THE STRESS INTENSITY FACTOR PROBLEM 15 SOLUTION

The questions so far in this tutorial have dealt with simple applications of Linear Elastic Fracture Mechanics (LEFM).  The more usual design case involves elastic-plastic fracture or so-called Yielding Fracture Mechanics (YFM), as tough ductile alloys are often operated at elevated temperatures.  Fracture assessment, therefore, generally involves a two parameter approach where the potential for fast fracture and yield dominated fracture (net-section yield) are assessed independently (see theory card).  It usually also involves more generally applicable fracture parameters like the crack tip opening displacement (CTOD, COD) of the J-integral which can cope with localised plasticity.

This problem illustrates the type of application where LEFM is not really suitable, although its predictions may still be conservative, and hence quite useful.

It should take around 15 minutes to work through.

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A particular chemical processing plant consists of a number of similar reactors which operate at temperatures ranging from -70ºC to 350ºC.   It is desired to use a single alloy to manufacture all the reactor vessels.   The fracture toughness and yield strength of this alloy vary with temperature as shown below:

KC = (63 + T/10)   MPa m½ over the range of temperature -100ºC to +400ºC

 Temperature ºC -100 0 100 200 300 400 Yield Strength MPa 550 450 412 400 362 300

The vessels have a wall thickness of 15 mm and are to be designed on the basis of 'leak-before-break'.  The stress intensity factor can be calculated using:

a)    Based on material performance, rather than operating conditions, graphically determine the temperature at which yield becomes more likely than fracture, i.e. the design criterion changes from fracture to yield control.

b)    Is the use of LEFM valid up this temperature?  If not, determine the range of operating temperatures over which use of YFM would be preferable.

You may assume plane stress conditions in the vessel wall and can take the plastic zone size as being:

Answer:  In this simplistic analysis, the design is fracture dominated up to around 221ºC.  YFM should be used for this design over the complete operating range - in fact, a YFM two-parameter approach should be adopted.

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a)    This part of the question is very straightforward, as it simply entails plotting yield strength and fracture stress as functions of temperature.  The problem can be solved analytically, but graphical presentation is useful to practising engineers.  The variation in fracture toughness is linear, so we only need the two end point values (KC at -70ºC and 350ºC) to calculate the fracture stress from the K equation and plot the line.   Because the design criterion is leak-before-break, we can assume that the critical crack size at fracture will be given by the wall thickness of the vessel, i.e. 15 mm (remember that the vessel must tolerate a crack with 2a = 2t).  The calculation then gives values of fracture stress of 258 MPa at -70ºC, and 451 MPa at 350ºC, e.g.:

The plot is shown below and the intersection point is seen to be at about 221ºC.  Simplistically, fracture is more likely at temperatures below this point and yield is more likely above it.  The assumption of plane stress is critical to this interpretation, as a biaxial stress state does not lead to yield point elevation.  However, there remains the question of whether crack tip plastic zone sizes are small enough to use LEFM.  Exploring this aspect is the purpose of the seocnd part of the question.

b)    For this alloy, fracture toughness is increasing with temperature and yield strength is decreasing.   Examination of the crack tip plastic zone equation shows that both of these effects will increase plastic zone size as operating temperature increases.  Consider the case at 220ºC  -  KC = 85  MPa m½ and the yield strength = 395 MPa, hence:

This is the same as the wall thickness, which certainly means that plane stress conditions exist, but is also the same as the crack length.  YFM would provide a more reliable assessment of fracture.

To see if LEFM is applicable over any part of the operating temperature range, we can check the situation at -100ºC - here KC = 53  MPa m½ and the yield strength = 550 MPa, hence:

This is around 1/5 of the wall thickness and LEFM may be valid.  However, the best solution to this problem would be to use YFM and the two parameter Failure Assessment Diagram (FAD) approach to cover the conjoint possibility of yielding and fracture.