MCEN 2024, Spring 2008 The week of Apr 07 HW 9 with Solutions

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1 MCEN 2024, Spring 2008 The week of Apr 07 HW 9 with Solutions The Quiz questions based upon HW9 will open on Thursday, Apr. 11 and close on Wednesday, Apr 17 at 1:30 PM. References to A&J: Chapters 13, 14, 15 and 16 Topic: FRACTURE Overview: Fracture differs from plastic yielding in that it is defined by the propagation of a crack. Therefore, a pre existing microcrack, or a "flaw", is the precursor to fracture. If a material is free from flaws then the intrinsic strength of the bonds among the atoms determines the stress required to produce fracture: this is called the ideal fracture strength, and, conceptually, is equivalent to the ideal shear strength for plastic yielding that was discussed previously. The essential length scale in fracture is the dimension of the flaw, which can range from just a few micrometers to macroscopic flaws that can be seen with a naked eye. In this session we discuss two issues: (i) what material parameter should be used in engineering design to safeguard against fracture failure, and (ii) how do we design the microstructure of materials in order to enhance their resistance to fracture. The engineering design parameter is called the fracture toughness, is designated by K IC, having he units of MPa m 1/2. Note that MPa is related to the stress required for fracture, and, as we shalll see, the length scale of meter is related to the dimension, or as often called, the size of the flaw. Sometimes K C or simply K are also used as the nomenclature for the fracture toughness. The materials science of the fracture process is studied by considering the work of fracture, written as G IC or G C, and this has units of J m -2. Notice that the units have work done per unit area. The physical significance of these units is that the work of fracture is expressed as the work done to propagate the crack such that its surface area is increase by a unit area.

2 Explanation of the Figures and the Tables (These are now appended at the end of the HW)I (i) The Tables for the Elastic Constants are included again, since the elastic constants are critically important in the study of the work of fracture, G C. (ii) Tables for the fracture toughness, K C, and for the work of fracture, G C of various materials, taken from A&J are attached. (iii) Often, we are concerned with fracture in beams of various cross sections. Here equation for the tensile stress, and displacements in beams of various cross sections are important. These equations are given at the end of this HW. Comparison Chart for Fracture Toughness

3 Comparison Chart for the Work of Fracture (called as the Toughness - not fracture toughness - in the book) The equation for the maximum stress on the right in the table for three point bending loading, is useful for relating the applied load to fracture. Equations for other geometries of the beam, and loading, are given at the end of this HW.

4 1. The mechanical loading of a component for the purpose of analyzing fracture is defined by the stress intensity factor, K I, which depends on the applied stress as well as on the flaw size as illustrated by the following sketch and equation: (Eq. 1) Fracture criterion: A common way to measure the fracture strength of glass is by the three point bending method. The sample, in the shape of a glass slide is supported on two roller pins. The force on a pin place above the slide to produce fracture in the slide is measured. If this force is F, the spacing between the roller supports at the bottom is L, the width of the glass side is w, and its thickness is b, then the fracture strength is given by: (Eq. 2) The fracture stress of the glass is 50 MPa, and. (i) Calculate the force required to break the glass slide. (What will be the force for fracture if the glass slide is twice as thick?) 1) The force is calculated simply by substituting the values (in meters) into Eq.(2). The answer is 6.7 N or about 0.7 kg 2) If the thickness, b, is doubled then the force required to break the glass will increase by a factor of four. The fracture strength is very sensitive to the thickness of the specimens in three point bending experiments. (ii) Assuming that the fracture toughness of glass is, calculate the flaw size in the glass slide. The fracture stress and the flaw size are related to the fracture toughness by Eq. (1). Substituting into Eq. (1) we get the flaw size to be 127 µm.

5 (ii) Calculate the range in the flaw size in the glass slide if the fracture stress varies from 10 MPa to 100 MPa. The flaw size would range from 32 µm for the high value of the fracture stress to 3184 µm for the low value of the fracture stress. 2. The equation for the maximum tensile stress (also called the hoop stress in such pressure vessel problems) in the wall of a pressurized cylinder with an inner radius of R and a wall thickness of t (assuming that R>>t) is given by: Show that the above equation has the correct units. R and t have the same units, that is length. Pressure has the same units as stress. The beer bottle has a radius of 5 cm. The fracture strength of the glass can vary from 10 MPa to 100 MPa. What should be the wall thickness for safe design if the maximum pressure within the bottle is expected to be 3 bar (note that 1 kbar = 14,500 psi = 100 MPa), and allowing for an overall safety factor of 2. Safety factor of two means that the maximum pressure for the purpose of design is 1.5 bar. The wall thickness must be designed for the lowest value of the fracture strength, i.e. 10 MPa. Therefore we have that: t = pr σ = 1.5*105 *5* *10 6 = 7.5*10 4 meters; i.e. the wall thickness should be at least 0.75 mm. 3. The design of a pressure vessel (the maximum pressure that is safe) can be limited by either the yield strength or the fracture toughness of the material. Please explain the following design curves for a pressure vessel made from aluminum:

6 (i) Explain the shape of the curves (why one is flat and the other curves downwards). The horizontal line signifies the yield strength, which is unrelated to the flaw size. The fracture limited design curve is dependent on the flaw size since the fracture stress is related to the fracture toughness of the materials and the flaw size by Eq. (1). (ii) Derive the transition from the yield limited design to fracture limited design. The transition occurs where the yield stress is equal to the fracture stress. Therefore the transition is given by: σ Y = σ f = K IC πa 3.

