Review Final UTeM 2015

CE 60Instructor: Paulo Monteiro

Polymers: Classification

• A) Thermoplastics such as polyethylene, whichsoften on heating.

• B) Thermosets or resins such as epoxi whichharden when two components are heatedtogether.

• C) Elastomers or rubbers• D) Natural polymers such as cellulose, lignin

and protein, which provide the mechanical basisof most plant and animal life

From: Asby& Jones


Engineering Thermoplastics

• This term was first introduced by the General ElectricCo. in the 1960’s, & they defined it as a polymer alloywhich could replace metals in many applications.

• Polyethylene is the most common of them. It is a linear polymer. That is why they soften when heated.

• Thermoplastics are made by adding together(polymerizing) sub-units (“monomers”) to form longchains.

• Example:-C - C

H

R

H

H

R may be hydrogen (polyethylene), orCH3 (polypropylene) or –Cl (polyvinylchloride)

mass of the polyethylene mer : (4 hydrogen atoms × 1 g/mol) + (2 carbon atoms × 12 g/mol) = 28

g/mol.

molecular weight of polymer (g / mol)DPmolecular weight of mer (g / mol / mer)410,000 g / mol28 g / mol / mer

=

=

= 14,643 mers

A high-molecular-weight polyethylene has an average molecular weight of 410,000 g/mol. What is its average degree of polymerization?

H H

C C

⎡ ⎤⎢ ⎥⏐ ⏐⎢ ⎥⎢ ⎥⏐ ⏐⎢ ⎥Η Η⎣ ⎦


Thermoplastics

• Nylons are one example of an engineeringthermoplastic.

• Polycarbonates have a “ring” structure in the chain which makes it very “stiff”molecule which translates into a highmelting point.

What type of bonding exists within the molecular chains of

thermoplastics?

• Within thermoplastic molecular chains, covalent bonds exist.


Thermosetting Plastics• Thermoplastics are usually easier to mold into complex shapes. The

polymer is heavily cross-linked• but thermosetting polymers offer more of the following properties:

– High thermal stability– High rigidity– High dimensional stability– Resistance to creep & deformation under load– Light weight (as compared to metals)– High electrical & thermal insulating properties

• Today many thermosetting “resins” are available which havesuperior properties. [See p. 330-340 in the Smith textbook].

Describe the atomic structural arrangement of thermosetting

plastics.

• Most thermosetting plastics consist of three-dimensional networks of covalently bonded atoms, as compared to the long chain-like molecules of thermoplastics.


Elastomeric Materials

• Elastomers are linear polymers withoccasional-cross links. These cross-linksprovide a memory so it returns to itsoriginal shape on unloading.

• Polymers which show “rubbery” behaviorat their operating temperature are called“elastomeric” [See the Smith textbook].

• Some elastomeric polymers are thermoplastics & others are thermosetting.

• The prototype is “natural rubber”.


E

Temperature

Glassy plateau

Glass transition

Rubbery plateau

Viscous flow


PortlandPortland CementCement

A hydraulic cement capable of setting, hardening and remaining stable under water. It consists essentially of hydraulic calcium silicates, usually containing calcium sulfate.


Manufacture

Raw Materials:

2/3 calcareous materials (lime bearing) - limestone

1/3 argillaceous materials (silica, alumina, iron)- clay


Based on the following notation:Based on the following notation:

C CaO

S SiO2

A Al2O3

F Fe2O3

H H2O


Cement MineralsCement Minerals

C3S : 3CaOSiO2C2S : 2CaOSiO2

C3A : 3CaOAl2O3C4AF : 4CaOAl2O3Fe3O4


CHEMICAL REACTIONSCHEMICAL REACTIONS

2C3S + 6H --> C3S2H3 + 3CH + 120 cal / g2C2S + 4H --> C3S2H3 + CH + 62 cal / gC3A + CSH2 --> Ettringite + 300 cal / g


SOLIDS IN CEMENT PASTESOLIDS IN CEMENT PASTE-Calcium Silicate Hydrate

Notation: C-S-HC/S Ratio: 1.5 to 2.0

Main Characteristics: High Surface (100 to 700 m2/ g) ----> High Van der Walls Force -----> Strength.

