Equipments when subjected to cyclic loads may fail due to fatigue. As an illustration consider an example of a wire being bent back and forth until the plastic deformations grow so that it ruptures. In ASME fatigue analysis typically the cyclic loads are pressure and/or temperature.

Here an ASME PTB-3 Validation for “Example E5.5.3 – Elastic Stress Analysis, and Equivalent Stresses” to check for protection against failure from cyclic loadings per ASME Sec VIII Div 2 is carried out using ANSYS.

Problem Statement:

Evaluate the vessel top head and shell base metal regions given in Example E5.2.1 in accordance with the fatigue methodology provided in paragraph 5.5.3. Note that the nozzle to head weld is machined and subjected to full volumetric examination and both ID and OD surfaces receive MT/PT and VT. The shell to head weld is in the as-welded condition and the OD surface receive the same inspection as above. The ID surface receives only full volumetric examination. The cyclic loading design requirements given in the Users’ Design Specification are provided below:

Operating pressure = 380 psig at 125^{o}F

Corrosion Allowance = 0.125 inches

Cyclic Life Requirement = 20000 full pressure cycles

Number of Shutdowns/Startups = 20

Solution:

Typically before performing a detailed fatigue analysis the equipment is checked to see if it meets fatigue screening criteria per sec 5.5.2. Three methods for fatigue screening criteria are provided in the code. We’ll see each one of them.

Fatigue Analysis Screening Based on Experience With Comparable Equipment.

We’ll assume no such data is available and move to next method.

Fatigue Analysis Screening, Method A.

- The expected (design) number of full-range pressure cycles including startup and shutdown, N
_{ΔFP }= 20000 + 20 = 20020 cycles - Operating pressure cycles (cycles for which range of pressure variation exceeds 20% of the design pressure for integral construction), N
_{ΔPO }= 0 cycles - The effective number of changes in metal temperature difference between any two adjacent points, N
_{ΔTE}= 0 cycles - The number of temperature cycles for components involving welds between materials having different coefficients of thermal expansion that causes the value of thermal strain to exceed 0.00034, N
_{ΔTα}= 0 cycles - Sum total of above calculated cycles N
_{ΔFP}+ N_{ΔPO}+ N_{ΔTE}+ N_{ΔTα}= 20020 which is greater than 1000 (Refer Table 5.9). Hence, Screening Method A is not met. So we proceed to Method B.

Fatigue Analysis Screening, Method B.

- For integral construction with no nozzles or attachments in the knuckle region criteria factors C
_{1}= 3 and C_{2}= 2 (Table 5.10) - N
_{ΔFP }= 20020 cycles as in Method A.

The allowable number of cycles can be obtained as N(C_{1}S) = 10^{X} (3-F.21) where X is obtained from Annex 3-F. A summary is shown in the above table.

Since N_{ΔFP }> N(C_{1}S) for both components, fatigue evaluation for both are required.

Fatigue Evaluation – Elastic Stress Analysis, and Equivalent Stresses

Finite Element Model from Example E5.2.1 is used in this analysis. The only cyclic load here is the pressure load which varies from 0 to 380 psi. Typically to find alternating stress amplitude one needs to solve two load cases:

Case 1: The maximum cyclic load (in our case pressure of 380 psi)

Case 2: The minimum cyclic load (in our case pressure of 0 psi)

Finally stress range is obtained as Case 3 = Case 1 – Case 2.

However for a linear elastic analysis like the one which we would be doing, the response (deformations, stresses) will be directly proportional to the applied loads. Hence Case 3 can be directly obtained by applying ΔP load = P_{max} – P_{min} = 380 – 0 = 380 psi in a single load case. Note, this approach would work even if we had a non zero P_{min}.

ANALYSIS SUMMARY

LOADS & BOUNDARY CONDITIONS

- Internal Pressure of ΔP = 380 psi was applied on the inner surfaces of the model.

- Pressure Thrust of 888.75 psi was applied on the flange face as negative pressure.

- Axial displacement was arrested at the shell base.

ANALYSIS RESULTS

Total Deformation Plot

Von-Mises Stress

Von-Mises Stress from PTB-3 example

SCL Locations

SCL Locations from ASME PTB-3 example

Membrane + Bending stress @ SCLs (The red values are from this analysis, the black values are from ASME PTB-3)

Allowable Primary + Secondary stress, S_{PS} is calculated as max(3S, 2S_{y}) as shown in the table below.

Since Membrane + Bending Stresses are less than S_{PS}, fatigue penalty factor K_{e} = 1

Computation of alternating stresses @ SCLs

S_{alt} = K_{f}*K_{e}*Total Von-Mises Stress Range/2

Here K_{f}, the fatigue strength reduction factor is taken from Tables 5.11 and 5.12.

The below table summarizes the results (values in red are from this analysis and those in black are from ASME PTB-3.

Finally allowable number of cycles, need to be calculated for S_{alt} using Annex 3-F. Calculation for allowable number of cycles are not being done for this analysis, however reference table from ASME PTB-3 is produced below.

Knowing the allowable number of cycles N and the design cycle life of the component n fatigue damage factor can be calculated as D = n/N. In case multiple cyclic loads are present, the miner’s rule D = n_{1}/N_{1} + n_{2}/N_{2} … can be used.

This results in calculated fatigue damage for the limiting region (nozzle outside radius) of 0.073 (from ASME PTB-3). Similarly, a fatigue damage of 0.039 (from ASME PTB-3) is calculated for the head knuckle.

Since D<1, the equipment is safe w.r.t fatigue damage.

What is the diffierence between Delta Sn.k (primary + Secondary equivalent stress range) & Delta Sp,k (range of primary + secondary + peak equivalent stress). As we linearize all stress components, only M+B will be obtained. So we can say both are same. Delta Sp,k means total raw stress from FEA software?

Linearization will give Delta(Sn), which is M+B stress. Delta(Sp) is the non linear total stress i.e without linearization. So Sn and Sp are not same.

Noted. I have one more doubt on this subject, Mr. Sandip. As per ASME Sec VIII, Div 2, Poisson correction factor Kv.k and Fatigue penalty factor Ke,K is equated. Code says, if the Fatigue penalty factor Ke,K is used for entire stress range, it can replace the use of Poisson correction factor Kv.k. How it is been related with each other. Being steel is used in pressure vessel industry, there won’t be vast change in poisson value (roughly it would be 0.3 to 0.35). Kindly provide your opinion on this?

Kv ranges from 1 to 1.4, whereas Ke ranges from 1 to 5. Directly applying Ke to the total stress range is a conservative but a simplified approach. Otherwise for a more accurate analysis the total stress range needs to be split up into thermal and non-thermal components and Kv and Ke can be respectively applied on them. You can find information on this in ASME PTB-1.