Task 2

This task involves using laboratory equipment to carry out one destructive and one non-destructive test on a metal and a non-metal. The type of tests carried out will depend on the equipment that you have to hand.

1) Destructive test: Tensile Test

Set up the equipment and take a photograph of it.

Carry out the test and make a record of results taken and observations made.

In following graph X-axis is the stress value and Y-axis are the strain value.

Put together the results so that you have a permanent record that can be used for future analysis

Area = 7.065 Sq.mm Length = 30.00 mm

S.N Load Displacement Stress Kn/mm2

Strain Young’s Modulus

1 0.11 0.06 5.2 x 10-03 1.66 x 10-03 3.13 x 10-06

2 0.13 0.06 6.57 x 10-03 2 x 10-03 3.28 x 10-06

3 0.14 0.07 7.89 x 10-03 2.33 x 10-03 3.38 x 10-06

Average Young’s modulus = E1 + E2 + E3 / 3

Answer = 3.26 x 10-06

Make a transcript of any discussion you may have had with your tutor about conducting the test.

I talked to my tutor and asked that what is Elastic Deformation?

My tutor replied: “The Engineered structures and components are subjected to

various loads and displacements during service. When the applied forces (or displacements) are relatively low, unloading causes the structure to return to its original size and shape. This type of mechanical response is described as ‘elastic’ (elastic deformation)”.

2) Non-Destructive test: Dye Penetrant Inspection

Set up the equipment and take a photograph of it.

Carry out the test and make a record of results taken and observations

made Below are the main steps of Liquid Penetrant Inspection: 1) Pre-cleaning: 2) Application of Penetrant: 3) Excess Penetrant Removal: 4) Application of Developer: 5) Inspection: 6) Post Cleaning:

Put together the results so that you have a permanent record that can

be used for future analysis

Dye Penetrant Inspection - Typical Minimum Penetration Times

Material Form Type of Discontinuity Water-Washable Penetration Time*

Aluminium Castings Porosity, Cold Shuts 5 to 15 min

Aluminium Extrusions, Forgings Laps NR**

Aluminium Welds Lack of Fusion, Porosity 30

Aluminium All Cracks, Fatigue Cracks 30, not recommended for fatigue crack

Magnesium Castings Porosity, Cold Shuts 15

Magnesium Extrusions, Forgings Laps not recommended

Magnesium Welds Lack of Fusion, Porosity 30

Magnesium All Cracks, Fatigue Cracks 30, not recommended for fatigue crack

Steel Castings Porosity, Cold Shuts 30

http://www.i/nspection-for-industry.com/dye-penetrant inspection.html#sthash.ZHXwjF8a.dpuf

(Accessed: 8/12/14)

Make a transcript of any discussion you may have had with your tutor about conducting the test.

I talked to my tutor and asked that “What are advantages and disadvantages of Dye

penetrant inspection”?

My tutor replied: “The main advantages of DPI (Dye Penetrate Inspection) are the

speed of the test and the low cost. Disadvantages include the detection of only surface

flaws, skin irritation, and the inspection should be on a smooth clean surface where

excessive penetrant can be removed prior to being developed. Conducting the test on

rough surfaces, such-as "as-welded" welds, will make it difficult to remove any

excessive penetrant and could result in false indications. Water-washable penetrant

should be considered here if no other option is available. Also, on certain surfaces a

great enough color contrast cannot be achieved or the dye will stain the work piece”.

Task 3

Describe a different process of degradation associated with each of metals, polymers and ceramics

Metal Oxidation process:

According to encylopeadia.com: Oxidation is one of the main constituents in the

atmosphere at low earth orbit. The atomic oxygen oxidation of metals has been the

topic of several studies.

The relatively high capture efficiency for atomic oxygen of silver makes it a suitable

detector for direct measurements in the upper atmosphere. These detectors depend on

the formation of silver-oxide and an increase in the resistance is observed. Although the

formation of A90 is essential in this practice, its effect on silver when used for example

as a structural or electrical item is undesirable, since it results in losses in both electrical

conductivity and mechanical strength. The major use of silver, exposed in the LEO

environment, is as interconnector material in solar arrays. These interconnectors serve

as the conductive path between cells forming the array.

As a result of the differences in thermal expansion coefficients of the several components

of the solar array, high stresses are found in the interconnectors. If the performance of

these interconnectors deteriorate by some action, loss of electrical power is observed. In

this investigation the effect of LEO exposure on silver was examined and several different

coatings were applied to the silver to examine their protective effect, also an alternative

interconnector material was tested.

