curso nptel - mechanics of laminated composite structure

Mechanics of Laminated Composite Structures

Nachiketa Tiwari

Indian Institute of Technology KanpurIndian Institute of Technology Kanpur

Lecture 1

IntroductionIntroduction

Lecture OverviewLecture Overview

• What are “composites”?What are composites ?

d f li i• Importance and areas of application

• Classification

• Advantages of fiber‐reinforced composites

What are “composites”?

• Composite: Two or more chemically different tit t bi d i ll tconstituents combined macroscopically to

yield a useful material

• Examples of naturally occurring composites– Wood: Cellulose fibers bound by lignin matrix

– Bone: Stiff mineral “fibers” in a soft organic matrix permeated with holes filled with liquids

– Granite: Granular composite of quartz, feldspar, and mica

What are “composites”?What are composites ?

• Some examples of man‐made compositesSome examples of man made composites– Concrete: Particulate composite of aggregates (limestone or granite) sand cement and water(limestone or granite), sand, cement and water

– Plywood: Several layers of wood veneer glued togethertogether

– Fiberglass: Plastic matrix reinforced by glass fibers

– Cemets: Ceramic and metal compositesCemets: Ceramic and metal composites

– Fibrous composites: Variety of fibers (glass, kevlar, graphite, nylon, etc.) bound together by agraphite, nylon, etc.) bound together by a polymeric matrix

These are not composites!These are not composites!

• Plastics: Even though they may have severalPlastics: Even though they may have several “fillers”, their presence does not alter the physical properties significantlyphysical properties significantly.

• Alloys: Here the alloy is not macroscopically heterogeneous especially from in terms ofheterogeneous, especially from in terms of physical properties.

M l i h i i i Th f• Metals with impurities: The presence of impurities does not significantly alter physical

i f h lproperties of the metal.

Where are composites used?Where are composites used?

• Automotive industry: Lighter stronger wearAutomotive industry: Lighter, stronger, wear resistance, rust‐free, aesthetics– Car body– Car body

– Brake pads

Drive shafts– Drive shafts

– Fuel tanks

H d– Hoods

– Spoilers


• Aerospace: Lighter stronger temperatureAerospace: Lighter, stronger, temperature resistance, smart structures, wear resistance– Aircraft: Nose doors struts trunnion fairings– Aircraft: Nose, doors, struts, trunnion, fairings, cowlings, ailerons, outboard and inboard flaps, stabilizers, elevators, rudders, fin tips, spoilers, , , , p , p ,edges

– Rockets & missiles: Nose, body, pressure tanks, frame, fuel tanks, turbo‐motor stators, etc.

– Satellites: Antennae, frames, structural parts


• Sports: Lighter stronger toughnessSports: Lighter, stronger, toughness, aesthetics, damping– Tennis– Tennis

– Bicycles

Badminton– Badminton

– Boats

H k– Hockey

– Golfing

l– Motorcycles …


• Transportation & Infrastructure: LighterTransportation & Infrastructure: Lighter, stronger, toughness, damping– Railway coaches– Railway coaches

– Bridges

Ships and boats– Ships and boats

– Dams

T k b di d fl– Truck bodies and floors

– RV bodies


• And many more industry sectorsAnd many more industry sectors– Biomedical industry

Consumer goods– Consumer goods

– Agricultural equipment

H hi– Heavy machinery

– Computers

H l h– Healthcare

Classification of CompositesClassification of Composites

• Chart ‐ BDA p4Chart BDA p4


• Particulate composites have one or moreParticulate composites have one or more material particles suspended in binding matrix. A particle by definition is not “long” vis‐à‐vis its own dimensions.

• Fibrous composites have fibers of reinforcing material(s) suspended in binding matrix. Unlike particles, a fiber has high length‐to‐diameter ration, and furthers, its diameter

b l i l imay be close to its crystal size.


• Particulate composites:– Random orientation: Orientation of particle is randomly distributed in all

directions (ex: concrete)

– Preferred orientation: Particle orientation is aligned to specific directions– Preferred orientation: Particle orientation is aligned to specific directions (ex: extruded plastics with reinforcement particles)

Note: Particulate composites in general do not have high fractureNote: Particulate composites in general do not have high fracture resistance unlike fibrous composites. Particles tend to increase stiffness of the materials, but have not so much of an influence on strength. In several cases, particulate composites are used to enhance performance at high temperatures In other case these compositesperformance at high temperatures. In other case, these composites are used to increase thermal and electrical properties. In cemets, which are ceramic‐metal composites, the aim is to have high surface hardness so that the material can be used to cut materials at high speeds, or resist wear.

Classification of CompositesClassification of Composites• Fibrous Composites: In general, materials tend to have much better thermo‐

mechanical properties at small scale than at macro scale This is shown inmechanical properties at small scale than at macro‐scale. This is shown in following table.

Material Fiber Tensile Strength (GPa) Bulk Tensile strength( GPa)g ( ) g ( )Glass 3.5 to 4.6 0.7 ‐ 2.1Tungsten 4.2 1.1 ‐ 4.1Beryllium 1.3 0.7Graphite 2 1 to 2 2 5 Very lowGraphite 2.1 to 2.2.5 Very low

At macro‐scale, imperfections in material have an accumulated effect of , pdegrading bulk mechanical properties of materials significantly. This is one reason why fibrous composites have been developed to harness micro‐scale properties of materials at larger scales. Man‐made fibers, have almost no flaws in directions perpendicular to their length, and hence are able to bear p p g ,large loads compared to bulk materials.


• Fibrous Composites:– Single‐layer: These are actually made of several layers of fibers,

but in the same direction. Hence they are considered as “single‐layer” composites. These composites may either have continuous and long fibers (filament wound shells) or they may be reinforcedand long fibers (filament wound shells), or they may be reinforced by discontinuous and short‐fibers (fiber glass bodies of cars). Further, the continuous long fiber composites may either have unidirectional reinforcement, or bi‐directional reinforcement. i ll i f i di i d h fibFinally, reinforcements in discontinuous and short‐fiber composites may either be randomly oriented or in preferred directions.

– Multi‐layer: Here, reinforcement is provided, layer‐by layer in different directions. If constituent material in all layers is same, then it is simply a laminate. Hybrid laminates have more than one p y yconstituent materials in the composite structure.

Advantages of CompositesAdvantages of Composites• Composites are engineered materials. Hence, we

i h ifi ll dcan engineer them specifically to meet our needs on a case‐to‐case basis. In general, following properties can be improved by using composite materialscan be improved by using composite materials.– Strength Electrical conductivity

M d l Th l d ti it– Modulus Thermal conductivity

– Weight Behavior at extreme temps.

Fatigue Acoustical insulation– Fatigue Acoustical insulation

– Vibration damping Aesthetics

Resistance to wear Resistance to corrosion– Resistance to wear Resistance to corrosion

Limitations of CompositesLimitations of Composites• Like all things in nature, composites materials have their limitations as well. Some of the important ones are:limitations as well. Some of the important ones are:– Anisotropy: A large number of composites have direction dependent material properties. This makes them more difficult to understand, analyze and engineer, vis‐à‐vis , y g ,isotropic materials.

– Non‐homogenous: Further, these materials, by definition are not homogenous. Hence their material properties vary from g p p ypoint‐to‐point. This too makes them difficult to model, and analyze.

– Costly: Composite materials are in general, expensive. Thus, they are used only in applications where their benefits outweigh the costs.

– Difficult to fabricate: Further, structures these materials are diffi l l d i f b idifficult, slow and expensive to fabricate.

What you learnt in this lecture?What you learnt in this lecture?



• Classification



Nachiketa Tiwari


Lecture 2





• Classification



• Particulate composites have one or moreParticulate composites have one or more material particles suspended in binding matrix. A particle by definition is not “long” vis‐à‐vis its own dimensions.

• Fibrous composites have fibers of reinforcing material(s) suspended in binding matrix. Unlike particles, a fiber has high length‐to‐diameter ration, and furthers, its diameter

b l i l imay be close to its crystal size.


• Particulate composites:– Random orientation: Orientation of particle is randomly distributed in all

directions (ex: concrete)

– Preferred orientation: Particle orientation is aligned to specific directions– Preferred orientation: Particle orientation is aligned to specific directions (ex: extruded plastics with reinforcement particles)

Note: Particulate composites in general do not have high fractureNote: Particulate composites in general do not have high fracture resistance unlike fibrous composites. Particles tend to increase stiffness of the materials, but have not so much of an influence on strength. In several cases, particulate composites are used to enhance performance at high temperatures In other case these compositesperformance at high temperatures. In other case, these composites are used to increase thermal and electrical properties. In cemets, which are ceramic‐metal composites, the aim is to have high surface hardness so that the material can be used to cut materials at high speeds, or resist wear.

Classification of CompositesClassification of Composites• Fibrous Composites: In general, materials tend to have much better thermo‐

mechanical properties at small scale than at macro scale This is shown inmechanical properties at small scale than at macro‐scale. This is shown in following table.

Material Fiber Tensile Strength (GPa) Bulk Tensile strength( GPa)g ( ) g ( )Glass 3.5 to 4.6 0.7 ‐ 2.1Tungsten 4.2 1.1 ‐ 4.1Beryllium 1.3 0.7Graphite 2 1 to 2 2 5 Very lowGraphite 2.1 to 2.2.5 Very low

At macro‐scale, imperfections in material have an accumulated effect of , pdegrading bulk mechanical properties of materials significantly. This is one reason why fibrous composites have been developed to harness micro‐scale properties of materials at larger scales. Man‐made fibers, have almost no flaws in directions perpendicular to their length, and hence are able to bear p p g ,large loads compared to bulk materials.


• Fibrous Composites:– Single‐layer: These are actually made of several layers of fibers,

but in the same direction. Hence they are considered as “single‐layer” composites. These composites may either have continuous and long fibers (filament wound shells) or they may be reinforcedand long fibers (filament wound shells), or they may be reinforced by discontinuous and short‐fibers (fiber glass bodies of cars). Further, the continuous long fiber composites may either have unidirectional reinforcement, or bi‐directional reinforcement. i ll i f i di i d h fibFinally, reinforcements in discontinuous and short‐fiber composites may either be randomly oriented or in preferred directions.

– Multi‐layer: Here, reinforcement is provided, layer‐by layer in different directions. If constituent material in all layers is same, then it is simply a laminate. Hybrid laminates have more than one p y yconstituent materials in the composite structure.

Advantages of CompositesAdvantages of Composites• Composites are engineered materials. Hence, we

i h ifi ll dcan engineer them specifically to meet our needs on a case‐to‐case basis. In general, following properties can be improved by using composite materialscan be improved by using composite materials.– Strength Electrical conductivity

M d l Th l d ti it– Modulus Thermal conductivity

– Weight Behavior at extreme temps.

Fatigue Acoustical insulation– Fatigue Acoustical insulation

– Vibration damping Aesthetics

Resistance to wear Resistance to corrosion– Resistance to wear Resistance to corrosion

Limitations of CompositesLimitations of Composites• Like all things in nature, composites materials have their limitations as well Some of the important onestheir limitations as well. Some of the important ones are:– Anisotropy: A large number of composites have directionAnisotropy: A large number of composites have direction dependent material properties. This makes them more difficult to understand, analyze and engineer, vis‐à‐vis isotropic materialsisotropic materials.

– Non‐homogenous: Further, these materials, by definitionNon homogenous: Further, these materials, by definition are not homogenous. Hence their material properties vary from point‐to‐point. This too makes them difficult to model and analyzemodel, and analyze.

Limitations of CompositesLimitations of Composites– Costly: Composite materials are in general, expensive. Thus, they

are used only in applications where their benefits outweigh the costs.

– Difficult to fabricate: Further, structures these materials are difficult, slow and expensive to fabricateslow and expensive to fabricate.

– Sensitivity to temperature: Laminated composites are particularly sensitive to temperature changes They come in with residualsensitive to temperature changes. They come in with residual thermal stresses, because they get fabricated at high temperatures, and then cooled. Such a process locks in thermal stresses.

– Moisture effects: Laminated composites are also sensitive to moisture, and their performance varies significantly when exposed to moisture for long periods of time.


Nachiketa Tiwari


Lecture 2



• Fibers and whiskers

• Matrices

• Mechanical Behavior of Composites Materials

• Basic terminology

• Problem Set

Fibers and Whiskers• A fiber has high length‐to‐diameter ratio, and also its

diameter approximates its crystal size.

• Modern composites exploit the fact that small scale samples of most of the materials are much more stronger p gthan bulk materials. Thus, thin fibers of glass are 200‐500 times stronger than bulk glass.

• Several types of fibers are available commercially. Some of the more commonly used fibers are carbon, glass, kevlar, steel, and other metals.steel, and other metals.

• Glass is the most popular fiber used in composites since it is relatively inexpensive It comes in two principal varieties; E‐relatively inexpensive. It comes in two principal varieties; E‐glass, and S‐glass. The latter is stronger than the former.

Fibers and Whiskers• Fibers are significantly stronger than bulk materials because:– They have a far more “perfect” structure, i.e. their crystals are aligned along the fiber axis.

– There are fewer internal defects, especially in direction normal to fiber orientation, and hence there are lesser number of dislocations.

• At larger scales, the degree of structural perfection within a material sample is far less that what is presentwithin a material sample is far less that what is present at small (micro and nano) scales. For this reason, fibers of several engineering materials, are far more strong than their counterpart bulk material samplesthan their counterpart bulk material samples.

Fibers and Whiskers• The following table lists bulk as well as fiber properties for

different materials. It is seen from the table that the diff b b lk d fib h i i ifidifference between bulk and fiber strengths is significant.

Table 2 1: Properties of Some Common Engineering Materials in Bulk and Fiber FormsTable 2.1: Properties of Some Common Engineering Materials in Bulk and Fiber Forms

Fiber Specific GravityYoung's Modulus

(GPa)Bulk Tensile Strength

(MPa)Fiber Tensile

Strength (MPa)

Aluminium 2 7 78 140 620 620Aluminium 2.7 78 140‐620 620

Titanium alloy/fiber 4.5 115 1040 1900

St l 7 8 210 340 212 4100Steel 7.8 210 340‐212 4100

E‐Glass 2.54 72 70‐210 350

S Gl 2 48 86 70 210 460S‐Glass 2.48 86 70‐210 460

Carbon 1.41 190 very low 1600

Fibers and Whiskers• Whiskers are similar in diameter to fibers, but in general, they are short and have low length‐to‐g , y gdiameter ratios, barely exceeding a few hundreds.

• Thus, difference in mechanical properties of a whisker vis‐à‐vis bulk material is even morewhisker vis à vis bulk material is even more acute. This is because, the degree of perfection in whiskers is even higher vis‐à‐vis that in fibers.

hi k d d b lli i i l– Whiskers are produced by crystallizing materials on a very small scale.

– Internal alignment within each whisker is extremely g yhigh.

Whiskers• The following table lists bulk as well as whisker properties for

different materials. It is seen from the table that the difference between bulk and whisker strengths is very significant.

Table 2.2: Properties of Some Common Engineering Materials in Bulk and Whisker Forms

Bulk Tensile Strength Whisker Tensile Fiber (MPa) Strength (MPa)

Alumina (Al2O3) 105‐107 19000

Silicon Carbide 3440 11000Copper 220 3000Iron whisker v/s bulk steel 525‐700 13000Boron carbide 155 6700Carbon very low 21000

• Modern composites derive much of their desired properties by using fibers and whiskers as one of the constituent materials.

• Carbon, E‐glass, S‐glass, and Kevlar are some of the commonly used fibers used on modern composite structures.

Problem Set

• Explore different types of fiber materials. What fibers would you used with an objectiveWhat fibers would you used with an objective to:

Improve thermal conductivity– Improve thermal conductivity

– Improve electrical conductivity

Improve mechanical strength– Improve mechanical strength

– Improve toughness


• Fibers and whiskers

• Matrices

• Mechanical Behavior of Composites Materials

• Basic terminology

• Problem set


Nachiketa Tiwari


Lecture 4


Matrix• Fibers and whiskers in composites are held together by a binder

known as matrix.

• Matrix serves several functions:– Transfers loads and stresses within the composite structure.– Support the overall structurepp– Protects the composite from incursion of external agents such as

humidity, chemicals, etc.

• Most of the matrix materials are relatively lighter, more compliant, and weaker vis‐à‐vis fibers and whiskers.

However the combination of fibers/whiskers and matrix can beHowever, the combination of fibers/whiskers and matrix can be very stiff, very strong, and yet very light.– Thus most of modern composites have very high specific strengths, i.e.

very high strength/density ratios.y g g y– This makes them very useful in aerospace applications, where weight

minimization is a key design considerations.

Matrix• Examples of commonly used matrix materials.

– Polyester resin: Yellowish in tint, it is commonly used for low tech applications It is inexpensive and easily tolow‐tech applications. It is inexpensive, and easily to handle. It is sensitive to UV and degrades over time. Thus it is coated for UV protection. Its hardner is MEKP, i.e. Methy‐Ethyl‐Ketone‐Peroxide.Methy Ethyl Ketone Peroxide.

– Vinylester resin: It has a purple/bluish/greenish tint. It is les viscous than polyester and more transparent It is fuelles viscous than polyester, and more transparent. It is fuel resistant, but melts when in contact with petrol. The resin is relatively more immune to UV vis‐à‐vis polyester, and also more flexible. Cost‐wise, it is as good as polyester., g p y

– Epoxy resin: When cured, it is almost transparent. It is very frequently used as a structural material in aerospacevery frequently used as a structural material in aerospace applications.

Matrix• Examples of commonly used matrix materials (contd.):

– Shape memory polymer resins: These resins may be epoxy based (used in automotive and outdoor applications) cyanate‐based (used in automotive and outdoor applications), cyanateester based (used in space applications), or acrylate based (used in low temperature applications). These have have shape memory, and thus they can be shaped and reshaped repeatedly ith t ff i t i l d d ti ti Thwithout suffering any material degradation properties. These

are also used in fabrication of shape‐memory composites.

PEEK: Polyether ether ketone is a thermoplastic matrix material– PEEK: Polyether‐ether ketone is a thermoplastic matrix material which retains its excellent thermo‐mechanical properties at high temperatures (120 C).

– Metals: Metals such as aluminum, magnesium, titanium are used in metal matrix composites as matrix materials. For high temperature applications, cobalt, and cobalt‐nickel alloys are

dused.

Mechanical Behavior of Composite Materials

• Most of the composite materials are neither homogeneous nor isotropic.– A homogeneousmaterial is where properties are uniform throughout, i.e. they do not depend on position in body.

– An isotropicmaterial is one where properties are directionAn isotropicmaterial is one where properties are direction independent.

• Composites are inhomogeneous (or heterogeneous) as well as non‐isotropic materials.– An inhomogeneous (or heterogeneous) material is one– An inhomogeneous (or heterogeneous) material is one properties of material vary from point‐to‐point.

– A non‐isotropic material is one, where material properties d d f di ti Th t i l’ d l bdepend of direction. Thus, a material’s modulus may be different in x, y, and z directions.


• A non‐isotropic material may be:– Orthotropic: Here, the composite has three mutually

di l l f i lperpendicular planes of material symmetry. • When a sample is pulled in tension along either of these planes, it produces only normal strains (tensile in pulling direction and compressive in transverse directions)direction, and compressive in transverse directions).

• Similarly, shear stress when applied along material symmetry planes will only generate shear strains.

• In contrast an isotropic material if pulled in tension in any• In contrast, an isotropic material, if pulled in tension in any direction, will only produce normal strains. And an isotropic material when applied shear force along any plane, will yield only shear strains.

– Anisotropic: Here the composite has no planes of material symmetry. y y

• Hence, tensile stress in any direction will produce shear strains as well.

Basic Terminology• Micromechanics: Study of composite materials by understanding interaction between constituent materials on a microscopic scalematerials on a microscopic scale. – Such an approach helps theoretically compute material properties, and failure mechanisms.S h h i diffi lt t f l t t– Such an approach is difficult to use for large structures because of computational limitations.

• Macro‐mechanics: Study of composite materials presumed to be homogeneous. In such an approach of study, averaged (smeared) properties of composite y, g ( ) p p pmaterial are used to account for the effects of constituent materials.– Such an approach works well for large structures.Such an approach works well for large structures.– Prediction of stresses at micro‐level is not accurate.

Basic Terminology• Lamina: A flat (sometimes curved as well) sample of unidirectional/woven fibers held together by a matrix.

If fib b k i l i it i th t i hi h t f– If a fiber breaks in a lamina, it is the matrix, which transfers load from one broken end to other broken end of fiber through shear mechanism. Plural of lamina is laminae– Plural of lamina is laminae.

– Also known as layer, and ply.

• Laminate: A laminate is stack of several laminaeoriented in different directions, glued together. – There may be significant shear stresses between twoThere may be significant shear stresses between two laminae, due to tendency of each layer to deform independently.

– These stresses are maximum at edges of laminate, andThese stresses are maximum at edges of laminate, and may cause delamination at such locations.

Basic Terminology

Problem Set

• Find out temperature of a missile nose, as it re‐enters earth’s atmosphere throughre enters earth s atmosphere through literature survey. What kind of matrix and reinforcement materials may be used toreinforcement materials may be used to protect the nose from such high temperatures?temperatures?


Nachiketa Tiwari


Lecture 5

Behavior of Unidirectional CompositesBehavior of Unidirectional Composites


• Material axes in unidirectional composites

• Constituent volume fraction and its relationship with composite densitycomposite density

• Importance of predictive methodologies used for p p gcomposite material properties

• Predictive model for longitudinal stiffness• Predictive model for longitudinal stiffness

• Predictive model for longitudinal strengthg g

Material axes in unidirectional compositesp• A lamina is the building block of modern composite laminated structures.

• Each lamina may have in itself have more than one types of fibers, and these fibers may be oriented in different directions.

• A laminate has several layers, or laminae.

E h l i h diff t• Each lamina may have different:– Thickness

– Fiber orientation angle

b– Fibers

– Matrix material

U d di h h i l b h i f l i i h fi i• Understanding the mechanical behavior of a lamina is the first step in understanding mechanics of laminated composite structures.

Material axes in unidirectional composites

• A lamina is also known as a ply, or a layer.

• Each layer in a laminate has, in general three planes of material symmetry. Because of this, they exhibit orthotropic behavior. – A lamina cut across these planes of symmetry will exhibit same mechanical– A lamina cut across these planes of symmetry will exhibit same mechanical

properties.

• Normals to these planes of material symmetry are called material axesNormals to these planes of material symmetry are called material axes, and are quite often designated as 1, 2 and 3 axes. These are shown in Fig. 5.1.– Axis‐1 runs parallel to the direction of fibers, and its direction is calledAxis 1 runs parallel to the direction of fibers, and its direction is called

longitudinal direction.

– Axis‐2 runs normal to axis‐1, but in the plane of lamina. Direction associated with axis‐2 is called transverse direction. Axis‐3 runs normal to axis‐1 and axis‐2. This is also transverse direction.

– 1, 2 and 3 are also known as principal material directions.


• Given the fact that fibers’ strength and stiffness is significantly larger than that of matrix, a lamina is stiffest and strongest inlarger than that of matrix, a lamina is stiffest and strongest in longitudinal direction.

• Further, in 2 and 3 directions, its mechanical properties are hl h f l ’ h lroughly the same. In fact, a lamina’s mechanical properties in

any direction lying in 2‐3 plane are quite the same. – For this reason, a unidirectional lamina is considered as transversely , y

isotropic, i.e. it is isotropic in 2‐3 plane.

• The thickness of a typical carbon or glass fiber ply is 0.127 mm In such plies fiber diameter may be approximately 10mm. In such plies, fiber diameter may be approximately 10 microns. Each ply may be constructed of yarns or rovings. These are shown in Fig. 5.2.– A yarn is a collection of long continuous and interlocked fibers.

– A roving is a narrow and long bundle of several fibers.


Fig. 5.2Roving (left) and Yarn (right)

Failure in Isotropic v/s Transversely Isotropic MaterialsMaterials• In isotropic materials, failure prediction requires calculating principal

stresses or strains and comparing them to allowable stress/strain limits for the materialthe material.

• In non‐isotropic materials (e.g. transversely isotropic materials), this approach does not workapproach does not work.– The notion of principal stress makes no sense for these materials, as material

strength changes with direction, and direction of principal stress may not in most of the cases coincide with direction of maximum strength.most of the cases coincide with direction of maximum strength.

• Thus, for unidirectional materials, we evaluate allowable stress field in context of different strengths of material in principal material directionscontext of different strengths of material in principal material directions. These are:– Longitudinal tensile strength Lateral tensile strength

– Longitudinal compressive strength Lateral compressive strength– Longitudinal compressive strength Lateral compressive strength

– In‐plane shear strength

Failure in Isotropic v/s Transversely Isotropic MaterialsMaterials

• These five material strength parameters for unidirectional it f d t l t i l ti f l icomposites are fundamental material properties of a lamina.

• Experimental data shows that these material strengthExperimental data shows that these material strength properties of a unidirectional lamina are mutually independent, particularly at macro‐scale.

• Hence, if we are able to calculate stress‐field in a unidirectional lamina using 1‐2‐3 axes as reference frameunidirectional lamina using 1‐2‐3 axes as reference frame, then we can predict failure in such lamina.

Volume and Mass Fraction• The relative proportions of fiber and matrix have a significant

influence on the mechanical properties of composite lamina.

• These proportions can be expressed either as volume fractions• These proportions can be expressed either as volume fractions, or as mass fractions. While mass fractions are easier to obtain during fabrication of composites, volume fractions are handier

h l l fin theoretical analyses of composites.

• If vc, vm, and vf, are volumes of composite, matrix, and fiber, respectively, then volume fraction of matrix (V ) and fiber (Vf)respectively, then volume fraction of matrix (Vm) and fiber (Vf) are defined below.– Vm= vm/vc and Vf = vf/vc where, vc= vm + vf

• And if mc, mm, and mf, are volumes of composite, matrix, and fiber, respectively, then volume fraction of matrix (Mm) and fiber (Mf) are defined below.( f)– Mm = Mm/Mc and Mf = Mf/Mc where, mc= mm + mf

Volume and Mass Fraction• Using volume fractions, we can thus, calculate the overall

density of the composite. If ρm, and ρf are densities of matrix, fiber and composite, respectively, then density of compositefiber and composite, respectively, then density of composite (ρc) can be calculated as shown below.– mc= mm + mf

– ρc vc= ρm vm + ρf vf

• Dividing this relation by volume of composite, we can write:Dividing this relation by volume of composite, we can write:– ρc = ρm vm / vc + ρf vf/ vc– ρc = ρm Vm+ ρf Vf (Eq. 5.1)

• Similarly, we can also develop a relation for composite’s density in terms of weight fractions and densities of fiber and matrix.


• Material axes of a lamina, and the notion of transverse isotropy

• How to calculate density of a composite, using volume and mass fraction values.

