MIT OpenCourseWare
http://ocw.mit.edu
16.982 Bio-Inspired Structures
Spring 2009
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How strong is it?
Under what sort of deformation?
F
Tension
F
l
l0
A0
Compression
A0
l0
l
F
F
A
B
T
φ
A0
Shear
T
Torsion
F
F
θ
F
C
D
Figure by MIT OpenCourseWare.
Reduced section
2 1"
4
3"
Diameter
4
0.505" Diameter
2"
Gauge length
3"
8 Radius
Figure by MIT OpenCourseWare.
Load cell
Extensometer
Specimen
Moving
crosshead
Figure by MIT OpenCourseWare.
• Basic mechanical behavior
of material
pulled in
• Specimen is “pulled”
tension at a constant rate
• Load (F) necessary to
produce a given elongation
(∆L) is monitored
• Load vs elongation curve
• Converted to stress-strain
curve
Engineering Stress, σ
F

A0
Reduced section
1"
2
4
3"
Diameter
4
0.505" Diameter
2"
Gauge length
3"
8 Radius
Figure by MIT OpenCourseWare.
where F is the applied load and A0 is the original
cross-sectional area. Units: N/m2 or lb/in2
E i
Engineering
i Strain,
St i ε
Li  L0 L


L0
L0
Reduced section
2 1"
4
3"
Diameter
4
0.505" Diameter
2"
Gauge length
3"
8 Radius
where L0 is the original length and Li is the
iinstantaneous.
t t
U
Unitless.
itl
Figure by MIT OpenCourseWare.
(Maximum) Tensile Strength
Fracture
Classification of
the material
B
Ductile
Stress
Brittle
B'
Figure by MIT OpenCourseWare.
A
C
Figure by MIT OpenCourseWare.
Strain
C'
Yielding occurs
• Elastic means reversible
Elastic Plastic
• Plastic means permanent
sy
Yield streng
gth
Stress
P
Unload
Strain
Stress
Modulus of Elasticity or
Youngs Modulus, E
Slope = modulus
of elasticity
Load
0
0
Strain
0.002
Figure by MIT OpenCourseWare.
Figure by MIT OpenCourseWare.
1. Initial
2. Small load
3. Unload
bonds
stretch
return to
initial

F
Elastic means reversible!
Hooke’s Law
σ = Eε
F
Linearelastic

Non-Linearelastic
1. Initial
2. Small load
bonds
t t h
stretch
& planes
shear
3. Unload
planes
still
sheared
elastic + plastic
F
Plastic means permanent!
plastic
p
F
linear
elastic
linear
elastic
plastic

Slipped
Slip
Plane
Do we break all bonds at once?
Recall force is d(energy)/da
Dislocation Motion Along Slip Plane
Image removed due to
copyright restrictions.
• Requires only one broken bond at
a given time
• Requires minimum energy
• Most plastic deformation occurs via
dislocation motion along slip plane
for metals and their alloys
alloys.
Ductility: degree of
plastic deformation at
fracture
B
Ductile
Stress
Brittle
B'
A
C
Strain
C'
Figure by MIT OpenCourseWare.
 A0  AF
%RA  
 A0

 100

 L f  L0 
 100
%EL  
 L0 
Brittle Materials have a
fracture strain of less
than approx. 5%
• Resilience is the capacity of
material to absorb energy as
elastic deformation and recover
the energy
sy
Stress
• Characterized by modulus of
y
resilience, Ur
U r    d
• If linear elastic region,
0.002
ey
Strain
Figure by MIT OpenCourseWare.
0
1
U r   y y
2
• These materials tend to have high
yield strength and low modulus of
elasticity
Fracture Toughness
Energy required to fail → area under stress-strain
curve. Material must be strong and ductile.
Engineering
tensile
stress, 
smaller toughness (ceramics)
l
t
h
larger
toughness
(metals, PMCs)
smaller toughnessi f
d
unreinforced
polymers
Engineering tensile strain, 
120
800
-100oC
80
60
400
40
0
25oC
0
0.1
0.2
20
0.3
0.4
o
Temperature ( F)
0
0.5
-400
0
800
400
1200
1600
Strain
E, yield
E
i ld and
d ttensile
il strength
t
th
decrease with temperature
Ductility usually increases
with temperature
60
400
Modulus of elasticity (GPa)
Figure by MIT OpenCourseWare.
70
Tungsten
50
300
40
Steel
200
30
20
Aluminum
100
10
0
-200
0
200
400
Temperature (oC)
Figure by MIT OpenCourseWare.
600
800
0
Modulus of elasticity (106 psi)
200
Stress (103 psi)
Stress (MPa)
600
Think of a rose or
banana dipped in
liliquid
id nitrogen
it
100
-200oC
400
Stress (MPa)
70
Tensile strength
450 MPa (65,000 psi)
60
103 psi
A
Yield strength
250 MPa (36,000psi)
MPa 40
300
200
200
40
30
30
20
100
0
0
20
10
100
0
0.10
0
50
Stress (103 psi)
500
10
0
0.005
0.20
Strain
0.30
0
0.40
Figure by MIT OpenCourseWare.
Do you notice
anything strange
about the stressstrain curve after a
material exceeds its
Tensile Strength?
σ = F/A0
ε = ∆L / L0
σT = F/Ai
εT = ln(Li / L0)
• In elastic region, change in crosssectional area and length are
negliligible
ibl
• As material deforms plastically, the
initial cross-sectional area and
length changes instantaneously
• True stress and true strain are
based on instantaneous crosssectional area and instantaneous
length
True
Stress
M'
M
Corrected
Engineering
If no volume
occurs during
deformation:
σT = σ(1+ε)
εT = ln(1+ε)
Strain
Figure by MIT OpenCourseWare.
After plastic deformation until necking:
σT = K εnT where K and n (strain hardening) are constants

isotropic material
y
x
 
z
z
σz
∆lz
2
l0x
∆lx
2
l0z
z
σz
y
x
• Elastic strain in compression
perpendicular to extension
caused by tensile stress
• Cannot be directly obtained from
stress-strain curve
• =0
0.26
26 to 0.35
0 35 for common
metal alloys
•  < 0.25 for ceramics
•  has a maximum value of 0.50
(no volume change)
∆lz/2
εz
=
2
l0z
∆lx/2
εx
=
2
l0x
Figure by MIT OpenCourseWare.
Shear Stress, 
Ft
Fs

A0
Fs
Shear Strain, 
A0
 = tan
Shear Modulus
Modulus, G
 = G
Fs
Area, A
where Fs is the shear load
and A0 is the initial crosssectional area parallel to
the loading direction
F
F
Ft
F
F
θ
F
Figure by MIT OpenCourseWare.
Tensile Stress and Strain
Shear Stress and Strain
Fs

A0
F

A0
Li  L0 L


L0
L0
 = tan
=G
G
σ = Eε
E = 2G((1+))
D
σyi
σyo
Stress
Unload
Reapply
load
Strain
Elastic strain
recovery
Figure by MIT OpenCourseWare.
• Up
pon release of load
during a stress-strain
test, some fraction of the
strain is recovered as
elastic strain
• Reapplication of stress
will traverse the same
curve
• Note the increase in the
yield stress – strain
hardening
• Ductility decreases