Polymer ageing process:

According to encyclopeadia.com: The irreversible change in the properties of polymers

under the action of, among other factors, heat, oxygen, sunlight, ozone, and ionizing

radiation. Depending on the factor, the aging will be of the thermal, oxidative, light, ozone,

or radiation type. Aging occurs during the storage and treatment of polymers, as well as

during the storage and use of articles made from polymers. Under actual conditions,

several factors act on polymers concurrently; for example, in atmospheric aging, oxygen,

light, ozone, and moisture all affect the polymer. An important factor accelerating aging

is mechanical stress that develops during the treatment of polymers and under certain

operating conditions of polymer articles.

Aging is caused by chemical transformations of macromolecules, which lead to the

degradation of the macromolecules and the formation of branched or three-dimensional

structures (cross-linking). There are various mechanisms of aging. For example,

degradation in oxidative aging is related to a chain oxidation reaction of the polymer

accompanied by the formation and decomposition of hydro peroxides. The rate of aging

depends on the sensitivity of the polymer to the factors mentioned above, on the intensity

of these factors, and on the composition of the polymer material. Carbon-chain polymers

whose macromolecules contain unsaturated bonds, in particular, certain rubbers (natural

rubber, synthetic isoprene rubbers), are most susceptible to aging.

Aging manifests itself in a deterioration of the mechanical characteristics of polymers, in

the appearance and growth of cracks on the surface that sometimes result in destruction

of the polymer, and in a change in color. The resistance of polymers to aging in many

cases determines the storage period and, sometimes, the service life of polymer articles.

Stabilization is an effective method for preventing aging in polymers.

Ceramic thermal shocking process:

According to www.its-inc.com: Conventional ceramics, including bricks and tiles, are well known for their ability to withstand high temperatures. Nonetheless, Fine Ceramics (also known as “advanced ceramics”) are more heat resistant than these materials by far. While aluminum begins to melt at approximately 660oC (approx. 1,220oF), alumina Fine Ceramics only begin to melt or decompose at temperatures above 2,000oC (approx. 3,632oF).The heat resistant properties of Fine Ceramics are measured by the temperatures at which they begin to melt, and by their levels of thermal shock resistance.

http://its-inc.com/upload/Photos/Service/Coatings/FCT%20Blade%20at%20Temp.jpg (Accessed: 8/12/14)

Thermal shock resistance refers to a material's ability to withstand rapid changes in temperature. Silicon nitride, a particularly heat tolerant material, displays superior resistance to thermal shock, as tested by heating the material to 550oC (1,022oF) and then rapidly cooling it by dropping it into water. Silicon nitride is thus suitable for applications involving extreme temperature variations, and in high-temperature industries such as metal manufacturing and energy generation.

Task 4

This task builds on Task 2 – you now need to explain how test procedures produce results of use to an engineer. Think about what sort of information they will want if they are going to use the results of the tests to inform the design of a product, e.g. in the case of a tensile test, the final breaking load is only one of at least five other pieces of data that can found if the test procedure is correct. Write an explanation (about 500 words) of how one destructive and one non-destructive test procedure can be carried out so that they produce results that are useful to an engineer.

Non-destructive Test: DPI (Dye Penetrant Inspection)

According to ndt-ed.org: The principle of DPI is DPI is based upon capillary action, where surface tension fluid low penetrates into clean and dry surface-breaking discontinuities. Penetrant may be applied to the test component by dipping, spraying, or brushing. After adequate penetration time has been allowed, the excess penetrant is removed and a developer is applied. The developer helps to draw penetrant out of the flaw so that an invisible indication becomes visible to the inspector. Inspection is performed under ultraviolet or white light, depending on the type of dye used - fluorescent or non-fluorescent (visible).

Evaluation for DPI: Test objects are coated with visible or fluorescent dye solution. Excess dye is then removed from the surface, and a developer is applied. The developer acts as blotter, drawing trapped penetrant out of imperfections open to the surface. With visible dyes, vivid color contrasts between the penetrant and developer make "bleed out" easy to see. With fluorescent dyes, ultraviolet light is used to make the bleed out fluoresce brightly, thus allowing imperfections to be readily seen.

Destructive test: Tensile Test Procedure: The following procedure should be adopted in ensuring that the data recorded from tensile test specimens will be taken in an organized and consistent manner.

