stress intensity factor and fracture toughness testing

Stress Intensity Factor and Fracture Toughness Testing

Theory

The stress intensity factor is defined from the elastic stress field equations for a stressed element near the tip of a sharp crack under biaxial (or uniaxial) loading in an infinite body. The situation is illustrated below:

The direct and shear stresses on the red element are given, to a first order approximation, by:

Only these first terms in the series expansion have the 1/r½ dependency, which causes a stress singularity at the crack tip (i.e. the stresses go to infinity as r goes to zero). Thus in the near-tip region, which is where fracture processes occur, the stress field is dominated by the singularity. Along the critical plane for cracking ahead of the crack tip (where the angle is zero), the equations reduce to the simple form of:

The numerator in these equations essentially gives gives a measure of the magnitude, or intensity, of the near-tip elastic stress field. Irwin defined the numerator as the stress intensity factor, K, and postulated that fracture would occur at critical values of K. Both numerator and denominator are multiplied by pi for expediency in showing that K and G (the critical strain energy release rate) are related. Thus critical values of K meet both the critical stress and the 'energetically favourable' criteria for crack growth. Generally speaking, finite geometry and crack shape correction factors have to be included in the expression for stress intensity factor, i.e.

where Y can be a relatively involved compliance-based function. This expression can be used to find the stress intensity factor corresponding to any combination of remote stress and crack length, but critical values of K at which fast fracture occurs are denoted K1C when conditions of plane strain apply, and KC otherwise.

Extensive experience has indicated that LEFM can still be applied in the presence of crack tip plasticity, provided that the ratio of plastic zone size to crack length is < 1/15.

Further information can be found in the following reference:

H L Ewalds and R J H Wanhill, Fracture Mechanics, Edward Arnold, London, 1989, pp. 28-42.

Question 1

For a through-thickness crack in an infinite plate, the tensile stress distribution ahead of the crack tip is accurately described by the equation:

where x is distance along the crack from its centre-line, as shown in the diagram below.

In terms of the stress intensity factor, the stress very near to the crack tip can be approximated as:

where r is distance ahead of the crack tip.

What is the difference between the stress predictions of these two equations at a distance r = 0.02a ahead of the crack tip?

Scientific Calculator | RPN Calculator | Theory

Answer: 1.5%

To solve this problem, all we need do is substitute into the first equation with x = 1.02a, i.e. x = a + r;

and then substitute r = 0.02a into the approximate K-based solution.

This represents a difference of 1.5%.

Question 2

A noted philanthropist offers you the chance to earn £ 50 000 by simply hanging from a rope for just one minute. The rope is attached to a sheet of glass which is 300 cm long by 10 cm wide and 0.127 cm thick.

Complicating the situation are:

1. The glass sheet contains a central crack with a length of 1.62 cm that is orientated parallel to the ground and perpendicular to the longest side of the glass sheet. The fracture toughness of the glass is known to be 0.83 MPa m½.

2. The rope is suspended over a deep pit containing rather annoyed green mamba snakes.

Would you try for the pot of gold at the end of the rainbow?

The stress intensity factor for a through-thickness crack is given by:

where:


Answer: For a 60 kg person, K = 0.75 MPa m½

The size of the glass sheet is necessary to calculate the stress in the sheet due to your weight hanging on the rope. Remember that this stress is calculated on the full cross-sectional area of the sheet, i.e. as though the crack is absent. This is clearly a finite sheet of glass, so we need to correct the stress intensity equation for this.

Essentially:

where the finite geometry correction function Y is given by:

Hence as 2a = 16.2 mm, a = 8.1 mm and W = 100 mm, which gives Y = 1.0196. The applied stress is simply obtained from:

Hence K = 1.0196 x 4.63√(π x 8.1 x 10-3) = 0.75 < 0.83 MPa√m

This calculation is based on my own weight and, if I was feeling brave, I might try for the money. In general terms, the margin of safety is somewhat low - it might be better to let a younger and lighter sibling earn the money for you!

Question 3

This question illustrates the application of constraints on determining valid K1C values from fracture toughness tests, according to BS 7448 : Part 1 : 1991. It should take about 15 minutes to complete.

