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Mechanics of Flexible Materials
By Hammad Mohsin
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Course Outline
A)FUNDAMENTALS & POLYMERS
Module 1 Introduction to Mechanics of Materials
• Role of Mechanics of Materials in Engineering
Role of Mechanics of Materials in Engineering,
Stresses and Deformations, True Stress and True
Strain
Module 2 Study of Stress and Strain
• Stress ‐ Strain Diagrams of Ductile and Brittle
g
Materials, Isotropic and An‐isotropic Materials,
Modulus of Elasticity, Modulus of Rigidity, Elastic and
Plastic Behavior of Materials, Non Linear Elasticity,
Linear Elasticity, 2
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• Stress and Strain in Changed Thermal Conditions,
Repeated Loading, Bending of Elasto‐plastic
Materials, Analysis of Stresses and Deformations
Module 3 Molecular basis of Rubberlike elasticity
• Structure of a Typical Network, Elementary
Molecular Theories, More Advanced Molecular
Theories Phenomenological Theories and
Molecular Structure, Swelling of Networks and
Responsive Gels Enthalpic and Entropic
p p p
Contributions to Rubber Elasticity: Force‐
Temperature Relations , Direct Determination of
Molecular Dimensions
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Module 4 Strength of Elastomers
• Initiation of Fracture, Threshold Strengths and
Extensibilities, Fracture Under Multiaxial
Stresses , Crack Propagation, Tensile Rupture,
Repeated Stressing: Mechanical Fatigue,
Repeated Stressing Mechanical Fatigue
Surface Cracking by Ozone, Abrasive Wear.
Module 5 Failure Prevention
• Analysis of polymer product failure Design
Analysis of polymer product failure, Design
aids for preventing brittle failure, Defect
analysis HDPE pipe durability.
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B) TEXTILE MATERIALS
Module 6 Mechanical Properties of Textile Fibres
• Tensile Recovery, Elastic Performance Coefficient in
Tension, Inter Fibre Stress and its Transmission,
, ,
Stress analysis of stable fibre, filaments, influence
of twist on yarn modulus
• Plasticity of textile fibers based on effect of load,
time, temperature superposition.
Module 7 Mechanics of Yarns:
• Mechanics of Bent Yarns, Flexural Rigidity, Fabric
Wrinkling, Stiffness in Textile Fabrics. Creasing and
Crease‐proofing of Textiles 5
• Module 8 Compression of Textile Materials
• Study of Resilience, Friction between Single
Fibres, Friction in Plied Yarns
• Module 9 Mechanical Properties of Non
Wovens and composite materials
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Books
• Neilsen L., Landel R.,“Mechanical Properties of Polymers
and Composites” (1994)
• Moalli J “Plastic Failure: Analysis and Prevention” (2001)
Moalli J Plastic Failure: Analysis and Prevention (2001)
• Mark J, Erman B., Elrich F “Science and Technology of
Rubbers” (2005)
• Ferdinand P Beer, E Russell Jhonston Jr., Jhon T Dewolf
“Mechanics of Materials” (2004)
• Jinlian Hu “Structure and Mechanics of Woven Fabrics”
Jinlian Hu Structure and Mechanics of Woven Fabrics
(2004)
• AE Bogdanovich, C M Pastore “Mechanics of Textile and
Laminated Composites” (1996)
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Assessment
• Quizzes: 10%
• Class participation & Discussion 10%
• Assignments : 10%
• Midterm: 30%
• Final: 40%
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Cause
& effect
model
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Material Properties
1. Elastic
a. Elastic Behavior causes a
a. Elastic Behavior causes a
material to return to its
original shape after being
deformed.
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b. Completely elastic behavior
F = kx
Force
k is called the elastic
modulus
(F)
Distance (x)
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2. Viscous
a. Viscous behavior is related to the rate of
deformation.
