Poisson’s Ratio (υ) = - lateral strain / Axial strain = - εx/εz = -εy/εz
= - [∆d / do] / [∆l/ lo]
εx =εy = - υ εz
The negative sign is added to
yield a positive value for υ
For many metals and
alloys the poisson’s
ratio range between
0.25 and 0.35
Shear stress – strain diagram
Elastic behavior :
τ=G γ
G = E / 2(1+ υ) for isotropic material, for which the
properties are the same in all directions.
True stress and strain
σT = F / Ai , load divided by the instantaneous cross section area
εT = ln (li / lo), li: instantaneous length ---- lo: original length
For plastic deformation (σ>σy) there is conservation of
volume: Aolo = Ai li
Ao / Ai = li/lo
Relations between true and engineering stress and strain:
σT = F / Ai = F / Ao x Ao/ Ai= σ x Ao/Ai = σ x li/lo
ε = li-lo / lo = li/lo –1 Æ li/lo = 1+ ε
Thus σT = σ (1+ ε) This equation is valid from yielding to the
on set of necking σy < σ < σu
εT = ln (li / lo) = ln (1+ε), This equation is valid from yielding to
the on set of necking σy < σ < σu
For some metals and alloys the region of the true stress – strain curve
from the onset of plastic deformation to the point at which necking
begins (σy < σ < σu) may be approximated by:
( )
σ T =KC ε T
“true” stress (F/A)
n
hardening exponent:
n=0.15 (some steels)
to n=0.5 (some copper)
“true” strain: ln(L/Lo)
K and n are constants that vary from alloy to alloy.
Taking logarithm of both sides yields a straight line:
Log σT = n log εT + log K (y = mx +c)
K: strength
coefficient
n (strain hardening exponent) defines the slope of the
straight line
HARDENING
• An increase in σy due to plastic deformation.
σ
High values of n
large hardening
σy
1
σy
small hardening
reload
unloa
d
0
ε
Since n is the slope of the straight line higher values of n
corresponds to higher strain hardening rate.
In case where n and K are to be found for a particular alloy, two
points on the stress strain curve that defines σ and ε in the
region where σy < σ < σu has to be given.
Log σT1 = n log εT1 + log K
Log σT2 = n log εT2 + log K
By subtracting the two equations:
n = (Log σT2 - Log σT1 ) / (log εT2 - log εT1 )
Then substitute in any equation to get K.
Construct a table of F , ∆L, ε and σ
F
∆L(l-lo)
0
0
7330
50.851-50.8
15100 50.902-50.8
ε = ∆L/lo
0.051/50.8
σ=F/Ao
7330/ [(π/4) d2]
0.05
0.165
Εlastic
strain
recovered
≅ 0.005
To Generate a True Stress – Strain Response
σ<σy
σ=σT, ε= εT [elastic deformation]
σy<σ<σut
σT = σ (1+ ε) , εT= ln(1+ε)
Yield > necking
σut < σ <σf
σT= F / Ai
Necking > failure ,
εΤ=ln li/lo = ln Ao/AiÆ this equation is more valid in the
neck region
N.B. To generate a true stress – strain from engineering
stress strain, the instantaneous diameters should be given in
the neck region (σut < σ <σf)[ from necking to failure ]