• The rationale underlying development of predictive methodologies for estimating composite material properties

• Relations to predict longitudinal modulus of a unidirectional lamina

R l i di h f idi i l l i• Relations to predict strength of unidirectional lamina

• Different failure criterion for a unidirectional lamina which undergoes pure uniform tensionundergoes pure uniform tension


Nachiketa Tiwari


Lecture 6



• Material axes in unidirectional composites

• Constituent volume fraction and its relationship with composite densitycomposite density

• Importance of predictive methodologies used for p p gcomposite material properties

• Predictive model for longitudinal stiffness• Predictive model for longitudinal stiffness

• Predictive model for longitudinal strengthg g

The Need for Predictive Models for Determining Composite Propertiesfor Determining Composite Properties

• Mechanical properties of a composite material depend on:Properties of constituent materials– Properties of constituent materials

– Orientations of each layer

– Volume fractions of each constituent

Thickness of each layer– Thickness of each layer

– Nature of bonding between adjacent layer.

Th ti b d t i d b d ti it bl• These properties may be determined by conducting suitable experiments as per industry standards.

• However, a specific set of experiments can only inform us about a specific fiber‐matrix system. Hence, if we want to design composite system by tuning its volume fraction, or fiber‐matrix combination, or orientation, then a very large number of experiments may have to be conducted.

The Need for Predictive Models for Determining Composite Propertiesfor Determining Composite Properties

• Such a process for material property determination is extremely tedious, prohibitively expensive and time consumingprohibitively expensive, and time consuming.

• Still further, exact fiber‐matrix combinations may not be always easily available for testingavailable for testing.

• Hence, there is a need for developing mathematical models, which can li bl d ifi bl di diff h h i l i freliably and verifiably predict different thermo‐mechanical properties of

composite materials.

• Such approaches are very useful for engineers since they provide significant savings in time and cost.

Predicting Longitudinal Modulus of Unidirectional Lamina

• Consider a unidirectional composite lamina with fibers which are continuous, uniform in geometric and mechanical properties and parallel throughout the length of lamina Weproperties, and parallel throughout the length of lamina. We also assume that the bonding between fiber and matrix is perfect, and thus the strains experienced by fiber (εf), matrix ( ) d ( ) l d l d ((εf ) and composite (εc)are same in longitudinal direction (1‐direction). For such a composite, when loaded in 1‐direction, the total external load Pc, will be shared partly by fibers, Pf, c p y y fand remaining by matrix, Pm. This is shown in Fig. 6.3.

Fig. 6.3: Loads on Composite, Fibers, &

in a Unidirectional Lamina.

Predicting Longitudinal Modulus Unidirectional Lamina

• We further assume, that fibers and matrix behave elastically. Thus, the stress in fibers, and matrix can be written in terms of their modulli (Ef, and Em) and strains as:

σf = Ef εf and

σm = Em εm

• Further, if total cross‐sectional area of fibers & matrix is Af and Am, respectively, then:

P = A σ = A E εPf = Afσf = AfEf εf,

and

Pm = Amσm= AmEm

Predicting Longitudinal Modulus Unidirectional Lamina

• Further we know that load on composite P is sum of Pf and P ThusFurther, we know that, load on composite, Pc, is sum of Pf and Pm. Thus,

Pc = Acσc = Afσf + Amσm, or

σ = (A /A )σ + (A /A )σσc = (Af/Ac)σf + (Am/Ac)σm

• But, for a unidirectional composite, Af/Ac and Am/Ac are volume fractions for fib d t i ti l Hfiber and matrix, respectively. Hence,

σc = Vfσf + Vmσm = (VfEf + VmEm) X ε (Eq. 6.2)

• And, if Eq. 4.2 is differentiated w.r.t strain (which is same in fiber and matrix), then.

dσc /dε = Vf(dσf /dε) + Vm(dσm/dε), or c f f m m

Ec = VfEf + VmEm (Eq. 6.3)

Predicting Longitudinal Modulus of Unidirectional Lamina

• Equations 6.2 and 6.3 show that contributions of fibers and matrix to average composite tensile modulus and stress are proportionately dependent on their respective volume fractions.

• In general, matrix material has a nonlinear stress‐strain response curve. For g pcomposites having nonlinear matrix materials, Eq. 6.2 does a good job in predicting stress‐strain relations for a unidirectional lamina. However, the stress‐strain response curve in such materials may not show up as strongly nonlinear, since fibers, especially when their volume fractions are high, dominate the stress‐strain response.

• The higher the fiber volume fraction, the closer is the stress‐strain curve for unidirectional lamina to that for fiber.

Predicting Longitudinal Modulus of Unidirectional Laminag g

• Experimental data pertaining to tensile test specimens agree very well with Eqs 6 2 and 6 3 However the results for compressive tests are not all thatEqs. 6.2 and 6.3. However, the results for compressive tests are not all that agreeable.

h b f b b kl h• This is because fibers under compression tend to buckle, and this tendency is resisted by matrix material. This is analogous to a structure with several columns on an elastic foundation. For a unidirectional composite, the compressive response is strongly dependent on shear stiffness of matrixcompressive response is strongly dependent on shear stiffness of matrix material.

h h h l d h d b f b l l b• Further, Eq. 6.2 shows us that load shared by fibers increases linearly by increasing fiber stiffness, and also by increasing its volume fraction. Experimental data shows that it becomes impractical to aim for fiber volume fractions exceeding 80% due to issues of poor fiber wetting and insufficientfractions exceeding 80% due to issues of poor fiber wetting and insufficient matrix impregnation between fibers.

Predicting Longitudinal Strength of Unidirectional Lamina

T di t l it di l t th f idi ti l l i t• To predict longitudinal strength of a unidirectional ply, requires one to understand the nature of deformation of such a ply as load increase. In general, stress‐strain response of unidirectional plies under tension has four stages These are:stages. These are:– In first stage, both the fiber and matrix exhibit elastic behavior.

– Subsequently, matrix starts becoming plastic, while most of the fibers continue to extend elasticallyto extend elastically.

– In the third stage, both fibers and matrix deform plastically. This may not happen in case of glass or graphite fibers, as they are brittle in nature.

– Finally, the fibers fracture leading to sudden rise in matrix stress, which in turn y, g ,leads to overall composite failure.

• A unidirectional lamina starts to fail in tension, when its fibers are stretchedA unidirectional lamina starts to fail in tension, when its fibers are stretched to their ultimate fracture strain. Here is assume that all of its fiber fail at the same strain level. If at this stage, the volume fraction of matrix is below a certain threshold, the it will not be able to absorb extra stresses transferred to it due to breaking of fibers. In such a scenario, entire composite lamina fails.


Th l i il h f idi i l l b• Thus, ultimate tensile strength of a unidirectional ply can be calculated as:σuc = Vfσuf + (1‐Vf)σm (Eq. 6.4)

where σuc and σuf are ultimate tensile strengths of ply and fiber, respectively,

and σm is stress in matrix at a strain level equaling fracture strain in fiber.

• If the fiber volume fraction is small, and below a certain threshold (Vmin), then even if all the fibers break, the matrix by itself will be able to bear the additional load transferred to itself from broken fibers,to bear the additional load transferred to itself from broken fibers, and may take additional load itself. In such a condition, the ultimate tensile strength of composite may be written as:σ = V σ = (1‐Vf)σ (Eq 6 5)σuc = Vmσum= (1 Vf)σum (Eq. 6.5)

• The relation for Vmin, can be developed by equating Eqs. 6.5 and 6.5, replacing V by V and solving for the latter This is shown in Eq 6 6replacing Vf by Vmin, and solving for the latter. This is shown in Eq. 6.6.Vmin = (σum ‐ σm )/(σuf + σum ‐ σm ) (Eq. 6.6)


F h ll d i d idi i l l i i h i• Further, a well designed unidirectional lamina requires that its ultimate tensile strength should exceed that of matrix. This can happen only when;σuc = Vfσuf + (1‐Vf)σm ≥ σum , where σum is ultimate tensile strength of matrix.

• This equation is satisfied only if fiber volume fraction exceeds a certain critical value, which is defined as:Vcrit = (σum ‐ σm )/(σuf ‐ σm ) (Eq. 6.7)

• Thus, if:– Vf < Vmin, then failure of matrix will coincide with failure of composite, while fibers will fail

prior to failure of matrix.prior to failure of matrix.

– Vf = Vmin, then failure of matrix, fiber and composite will happen at the same time.

– Vf > Vmin, then failure of fiber, will immediately lead to failure of matrix as well as composite.

– Vf > Vcrit, then failure of fiber, will immediately lead to failure of matrix as well as composite, and also the strength of unidirectional composite will exceed that of matrix alone.


• Material axes of a lamina, and the notion of transverse isotropy

• How to calculate density of a composite, using volume and mass fraction values.

• The rationale underlying development of predictive methodologies for estimating composite material properties

• Relations to predict longitudinal modulus of a unidirectional lamina

R l i di h f idi i l l i• Relations to predict strength of unidirectional lamina

• Different failure criterion for a unidirectional lamina which undergoes pure uniform tensionundergoes pure uniform tension


Nachiketa Tiwari


Lecture 7



• Predictive models for transverse stiffness

• Shear modulus and Poisson’s ratio

• Estimates for transverse strength

• Predictive models for coefficient of thermal expansion

• Thermal conductivityThermal conductivity

• Failure mechanisms in unidirectional composites

Predicting Transverse Modulus of Unidirectional Lamina

• Figure 6.1 shows a very simple model for predicting transverse modulus of unidirectional lamina. Here, the model constitutes of two “slabs” of materials, fiber and matrix, of thicknesses tf and tm, respectively. The

ll hi k f i l b i hi h i f d Ioverall thickness of composite slab is tc, which is sum of tf and tm. It may be noted here that these thicknesses of fiber and matrix are directly proportional to their respective volume fractions.

• In such a system, externally imposed stress on the composite (σc) is assumed to be same as that seen by fiber (σf) and also by matrix (σm).

• This is contrast to the model developed for predicting longitudinal• This is contrast to the model developed for predicting longitudinal modulus, where we had assumed that strain, and not stress, in composite, fiber and matrix are all the same.


• Further, in such a model, which is akin to springs in series, the overall displacement in composite (Δc) in transverse direction due to external load is a sum of displacement in fiber (Δ ) and displacement in matrix (Δ )is a sum of displacement in fiber (Δf) and displacement in matrix (Δm).

Δc = Δf + Δm

• Further, recognizing the relation between strains in each constituent, and their thickness above equation can be rewritten astheir thickness, above equation can be rewritten as:

εc tc = εm tm + εf tf

• Dividing above equation by thickness of composite (tc), and realizing that tf/tc, and tm/tc equal Vf and Vm, respectively, we get:

εc = εm Vm + εf Vf

• In linear‐elastic range, strain is a ratio of stress and modulus. Hence, above equation can be further re‐written as:q

(σ c/Ec)= (σ m/Em)Vm + (σ f/Ef)Vf


• But we had earlier assumed that externally imposed stress on the composite (σc) is same as that seen by fiber (σf) and also by matrix (σm). Thus previous equation can be rewritten as:Thus, previous equation can be rewritten as:

1/Ec= Vm/Em + Vf/Ef (Eq. 7.1a)

Or alternatively,

Ec = ( EfEm)/([(1‐Vf)Ef + VfEm] (Eq. 7.1b)

• Equation 5.1 gives us an estimate for transverse modulus of unidirectional lamina. The relation shows that significant increase in fiber volume fraction is required to raise overall transverse modulus in moderate amounts. This is in stark contrast with longitudinal modulus, which is linearly dependent on fiber volume fraction.

• Equation 6.1, even though based on a simple model, is not borne out well be experimental data, and also lacks mathematical rigor.


• However, in this lecture, we will use simple and generalized expressions for transverse modulus as developed by Halpin and Tsai. These are relatively simple relations and hence easy to use in design practice Therelatively simple relations, and hence easy to use in design practice. The results from Halpin and Tsai are also quite accurate especially if fiber volume fraction is not too close to unity.

• As per Halpin and Tsai, transverse modulus (ET)can be written as:

ET/Em = (1 + ξηVf)/(1 ‐ ηVf) (Eq. 7.2)

where,

η = [(Ef/Em) ‐ 1] / [(Ef/Em) + ξ]

Here, ξ is a parameter that accounts for packing and fiber geometry, and loading conditions. Halpin and Tsai suggest ξ to be 2 for square and round fibers, and for rectangular fibers, ξmay be calculated as 2a/b, where a is the dimension of fiber in direction of loading, and b is the other dimension of cross‐section.

Shear Modulus and Poisson’s Ratio

• A perfectly isotropic material has two fundamental elastic constants, E and ν. Its shear modulus and bulk modulus can be expressed in terms of these two elastic constantstwo elastic constants.

• Likewise, a transversely isotropic composite ply has four elastic constants. These areThese are:

– EL, i.e. elastic modulus in longitudinal direction.

– ET i.e. elastic modulus in transverse direction.

– GLT i e longitudinal shear modulusGLT i.e. longitudinal shear modulus.

– νLT i.e. Poisson’s ratio

• A detailed discussion on the mathematical logic underlying existence of g y gfour such constants will be conducted in a subsequent lecture.

• Till so far we have developed relations for EL and ET Now we will learnTill so far, we have developed relations for EL, and ET. Now we will learn about similar relationships for GLT and νLT.

Shear Modulus and Poisson’s Ratio

• Halpin and Tsai have developed relations similar to Eq. 6.2 which can be used to predict longitudinal shear modulus, GLT. This is shown below.

GLT/Gm = (1 + ηVf)/(1 ‐ ηVf) (Eq. 7.3)

where,

η = [(Gf/Gm) ‐ 1] / [(Gf/Gm) + 1]

• For predicting Poisson’s ratio νLT, we exploit the fact that a longitudinal p g LT, p gtensile strain in fiber direction, will generate Poisson contraction in transverse direction in both, matrix and fiber materials.

• In this context, we also use the fact that relative strain values for such a ,contraction will be proportion to each constituent material’s volume fraction. Thus, overall Poisson’s ratio νLT for the composite can be written as:

νLT = νfVf + νfVm (Eq. 7.4)

Transverse Strength

• We have seen that a unidirectional ply, when put to tension in fiber direction tends to break at stress values which exceed matrix tensile strength This is particularly true when fiber volume fraction exceeds Vstrength. This is particularly true when fiber volume fraction exceeds Vcrit. Similarly, fibers play a central role in significantly enhancing the stiffness of the ply in fiber direction, and the overall stiffness of the system tends to far surpass that of pure matrix.p p

• This occurs because fibers, which are stronger and stiffer vis‐à‐vis matrix, carry a major portion of external load thereby enhancing composite’scarry a major portion of external load, thereby enhancing composite s stiffness and strength.

• However the same may not be said for a unidirectional ply loaded in• However, the same may not be said for a unidirectional ply loaded in tension in the transverse direction. This is because, no load sharing, as it occurs in fiber direction, occurs in transverse direction.

• When a unidirectional load is subjected to tension in transverse direction, the stiff fibers act constrain the deformation of the matrix.

Transverse Strength• Such a constraint on matrix deformation tends to increase ply’s transverseSuch a constraint on matrix deformation, tends to increase ply s transverse

modulus, though only marginally (unless fiber volume fraction is high).

• However, the story is even more starkly different in case of transverse strength. o e e , t e sto y s e e o e sta y d e e t case o t a s e se st e gt .The deformation constraints imposed on matrix by fibers tend to generate strain and stress concentrations in matrix material.

• These stress and strain concentrations cause the matrix to fail at much lesser values of stress and strain, than a sample of matrix material which has no fibers at all. Thus, unlike longitudinal strength, transverse strength tends to get reduced for composites due to presence of fiberscomposites due to presence of fibers.

• This reduction in transverse strength of a unidirectional ply is characterized by a factor, S, the strength‐reduction‐factor. The exact value of this factor can befactor, S, the strength reduction factor. The exact value of this factor can be calculated by using a combination of advanced elasticity formulations and numerical solution techniques.

• The strength of unidirectional ply in transverse direction, σuT, can be written as:

σuT = σuf /S (Eq. 7.5)

Some Other Properties of Unidirectional Plies

• Using approaches as described earlier, thermal conductivity in L (kL) direction can be written as:

k V k + V k (E 7 6)kL = Vfkf + Vmkm (Eq. 7.6)

• Similarly, transverse conductivity, kT, can be written as:

kT/km = (1 + ξηVf)/(1 ‐ ηVf) (Eq. 7.7)

where,

η = [(kf/km) ‐ 1] / [(kf/km) + ξ], where, log ξ = 1.732 log(a/b)η [( f/ m) ] / [( f/ m) ξ], , g ξ g( / )

• Finally, longitudinal and transverse thermal expansion coefficients have been shown in engineering literature to be:been shown in engineering literature to be:

αL = (EfVf α f + EmVm α m)/EL (Eq. 7.8)

( ) ( ) ( ) ( )αT = (1+νf)Vf α f + (1+νm)Vm α m ‐ (1+νf) νLT (Eq. 7.9)


• Predictive models for transverse stiffness

• Shear modulus and Poisson’s ratio

• Estimates for transverse strength

• Predictive models for coefficient of thermal expansion

• Thermal conductivityThermal conductivity

• Failure mechanisms in unidirectional composites


Nachiketa Tiwari


Lecture 8

Analysis of an Orthotropic PlyAnalysis of an Orthotropic Ply


• IntroductionIntroduction

k ’ f O h i i l• Hooke’s Law for Orthotropic Materials

• Relations between compliance and stiffness matrices.

Introduction

• Most of the composite materials are neither homogeneous nor isotropic.– A homogeneousmaterial is where properties are uniform throughout, i.e. they do not depend on position in body.

– An isotropicmaterial is one where properties are directionAn isotropicmaterial is one where properties are direction independent.

• Composites are inhomogeneous (or heterogeneous) as well as non‐isotropic materials.– An inhomogeneous (or heterogeneous) material is one– An inhomogeneous (or heterogeneous) material is one properties of material vary from point‐to‐point.

– A non‐isotropic material is one, where material properties d d f di ti Th t i l’ d l bdepend of direction. Thus, a material’s modulus may be different in x, y, and z directions.

Introduction• Consider a rectangular slab of isotropic material• Consider a rectangular slab of isotropic material.

– It this slab is pulled in tension, as shown in Fig. 8.1a, then it only produces normal strains. These strains are tensile in loading direction, and compressive (due to Poisson’s effect) in transverse direction. ( )

– Further, if this slab is subjected to pure shear stresses, then as shown in Fig. 8.1b, the slab undergoes pure shear strain in X‐Y plane.

– These are important characteristic of isotropic materials, i.e. normal stresses produce pure normal strains, and shear stresses produce pure shear strains.

Introduction• Next we consider a rectangular slab of fully anisotropic material• Next, we consider a rectangular slab of fully anisotropic material.

– It this slab is pulled in tension, as shown in Fig. 8.2a, then it not only produces normal strains, but also shear strains. Further, if this slab is subjected to pure shear stresses, then as shown in Fig. 8.2b, the slab exhibits not only shear , g , ystrain in X‐Y plane, but also normal strains.

– This is a very important characteristic of anisotropic materials, i.e. normal stresses produce normal as well as shear strains, and shear stresses produce normal strains in addition to shear strains.

Introduction• Finally, we consider a rectangular slab of orthotropic material.

– In general, this material behaves in ways very similar to anisotropic materials. Thus when subjected to normal stresses it will not only exhibit normalThus, when subjected to normal stresses, it will not only exhibit normal strains, but also shear strains. This is shown in Fig. 8.3a‐b.

– However, the response of these materials mimics that of isotropic material, ifHowever, the response of these materials mimics that of isotropic material, if the edges of slab are parallel to a special set of three mutually perpendicular axes. This is shown in Fig. 8.4a‐b.

– The exact orientation of these three mutually perpendicular axes depends on the internal material structure, and in case of unidirectional composites, on the direction of fibers.

– These axes are known as natural material axes. Also, the planes for which these axes act as normals are known as planes of material symmetry. In case of unidirectional composites the direction of fiber is one such material axisof unidirectional composites, the direction of fiber is one such material axis, and is called longitudinal axis, while the direction normal to it, and in plane of lamina is called transverse axis.

Introduction• Fig. 8.3a‐b

• Fig. 8.4a‐b

Hooke’s Law for Orthotropic LaminaU d di h i f lid i k l i hi b• Understanding mechanics of a solid requires one to know relationships between strains and stresses. For isotropic solids, this relationship is simple and straightforward. For an isotropic sample under pure tensile stress the relationship between stresses and strains is given below.σt = Eεtwhere, σt and εc are stress and strain in direction of tension, and E is Young’s modulus of the material.

• We also know, that due to such a tensile stress, the material sample experiences contraction in transverse directions, and the consequent lateral strain can be expressed as:

ν

expressed as:εc = ‐νεt, where ν is Poisson’s ratio for the material.

• We also know that relation between shear stress, τ, and shear strain, γ, forWe also know that relation between shear stress, τ, and shear strain, γ, for isotropic solids is: τ = Gγ, where G is material’s shear modulus, and it can be expressed in terms of E and ν.

• Hence, an isotropic material has two fundamental elastic constants, which relate stresses and strains.

Hooke’s Law for Orthotropic Lamina• Similarly, we have to develop mathematical relationships between stresses

and strains present in an orthotropic lamina.

• Figure 8.5 shows different types of strains which can act on an infinitesimal material element.

• As shown in Fig. 8.5, there are a total of nine different types of stresses; σ11, σ22, σ33, τ12, τ13, τ21, τ23, τ31, τ32. Here,

– σ11, σ22, and σ33 are normal stresses. They can be compressive or tensile in nature. Their first subscript indicates the plane on which they are acting, and the second subscript indicate the direction in which they point to.

– τ12, τ13, τ21, τ23, τ31, τ32 are shear stresses. Their first subscript indicates the plane on which they are acting, and the second subscript indicate the direction in which they point to. y g, p y pThus, shear stress, τ12, acts on plane 1, and it points in 2‐direction. In contrast, shear stress τ21, acts on plane 2, and it points in 1‐direction.

Si il l th i diff t t f t i t Th• Similarly, there are nine different components of strain tensor. These are: ε11, ε22, ε 33, ε12, ε13, ε21, ε23, ε31, ε32. Here, first three are normal strains, while the remaining six are tensorial shear strains.

Hooke’s Law for Orthotropic Lamina• Now, by factoring into thermodynamic considerations pertaining to strain‐

energy density (details not discussed here), we can show the an anisotropic solid require only 21 independent elastic constants without any loss of

litgenerality.

• At this stage, we introduce the notion of orthotropy, which requires i t f th l f t i l t Thi i li th t t f 21existence of three planes of material symmetry. This implies that out of 21

elastic constants, 12 have to be zero, thereby reducing total number of elastic constants to 9. Using these constants, we can write the stress strain relation as:relation as:

σi = Qij∙εj i, j = 1, 2, 3, 4, 5, 6. (Eq. 8.1)

where,

Qij is the stiffness matrix, σi represents six different stress components, and εj represents engineering strain vector.

• Equation 8.1 is a generalized Hooke’s Law for orthotropic solids.

Hooke’s Law for Orthotropic Lamina• Equation 8.1 may also be represented as:

• In Eq. 8.2, subscripts 1, 2 and 3 coincide with orthotropic material axes.

• Now a lamina may be assumed to have only two‐dimensions as its thickness is very small compared. Hence, all the terms related to thickness direction may be dropped. The stress‐strain relationship for such a lamina is:

Hooke’s Law for Orthotropic Lamina• Finally, for the case when stresses are know in an orthotropic lamina, and

we wish to know strains, we can simply multiply both sides of Eq. 8.3 by inverse of stiffness matrix [Q]. This yields us expressions which may be used t l l t t i i t f t d li t tto calculate strains in terms of stresses and a compliance constants represented by [S]. The general form of such a relation is shown in Eq. 8.4.

• In above equation, compliance terms relate stress to strain. These terms may be expressed in terms of stiffness coefficients as shown in Eqs. 8.5.

S11 = Q22/(Q11Q22 ‐ Q212), 11 Q22/(Q11Q22 Q 12),

S22 = Q11/(Q11Q22 ‐ Q212),

S12 = Q12/(Q11Q22 ‐ Q212), (Eq. 8.5)

andand

S66 = 1/Q66.

Hooke’s Law for Orthotropic Lamina• Similarly, following equation may be used to find out stiffness constants of

an orthotropic lamina, if its compliance coefficients were know.

Q11 = S22/(S11S22 ‐ S212),

Q22 = S11/(S11S22 ‐ S212),

Q12 = S12/(S11S22 ‐ S212), (Eq. 8.6)

anda d

Q66 = 1/S66.

• It needs to reiterated here that Eqs 8 3 8 4 8 5 and 8 6 relate to a two• It needs to reiterated here that Eqs. 8.3, 8.4, 8.5 and 8.6 relate to a two‐dimensional orthoropic lamina. Such materials require only four independent elastic constants. For a three‐dimensional orthotropic lamina, nine elastic constants are needednine elastic constants are needed.

• Thus, understanding three‐dimensional orthotropy involves more complexity compared to that of isotropy or two dimensional orthotropycomplexity compared to that of isotropy or two‐dimensional orthotropy.

Problem Set

• dwdw



k ’ f O h i i l• Hooke’s Law for Orthotropic Materials

• Relations between compliance and stiffness matrices.


Nachiketa Tiwari


Lecture 9



• Introduction

• Engineering constants for an 2‐D orthotropic lamina

• Relationship between engineering constants and li d tiff t icompliance and stiffness matrices

• Restrictions on elastic constants• Restrictions on elastic constants

• Problem setProblem set

Introduction• In previous lecture, we have developed stress‐strain relationship, for a

two‐dimensional orthotropic material.

• Also, we defined four stiffness constants [Q], which can be used to evaluate stress in terms of strains.

Additi ll d fi d f li t t [S] hi h l t t i• Additionally, we defined four compliance constants [S], which relate strain to stress. Relationships between components of [S] and [Q] have also been defined.

• However, neither components of [S] nor those of [Q] are easy to measure. Hence, there is a need to define a third set of constants, known as “engineering constants”, which are relatively easy to measure. For isotropic materials, Young’s modulus (E), and Poisson’s ratio (ν) are such p , g ( ), ( )constants.

• Similarly, we need four, easy to measure engineering constants for two‐dimensional orthotropic materials which may be subsequently used todimensional orthotropic materials, which may be subsequently used to calculate elements of [Q] and [S] matrices.