d f ti
⎛ Δx ⎞
F = η⎜ ⎟
⎝ Δt ⎠
Viscosity Rate of deformation
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fast
f t
force, F
slow
distance, x
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3. Viscoelastic
a. Fibers exhibit viscoelastic behavior
b. force required to deform a material
dependents amount of deformation and rate
at which the material is deformed
fast
F viscous
slow
elastic
x 14
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B. Internal Structure
1. Chemical Composition
Sequence and kind of atoms in structure
Sequence and kind of atoms in structure
2. Crystallinity
Polymer chains or sections packed together
3. Orientation
Alignment of chains along fiber axis
C. Thermal Properties
C Thermal Properties
Melting Temperature
2. Glass Transition Temperature
Most polymers are thermoplastic – they soften
before melting
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D. Physical Properties
Breaking Strength
Force required to break a fiber
Force required to break a fiber
2. Breaking Elongation
Amount of stretch before breaking
3. Modulus
Resistance to deformation
4. Toughness
Amount of energy absorbed
5. Elasticity
Ability to recover after being deformed
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Structural factors=> Mechanical
Behavior
l. Molecular weight
2. Cross‐linking and branching
2 Cross‐linking and branching
3. Crystallinity and crystal morphology
4. Copolymerization (random, block, and graft)
5. Plasticization
6. Molecular orientation
7. Fillers
7 Fillers
8. Blending
9. Phase separation and orientation in blocks,
grafts, and blends
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External Factors Mechanical
Properties
1. Temperature
2. Time, frequency, rate of stressing or straining
3. Pressure
4. Stress and strain amplitude
5. Type of deformation (shear, tensile, biaxial, e tc. )
6. Heat treatments or thermal history
6 Heat treatments or thermal history
7. Nature of surrounding atmosphere, especially
moisture content
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5 assumptions ‐> Mechanical Behavior
1) Linearity: Two types of linearity are normally
assumed: A) Material linearity (Hookean stress
assumed: A) Material linearity (Hookean stress‐
strain behavior) or linear relation between stress
and strain; B) Geometric linearity or small strains
and deformation.
2) Elastic: Deformations due to external loads are
completely and instantaneously reversible upon
load removal.
load removal.
3) Continuum: Matter is continuously distributed
for all size scales, i.e. there are no holes or voids.
4) Homogeneous: Material properties are the same
at every point or material properties are invariant
upon translation. 20
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5) Isotropic: Materials which have the same
mechanical properties in all directions at an
h i l ti i ll di ti t
arbitrary point or materials whose properties
are invariant upon rotation of axes at a point.
Amorphous materials are isotropic.
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Stress‐ Strain > Definations
• Dog Bone is used and material properties such as
• 1) Young’s modulus, 2) Poisson’s ratio, 3) failure (yield) stress and
) g , ) , ) (y )
strain.
• The specimen may be cut from a thin flat plate of constant
thickness or may be machined from a cylindrical bar.
• The “dogbone” shape is to avoid stress concentrations from
loading machine connections and to insure a homogeneous state
of stress and strain within the measurement region.
• The term homogeneous here indicates a uniform state of stress
or strain over the measurement region, i.e. the throat or reduced
central portion of the specimen.
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• The engineering (average) stress can be
calculated by dividing the applied tensile
l l t d b di idi th li d t il
force, P, (normal to the cross section) by the
area of the original cross sectional area A0 as
follows,
Stress
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Strain
• The engineering (average) strain in the direction
of the tensile load can be found by dividing the
change in length, ∆L, of the inscribed rectangle by
the original length L0,
• The term lambda in the above equation is called
the extension ratio and is sometimes used for
large deformations e.g., Low modulus rubber
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True Stresses and Strain
• True stress and strain are calculated using the
instantaneous (deformed at a particular load)
i t t (d f d t ti l l d)
values of the cross‐sectional area, A, and the
length of the rectangle, L,
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Young Modulus
• Young’s modulus, E, may be determined from
the slope of the stress‐strain curve or by
th l f th t t i b
dividing stress by strain,
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• the axial deformation over length L0 is,
• Poisson’s ratio, , is defined as the absolute
value of the ratio of strain transverse, єy, to
the load direction to the strain in the load
direction, є x ,
Where strain transverse
‐ve for Applied tensile load, 27
Shear
• L = length of the cylinder,
• T = applied torque,
• r = radial distance,
• J = polar second moment of area
• G = shear modulus.
• =shear stress, = angle of twist,
• =shear strain, 28
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• The shear modulus, G, is the slope of the shear
stress‐strain curve and may be found from,
where the shear strain is easily found by measuring
only the angular rotation, , in a given length, L.
The shear modulus is related to Young’s modulus
• As Poisson’s ratio, , varies between 0.3 and 0.5 for
most materials, the shear modulus is often
approximated by, G ~ E/3. 29
Typical Stress Strain Properties
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Yield point
• if the stress exceeds the proportional limit a
residual or permanent deformation may remain
when the specimen is unloaded and the material is
p
said to have “yielded”.
