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Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
1
Outline
• Example• Torsion - introduction• Torsion test• Torsional Failure Modes• Plastic deformation• Dislocations – introduction• Edge Dislocation• Dislocation movements
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
- From observation, the angle of twist of the
shaft is proportional to the applied torque andto the shaft length.
T^ L
∝ ∝
φ φ
- When subjected to torsion, every cross-section
of a circular shaft remains plane andundistorted, because a circular shaft isaxisymmetric.
- Cross-sections of noncircular (non-
axisymmetric) shafts are distorted whensubjected to torsion.
Torsion -
Introduction
- Torsion is a variation of shear occurring in
machine axles, drive shafts and twist drills
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
3
Torsion -
Theory
J
r
M
T
r
J
M
T
→
M
T
= Torsional moment
τ
= shear stress
r = radial distance from centreJ = Polar moment of inertia
4 Often tests are done ontubular cross sections
12
c
J
π
=
r
Shear stress is zero at centre of bar increasing linearly to max at surface.
3
max
2
c M
T
π
τ
=
(^41)
(^42)
12
c
c
J
−
=
π
r
)
2 (
(^41)
42
2
max
c
c
c
M
T −
=
π
τ c
2
= outer radius
c
1
= inner radius
L
r
tan
Shear Strain:
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
Torsion Test
Not as common in testing as tensile test.
Torsion test samples (similar to tensile samples).
But also used on full sized parts such as shafts, axles, drills etc.
Troptometer
Torsion machine
Torsion machines use an electricalmotor and gear drive to apply a torqueto the specimen
The specimen is gripped on both ends,with one end remaining stationary andthe other rotated by the motor
- Troptometers are used to measure howmuch the specimen has been twisted.• Combining this twisting information withthe applied torque, we are able todetermine the mechanical properties of thespecimen.
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
5
During test, measure angle of twist,
(in radians) and plot against M
T
L
r
tan
Torsion Test
Shear stress
in plastic region can be
calculated using diagram of M
T
vs.
In elastic region, we canmeasure shear modulus, G:
T J
L
M
G
θ
=
J
r
M
G
T
γ
τ
=
=
→
/L)
Also the
ultimate torsional shear
strength
Modulus of Rupture
3
max
a
M
u
π
τ
3
CD
BC
a
a
If r = a, then:
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
- Elements with faces parallel and perpendicular
to the shaft axis are subjected to shear stressesonly. Normal stresses, shearing stresses or acombination of both may be found for otherorientations.
max
0
0
max
45
0
max
0
max
2
2
2
45
cos
2 o
τ
τ
σ
τ
τ
=
=
=
=
=
A
A
F A
A
A
F
a
is in ……… shear.
- Note that all stresses for elements
a
and
c
have
the same magnitude
c
is subjected to a tensile stress on
two faces and compressive stress on the othertwo.
- Consider an element at 45
o
to the shaft axis,
Torsional Failure Modes
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
7
- Ductile materials generally fail in
shear. Brittle materials are weakerin tension than shear.
- When subjected to torsion, a ductile
specimen breaks along a plane of maximum shear
, i.e., a plane
perpendicular to the shaft axis.
- When subjected to torsion, a brittle
specimen breaks along planes………………. to the direction inwhich tension is a maximum, i.e.,along surfaces at 45
o
to the shaft
axis.
Torsional Failure Modes
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
Why metals could be plastically deformed?
Why the plastic deformation properties could be changed
to a very large degree by forging without changing thechemical composition?
Why plastic deformation occurs at stresses that are much
smaller than the theoretical strength of perfect crystals?
Plastic deformation – the force to break all bonds in the
slip plane is much higher than the force needed to cause thedeformation. Why?
These questions can be answered based on the idea proposed in1934 by Taylor, Orowan and Polyani:
Plastic deformation is
due to the motion of a large number of ……………..
Plastic Deformation
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
13
dislocations are
intrinsic
defects
like vacancies
dislocation density
is the total
dislocation length/unit volume
units: mm/mm
3
or mm
annealed metal: 10
5
6
mm
deformed: 10
9
10
mm
atoms above slip plane are incompression
atoms below slip plane are intension
creates a
strain field
around the
dislocation
dislocations contain
stored energy
Regions of compression (dark) and
tension (colored) located around an edge
dislocation.
Dislocations
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
dislocations ………. duringplastic deformation
dislocations can either
repel
or
attract
one another
depends on orientation or
sign
(positive or negative)
important since deformationincreases dislocation density →
work hardening
this is a strengtheningmechanism
+ve
+ve
-ve
→
0
Dislocation Interaction
repulsion
attractive
Two extra half-planes will align and become a complete plane
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
15
When compared to experimental shear yield strengths, commonmetals are 1000 to 10,000 times weaker than theory predicts. Theoretical Shear Strength,
τ
TH
G/
π
to
G/30 depending on method.
Theoretical vs. Experimental Mech properties
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
- The combination of C-P plane (the slip plane) and C-P direction(the slip direction) is called a
…………..
.
Recall: SLIP SYSTEMS DEPEND ONTHE CRYSTAL STRUCTUREOF THE MATERIAL!
Under applied shear stress, dislocations can move by breaking bonds CONSECUTIVELY
(rather than simultaneously).
Requires less energy, (reason why expt. Shear strength is lower).Deformation by dislocations movement is called
SLIP.
Movement of Dislocations
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
17
The more slip systems available, the easier it is for dislocations tomove, which is why (on the average) FCC and BCC metals aremore ductile than HCP metals.
number of slip systems
with temperature e.g. HCP
metals
more ductile at high temperature
Slip Systems
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
Edge dislocations can move “out” of theslip plane by non-conservative motion.Requires diffusion of vacancies to bottomof extra 1/2 plane thus dislocationCLIMBS to a higher SLIP plane.Thermally activated process (diffusion +number of vacancies) so usually onlyimportant at high temps. > 0.5 T
m
(K)
Dislocation climb involvingvacancy ( ) diffusion to edgedislocation allowing its movementto climb from plane A to plane B.
Movement of Dislocations
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
19
when atoms leave the
dislocation line to create
interstitials or to fill vacancies
Movement of Dislocations
Can dislocations climb?
or when atoms are attached to
the dislocation line by
creating vacancies or
eliminating interstitials
Dr. M. Medraj
Mech. Eng. Dept. - Concordia University
MECH 321 lecture 3/
If we have a material
without
dislocations (i.e. SLIP
cannot occur)!!
Is the strength closer to the theoretical value?
Quite close; onlyx10 not
x10,000.
Strength of a perfect Crystals