Engineering Constants for a 2‐D Orthotropic Lamina

A t di i l th t i li l l ti l i h f i i• A two‐dimensional orthotropic linearly elastic lamina has four engineering constants. These are:– Longitudinal modulus (EL): If a unidirectional lamina is pulled in tension along

its fiber length then its longitudinal modulus is defined as:its fiber length, then its longitudinal modulus is defined as:

EL = σL/εL. (Eq. 9.1)

Transverse modulus (E ): If a unidirectional lamina is pulled in tension across– Transverse modulus (ET): If a unidirectional lamina is pulled in tension across its fiber length, i.e. transversely, then its transverse modulus is defined as:

ET = σT/εT. (Eq. 9.2)

– Shear modulus (GLT): If a unidirectional lamina is subjected to pure shear in L‐T plane, then its shear modulus is defined as:

GLT = τLT/γLT. (Eq. 9.3)LT LT/γLT ( q )

– Major Poisson’s ratio (νLT): It is the ratio of negative of transverse strain and longitudinal strain for a unidirectional lamina pulled in fiber direction. Mathematically:

ν LT = ‐ εT/εL. (Eq. 9.4)

Relationship between Engineering Constants and Elements of [Q] & [S] Matrices

• Consider an orthotropic lamina under arbitrary state of stress such that:– Stress in fiber direction is σL– Stress in transverse direction is σTT– Shear stress in L‐T plane is τLT

• Such a lamina will undergo strains due these stresses and the stress• Such a lamina will undergo strains due these stresses, and the stress state can be defined as:– Strain in fiber direction is εL

S i i di i i– Strain in transverse direction is εT– Shear strain in L‐T plane is γLT

• If, the lamina was applied only longitudinal stress (i.e. σL, and τLT are zero) then, using Eq. 8.2, we can write:– σL = Q11 εL + Q12 εLL Q11 L Q12 L

– σT = 0 = Q12 εL + Q22 εT

Relationship between Engineering Constants and [Q], [S] Matrices

• Solving for strains we get:• Solving for strains we get:εL = Q22 /(Q11 Q22 ‐ Q2

12) σLεT = Q12 /(Q11 Q22 ‐ Q2

12) σL

• Rearranging these equations we get:σL/εL = (Q11 Q22 ‐ Q2

12)/Q22

σL /εT = ‐(Q11 Q22 ‐ Q212)/Q12

• Recognizing the definitions of longitudinal modulus, and major Poisson’s g g g , jratio from Eqs. 9.1, and 9.4, respectively, and comparing those with above equations, we can write:EL = (Q11 Q22 ‐ Q2

12)/Q22 (Eq. 9.5)L 11 22 12 22

ν LT = Q12/Q22 (Eq. 9.6)

• And similarly, we can also write:And similarly, we can also write:ET = (Q11 Q22 ‐ Q2

12)/Q11 (Eq. 9.7)

GLT = Q66 (Eq. 9.8)

Relationship between Engineering Constants and Elements of [Q], [S] Matricesand Elements of [Q], [S] Matrices

• Finally, the engineering constants can also be used to define elements of compliance matrix. The methodology for developing these relations is very similar to one used earlier to develop Eq. 9.9 ‐ 9.14.

• Here we directly write the resulting equations.S11 = 1/EL (Eq. 9.15)

S22 = 1/ET (Eq. 9.16)

S12 = ‐νLT /EL = ‐νTL /ET (Eq. 9.17)

S66 = 1/GLT (Eq. 9.18)

Constraints on Values of Elastic Constants

• An isotropic material has several elastic constants; E, ν, G, K, etc. However, out of these, only two are mutually independent. The remaining constants

b d i t f th t F i t if E dcan be expressed in terms of other two. For instance, if E, and ν, are assumed to be mutually independent, then shear and bulk moduli of the material can be expressed as:

G E/[2(1 )] d K E/[3(1 2 )]G = E/[2(1+ ν)], and K = E/[3(1‐2ν)]

• These equivalence relations have implications on the values of elastic constants. Thus; the relation K = E/[3(1‐2ν)], implies that Poisson’s ratio cannot exceed 0.5, for in that case, K would be negative, implying that application of an external inward pressure would cause the material to bulge

t d hi h ld b i i t t ith h i l l f toutwards, which would be inconsistent with physical laws of nature.

• Similarly, there are restrictions on values of elastic constants for orthotropic, and transversely isotropic materials.


• An isotropic material has several elastic constants; E ν G K etc However• An isotropic material has several elastic constants; E, ν, G, K, etc. However, out of these, only two are mutually independent. The remaining constants can be expressed in terms of other two. For instance, if E, and ν, are assumed to be mutually independent, then shear and bulk moduli of theassumed to be mutually independent, then shear and bulk moduli of the material can be expressed as:G = E/[2(1+ ν)], and K = E/[3(1‐2ν)]

• These equivalence relations have implications on the values of elastic constants. Thus; the relation K = E/[3(1‐2ν)], implies that Poisson’s ratio cannot exceed 0.5, for in that case, K would be negative, implying thatcannot exceed 0.5, for in that case, K would be negative, implying that application of an external inward pressure would cause the material to bulge outwards, which would be inconsistent with physical laws of nature. Similarly, Poisson’s ratio cannot be less than ‐1.0, or else, a positive shear y pstrain would cause negative shear strain, which is not possible in reality.

• Similarly, there are restrictions on values of elastic constants for otherSimilarly, there are restrictions on values of elastic constants for other classes of materials, including orthotropic, and transversely isotropic materials, as well.


• For a transversely isotropic material there are five independent elastic• For a transversely isotropic material, there are five independent elastic constant. If its natural material axes are designated as L, T, and T’; where T’ is an axis normal to both, L and T, axes, then its five independent elastic constants are, EL, ET, GLT, νLT, and νTT’ . For such a material to be transverselyconstants are, EL, ET, GLT, νLT, and νTT . For such a material to be transversely isotropic implies that following conditions must be satisfied.ET = ET’ (Eq. 9.19a)

GLT = GLT’ (Eq. 9.19b)LT LT ( q )

νLT = νLT’ (Eq. 9.19c)

GT’T = GTT’ = ET/[2(1+ νT’T)] (Eq. 9.19d)

• Finally, to ensure fundamental physical laws are not violated, the following equations acts as restraints on values of elastic constants.EL, ET, ET’, GLT, GLT’, and GTT’ > 0 (Eq. 9.20a)EL, ET, ET’, GLT, GLT’, and GTT’ > 0 (Eq. 9.20a)

(1‐ νLTνTL), (1‐ νLT’νT’L), (1‐ νTT’νT’T) > 0 (Eq. 9.20b)

1 ‐ νLTνTL ‐ νLT’νT’L ‐ νTT’νT’T ‐ 2νLT νT’L νTT’ > 0 (Eq. 9.21c)

• Details of above equations can be found in Lempriere’ paper: Poisson’s ratio in Orthotropic Materials, AIAA Journal, Nov. 1968.

Problem Set

• dwdw


• The need for engineering constants

• Definitions of engineering constants for an 2‐D orthotropic laminaorthotropic lamina

• Relationship between engineering constants and p g gcompliance and stiffness matrices

• Restrictions on elastic constants• Restrictions on elastic constants

• Problem set


Nachiketa Tiwari


Lecture 10



• Transformation of stress and strainsTransformation of stress and strains

• Stress‐strain relations for a lamina with any• Stress‐strain relations for a lamina with any orientation

• Strength of an orthotropic lamina

• Problem set

Introduction• Earlier, while discussing the stress state in 2‐D orthotropic materials, it was , g p ,

assumed that reference axes for measuring stresses and strains were coincident with material axes. In reality, that may not be the case.

• Hence, there is a need to develop stress‐strain relations in a 2‐D orthotropic lamina oriented arbitrarily. Towards this goal, as a first step, we have to transform stresses and strains from material axes to arbitrary ystresses, and vice‐versa.

• Consider a tetrahedron with vertices ABCP Its face ABC with surface areaConsider a tetrahedron with vertices ABCP. Its face ABC, with surface area A and, normal n (with direction cosines, nx, ny, and nz), experiences is subjected to stress vector T, such that the total external force on face ABC is T∙A. This is shown below.

• Fig. 10.1


F th th t th b d i i ilib i d th th th• Further, we assume that the body is in equilibrium, and thus, other three faces experience normal and shear stresses as shown in Fig. 10.1.

• Given that the body is in equilibrium, we can write the following three equilibrium equations.σxxAnx + τyxAny + τzxAnz = (Tnx).A

τxyAnx + σyyAny + τzyAnz = (Tny).A

τxzAnx + τyzAny + σzzAnz = (Tnz).A,

• Cancelling A, we get Cauchy stress‐formulae, as follows.

σxxnx + τyxny + τzxnz = Tnx = Txτxynx + σyyny + τzynz = Tny = Ty (Eq. 10.1)

τxznx + τyzny + σzznz = Tnz = Tz

• In Eq. 10.1, Tx, Ty, and Tz, are x, y, and z, components of stress‐vector T. Further, T, can also be resolved in terms of its normal and tangential component, with respect to surface A. The normal component can be expressed as:

σn = Txnx + Tyny + Tznz (Eq. 10.2)

Engineering Constants for a 2‐D Orthotropic Lamina• Combining 10 1 and 10 2 we can write the relation for normal stress as:• Combining 10.1 and 10.2, we can write the relation for normal stress as:

σn = σxxnx2 + σyyny2 + σzznz2 + 2τxynxny + 2τyznynz + 2τzxnznx (Eq. 10.3)

E 10 3 b d f l f f• Eq. 10.3 can be used to transform normal stress from one set of axes to another set of axes.

• Now, let us consider Fig. 10.2. Here, we assume that the stress state at a point is known, with respect to an arbitrary set of axes, x, y, and z. From this stress‐state, we are interested in finding the state of stress with

t t th t f hi h t l t i lrespect to another set of axes, which are natural material axes.

• Fig. 10.2.


F Fi 10 2 it i th t th t i l i t i ti ll• From Fig. 10.2, it is seen that the material axis system is essentially a rotation of x‐y axes, around z axis by an angle θ. Thus the direction cosines for the material axis system (L‐T‐T’), with respect to x‐y‐z system are, cos θ sin θ and 1θ,sin θ and 1.

• For such a reference system, normal stresses σxx, and σyy can be written as:

σL = σxx cos2 θ + σyy sin2 θ + 2τxy cos θ sin θ (Eq. 10.4a)

σT = σxx sin2 θ + σyy cos2 θ + 2τxy cos θ sin θ (Eq. 10.4b)

• Using similar approach, we can also write the equation for shear stress as:

τLT = ‐σxx cos θ sin θ + σyy cos θ sin θ + τxy cos2 θ sin2 θ (Eq. 10.4c)

• Eqs. 10.4a‐c, can also be written in matrix form as:


Si il ti l b d t t f t i f• Similar equations can also be used to transform strains from one coordinate system to another one. The strain transformation equations are:

• In Eqs. 10.5 and 10.5, [M] is transformation matrix, and is defined as:

It b t d h th t lik t t f ti ti t i• It may be noted here, that unlike stress transformation equations, strain transformation equations have a factor of ½ within strain vectors. This is because during transformation of strains, tensor strains (and not engineering strains) have to be used While mathematical definitions ofengineering strains) have to be used. While mathematical definitions of normal tensor strain and normal engineering strains are identical, tensorial shear strain is one‐half times that of engineering shear strain.

Transformation of Engineering Constants

N th t h l ti hi h b d t t f t i f• Now, that we have relations which can be used to transform strains from one system to other, we proceed to develop relations which will help us transform engineering constants. Pre‐multiplying Eq. 10.5 by [M]‐1 on either sides we get:either sides, we get:

[M]‐1{σ}L‐T = [M]‐1 [M] {σ}x‐y or {σ}x‐y = [M]‐1{σ}L‐T (Eq. 10.8)

where {σ} and {σ} are stresses measured in x y and L T referencewhere, {σ}L‐T and {σ}x‐y are stresses measured in x‐y, and L‐T reference frames, respectively.

F th f E 8 3 it• Further, from Eq. 8.3, we can write:{σ}L‐T = [Q] {ε}L‐T (Eq. 10.9)

• Putting RHS of Eq. 10.9 in RHS of Eq. 10.8, we get:

{σ}x‐y = [M]‐1 [Q] {ε}L‐T (Eq. 10.10)


A d fi ll i { } i t f { } i i t• And finally expressing {ε}L‐T in terms of {σ}x‐y , using appropriate transformations, in Eq. 10.10, we get:

{σ}x‐y = [M]‐1 [Q] [M]{ε}x‐y or,

{σ}x‐y = [Q]{ε}x‐y (Eq. 10.11)

• Equation 10.11 helps us compute stresses measured in x‐y coordinate system in terms of strains measure in the same system. Here, [Q] is the transformed stiffness matrix, and its individual components are:

Q11 = Q11 cos4θ + Q22 sin4θ + 2(Q12+2Q66) sin2θ cos2θ

Q22 = Q11 sin4θ + Q22 cos4θ + 2(Q12+2Q66) sin2θ cos2θ

Q12 = (Q11 + Q22 ‐ 4Q66)sin2θ cos2θ + Q12 (cos4θ + sin4θ)

Q66 = (Q11 + Q22 ‐ 2Q12 ‐ 2Q66)sin2θ cos2θ + Q66 (cos4θ + sin4θ)

Q16 = (Q11 ‐ Q22 ‐ 2Q66)sinθ cos3θ ‐ (Q22 ‐ Q12 ‐ 2Q66 )sin3θ cos θ

Q26 = (Q11 ‐ Q22 ‐ 2Q66)sin3θ cosθ ‐ (Q22 ‐ Q12 ‐ 2Q66 )sinθ cos3θ

(Eq. 10.12)


F ll i b ti b d f E 10 12• Following observations can be made from Eq. 10.12.– Unlike, [Q] matrix, [Q] matrix is fully populated.

f– In case of [Q] matrix, terms Q16, and Q26, were exactly zero. However, terms Q16, and Q26, are non‐zero entities, and linear combinations of other elements of four basic elements of [Q] matrix.

– For a specially orthotropic lamina, where loading direction coincide with material axes, normal stresses produce only normal strains, and shear stresses produce pure shear strains.

– For a generally orthotropic lamina, i.e. when loading direction and material axes are not coincidental, normal stresses produce normal as well as shear strains. This occurs because of non‐zero values for terms Q16, and Q26, which couple normal and shear responses. These terms are also known as cross‐coupling stiffness coefficients.


F ll i th t f ti d f [Q] li t i [S]• Following the transformation procedure for [Q], compliance matrix [S] can also be transformed to an arbitrary coordinate system. The elements of transformed compliance matrix [S] are defined below.

S11 = S11 cos4θ + S22 sin4θ + (2S12 + S66) sin2θ cos2θ

S22 = S11 sin4θ + S22 cos4θ + (2S12 + S66) sin2θ cos2θ

S12 = (S11 + S22 ‐ S66)sin2θ cos2θ + S12 (cos4θ + sin4θ)

S66 = 2(2S11 + 2S22 ‐ 4S12 ‐ S66)sin2θ cos2θ + S66 (cos4θ + sin4θ)

S16 = 2(2S11 ‐ 2S22 ‐ S66)sinθ cos3θ ‐ 2(2S22 ‐ 2S12 ‐ S66 )sin3θ cos θ

S26 = 2(2S11 ‐ 2S22 ‐ S66)sin3θ cosθ ‐ 2(2S22 ‐ 2S12 ‐ S66 )sinθ cos3θ

(Eq. 10.13)

Strength of an Orthotropic Lamina

I i i i l f il di i i l l i• In isotropic materials, failure prediction requires calculating principal stresses or strains and comparing them to allowable stress/strain limits for the material.

• In non‐isotropic materials (e.g. transversely isotropic l ) h h d kmaterials), this approach does not work.

– The notion of principal stress makes no sense for these materials, as material strength changes with direction, and direction of principal stress may not in most of the cases coincide with direction of maximum strength.

• For an isotropic material, we can fully describe allowable stress field by knowing the material’s tensile, compressive and h t thshear strength.

Failure in Isotropic v/s Transversely Isotropic MaterialsMaterials• Similarly, for 2‐D orthotropic materials, we evaluate allowable

stress field in context of five different strengths of material measured with respect to its principal material directions. These are:– Longitudinal tensile strength (σLU)g g ( LU)

– Lateral or transverse tensile strength (σTU)

– Longitudinal compressive strength (σ’LU)

Lateral or transverse compressive strength (σ’ )– Lateral or transverse compressive strength (σ’LU)

– In‐plane shear strength (τLTU)

• These five material strength parameters for an orthotropic lamina are its fundamental material properties.

Failure in Orthotropic Materials• Similar to isotropic materials, several theories have been developed to

predict failure in orthotropic materials. Some of the more widely used theories are based on maximum stress maximum strain and maximumtheories are based on maximum stress, maximum strain, and maximum work.

• Maximum Stress Theory As per this theory failure will occur once stresses• Maximum Stress Theory: As per this theory, failure will occur once stresses measured with respect to principal material axes, exceed their respective allowable limits. Thus, for failure, at least one of the following conditions must be violatedmust be violated.

σL < σLU, σT < σTU, τLT < τLTU.

d f l hAnd if normal stresses are compressive, then:

σL < σ’LU, σT < σ’TU. (Eq. 10.14)

• One limitation of this theory is that different modes of potential failure do not interact with each other.

Failure in Orthotropic Materials• Maximum Strain Theory: As per this theory, failure will occur once strains y p y,

measured with respect to principal material axes, exceed their respective allowable limits. Thus, for failure, at least one of the following five conditions must be violated.εL < εLU, εT < εTU, γLT < γLTU.

And if normal strains are compressive, then:And if normal strains are compressive, then:εL < ε’LU, εT < ε’TU. (Eq. 10.15)

• If material is linearly elastic then Eq 10 15 can be re written as:• If material is linearly elastic, then Eq. 10.15 can be re‐written as:εL < σLU /EL, εT < σTU /ET, γLT < τLTU/GLT.

εL < σ’LU /EL, εT < σ’TU /ET. (Eq. 10.16)

• Predictions from max. stress and strain theories are very similar, with minor differences being attributable to role of Poisson’s ratio. This is true f li l ti t i l F li l ti t i l E 10 16for linear elastic materials. For non‐linear elastic materials, Eq. 10.16 should not be used, and significant difference should be expected between results from two theories.

Failure in Orthotropic Materials• Tsai‐Hill or Maximum Work Theory: As per this theory, failure occurs when the y p y,

following inequality condition is violated.

(σL /σLU)2 ‐ (σL /σLU) (σT /σTU) + (σT/σTU)2 + (τLT /τLTU)2 < 1 (Eq. 10.16)( L / LU) ( L / LU) ( T / TU) ( T/ TU) ( LT / LTU) ( q )

• Here, if normal stresses are compressive, then compressive strength should be used in the equation. Also, if the lamina is subjected to uni‐directional normalused in the equation. Also, if the lamina is subjected to uni directional normal stress, then above equation can be simplified as:

(cos2θ/σLU)2 ‐ (cosθ sinθ/σLU) 2 + (sin2θ /σTU)2 + (cosθ sinθ/τLTU)2 < (1/ σx)2 (Eq. 10.17)

• Unlike maximum stress and strain theories, Eq. 10.16 provides a single criterion for predicting failure. It also accounts for interaction between p gdifferent strengths of the material. Predictions of strength from this theory are slightly lesser than those from max. stress and strain theories.

• All the theories discussed till so far work only for a lamina subjected to bi‐axial stress state and not for tri‐axial stress state.

Problem Set

• dwdw


• Transformation of stress and strainsTransformation of stress and strains

• Stress‐strain relations for a lamina with any• Stress‐strain relations for a lamina with any orientation

• Strength of an orthotropic lamina

• Problem set


Nachiketa Tiwari


Lecture 11

Analysis of a Laminated CompositeAnalysis of a Laminated Composite



S i fi ld i l i• Strain‐field in a laminate

• Stress‐field in a laminate

• Problem set

Introduction

• Composites offer a very significant advantage: Their properties can be tailor‐made, layer‐by‐layer to meet specific functional requirements.

• Further, each layer can be itself engineered by altering selection of fibers, having a mix of fibers, changing their orientation, using matrix material with appropriate properties, and controlling fiber volume fraction.pp p p p , g

• Analytical models developed thus far help us calculate, with fair levels of accuracy mechanical properties of each lamina These models allowaccuracy, mechanical properties of each lamina. These models allow variability of properties of fibers and matrices, volume fractions, and orientation of fiber angles.

• The next step in this journey is to develop a theoretical construct which will help us predict the mechanical response of a laminate, i.e. a collection of laminae, stacked up and bonded together. Each lamina in this stack‐upof laminae, stacked up and bonded together. Each lamina in this stack up may have different properties (i.e. [Q] matrices and thicknesses).

Strain‐Field in a Laminate

• Before developing an understanding about variation of strains in a laminate, we will make certain assumptions about it. These are:– Laminates are manufactured so that they act as single‐layermaterials. In most

of applications, such a response from the laminate is required so that its overall strength and stiffness can be maximized.

– The requirement of “single‐layer materials” necessitates that the adhesive bond between two adjacent layers is perfect in the sense it has:

• Almost zero thickness

• No shear deformation Thus adjacent lamina cannot slip over each other• No shear deformation ‐ Thus, adjacent lamina cannot slip over each other.

– The assumption of “single‐layer material” also implies that displacements are continuous across the bond between two adjacent layerscontinuous across the bond between two adjacent layers.

– Laminates are thin in the sense their overall thickness is significantly smaller other dimensions of the laminate.ot e d e s o s o t e a ate


• Consider Fig. 11.1. The figure shows how a section of laminate, taken in x‐zdirection, appears after deformation due to application of forces. Here, z, is the thickness direction on reference coordinate system.


• In Fig. 11.1, the left side is a view of un‐deformed laminate. The right portion of Fig. 11.1 shows the deformed state of laminate’s section.

• In the un‐deformed section, line ABCD, is perfectly straight and normal to mid‐plane of the laminate. This line is assumed to remain straight and normal to mid‐plane even after deformation implying:p p y g– Out‐of‐plane shears strains γxz, and γyz, are zero.

– There is no inter‐laminar shear or slipping.


• Further, it is assumed that the length of line ABCD remains same after deformation. This in turn implies that strain in z direction, εzz, is zero.

• The assumptions of normality, in‐extensibility and straightness for line ABCD are together known as Kirchoff‐Love assumptions for shells, and Kirchoff’s assumptions for plates.p p

• Now, point B, due to deformation undergoes translation uo, vo, and wo, in x y and z directions Also line ABCD rotates about B by an angle α in the zx, y, and z directions. Also, line ABCD rotates about B by an angle α in the zplane. Figure 11.1 does not show vo displacement explicitly because the figure is a side view of the laminate undergoing deformation.

• Thus, displacement of point C, which is z distance away from mid‐plane is:

u(z) = u ‐ z∙α = u ‐ z (Eq 11 1)u(z) = uo z α = uo z (Eq. 11.1)

• In Eq. 11.1, we use the fact that α is partial differential of wo, w.r.t, x.


• Similarly, we can also write the relation for displacement v(z), as:

(Eq. 11.2)

• Also, for small displacements, following relations hold for strains.

(Eq. 11.2a)

i d fi i i f d i b i d fi i i• Now, using definitions for u, and v, in above strain definitions, we get:

(Eq. 11.3)

• Also, as mentioned earlier, εzz , γxz, and γyz are zero.


• As per Eq. 11.3, normal and shear strains at a point in a laminate can be decomposed into mid‐plane component, and curvature component. Thus, Eq. 11.3 may be re‐written as:

• ( (Eq. 11.4)

where the mid‐plane strains are defined as:

(Eq. 11.5)


• And, mid‐surface curvatures are defined as:

(Eq. 11.6)(Eq. 11.6)

• In Eq. 11.6, the last term represents twist curvature of mid‐surface of composite laminate.composite laminate.

• Equations 11.3‐11.6 are valid only for plates and not for shells. This is because our strain definitions as per Eq 11 2a are valid only for platesbecause our strain definitions, as per Eq. 11.2a are valid only for plates.

• Equation 11.4 shows that strains vary linearly over the thickness of a it l t ith th t i t d l t ’ thi kcomposite plate, with the average strain computed over plate’s thickness

equaling mid‐plane strain.

Stresses in a Laminate

• If one were able to compute mid‐plane strains and curvature of the plate, then predicting stresses over the thickness is simply a matter of multiplying these strains, using strain‐stiffness relations, on a layer‐by‐layer basis.

• Thus, stresses in kth layer of the laminate may be calculated using , y y gfollowing relations.

(Eq 11 7)(Eq. 11.7)

• Since [Q] matrix for each lamina may vary abruptly between two adjacent layersSince [Q] matrix for each lamina may vary abruptly between two adjacent layers, variation of stresses between two layers need not be linear, or even continuous. Instead, they will be expected to vary in a discontinuous manner, even though strain will vary in a linear manner across entire laminate thickness. However, over the thickness of a single lamina, stress variation will be linearly continuous.

Problem Set



S i fi ld i l i• Strain‐field in a laminate

• Stress‐field in a laminate

• Problem set


Nachiketa Tiwari


Lecture 12




• Resultant Forces and Moments• Resultant Forces and Moments

Pi i i t ti f l l ti lt t• Piece‐wise integration for calculating resultant forces and moments.

• Problem set

Introduction

• In previous lecture, mathematical relations have been developed, which define:– Variation of strains over thickness of a laminate

– Variation of stresses over thickness of a laminate

• These relations are useful in predicting stresses in a plate, if mid‐planeThese relations are useful in predicting stresses in a plate, if mid plane strains and mid‐plane curvatures were known.

• In most of real applications we may not be necessarily know mid plane• In most of real applications, we may not be necessarily know mid‐plane strains and curvatures for a composite, and sans their knowledge, predicting stresses in composite laminates is not possible just by using equations developed earlier.equations developed earlier.

• However, in a several cases, we do know the value of externally applied loads and moments on plates Thus there is a need to develop relationsloads and moments on plates. Thus, there is a need to develop relations which connect mid‐plane strains, mid‐plane curvatures, stresses, and external forces and moments.

Resultant Forces and Moments• Consider a small part of composite plate, which is acted upon by forces

and moments on its edges as shown in Fig. 12.1 due to different stresses. Here, Nx, Ny and Nxy are resultant forces per unit length acting on the y yedges of the composite plate in directions as shown in Fig. 12.1. Similarly, Mx, My and Mxy are resultant moments per unit length acting on the edges of the composite plate in directions as shown in Fig. 12.1.

Fig. 12.1: Forces and Moments at Mid‐Plane of a Plate

Resultant Forces and Moments• Using principles of equilibrium, we can relate stresses to force resultants

by integrating appropriate stress components through the plate thickness. Thus, we get.

(Eq. 12.1)( q )

• In above equation, t, represents overall composite plate thickness.

Resultant Forces and Moments• Similarly, using principles of equilibrium, we can relate stresses to moment

resultants by integrating appropriate products of stress components and distance from mid‐plane, through the plate thickness. Thus, we get.

(Eq. 12.3)

• In these equations, unit of force resultants (Nx, Ny and Nxy) is N/m, while that for moment resultant (Mx, My and Mxy) is N. Also, Fig. 12.1 depicts conventions used for positive resultant forces and resultant moments.p

Resultant Forces and Moments

Resultant Forces and Moments• Consider Fig. 12.2. Looking at it, and also realizing from previous analysis,

we realize that variation of stress is:– Discontinuous over the thickness of the whole laminate.

– Linearly continuous over the thickness of a single layer. Such a variation of stresses over laminate thickness.

• Thus, the integrands for resultant forces and moments, as defined in Eqs. 12.1 and 12.2 are not continuous functions of z.

• Rather, they are piece‐wise continuous over the thickness of composite plate. Hence, their integration over entire thickness requires one to:– Piece‐wise integrate the function over each lamina’s thickness.Piece wise integrate the function over each lamina s thickness.

