Secondary stress is developed in a piping system owing to constraint of the system against displacements, either due to thermal expansion or imposed anchor and restraint movements. As per ASME B31.3 section 302.3.5, the computed displacement stress range SE in a piping system shall not exceed the allowable displacement stress range SA calculated by the following equation:

SA = f(1.25Sc + 0.25Sh)...........................1a

## Liberal Allowable Stress Range

When the basic allowable stress Sh (as per Table A-1) is greater than the stresses due to sustained loads SL, the difference Sh-SL may be added to the above giving the following equation:

SA =  f[(1.25Sc + 0.25Sh + (Sh - SL)]

SA  = f[1.25(Sc + Sh) -SL)]...........................1b

where Sc and Sh are the basic allowable stresses in the cold and hot conditions and 'f' is the stress-range reduction factor calculated by the equation 1c below.

f = 6.0(N-0.2) ≤ 1.2...........................1c

where N is the equivalent number of full displacement cycles during the expected service life of the piping system.

This addition of the unused portion of basic allowable stress (Sh - SL) gives us a higher value of allowable displacement stress range often referred to as liberal allowable stress range. It may be noted that the terminology 'Liberal Allowable Stress Range' is not referred to in the ASME B31.3 code and is derived from its usage in the Caesar II program for stress analysis calculations. The application of equation 1b to take the benefit of unused portion of sustained loads is not mandatory as per the code and is at the discretion of the piping engineer.

## Stress Range Reduction Factor

The stress-range reduction factor provides a basis for reduction in allowable displacement stress range SA depending on the number of secondary stress cycles the plant undergoes during its design life. It is important to understand that the factor 'f' is not intended to take care of vibration cycles induced in the piping system due to other excitation mechanisms.

The 2002 edition of ASME B31.3 limited the value of 'f' to a maximum of 1.0. The later editions including the 2018 edition allows maximum 'f' value to equal 1.2 for piping systems.

Substituting the value of N in equation 1c would give us the following values of stress range reduction factor 'f'.

Sr.No Cycles N From/To Max Cycles N Factor 'f'
for maximum N
1 Up to 7000 7000 1.02
2 7000 < N ≤ 14000 14000 0.89
3 14000 < N ≤ 22000 22000 0.81
4 22000 < N ≤ 45000 45000 0.7
5 45000 < N ≤ 100000 100000 0.6
6 100000< N ≤ 200000 200000 0.52
7 200000< N ≤ 700000 700000 0.41
8 700000< N ≤ 2000000 2000000 0.33

The minimum cycles of 7000 in the code is considered to be based on approximately 1 cycle per day for a 20 year life of the facility. However, very few process facilities are likely to face this magnitude of thermal cycling.

ASME 31.3. edition 2002 contained the following table which closely approximates to the values calculated above using equation 1c. This table has been deleted from the recent version of the code.

Sr.No Cycles N Factor 'f'
for max N
1 7000 and less 1.0
2 Over 7000 to 14000 0.9
3 Over 14000 to 22000 0.8
4 Over 22000 to 45000 0.7
5 Over 45000 to 100000 0.6
6 Over 100000 to 200000 0.5
7 Over 200000 to 700000 0.4
8 Over 700000 to 2000000 0.3

A quick review of the above table reveals that all the values of factor 'f' do not match the value of N. For example the value of N should be 7774 to get a value of f = 1. Further, the above table stipulates same value of 'f' for a range of cycles which is not consistent with equation 1c. Removal of this table from the code appears to be justified to eliminate the above inconsistencies.

Also it is noted that the validity of equation 1c was limited to 2 x 106 cycles for a minimum value of f = 0.3 in 2002 edition of code. However, the minimum value of factor 'f' as per ASME B31.3 - 2018 edition is 0.15 for an indefinitely large number of cycles. From equation 1c, the value of N for f = 0.15 works out to 1.006 x 108 cycles.

Figure below shows the plot of Cycles N versus the Stress range reduction factor for values up to 0.4. The secondary axis shows the plant cycles per day considering a 20 year design life of the piping system.

## Basis of Equation 1a for calculating Allowable Displacement Stress Range

As per Tabe A-1 of ASME B31.3 the allowable stress of materials below the creep range is 1.5 times the yield stress of the material. For example the yield stress of ASTM A106 Grade B is 35000 psi and allowable stress up to 100°F is 23300 psi. The ratio (yield stress/allowable stress) = (35000/23300) = 1.5. Thus the bending stress at which plastic flow will start in the hot condition is 1.5Sh and the bending stress at which plastic flow will start in the cold condition is 1.5Sc. Hence the sum of these stresses would represent the maximum stress range to which a piping system could be subjected to without plastic flow occuring either in the hot or cold condition.

