an object in motion will maintain its state of motion and the presence of an unbalanced force results in a change in its velocity.
1.3.2 Engineering stress
Engineering stress S is the force per unit area of the metal cross section. A stress may be normal, shear or torsion, leading to corresponding deformations. While stress cannot be measured directly, deformations can be measured.
Units for engineering stress:
N/m^{2} (Pascal, Pa)
lb_{f}/in^{2 }(psi)
kg_{f}/cm^{2}
Commonly used units for stress:
Kilo pounds per square inch (ksi) = 10^{3 }psi
Megapascals (MPa) = 10^{6} Pa
Commonly used conversion factors for stress:
1 lb_{f}/in^{2} (psi) = 0.0703 kg_{f}/cm^{2} = 6.896 kPa
1 lb_{f}/in^{2} (psi) = 6.896 kPa
1 MPa = 145 psi
1 ksi = 6.88 MPa
1.3.3 Deformation of materials and engineering strain
When the elements of materials are subjected to tensile or compressive loads, they undergo small deformations. These deformations can be “elastic” or “plastic”. Within the elastic limit, the deformation is “elastic”, i.e. the material springs back to its original shape when the load is removed. Thus, elastic deformation is temporary in nature and exists only when the load is present. After the material begins to yield, the deformation is permanent and remains even after the load is removed. This is called “plastic” deformation. Figure illustrates the manner in which deformation of a material occurs, when it is subjected to tensile and compressive loads.
Tension Compression
Figure 1.1
Deformation of a material when subjected to tensile and compressive loads
Engineering strain e is the change in length divided by the original length, i.e.
Where
DL is the change in length
L_{o} is the original length
Units of strain: in/in or mm/mm.
While an object in tension has resulting tensile strain, an object in compression has resulting compressive strain. The above equation for strain is only valid if the deformation of the object is uniform throughout its volume.
1.3.4 Modulus of elasticity (E)
Modulus of Elasticity E is a material property that is indicative of the strength of the material. The modulus of elasticity values for steel and aluminum are given here. The values indicate that steel is about three times stronger than aluminum.
E_{steel} = 30 x 10^{6} psi = 2.07 x 10^{5} MPa
E_{aluminum} = 10 x 10^{6} psi = 0.70 x 10^{5} MPa
The modulus of elasticity of materials decreases with increase in temperature. This is due to the thermal expansion of materials. At higher temperatures, thermal expansion results in a lesser force being required to cause a given amount of strain, resulting in a lower modulus of elasticity. The modulus of elasticity for different materials and at various temperatures is listed in Table 1.1. Modulus of elasticity is also referred to as “Young’s Modulus”.
Table 1.1
Modulus of Elasticity of Different Materials at Various Temperatures
(Modulus of Elasticity is given in 10^{5} MPa. The values in parenthesis are in 10^{6} psi)
Material 
130°C (203°F) 
20°C (68°F) 
260°C (500°F) 
540°C (1004°F) 
810°C (1490°F) 

Carbon steels (<3% C) 
2.03 (29.5)

1.92 (27.9)

1.82 (26.4)

1.06 (15.4)

 


Low, Intermediate alloy steels

1.96 (28.5)

1.88 (27.4)

1.79 (26.0)

1.57 (22.8)

 


Austenitic stainless steels

2.06 (29.9)

1.95 (28.3)

1.80 (26.1)

1.56 (22.7)

1.23 (17.9)


Monel (67Ni, 30Cu)

1.83 (26.6)

1.79 (26.0)

1.75 (25.4)

1.10 (16.0)

 


CuproNickel (70Cu, 30Ni) 
 

1.49 (21.6)

1.40 (20.3)

 

 


Aluminum Alloys
Copper
Brass (66Cu, 34Zn)
Bronze (88Cu, 6Sn, 4.5Zn, 1.5Pb) 
0.750 (10.9)
1.15 (16.7)
1.01 (14.7)
0.945 (13.8)

0.695 (10.1)
1.10 (16.0)
0.963 (14.0)
0.894 (13.0)

0.530 (7.7)
1.01 (14.7)
0.874 (12.7)
0.805 (11.7)

 
 
 
 

 
 
 
 