– Sum up all lamina specific integrals for all the layers.

This is accomplished in subsequent sectionThis is accomplished in subsequent section.

Resultant Forces and Moments• Consider Fig. 12.3, which shows cross‐section of stack up of a n‐layered

orthotropic laminate. Here, the z coordinate of top and bottom surfaces of kth laminate is zk, and zk‐1. Further, as per the convention in this schematic, top most layer, with a z‐coordinate of t/2 is considered the 1st layer, while the bottom most layer is considered the nth layer.

Resultant Forces and Moments• For such a schematic, the relations of resultant forces and moments, using

Eqs. 12.1 and 12.2 can be written as:

(Eq. 12.4)

and,

(Eq. 12.5)

Resultant Forces and Moments• Equations 12.4 and 12.5 are summations of integrals. If there are n layers

in the composite, then there will be n summations.

• In this way, contribution of each layer is summed up while calculating resultant forces and moments.

• After integration and summation, coordinate z, no longer appears in expressions for resultant forces and moments.

• At this stage, i.e. after summation and integration, the resultant forces and moments numbering six, can be conceptualized as getting applied on a composite plate at its mid plane to generate stresses and strainscomposite plate at its mid‐plane, to generate stresses and strains.

• However, even after summation and integration, the consequential lt t f d t ith t t dresultant forces and moments may vary with respect to x and y

coordinates.

Problem Set



• Resultant Forces and Moments• Resultant Forces and Moments

Pi i i t ti f l l ti lt t• Piece‐wise integration for calculating resultant forces and moments.

• Problem set


Nachiketa Tiwari


Lecture 13



• Introduction

• Stiffness Matrices for Laminate

• Discussion on Stiffness Matrix Elements

• Special Lamination Sequences

• Problem set

Introduction

• Earlier, it resultant stress and moments have been defined as summations of integrals of stress, and stress times distance from mid‐plane, on a layer‐by‐layer basis.

• Substituting in Eq. 12.5 definitions of stress field using Eqs. 11.7, we get:

• This can be simplified as:

• But, we know that mid‐plane strains and curvatures are independent of z.But, we know that mid plane strains and curvatures are independent of z. Hence, the strain and curvature terms can be removed from under the integral and summation operations.

Laminate Stiffness Matrices• Also, [Q] is a constant over the thickness of each layer, but changes from

one layer to the other. Thus, the relation for resultant stresses becomes:

• This equation can be simplified as:

(Eq. 13.1)

• where,

(Eq. 13.2)

Laminate Stiffness Matrices• Similarly, the relation for moment resultants can be developed in the

following form:

(Eq.13.3)

where,

(Eq. 13.4)( q )

• Matrix [A] is called extensional stiffness matrix.

• Matrix [B] is called bending‐extension coupling matrix.

M i [C] i ll d b di iff i• Matrix [C] is called bending stiffness matrix.

Laminate Stiffness Matrices

• Extension Stiffness Matrix: This matrix influences extensional strains in the laminate. – For a given resultant force, mid‐plane strains decrease as elements of this

matrix increase in magnitude.

– Magnitude of extensional stiffness increases directly in proportion to the thickness of each layer, since zk ‐ zk‐1, equals thickness of kth lamina.

T A d A l h d l f h l i If– Terms A16 and A26 couple shear and normal responses of the laminate. If either of these terms is non‐zero, then:

• An extensional resultant force will generate not only extensional strain, but shear strain as well.

• Similarly, a shearing resultant force will generate not only shear strain in the laminate, but extensional strain as well if these terms are non‐zero.


• Extension‐Bending Coupling Matrix [B]: This matrix couples extensional response to the bending response of the laminate.– Magnitude of [B] matrix depends on the square of the distance of each

lamina’s surface from the mid‐plane.

– If the magnitude of this matrix is non‐zero, then:• A composite laminate will exhibit bending and twisting, even if external moment on

it is perfectly zero.

• A composite laminate will exhibit extensional and shear strains even if externalA composite laminate will exhibit extensional and shear strains, even if external resultant forces on it are perfectly zero.

– In a very large number of applications, coupling of bending and extensional responses is cautiously avoided, as it can trigger early material failure. In such cases, laminates are carefully engineered such that all elements of [B] are identically set to zero.


• Bending Matrix [D]: This matrix significantly influences the bending response of a laminate.– Magnitude of [D] matrix depends on the cube of the distance of each lamina’s

surface from the mid‐plane.

– A [D] matrix with large magnitude implies that a unit bending moment will generate very little curvature in the laminate and vice‐versa.

Terms D and D couple bending and twisting responses of the laminate If– Terms D16 and D26 couple bending and twisting responses of the laminate. If either of these terms is non‐zero, then:

• A pure bending moment will generate not only bending curvature kx, and ky, but twist curvature, kxy, as well.y

• Similarly, a pure torque will generate not only twisting curvature in the laminate, but bending curvatures as well.


• Equations 13.1 and 13.3 can also be expressed in one single set of equations expressed in matrix format. This is shown below.

(Eq. 13.5)

where,

N is a vector for resultant stresses.

M is a vector for resultant moments,

εo is a vector for mid‐plane strains, and,

κ is a vector for mid plane curvaturesκ is a vector for mid‐plane curvatures.

Special Lamination Sequences

• Here, we list certain special lamination sequences, and associated terminologies, which are frequently used in composite industry, and cater to specific functional requirements.– Symmetric Laminates: Such laminates are symmetric with respect to geometry

(orientation of fibers) and material properties. Such laminates:l b b d d l h ll l• Have no coupling between bending and extensional responses. Thus, all elements

for such laminates’ bending matrix are zero.

• When such elements are cooled after fabrication, the induced thermal stresses do not cause twisting and bending of laminate, unlike unsymmetrical laminates.

– Cross‐ply laminates: Here, all layers in the laminate have an orientation of either 0 or 90 degrees.

• For such laminates, A16, A26, B16, B26, D16, and D26 terms are zero.

• Also, shear and extensional responses are decoupled.

• And, bending and twisting responses are decoupled.


– Angle‐Ply Laminates: All layers in such laminates have an orientation of either θ, or ‐θ. For such laminates, if:

• The laminate has an equal number of plies oriented in either directions, then terms A16, A26 are zero.

• The laminate has a large number of plies totaling an odd number of plies, and the ply sequence alternating between θ, or ‐θ, then, coupling matrix [B] is zero, and terms A16, A26, D16, and D26 are very small.

– Anti‐Symmetric Laminates: Here, the material orientation of layers above mid‐plane is negative of those below the mid plane For such laminates:plane is negative of those below the mid‐plane. For such laminates:

• Terms A16, A26, D16, and D26 are zero.

• However, matrix [B] is non‐zero.

– Balanced‐Ply Laminates: For every ply with an orientation θ, the laminate also has a corresponding ply of same thickness, but orientation ‐ θ in the laminate. For such cases, terms A16 and A26 are zero


– Quasi‐istotropic Laminate: Such a laminate is widely used. An important property of such a laminate is that its extensional stiffness matrix is similar to th t f i t i t i l Th f h t i lthat for isotropic materials. Thus, for such materials:

• A11 = A22

• A66 = (A11 ‐ A12)/2

• A16 = A26 = 0A16 A26 0.

Such a laminate can be constructed if following conditions are met.• The laminate should have more than two layers.

• All layers should have identical stiffness matrices [Q], and thicknesses.

• The difference in orientation angle of two adjacent layers must be equal. Thus, the angle between two adjacent layers should be 2π/n, where n is total number of layers in the composite platelayers in the composite plate.

• Two examples of quasi‐isotropic laminates are:– [0/±60] is a quasi‐isotropic three‐layer laminate.

– [0/±45/90] is a quasi‐isotropic four‐layer laminate.

Problem Set


• Introduction

• Stiffness Matrices for Laminate

• Discussion on Stiffness Matrix Elements

• Special Lamination Sequences

• Problem set


Nachiketa Tiwari


Lecture 14

Equilibrium Equations for PlatesEquilibrium Equations for Plates



ilib i i• Equilibrium Equations– Assumptions

– Equations for force

– Equations for moments

Introduction

• Earlier, the following equations have been developed, which govern stress‐strain relations in a composite laminate.

• Such an equation works well as long as {N} and {M} are known. But if they are unknown, or vary with respect to x, and y coordinates, then it is important to understand the laws which govern such a variation.

• There are two approaches which can be used to understand such pprelationships. These are:– Equilibrium approach, which is based on Newton’s 1st Law of Motion.

– Variational approach, which is based on energy equilibrium.pp , gy q

• Here, we start our discussion with the Newtonian approach.

Equilibrium Equations

• As per Newton’s 1st Law of Motion, for a body to remain in equilibrium, the sum of external forces and moments on it, should be zero. This can be mathematically expressed as: – ∑Fi = 0

– ∑Mi = 0,

where, i, is an index which can assume values x, y, or z.

• Thus, there are a total of six equilibrium equations; three for force, and three for moment. Consider an infinitesimal portion of a plate, as shown below in Fig. 14.1. Here, only forces in x direction are shown.

Fig 14 1: x direction forcesFig. 14.1: x‐direction forces acting on the mid‐plane of a plate (origin assumed to be at geometric center of plate element)

Equilibrium Equations

• In Fig. 14.1, the geometric center of the plate coincides with the origin. Also, the dimensions of this plate are Δx, Δy, Δz. Further, we assume that force and moment resultants at the origin are Nx, Ny, Nxy, Mx, My, and Mxy. y y y yFigure 14.1 shows force resultants on different faces of the plate, acting in x‐direction.

• For equilibrium in x direction, the following equation can be written as shown below.

(Eq. 14.1)

• Similarly, the equilibrium equation for y direction can be written as:y, q q y

(Eq. 14.2)

Equilibrium Equations• Here, it should be noted that resultant forces Nz, Nxz, and Nyz are zero, and

so are moments Mxz, and Myz. This is an outcome of Kirchhoff’s assumptions for a plate due to which out‐of‐plane shear and normalassumptions for a plate due to which out‐of‐plane shear and normal strains are zero.

• Next we develop an equilibrium equation for z direction Consider an• Next, we develop an equilibrium equation for z direction. Consider an infinitesimal portion of a plate subjected to normal load per unit area, q. Figure 14.2 shows force resultants on different faces of the plate, acting in z‐directionz direction.

Fig. 14.2: Equilibrium schematic for z‐direction on an infinitesimal portion of a composite plate

Equilibrium Equations• As per Kirchhoff’s assumptions, out of plane shear stresses (τxz, and τyz)

should be zero. This in turn implies that the equilibrium equation for zdirection should be:q∙dx∙dy = 0 or q = 0.

• Above relation implies that for a plate to be in equilibrium in z direction, it p p q ,should not be subjected to any external load acting in out‐of‐plane direction.

• However, this is not consistent with reality, where plates do take up significant out‐of‐plane loads. This forces us to revisit Kirchhoff’s assumptions concerning out‐of‐plane shears, and it is not true that these entities are zero in platesentities are zero in plates.

• Thus, we apply a correction to Kirchoff’s plate as suggested above.

Equilibrium Equations• As per this correction we do not assume out‐of‐plane shear stresses to be• As per this correction, we do not assume out‐of‐plane shear stresses to be

zero, and thus define two non‐zero entities as follows.

• Thus, the equation for force equilibrium in z direction is:

(Eq. 14.3)

• Next we look at moment equilibrium about origin. Consider Fig. 14.3, which depicts various moments acting about y‐axis.

Fig. 14.2: Depiction of moments and forces needed to develop momentforces needed to develop moment equilibrium equation about y‐axis (moments and forces shown only on positive faces of the plate element).

Equilibrium Equations• Taking moments about y‐axis, we get:

• Neglecting higher order terms we get moment equilibrium about y axis• Neglecting higher order terms, we get moment equilibrium about y axis.

Equilibrium Equations• Finally, we consider moment equilibrium about z axis. Here, we note the

following:– Nx, and Ny, do not contribute to moments about z axis, as they per definition,

act on the mid‐plane.

– q(x, y) also does not contribute about z axis, sd its contribution would be of a higher order.

O l N d N t ib t t t b t i B t th i t ib ti– Only Nxy, and Nyx, contribute to moments about z axis. But their contributions add up to exact zero. Thus, moment equilibrium equation for z direction given below, is identically satisfied.

N ‐ N = 0 (Eq. 14.6)Nxy Nyx 0 (Eq. 14.6)

• Equations 14.1, 14.2, and 14.3 represent force equilibrium in x, y, and z, directions, respectively.p y

• Equations 14.4, 14.5, and 14.6 represent moment equilibrium in x, y, and z, directions, respectively.

Equilibrium Equations• Differentiating Eq. 14.4 and 14.5 with respect to x, and y, respectively,

summing them up, and then substituting Eq. 14.3 in this “summed up” relation, we get:

• Thus, we have a total of four equilibrium equations. These are:

• It may be note that embedded within Eq. 14.10 are three conditions for ilib i f b l i di i d b l i dequilibrium; force balance in z direction, and moment balance in x and y

directions.



ilib i i• Equilibrium Equations– Assumptions

– Equations for force

– Equations for moments


Nachiketa Tiwari


Lecture 15

Boundary Conditions forBoundary Conditions for

Rectangular Plates



d C di i f l• Boundary Conditions for Rectangular Laminates

Boundary Conditions• Equations 14.7 to 14.10 are valid for any composite plate of arbitrary

shape or dimensions. These equations have to be solved to get displacement‐field, and stress‐field information for the plate.

• For such a plate, there are four sets of boundary conditions (BCs) for each edge of the plate.

• These four sets of BCs correspond to four equations of equilibrium. – The first two BCs correspond to inplane equilibrium equations, i.e. Eqs. 14.7The first two BCs correspond to inplane equilibrium equations, i.e. Eqs. 14.7

and 14.8.

– The second set of two BCs correspond to out‐of‐plane equilibrium equation, i.e. Eq. 14.9.

• Each edge of a rectangular plate could have five different resultant forces, and moments. For instance, the x = a/2 edge could be subjected to g jexternal force/moments as, Nx

+, Nxy+, Ny

+,Mx+,Mxy

+ and Qx+. However,

there are only four BCs per edge.

Boundary Conditions• Boundary Conditions at x = +a/2.

(Eq. 15.1)

Boundary Conditions• Boundary Conditions at x = ‐a/2.

(Eq. 15.2)

Boundary Conditions• Boundary Conditions at x = +b/2.

(Eq. 15.3)

Boundary Conditions• Boundary Conditions at x = ‐b/2.

(Eq. 15.4)

Boundary Conditions• In Equations 15.1 through 15.4, the BC on equality of out‐of‐plane shears is

note‐worthy.

• As per this condition, either displacement in z direction should be known, or the sum of Q and partial derivative of Mxy should be known.

• This condition does not require knowledge of either Q, or partial derivative of Mxy, individually. Rather, what is needed to be known is their sum. It is for this reason, five variables on each face of plate, are linked by only four , p , y yboundary conditions.

• Overall there are three displacements three strains three stresses and sixOverall, there are three displacements, three strains, three stresses, and six force and moment resultants; a total of 15 unknown variable. These can be calculated by solving 15 simultaneous equations; 3 kinematic relations, six equations of equilibrium, and 6 relationships between force and momentum q q , prelations and strains using A, B and Dmatrices. The integration constants/functions may be determined from boundary conditions.

Boundary Conditions• Finally, we note that for the edge, x =+a/2, one of the boundary condition is:

or, mid‐plane displacement wo, should be known.

• The term represents net out of plane shear force acting at edge• The term represents, net out‐of‐plane shear force acting at edge, x=+a/2. It is sometimes also referred as Q+

x_eff.

Si il b d di i i l i f h d f h• Similar boundary condition requirements also exist for other edges of the plate.


Nachiketa Tiwari


Lecture 16

Semi‐Infinite PlatesSemi‐Infinite Plates



i• Assumptions

• Different Cases to be Studied

• Solutions for Some Cases

Introduction

• Here, we will study how governing equations for composite plates as developed earlier, can be used to calculate the response of semi‐infinite l tplates.

• We start with semi‐infinite plates because of the following reasons:– Closed form solutions for such plates are easy to develop. The same may not

be true for more “common” geometries and boundary conditions.

– Such “idealized” analyses provide us with useful insights related to role of t i l ti b d diti d t l l d thmaterial properties, boundary conditions and external loads on the response

of plates.

– The closed form solutions developed here, may be used to check validity of other solution approaches, e.g. finite element method.other solution approaches, e.g. finite element method.

Assumptions about Semi‐Infinite Plates

• Consider a plate which is infinitely long in one direction. This plate is simply supported along both of its short‐edges as shown in below in Fig. 16 1 Th l t i l d d i di ti ith l d i t it16.1. The plate is loaded in z direction with load intensity q.

Fig. 16.1 Fig. 16.1

y

x

• In reality, a very long plate, with dimensions a, and b, such that a >> b, will approximately mimic the response of an “infinitely” long plate. Here, a is

’plate’s dimension in x direction.

• Given that the plate is very long in x direction, it can be assumed that partial derivatives of all entities in y direction for such a plate will be zero. This also implies, that for such a plate, partial derivative of an entity in xdirection will equal total derivative in x direction. Thus for such a plate:

Different Cases to be Studied

• Finally we define the boundary conditions and lamination sequence of the plates to be studied. Overall, four cases will be studied. The BCs and l i ti f th li t d i Fi 16 2lamination sequences for these cases are listed in Fig. 16.2.

CASE A: [0/90]s

CASE B: [0/90]s

/

[ / ]s

CASE C: [02/902]T

CASE D: [02/902]s

Fig. 16.2: Boundary Conditions for Semi‐Infinite Plates

Solutions for Response of Infinitely Long Plates

• Kinematic Equations: Rewriting Eqs. 11.4, 11.5 and 11.6 we get:

• Given that partial derivatives with respect to are zero for such plates, we can write these relations as:

(Eq. 16.1)


• Equilibrium Equations: Next, we simplify equilibrium equations, in the same way as we did kinematic equations, i.e. by using the fact that partial d i ti ith t t f h l tderivatives with respect to are zero for such plates.

• Thus, Eqs. 14.7, 14.8 and 14.9 may be simplified as:

(Eq. 16.2)


• Stiffness Matrices: For cases A and B, as defined in Fig. 16.2, the laminate is [0/90/]s. This implies that the laminate is symmetric, and also cross‐ply. F h l i tFor such a laminate:– The condition of symmetry implies that the coupling matrix, [B], for such a

laminate is zero.

And because all the plies in the laminate are oriented either at 0 or 90– And because all the plies in the laminate are oriented either at 0 or 90 degrees, it implies that following terms are zero.

• A16 = A26 = 0

• D16 = D26 = 0 (Eq. 16.3)16 26 ( q )

• For cases C and D, as defined in Fig. 16.2, the laminate is [02/902]T. This is a non‐symmetric but cross‐ply laminate. For such a laminate:– All plies in the laminate are oriented either at 0 or 90 degrees, it implies that

following terms are zero.• A16 = A26 = 0

• B16 = B26 = 0

• D16 = D26 = 0 (Eq. 16.4)

Solutions for Case A

• For Case A, we now combine Eqs. 16.1 and 16.2 using material constitutive relations, and also using Eqs. 16.3. Thus, we get:

(Eq. 16.5)

( (Eq. 16.6)

(Eq. 16.7)

(Eq. 16.8)


• Also, considering the momentum equation, and assuming that external normal load per unit area q, is a constant, we get from Eq. 16.2:

2/Mx = ‐qx2/2 + C3x + C4 (Eq. 16.9)

• And expressing Mx in terms of strains and curvatures, we write:Mx = D11(d2wo/dx2) (Eq. 16.10)

because terms D16, D26, and Bij are zero.

• Combining these two equations, and integrating it twice, we get:wo(x) = {‐qx4/24 + C3x3/6 + C4x2/2 + C7x + C8}/D11 (Eq. 16.11)

• Also, writing expression for My, and using Eq. 16.8, we get:My = D12(d2wo/dx2) = (D12/ D11)Mx (Eq. 16.12)


To be continued in next lecture.


Nachiketa Tiwari


Lecture 17



• Solution for Case ASolution for Case A

S l i f C• Solution for Case B


• It may be noted here, that Eqs. 16.5 through 16.12 are valid for Cases A and B, because the problem definition for either cases is same sans b d diti A h t t li d b d ditiboundary conditions. As we have not yet applied boundary conditions, these equations are applicable to Case B as well.

• Boundary Conditions for Case A– As shown in Figure 16.2, the BCs for this plate are:

BCs at x = ‐a/2 BCs at x = +a/2BCs at x a/2 BCs at x a/2

uo‐ = 0 Nx+ = 0

vo‐ = 0 Nxy+ = 0

M 0 M + 0Mx‐ = 0 Mx

+ = 0

wo‐ = 0 wo+ = 0


• Plugging 1st set of BCs, i.e. uo‐ = 0 and Nx+ = 0 in Eqs. 16.5 and 16.6, yields:

C1 = C5 = 0.

Thus,Nx(x) = 0 uo(x) = 0 Ny(x) = 0 (Eq. 17.1)

• Similarly, second set of BCs, i.e. vo‐ = 0 and Nxy+ = 0 yields:

C2 = C6 = 0.

Thus,

Nxy(x) = 0 vo(x) = 0 (Eq. 17.2)

• Further, the 3rd set of BCs , i.e., M0± = 0, at x = ±a/2 implies (from Eq. 16.9):, , , , / p ( q )0 = ‐qa2/8 + C3a/2 + C4 (for x = +a/2)

0 = ‐qa2/8 ‐ C3a/2 + C4 (for x = ‐a/2)

Thus,Thus,

C3 = 0, C4 = qa2/8, and Mx = q(a2/4 ‐ x2)/2 (Eq. 17.3)


• Finally, plugging 4th set of BCs, i.e. w0± = 0, at x = ±a/2 in Eqs. 16.11,and also using Eq. 17.3, yields:

4/ 3/ 2/ / /0 = {‐qa4/384 + C3a3/48 + C4a2/8 + C7a/2 + C8}/D11

0 = {‐qa4/384 ‐ C3a3/48 + C4a2/8 ‐ C7a/2 + C8}/D11

Thus,

C7 = 0, C8 = ‐5qa2/384, and

wo(x) = {‐16(x/a)4/24 + 24(x/a)2/16 ‐ 5}(qa4)/(384D11) (Eq. 17.4)

Also,

My = (D12/ D11)Mx and Mxy = 0 (Eq. 17.5)

• Equations 17.1 through 17.5 constitute the solution for Case A.

Comments on Solution for Case A

• It is seen from Eqs. 17.1 through 17.5, that the solutions for mid‐plane displacements, resultant forces, and resultant moments is symmetric with

t t irespect to x axis.

• However, as seen in Fig. 16.2, the boundary conditions for Case A are not symmetric about the same axis. Further, even though the beam is free to move in x direction at x = +a/2, the solution for displacement at this end, uo, is zero. There are two reasons for this.

f l h h l b– Presence of symmetric laminate ensures that there is no coupling between u(x) and Mx. Hence, the moment imposed by external load intensity q, cannot create inplane resultant force Nx, and thus, no inplane displacement in x‐direction is getting generated.d ect o s gett g ge e ated.

– Further, we have assumed linearity of strains our strain‐displacement formulation.


• Next, consider Fig. 17.1(a), which shows the un‐deformed shape of a small portion of the plate being currently analyzed.

Fig. 17.1(a): Undeformed Shape of a Small Plate Under Being Subjected to Various Loads and Moments which may Changes Its Curvature


Fig. 17.1(b): Anti‐clastic curvature in plate under loads and moments

Comments on Solution for Case AThi l t h l th f ti th l i t Th M• This plate, has only three forces acting on the laminate. These are, q, Mx, and My. All other forces have been found to be zero as per our calculations.

• It may be noted that surface ABCD is the mid‐plane for the plate in Fig. 17.1a. Now, because in‐plane displacements at midplane (uo, and vo ) and

t d o/d ( E 17 1 d 17 2) li AB d CDcurvature dwo/dy are zero (see Eqs. 17.1 and 17.2), lines AB, and CD, which were straight prior to deformation, will remain straight after deformation as well. This is counter‐intuitive, as external load q, would tend to induce curvature on these line‐segmentstend to induce curvature on these line‐segments.

• However, presence of non‐zero My, ensures that these line segments i t i ht Thi i l i d f llremain straight. This is explained as follows.

Comments on Solution for Case A• Vertical force due to q is balanced byM andM However given that• Vertical force due to q is balanced by Mx, and My. However, given that

My/Mx equals D12/D11, and typically, D12 << D11, q is predominantly balanced by Mx.

• However, in absence of My, plate would have curved as shown in Fig. 17.1b, and a straight line AB, would have curved upwards as A’B’.

• This curvature which would have existed in absence ofM is called• This curvature, which would have existed in absence of My, is called anticlastic curvature.

• But, since My is non‐zero, it acts to make segment A’B’ once again ystraight.

• Also, note that My is non‐zero, because D12 is non‐zero, and the latter term is directly proportional to Poisson’s ratio ν Hence presence ofM andis directly proportional to Poisson s ratio ν12. Hence, presence of My, and straightening of AB is a demonstration of Poisson’s effect.

• Finally, absence of My, would have caused a non‐zero curvature in y direction, hi h ld h i l t d th diti th t ti l d i ti i di tiwhich would have violated the condition that partial derivatives in y direction

should be zero. Hence, we see that it is the material, which drives the condition of that partial derivative in y‐direction should be zero.

Solution for Case B• Case B is very similar to Case A except changes in some boundary• Case B is very similar to Case A, except changes in some boundary

conditions. The BCs for Case B are given below.

BCs at x = ‐a/2 BCs at x = +a/2

uo‐ = 0 uo+ = 0

vo‐ = 0 Vo+ = 0

Mx‐ = 0 Mx

+ = 0

• Plugging 1st set of BCs, i.e. uo± = 0 at x = ±a/2 in Eqs. 16.5 and 16.6, yields:C1 = C5 = 0. Thus,

wo‐ = 0 wo+ = 0

1 5 ,Nx(x) = 0 uo(x) = 0 Ny(x) = 0 (Eq. 17.6)

• Similarly, second set of BCs, i.e. vo± = 0 at x = ±a/2 yields:C2 = C6 = 0. Thus,Nxy(x) = 0 vo(x) = 0 (Eq. 17.7)

• Since out of plane problem is identical to Case A and is decoupled from• Since out of plane problem is identical to Case A, and is decoupled from inplane problem (B matrix is zero), thus, solution for out of plane problem is same as that for Case A.


Nachiketa Tiwari


Lecture 18



• Solution for Case CSolution for Case C

S l i f C• Solution for Case D

Solution for Case C

• Firstly, we reproduce equilibrium relations (Eq. 16.2), which apply to all cases. These are:

(Eq. 16.2)

• Further, the kinematic relations for Case C, are being reproduced from Eq. 16.1 as:16.1 as:

(E 16 1)(Eq. 16.1)

Solution for Case C

• For cases C and D, the laminate is [02/902]T. For such a laminate, we reproduce results from Eq. 16.4.