The figure below shows the load line displacement trace recorded from a standard sized compact tension fracture toughness specimen. Specimen thickness was 25 mm, the crack length at fracture was 25 mm and the steel alloy had a yield strength of 650 MPa.

a) Calculate FQ and hence KQ. Apply the required checks on plastic zone size, stress state and plasticity during the test to determine whether KQ is a valid plane strain fracture toughness value.

b) What is the maximum K1C value that can be determined for this steel using 25 mm thick specimens?

Note that:


Answer:

a) FQ = 19 kN, KQ = 32.8 MPa m½

b) 91.9 MPa m½

a) In order to find FQ all we need to do is construct a line from the origin with a slope 5% less than that of the tangent to the initial straight line part of the load-displacement record. This line is shown in the figure above and is simple positioned from the definition of slope of a line, by drawing the line through the origin and the point (0.34 0.95x17), i.e. the point (0.34 16.15).

This line intersects the load-displacement trace at 19 kN, and there is no previous higher value of load on the trace. Hence FQ = 19 kN. To find KQ we need only to calculate the function of a/W (which is the finite geometry correction factor) and substitute these values into the given equation. As the specimen is standard sized, we know that W = 2B = 50 mm, hence a/W = 25/50 = 0.5:

To determine whether this value of KQ can be taken as a valid K1C, we must check that:

Essentially, the requirement on a ensures that the crack tip plastic is a small enough percentage of crack length that LEFM is valid, the constraint on B helps to ensure that plane strain conditions prevail, while the condition on (W - a) ensures that a plastic hinge does not develop ahead of the crack.

Clearly, the specimen dimensions all meet this inequality and hence we appear to have found a valid K1C. The final check to be implemented is the constraint that the ratio Fmax/FQ < 1.10, i.e. that there is not too much crack tip plasticity prior to fracture. As Fmax = 21.0 kN, the ratio is 21/19 = 1.105 which is close enough.

b) The maximum K1C value that can be determined from this thickness of specimen in steel of this grade, can be found simply by finding when the inequality between B and the ratio KQ/yield

strength is no longer met.

Question 4

This question illustrates the potential effect of quenching and residual stresses on fast fracture. It should take about 15 minutes to complete.

During water quenching of steel components with a section thickness of 30 mm, heat transfer calculations indicate that a peak stress of 130 MPa is generated in the section. Prior to heat treatment, the components were ultrasonically inspected to detect defects. The inspection technique has a minimum detection size of 0.5 mm.

a) What type of defect will be most critical?

b) Calculate the size of defect which would cause fracture of the component during the quenching operation, given that the aspect ratio of the crack is 2c/a = 10.

c) Would this inspection procedure guarantee integrity of the component if the quenching stresses approached the proof stress of the steel?

Note that the value of the plane strain fracture toughness K1C = 30 MPa m½ and the proof stress = 620 MPa. The stress intensity calibration for this component and crack geometry is given in the figure below (note that the subscript y indicates proof stress).

Where, for surface flaws:

and for embedded flaws:


Answer: a) surface defects b) 15.4 mm

c) Procedure does not guarantee integrity - ac = 0.54 mm

a) From inspection of the stress intensity solutions for surface and embedded flaws we can see

that, because of the factor of 1.1 for surface defects, they will become critical at smaller values of a than embedded defects.

b) This part of the problem merely requires substitution into the K equation for surface defects, but we need the appropriate value for Q. From the graph, as:

This flaw is very much bigger than the NDT detection limit and there should be very little risk of failure during the quenching.

c) If the quenching stresses approached the proof stress of the material, the situation changes dramatically. Surface defects are still critical, but the value of Q has changed:

As the critical size of defect is around the NDT detection limit, the inspection would not guarantee integrity. A change to the quenching procedure would have to be implemented, e.g. using a slower quenchant.

Question 5

Following the failure of a rocket motor casing at an applied stress of 1260 MPa during hydraulic proof testing, investigation revealed an internal elliptical flaw which had extended to 4.0 mm by 1.6 mm prior to fracture. The material had been heat treated to a proof strength of 1645 MPa and had a K1C value of 60 MPa m½.