Volume % : 50 a 60


SOLIDS IN CEMENT PASTESOLIDS IN CEMENT PASTE

-Calcium Hydroxide ( portlandite)

Ca(OH)2Volume % : 20 to 25

low Van der Walls forceproblems with durability and strength


SOLIDS IN CEMENT PASTESOLIDS IN CEMENT PASTE

-Calcium Sulfoaluminate HydratesVolume % : 15 to 20

first : ettringiteafter : monosulfate hydrated.


Hydration process – Initial Condition

Let’s study a cement paste with w/c= 0.63

Start with 100 cm3 of cement.

Compute the mass of cement: Mc = 3.14* 100 = 314 g

Compute the mass of water: Mw = 0.63 * 314 = 200 g

Vc= 100 cm3

Vw= 200 cm3


ASTM Portland CementsASTM Portland Cements

Type I- General Purpose

Type II- moderate heat of hydration and sulfate resistance (C3A < 8%) : general construction, sea water, mass concrete

Type III- high early strength (C3A < 15%) : emergency repairs, precast, winter construction.

Type IV- low heat ( C3S < 35%, C3A < 7%, C2S > 40%) : mass concrete

Type V- sulfate resistant ( C3A < 5%) : sulfate in soil, sewers.


Aggregates

• cost• provide dimensional stability• influence hardness, abrasion resistance,

elastic modulus


Aggregate Type

•Coarse aggregate ( > 3/16 in. - 4.75 mm of No. 4)•Fine aggregate < 3/16 in. and > 150 (No. 200)


Aggregate Type -mineralogy

•Sedimentary Rocks (cost effective - near the surface), •about 80% of aggregates•Natural sand and gravel•Sandstone, limestone (dolomite), chert, flint, graywacke

•Metamorphic Rocks: slate, gneiss : excellent to poor



• Fineness modulus is the sum of the total percentages retained on each of thespecified sieve divided by 100. Thespecified sieves are 3, 1 1/2, 3/4 and 3/8 in and Nos. 4, 8, 16, 30, 50 and 100.


Characteristics of coarse aggregate Characteristics of fine aggregateType Used:________________ Type Used: ______________

Max. Size:______1______ inch F.M. _____2.93______________B.S.G: 168 ______lb/ft3 B.S.G: 167 ______lb/ft3

Moisture deviation from S.S. D.=_-0.4%__ Moisture deviation from S.S. D.=0.7%__Dry-rodded unit wt.__104_lb/ft3____

B.S.G of cement = 196 lb/ft3


Effect of moisture


TestingTypes of Elastic Modulus

ASTM Testing


Creep and Shrinkage


Importance


Compressive Strength

• Fundamental relationship

• S = So exp (-kp)

• Where So is the strength at zero porosity, p is the porosity and k a constant.


Interfacial Transition Zone


REASON


Microstructural improvement

• Use of silica fumereduce the porosity of the ITZgeometrical effect (no space)reduces the amount of CH due to pozzolanic reaction


Humidity

• Great importance of moist curing.


Temperature

• Cast and cured at the same temperature• Cast at different temperature but cured at

the same temperature• Cast at normal temperature but cured at

different temperatures.


Testing parameters

• Specimen Size: Fracture mechanics will explain the importance of size effect.

• Loading Rate: Increasing rates lead to increasing strength.



Thermal stresses

where:σt: tensile stressKr: degree of restraintE: elastic modulusα: coefficient of thermal expansionΔT: temperature changeϕ: creep coefficient

σ t =K r

E1 + ϕ

αΔT


Temperature Evolution

ΔT = placement temperature of fresh concrete + adiabatic temperature rise - ambient or service temperature - heat losses.


Durability Durability

•Durability of concrete: ability to resist weathering action, chemical attack, abrasion, or any process of deterioration


Water Structure


Abrasion - Erosion

•Note: the deterioration starts at the surface, therefore special attentions should be given to quality of the concrete surface.•Avoid laitance (layer of fines from cement and aggregate).


The problem

The transformation of ice from liquid water generates a volumetric dilation of 9%. If the transformation occurs in small capillary pores, the ice crystals can damage the cement paste by pushing the capillary walls and by generating hydraulic pressure.