A = A = 0– A16 = A26 = 0– B16 = B26 = B12 = B66 = 0– D16 = D26 = 0 (Eq. 16.4)

• Finally, BCs for Case C are:Finally, BCs for Case C are:


uo‐ = 0 Nx+ = 0u 0 x 0

vo‐ = 0 Nxy+ = 0

Mx‐ = 0 Mx

+ = 0o 0 o+ 0

• Since Nx is zero at +a/2, we get from Eq. 16.2:C = N = 0 = A (duo/dx) + B (d2wo/dx2) or

wo‐ = 0 wo+ = 0

C1 = Nx = 0 = A11(duo/dx) + B11(d2wo/dx2), or,

uo(x) = [C1x + C5 + B11(dwo/dx)]/A11 (Eq. 18.1)

Solution for Case C

• Since Nxy is zero at +a/2, we get from Eq. 16.2:C2 = Nxy = 0 = A66(dvo/dx), or,

vo(x) = [C2x + C6]/A66 (Eq. 18.2)

• Also, we know that Mx = B11(duo/dx) + D11(d2wo/dx2). Integrating this equation after substituting the definition of Mx from Eq. 16.2, we get:

B11uo(x) + D11(dwo/dx) = ‐qx3/6 + C3x2/2 + C4x + C7 (Eq. 18.3)

• Equations 18.1‐3 can be used to solve for uo, vo, and wo. Here, term B11 couples in‐plane and out‐of‐plane displacements. Solving them yields:

(A11‐B211/D11) uo(x) = (B11/D11)(qx3/6) ‐ (B11/A11) (C3qx2/2) + (C1‐C4B11/A11)x + (C5‐C7B11/A11)

(D11‐B211/A11)wo(x) = qx4/24 ‐ C3x3/6 ‐ (C1B11/A11 ‐ C4)x2/2 + (C5B11/A11 ‐ C7)x + C8(D11 B 11/A11)w (x) qx /24 C3x /6 (C1B11/A11 C4)x /2 + (C5B11/A11 C7)x + C8(Eq. 18.4a, b)

Solution for Case C

• Also, My = D12(d2wo/dx2), and Mxy = 0 (Eq. 18.5)

• Equations 18.1‐18.5 are applicable to Case C and Case D, as BCs yet remain to be applied. At this stage, we apply boundary conditions for Case C.

• Since, Nx, Nxy, are zero at x = +a/2, we get:C1 = C2 = 0.

• Further, since vo is zero at x = +a/2, we get from Eq. 18.2:, / , g qC2 = 0.

• Further, since moment Mx is zero at both ends, we get from Eq. 16.2:C = 0 and C = qa2/8C3 = 0 and C4 = qa /8.

• Now, we are left with C5, C7, and C8. To find them, we use two BCs for wo

condition, and one BC for uo.

Solution for Case C

• Applying two BCs for wo in Eq. 18.4b, and one BC for uo in Eq. 18.4a, we get values for C5, C7, and C8. From these, we get relations for wo and uo:

uo(x) = (B11qa3)[4(x/a)3 ‐ 3(x/a) ‐ 1]/ [24D11(A11‐B211/D11)] (Eq. 18.6)

vo(x) = 0 (Eq. 18.7)

wo(x) = (‐5qa4)[16(x/a)4 ‐ 24(x/a)2 +5]/ [384 (D11‐B211/A11)] (Eq. 18.8)

• Equations 18.6‐8, constitute solution for displacement field for Case C.

• Next, we consider Case D. Here, the general solution as expressed through Eqs. 18.1‐5 remains same as Case C. However, integration constants are different, due to difference in boundary conditions.

Solution for Case D

• The boundary conditions for Case D are given below.


uo‐ = 0 uo+ = 0

vo‐ = 0 vo+ = 0

M ‐ = 0 M + = 0Mx = 0 Mx+ = 0

wo‐ = 0 wo+ = 0

• Using different boundary conditions we get the final solutions as:uo(x) = (B11qa3)[4(x/a)2 ‐ 1]/ [24D11(A11‐B211/D11)](x/a) (Eq. 18.9)

vo(x) = 0 (Eq. 18.10)

wo(x) = (‐qa4)[16(x/a)4 + 48{B11/(3A11D11) ‐ 1/2}(x/a)2 +

{5 ‐ 4B211/(A11D11)}]/ [384 (D11‐B211/A11)] (Eq. 18.11)

Comments on Cases C & D

• In Cases C and D, the direction of uo(x) depends on sign of B11.– The sign for B11 for a [02/902]T laminate is negative of that for a [902/02]T

laminatelaminate.– Thus, direction of uo(x) for a [02/902]T laminate is negative of that for a

[902/02]T laminate.

• In either case, uo(x) and wo(x) are odd and even functions of x. This observation is consistent with intuition.

• The term (D11‐B211/A11) is known as reduced bending stiffness. It appears as denominator in expressions for wo(x). The out of plane displacement is inversely proportional to this term, and not just D11. For symmetric laminates this term is identical to D Also higher the value of B lowerlaminates, this term is identical to D11. Also, higher the value of B11, lower reduced bending stiffness.

• The term (A11‐B211/D11) is known as reduced extensional stiffness. ItThe term (A11 B 11/D11) is known as reduced extensional stiffness. It appears as denominator in expressions for uo(x). The in‐plane displacement is inversely proportional to this term, and not just A11.


Nachiketa Tiwari


Lecture 20

Thermal Stresses in PlatesThermal Stresses in Plates



h i l d h l S i• Mechanical and Thermal Strains

• Stiffness Matrix for a Lamina

• Thermal Forces and Moments

Introduction

• A change in a body’s temperature causes its dimensions to grow or contract, thereby generating thermal strains.

• For a linear system, i.e. where thermal strain is directly proportional to change in temperature, the expression for thermal strain (εT ) for isotropic material is:

εT = αΔT

where, α is coefficient of thermal expansion, and ΔT is rise in body’s temperature.

• For an orthotropic unidirectional lamina, thermal strains in longitudinal and transverse directions are defined as:εTL= αLΔT

εTT= αTΔT

where, αL and αT are coefficient of thermal expansion in longitudinal and transverse directions, respectively. Also, it should be note that there are no‐shear strains in L‐T plane associated with thermal expansion. Hence, γTLT = 0.

Introduction

• Looking at expressions for thermal strains with respect to material axes, it can be stated that αL and αT are identical to thermal strains corresponding t t t i t f it H th ffi i t f ll thto temperature increment of unity. Hence, these coefficients follow the same transformation laws as strain vector.

• Thus, coefficient of thermal expansions measured with respect to an arbitrary coordinate system (x‐y) can be written as:

• Here, [T]‐1 is inverse of [T] as defined in Eq. 10.7. Using these relations, e e, [ ] s e se o [ ] as de ed q 0 Us g t ese e at o s,thermal strains in arbitrary coordinate system can be written as:

Mechanical and Thermal Strains

• Thermal strains by themselves cannot generate a resultant force or moment, unless the body is not completely free to deform due to t t Th t th l l f h l l i t th lt ttemperature. Thus, at the level of a whole laminate, there are no resultant forces and moments due to temperature alone.

• However, at the level of a lamina, the same may not be true. This is because a lamina by itself is not entirely free to bend or twist or expand due to temperature rise or fall.

• These stresses in a lamina are attributable to those strains which are in excess of thermal strains as defined in Eq. 19.2. These excessive strains are known as mechanical strains, and are denoted by a superscript M. Thus:

Mechanical and Thermal Strains

• In Eq. 19.3, we see that mechanical strain is the difference of total strain, and thermal strain.

• Further, thermal stresses at individual ply level may be calculated by multiplying mechanical strain vector with lamina stiffness matrix. Thus:

• However, in Eq. 19.4, mid‐plane strains and mid‐plane curvatures are not known.

• Equation 19.4 may be integrated over thickness (assuming temperature is constant over thickness) as per Eq. 12.1 to give resultant force vector. Similarly, Eq. 19.4 may be multiplied with z, and then integrated over y, q y p , gthickness as per Eq. 12.2, to give resultant moment vector.

Stiffness Matrices

• Thus, we get following relations for resultant forces.

Thermal Forces and Moments

• Here, thermal force and moment vectors are defined as:

(Eq. 19.6)

• Finally, the equilibrium equations for the plate are essentially the same as defined in Eqs. 14.7, 14.8, and 14.9. This is so, since resultant force and

t ti l d i l d th l ff t d t th i i dmoment equations already include thermal effects due to their revised definitions.

Thermal Forces and Moments

• Thermal stresses get induced in a laminate whenever the temperature of laminate differs from that of its free stress state.

• During the fabrication of composite plates, plies are stacked together at elevated temperatures. At these temperatures, matrix material permeates and different layers, and binds them because it gets cured. Later, the cured laminate is cooled to room temperature.

• As a result of this process, thermal stresses get induced in a laminate between different layers, because individual layers are no‐longer, after curing, free to respond to cooling. These stresses, which get induced in the laminate due to such cooling are known as residual or curing stresses.

• Residual stresses, if not properly managed, may lead to failure of laminate.


Nachiketa Tiwari


Lecture 20

Thermal Stresses in PlatesThermal Stresses in Plates


• Analysis of Thermal Stresses in Semi‐InfiniteAnalysis of Thermal Stresses in Semi Infinite Plate with Unsymmetric Lamination Sequence– Boundary Conditions– Boundary Conditions

– Equilibrium Equations

Constitutive Relations– Constitutive Relations

– Solution of Problem

C t– Comments

Free‐Free Unsymmetric Laminate

• Consider a free‐free unsymmetric semi‐infinite plate. Alternatively, if the dimensions of the plate are very large in one direction vis‐à‐vis other di ti it b h i i t th t f i i fi it l t L t thdirection, its behavior approximates that of semi‐infinite plate. Let the dimensions of such a plate be a X b.

• The plate undergoes a uniform temperature rise of ΔT.

• The lamination sequence for such a plate be [02/902]T.2 2 T

• Since the plate is unrestrained at all of its edges, its BCs can be written as:

BC t /2 BC t + /2BCs at x = ‐a/2 BCs at x = +a/2

Nx‐ = 0 Nx

+ = 0

Nxy‐ = 0 Nxy

+ = 0

Mx‐ = 0 Mx

+ = 0

Q‐x_eff= 0 Q+

x_eff= 0


• Also, because a >> b, we can state that:

• Applying these conditions in equilibrium equations, we get:


• Next, using BCs for Nx, Nxy, we get:

c1 = 0

c2 = 0

• Also, since there is no external load, q, and external moments at bothAlso, since there is no external load, q, and external moments at both ends of the plate are zero, we get from moment equilibrium equation:

c3 = 0

c = 0c4 = 0

• Thus, the solution for the plate for forces and moments is:

Nx = Nxy = Mx = 0. (Eq. 20.1)


• Next, we solve for displacement field for this plate.

• Here, we realize that the plate has a cross‐ply lamination sequence. Thus:A16 = A26 = 0

B12 = B16 = B26 = B66 = 0

D16 = D26 = 0.

• Also, for a cross‐ply laminate it can be easily shown that:TNTxy = 0

MTxy = 0

• Thus relations for N N and N can be written down using Eq 19 4 as:• Thus, relations for Nx, Nx and Nxy, can be written down using Eq. 19.4 as:


• Similarly, relations for Mx, My, Mxy, can be written as:

• In these relations, there five unknowns (uo’, vo’, wo’’, Ny, andMy), and fiveIn these relations, there five unknowns (u , v , w , Ny, and My), and five non‐trivial equations (three from Eq. 20.2, and two from Eq. 20.3). Here, superscript ‘ refers to total derivative of an entity with respect to x.

• Thus, these five equations can be solved, and integrated to get expressions for uo, vo, wo, Ny, and My. Out of these, the expressions for displacements are:d sp ace e ts a e


• In Eq. 20.4;

• These expressions may be used to further calculate force and momentThese expressions may be used to further calculate force and moment resultants using Eqs. 20.2 and 20.3.

• Thus the curvature of a free‐free semi‐infinite plate may be predicted• Thus, the curvature of a free‐free semi‐infinite plate may be predicted.

• It may be noted that term B11, changes sign when lamination sequence h f [0 /90 ] t [90 /0 ] S h hchanges from [02/902]T to [902/02]T. Such a change:– Has no influence on the sign of uo, as it is dependent on square of B11.

– However, the sign of wo, gets inverted because sign of B11as well as MTx flip as

lamination sequence gets invertedlamination sequence gets inverted.


Nachiketa Tiwari


Lecture 21

Finite Rectangular PlatesFinite Rectangular Plates



C l d i lifi i h• Commonly used simplification approaches while solving problems related to finite plates– Stiffness matrix elements

– Boundary conditions

• Out‐of‐plane BCs for different end conditions.p

Introduction

• Consider a rectangular composite plate with dimensions a and b. Solving for displacement field for such a plate is significantly different than that f i i fi it l tfor a semi‐infinite plate.

• In the semi‐infinite plate, we had assumed that all partial derivatives in one direction (e.g. y direction) were zero. Such a simplification renders equilibrium equations, which are originally Partial Differential Equations (PDEs), to Ordinary Differential Equations (ODEs).

• Such a simplification helped us develop closed form solutions for the problem at hand.

• The same simplification can not be used for finite plates, since the displacement field exhibits variation in both, x, and y directions.

Introduction

• Hence, only in very special cases, closed form solutions for finite plates are possible. In general, finite plate problems are solved using approximate

th d S f th l th d d t dd fi it l tmethods. Some of the more popular methods used to address finite plate problems are:– Galerkin method

R l i h Rit th d– Rayleigh‐Ritz method

– Finite element method

– Series methods

• The last approach, i.e. the series method, works well in some special cases, and yields solutions which are of “semi‐closed” nature.

• The overall strategy to solve a finite plate problem starts with simplification of the problem to the maximum possible extent, without compromising on the physics of the problem. Such simplifications can be applied to governing equations, kinematic relations, material properties, and BCs.

Common Simplification Strategies

• Kinematics: Look out for specific displacements which may be constant or zero throughout the field.

• Equilibrium: Some force resultants or moment resultants may be known or zero throughout the field.

• Stiffness matrices:– Symmetric laminates: Bij = 0. y ij

Such a simplification decouples resultant force {N} and moment vectors {M}. Hence the third equilibrium equation for z direction involving wo, gets decoupled with those for inplane displacement variables, i.e. uo and vo, when infinitesimal

i id dstrains are considered.

– Balanced and symmetric laminates: Bij = 0, and A16 = A26 = 0.

h l f l d l h f l f h lSuch a simplification not only decouples the out‐of‐plane response of the plate with in‐plane response, but also extensional strains with shear strains.


• Stiffness matrices (contd.):– Symmetric and specially orthotropic laminates: Bij = A16 = A26 = D16 = D26 = 0

Such a simplification not only decouples the out‐of‐plane response of the plate with in‐plane response, but also extensional strains do not cause shear strains. Further, there is no twist curvature in the system, unless it is caused by externally applied twisting moments In such a case we can solve the inplane part of theapplied twisting moments. In such a case, we can solve the inplane part of the problem independently vis‐à‐vis out‐of‐plane part of the problem.

– Non‐symmetric but specially orthotropic laminates: A16=A26 = B16=B26 =D16=D26 =0y p y p 16 26 16 26 16 26

– Balanced and anti‐symmetric laminates

– Cross‐ply and anti‐symmetric laminates


• Boundary Conditions: Equation 15.2 lists four sets of boundary conditions for each edge of the plate. These are relisted below for edge x = a.


• Boundary Conditions (contd.): Out of these four BCs listed earlier:– Conditions (i) and (ii) relate to inplane variables.

– Conditions (iii) and (iv) relate to out‐of‐plane variables.

– If the laminate is symmetric and we are only interested in out‐of‐plane response of the plate [i.e. only wo(x,y)], then only BCs (iii) and (iv) may be considered, and appropriate simplifications may be made.

• Next, we look at different edge conditions for rectangular plates, and their associated boundary conditions. Here we are addressing only out‐of‐plane boundary conditionsboundary conditions.

Different Edge Conditions for Plates

• Simply Supported Condition: For a plate simply supported along edge x = a/2, the out‐of‐plane BCs along this edge will be:– wo = 0 and M+

x = 0.

– The condition for moment can be further elaborated as shown below.

• Clamped Support: Here, displacement and slope are zero along the edge.


• Traction Free Edge: For a plate with traction free edge at x = a/2, the relevant BCs will be Q+

x = M+x = M+

xy = 0. Substituting this in BC (iii) in Eq. 15.2, we get,

(Equivalent BC for traction free edge)

since Q+x and partial derivative of Mx w.r.t. to y are zero along edge x = a/2. x x

Also from Eq. 15.2‐iv, we set the other BC for traction free edge as:

Mx = 0.

• Now for a symmetric laminate,


• Traction Free Edge (contd.): Thus, out‐of‐plane BCs for such plates at x=a/2 are:

• For specially orthotropic case, when D16 = D26 = 0, we get:


Nachiketa Tiwari


Lecture 22Simply Supported Plates with Normal LoadSimply Supported Plates with Normal Load


• Analysis of a Simply Supported Plate on AllAnalysis of a Simply Supported Plate on All Four Sides– Governing equations– Governing equations


Stiffness matrices– Stiffness matrices

– Series solution

• Assessing Convergence of Solution

Simply‐Supported Plate on All Sides

• Consider a plate of dimensions a X b in x and y directions, respectively, and which is simply supported on all four sides.

• Further, we assume that this plate has a symmetric and specially orthotropic lamination sequence. Thus,– [B] = [0] due to symmetry

– A16 = A26 = D16 = D26 = 0

• Finally we assume that the plate is normally loaded as shown in Fig. 22.1 with a load intensity of q(x,y).

q(x y)

x

q(x,y)

y

Fig. 22.1


• For such a plate, the out‐of‐plane boundary conditions are:– w± = 0 on all four sides.

– Mx± = 0 at x = 0, a.

– My± = 0 at y = 0, b.

• As explained earlier, the governing equation for equilibrium for out‐of‐plane direction for such a plate is decoupled with in‐plane equations, because the plate’s lamination sequence is symmetric. This equation is

d d b lreproduced below.

• Expressing this equation in terms of derivatives of wo, we get:

(Eq. 22.1)(Eq. 22.1)


• Equation 22.1 has to be solved for wo. At this stage, we assume a series function for wo, which satisfies the boundary conditions listed earlier. Let thi d f ti bthis assumed function be:

where, m, and n are integers which can vary between 1 and infinity.

• Such a series expression for wo satisfies out‐of‐plane boundary conditions.Such a series expression for w satisfies out of plane boundary conditions. If this expression also satisfies Eq. 22.1, then it is a valid solution. For this expression to satisfy Eq. 22.1, we put it into the equation, and get the following condition, which has to be satisfied for getting a valid solution for wo.

(Eq. 22.2)


• Now, if q(x,y) can also be expressed as a double Fourier series, as shown below:

then, for Eq. 22.2 to be true, the following condition must hold.

• Hence, if we know qmn, we can find out‐of‐plane deflections for such a plate.


• The values for qmn can be gotten by decomposing q(x,y) into a double Fourier series as shown below.

• While solving for qmn, following identities may be used.mn

• If functions are more complex, then Gaussian integration may be used.

• Also, if there are point loads, then Dirac‐Delta functions δ[(x‐xo), (y‐yo)] may be used while computing qmn using above formula.

Convergence of Solution

• While computing the out‐of‐plane response for the plate, one has to estimate the number of terms to be used in serial expansion of q(x,y). Th ti ll th l th b f t th t (Theoretically, the larger the number of terms the more accurate (or converged) is the solution.

• But then, one question that requires careful consideration relates to the context in which accuracy/convergence is being assessed. Are we looking for convergence of displacement, or stress, or load, or … ?

• In this context, consider that q(x,y) is constant over the plate, and has a value qo. For such a load, the Fourier series expansion is:

where, m = 1, 3, 5, …. and n = 1, 3, 5, …, , , , , , ,


• Here, for a two‐term solution (i.e. m, and n can assume two distinct values):

/ 2 / 2 / 2q1,1 = (16qo/π2) q3,3 = (16qo/π2) X (1/ 32)

• Similarly, for a 20‐term solution:q1,1 = (16qo/π2) q39,39 = (16qo/π2) X (1/ 392)

• Hence, we see that load converges in a 1/m2 way.

• Next, we look at convergence of deflection. Earlier, for this plate, it has been shown that:wmn = qmn/dmn

• But, we know that d and q are directly proportional tom4 and 1/m2,But, we know that dmn and qmn are directly proportional to m and 1/m , respectively. Thus, deflection converges in a 1/m6 fashion.


• Similarly, we can find the pace of convergence for other parameters. The following table lists the rate of convergence for different mechanical

t llparameters as well.Load m‐2

Out‐of‐plane shear m‐3

4Curvature m‐4

Resultant Moment m‐4

Stress m‐4

l ( ) 6Displacement (w) m‐6

• Hence, we find that among the unknowns, out‐of‐plane shear converges h l l h f l d lmuch more slowly than out‐of‐plane displacement.


Nachiketa Tiwari


Lecture 23Rectangular Plates with SS on Two SidesRectangular Plates with SS on Two Sides


• Analysis of a Normally Loaded Plate withAnalysis of a Normally Loaded Plate with Simple Supports on Two Sides– Governing equations– Governing equations


Stiffness matrices– Stiffness matrices

– General Approach to Solution

• Assessing Convergence of Solution

Rectangular Plate with SS on Two Edges

• Consider a plate of dimensions a X b in x and y directions, respectively, and which is simply supported on two sides (y=0, b). On other two edges, there are some other support conditionsthere are some other support conditions.

• Further, we assume that this plate has a symmetric and specially orthotropic lamination sequence Thusorthotropic lamination sequence. Thus,– [B] = [0] due to symmetry– A16 = A26 = D16 = D26 = 0 due to special orthotropy


q(x y) SS

x

q(x,y) SS

Some other support condition


y

Fig. 23.1SS

Rectangular Plate with SS on Two Edges• For such a plate, the out‐of‐plane boundary conditions are:

– wo± = 0 at y = 0, b.

– My± = 0 at y = 0, b.

• Consider the BC on My as stated above.

• Thus, the out‐of‐plane BCs for edges y=0 and y=b, can be rewritten as:

Rectangular Plate with SS on Two Edges• Also, the out‐of‐plane BCs for edges x=0 and x=a, could be anything

(clamped, free or elastic foundation ...). Such a problem has been approached by Whitney in a very unique way. He breaks this problem, i i i l f li i i h i Fi 23 2using principles of linear superposition as shown in Fig. 23.2.

SS SS

Correction

SS

Original BC

1SS SS

2

CorrectionCorrection

3

Original BCOriginal BC

+ =Fig. 23.2

• In such an approach, the original problem (3) is decomposed into a sum of two problems (1, and 2). We have already developed solution for Problem

SS SS SS

p ( , ) y p1. If we can figure out corrections to this, (Problem 2), then, the sum of these two solutions will give us answer to Problem 3.

• Such an approach works only if equilibrium, kinematic, and material constitutive relations are linear in nature.

Rectangular Plate with SS on Two Edges• It has been shown earlier, that the out‐of‐plane solution for Problem 1 is:

• Now, let us develop a general correction in context of Problem 2, as defined in Fig. 23.2.defined in Fig. 23.2.

Rectangular Plate with SS on Two Edges• The general for of this correction could be:

• Such a correction function satisfies all the out‐of‐plane boundary conditions for edges y=0 and y=b, regardless of our choice for n(x).

• Hence, the overall solution of our original problem, which is Problem 3 as per Fig. 23.2, is:

• This can be rewritten as:

Rectangular Plate with SS on Two Edges• Now, Eq. 23.1 should satisfy the governing equation, which is:

(Eq. 23.2)

• Putting in this equilibrium eqn. the assumed solution for Problem 3, we get:

(Eq. 23.3)

• Here, the terms in red add up to zero, as they collectively represent the solution for Problem 1 i e for a plate which is simply supported on all foursolution for Problem 1, i.e. for a plate which is simply supported on all four sides. Hence, above equation can be simplified and rewritten as:

(Eq. 23.4)

Rectangular Plate with SS on Two Edges• Equation 23.4 may be simplified as:

(Eq. 23.5)

• Equation 23.5 is an ordinary differential equation of 4th degree. It has no forcing term. Hence, it is also homogenous in nature. We can assume its solution to be an exponential function as shown below:

φn(x) = Serx

• Substituting this assumed expression for φn(x) in Eq. 23.5, we get:

(Eq. 23.6)

• For Eq. 23.1 to be a valid solution for w, above condition must be satisfied.

Rectangular Plate with SS on Two Edges• One of the solutions for Eq 23 6 is S = 0• One of the solutions for Eq. 23.6 is S = 0.

– This is the case for a plate which is simply supported on all four edges. In such a case, no correction, as defined in Fig. 23.2, is required.

• For other BCs’ along edges x=0, a, the following condition must be satisfied.

• This equation can be rewritten as:

• This is a bi‐quadratic equation with following four roots λ1,2,3,4.

• Thus, we get four different values of r, which are:


• Thus the complete solution for the problem is:

(Eq. 23.7)

• This is the general solution for a plate which is:– Simply supported along edges y = 0 and y = b.

– Has some other conditions along edges x = 0 and x = a.

• In Equation 23.7, constants S1, S2, S3 and S4 can be determined by applying actual boundary conditions along edges x = 0 and x = a.

• In subsequent lecture, we will find solutions for such plates, when actual conditions along edges x = 0 and x = a are known.


Nachiketa Tiwari


Lecture 24Rectangular Plates with SS on Two SidesRectangular Plates with SS on Two Sides


• In previous lecture, the solution for a plate with dimensions a X b in x and y directions, respectively, and which is simply supported on two sides (y=0 b) was developed(y=0, b) was developed.

• Such a solution was specific to a plate with following two conditions.– The plate’s lamination sequence is symmetric and also specially orthotropicThe plate s lamination sequence is symmetric and also specially orthotropic

lamination sequence. Thus,– [B] = [0] due to symmetry– A16 = A26 = D16 = D26 = 0, due to special orthotropy.– The plate is normally loaded as shown in Fig. 23.1 with a load intensity of

q(x,y).

q(x y) SS

x

q(x,y) SS



y

Fig. 23.1SS

Rectangular Plate with SS on Two Edges• For such a plate, using principles of linear superposition, we developed a

general solution for out‐of‐plane displacement in Eq. 23.7. This has been reproduced below.

where,

(Eq. 23.7)

• It can be seen from above relations that the values of λ are real, because ,the term within [] brackets is always positive. Hence, we state four values of λ as:λ 1 = +λIλ 2 = ‐λI[[λ 3 = +λIIλ 4 = ‐λII4 II

where, λI and λII are real and positive numbers.


• Thus the complete solution for the problem is:

• Now we know that:


• In this expression for n(x):– An, Bn, Cn, and Dn, depend on boundary conditions.

b d d l t ’ i– a, b, depend on plate’s size.

– λI and λII depend on plate’s bending stiffness matrix elements, D11, D12, and D66.

Th i f ( ) b b i d b k i E 23 7• The expression for n(x) can be substituted back in Eq. 23.7 to get true solution for a specially orthotropic plate which is simply supported on two opposite sides, and has a different set of boundary conditions on its remaining two edgesremaining two edges.