Calculate the applied stress necessary to cause fracture, using the K-calibration shown below. Does the calculated value agree with the observed value of failure stress?

Where, for surface flaws:

and for embedded flaws:


Answer: 1220 MPa - Yes, to a first order

This problem merely requires substitution into the K equation for embedded defects, but we need the appropriate value for Q. From the graph, as:

This agrees pretty well with the observed value, bearing in mind that this is a first order analysis.

Question 6

A thin plate of steel contains a central through-thickness flaw of length 16 mm, which is subjected to a stress of 350 MPa applied perpendicularly to the flaw plane. The 0.2% flow stress of the material is 1400 MPa.

Calculate the plastic zone size and the effective stress intensity level at the crack tip, making

reasonable assumptions about the state of stress.

If, after heat treatment, the flow stress of the steel dropped to 385 MPa, what would the plastic zone size be under the applied stress of 350 MPa, and what conclusions would you draw about the use of LEFM?


Answer: For the 1400 MPa strength level, rp = 0.25 mm and Kmax = 56.35 MPa m½.

For the 385 MPa strength level, rp = 3.31 mm, and the use of LEFM is becoming dubious.

This problem requires two simple assumptions to be made and then is a simple case of substitution into standard formulae. The first assumption is that the plate is large compared to the size of the crack; this allows us to use the simple infinite plate formula for stress intensity factor, i.e<

We must remember that the length of a central through-thickness crack is defined as 2a, hence we use half the length (8 mm) in the stress intensity equation.

Our second assumption is that the steel plate is in a state of plane stress - this is reasonable, as it is stated to be 'thin'. An accurate assessment of stress state would require a comparison to be made between plate thickness and plastic zone size - if this ratio tends towards 1, plane stress prevails, while if it tends towards 15 we are dealing with plane strain. Plane stress is also a conservative assumption, in that K values are higher (through Irwin's plastic zone correction) when plastic zones are bigger.

Irwin's plastic zone correction factor to crack length is given by:

This is small compared with the crack length and its effect on K will be correspondingly small:

This is around a 1.5% change and thus a single iteration of the calculation is sufficient.

However, if the flow stress drops to 385 MPa after heat treatment, the situation is quite different. The plastic zone size now becomes:

This represents a correction of around 18.9% and the use of LEFM becomes dubious. This is confirmed by the fact that the applied stress (350 MPa) is now some 91% of the flow stress. A yielding fracture mechanics parameter should be used to characterise the propensity for fracture.

Question 7

This question illustrates the dichotomy in fracture mechanics between high strength and high toughness. Generally speaking, as strength level increases in an alloy class, the toughness decreases. This has led to some problems for designers who have been schooled in yield orientated design. Certain alloys, such as the maraging steels, combine high strength and high toughness through careful compositional and microstructural design. Maraging steels are used in applications like undercarriage legs for aircraft, missile cases, cannon recoil springs and fan shafts in jet engines. They can achieve strength/toughness combinations ranging from about

1500 MPa/120 MPam½ to 2000 MPa/60 MPam½. Such steels typically have very high nickel, cobalt and molybdenum contents and very low carbon. Their high strength and toughness derives from age hardening (precipitation of intermetallic compounds) of a low carbon, iron-nickel lath martensite matrix. Further information on these steels can be found in:

ASM Handbook, Vol. 1 10th edition (1990), Properties and Selection: Irons, Steels and High Performance Alloys, American Society for Materials, Materials Park, Ohio, pp.793-800.

This problem is the first one that uses hoop stress, calculated using stress analysis for a thin walled pressure vessel. Pressurised vessels are commonly found in engineering practice and hence turn up regularly in fracture mechanics eaxmples. Thin walled theory is approximate, and can be used for ratios of diameter/wall thickness of > 10. For ratios lower than this, the thick walled theory should be used. The two relevant stress equations are given in example 4.4.1, which deals with fatigue crack growth. In pressurised vessels with internal surface-breaking cracks, one has to combine the stress intensity factors arising from the hoop stress and the internal pressure (see the Theory card in example 2.16 for information on superposition of K values).

Thin and thick walled pressure vessel theory is given in the following website:

http://www.engin.brown.edu/courses/En175/Elasticity2/Elasticity2.htm

The problem should take around 15 minutes.