Deterioration by fire

•Concrete is able to retain sufficient strength for a reasonably long time.


Effect of High Temperature on Cement Paste

•(a) degree of hydration•(b) moisture state•de-hydration:•ettringite > 1000C•Ca(OH)2 500-6000C•CSH ~ 9000C


Electrochemical process of steel corrosion in concrete


Volumetric change


The chemistry is simple

1) The high pH in the cement paste promotes the hydrolysis of silica

Si-OH + Si-OHSi-O-Si + H OH aggregate paste

2) Si-OH react with the paste to form Si-O-

3) Si-O-, adsorbs Na, K, and Ca to form a gel.


Expansive Reaction

• C3A + gypsum C3A.3C$.H32 (ettringite)

C3A.C$.H18 (monosulfate)

In the presence of sulfates


Sodium sulfate attack:

• Na2SO4 +Ca(OH) 2 +2H2O CaSO4.2H2O + 2NaOH

the formation of sodium hydroxide as a by-productof the reaction ensures the continuation of highalkalinity in the system, which is essential for thestability of the cementitious material C-S-H.


Magnesium sulfate attack

• MgSO4 +Ca(OH) 2 +2H2O CaSO4.2H2O + Mg(OH) 2• 3 MgSO4 + 3CaO .2SiO2 .3H2O + 8 H2O 3 CaSO4.2H2O + 3

Mg(OH) 2 + 2SiO2.H2O

• the conversion of calcium hydroxide to gypsum is accompaniedby the simultaneous formation of relatively insoluble magnesiumhydroxide.

• In the absence of hydroxyl ions in the solution C-S-H is no longer stable and is also attacked by the sulfate solution.

• The magnesium sulfate attack is, therefore, more severe onconcrete.


Factors influencing sulfate attack

• amount and nature of the sulfate present, • level of the water table and its seasonal

variation, • flow of groundwater and soil porosity, • form of construction, • quality of concrete.


Determine the lattice points per cellin the cubic system

Simple cubic:Lattice points are located only at the corners of the cube

8 corners (1/8) = 1

In BCC unit cells, lattice points are located at the corners and the center of the cube:

8 corner (1/8) + 1 center (1) = 2

In FCC unit cells, lattice points are located at the corners and faces of the cube:

8 corners (1/8) + 6 faces (1/2) = 4


Calculate the radius of an atom thatwill just fit into a cubic site

2R + 2r= 2R sqrt(3)

r/R = 0.732 R

r

2R + 2r= 2R sqrt(3)

2R


Problem

• Calculate the change in volume thatoccurs when BCC iron is heated andchanges to FCC iron. The latticeparameter of BCC iron is 2.863 A and ofFCC iron is 3.591 A

Volume of BCC cell = a3 = 2.863 = 23.467 Volume of FCC cell = a3 = 3.591 = 46.307

But the FCC unit cell contains four atoms and the BCC unit cell contains only twoatoms. Two BCC unit cells with a total volume of 46.934 will contain 4 atoms.

Volume change/atom = (46.307 -46.934)/46.934 = -1.34%Steel contracts on heating!!


Hypoeutectoid Phase Diagram

• If a steel with a composition x% carbon is cooled from the Austenite region at about 770 °C ferrite begins to form. This is called proproeutectoid (or prepre--eutectoid) ferrite since it forms before the eutectoid temperature.


Problem


An Example (Assume a Eutectoid Low Carbon Steel)

(a) Water-quench to room Temperature.(b) Hot-quench at 690°C & hold 2 hr; water-quench

(c) Hot-quench at 610°C & hold 3 min; water-quench

(d) Hot-quench at 580°C & hold 2 sec; water-quench

(e) Hot-quench at 450°C &hold 1 hr; water-quench

All martensite

Pearlite

Pearlite

50% pearlite + 50 martensite

Bainite


Types of Atomic & Molecular Bonds

• Primary Atomic BondsIonic Bonds

Covalent BondsMetallic Bonds

• Secondary Atomic & Molecular BondsPermanent Dipole BondsFluctuating Dipole Bonds

Documents

Review Final UTeM 2015