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

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.
Topic: FRACTURE
References to A&J:
Chapters 13, 14, 15 and 16
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 KIC, having
he units of MPa m1/2. 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 KC 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 GIC or GC, 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.
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, GC.
(ii) Tables for the fracture toughness, KC, and for the work of fracture, GC 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
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.
1. The mechanical loading of a component for the purpose of analyzing fracture is defined by the
stress–intensity factor, KI, 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.
(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:
pR 1.5 *10 5 * 5 *10 −2
=
= 7.5 *10 −4 meters; i.e. the wall thickness should be at least 0.75 mm.
t=
6
σ
10 *10
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:
(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:
K
σ Y = σ f = IC
πa
3.
(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 KIC=0.5 MPa m0.5, 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.
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 m0.5. 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.
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 m1/2. 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 m0.5, 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).
a) Show that GC
has units of J m-2.
KC has units of stress times square root of length. E has units of stress. Therefore GC 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 GC and KC 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.
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.