# HW6FinalwSolutionsl

```MCEN 2024, Spring 2008
The week of Feb 25
HW 6–Final
Quiz #6 will be based upon this HW, and will be open from Saturday morning (Feb 29) until 1:30
PM, Monday, Mar 03. Please have this homework by you side when you open the quiz.
References to A&amp;J:
The reading list for this week:
Chapter 8: The Yield Strength, Tensile Strength, and Ductility
Overview:
We now move over to learn about plastic deformation in solids. Such deformation is
characterized by permanent deformation, that is, a change of shape. The topic is studied from the
point of view of bridging the length scales, from the atomistic to the macroscopic.
For example the design of a diving board so that it does not “yield” when a person stands on
the edge is related to the yield stress of the material. The yield strength is then related to the atomic
structure. The objective is to model the yield stress in terms of atomistics (atoms spacings and
bonding), and then to relate the yield stress to component design.
1. Draw a distinction between the two phenomena expressed in each of the sub–questions given
below:
(a) Bending a copper tube, and hot–forging of a horse–shoe.
(b) Hot–forging of a horse–shoe and forming of glass into a chandelier.
(c) Deformation of a ductile metal versus the deformation of a &quot;brittle material&quot;.
(d) Deformation of a metal vs. the time dependent deformation of a polymer.
(a) Because the yield stress decreases with increasing temperature. The yield
stress of iron–alloys is too high at room temperature to forge into shape
with a hammer.
(b) Deformation of metals involves a distinct yield stress. Glass, is different
it flows like a liquid albeit with a very high viscosity.
(c) A brittle material breaks when stressed to its elastic limit, while a
ductile deforms plastically at the elastic limit.
(d) The yield stress of metals is essentially time independent, while
deformation of polymers is time dependent, for example they continue to deform
slowly if loaded for an extended period of time.
2. Make a plot of the yield stress (vertical axis) and the elastic modulus of the following materials,
using the data given in the tables above: aluminum alloys, mild steel, stainless steels, cast iron,
nickel alloys, beryllium alloys, copper alloys, and titanium alloys. Show the range in the data with
error bars.
(a) Why are the error bars greater for the yield stress than for the elastic modulus?
(b) Draw two straight lines through the data which border the upper bound and the lower bound for
the data. The slope of these lines gives the yield stress as a fraction of the elastic modulus. What are
these fractions?
(c) Show that the fractions given above are equal to elastic strain at the onset of plastic yielding.
The plot is given on the next page:
(a) The error bars are greater for the yield stress because alloying
and heat–treatments are used to &quot;engineer&quot; the yield behavior, while
the elastic constants depend only on the composition and can be roughly
calculated using the rule–of–mixtures of the values for E for the
constituents.
(b) The slopes give 1.7*10-3 &lt; σ y / E &lt; 6.7*10-3.
(c) The elastic strain at the onset of yielding is the same as σ y / E ,
that is 0.17% to 0.67%.
3. Plastic strains are often much larger than elastic strains. In some situations, like in forging and
rolling the strains can be very large indeed. Therefore the stresses, and strains in large deformations
have to be defined more carefully than the stresses and strains for elastic deformation. For large
deformations, these quantities are described as &quot;true stress&quot; and &quot;true strain&quot;, written as σ and ε .
The nominal stresses and strains, σ n and ε n , are defined with respect to the initial cross–section,
Ao, and initial length, Lo of the specimen. If the force at any point of deformation is written as P,
and the length as L, then:
σn =
P
L − Lo
, εn =
.
Ao
Lo
(1)
For plastic deformation, the strain is defined by the incremental deformation, so that:
dε =
dL
L
Show that, for a simple uniaxial deformation:
(2)
ε = n(1 + ε n )
(3)
Then using the condition of constant volume for plastic deformation show that:
σ = σ n eε
(4)
The volume of the specimen at any state of plastic deformation is given
by the product of the cross sectional area and the gage length of the
specimen. Since the volume remains constant we have that:
AL = Ao Lo
(A)
Now, substituting from Eq. (1) into Eq. (3), and then again from Eq.
(A) we have:
ε = n(1 +
L − Lo
L
A
) = n( ) = n( o )
Lo
Lo
A
(B)
Raising the left most and the right most terms in (B) to the power of
&quot;e&quot;, we have:
Ao
= eε
A
(C)
Now,
σ=
P P Ao
=
.
