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True stress strain curves Lecture 5 October 25, 2010 10/25/2010 1 Production porcesses, Dr. B. Gharaibeh

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Page 1: True Curve

True stress strain curves

Lecture 5

October 25, 2010

10/25/2010 1Production porcesses, Dr. B. Gharaibeh

Page 2: True Curve

True stress strain curve

10/25/2010 Production porcesses, Dr. B. Gharaibeh 2

K : Strength coefficient

n : Strain hardening exponent

0 < n < 1

Y = a + b X

nK

Ln nLnKLn

K = Strength coefficient N: Strain hardening exponent

K is the true stress at true strain = 1n > 0Elastic and plastic are proportional

Page 3: True Curve

10/25/2010 Production porcesses, Dr. B. Gharaibeh 3

(b) True stress-true strain curve plotted on alog-log scale.

slope is always positive and the slope decreases with increasing strain.

Y : yield strength Yf : flow stress is the true stress to complete plastic deformation at a particular true strain ε1Yf increases with increasing strain

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Plastic instability point

• The point of instability satisfies the relation

10/25/2010 Production porcesses, Dr. B. Gharaibeh 4

d

d

nK

.......(2)

max.pat 0

..

.......(1) .P

A

dAd

dp

dAAddp

A

0.

... 00

LdAdLA

constLALA.......(4)

2 into 3 sub.

.......(3) d

0.

... 00

d

d

A

dA

L

dL

LdAdLA

constLALA

Page 5: True Curve

Plastic instability point in simple tension

1nn

1n

KnK

d

d but

Knd

d

nK

n

The higher the strain hardening exponent the greater the strain a material can be stressed at before necking

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Example:

A material has a true stress strain curve given by

Calculate the true ultimate tensile strength stress (UTS) and the engineering UTS of this material.

psi 000,100 5.0

Page 7: True Curve

• These curves are used to simplify the numerical analysis associated computer simulation

idealized stress strain curves

Page 8: True Curve

Types of Stress-Strain Curves (the idealized stress strain curves)

Various types of idealizedstress-strain curves.

(a) Perfectly elastic.(b) Rigid, perfectly plastic.(pure lead)(c) Elastic, perfectly plastic.(d) Rigid, linearly strain hardening. (e) Elastic, linearly strain hardening

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Page 9: True Curve

Characteristics of the elastic region

► There is change in volume in this region while the material is loaded.► Strain in this region is very small.► when the loads are released the specimen will return to itsoriginal, un-deformed configuration.

► Poisson's ratio (ν) = [ - Lateral strain \ axial strain ] takes a value 0< ν<0.5 but not 0.5

► in this region tanθ gives the Modulus of elasticity, E ( where θis the angle between the straight line of the stress-strain curve and the horizontal line)

limit alproportion at Strain

limit alproportion at StressE

10/25/2010 9Production porcesses, Dr. B. Gharaibeh

Page 10: True Curve

Characteristics of the plastic region

► In this region there is no change in volume.

► The strains in this region are very large compared with strains in elastic region.

► the material does not return to its initial (original) un-deformed state

after unloading.

► Poisson's ratio (ν) for all materials in this region equals 0.5

► In this region [ε1 + ε2 + ε3 = 0] (in length, depth and width).

►In this region the approximate slope of the plastic line gives the modulus of plasticity P, (P<E) always.

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Page 11: True Curve

Rigidity

In the elastic region the relationship between G, E, and ν is given by

Where G: Modulus of rigidityν: Poisson's ratio

E: Modulus of elasticity

)1(2

EG

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Page 12: True Curve

• The energy per unit volume that has been dissipated up

to fracture.• Approximate by the area under the true stress-strain curve.

Toughness

T

f

0dT

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Page 13: True Curve

• The ability of material to show plastic deformation before

fracture. And it is measured by two quantities

Ductility

(2) % Reduction of area (RA)

100xA

AARA%

o

fo-

=

x 100L

LLEL%

o

of

Lf

AoAf

Lo

(1) % elongation (EL)

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Page 14: True Curve

Example:

A tensile test specimen has an original gauge length of 50 mm and a diameter of 4 mm. The maximum load during the test is 10 kN. The final gauge length is 80 mm and the diameter at fracture is 3 mm.Calculate :a)The Ultimate tensile stress (UTS).b)Percent of elongation.c)Percent reduction of area.