• Next, we look at the solution for a plate when it is clamped on two edges, d i l t d th t dand simply supported on other two edges.

Rectangular Plate with SS and CC Edges

• For a plate which is clamped on edges x=0, and x=a the boundary conditions are:

wo 0 at edges x = 0 and x = a– wo = 0 at edges x = 0, and x = a

– at x = 0, and x = b.

• These four boundary conditions give us four sets of equations which may be• These four boundary conditions give us four sets of equations, which may be used to solve for constants, An, Bn, Cn, and Dn.

Th l l i f hi l d l d i E 23 7 i• The general solution for this plate, as developed in Eq. 23.7 is:

(Eq. 24.1)


• Applying the 1st boundary condition, wo = 0 at x=0 edge, we get:

(Eq. 24.2)

• Applying the 2nd boundary condition, wo = 0 at x=a edge, we get:

(Eq. 24.3)

• The 3rd and 4th boundary conditions involve derivative of w with respect to xThe 3 and 4 boundary conditions involve derivative of w, with respect to x. The partial derivative of w with respect to x is:


• This derivative is zero at x=0, and x=a. Forcing this derivative to be 0 at x=0 we get:

(Eq. 24.4)

ll l h f h b d d h h l• Finally, we implement the fourth boundary condition, which is a partial derivative of w w.r.t. x, is zero at x = a.

(Eq 24 5)(Eq. 24.5)

Rectangular Plate with SS and CC Edges• Equations 24 2‐24 5 are four equations which involve summations over• Equations 24.2‐24.5 are four equations, which involve summations over

index n. For these equations to be always valid, the following equations have to be satisfied.

• Thus, for each value of n, we have four simultaneous linear equations, with A B C and D as unknowns All other parameters in these equations areAn, Bn, Cn, and Dn, as unknowns. All other parameters in these equations are known. Hence these equations can be solved for the unknowns.


Nachiketa Tiwari


Lecture 25Virtual Work ApproachesVirtual Work Approaches

Introduction

• Till so far, we have solved problems using exact approaches. Two general approaches have been used in this regard:– Simplification of the problem by converting PDEs into ODEs, and then

integrating them to get exact solutions.

– Simplification of the problem, and then using series approach to satisfy PDEs , and through such a process extracting constants used in the seriesand through such a process extracting constants used in the series.

• However, such approaches, even though they provide exact solutions have a limited range of applicability In a very large number of problems sucha limited range of applicability. In a very large number of problems such approaches are difficult to implement because:– Either closed form solutions do not exist.

– Or indentifying them is not very much obvious– Or indentifying them is not very much obvious.

• Hence, a slew of other solution approaches have been developed to address relatively more complex problemsaddress relatively more complex problems.

Introduction

• Consider a rectangular plate of dimensions 4a X 4b, which is clamped on all four edges. This is shown below in Fig. 25.1.

x

q(x,y) Clamped

Clamped x

Fig. 25.1

Clamped Clamped

yClamped

• For such a problem the boundary conditions require w and its slope to be zero at all four edges of the plate.

Introduction• Thus, we assume a 1‐term solution, which satisfies these boundary

conditions. This 1‐term solution and its derivatives are given below.

• Such a function satisfies all the boundary conditions which are:Such a function satisfies all the boundary conditions, which are:

Introduction• Even though the assumed function satisfies all the BCs, it does not satisfy

the equilibrium equation for wo. This is shown below.

• Assume that the plate is specially orthotropic, and also symmetric. Thus, the out‐of‐plane equilibrium equation for such a plate is:

• Substituting the expression for wo in this relation, we get:

• We see here, that the LHS of the equation does not become zero. Thus, there is an error [E(x y)] in the equation and hence our assumed solutionthere is an error [E(x,y)] in the equation, and hence, our assumed solution is not a valid one. The units of this error N/m2.

Virtual Work Approaches• It may noted here that the fact statement error [E(x,y)] is not zero means:

– There is an imbalance between external and internal forces in z direction because in this specific case, the error E(x,y) relates equilibrium equation for p , ( ,y) q qz‐direction.

– The value of error changes from point to point over the surface of the plate. Hence, the equilibrium equation is not satisfied in a point to point‐wise sense.

• However, if we chose to equate virtual work over the entire domain (in this case the area of the plate), then we do have a solution which may be p ), ygood enough, though not the best possible.

• “Virtual work” is the dot product of forces and virtual displacements.p p

Virtual Work Approaches• A virtual displacement is an assumed infinitesimal change of system

coordinates occurring while time remains stationary. Such displacements are called “virtual” and not “real” since no actual displacement can occur pwhile time is held stationary.

• Virtual displacements should be such that all kinematic conditions of theVirtual displacements should be such that all kinematic conditions of the problem are satisfied. This in turn implies that:– Boundary conditions on displacements and slopes must be satisfied.

– Continuity within boundaries must be maintainedContinuity within boundaries must be maintained.

• Finally we note that imposition of virtual displacements does not lead to change of forceschange of forces.

Galerkin Method• Consider wo(x,y) as the actual displacement field for a plate.

– From this reference condition the plate may be virtually perturbed by some virtual displacement field εwo

1 (x,y). Such a virtual displacement field should p 1 ( ,y) psatisfy kinematic conditions.

– Here, ε is small scale parameter. It gives “smallness” to virtual displacements. Thus,

– Total displacement in z‐direction = wo(x,y) + εwo1 (x,y).

• In Galerkin’s method we set virtual work to zero. Thus, for out‐of‐plane pequation we can write virtual work statement as:– Virtual work = E[wo(x,y)]∙ εwo

1 (x,y) dxdy = 0,

– where,

– E[wo(x,y)]∙ dxdy is residual force.

• However, above statement of virtual work cannot be valid, becauseHowever, above statement of virtual work cannot be valid, because none of three terms in it, E[wo(x,y)], ε and wo

1 (x,y) are zero.

Galerkin Method• However, our earlier statement of virtual work cannot be valid because

none of the three terms in it, E[wo(x,y)], ε and wo1 (x,y) are zero.

• Hence, we satisfy virtual work statement in an integral sense.

h ll h b d• Mathematically, this may be expressed as:

∫ΩE[wo(x,y)]∙ εwo1 (x,y) dxdy = 0 (Eq. 25.1)

• Equation 25.1 is the General Galerkin Equation corresponding to out‐of‐plane equilibrium equation. p q q

• Similar relations may be developed corresponding to other equilibrium equations as wellequations as well.

Comments on Galerkin Method• We are able to satisfy the equilibrium equation only in an integral sense,

and not in a point‐by‐point sense.

• Since as the differential equation is satisfied only in an integral sense (i.e. over the entire domain), we are actually forcing the integral of error to become zero Hence Galerkin solution is stiffer vis‐à‐vis exact methodbecome zero. Hence Galerkin solution is stiffer vis à vis exact method solution.

• In this method there is no restriction on the choice of virtual displacement• In this method there is no restriction on the choice of virtual displacement field except that it has to satisfy kinematic conditions. Hence, it is known as General Galerkin Method.

• In Special Galerkin Method, we choose virtual displacement field such that it is of the same form as that of wo(x,y). In such a method, statement of virtual work corresponding to the 3rd equilibrium equation is:of virtual work corresponding to the 3rd equilibrium equation is:

∫ΩE[wo(x,y)]∙ εwo (x,y) dxdy = 0 (Eq. 25.2)

Comments on Galerkin Method• There are several interpretations of Galerkin method. One of the

interpretations has already been discussed earlier, i.e. total virtual work done over the domain is zero.

• Another common interpretation relates to the notion of orthogonality. We know that work can be defined as:know that work can be defined as:dW = F∙dU,

i.e. work is a scalar, which is inner product of force and displacement vectors.

• Thus, if F and dU are mutually orthogonal, then their inner product, i.e. dW will be zero. Now in Eq. 25.1, we can state that in an average sense:(Error in force) Virtual Displacement = 0 is the condition for equilibrium(Error in force) ∙ Virtual Displacement = 0 is the condition for equilibrium.

• This implies that in an average sense, error forces and virtual displacements are mutually orthogonaldisplacements are mutually orthogonal.


Nachiketa Tiwari


Lecture 26Application of Galerkin MethodApplication of Galerkin Method

Application of Galerkin Method: Example 1

• Here, we consider a beam which is simply supported at both ends, as shown in Fig. 26.1. The overall length of the beam is L.

x

Fig. 26.1

• Here we assume:

xz

• Here, we assume:– Material is isotropic.

– Normal uniform load of intensity q0 is applied over the length of the beam.

Strain displacements relations are linear– Strain‐displacements relations are linear.

• For such a system, the governing equation for normal deflection is:EI(d4w/dx4) ‐ q0 = 0. (Eq. 26.1)


• In this example, we will compare the accuracy of Galerkin solution for Eq. 26.1 with exact solution.

• In such a case, the boundary conditions are:– Displacement at ends of the beam is zero, i.e. w = 0 at x = ±L/2.

– Moment at both beam ends is zero.

• At this stage, we select a displacement function, which satisfies the kinematic boundary conditions. Thus, we assume:

wo(x) = A cos (πx/L)

• Such a displacement function satisfies the displacement BC at both ends of beam. Substituting this function in governing equation gives us error in the force. The relation for this error is:

E[w(x)] = AEI∙(π/L)4∙ cos (πx/L) ‐ q0 (Eq. 26.2)


• The virtual work done by this error force as defined in Eq. 26.2, when integrated over the length of beam should be zero. Thus, we get:

∫ΩE[w(x)]∙ εw1(x) dx = 0, where integration limits are ‐L/2 and L/2.

or,

∫Ω [AEI∙(π/L)4∙ cos (πx/L) ‐ q0 ]∙ εw1(x) dx = 0. (Eq. 26.3)Ω 0 1

• At this stage, we chose w1 as defined below:

w (x) = A cos (πx/L)w1(x) = A1 cos (πx/L)

• Thus, Eq. 26.3 can be rewritten as:

/ 4 / /∫Ω [AEI∙(π/L)4∙ cos (πx/L) ‐ q0 ]∙ ε A1 cos (πx/L) dx = 0. (Eq. 26.4)

• From Eq. 26.4, we get:

A = 4q0L4/(π5EI) (Eq. 26.5)


• Thus, the approximate solution as per Galerkin method is:

w(x) = [4q0L4/(π5EI)] cos (πx/L)

• At x = 0, the beam deflection is:

wGalerkin(0) = 0.01309[q0L4/(EI)]wGalerkin(0) 0.01309[q0L /(EI)]

• Also, the exact solution for beam deflection is:

(0) 0 1302 [ L4/(EI)]wExact(0) = 0.1302 [q0L4/(EI)]

• Comparing exact and approximate values (as per Galerkin method), we see the two answers are fairly close to each other.


• Now through Example 1, that the Galerkin solution for displacement is slightly less than as predicted by exact solution. This is because Galerkin

th d di t hi h tiff f tmethod predicts higher stiffness for a system.

• To improve the accuracy of the solution we may use more than one term in the assumed solution form. Thus, we assume:

wo(x) = A1 cos (πx/L) + A3 cos (3πx/L)

• Such an assumed expression for w(x) satisfied the kinematic boundary conditions at both ends of the simply supported beam. At this stage, we also assume an expression for virtual displacement. Thus,

w1(x) = B1 cos (πx/L) + B3 cos (3πx/L)

• Like the expression for w(x) the expression for virtual displacement w (x)Like the expression for w(x), the expression for virtual displacement, w1(x), also satisfies kinematic boundary conditions.

Application of Galerkin Method: Example 2• At this stage, we develop an expression for virtual work over the entire

domain, and equate it to zero. Thus:

• But we know that B1, and B2 can have arbitrary magnitudes. Hence,But we know that B1, and B2 can have arbitrary magnitudes. Hence, integral of virtual work over domain can only be zero, if both terms in [ ] are individually zero.

Application of Galerkin Method: Example 2• Thus we get two parallel equations in A1, and A3. Solving for these

equations gives us:– A1 = 4q0L4/(π5EI)

– A3 = ‐A1/35

• Thus,

w(x) = [4q0L4/(π5EI)] [cos (πx/L) ‐ (1/35) cos (3πx/L)].

• At the center i e when x = 0 the value of displacement of the beam is:At the center, i.e. when x 0, the value of displacement of the beam is:

wGalerkin(0) = 0.01309[q0L4/(EI)] using 1‐term solution

wGalerkin(0) = 0.01302[q0L4/(EI)] using 2‐term solution

(0) 0 1302 [ L4/(EI)] l iwExact(0) = 0.1302 [q0L4/(EI)] exact solution.

• Thus, we see that in this case, a 2‐term solution brings us remarkably close to the exact solution.

Application of Galerkin Method: Example 2• Based on this analysis, we make following observations.

– As we increase the number of terms in an assumed form of Galerkin solution, it approaches the exact value monotonically.

– A solution with lesser number of terms represents a stiffer system vis‐à‐vis a system which has more terms.

– As shown later, this asymptotic convergence on stiffness (and displacements) occurs because Galerkin method is similar to a “least squares approach”. As number of terms increase so does the flexibility of the system therebynumber of terms increase, so does the flexibility of the system, thereby reducing its overall energy. Hence, accuracy of solution increases with number of terms.

– Galerkin method tends to help itself.

– This method can be easily extended for plates.


Nachiketa Tiwari


Lecture 27The Galerkin MethodThe Galerkin Method

Example 3: Simply‐Supported Plate on All Sides

• Consider a plate of dimensions a X b in x and y directions respectively, which is simply supported on all four sides.


– A16 = A26 = D16 = D26 = 0 due to special orthotropy.

• Finally we assume that the plate is normally loaded as shown in Fig. 27.1 with a constant load intensity of q0.

q(x y)

x

q(x,y)

y

Fig. 22.1



– Mx± = 0 at x = 0, a.

– My± = 0 at y = 0, b.



• Expressing this equation in terms of derivatives of wo, we get:

(Eq. 22.1)(Eq. 22.1)

Example 3: Simply‐Supported Plate on All Sides• For such a plate we have already developed an exact solution Here we• For such a plate, we have already developed an exact solution. Here we

develop Galerkin solution and compare it with the exact solution.

F thi th t th l ti i f f• For this, we assume that the solution is of form:

• Such an assumed solution form satisfies all the kinematic BCs for the problem. Using this, we compute the error in PDE.

• Multiplying this error with virtual displacement and integrating the product over plate’s area we get:p p g

(Eq 27 1)(Eq. 27.1)


• From Eq. 27.1, we get:

• This value of w11 is exactly the same as that determined in the exact solution. In this case, we get same results because our choice of w(x,y) coincides with that of the exact solution.

• In this Example, we have assumed the plate to be specially orthotropic. Thus the role of terms D16 and D26 was not present Later we will exploreThus, the role of terms D16 and D26 was not present. Later, we will explore this role by solving another problem which is slightly different than Example 3.

Another Interpretation of Galerkin Method

• Consider the 1‐D beam equation as discussed earlier. For such a beam, the error in force equilibrium equation using an assumed solution is:

E[w(x)] = EI∙(d2w/dx2)‐ q0

• The value of such an error changes with position, both in magnitude and isign.

• Also, the integral of square of this error over the domain (i.e. beam length) can be expressed as:

∫Ω {E[w(x)]}2 dx,

or,

∫Ω [EI∙(d2w/dx2)‐ q0]2 dx


• A “good enough” solution for w(x), would be when the integral of this squared error is minimized.

• The condition for a minima of a function is when its 1st derivative is zero. Using this, we get the condition for minima of integral of square‐error as:

• At this stage, we assume a form for w(x), and introduce it in above equation. We assume,


• Now the integral of square error involves E, I, A, L, qo, but not x. Thus, when this error is minimized we have to differentiate it w.r.t. A, because E, I d L k t tI, and L are known constants.

• Thus, the condition for minima of integral of square error is:Thus, the condition for minima of integral of square error is:


• However, since the integral is over x, we can rewrite above expression as:

Thi ld b itt• This could be re‐written as:


• Thus we see that such an approach is equivalent to Special Galerkinmethod, where mathematical expressions for virtual displacement and t l di l t i h thactual displacement expressions are very much the same.

• As Galerkin method involves a least squares approach, multiple terms increase the flexibility of the system so that its energy is reduced.

• Hence, the accuracy of Galerkin method increases with number of terms used in the solution.

• Such an interpretation can also be generalized in context of partial Suc a te p etat o ca a so be ge e a ed co te t o pa t adifferential equations.


Nachiketa Tiwari


Lecture 28The Role of D in a Simply‐The Role of D16 in a Simply‐

Supported Rectangular Plate

Influence of D16 on Response of a SS‐SS Plate


• Further, we assume that this plate has a symmetric.– [B] = [0] due to symmetry

• Here, we do not assume that the plate is specially orthotropic. Hence terms D16, and D26, are not zero.


q(x y)

x

q(x,y)

y

Fig. 28.1



– Mx± = 0 at x = 0, a.

– My± = 0 at y = 0, b.



(Eq. 28.1)

• Also, writing expressions for Mx, My, and Mxy, we get:


• Also, writing expressions for Mx, My, and Mxy, we get:

(Eq. 28.2)(Eq. 28.2)

• Substituting these equations in Eq. 28.1 yields:

(Eq. 28.3)


• At this stage, we use Galerkin approach to solve this equation. Here we assume that out‐of‐plane displacement is of the form:

(Eq. 28.4)

• We chose such a form for wo, because it satisfies all the kinematic boundary conditions of the system. Inserting Eq. 28.4 in Eq. 28.3, we get the error in the equation as:

(Eq. 28.5)


• As per Galerkin method, the virtual work performed by this force error, when integrated over plate’s area, should be zero. This is shown in Eq. 28.6.

(Eq 28 6)(Eq. 28.6)

• Equation 28.6 involves summation over indices m, and n. In this equation constants w are unknown When this equation isequation, constants wmn, are unknown. When this equation is integrated over the domain, we will get several terms multiplied by wij coefficient of virtual work function, where i and j, are indices with

i i t lvarying integer values.


• Equation 28.6 gives a set of simultaneous equations which equal m X n in number, where values of wmn are not known. Solving these equations gives

th d i d l tius the desired solution.

• We have seen that in Galerkin approach, our choice of displacement function requires only kinematic boundary conditions to be satisfied. However, in this problem, the displacement as well as moment along all four edges of the plate is zero.

• While our choice displacement function (as defined in Eq. 28.4) ensures satisfaction of displacement boundary condition, the same may not be true of the following boundary conditions:– Mx

± = 0 at x = 0, a.

– My± = 0 at y = 0, b.


• Consider the expression for Mx.

• Substituting Eq. 28.4 in this equation we get:

M h ld b t l d 0 At 0 th t i t th At th t• Mx should be zero at along edges x=0,a. At x=0, that is not the case. At that edge, Mx, as calculated from above equation, is:


• Thus we see that usage of approximate methods such as Galerkin approach, generates moments at edges for certain lamination sequences, even if the d i i l t dedge is simply‐supported.

• This phenomenon is called the “Edge Effect”.

• Thus, while using such approaches, care has to be taken that these residual moments due to edge effect are not significantly large. For this the following is recommended.– Find the maximum value of Mx over the plate’s area. For a rectangular plate

which is simply supported on all edges and which is loaded with uniform load ll h f h lintensity, Mmax, may typically exist at the center of the plate.

– Ensure that the ratio of “residual moments” at edges, and Mmax, is small. Typical upper threshold for this ratio should be 5%.

Influence of D16 on Response of a SS‐SS Plate16

• Like D16, D26 causes generation of edge effects along edges y=0,b.

• This is a limitation of all approximate methods. The extent of this problem may be minimized by increasing the number of terms in approximate lsolution.

• Such an approach drives edge‐moments closer to zero.

• Similar issues are also encountered in other approximate methods as well, such as the finite element method.

• A reasonably accurate and converged solution requires that these edge moments must be minimized to the extent possiblemoments must be minimized to the extent possible.


Nachiketa Tiwari


Lecture 29Energy MethodsEnergy Methods

Introduction

• Till so far, we have deduced solutions for laminated composite plates by solving equilibrium equations.

• These equations are derived from Newton’s Laws of Motion, as per which “for an equilibrium state there must be no net force acting on the body” (or system).

• There is an alternative approach to solve these problems without using Newton’s method. This method was originally developed by John Bernoulli (1667‐1745) and later pefected by Lagrange (1736‐1783).

• Bernoulli conceived the notion of “virtual work” to solve the same problem. As per this method, “for all possible displacements (of a system), the sum of the products of force and initial displacement in the direction of the force (i.e. the virtual work) must balance for equilibrium.

Introduction

• Comments on Newton’s and Bernoulli approaches:– These are totally different concepts.

– Newton’s method is force oriented, and free‐body‐diagrams have to be constructed to develop equilibrium equations.

– Bernoulli’s method is displacement oriented (or configuration oriented).

B h h h d id fi l f ilib i– Both these methods provide us final state of equilibrium.

– These methods are mutually independent.

• The virtual work method is applicable only for static problems. However, its dynamic analogue is d’Alembert’s Principle, which is also known as Lagrange‐d’Alembert’s Principle.

i l “ i ” f hi h d i k “ il ’ i i l ”– A special “version” of this method is known as “Hamilton’s Principle”.

– Also, it can be mathematically shown that this principle maps to the Principle of Virtual Work in static systems.

Introduction• Here we will understand how these methods can be used to address• Here, we will understand how these methods can be used to address

problems of statics and dynamics.

• Specifically we will use:• Specifically, we will use:– Principle of minimum potential energy (similar to virtual work principle) to

address static problems.– Hamilton’s method to solve dynamic problems.

• There are several significant advantages of energy methods over Newton’s method. These are:– Problem formulation is relatively straightforward.– Such formulations can be expressed in “weak” and “strong” versions. Weak

formulations, when automated in FEA lead to symmetric stiffness matrices for linear systems.linear systems.

– It is easy to discern boundary condition requirements using such methods.

• However, Newton’s approach is more intuitive vis‐à‐vis energy based pp gymethods.

Total Potential Energy Principle

• The total potential energy in a structural system can be expressed as:

(Eq. 29.1)

• Here, W, is the strain‐energy density function, Bi represents body force per iunit volume in i‐direction, and Ti represents traction force per unit area in i‐direction. For purposes of brevity only, we will omit body forces going further.

• Now, we know that:

W = [σxxεxx+ σyyεyy+ σzzεzz+ τxyγxy + τyz γyz + τzx γzx ]/2[σxxεxx σyyεyy σzzεzz τxyγxy τyz γyz τzx γzx ]/

• For a laminated plate, due to assumption of normality, we can discard shear strains in y‐z and z‐x plane Also using Eq 10 11 in above relation we get:strains in y‐z and z‐x plane. Also, using Eq. 10.11 in above relation we get:

W = [Q11ε2xx+ 2Q12 εxxεyy + 2Q16 εxx γxy + 2Q26 γxyεyy + Q22ε2yy+ 4Q66 γ2xy]/2

(Eq. 29.2)


• Putting Eq. 29.2 in 29.1, we get:

(Eq. 29.3)

• Using strain definitions from Eq. 11.3 above, and integrating the strain energy term over thickness of plate, we get:

(Eq. 29.4)

• In Eq. 29.4, H(uo, vo, wo) is defined in Eq. 29.5. Also, it should be noted q ( ) qhere that the domain of integration for H is midplane area of the plate, while the domain of integration of tractions constitutes six external surfaces of a rectangular plate.


(Eq. 29.5)

• According to Theorem of Minimum Potential Energy, “of all possible displacement fields which satisfy compatibility and prescribed boundary conditions, the displacement field which satisfies the equilibrium equation makes the total potential energy a minimum”. Mathematically, this implies that the condition for equilibrium is that the 1st variation of total potential

h ld b Thenergy should be zero. Thus,

δΠ = 0. (Eq. 29.6)


• Thus, setting the 1st variation of Eqn. 29.4 to zero, we get:

• Plugging definitions of Nx, Ny, etc. from Eq. 13.1 above, we get:

(Eq 29 7)(Eq. 29.7)


• Till so far we have not addressed forces attributable to known tractions Ti. These forces may exist on various surfaces of the plate. For a rectangular plate, there are six such surfaces. The second integral (over surface S)

b l d ll f h i fmust be evaluated over all of these six surfaces.

• We start this by assuming that the origin is located at geometric center of the plate, and the plate dimensions are a x b. Also, the plate thickness is assumed to be t.

• Now consider 1st surface (or edge) of the plate, x = a/2. At this edge, dSequals dydz. If we integrate the second integral on this surface, we get:


• Integrating each component over the thickness, we get the following expression for second integral on surface x = a/2.

(Eq. 29.8)

where,

where,

(Eq. 29.9)( q )


• It is seen in Eq. 29.8 and 29.9, that Kirchhoff assumption along with the surface integral for tractions shifts emphasis from known stresses to known stress‐resultants.

• Hence, developing the solution of plate does not necessarily require us to know stresses on the boundary on a point‐to‐point basis. Rather, what is needed are overall integrals of those stresses over the edge of the plate.

• This in turn means, that several stress distributions can lead to same integrated value. Thus, the plate theory under discussion does not provide us with unique answers for a given stress distribution over a plate’s boundary, as some other stress distribution may also lead to same stress resultants. This is a limitation of several structural level theories of laminated plates.

• However, these theories provide sufficient detail as we move away from the edge of the plate, as per St. Venant’s principle.


• Using a similar approach as discussed earlier, we can evaluate surface integrals (2nd integral) for other edges of the plate as well, i.e. x = ‐a/2, y=‐b/2, and y=+b/2, and get results similar to that in Eq. 29.8.

• Finally, we evaluate integrals for top and bottom surfaces of the plate, i.e. z=+t/2. On these surfaces, typically, Tx, and Ty, are zero, and dS is dxdy. Thus, for the top surface of the plate, this integral can be expressed as:

(Eq. 29.10)

• Similarly, the integral for the bottom surface is ‐q‐(x,y).Similarly, the integral for the bottom surface is q (x,y).

• We add them together such that:

( ) + ‐q(x,y) = q+ ‐ q‐


• Putting all these results for surface integral terms corresponding to six surfaces in Eq. 29.7, we get:

(Eq. 29.11)


Nachiketa Tiwari



• Earlier, using total potential energy principle, we developed Eq. 29.11. This is shown below.

(Eq. 29.11)


• Equation 29.11 is an important form of the virtual work equilibrium equation. It has been derived using the total potential energy principle. Here total potential energy has been minimized by setting its 1st variation f bof to be zero.

• We see from Eq. 29.11, that the formulation by itself generates requisite boundary conditions in an appropriate form.

• Equation 29.11 is widely used since it serves as the basis of classic q yRayleigh‐Ritz formulations. These formulations are used for obtaining approximate solutions, and are very widely used in finite element formulations.

• Equation 29.11 is also known as “weak form”. Later, we will also develop a “strong form” for equilibrium. The rationale underlying such terminologies will be discussed later.

Generation of Equilibrium Equations

• Equation 29.11 can be further processed to generate equilibrium equations, which we have already developed through 1st principles earlier. We will see that such an approach not only generates requisite

ilib i i i l id i h b d di i iequilibrium equations, it also provides us with boundary conditions in appropriate form.