The different physical quantities of force, stress, strain, modulus of elasticity and their respective units have already been discussed. Let us go ahead and discuss some of the other physical quantities used in pipe stress analysis.
The density of a substance is its mass per unit volume. It is represented by the symbol “r”. Density for a given substance can be calculated from the following equation,
Density (r) = Mass of the substance (m) / Volume of the substance (V)
Density has the units, lbm/ft^{3} or kg/m^{3}.
If equal masses of cotton and lead are taken (say 1 kg each), we will find that the volume of cotton is much larger than the volume of lead. This is because lead is heavier (denser) than cotton. The particles of lead are closely packed while those of cotton are more diffused.
Density tends to change with change in temperature.
The specific gravity of a substance is the ratio of the density of a substance to the density of some standard substance. The standard substance is usually water (at 4°C) for liquids and solids, while for gases it is usually air. Specific gravity is also known as Relative Density.
Relative density for liquids and solids (s) = Density of substance
Density of water at 4°C
Relative density for gases (s) = Density of substance
Density of air
Density of substance = Density of water at 4°C ´ Relative density of liquid or solid
i.e.: r (for liquids and solids) = 1000 ´ s and
r (for gases) = 1.29 ´ s
Specific gravity is a dimensionless number.
The specific weight of a substance is the weight per unit volume. It has units of kN/m^{3} or kg_{f}/m^{3} or lb_{f}/ft^{3&a }
World Malaria Day takes place on the 25th of April. As we look towards the future, engineers will continue to be at the forefront when it comes to finding unique ways of malaria prevention. Continue reading How Engineers Can Help in the Fight Against Malaria at EIT  Engineering Institute of Technology.
This year two student ambassadors were selected to represent the Engineering Institute of Technology (EIT), Aaron Brook and Douglas Mugweni. In this profile piece, we are proud to introduce Aaron Brook, who is currently studying EIT’s 52708WA  Advanced Diploma of Industrial Automation. An accomplished Instrument Reliability Specialist, Aaron has spent several years combining his career with continued education. With previous posttrade qualifications earned in a range of fields including leadership and management, project management, work health and safety, hazardous area installation, maintenance and inspection and audio engineering, EIT is just the latest addition to Aaron’s learning ventures. Continue reading Meet EIT Student Ambassador: Aaron Brook at EIT  Engineering Institute of Technology.
No one knows the value of training better than EIT’s International Engineering Education Manager, Kevin Baker. This year, he celebrates his 21st anniversary in charge of the institute’s group training department, a stewardship that has seen the company run courses for everyone from; CocaCola to NASA, the major oil and gas players through to the United Nations. While EIT prides itself on providing inperson training for any class anywhere in the world, for the past 13years, there has been a steady migration to the relatively new territory of remote (virtual) teaching. Continue reading EIT Celebrates Two Decades of Group Training at EIT  Engineering Institute of Technology.
As energy demand continues to increase in parallel with the growth of the world population, our energy systems need to evolve to be more flexible, sustainable, and distributed. It was only a month ago that extreme weather events highlighted the fragility of the Texas electrical grid. As Texans’ demand for electricity soared, utility operators revealed just how poorly prepared they were for the cold weather. The Texas electricity grid failed in spectacular fashion, and there are chances it will happen again. Continue reading Making Smart Grids Smarter with Machine Learning at EIT  Engineering Institute of Technology.
When you decide to enrol in a program and start studying you are investing time and money into your future goals. While engineers are currently in highdemand, professional selection criteria is becoming increasingly competitive. Hard work and good grades are no longer enough to set yourself apart in the industry. As one of Dyson’s successful engineers commented ‘It’s not always the most intelligent student that makes the best engineer’. This month we came up with a few points that will help you stand out from the crowd: 1 Create a competitive CV As an engineer, your résumé becomes a marketing tool that will emphasize your assets. Continue reading How Successful Engineers Stand Out from the Crowd at EIT  Engineering Institute of Technology.
The Internet of Things (IoT) is enabling manufacturers to improve efficiencies, reduce waste and increase profits. Continue reading How IoT Applications are Changing the Future of Manufacturing at EIT  Engineering Institute of Technology.
Originally from Nigeria, Felix Okoh is a current oncampus Master of Engineering (Industrial Automation) student who began his EIT learning journey online. He is now based in Perth in Western Australia and this month we had the pleasure of hearing about his experience from starting his studies online before transferring to oncampus. Why did you select EIT as your education provider? Before joining I did extensive research on different institutes. Continue reading Felix’s Experience: Transition from Online to OnCampus Study at EIT  Engineering Institute of Technology.
EIT’s courses are specifically designed to balance academic excellence and preparing our students for the competitive engineering jobs market. One critical aspect of preparing for a career in the sector is the practical skills earned in undertaking an internship. This month we sat down to map out the top five reasons for doing an engineering internship… 1. There’s only so much you can learn in the classroom No matter how much time you spend in lectures and workshops, there’s no substitute for putting what you’ve learned into practice. Continue reading Practical Positives of the Engineering Internship at EIT  Engineering Institute of Technology.
The Texas blackouts left millions without power as freezing weather pummeled the state. The blackout is among the largest in US history. Continue reading Texas Blackouts: How a Cold Snap Left the State in Darkness at EIT  Engineering Institute of Technology.
The Engineering Institute of Technology is proud to announce Douglas Mugweni and Aaron Brook as EIT Student Ambassadors for 2021. Every year, the Engineering Institute of Technology (EIT) selects an applicant to fill the role of Student Ambassador. For 2021, however, the number of applications received was greater than ever before, and the standard tremendously high – too high, in fact, to choose only one candidate. As a result, the committee has elected to create two positions for the coming year. Continue reading EIT Student Ambassadors 2021 Announcement at EIT  Engineering Institute of Technology.
IDC Technologies © 1991  2021