lecture 15 true stress strain hardening
TRANSCRIPT
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Mechanical Response of
Engineering Materials:
EMch 315
Plastic Deformation & Ductile Failure II
Lecture 15
Chapter 6
(Mechanical Response of Engineering Materials)
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Midterm Exam Format
Chapters 1 and 2: Mohrs circle (either stress or strain)
Chapter 4: Tensile/Compressive response
Chapter 3: Thermal Strain/Stress problem
Chapter 2: Stress concentration
Chapter 5: Yield/Safe design Theories
All the above questions: 20 points each.
Bonus Points (5): Short question on Hookes law (Constitutive equations)or definitions or true/false statements.
2
Closed book, one 3x5 inch card with equations is allowed
March 13, Tuesday, 6:307:45PM, Room 100, Thomas Bldg
No class on Tuesday, March 136 (5+1) Questions: The main topics will include:
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Is plastic deformation useful? _________________________________
Plastic region is also characterized by_________________, only shape
changes.
Small ______________________occurs in elastic region due to stretching
of bonds (spring action)
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True Stress and True Strain
Plastic Deformation and Ductile Behavior
etn= sot
= snomtTrue stress-strain curve
is same as engineering
curve up to yield stress
Parabolic law for
ductile materials
e.=
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etn= sotThe strength coefficient, , determines the magnitude of the true stress in the large strain region
(necking), so it is included as a measure of strength and hence the name. n, the strain hardening
exponent is a measure of the rate of strain hardening for the true stress-train curve. For
engineering metals, values above n = 0.2 are considered relatively high, and those below 0.1 are
considered relatively low. Al has higher n than steel, and hence easier towards strain hardening.
so
so
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s
ea b c d
X
smax
1 2 3 4
Necking
Strainhardening
X
True Stress and True Strain
st = P/Ai snom = P/A0st = snom A0/Ai
Recall et = AoAilnet = ln (enom + 1)Ao
Ai= (enom + 1)
(enom + 1)= snom
st (enom + 1)= snom
snomA0 = P
et = ln (enom + 1)
Instability
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Necking Region:Under tensile load it
is highly unstable dueto continuous
formation of
micro-voids,
conversion into
micro-cracks,
propagation and
growth of micro-
cracks into large
elliptical crack
finally causing the
physical
failure/fracture
Internal cracking in the necked
region of a polycrystalline
specimen of high-purity copper.
(Magnification 9x.)
TS/UTS Necking MicrovoidsMicrovoids
coalescence
Fracture/failure
Load Instability in Tension
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Load Instability in Tension
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etn= sotThis correlates true-stress and true-strain
behavior of a ductile material, but it doesnot contain any information of the
failure/fracture.
In the case of engineering stress-strain
response the instability occurs at TS/UTS.
Physically, ductile materials do not really
show any load instability in tension.
Instead, the true-stress curve continues to
increase even as the post instability load
decreases because the cross sectional area is
shrinking at a faster rate.
However, a measure of instability based on
true stress and strain concepts can be found
by examining changes in stress in the
vicinity of maximum load.
s
ea b c d
X
smax
X
Pmax = P*
Instability
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Load instability in Tension
sn
enst
et
Material still
exhibits instability
We know that there is a maximum
load that the material can carry
from engineering stress-strain
curve, however, we want to know
what is it using true stress-strain
curve
Want _____ whichcorresponds to Pmax.
We will call this
stress:_____
____ = sonnt*e
t
n= sot
Engineering/nominalStress-strain curve
Instability begins
Pmax
P*
*
*
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P = P* = Pmax
dP = 0
dP = = = 0
st = so etnst *= so et *n
;
sn
en
Xwant
at maximum load, slope is zero and
dAi *Ai *
dst*st * = dst*det * st*=
dst*det * n so et *(n-1)=
We want a quantitativemeasure of stress when
load becomes unstable.
or
recall
so at maximum load
=so et*n
dAi *Ai *
= det *det*
d(stAi) stdAi + Aidst
n so et *(n-1) = n so =so et*nt *n
et *
Pmax = P*
P = stAiAt unstable point
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et * = n
True strain at maximum load or strain corresponding to load
instability = strain hardening coefficient Onset of necking
st *= so et *net *= ln (enom* + 1) = n
et* = nValid only at the max. load Onset of neckingn so et *(n-1) = n so =so et*nt *net *
= sonnet = ln (enom + 1)
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Engineering/True Stress-Strain Relationships and
Limitations
s= P/Ao
st= P/Aiet
st=t=et= ln(Ai/Ao)t=
e= DL/LoOnly average values,
not accurate
Both are almost
same
Good only in constant
volume region
An arbitrary limit below
which true and engineering
stress are almost same
y
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V = 0 in plastic deformation
What is ?