You are involved in the design and manufacture of 6.6m diameter (D) rocket motor cases with a wall thickness (t) of 18.5 mm, and an operational pressure (p) of 6.6 MPa. These components are presently manufactured from a Grade 200 maraging steel with a yield strength of 1515 MPa and K1C = 136.5 MPam½. In order to save weight, a design engineer has proposed changing to a Grade 250 maraging steel with a yield strength of 1650 MPa and a plane strain fracture toughness value of 72.5 MPam½, and has requested an fracture analysis of allowable defect size against failure stress.

Failure of the motor case can be assumed to occur from embedded elliptical defects orientated perpendicular to the hoop stress. Typical elliptical defects resulting from the welding process can be detected by NDT, and are known to occur with sizes up to 5.5 mm by 35.5 mm. Determine design data for these alloys of fracture stress against allowable defect size, over a range of major axis lengths from 20 mm to 50 mm.

Based on this data, make recommendations to the designer as to the appropriate material to use when welding defects are likely to be present. Also determine whether the use of LEFM, based on K1C, is likely to be conservative or not for the given dimensions of defects and motor case, i.e. whether plane strain conditions exist.

Hoop stress is given by pD/2t, and the stress intensity solution for an embedded crack, at the

semi-minor axis position (crack depth a), is given below:


Answer: The Grade 200 steel is the best choice, and use of K1C is conservative.

To provide the fracture analysis requested by the designer, all we need do is calculate a table of fracture stress against defect size for the two materials. Although the calculations are based on the length of the semi-minor axis (a), it is useful to show the full length of the minor axis (2a) in the table, as this would be the parameter obtained from NDT. The fracture stress corresponding to likely weld defects can then be compared with the design (hoop) stress and the appropriate recommendation made.

The hoop stress in this component is:

As the defect is embedded, no additional contribution to stress intensity arises from the internal pressure in the motor case (see Theory Card in this question for information on superposition of K values). Thus the design stress is synonomous with the hoop stress. The table below gives fracture stress for various values of 2a corresponding to the range of defect sizes of interest (20 mm < 2c < 50 mm).

Alloy Defect Size (2a) mm

3.4 4.5 5.0 5.5 6.0 6.5 7.0 7.5 12.1

Grade 200 1931 1832 1747 1673 1607 1549 1496 1177

Grade 250 1177 1026 973 928 888 854 822 795

This column in bold indicates the fracture stresses corresponding to the presence of typical weld defects. The Grade 200 steel can tolerate cracks of up 2a = 12.1 mm at the design stress, while the fracture stress for a crack with 2a = 5.5 mm is greater than the yield strength. (Note, however, that we do not know what temperature the given material property data refer to, or what the operating temperatures of the motor case are - this is important information, as fracture toughness and yield strength are functions of temperature). The Grade 250 steel, however, can only tolerate cracks with 2a = 3.4 mm and would suffer fracture at the design stress.

The recommendation to the design engineer would be to continue with the use of the Grade 200

steel.

The plane strain fracture toughness, K1C, is a lower bound toughness value that applies to conditions where full triaxial constraint is developed. If conditions in the component are less constrained, i.e. tending towards plane stress, then use of K1C as a design criterion will be conservative. Estimation of the stress state of typical weld defects in the component can be done by checking the following ratio inequalities, which must all be < 1/15:

crack tip plastic zone size (rp)/motor case thickness (t) rp/crack length (a) a/t a/uncracked ligament

For an embedded crack, one can probably use the plane strain plastic zone in these checks, although for a surface breaking crack, the plane stress zone (which is 3 times larger) may be more relevant. In the present ase, however, the Grade 200 steel is tough enough that fracture would probably occur under conditions of plane stress or net section yield (the above inequalitites would not be met for a crack with 2a = 12.1 mm). Hence the K1C approach is conservative.

Question 8

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 fracture-safe design process. This is demonstrated graphically in the question.

It should take around 15 minutes to complete.

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, KC = (150 + 0.05T) MPa m½ 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 KC = (100 + 0.25T) MPa m½, 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.

