E7.19. If the crash barrier of the last
exercise is part of a vehicle fuel is
saved if it is light. We then want the
materials with the largest value of
W pl / where is the density.
These are found by using a selection
line of slope 1 on the chart made in
E8.10. Suppose that the resolution limit of the NDT facility available to you is 1mm, meaning that it
can detect cracks of this length or larger. You are asked to explore which materials will tolerate
cracks equal to or smaller than this without brittle f
Whether polycarbonate (PC) is has a higher fracture toughness K 1c toughness than glass.
Answer.
CFRP has lower fracture toughness K 1c than aluminum alloys.
Polypropylene (PP) has a higher toughness G c than aluminum alloys.
Polycarbonate (PC) is has a m
E7.17. The text showed that the power
required to roll a metal is proportional to
its yield strength. Make a bar-chart of
yield strength, y for metals. Open the
record for low-carbon steel, find the range
of it yield strength and take the average.
Normali
Chapter 8
E8.1. What is meant by toughness? How does it differ from strength?
Answer. Strength is resistance to plastic flow and thus is related to the stress required to move
dislocations through the solid. The initial strength is called the yield streng
E7.14. A material is required for a
spring that must be as light as possible.
To be stiff enough it must also have a
Youngs modulus E > 20 GPa. Make a
bar chart with the index 2 / E for
y
selecting light springs derived in
exercise E7.10 (you will need to
C m . Thus the material cost of the panel is minimized, while meeting the constraint on strength, by
choosing the material with the largest value of the index
1/ 2
y
Cm
.
E 7.6. A centrifuge has a solid rotor of diameter of 200 mm made of a material of
(a) the nominal stress
(b) The maximum stress in the plate?
Will the plate start to yield? Will it collapse completely?
Answer. The nominal stress is the load divided by the nominal area, 500 x 15 mm, giving
nom = 6.67 MPa
The maximum stress is the nomin
E7.12. Use the Search facility in CES to search for materials that are used for light springs. Report
what you find.
Answer. Here the search gives a smaller number of returns than in Exercise 7.11. They are
CFRP, epoxy matrix (isotropic)
Titanium alloys
P
E7.3. Derive the index for selecting materials for a light, strong beam with a square cross section,
equation (7.18).
Answer. The objective is that of minimizing mass the same as in Exercise 7.2. The constraint is that
of strength: the beam must support F
E6.14. Work hardening causes dislocations to be stored. Dislocations disrupt the crystal, and have
associated energy. It has been suggested that sufficient work hardening might disrupt the crystal so
much that it became amorphous. To do this, the energy a
E6.9. Use a Limit stage to find materials with a yield strength y , greater than 100 MPa and the
density, , less than 2000 kg/m3. List the results.
Answer. The Limit stage gives two classes of materials, alloys of magnesium, and composites:
CFRP, epoxy ma
E6.12. Apply the same procedure as that of the last exercise to explore copper and its alloys. Again
use your current knowledge to make your best judgement about the origins of the trends.
Answer. The chart shows that
All alloying increases the strength b
Chapter 7
E7.1. What is meant by the elastic section modulus, Z e ? A beam carries a bending moment M .
What, in terms of Z e is the maximum value M can take without initiating plasticity in the beam?
Answer. The basic equation of elastic beam-theory is
y
E5.17. Make a chart of the sound velocity (E / )1 / 2 for the elements. To do so, construct the
quantity (E / )1 / 2 on the y-axis using the Advanced facility in the axis-choice dialog box, and plot it
against atomic number An . Use a linear scale for An
E6.5. A metal matrix composite consists of aluminum containing hard particles of silicon carbide SiC
with a mean spacing of 3 microns. The composite has a strength of 180 MPa. If a new grade of the
composite with a particle spacing of 2 microns were devel
Exploring design with CES.
E5.10. Use a Limit stage to find materials with modulus E > 180 GPa and price C m < 3 $/kg.
Answer. All are ferrous alloys:
Cast iron, ductile (nodular)
High carbon steel
Low alloy steel
Low carbon steel
Medium carbon steel
Stai
E 1/2 /C m
E5.8. Devise an elastic mechanism that, when compressed, shears in a direction at right angles to the
axis of compression.
Answer. The molding shown here has the required
response.
E5.9. Universal joints usually have sliding bearings. Devise a
E5.14. A material is required for a tensile tie to link the front and back walls of a barn to stabilize both.
It must meet a constraint on stiffness and be as cheap as possible. To be safe the material of the tie
must have a fracture toughness K 1c > 18 M
Chapter 6
E6.1. Sketch a stress-strain curve expected of a
metal. Mark on it the yield strength, y , the
tensile strength ts and the ductility f . Indicate
on it the work done per unit volume in deforming
the material up to a strain of < f (pick your
own
Answer. The figure shows the chart of Figure 4.6 onto which contours of
E / are plotted
(remember to multiply E by 10 and in the units of this chart by 10 in order to get the velocity in
m/s). Tungsten, titanium, nickel, aluminum, magnesium and steel all
Chapter 5
E5.1. Distinguish tension, torsion, bending and buckling.
Answer. This is best answered by a sketch.
E5.2. What is meant by a material index?
Answer. A material index measures the performance of a material in a given application. Material
indice
further physical properties involved such that the correlation coefficient is dimensionless. We return
to this correlation in a later chapter, where these other properties are identified. The correlation is then
a much better one.
E4.19. The force require
Answer. A line of slope 1, sketched through the data for Youngs modulus E (GPa) plotted against
absolute melting point Tm (K) for metals, gives the relationship
E / Tm 0.1
(Change the units of temperature from C to K via Tools / Options / Units / Use abso
Answer. The chart shows that there is an approximately linear relationship between the left and right
hand sides of the equation. The red line is a plot of the equation above. The correlation is pretty good.
specific modulus, E / , of the composite greater than that of unreinforced aluminum? How much
larger is the specific modulus if the same volume fraction of SiC in the form of continuous fibers is
used instead? For continuous fibers the modulus lies very c
E 4.18. The cohesive energy H c is the energy that binds atoms together in a solid. Youngs modulus
E measures the force needed to stretch the atomic bonds and the melting point, Tm , is a measure of
the thermal energy needed to disrupt them. Both derive f
a crystalline solid
an amorphous solid
a thermoplastic
a thermoset
an elastomer?
Answer.
A crystalline solid is one in which the atoms or molecules are arranged in an ordered way that
can be described by a geometric lattice.
An amorphous solid (or glassy
Wind turbine blade: bending, caused by the self-weight of the blades and wind loads.
Climbing rope: tension.
Bicycle forks: bending and compression
Aircraft fuselage: bi-axial tension in the fuselage wall caused by the hydrostatic pressure of
pressurized