THE STRESS INTENSITY FACTOR 
PROBLEM 14 
SOLUTION 
This question illustrates some aspects of material selection in the design of a pressure vessel, which experiences a range of operating temperatures. Hence material toughness, as a function of temperature, forms an important part of the fracturesafe design process. This is demonstrated graphically in the question.
It should take around 15 minutes to complete.
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Power generation pressure vessels are usually thick walled, operate at a range of temperatures from ambient to elevated and are required to be failsafe (leakbeforebreak). In one particular case, a spherical pressure vessel is proposed which will operate at an internal pressure (p) of 40 MPa and at temperatures from 0ºC to 300ºC. The proposed wall thickness (t) is 100 mm and the diameter (D) is 2 m. Two candidate steel alloys have been suggested:
Steel A: For this steel, K_{C} = (150 + 0.05T) MPa m^{½} where T is operating temperature in degrees centigrade, and the yield strength varies in a linear fashion from 549 MPa at 0ºC to 300 MPa at 300ºC.
Steel B: Here K_{C} = (100 + 0.25T) MPa m^{½}, and the yield strength varies linearly from 650 MPa at 0ºC to 500 MPa at 300ºC.
Graphically determine, by inspection, the range of temperatures over which each of these alloys would have the highest safety factor with respect to fast fracture.
Throughthickness cracks can be assumed to be critical and the stress intensity factor for such cracks in this geometry is given by:
The membrane stress in the pressure vessel wall may be taken as pD/4t.
Answer: Steel A is best below approximately 212ºC, Steel B is better above this temperature.
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Solving this question is simply a matter of calculating the required values of fracture toughness K_{C} to avoid fracture at various temperatures in the operating range, say 0ºC, 100ºC, 200ºC and 300ºC for both steels. These can be plotted against the actual values of alloy toughness at the same temperatures, and the alloy with the highest toughness, and the highest margin between actual and required toughness values, assessed visually. Graphical presentation of engineering data is often more useful, and accessible, than presenting the same information analytically, or in a table. The problem could be solved analytically for the crossover temperature but at the loss of easy visualisation of safety margins in fracture performance.
The membrane stress is simply found from:
Although the design case is based on leakbeforebreak, the amount of pressure relief caused by a throughthickness crack is unknown. It is therefore conservative to assume that the internal pressure will load the crack surfaces, hence the total stress intensity factor will be calculated using the sum of the membrane stress and the internal pressure, i.e. 240 MPa. Leakbeforebreak design requires the pressure vessel to tolerate a throughthickness crack of total length 2a = the surface length (2c) of the precursor semielliptic crack. As we have no information regarding crack ellipticity, we will have to assume that it was semicircular and hence 2a = 2t, where t is the wall thickness. Therefore the required values of toughness are found from:
The table below gives required and available toughness values for the two alloys.
0ºC 
100ºC 
200ºC 
300ºC 

Steel A  Yield Strength MPa 
540 
460 
380 
300 
Required K_{C} MPa m^{½} 
141.7 
144.7 
150.3 
163.1 

Actual K_{C} MPa m^{½} 
150 
155 
160 
165 

Steel B  Yield Strength MPa 
650 
600 
550 
500 
Required K_{C} MPa m^{½} 
139.4 
140.2 
141.4 
143.0 

Actual K_{C} MPa m^{½} 
100 
125 
150 
175 
This data is plotted in the graph below. By inspection, the toughness values for Steel A are highest and therefore the safety margin greatest, up to about 212ºC. Above that temperature, Steel B has become the best choice because of its very steep increase in toughness with temperature.