September 9, 2011Page 1
MECH 321 Solid Mechanics II
Week 1, Lecture 2
Review of CIVL 220 (cont’d)
September 9, 2011Page 2
The Stress–Strain Diagram
Conventional Stress
Conventional Stress
–
–
Strain Diagram
Strain Diagram
•Nominal or engineering stress is obtained by dividing
the applied load P by the specimen’s original cross
sectional area.
•Nominal or engineering strain is obtained by dividing the
change in the specimen’s gauge length by the specimen’s
original gauge length.
0
A
P
=
σ
0
L
δ
ε
=
September 9, 2011Page 3
The Stress–Strain Diagram
Conventional Stress
Conventional Stress
–
–
Strain Diagram
Strain Diagram
•Elastic Behaviour
Stress is proportional to the strain.
Material is said to be
linearly elastic.
{
Proportional, linearly
elastic range
September 9, 2011Page 4
The Stress–Strain Diagram
•Yielding
Increase in stress above elastic limit will cause material
to deform permanently.
September 9, 2011Page 5
The Stress–Strain Diagram
Strain Hardening.
After yielding a further load will reaches a ultimate stress.
September 9, 2011Page 6
The Stress–Strain Diagram
•Necking
At ultimate stress, crosssectional area begins to decrease
in a localized region of the specimen.
–Specimen breaks at the fracture stress.
September 9, 2011Page 7
True Stress–Strain
The values of stress and strain computed from these
measurements are called true stress
and true strain
.
Use this diagram since most engineering design is done within
the elastic range.
September 9, 2011Page 8
Stress vs. Strain –Ductile Material
Ductile material –Low Carbon Steel.
σu
= Ultimate Strength
(450 MPa)
σY
= Yield Strength
(265 MPa)
Strain hardening 
stress increases
Necking –stress decreases
September 9, 2011Page 9
Ductile Material Aluminum
Length increases linearly at a slow rate.
σY
–yield stress point occurs after a large amount of
deformation largely due to shear stresses (internal slippage
along grains).
Rupture
σB
σY
σu
(450 MPa)
300
0.002
Necking
0.2
Aluminum Alloy
ε
September 9, 2011Page 10
Brittle Material
σu
= σB
Rupture
ε
Cast Iron
September 9, 2011Page 11
Hooke’s Law
σ= E εor E = σ/ ε
E= modulus of elasticity (same units as stress, σ)
σ
ε
Tempered steel
HighC steel
LowC steel
Iron
Different: σY, σU, σB, εB
Same: E
September 9, 2011Page 12
Hooke’s Law
For the straight section of stressstrain (σ–ε) curve, stress
is directly proportional to strain.
Proportional Limit: limit of linear stressstrain (σ–ε)
relationship.
Alloys of the same material will have different properties
(strength, ductility, corrosion resistance, etc), but will often
have the same E. Same stiffness within their linear range.
σ= E εor E = σ/ ε
September 9, 2011Page 13
Hooke’s Law
Strain Hardening
Strain Hardening
•When ductile material is loaded into the plastic region and then
unloaded, elastic strain is recovered.
•The plastic strain remainsand material is subjected to a permanent
set.
September 9, 2011Page 14
Strain Energy (new)
•When material is deformed by external loading, it will store
energy internally throughout its volume.
•Energy is related to the strains called strain energy.
Modulus of Resilience
Modulus of Resilience
•When stress reaches the
proportional limit (σpl), the
strainenergy density is the
modulus of resilience, ur.
E
u
pl
plplr
2
2
1
2
1
σ
εσ
==
September 9, 2011Page 15
Strain Energy
Modulus of Toughness
Modulus of Toughness
•Modulus of toughness, ut, represents the entire area
under the stress–strain diagram.
•It indicates the strainenergy density of the material just
before it fractures.
September 9, 2011Page 16
Example 3.2
The stress–strain diagram for an aluminum alloy that is used for making aircraft
parts is shown. When material is stressed to 600 MPa, find the permanent strain
that remains in the specimen when load is released. Also, compute the modulus of
resilience both before and after the load application.
September 9, 2011Page 17
Poisson’s Ratio
•Poisson’s ratio, v(nu), states that in the elastic range, the
ratio of these strains is a constantsince the deformations
are proportional.
•Negative sign since longitudinal elongation (positive strain)
causes lateral contraction (negative strain), and vice versa.
long
lat
v
ε
ε
−=
Poisson’s ratio is dimensionless.
Typical values are 1/3 or 1/4.
September 9, 2011Page 18
Poisson’s Ratio
L
long
δ
ε
=
r
lat
δ
ε
′
=
long
lat
ε
ε
υ
=
September 9, 2011Page 19
Poisson’s Ratio
Unloaded Loaded
September 9, 2011Page 20
Example 3.4
A bar made of A36 steel has the dimensions shown. If an axial force of is applied to
the bar, determine the change in its length and the change in the dimensions of its
cross section after applying the load. The material behaves elastically.
September 9, 2011Page 21
The Shear Stress–Strain Diagram
•For pure shear, equilibrium
requires equal shear stresses
on each face of the element.
•When material is
homogeneous and isotropic,
shear stress will distort the
element uniformly.
September 9, 2011Page 22
The Shear Stress–Strain Diagram
•For most engineering materials the elastic
behaviour is linear, so Hooke’s Law for shear
applies.
•3 material constants, E, and G are actually
related by the equation
γ
τ
G
=
G= shear modulus of elasticity
or the modulus of rigidity
()
v
E
G
+
=
12
September 9, 2011Page 23
Example 3.5
A specimen of titanium alloy is tested in torsion and the shear stress–strain diagram
is shown. Find the shear modulus G, the proportional limit, and the ultimate shear
stress. Also, findthe maximum distance d that the top of a block of this material
could be displaced horizontally if the material behaves elastically when acted upon
by a shear force V. What is the magnitude of V necessary to cause this
displacement?
September 9, 2011Page 24
Next Time
Axial Loading (review) and the
Principle of Superposition
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