• The exact yield point may not be the same as the
proportional limit and if this is the case the location
is difficult to determine.
• As a result, an arbitrary “0.2% offset” procedure is
often used to determine the yield point in metals
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• That is, a line parallel to the initial tangent to
the stress‐strain diagram is drawn to pass
th t t i di i d t
through a strain of 0.002 in./in.
• The yield point is then defined as the point C
of intersection of this line and the stress‐strain
diagram.
g
• This procedure can be used for polymers but
the offset must be much larger than 0.2%
definition used for metals.
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• If the strain scale of Fig. (a) is expanded as
illustrated in Fig. (b),
• the stress‐strain diagram of mild steel is
approximated by two straight lines;
• i) for the linear elastic portion and
• ii) is horizontal at a stress level of the lower
yield point. 35
• This characteristic of mild steel to “flow”,
“neck” “d
“ k” or “draw” without rupture when the
” ith t t h th
yield point has been exceeded has led to the
concepts of plastic, limit or ultimate design.
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Idealized Stress‐ Strain
• a linear elastic perfectly brittle material is assumed
to have a stress‐strain diagram fig (a)
• a perfectly elastic‐plastic material with the stress‐
strain diagram Fig (b) mild steel or Poly C 37
• Metals (and polymers) often have nonlinear
• stress‐strain behavior as shown in Fig. (a). These
are sometimes modeled with a bilinear diagram
as shown in Fig. (b) and are referred to as a
perfectly linear elastic strain hardening material.38
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Mathematical Definitions
Definition of a Continuum: A basic assumption
of elementary solid mechanics is that a
f l t lid h i i th t
material can be approximated as a continuum.
That is, the material (of mass M) is
continuously distributed over an arbitrarily
small volume, V, such that,
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Mathematical/ Physical Def. of
Normal and Shear Stress
• Consider a body in
equilibrium under the
ilib i d th
action of external
forces
• F1, F2, F3, F4 = Fi as
shown in Fig.g
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• If a cutting plane is
passed through the
d th h th
body as
• shown in Fig,
equilibrium is
maintained on the
remaining portion
by internal forces
distributed over the
surface S. 41
• At any arbitrary point p,
• the incremental resultant force, ∆Fr, on the
cut surface can be broken up into a normal
force in the direction of the normal, n, to
surface S and
• a tangential force parallel to surface S.
• The normal stress and the shear stress at
point p is mathematically defined as,
i i h i ll d fi d
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Alternatively, the resultant
force, ∆Fr, at point p can
be divided by the area, ∆
A,
and the limit taken to
and the limit taken to
obtain the stress resultant
σr as shown in Fig. Normal
and tangential
components of this stress
resultant will then be the
normal stress σn and
shear stress τs at point p
on the area A.
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• If a pair of cutting planes a differential distance
apart are passed through
• the body parallel to each of the three coordinate
yp
planes, a cube will be identified.
• Each plane will have normal and tangential
components of the stress resultants.
• The tangential or shear stress resultant on each
plane can further be represented by two
components in the coordinate directions.
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• The internal stress
state is then
represented by
three stress
components on
components on
each coordinate
plane as shown in
Fig. Therefore at
any point in a body
there will be nine
h ill b i
stress components.
These are often
identified in matrix
form such that, 45
Using equilibrium, it is easy to show that the
stress matrix is symmetric,
t ti i ti
or
• leaving only six independent stresses existing at
a material point. 46
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Physical and Mathematical Def. of
Normal & Shear Strain
• If there is stress acting on the body. For
example
l
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• Both shearing and normal deformation may occur
with displacements.
• u is the displacement component in the x
direction and v is the displacement component in
the y direction.
the y direction
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• The unit change in the x dimension will be the
strain єxx and is given by,
• If we apply similarly for y and z direction, and
assume that change of angle is very small then
∆u will be ignored. Then in 3 co‐ordinate
system normal strains are defined as:‐
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Shear strains
• Shear strains are defined as the distortion of
the original 90º angle at the origin or the sum
of the angles Ѳ1 + Ѳ2. That is, again using the
small deformation assumption,
• After solving in all 3 directions shear strain is
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• Like stresses, nine components of strain exist
at a point and these can be represented in
matrix form as,
• Again, it is possible to show that the strain
g , p
matrix is symmetric or that,
• Hence there are only six independent strains. 51
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