1. Choose specimen. Care is to be taken to ensure that the specimens did not have any

notching or cracks from manufacturing or any surface defects that would adversely affect

the tensile tests.

2. Before loading the specimens in the Tensometer, the computer system connected to

the machine should be set up by inputting the necessary information of gauge length and

width of the specimen. The computer system will then be prepared to record data and

output necessary load-deflection graphs.

3. The specimens then be loaded into the Tensometer, and a tensile test will be

performed. The data which will be recorded electronically in text files and the load-

deflection curve will be shown on the computer screen as a visual representation.

Evaluation:

Selection of material is performed by matching its mechanical properties to the required

service conditions. Once the required properties are known, the material selection can be

made. Obtaining the physical properties through an experiment reflective of the

application is one of the key steps that must be done before a selection can be made,

since properties from idealized tests do not necessarily reflect the application

environment of the material.

Tensile tests help characterize the strength, stiffness, and plasticity of a polymer. By

obtaining the stress-strain curve of a polymer the yield strength, necking point, nonlinear

elastic deformation, linear elastic deformation, plastic deformation, and tensile strength

can be identified.

The tested specimens were of an amorphous structure, also known as “glassy” polymers.

Plastic deformation in amorphous polymers occurs inhomogeneously in two main forms:

crazing, and shear banding.

Crazing consists of narrow zones of highly deformed polymer containing voids. The zones

are oriented perpendicular to the stress axis. In the crazed zone, the molecular chains

get aligned along the stress axis, but are “interspersed” with voids. The typical craze

thickness in glassy polymers is in the order of 5µm or less. Since the lateral surface

contraction associated with the craze is negligible by comparison, the density of the craze

decreases. Generally, crazing occurs in brittle glassy polymers such as the specimens

tested.

Shear bands form at 45° to the largest principle stress. The polymer molecular chains get

oriented within the shear bands without any accompanying change in volume. Shear

yielding can take two forms: diffuse shear yielding and localized shear band formation. In

localized shear, the shear is concentrated in thin planar regions, and the process involves

a “cooperative” movement of molecular chains, resulting in a band formation of about 45°

to the stress axis. Shear yielding is usually the dominant mode of deformation in ductile

amorphous thermoplastics. Since Acrylic and Lexan are brittle amorphous

thermoplastics, shear yielding was not the dominant form of deformation.

Task 5

In Task 3 you described how materials degrade over a period of time. From the list choose two degradation processes and explain how they affect the behaviour of engineering materials. Include in your answer reference to specific materials and give examples of where they are used in service and how their behaviour is affected by the degradation. Write around 600 words and include images, diagrams and web links to examples of products that have suffered as a result of the degradation process. Oxidation process: How does it affects the engineering materials and how its properties are affected by degradation process? To explain oxidation process of engineering materials I have taken an example of iron. According to physicsforums.com: In oxidation of iron when impure (cast) iron is in contact with water, oxygen, other strong oxidants, or acids, it rusts. If salt is present, for example in seawater or salt spray, the iron tends to rust more quickly, as a result of electrochemical reactions. Iron metal is relatively unaffected by pure water or by dry oxygen. As with other metals, like aluminium, a tightly adhering oxide coating, a passivation layer, protects the bulk iron from further oxidation. The conversion of the passivating ferrous oxide layer to rust results from the combined action of two agents, usually oxygen and water. Other degrading solutions are sulphur dioxide in water and carbon dioxide in water. Under these corrosive conditions, iron hydroxide species are formed. Unlike ferrous oxides, the hydroxides do not adhere to the bulk metal. As they form and flake off from the surface, fresh iron is exposed, and the corrosion process continues until either all of the iron is consumed or all of the oxygen, water, carbon dioxide, or sulfur dioxide in the system are removed or consumed When iron rusts, the oxides take up more volume than the original metal; this expansion can generate enormous forces, damaging structures made with iron.

http://upload.wikimedia.org/wikipedia/commons/8/80/Rust_wedge.jpg Outdoors Rust Wedge display at the Exploratorium shows the enormous expansive force of rusting iron. (Accessed 8/9/14)