Unit cube, loading. L0 one side
Lx = new length in x-direction; Ly, Lz
V0 = L03 ; VF = Lx Ly Lz
but Ly = Lz VF = Lx Ly2
dV = Ly2 dLx + Lx dLy (2)Ly = 0
Ly
dLx
= 2 Lx
dLy
(dLx / Lx) =2 (dLy/ Ly)
x =2yn =y /x = 1/2
Loading
X
y
z
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.001 .003.002 .006.004 .007.005
4000
2000
0
10000
5000
1.12
0
Deformation, inches
Deformation, inches
Load,pounds,
F
Load,pounds,
F
x
UTS
Fracture
E
spl3000
L0 = 2 inch
D = 0.505 inch
A0 = p(0.505/2)2 =sy
0.2% of L0
E = Ds/De = (F/A0)/(DL/L0)
4025
ur =spl)22ET=(sy + TS)ef 2
0.2 in2
LfL01.08
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In-class Problem 2
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Tensile load-deformation curves for cartridge brass 70% Cu30% Znin the 0525
temper are shown. The top graph has a magnified deformation axis to allow for
greater resolution in the elastic and yield regions: the bottom graph is the entire
response to fracture. Find the following quantities, indicating points used on theappropriate, and be sure to give correct units. The round specimen tested had an
original diameter of 0.505 inches and a gage length of 2.0 inches.
a. Modulus of Elasticity ____________________________________________
b. Proportional limit of stress _________________________________________
c. 0.2% Offset yield strength ___________________________________________
d. Ultimate tensile Strength _____________________________________________
e. Percent Elongation __________________________________________________
f. Strain Hardening Exponent ___________________________________________
g. Modulus of Resilience _______________________________________________
h. Modulus of Toughness ______________________________________________
E = Ds/De = [3000 lbs/p(0.505/2)2]/(0.002/2) = 15x106 psis
pl
= [3000 lbs/p(0.505/2)2] = 15000 psisy = [4025 lbs/p(0.505/2)2] = 20,125 psi
TS or UTS = [10,000 lbs/p(0.505/2)2] = 50,000 psi
ef(100)= (LfL0)/L0= 1.12 / 2 = 0.56 = 56%
ur =(spl)2 2E = (15000)2 2(15x106) = 75 in.lbs/in3
T=(sy + TS)ef 2 = (20,125 + 50,000)0.56 2 = 19,635 in.lbs/in3
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Example Problem
For the 6061-T6 Aluminum alloy shown, determine the following: (1) Strain
exponent and strength coefficient and (2) the nominal tensile strength.
n = 7/55 = 0.127
s0 = 74,000 psist *= so et *n = sonn
st *= sonn
68
41
0.007 0.4
n = (log68log41 ) / (log0.4log0.007 ) = 0.125
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No volume change during plasticity!
But, at
instability:
Hence:
Note: TSnom
< st* because A
o
> Ai
*
56,940Ai* = TSnomAo
TSnom = 56,940Ai
*Ao
AoAi*ln = et* = nAi
*
Ao= e-n = e-0.127 = 0.88
Ai*
AoTSnom = 56,940 = 56,940 (0.88) = 50,100 psi
st *= sonn = 74,000 (0.127)0.127 = 56,940 psi
P* = Pmax
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(so)n
Failure
True Strain et
True
Stress
st
x
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20From Mechanical Response of Engineering Materials, pages 127-128.
Homework ProblemsReading Assignment: chapter 6
VI.
VI.
%1001%100%0
0=
=
F
teA
AARA
F e
There is a typo in the answer for VI.3
given in the back of the book
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21From Mechanical Response of Engineering Materials, pages 127-128.
Homework Problems
Reading
Assignment:
Chapter 6
VI.
VI.
VI.