Lecture 15 True Stress Strain Hardening

8/2/2019 Lecture 15 True Stress Strain Hardening

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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)

3

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.

Documents

Lecture 15 True Stress Strain Hardening