Therefore Maximum Stress Range = 1.5(Sc + Sh)

The Maximum Displacement Stress Range allowed by ASME B31.3 = 1.25(Sc + Sh)

From this total stress range 0.5Sh is reserved for pressure stress (Note that longitudinal stresses due to pressure are half the circumferential or hoop stresses) and 0.5Sh is reserved for weight and other sustained load stresses. Thus 0.5Sh0.5Sh = Sh is deducted from above equation resulting in an allowable stress range of

SA = 1.25Sc + 1.25Sh - Sh

Thus SA = 1.25Sc + 0.25Sh

## Example calculation of Allowable Stress Range

Problem: Calculate the allowable stress range for ASTM A106 Grade B pipe material for design temperature of 300°C and for 18,000 thermal cycles of service. Consider the installation temperature of pipe as 20°C and longitudinal stresses due to sustained loads SL as 60 MPa.

Solution: From ASME B31.3 Table A-1 for ASTM A 106 Grade B

Sc =  138 MPa (20,000 psi) for temperature up to 40°C.

Sh = 126 MPa (18275 psi) for temperature of 300°C.

f = 6.0(18000-0.2) = 0.845

Allowable Stress Range SA = 0.845(1.25*138 + 0.25*126) = 172.4 MPa or

0.845(1.25*20000 + 0.25*18275) = 24986 psi............................as per equation 1a

The above piping system will operate safely provided the displacement stress range does not exceed 172.4 MPa (24986 psi) and the number of thermal cycles does not exceed 18,000.

In this example, since Sh = 126 MPa is greater than SL = 60 MPa, we can add the difference to the above equation 1a. The difference Sh - SL = 126 - 60 = 66 MPa represents the unused portion of allowable stress for sustained loads. This should be multiplied by factor f = 0.845 to account for the thermal cycles. Thus the unused portion of allowable stress considering thermal cycles = 0.845*66 = 55.8 MPa

Thus the new value of allowable stress range SA or liberal allowable stress range is

SA = 172.4 MPa + 55.8 MPa = 228.2 MPa

The above value can also be calculated directly using equation 1b.

The allowable stress range is increased by approximately 32% by including the unused portion of the allowable stress for pressure and other sustained loads.

At 300°C, the approximate value of yield strength of ASTM A106 Grade B is (3/2)*126 MPa = 189 MPa. It is noted that the allowable stress range using equation 1b is 228.2 MPa and is larger value than the yield strength of ASTM A106 Grade B at the design temperature. This brings us to an understanding of a very important concept.

Can the allowable stress range exceed the yield strength of the material at the temperature. The answer is yes. This is because the allowable stress range is for expansion stresses which are self-limiting by nature. This means the expansion stresses will diminish in time through local yielding of the components at locations where the stresses exceed the yield strength. When the pipe cools and returns to its original position, it will retain a residual in the reverse direction (compressive stress). On each subsequent loading cycle, this residual compression must be overcome before the pipe can go into tension. This retains the elastic range of the pipe material and thus the allowable displacement stress range.

In summary, the allowable stress range can exceed the yield strength but not the elastic range of the material. (SA can be as large as 1/3-2/3 times the yield strength of the material. The code allows the piping material to yield during its initial cycle provided it afterwards stays within its elastic range.

## Calculation of Equivalent N when the computed stress range varies

Let us consider a scenario where the plant operating conditons have changed due to change in process parameters or plant modifications and the plant piping now undergoes thermal cycling at the new operating conditions. This will result in change of computed displacement stress range for the piping. Let us assume this displacement strange as Si. Let us associate the number of cycles associated with this displacement stress range as Ni.

Let NE be the number of thermal cycles associated with the maximum computed displacement stress range SE.

If the plant undergoes several upgrades and thermal cyclings due to changing operating conditions, the stress range reduction factor will not remain the same as the original conditon. To calculate the stress range reduction factor we have to calculate the equivalent number of cycles.

The equivalent N in such cases can be calculated using equation 1d

N = NE + ∑(ri5Ni) for i = 1,2,3,---,n...........................1d

where ri = Si/SE

Problem: Table below provides data for a plant piping system that undergoes the a number of thermal cycles. The corresponding computed displacement stress range for each thermal cycle is provided. What is the equivalent number of thermal cycles.

Number of Thermal Cycles (N) Computed Displacement Stress Range (Si)
8000 20000 psi
6000 12000 psi
5000 10000 psi
3000 8000 psi
2000 6000 psi
1000 4000 psi

Solution: The number of cycles associated with maximum displacement stress range SE = 20000 is NE = 8000.

N1 = 6000 and r1 =  S1/SE = 12000/20000 = 0.6

N2 = 5000 and r2 =  S2/SE = 10000/20000 = 0.5

N3 = 3000 and r3 =  S3/SE = 8000/20000 = 0.4

N4 = 2000 and r4 =  S4/SE = 6000/20000 = 0.3

N5 = 1000 and r5 =  S5/SE = 4000/20000 = 0.2

N = NE + N1r1 + N2r2 + N3r3 + N4r4 + N5r5

N = 8000 + 6000*0.6 + 5000*0.5 + 3000*0.4 + 2000*0.3 + 1000*0.2

N = 8000 + 3600 + 25000 + 1200 + 600 + 200 = 16100 cycles

Conclusion: Though the plant has gone through 25000 cycles in total at different computed displacement stress range, the equivalent number of cycles as per equation 1d of the code is 16100 cycles.