A Ao A
(D)
Substituting from (C) into (D) gives the desired result.
4. Show that the area under the stress–strain curve, where the y–axis is the true stress, and the
x–axis is the true strain, is a measure of the total work done per unit volume of the material that is
deformed in a uniaxial test. (In class we considered nominal stress and nominal strain. Here you
need to extend that integral to true stress and strain. Use the constant volume condition that
Ao Lo = AL ).
The total work done on the specimen is the integral of the force times
the displacement, which can be written as:
WTOT = ∫ Pd(ΔL)
(E)
Dividing both sides by the volume of the specimen which is equal to
AL, we get:
WTOT
P d(ΔL)
(F)
=∫
= ∫ σdε
AL
A L
which is equal to the area under the true-stress true–strain curve.
W perVOL =
5. A plate of aluminum which is 5 mm thick and 25 cm wide is rolled in one pass to a sheet of the
same width but a thickness of only 0.1 mm. The yield stress of the aluminum (in uniaxial
compression) may be assumed to have a constant value of 300 MPa. The work capacity of the
roller–system is 100 HP (horse–power). Calculate the maximum speed at which the plate can be put
through the roller machine. (Hints: 1 HP = 746 watts, 1 watt = 1 J s-1. First calculate the true strain
in the sheet, and then the work done per unit volume. This will tell you how much volume of the
plate can be put through the machine in one second).
Work done per unit volume of the material is simply equal to the product
of true-stress and true–strain (since the stress–strain curve is assumed
to be square shaped, that is, the yield stress is assumed to be constant).
True strain in the plate is given by Eq. (B):
ε = n(
L
0.1
) = n( ) = −3.9
Lo
5
(G)
Therefore the work done per unit volume of throughput will be given
by: 300*106*3.9 J m-3 (the minus sign means that the strain is negative,
not that the work done is negative).
If the feed velocity for the plate is V m/s, and the input cross section
is 25cm x 5mm, then the work done per second on the metal, which must
be equal to the HP capacity of the machine, leads to the following
equation (keep track of units - use J and m):
(300 *10 6 * 3.9) * (25 *10 −2 * 5 *10 −3 *V ) = 100 * 746
which gives: 0.05 m/s
of 10)
(the answer in the book is too large by a factor
6. and 7.
H (MPa)
TrueStress
NomStrain
NomStrain+H TrueStrain NomStress
423
141
0.01
0.09
0.09
129.36
606
202
0.1
0.18
0.17
171.19
756
252
0.2
0.28
0.25
196.88
870
290
0.3
0.38
0.32
210.14
957
319
0.4
0.48
0.39
215.54
1029
343
0.5
0.58
0.46
217.09
1080
360
0.6
0.68
0.52
214.29
1116
372
0.7
0.78
0.58
208.99
1170
390
1
1.08
0.73
187.50
Notes:
• Hardness measures the true value of the yield stress, and is three
times the tensile yield stress
• The fourth column is simply the nominal strain +0.08, as said
in the problem
• The true strain is calculated from Eq. (3) above.
• The nominal stress can now be calculated from Eq. (4).
The nominal stress-strain plot is obtained from the 3rd and the 6th
columns, and is given below.
The solutions to problems (a), (b), and (d) are explained by the
above figure. For the reduction in area at the point of tensile strength,
problem (c), we proceed as follows:
RA =
Ao − A
A
L
= 1−
= 1− o
Ao
Ao
L
(H)
But we know that:
εn =
L − Lo L
=
−1
Lo
Lo
Therefore, RA = 1−
8. 9. and 10.
(I)
1
= 0.38
1 + εn
8. Copper can be deformed in rolling because strain localization in
the neck which leads to failure is prevented in compressive
deformation.
9. The tensile strength is achieved where the applied load is a maximum,
which occurs at 101kN. The answer to (a) therefor is
0.25*101E+3/(160E-6) Pa = 158 MPa.
The extension at 0.1% strain will be equal to 0.001*50=0.05 mm. The
load at this extension is 12kN which gives a nominal stress of 75
MPa. With a safety factor of 0.6 the stress is 45 MPa (the answer
in the book appears to be wrong).
10. The area of the footprint for a sphere, of radius r indented into
a surface to a depth of h is equal to 2πrh , hence the answer.
```