Upload
anees-ur-rahman
View
2.070
Download
0
Embed Size (px)
Citation preview
4/29/2011
1
Mechanics of Flexible Materials
By Hammad Mohsin
1
Course Outline A)FUNDAMENTALS & POLYMERS
Module 1 Introduction to Mechanics of Materials
• Role of Mechanics of Materials in EngineeringRole 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 gMaterials, Isotropic and An‐isotropic Materials, Modulus of Elasticity, Modulus of Rigidity, Elastic andPlastic Behavior of Materials, Non Linear Elasticity, Linear Elasticity,
2
4/29/2011
2
• 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 pContributions to Rubber Elasticity: Force‐Temperature Relations , Direct Determination of Molecular Dimensions
3
Module 4 Strength of Elastomers
• Initiation of Fracture, Threshold Strengths and Extensibilities, Fracture Under Multiaxial Stresses , Crack Propagation, Tensile Rupture, Repeated Stressing Mechanical FatigueRepeated 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.
4
4/29/2011
3
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
6
4/29/2011
4
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)
7
Assessment
• Quizzes: 10%
• Class participation & Discussion 10%
• Assignments : 10%
• Midterm: 30%
• Final: 40%
8
4/29/2011
5
Cause & effect model
9
Material Properties
1. Elastic a. Elastic Behavior causes aa. Elastic Behavior causes a material to return to its original shape after being deformed.
10
4/29/2011
6
b. Completely elastic behavior
Force
(F)
kxF =k is called the elastic modulus
Distance (x)11
2. Viscous
a. Viscous behavior is related to the rate of d f tideformation.
⎟⎠⎞
⎜⎝⎛ΔΔ
η=txF
Viscosity Rate of deformation
12
4/29/2011
7
f t
force, F
fast
slow
distance, x
13
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
F viscous
fast
F
x
elastic
viscousslow
14
4/29/2011
8
B. Internal Structure 1. Chemical Composition
Sequence and kind of atoms in structureSequence and kind of atoms in structure
2. CrystallinityPolymer chains or sections packed together
3. OrientationAlignment of chains along fiber axis
C Thermal PropertiesC. Thermal Properties Melting Temperature2. Glass Transition Temperature
Most polymers are thermoplastic – they soften before melting
15
D. Physical Properties Breaking Strength
Force required to break a fiberForce required to break a fiber
2. Breaking ElongationAmount of stretch before breaking
3. ModulusResistance to deformation
4. ToughnessAmount of energy absorbed
5. ElasticityAbility to recover after being deformed
16
4/29/2011
9
17
Structural factors=> Mechanical Behavior
l. Molecular weight2 Cross‐linking and branching2. Cross‐linking and branching3. Crystallinity and crystal morphology4. Copolymerization (random, block, and graft)5. Plasticization6. Molecular orientation7 Fillers7. Fillers8. Blending9. Phase separation and orientation in blocks, grafts, and blends
18
4/29/2011
10
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 history6. Heat treatments or thermal history
7. Nature of surrounding atmosphere, especially moisture content
19
5 assumptions ‐> Mechanical Behavior
1) Linearity: Two types of linearity are normally assumed: A) Material linearity (Hookean stress‐assumed: A) Material linearity (Hookean stressstrain 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
4/29/2011
11
5) Isotropic: Materials which have the same h i l ti i ll di ti tmechanical properties in all directions at an
arbitrary point or materials whose properties are invariant upon rotation of axes at a point. Amorphous materials are isotropic.
21
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.
22
4/29/2011
12
• The engineering (average) stress can be l l t d b di idi th li d t ilcalculated by dividing the applied tensile
force, P, (normal to the cross section) by the area of the original cross sectional area A0 as follows,
Stress
23
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
24
4/29/2011
13
True Stresses and Strain
• True stress and strain are calculated using the i t t (d f d t ti l l d)instantaneous (deformed at a particular load) values of the cross‐sectional area, A, and the length of the rectangle, L,
25
Young Modulus
• Young’s modulus, E, may be determined from th l f th t t i bthe slope of the stress‐strain curve or by dividing stress by strain,
26
4/29/2011
14
• 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
4/29/2011
15
• 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
30
4/29/2011
16
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 psaid 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
31
• That is, a line parallel to the initial tangent to th t t i di i d tthe stress‐strain diagram is drawn to pass 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.
32
4/29/2011
17
the stress is nearly
linear with strain until it reaches the
upper yield point
stress which is also
known as the
elastic‐plastic
tensile instability
point. p
At this point the load (or stress) decreases as the deformation continues to increase. That is, less load is necessary to sustain continued deformation.
33
The region between the lower yield point
and the maximum stress is a region of strain hardening, Poly‐Carbonate shows the similar behavior
34
4/29/2011
18
• 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”,“ k” “d ” ith t t h th“neck” or “draw” without rupture when the yield point has been exceeded has led to the concepts of plastic, limit or ultimate design.
36
4/29/2011
19
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
4/29/2011
20
Mathematical Definitions
Definition of a Continuum: A basic assumption f l t lid h i i th tof elementary solid mechanics is that a
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,
39
Mathematical/ Physical Def. of Normal and Shear Stress
• Consider a body in ilib i d thequilibrium under the
action of external forces
• F1, F2, F3, F4 = Fi as shown in Fig.g
40
4/29/2011
21
• If a cutting plane is d th h thpassed through the
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 i i h i ll d fi dpoint p is mathematically defined as,
42
4/29/2011
22
Alternatively, the resultant force, ∆Fr, at point p can be divided by the area, ∆ A,
and the limit taken toand 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.
43
• If a pair of cutting planes a differential distance apart are passed through
• the body parallel to each of the three coordinate y pplanes, 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.
44
4/29/2011
23
• The internal stress state is then represented by three stress components oncomponents on each coordinate plane as shown in Fig. Therefore at any point in a body h ill b ithere will be nine stress components. These are often identified in matrix form such that, 45
Using equilibrium, it is easy to show that the t t i i t istress matrix is symmetric,
or
• leaving only six independent stresses existing at a material point. 46
4/29/2011
24
Physical and Mathematical Def. of Normal & Shear Strain
• If there is stress acting on the body. For lexample
47
• 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 directionthe y direction.
48
4/29/2011
25
• 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:‐
49
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
50
4/29/2011
26
• 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 , pmatrix is symmetric or that,
• Hence there are only six independent strains. 51