7 (above is problem #4) Prob. 3: The butt joint problem is like the three point bending problem described by Eq. (1). First you must calculate the fracture stress using Eq. (2), setting K IC =0.5 MPa m 0.5, and a=1 mm. This gives a fracture stress of 8.9 MPa. You can now calculate the maximum load from the three point bending problem by setting l=2m, and w and b=0.1m. You should get the answer 2967 N, or about 300 kg. Prob. 4: First you calculate the fracture stress using the same procedure as in Prob. 3. If the fracture stress is smaller than the yield stress then failure would occur by fracture rather than by yielding. 5. The following table gives the data obtained by Griffith in 1921 for the tensile strength of glass fibers as a function of their diameter. These data led to his theory of fracture which became the basis for modern fracture mechanics. The data are in inches (the diameter is in unites of mils ) and psi (recall that 14,500 psi, that is 1 kbar is equal to 100 MPa). Please convert these numbers into micrometers and MPa.

8 Make a plot of the data with the fiber diameter along the x axis and the tensile strength along the y axis. Estimate a range for the flaw size in the glass fibers in Griffith's experiments. The data in the table is plotted in the above figure. The vertical axis is the fracture stress normalized with respect to the elastic modulus of glass, which is assumed to be 100 GPa. Therefore the fracture stress ranges from about 3.5 GPa at the high end to 300 MPa at the low end. The fracture toughness for glass is 1 MPa m 0.5. Substituting these values into Eq. (2) gives the range for the flaw size as 26nm for the high stress and 3.5 µm for the low value of the fracture stress.

9 4. We wish to design the glass overhang for canyon viewing, is shown on the following page. The length and the width of the overhang are functional design parameters that can be estimated from the scale of the picture. The question is what should be the thickness of the glass platform, assuming that the fracture toughness of glass is 1 MPa m 1/2. How would you specify the size of the largest flaw in the design process? How would you design the glass platform taking into account both the physical weight of the glass and the force of the maximum loading from people standing on it? Here you have to use the cantilever beam equation. Substitute the weight of the glass as the force acting on the beam, which will give you an equation for the stress in the beam as a function of the beam thickness. The stress is then substituted into Eq. (2). Setting the fracture toughness to be equal to 1 MPa m 0.5, and the flaw size to be 1 cm, you can now solve for the thickness of the beam. 5. The work of fracture, defined as the mechanical work done to propagate the crack so that its surface area increases by a unit area, is given by: (Eqn. A) (Please note that there is often a discrepancy by a factor of 2 in the equations that you will find in the literature: this usually stems from considering fracture to be represented by two surfaces that are created when a crack propagates, or just one surface).

10 a) Show that G C has units of J m -2. K C has units of stress times square root of length. E has units of stress. Therefore G C has units of stress times length. Now stress has units of energy per unit volume, i.e. J m -3, which multiplied by length gives J m -2. b) The work of fracture for polymers is much higher than it is for a glass (like silica) why? yet the fracture toughness of glass and polymers are often the same. The work of fracture for polymers involves stretching of chains at the crack tip which involves a great deal of work, much more than fracture in glass where the bonds at the crack tip are broken to propagate the crack. However, the fracture toughness of these two classes of materials is similar because polymers have a much lower modulus than that of glass, since Eq. (A) the fracture toughness is related to the product of the work of fracture and the elastic modulus. 6. The Table for G C and K C given in the book and at the end of this problems set gives the values for different kinds of iron based alloys, including: Steel Rotor PressVess HighStren MildSteel CastIron Make a plot of versus and show that the relationship given in Eq (A) is reasonably correct. This plot is left to the students. 7. Why, in Problem 6, does the elastic constant of iron remain essentially unchanged while the fracture toughness and the work of fracture vary widely. The elastic constants of materials is related to the bonding between the atoms, which, on the average remains approximately unchanged even if small amounts of alloying agents, such as carbon, are added to the iron. However, the work of fracture is related to the yield strength because the work done to propagate a crack is strongly related to the work done in plastic deformation near the crack tip. The plastic work depends on the yield stress of the metal, which can be changed by alloying and heat treatment. Therefore the work of fracture is variable.

11 8. Explain why there is a trade off between the yield stress and the fracture toughness in the metallurgical design of steels (if the yield stress goes up, the fracture toughness goes down). A higher yield stress always means that the metal is less ductile. For example tool steels are semi brittle materials. The work of fracture therefore decreases as the yield strength increases since the metal fractures more easily at the crack tip. From Eq. (A) a lower work of fracture also means a lower fracture toughness, provided the elastic modulus does not change significantly. 9. Pick the following classes of materials from the Table for fracture toughness and work of fracture in the Table given the book (and appended here): Ice Soda Glass Silicon Nitride Polycarbonate Cast Iron Common Woods CFRPs Titanium Alloys Aluminum Alloys Rotor Steels Make a plot of to check whether or not Eq. (A) is valid for a wide variety of material. Since the data covers several orders of magnitude.. it would be better to plot the data on a log-log scale: then see if the general slope for the data in this plot is equal to one. This plot is left to the students.

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