Through-thickness 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. Scientific Calculator | RPN Calculator | Theory

Answer: Steel A is best below approximately 212ºC, Steel B is better above this temperature.

Solving this question is simply a matter of calculating the required values of fracture toughness KC 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 cross-over 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 leak-before-break, the amount of pressure relief caused by a through-thickness 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. Leak-before-break

design requires the pressure vessel to tolerate a through-thickness crack of total length 2a = the surface length (2c) of the pre-cursor semi-elliptic crack. As we have no information regarding crack ellipticity, we will have to assume that it was semi-circular 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 AYield Strength

MPa540 460 380 300

Required KC

MPa m½141.7 144.7 150.3 163.1

Actual KC

MPa m½150 155 160 165

Steel BYield Strength

MPa650 600 550 500

Required KC

MPa m½139.4 140.2 141.4 143.0

Actual KC

MPa m½100 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.

Question 9

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.

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.

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.

Question 10

The last question in this section of the tutorial illustrates the concept of superposition of stress intensity factors, which has already been invoked in dealing with pressure vessels. For internal defects in pressurised vessels, we had to take account of the stress intensity arising from the internal pressure on the crack faces, as well as that due to the membrane stresses in the shell.

This particular case also demonstrates the application of LEFM to ice, a glass-like material.

It should take around 10 minutes to complete.

The figure below shows stress intensity factors corresponding to two load cases for a vertical edge crack in an infinite body. Case A represents uniform tension, while Case B illustrates a linearly increasing tensile load with depth into the body. The two stress intensity equations are:

Consider a surface crevasse in a wide, thick glacier, which is subject to a uniform longitudinal stress of 200 kN/m2. Determine the depth to which this vertical crack can grow, given that the density of ice is 0.92 x 103 kg/m3 and the gravitational constant is g = 9.81 m/s2.


Answer: The crack will reach a depth of about 36.34 m.

This problem illustrates simple superposition stress intensity factors, but requires a bit of thought in relating the cases shown, with what is required by the physics of the problem. Both stress intensity solutions shown are tensile but, clearly, this circumstance cannot directly apply to the crack in the glacier, or it would not arrest. The crack in the glacier is stated to be subject to tensile loading (and this, presumably, is true or the glacier would not move forwards!).

For the crack to arrest, this tensile loading must be opposed by a compressive loading, which arises from hydrostatic pressure in the ice. Hydrostatic loading will induce a linear compressive stress gradient identical with the one shown in Case B, but opposite in sign. Although compressive K values have no physical meaning, positive and negative stress intensity factors can be algebraically added. Hence the crack will arrest when:<

KA - KB = 0

The answer is obtained by substituting the appropriate values into the two equations, noting that:

Hence we get:

Question 11

This question is straightforward, and shows an application of the concept of stress state being determined by the ratio of crack tip plastic zone size to specimen thickness. This concept is used in plane strain fracture toughness tests to BS 7448 : Part 1 : 1991 in choosing specimen thickness and checking that plane strain conditions existed in the test (see Problem 6). It should take about 10 minutes to complete.

Catastrophic fracture occurred in a thick steel plate during proof testing, at an applied stress of 700 MPa.& The initiating defect was an embedded sharp penny-shaped flaw with a radius of 2.5 cm. Calculate the fracture toughness of this steel.

It is desired to check this value by determining the plane strain fracture toughness from standard tests. The yield strength of the steel is 1100 MPa. A sheet of nominally similar steel, 7.5 mm thick, is available. Is this sufficiently thick to obtain a valid K1C value? If not, what thickness of steel should you order?


Answer: KC = 124.9 MPam½ - a valid K1C value requires a thickness > 35 mm.

Calulating KC requires simple substitution of values into the formula, i.e.

If this is assumed to be a valid plane strain fracture toughness value, then the minimum specimen thickness required is given by:

Substituting the given information into this equation gives the required thickness as:

Hence the thickness of steel in stock is insufficient to provide a valid K1C value. One would order the next standard plate thickness greater than 32.2 mm, probably 35 mm thick.

http://www.tech.plym.ac.uk/sme/interactive_resources/index.html

http://www.tech.plym.ac.uk/sme/interactive_resources/index.html

stress intensity factor and fracture toughness testing

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