• First, we apply integration by parts to all the terms which are integrated over the area of the plate, in Eq. 29.11. There are a total of eight such terms. Consider the 1st part:

• In above expression we have defined the domain of integration.– Thus, the area of integration is now a rectangle of size a X b.

– Also, dA, has been substituted by dxdy.Also, dA, has been substituted by dxdy.


• Next, consider following identities.

• Applying this identity to the 1st term, we get:

(Eq. 30.1)


• Next, we recall the component form of divergence (or gradient) theorem.

• We use this theorem to compute the area integral of ∂(Nxδuo)/∂x in Eq. 30.1, to get:

(Eq. 30.2)

• Here, nx is the x‐direction component of a unit vector normal to the boundary. Thus, its value is 1, 0, ‐1, and 0 along edges x=a/2, y=b/2, x=‐a/2, and y=‐b/2, respectively, of the composite plate being analyzed here.y p y p p g y

• Using these values of n along the closed boundary of composite plate we

Generation of Equilibrium Equations• Using these values of nx along the closed boundary of composite plate, we

calculate the contour integral as defined in Eq. 30.2 as:

(Eq. 30.3)

• Putting Eq. 30.3 in Eq. 30.2, the final form of 1st part of Eq. 29.11 is:g q q , p q

(Eq. 30.4)


• Similarly, we apply integration by parts to remaining seven terms in Eq. 29.11. Putting all these terms together, and reorganizing, them we get:

(Eq. 30.5)

• where, P5, P6, P7, P8, and P9 are defined as follows.


• and,


• To be continued in next lecture


Nachiketa Tiwari



• Till so far, we have used the process of integration of parts to transform Eq. 29.11 (weak form) into Eq. 30.5.

(Eq. 30.5)

h P5 P6 P7 P8 d P9 h b d fi d li• where, P5, P6, P7, P8, and P9 have been defined earlier.

• Equation 30.5 is called the “strong form” of equilibrium condition. It is termed as “strong” because the differentiability conditions for an assumed displacement function required for such a form are stronger.

• For instance, here the equation involves second derivatives of moments, while in Eq. 29.11 did not involve any derivatives of moments.


• Given that moments are directly proportional to second derivatives of wo(x,y), it implies that:– A valid function for wo(x,y) has to have at least 4th order differential continuity,

f h f l ( ) f l h blif we use the strong formulation (i.e. Eq. 30.5) for solving the problem.

– However, if we used Eq. 29.11 for solving the same problem, the formulation would require second order, i.e. C2, continuity for wo(x,y).

• Thus, we see that Equation 30.5 requires higher order of continuity for assumed displacement functions, vis‐à‐vis Eq. 29.11. It is for this reason that Eq 30 5 is also known as the “strong form” while Eq 29 11 is knownthat Eq. 30.5 is also known as the strong form , while Eq. 29.11 is known as the “weak form”.

Of i b h h d di l f i h ld• Of course, in both cases, the assumed displacement function should satisfy essential boundary conditions for the problem.

• Moving further we now extract equilibrium equations from Eq. 30.5.


• Looking at Eq. 30.5, we realize that it is a sum of area integrals, line integrals (P5‐P8), and point‐specific values (P9). We further note, that there are an infinite number of mathematically valid variations in di l fi ld ibl hi h b d i hi i ldisplacement field possible which may be used in this equation as long as they satisfy the following conditions.– Differentiability requirements

E i l b d di i– Essential boundary conditions

• Thus, Eq. 30.5 can be zero only if, area integrals, line integrals, and point‐ifi l (P9) i di id llspecific values (P9) are individually zero.

• Thus, from area integrals we get equilibrium equations, while terms P5‐P9 give us a mathematically consistent set of boundary conditions.


• Thus, the equilibrium equations are:

• These equilibrium equations are identical to the ones developed earlier (Eq. 14.7‐9) developed using the Newtonian approach.

• Next we look at boundary conditions. Terms P5, P6, P7, P8 correspond to boundaries x=a/2, x=‐a/2, y=b/2, and y=‐b/2, respectively.


• Looking at term P5, which corresponds to edge, x=a/2, we notice that there are four BCs along this edge. Further, for variational statement to be true, each of these BCs should be individually zero.

• Thus the boundary conditions along edge x=a/2 are:


• Similarly, BCs conditions for x=‐a/2 are:

• Likewise, BCs for other two edges of the plate, y=+b/2 may also be computed in a straight‐forward way.computed in a straight forward way.

• Finally we look at term in P9 as defined below

Generation of Equilibrium Equations• Finally, we look at term in P9 as defined below.

• These are 8 corner conditions.

Observations

• The equilibrium equations developed here and the associated BCs govern the plate behavior completely.

• Each edge of the plate is associated with four sets of BCs. Within each set, there is a pair of conditions of which, one must be satisfied.

• For each of these pairs, either a force or moment must be equated to externally applied force or moment, or a kinematic condition (displacement or slope) must be specified.

• Such a pairing of force/moment and kinematic conditions emerges Suc a pa g o o ce/ o e t a d e at c co d t o s e e gesautomatically, when variational process is used to minimize the total potential energy of the system.

Observations

• The boundary conditions developed here are the only valid form of BCs which can be enforced on a plate’s boundary, and also ensure problem’s

i t ith l i l l i ti thconsistency with classical lamination theory.

• If BCs are specified in some other way then it quite likely that we will not be able to solve the problem.

• Specification of BCs in ways different than laid out earlier leads to an ill‐posed problem.

• The strong form of variational formulation requires higher order e st o g o o a at o a o u at o equ es g e o decontinuity of displacement field. Such a formulation is used in Galerkinmethod. In contrast, the weak formulation is used in Rayleigh Ritz method.


Nachiketa Tiwari


Comments on Energy Methods

• Earlier, we have commented on two types of minimum potential energy formulation; weak and strong. Weak form is also known as “variational f ”form”.

• In contrast to weak formulation, the continuity requirements on displacement and weight functions are higher for strong formulation.

• Further, it should be noted that energy methods do not explicitly require displacement and weight functions to be chosen such that all the boundary conditions are satisfied.

• Instead, the requirement is only to satisfy essential boundary conditions, i.e. those BCs which involve primary variables. Thus, such formulations do not necessarily ensure boundary conditions relating to secondary variables are enforced.

Example : Differentiability Requirements



– A16 = A26 = D16 = D26 = 0

• Finally we assume that the plate is normally loaded as shown in Fig. 32.1 with a load intensity of q(x,y) equalling a constant, q0.

q(x y)

x

q(x,y)

y

Fig. 32.1

Example : Differentiability Requirement

• Since the plate’s lamination sequence is symmetric and thus, its [B] matrix is [0], the out‐of‐plane and in‐plane equilibrium equations are decoupled.

• Hence, we can solve the out‐of‐plane equilibrium equation directly. We have done this using Galerkin approach earlier.

• Earlier, we had assumed a sine‐sine displacement field for wo, and found it to be a good approximation.

• Here, we redo the same exercise, to solve the problem but we use a different approximation for wo. We assume that:d e e t app o at o o e assu e t at

•wo(x,y) = W11 (x2‐ax) (y2‐by), and

use Galerkin (strong form) as well as weak form to solve the problem.


• Solution Using Strong Form (Galerkin method)– Step 1: Verify that the assumed solution meets all essential boundary

diti Thi i d b lcondition. This is done below.

wo(x,y) = W11 (x2‐ax) (y2‐by).

By evaluating this function on boundaries of the SS‐SS plate, we indeed find that this function vanishesthat this function vanishes.

Hence, all essential BCs are satisfied.

Step 2: Check continuity requirement We omit this step here We will later– Step 2: Check continuity requirement. We omit this step here. We will later realize that omitting this step can be a serious mistake and lead to erroneous solutions.

– Step 3: Find the error function (or residue) by substituting this function in out‐of‐plane equilibrium equation.

• The out of plane equilibrium equation is:


• Solution Using Strong Form (Galerkin method ‐ contd.)– Step 3: Find the error function (or residue) by substituting this function in out‐

f l ilib i tiof‐plane equilibrium equation.• Since D16 = D26 = 0, the out of plane equilibrium equation is:

• Also, we note that for the assumed function,

• Thus the expression for residual error E(x y) is:• Thus, the expression for residual error E(x,y) is:


• Solution Using Strong Form (Galerkin method ‐ contd.)– Step 3: Multiply this error with a virtual displacement εδW11 (x2‐ax) (y2‐by),

i t t th d t l t ’ d t it t Th tintegrate the product over plate’s area and equate it to zero. Thus we get:

– Step 4: From above relation, we can calculate W11. We find that:


• We see from previous expression that W11 does not depend on terms D11, and D22. This is clearly incorrect, since in plate’s bending stiffness is d i t d b t D d Ddominated by terms D11 and D22.

• The reason for this error is that we ignored to check the continuity requirements on differentials of function wo(x,y) = W11 (x2‐ax) (y2‐by).

• Any assumed function for wo(x,y) should be at least four times differentiable in x and y.

• This is so, because the governing differential equation involves 4th order s s so, because t e go e g d e e t a equat o o es o departial derivatives of wo(x,y) in x and y directions. In our case the assumed function does not meet this differentiability requirement. Hence, it yields wrong results.

Example 1: Differentiability Requirement

• Next, we solve the same problem using the weak form, as per Eq. 29.11. We first simplify this equation, since we know that for a symmetric l i t t f l d fl ti t i fl d b i llaminate, out‐of‐plane deflections are not influenced by in‐plane displacements. Thus, we get:

(Eq. 32.1)


• Now let us consider edge x=a/2. For this edge, we have:– wo is known (actually zero) along the entire edge. Hence its variation, δwo

l th d ialong the edge is zero.

– The slope of wo in y‐direction [∂wo/ ∂x] is zero, since wo is zero along entire edge. And since this slope is known (does not have to be zero) along the entire edge its variation δ[∂wo/∂x] is zero along the edge Mathematically we canedge, its variation, δ[∂w /∂x] is zero along the edge. Mathematically, we can also say δ[∂wo/ ∂x] = [∂(δ wo)/∂x] = 0.

– Finally, given that this edge is simply supported, there is no external moment resultant (M+

x) along the entire edge.

– Hence, term P1 is zero.

• Similarly, we can show that terms P2, P3, and P4 are zero as well. Hence, y, , , ,Eq. 32.1 can be restated as:

(Eq. 32.2)


• Next, we develop expressions for Mx, My, and Mxy.

• Substituting these expressions, along with relations for wo and its variationSubstituting these expressions, along with relations for w and its variation in Eq. 32.2, integrating the equation, and then solving for W11, we get:

• Thus, we see that the weak form is able to give a “reasonable” value of W i à i th t f b i th f diff ti bilitW11 vis‐à‐vis the strong form because in the former differentiability requirements are less stringent.


Nachiketa Tiwari


Lecture 33Stability of Conservative SystemsStability of Conservative Systems

Introduction

• This lecture focuses on the static stability of structures.

• A system even though in equilibrium may or may not be stable. Consider Fig. 33.1. Here equilibrium exists in all three configurations, since the sum of forces and moments in all directions for all configurations is zero. gHowever, the same may not be said of each configuration’s stability.

A BC

Fig. 33.1

• If we disturb configuration by providing the ball infinitesimally small displacement or velocity then:

g

displacement or velocity, then:– Config. A is stable with respect to displacement & velocity disturbances.

– Config. B is unstable with respect to displacement & velocity disturbances.

Config C is stable with respect to displacement disturbances but unstable– Config. C is stable with respect to displacement disturbances, but unstable with respect to velocity disturbances.

Introduction

• As per the Law of Minimum Potential Energy, “a conservativemechanical system is in static equilibrium if and only if the value of potential energysystem is in static equilibrium if and only if the value of potential energy is at a relative minimum”. Mathematically, this implies:

δΠ = 0.

• However, such a condition for equilibrium is valid only for:– Conservative systems: In such systems, the forces involved are conservative

in nature This implies that systemic energies depend only on the state of thein nature. This implies that systemic energies depend only on the state of the system, and not on path the system took to get to its final state. A system which has dissipative effects (damping, or friction) is not conservative.

– Static system: The law of minimum P.E. holds only for static systems. It is not y y yvalid for dynamic systems.

Introduction

• The Law of Minimum P.E. is based on a more generic concept ‐ that of dynamic stability as per which the condition for equilibrium (dynamic ordynamic stability, as per which, the condition for equilibrium (dynamic or static) requires the system to be always in a configuration such that its total energy (E) is minimized.

• Such a criterion for dynamic equilibrium is valid only for conservative systems.

• However, a system which is in static equilibrium may or may not be stable. This has also been shown in Fig. 33.1. Going forward, we develop a criterion for stability of static systemsa criterion for stability of static systems.

Criterion for Static Stability

• Earlier, we had stated that the condition for minimum potential energy is that the 1st variation of P.E. is zero. Mathematically, this was stated as:

y

y,δΠ = 0.

• However, above condition does not guarantee a relative minima. TheHowever, above condition does not guarantee a relative minima. The limitations of above condition become apparent, especially in nonlinear systems. For purposes of analogy, a function f(x) is at a minima if following two conditions are held simultaneously.

/– df(x)/dx = 0and– d2f/dx2 > 0 at a value of x, when df(x)/dx = 0.

• In linear conservative systems, the condition δΠ = 0 is by itself sufficient to ensure stability, since other requirements (which will be described later) are satisfied identically. However, in nonlinear systems a more ) y , yexplicit definition of stability is needed.


• If a system’s total potential energy is at a minima, then, its potential energy can only increase if the system is gets perturbed by infinitesimally small

y

y y g p y ydisplacements to acquires some other configuration. Let this increase in its potential energy be ΔΠ.

L t th fi ti f th t b t d b t f f ti• Let the configuration of the system be represented by a set of functions qo, and the corresponding P.E. be Πo when the system is equilibrium. Now, when it is disturbed by εq1, (where, ε is a scalar of infinitesimal magnitude) then the change in potential energy, ΔΠ, may be expressed as:

(Eq. 33.1)

Criterion for Static Stability• Equation 33.1 can be rewritten in following abridged form.

(Eq. 33.2)

• Now, if Π has to be minimum it implies that ΔΠ has to be a positive number. Now, ΔΠ can be only positive if:– Π1 (which is the 1st variation of Π, and also earlier written as δΠ) is zero. If it

was not zero, then it could be negative or positive, depending on the sign of parameter ε. And a negative value of 1st variation would imply that P.E. is not at a minima.

– Π2 (which is 2nd variation of Π) has to be positive since it is multiplied by ε2, which is always positive. Hence, for ΔΠ to be positive, 2nd variation has to be positive.

– All other variations of PE (Π Π Π ) do not influence the sign of ΔΠ since– All other variations of P.E. (Π3,Π4, Π5,…) do not influence the sign of ΔΠ, since their contribution would be very small vis‐à‐vis that of Π2 as they are multiplied with higher powers of ε.


• Hence, following conditions must be satisfied for a conservative system to be statically stable:– Π1 = 0 (This condition will give us the equilibrium configuration q0.)

– Π2 > 0 (For determining stability of configuration q0, Π2 should be

evaluated for the system when it is in configuration q0).

• It could happen in certain cases that Π2 is itself zero. In such a case, we should look at higher order variations in P.E. For such cases following are the conditions for stability:– Π3 = 0

– Π4 > 0

• In such a situation all higher order variations of P.E. should be evaluated for a system when it is equilibrium state q0.

Example of a Rigid Column

• Consider a rigid column as shown below.

Fig 33 2: Rigid Bar Connected to Ground Through a Torsional SpringP

α

Fig. 33.2: Rigid Bar Connected to Ground Through a Torsional Spring

Length of bar: LInitial inclination with vertical axis at no‐load condition: αFinal inclination with vertical axis when loaded: q

Torsional stiffness: c

Vertical load applied at bar’s end: PTorsional stiffness of spring at bar’s base: c

• Here, the total potential energy of the system is:

Π = c(q‐ α)2/2 ‐ PL(1‐cosq) (Eq. 33.3)

• The state of the bar at any stage is defined by its single degree of freedom, q. Let qo be the state corresponding to equilibrium. We now perturb this state by providing a small disturbance of εq1 to find out conditions for 1 equilibrium and stability of the system.


• Putting Eq. 33.3 in Eq. 33.2, we get the value of ΔΠ in the vicinity of equilibrium state qo, as follows:

ΔΠ = εΠ1 + ε2Π2 + …

where,

Π1 = [c(qo‐ α) ‐ PL(sin qo)]q1Π1 [c(qo α) PL(sin qo)]q1Π2 = [c ‐ PL(cos qo)] q1 2

If t Π t diti f ilib i Th ti f• If we set Π1 as zero, we get condition for equilibrium qo. The equation of equilibrium is:

c(qo‐ α) ‐ PL(sin qo) = 0.

• If the bar was straight up to begin with, i.e. if α is zero, then above equation for equilibrium becomes:

cqo ‐ PL(sin qo) = 0. (Eq. 33.4)


• Equation 33.4 gives us two solutions for qo. These are:qo = 0. (Solution 1)q /(sin q )= PL/C (Solution 2)qo/(sin qo) PL/C (Solution 2)

• We have to figure out the stability of each of these solutions. For this, we explore the sign of second variation of P.E. as defined below:p g

Π2 = [c ‐ PL(cos qo)] q1 2

Since q1 2 is always positive, the sign of Π2 only depends on that of [c ‐ PL(cos qo)].

• For solution 1, qo = 0. If we put this solution in expression [c ‐ PL(cos qo)], we get:[c ‐ PL(cos qo)] = c‐PL.[c PL(cos qo)] c PL.This will be positive only if P < c/L.


• Hence Solution 1 (i.e. qo = 0) is stable as long as external load P, does not exceed c/L.

• Next, we look at Solution 2. i.e. qo/(sin qo)= PL/C. This solution implies that PL = cqo/(sin qo).qo qo

• If we put this solution in expression [c ‐ PL(cos qo)], we get:

[c PL(cos q )] = c cq (cos q ) /(sin q )[c ‐ PL(cos qo)] = c‐ cqo(cos qo) /(sin qo).

• This term is positive only if the following condition holds:

qo/L > Tan qo.

• In this way, are able to extract, from theorem of min. P.E. conditions for equilibrium, and also conditions for stability.


Nachiketa Tiwari


Lecture 34Geometric Nonlinearity in Composite PlatesGeometric Nonlinearity in Composite Plates

Introduction

• Till so far we have used an analysis paradigm where the deformed and original system is approximately the same from standpoints of material g y pp y pbehavior, and stiffness.

• While such an approximation may work in linear systems (since thereWhile such an approximation may work in linear systems (since, there the stiffness of deformed and original system do not differ significantly), the same may not be true in systems involving large deformation scales, and change of material properties. Either of these factors render the g p psystemic behavior nonlinear.

Introduction• In structures, there are three broad categories of nonlinearities. These

are:– Geometric nonlinearities: Here, the nonlinear effect is attributable to largeGeometric nonlinearities: Here, the nonlinear effect is attributable to large

or moderate displacement and rotations.

– Material nonlinearity: Here, the material may stiffen with increasing load y , y gsince at high stresses, fibers straighten and thus overall structure becomes stiffer. In other cases, plastic effects cause the stiffness to decrease.

– Nonlinear loading: In such a case, the magnitude and direction of load changes with increasing deformation of the system. Loads which change direction due to change in configuration of the structure are called follower loadsloads.

• Here, we will only address effects of geometric nonlinearities caused due to moderate rotationsto moderate rotations.

Geometric Nonlinear Plate Theory

• Consider a small line segment AB of length dso, such that the projections of its length along x, y, and z axes are dx, dy, dx. This line segment translates as well as d f I it fi l t t it i d t d b A’B’ L t th t thdeforms. In its final state it is denoted by A’B’. Let us assume that the coordinates of point A are (0,0,0), and Point A translates by distances u, v, and w, along x, y, and z, directions. Thus, we calculate the coordinates of points A, B, A’, and B’ as shown in following table.g

Point X‐Coordinate Y‐Coordinate Z‐CoordinateA 0 0 0A 0 0 0

B dx dy dz

A’ u v w

• We also note that the final length of A’B’ is ds. It should be noted here, that

B’ dx + u + (∂u/∂x)dx + (∂u/∂y)dy + (∂u/∂z)dz

dy + v + (∂v/∂x)dx + (∂v/∂y)dy + (∂v/∂z)dz

dz + w + (∂w/∂x)dx + (∂w/∂y)dy + (∂w/∂z)dz

g ,while calculating coordinates of B’, we have only taken 1st order derivatives and ignored higher order on account of their “smallness”.


• Thus, we develop expression for ds, and dso, as shown below:

(Eq. 34.1)


• Comparing this relation with Eq. 34.1, we get the following definitions of strains.

(Eq. 34.2)


• In plates, the nonlinear effects due to stretching (i.e. due to u and v) are very small compared to those due to bending displacement w. Thus, we simplify Eq. 34 2 di l t t f ll i t i di l t l ti34.2 accordingly to get following strain‐displacement relations:

(Eq. 34.3)


• At this stage, we make two more simplifications.

– First we drop off out‐of‐plane shear and extensional strains. We do this not because these terms are zero, but because in a plane‐stress problem out‐of‐plane stresses, are zero, and hence out‐of‐plane strains do not enter the picture.

Secondly we had earlier developed Eqs 11 1 and 11 2 using Kirchhoff’s– Secondly, we had earlier developed Eqs. 11.1 and 11.2 using Kirchhoff s assumptions for plates. These relations decompose u and v displacements into their midplane components (uo and vo), and also a rotational component. With these modifications, we get:

(Eq. 34.4)


• Equation 34.4 represent strain‐displacement relations for a plate, which experiences moderate rotation. These are a total of three equations. Each

ti h t t f t i t d i d d lequation has two sets of terms, printed in red and green colors.

– The terms in red are known as membrane strains, as they develop due to stretching effects. The source of geometric nonlinearity is embedded in membrane strains. These strains depend on 1st order derivatives of themembrane strains. These strains depend on 1 order derivatives of the displacement field.

– The terms in green are known as curvature strains. The more the curvature of a bent plate, the higher are these strains. These strains depend on 2nd

order derivatives of wo.

• Now let us explore the role of these components through an example. Consider a line segment AB, which is originally of length unity. This line segment experiences some stretching, and some bending. Here we assume:– vo = 0; and

– 1st and 2nd order derivatives of uo, and wo with respect to y direction are 0.


• Thus, such a line segment experiences just extensional strain in x‐direction, εxx defined as:

• Let us assume that the value of ∂uo/∂x is 0.001 when the slope ∂wo/∂x is 1°. It may be noted here that a strain of 0.001 is fairly large given that many engineering materials start yielding at such a strain level.many engineering materials start yielding at such a strain level.– When ∂wo/∂x = 1° = 0.01745 radians, (∂wo/∂x)2/2 = 0.000152

Thus, for such a value of ∂wo/∂x , nonlinear component of strain is very small compared to linear component of strain.p p

– However, when ∂wo/∂x = 5° = 0.087 radians, (∂wo/∂x)2/2 = 0.0038.

Thus, a mere 5 degree of slope produces nonlinear strains, which exceed , g p p ,linear component of strain by approximately 3.8 times.


• While in nonlinear plate theory, strain‐displacement relations have gotten modified to accommodate the nonlinear effect of moderate rotations, other relations remain unchanged.

• Specifically, equilibrium equations do not change.p y q q g

• Relations linking force and moment resultants to strains do not change as well.as well.

• Neither do definitions of [A], [B], and [D] matrices.

• Thus, we can use solution strategies discussed earlier to solve nonlinear composite plate problems.


Nachiketa Tiwari


Lecture 35Buckling and Geometric NonlinearityBuckling and Geometric Nonlinearity

Buckling of a Plate SS on Two Sides

• In this lecture, we shall explore the concept of buckling, and how it relates to concepts of minimum potential energy theorem and geometric nonlinearity.

• Consider a plate as shown in Fig. 35.1 of dimensions 2a X 2b, with the i i l d f lorigin located at center of plate.

SSSS

x

y

Fig. 35.1

• This plate has the following boundary conditions.No constraints on edges y = +b.wo = 0 on edges, x = +au = 0 on edge x = ‐a

y

u = 0 on edge x = aExternal constant force per unit length (‐N) applied at edge x = +a. It should be noted here

that loading direction is compressive, and hence load is negative.


• Further, we assume that the plate is:– Symmetric: Thus [B] = [0]y [ ] [ ]

– Specially orthotropic: Thus, A16 = A26 = D16 = D26 = 0.

• This plate will remain flat at very low values of N. At small values of N, the plate will demonstrate only in‐plane displacements, and no out‐of‐plane displacement. We call such a configuration of the plate its primary configuration.

• However, beyond a critical value of Ncr the plate will buckle and shift move to an alternative configuration, i.e. its secondary configuration.

• In this example, we want to find the stability of its primary configuration. We expect that initially this configuration will be stable, but once we cross a threshold Ncr, instability will set in, thereby causing plate to buckle.


• To conduct a stability analysis of plate’s primary configuration, we plan to:1. Solve displacement field for primary configuration. As this is an easy problem, we will p p y g y p ,

directly use equilibrium equations to solve it, and get the exact solution.

2. Evaluate total potential energy (Πinitial) in the system associated with primary configurationconfiguration.

3. Next, we will perturb this configuration with some infinitesimally small displacements, which are consistent with essential BCs of the problem.

4. Once the system is perturbed, we will calculate the new PE of the system, i.e. Πnew. While calculating Πnew, nonlinear effects due to moderate rotations will be incorporated in the definition of strains.in the definition of strains.

5. From these two values of potential energy, we will calculate ΔΠ.

6. Finally, we will develop conditions under which ΔΠ will no longer remain positive. This will be the point at which the plate will buckle.


1. Solve for displacement field for primary configuration, i.e. when plate remains flat under compression. This is done below:

(Eq. 35.1)

(Eq 35 2)(Eq. 35.2)

Given that this is a homogenous plate at a macroscopic level, and in uniformly load symmetric plate there is no bending, strains in the plate

ld b i i H b i f iwould be constant over its entire area. Hence, above expressions for strain are valid not only at the edge of the plate, but also over its entire area.


Thus, solving for displacement field for primary configuration, we get:

(Eq. 35.3)

Here, it should be noted that: (i) we have placed the origin of coordinate system at the geometric center of plate, (ii) plate’s dimensions are 2a and 2b, and (iii) u displacement at x=‐a is zero.2b, and (iii) u displacement at x a is zero.

2. Next we find the total P.E. associated with this configuration.

(Eq. 35.4)

Here, we use Eq. 35.3 in Eq. 35.4 to calculate Πinitial.


3. Next we apply perturbations to existing displacement field. The perturbation field is assumed to be:

δu(x,y) = δv(x,y) = 0, and δwo(x,y) = εW11(x2‐a2)(y2‐b2)

Such a perturbation field is chosen such that it satisfies essential boundary conditions, and also differentiability requirements. Since we are using weak formulation, we need C2 differentiability out‐of‐plane displacement field.

Adding this field to original displacement field (Eq. 35.3), we get:dd g t s e d to o g a d sp ace e t e d ( q 35 3), e get

(Eq. 35.5)


4. Now, using Eq. 35.4 we find the new PE of the system, i.e. Πnew in the vicinity of its original configuration. While finding Πnew, nonlinear effects d t d t t ti ill b i t d i th d fi iti fdue to moderate rotations will be incorporated in the definition of strains. These revised strain definitions are:

(Eq. 35.6)


Equations 35.6 provide us with definitions of “new” mid‐plane strains and curvatures. From these we now calculate expressions for “new” force and moment resultants. These are:

Finally, we now find revised P.E. (Πnew) of the system using the following expression.p

(Eq. 35.7)( q )

It should be noted here, that the external work done does not changeIt should be noted here, that the external work done does not change due to application of variation in the displacement field since variations in uo and vo fields are zero.


5. Next, we find the value of ΔΠ. It is computed to be:

(Eq. 35.8)

6. From above expressions, we make the following observations.– The plate is under equilibrium, since 1st variation of Π is zero.

– Π is positive for low values of N Once N exceeds a threshold N the second variationΠ2 is positive for low values of N. Once N exceeds a threshold, Ncr, the second variation of P.E. is no longer positive. This value of Ncr can be calculated by setting the second variation of P.E. to be zero. Thus,


Thus, the plate’s initial configuration as defined in Eq. 35.3 is:– Stable as long as N < Ncr.– Metastable at N = Ncr, because at this value of N, Π2= 0.cr 2

– Unstable for N > Ncr.

Also, when N = Ncr, we see from Eq.35.8 that:

Π1= 0 Π2= 0Π3= 0 Π4> 03 4

This implies, that when N equals Ncr, ΔΠ is a positive entity.

Hence, when N equals Ncr the plate’s initial configuration is stable.

It should be noted here that our observations are valid for a chosen set of trial functions.


Nachiketa Tiwari


Lecture 37Buckling of PlatesBuckling of Plates

Buckling of Plates

• In previous lecture, it has been shown that a plate buckles once external compressive load on it exceeds a certain value. We were able to predict its p pbuckling load by calculating the value of Ncr at which second variation of P.E. became zero.

• In these calculations we have used the energy method to calculate buckling load. However, such an approach though ideal for computers, is tedious for hand calculations.

• Hence, we will now develop relations similar to equations of equilibrium, which will help us identify buckling loads.which will help us identify buckling loads.

Buckling of Plates

• Consider a plate, which is symmetric and also specially orthotropic. Thus, such a plate will have:p

– [B] = [0]

– A16 = A26 = D16 = D26 = 0.

• If such a plate is loaded in compression then it will initially remain flat, but after a certain threshold load it will buckle under external compressive load. Since, the plate is symmetric its initial displacement field at its mid‐l ll b ( ) h d f f d hplane will be (uo, vo, 0), where uo and vo are functions of x and y. For such

a state, its initial P.E. can be gotten from simplified versions of Eq. 29.4. and 29.5. Thus,

(Eq. 36.1)

Buckling of Plates

• Next, we perturb the plate with some disturbance such that the final displacement is:p

(Eq. 36.2)

W E 36 2 t fi d l f PE f th t• We now use Eq. 36.2 to find new value of P.E. of the system.

(Eq. 36.3)

• In Eq. 36.3, expressions for “new” Nx, Ny, … have been calculated usingIn Eq. 36.3, expressions for new Nx, Ny, … have been calculated using revised definition of strains as given in Eq. 36.4 as shown further.

Buckling of Plates

(Eq. 36.4)

• Next, we find the difference between two energy values, i.e. ΔΠ. The expression for ΔΠ can be written as:

ΔΠ = εΠ1 + ε2Π2 + ε2Π2 + ε4Π4 + ... (Eq. 36.5)1 2 2 4 ( q )

• We look at 1st two terms of Eq. 36.5 in particular. The 1st variation of P.E., when equated to zero, gives us the condition for equilibrium.when equated to zero, gives us the condition for equilibrium.

Buckling of Plates

(Eq 36 6)(Eq. 36.6)

• This is identical to Eq. 29.7, except that it does not have terms involving Mx, My, Mxy, and Tz. This is only to be expected as:

– The plate is not subjected to normal loads Hence the value of T term is zeroThe plate is not subjected to normal loads. Hence, the value of Tz term is zero.

– We have assumed that the primary solution for this problem does not involve wo. displacement. Further the plate is symmetrically laminated. Hence there are no moment related terms in this equation.

• Now, we can conduct mathematical operations similar to the ones which we performed to develop Eq. 30.5. Thus, we get the following set of equilibrium equations which help us determine plate’s primaryequilibrium equations, which help us determine plate s primary configuration.

Buckling of Plates

(Eq. 36.7)

• Also, the boundary conditions for these equations are same as those described in lecture 31. We note however, that all BCs related to out‐of‐plane direction are identically satisfied.

• Next, we look at term Π2 in Eq. 36.5.

Buckling of Plates

• Reorganizing these terms, we get:

(Eq. 36.8)

where,

• Next, we integrate individual terms in Eq. 36.8 by parts. While performing such operations we assume that N N and N are constant as they havesuch operations, we assume that Nx, Nxy, and Ny, are constant as they have already been determined from equilibrium equations.

Buckling of Plates

• Thus we get:

• The condition for buckling is Π2 =0. Applying such a condition, we get following equations for buckling.g q g

(Eq 36 8)(Eq. 36.8)

Buckling of Plates

• In Eqs. 36.8, Nx1, Ny1, Nxy1,Mx1, My1, and Mxy1 represent variations in Nx, Ny, Nxy,Mx, My, and Mxy, respectively.y, xy, x, y, xy, p y

• The 3rd equation couples inplane force resultants with out‐of‐plane moment resultants Such a coupling is entirely attributable to themoment resultants. Such a coupling is entirely attributable to the presence of nonlinear terms in strain‐displacement equations.

• Further the values of N N and N have to be determined by solving the• Further, the values of Nx, Ny, and Nxy, have to be determined by solving the equilibrium equations (Eq. 36.7). These values are then introduced into buckling equations.

• Care has to be taken for signs of Nx, Ny, and Nxy. A tensile force resultant tends to stabilize the system, while a compressive force (negative in sign) does just the oppositedoes just the opposite.


Nachiketa Tiwari


Lecture 37Buckling of PlatesBuckling of Plates

Buckling of Infinitely Long Plate SS on Two Sides

• Consider a plate as shown in Fig. 37.1. This plate:– Is infinitely long in x‐direction.y g

– It’s width in y‐direction is b.

– An external load is applied in x‐direction, such that its per unit length value is No.

– Plate’s lamination sequence is specially orthotropic and also symmetric.

SSx

y

Fig. 37.1

NN

Fig. 37.1

SS


• For such a plate, the following stiffness terms are zero.– [B] = [0][ ] [ ]

– A16 = A26 = D16 = D26 = 0.

• Hence, we can use equilibrium equations as per Eq. 36.7. Solving these equations will give us strain and displacement fields for the plate. Thus we get:


• We now integrate expressions for strain to get the appropriate displacement field as defined below:

• If the value of N exceeds a certain threshold Ncr the plate will buckle. We will now estimate the value of Ncr. For this we have to solve the following buckling equations.

(Eq. 36.8)


• To solve these equations, we have to assume a set of perturbations in the displacement field. Here, we assume:

(Eq. 37.1)

• Here, λ is the wavelength of the buckling pattern in x‐direction. For the y‐direction, we are assume that the wave will have a half‐wave. Such a choice of shape function for w1 satisfies out‐of‐plane boundary conditions for the buckling equations. These conditions are:

– w1‐displacement perturbation is zero along long edges of the plate.

– There are no perturbations on Mx along long edges of the plate as Mx has to be zero along simply supported edges of the plate.

• The perturbation field chosen through Eq. 37.1 satisfies these BCs.


• Next, we apply this perturbation field to Eq. 36.8. If such a field satisfies these equations, then our choice of field is correct.

• Of these three buckling equations the 1st two are identically satisfied. In the third equation, we use the fact that:

• Thus the buckling equation is rewritten as:Thus the buckling equation is rewritten as:

(Eq. 37.2)


• But, Nx = N. Hence, Eq. 37.2 can be re‐written as:

(Eq. 37.3)

• The boundary conditions associated with this equation are also satisfied because of choice of our shape function for w1. Substituting 3rd of Eq. 37.1 in Eq. 37.3, we get:

(Eq. 37.4)

• Eq. 37.4 has to be satisfied for buckling to occur. Solving for N, we get:

(Eq. 37.5)


• Equation 37.5 defines the buckling load of an infinitely long specially orthotropic and symmetric composite plate subjected to inplane uniform compressioncompression.

• For such a plate, if: N N th l th λ i i b d th l t ’ i iti l fl t– N < Ncritical, the wavelength λ comes as an imaginary number and the plate’s initial flat configuration remains stable.

– N = Ncritical, the wavelength λ becomes real. However, at this stage, we can show by critical g g yexploring the sign of 4th variation of P.E., the configuration is still stable.

– N > Ncritical, the wavelength λ becomes real, and plate’s primary flat configuration becomes unstable If the plate is perturbed with some out of plane disturbance it willbecomes unstable. If the plate is perturbed with some out‐of‐plane disturbance it will buckle, thereby lowering its P.E.

Buckling of Finite Plates SS on All Edges

• Consider a plate which is simply supported on all of its edge as shown in Fig. 37.2.

xSSNN

KN

y

Fig. 37.2SSSS SS

KN• Here, the plate is loaded bi‐directionally. For such a plate the following

conditions hold.– If K <0, then the plate’s loading in y‐direction is tensile.

If K = 0 then the plate is in unidirectional compression

y

– If K = 0, then the plate is in unidirectional compression.– If K < 0, then the plate is in bidirectional compression.

• Here, the plate is simply supported on all four sides and has specially h i ll i l i iorthotropic as well as symmetric lamination sequence.


• Boundary conditions for the plate are:– uo = 0 at x = 0.

– vo = 0 at y = 0– v = 0 at y = 0.

– wo= 0 on all four edges.

– Mx = 0 on x = 0, a

– My = 0 on y = 0, by

• The pre‐buckling solution for the problem is:– Nx = ‐N

– Ny = ‐KN

– Nxy = Mx =My =Mxy = wo = 0.

h l f ll h b d d• Such a solution satisfies all the boundary conditions pertaining to equilibrium equations. Next, we assume a perturbation field defined as:


• Such a perturbation field satisfies all the BCs of buckling equations. Putting it in third buckling equation, we get:

• Hence, such a plate has infinite number of buckling shapes, each associated with a buckling load The minimum value of this load dependsassociated with a buckling load. The minimum value of this load depends on m, n, K, and aspect ratio of the plate. For a square plate loaded unidirectionally, minimum buckling load is associated with a buckling mode shape when m=n=1.p


Nachiketa Tiwari


Lecture 38

Behavior of Short‐Fiber CompositesBehavior of Short‐Fiber Composites


• Background information about short‐fiber compositesg p

• Load transfer mechanism in short‐fibers

• Longitudinal and transverse moduli for short‐fiber it ith idi ti l li tcomposites with unidirectional alignment

• Modulus of randomly oriented short fiber composites• Modulus of randomly oriented short‐fiber composites

• Problem setProblem set

About Short‐Fiber Composites

• We have seen earlier that unidirectional composites tend to be very stiff and strong in fiber direction, but very weak in the transverse direction. Th i k i t di ti i tt ib t bl t fTheir weakness in transverse direction is attributable to presence of significant stress concentration at the interface of matrix and fiber.

• Given these attributes, unidirectional composites are very useful in applications where state of stress is well known.– In such applications, lamination sequence of composite can be tailor‐made to

b t l l d ti llbear external loads optimally.

• However, if externally applied loads are omni‐directional, or if their direction can vary in time, then such laminates fabricated by stacking up unidirectional laminae may not necessarily meet our design needs.

About Short‐Fiber Composites

• We may still be able design a laminate for such cases (that is when loading is omni‐directional) which is equally strong in all directions, but even in such a design, the top and bottom layers will be weak in transverse directions, and failure could get initiated from here.

• Hence, in such applications, it is useful to have lamina which have in‐plane isotropy.

• One way to produce such lamina is by using short fibers which are randomly oriented. Such composites, in general are significantly less expensive than unidirectional composites. The fiber lengths in these are between 1 to 8 cm.

• Such composites are used very extensive in general purpose applications,Such composites are used very extensive in general purpose applications, such as car body panels, boats, household goods, etc. In most of such applications, glass fiber is used as reinforcing material for matrix.

Load Transfer Mechanism in Short‐Fiber Composites

• In composite materials, fibers are invariably surrounded by matrix material. Hence, external load is directly applied to matrix, and from here, it gets transferred to fibersit gets transferred to fibers.

• A part of this load gets transferred to fibers at their ends, while remaining part gets transferred through their external cylindrical surfacespart gets transferred through their external cylindrical surfaces.

• For unidirectional composites with continuous fibers, transfer of load at fiber ends may be very small vis‐à‐vis load transfer through fiber’s external surface.

• This is because fibers are very long, and hence their cylindrical surfaces, across which load gets transferred through shear‐mechanism, are sufficiently long. In such fibers, the effect of load transfer through fiber

d f l ff ll h f l d fends may not significantly affect overall mechanics of load transfer.


• However, in short‐fiber composites, the same may not be true. In such composites, the length of the fiber is not sufficiently long such that much of load transfer happens across cylindrical surfaces of fibersof load transfer happens across cylindrical surfaces of fibers.

• Thus, in such fibers, both the ends, as well as external cylindrical surfaces of fibers play a significant role in matrix to fiber load transferof fibers play a significant role in matrix‐to‐fiber load transfer.

• Hence, it is important to understand role of end‐effects in context of load transfer to fibers. Sans this understanding, our understanding of reinforcing effects in short‐fiber composites will be inaccurate and flawed.


• Consider a short‐fiber of length l, embedded in matrix, as shown in Fig. 38.1. The figure also shows the details of an infinitesimal portion of fiber of length which experiences normal stress in length direction and shearof length , which experiences normal stress in length direction, and shear stress, τ, along its cylindrical surface.

• Please note that while normal stress at one end of infinitesimal fiber is σ• Please note that while normal stress at one end of infinitesimal fiber is σf, it is σf +dσf at its other end.

• This variation in normal stress along the length of infinitesimally long fiber is because some of the load gets transferred from matrix to fiber due to application of shear stress on its cylindrical surface.


Fig. 38.1: Force Equilibrium of an Infinitesimal Portion of Discontinuous Fiber Which is Aligned to External Load


• Using Newton’s Laws, we can now write equation of force equilibrium for this infinitesimally‐sized fiber.

2 + (2 d ) 2 ( + d )π r2∙σf + (2 π r τ dz)∙σf = π r2∙(σf + dσf )

• Cancelling out term Πr2∙σf from both sides, and rearranging remaining terms we get:

dσf /dz = 2τ/r

• Integrating above equation yields,

σf = σfo+ (2/r)∫ τ dz, where the integral limits are 0 to z.

• Quite often, fiber separates from the matrix due to presence of large stress concentration. In other cases, matrix yields at the fiber end. The implication of either cases is that the integration constant for aboveimplication of either cases is that the integration constant for above equation, σfo, is zero. Thus, above equation can be rewritten as:

σf = (2/r)∫ τ dz


• The integral equation shown earlier can be evaluated if variation of shear stress, τ, with respect to coordinate z, is known. At this point, we make an assumption that shear stress at the interface of fiber and matrix isassumption that shear stress at the interface of fiber and matrix is constant, along fiber length, and equals matrix yield shear, i.e. τy. Such an assumption may be made, if we have a system where matrix material transmits maximum possible stress to fiber, which would be τy. For such a p , ycase, the integral equation can be simplified as:

σf = 2τy z/r (Eq. 38.1a)

• For short fibers, maximum fiber stress is expected to occur at mid‐length, i.e. z = l/2, while it will be zero at its extremities for reasons explained earlier Hence the equation written above will hold good only for values ofearlier. Hence, the equation written above will hold good only for values of z = 0 to l/2, and for the region z = l/2 to l, the equation will have to have a negative slope. Further, the maximum value of fiber stress will be, as per above equation:q

• σf_max = τy l/r, corresponding to z = l/2. (Eq. 38.1b)


• Equation 38.1 places no limit on the upper bound for fiber stress, and can approach very large values if l is made very large. However, in reality there will indeed be a limit which will correspond to the stress borne bywill indeed be a limit, which will correspond to the stress borne by continuous and infinitely long fibers in unidirectional plies. This stress, as calculated earlier is Ef/Emσc. Equating this value to maximum fiber stress in short‐fiber (as per Eq. 38.1) gives us a load‐transfer length, lt, which is ( p q ) g g , t,required to achieve maximum possible stress in fiber. This is shown below.

σf = τ l /r = Ef/E σ (Eq 38 2a)σf_max = τy lt/r = Ef/Emσc (Eq. 38.2a)

or,

lt/r = σf_max /τy = (Ef/Emσc) / τy (Eq. 38.2b)

• Thus, a fiber over length lt, develops maximum fiber stress (Ef/Emσc) as defined in 7.2a, when the externally applied stress is σc.

Load Transfer Mechanism in Short‐Fiber CompositesLoad Transfer Mechanism in Short Fiber Composites

• Hence, if we increase external stress σc, we will have to increase lt to ensure c tmaximum load in fiber, as σf_max, which equals Ef/Emσc,will also increase.

• However, there is a limit beyond which external stress σc cannot beHowever, there is a limit beyond which external stress σc cannot be increased.

• This limit corresponds to a point when the stress in fiber equals its ultimate• This limit corresponds to a point when the stress in fiber equals its ultimate strength (σuf ), – At this limit, any further stress in external stress will lead to failure of fiber.

– Thus the condition for maximum possible stress in fiber is:– Thus, the condition for maximum possible stress in fiber is:

σf_max = σuf = Ef/Emσc (Eq. 38.3)


• For such a limiting stress, there is a corresponding minimum fiber lengthwhich is required to support such a level of stresswhich is required to support such a level of stress.

• Mathematically, the value of minimum fiber length can be calculated from b b lEq. 38.2b and is given below.

lmin/r = σuf/τy (Eq. 38.4)

• Thus, any design of a short‐fiber composite should ensure that its fiber is at least lmin long, because in such a system the overall composite strength will be will maximized .

• If fibers are shorter than this critical length, then composite strength would not be at its maximum value, thereby adding mode weight and cost to the , y g gstructure. Finally, if l is very large compared to lmin, then composite increasingly behaves as one with continuous fibers.


Nachiketa Tiwari


Lecture 39

Behavior of Short‐Fiber CompositesBehavior of Short‐Fiber Composites


• Till so far, we have assumed that the matrix material in fiber‐matrix interface region is perfectly plastic. g p y p

• This is not entirely true. In reality, most of the matrix materials exhibit elasto‐plastic behavior For such systems performing a theoretical analysis isplastic behavior. For such systems, performing a theoretical analysis is difficult.

• Hence numerical methods may be used to solve such problems and get• Hence, numerical methods may be used to solve such problems, and get better understanding of load transfer mechanisms in short‐fiber composites.

S l h t di h h th t l d t f t fib d i t• Several such studies have shown that load transfer at fiber ends is not significant, and hence our earlier assumption that σfo is zero, stands validated, though in an approximate sense.

L d T f M h i i Sh t Fib C itLoad Transfer Mechanism in Short‐Fiber Composites

• Figure 39.1 is a plot of variation of fiber strength for three different fiber lengths.

Fig. 39.1


• Following observations can be made from Fig. 39.1.

• If fiber length is less than lt, then the normal stress in fiber is zero at both ends of fiber, and reaches a peak value at mid‐fiber length. In such a case, the longer the fiber (as long as it does not exceed lt), the longer is the value of peak normal stress at its mid lengthof peak normal stress at its mid‐length.

• If fiber length equals lt, then normal stress in fiber gets maximized. However, the shape of stress plot still remains triangular.

• Finally, if fiber length exceeds lt, then normal stress in fiber:– Rise from zero to a maximum value over part of the fiber length.

– Remains constant once it has maximized.

– Falls back to zero, over remaining part of fiber length.

• Utilization of fiber strength is maximized in the third configuration.

Modulus of Short‐Fiber Composites• Even though mathematically rigorous solutions can be developed to g y g p

compute moduli of short‐fiber composites, it is usually desirable to have relatively simple relations for estimating the same from a design perspective. Here, we directly cite results of Halpin and Tsai, which help us calculate with reasonable accuracy, longitudinal and transverse moduli of short‐fiber composites, with fibers aligned in a single direction.

EL/Em = [1 + (l/r)ηLVf)/(1 ‐ ηLVf) (Eq. 39.1)

where,

η = [(Ef/E ) ‐ 1] / [(Ef/E ) + l/r]ηL = [(Ef/Em) 1] / [(Ef/Em) + l/r].

Also,

/ [ )/( ) ( )ET/Em = [1 + 2ηTVf)/(1 ‐ ηTVf) (Eq. 39.2)

where,

ηT = [(Ef/Em) ‐ 1] / [(Ef/Em) + 2]

Modulus of Short‐Fiber Composites

• It may be noted from Eq. 39.3, that transverse modulus of unidirectional short‐fiber composites are not influenced by l/r ratio, and its value equals p y / , qthat for continuous unidirectional composite.

• Earlier we had discussed the need for randomly oriented short‐fiberEarlier, we had discussed the need for randomly oriented short fiber composites, since they are isotropic in a plane, and hence are appropriate for loads coming from all directions.

Modulus of Short‐Fiber Composites

• Such composites do not crack easily in their surface layers in the transverse direction.

• Predicting their modulus is analytically difficult. Hence, we cite a well known empirical result which predicts modulus of a randomly orientedknown empirical result, which predicts modulus of a randomly oriented short‐fiber composite.

E = (3E + 5E )/8Erandom = (3EL + 5ET)/8

Here, moduli EL and ET may be calculated from relations given earlier, or d i d i lldetermined experimentally.

• Above equation is a good engineering tool for designing composites with randomly oriented fibers.


• Background information about short‐fiber gcomposites

• Load transfer mechanism in short‐fibers

L it di l d t d li f h t fib• Longitudinal and transverse moduli for short‐fiber composites with unidirectional alignment

• Modulus of randomly oriented short‐fiber composites


Nachiketa Tiwari


Lecture 40

ClosureClosure

Summary

• This lecture series can be categorized into seven different modules.

• The 1st module is introductory in nature. Here, the basic nature of composites has been described. This module address:– Advantages of modern composites vis‐à‐vis traditional engineering materials– Advantages of modern composites vis‐à‐vis traditional engineering materials.

– Application areas of modern composites

– Qualitative introduction to why composites behave the way they do.

Key mechanical properties of different types of matrices fibers and whiskers– Key mechanical properties of different types of matrices, fibers, and whiskers.

– Classification of different composite materials.

– Important terminology relating to composite materials.

Introduction to terms like istotropy orthotropy and anisotropy– Introduction to terms like istotropy, orthotropy, and anisotropy.

Summary

• The second module addresses behavior of unidirectional continuous fiber composites in detail. This module addresses:p– Material axes for a unidirectional laminate

– Difference in failure mechanisms in composites vis‐à‐vis isotropic materials

– Key strength parameters of a unidirectional compositey g p p

– Concepts of volume and mass fraction for fibers and matrix materials

– Predictive models for longitudinal stiffness, lateral stiffness, longitudinal tensile and compressive strengths, etc.

– Predictive models for Poisson’s ratio, coefficient of thermal expansion in principlamaterial directions.

– Hooke’s Law for an orthotropic lamina

– Models for calculating engineering constants for an orthotropic lamina in arbitrary directions.

– Transformation of stress, strain, stiffness and compliance matrices

– Different failure criteria for composites

Summary

• The third module addresses behavior of alaminated composite plate details. This module covers:– Kirchhoff’s assumptions for plates.

– Variation of strain field in a plate over its thickness

– Through thickness variation of stress field in a plateg p

– Definition of mid‐plane strains and curvatures

– Definition of force and moment resultants and their variation across plate’s thickness

– Definition of [A], [B], and [D] matrices

– Role of different stiffness matrix components in influencing mechanical behavior of laminates

– Features of frequently used lamination sequences

Summary

• The fourth module addresses equilibrium equations and their solution for semi‐infinite plates using exact method. The module covers:p g– Equilibrium equations for a linearly elastic composite plate

– Importance of out‐of‐plane shears, and application of appropriate correction for such loads

– Different boundary conditions on four edges of a rectangular plate and their physical interpretation

– Closed form solutions for a semi‐infinite plate which is simply supported at two edges.

• Two sets of lamination sequences studied: Symmetric, and unsymmetric.

– Anti‐clastic curvature

R d d b di d t i l tiff– Reduced bending and extensional stiffness

– Response of semi‐infinite plates with non‐zero [B] matrices to inplane loads

– Modified Hooke’s Law for thermal stresses

Th l i i i fi i l– Thermal stresses in semi‐infinite plates

Summary

• The fifth module uses equilibrium equations to solve problems pertaining to finite plates. The module covers:p– Overview of commonly used simplification approaches to solve PDEs.

• Boundary conditions

• A, B, D, matrices

• PDE

• Kinematic relations

– Series solution method for a simply supported rectangular plate loaded normally.

– Convergence of different entities

– Rectangular plate simply supported on two edges• Application of corrections to solution for a plate SS on all four sides

Summary

• The sixth module details energy based methods to solve problems of composite plate mechanics. Specifically it covers:p p p y– Principal of virtual work

– Notion of residual error, and integral of residual errors.

– Galerkin method

– Special Galerkin method

– Examples using Galerkin method

– Strategies for improving accuracy of Galerkin based solution approachesg p g y pp

– Influence of D16 and D26 on out‐of‐plane response of composite plates subjected to normal loads

– Total potential energy principle

– Weak and strong formulations

– Differentiability requirements for weak and strong formulations

Summary

• The seventh module addresses two key concepts: nonlinearity, and stability. Further, in this module, the problem of buckling has been explained by using , , p g p y gconcepts of stability and nonlinearity. Specifically it covers:– Nonlinear strain‐displacement relations

– Notion of static stabilityy

– Criterion for equilibrium

– Criterion for stability

– Meta‐stable states, and assessing stability of these states using 4th order , g y gvariations

– Buckling equation for rectangular composite plates

– Solution of buckling equation for typical plate topologies:• Semi‐infinite plate

• Rectangular plate SS on two edges

• Rectangular plate SS on all edges, and which is loaded bi‐directionally

Summary

• The eight and final module addresses addresses behavior of short‐fiber composites. They have been dealt with in this module, because they are used p y , yin a very large number of applications. The module covers:– Background information about short‐fiber composites

– Load transfer mechanism in short‐fiber compositesp

– Longitudinal and transverse moduli for short‐fiber composites with unidirectional alignment

– Modulus of randomly oriented short‐fiber composites

– Load transfer mechanism in short‐fiber composites after incorporating plasticity of matrix material

– Improved models for predicting longitudinal and transverse moduli for short‐fiber icomposites

curso nptel - mechanics of laminated composite structure

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