Ageing Process: How does it affects engineering material and how its properties are affected by degradation process? To explain Ageing process of engineering materials I have taken an example of polymer. According to encyclopeadia2.com: The polymer ageing is a irreversible change in the properties of polymers under the action of, among other factors, heat, oxygen, sunlight, ozone, and ionizing radiation. Depending on the factor, the aging will be of the thermal, oxidative, light, ozone, or radiation type. Aging occurs during the storage and treatment of polymers, as well as during the storage and use of articles made from polymers. Under actual conditions, several factors act on polymers concurrently; for example, in atmospheric aging, oxygen, light, ozone, and moisture all affect the polymer. An important factor accelerating aging is mechanical stress that develops during the treatment of polymers and under certain operating conditions of polymer articles.

http://biofinagroup./com/wp-content/uploads/2013/07/peter1.jpg Degrading Polymer by ageing. (Accessed: 8/12/14) Aging is caused by chemical transformations of macromolecules, which lead to the degradation of the macromolecules and the formation of branched or three-dimensional structures (cross-linking). There are various mechanisms of aging. For example, degradation in oxidative aging is related to a chain oxidation reaction of the polymer accompanied by the formation and decomposition of hydro peroxides. The rate of aging depends on the sensitivity of the polymer to the factors mentioned above, on the intensity of these factors, and on the composition of the polymer material. Carbon-chain polymers whose macromolecules contain unsaturated bonds, in particular, certain rubbers (natural rubber, synthetic isoprene rubbers), are most susceptible to aging. Aging manifests itself in a deterioration of the mechanical characteristics of polymers, in the appearance and growth of cracks on the surface that sometimes result in destruction of the polymer, and in a change in colour. The resistance of polymers to aging in many cases determines the storage period and, sometimes, the service life of polymer articles. Stabilization is an effective method for preventing aging in polymers.

Task 6

This task builds on the work that you carried out for Tasks 2 and 4. Choose either the destructive test or the non-destructive test and evaluate the results that that you took. If, for example, you decide to use the results of a tensile test, and the material investigated was a coded specimen of known type, you could compare your numerical values with the published ones and evaluate how accurate your test procedure was. Write around 600 words and include diagrams if appropriate.

Destructive test: Tensile stress, comparing numerical values with Baxial method and mine Tensometer test:

Following are the numerical readings of Baxial test which are taken from different style of method: The plane biaxial test device using cruciform specimens developed at the Free University of Brussels has four independent servo-hydraulic actuators with an appropriate control unit to keep the centre of the specimen explicitly still. The device has a capacity of 100 kN in both perpendicular directions, but is restricted to tensile loads. As no cylinders with hydrostatic bearing were used, failure or slip in one arm of the specimen will result in sudden radial forces which could seriously damage the servo-hydraulic cylinders and the load cells. To prevent this, hinges were used to connect the specimen to the load cells and the servo-hydraulic cylinders to the test frame. Using four hinges for each loading direction results in an unstable situation in compression and consequently only tension loads can be performed.

A baxial method of measuring tensile test. http://www.sciencedirect.com/science/article/pii/S0020768306002848#gr1 (Accessed 8/12/14)

An extended database of experimental static and fatigue results on beamlike glass fibre reinforced epoxy specimens with a View the MathML source-lay-up has been set-up within the framework of a project that deals with the optimisation of the use of composite materials for wind turbine rotor blades. One of the aims of the project is to investigate the material behaviour under complex stress states. For the glass fibre reinforced composite laminate with the mentioned lay-up the average and standard deviation material parameter results of about 400 traditional beamlike tests are given in table below. No information about the shear modulus is available for this lay-up. Only for a single lamina, information exists from off-axis tests and tests on a (+45°/−45°) lay-up. Based on the ply data, the theoretically expected properties of the laminate can be calculated using classical laminate theory .

Following are the numerical readings of tensile test which are taken from my style of method:

Put together the results so that you have a permanent record that can be used for future analysis

Area = 7.065 Sq.mm Length = 30.00 mm

S.N Load Displacement Stress Kn/mm2

Strain Young’s Modulus

1 0.11 0.06 5.2 x 10-03 1.66 x 10-03 3.13 x 10-06

2 0.13 0.06 6.57 x 10-03 2 x 10-03 3.28 x 10-06

3 0.14 0.07 7.89 x 10-03 2.33 x 10-03 3.38 x 10-06

Average Young’s modulus = E1 + E2 + E3 / 3

Answer = 3.26 x 10-06

Procedure: The following procedure should be adopted in ensuring that the data recorded from tensile test specimens will be taken in an organized and consistent manner.

1. Choose specimen. Care is to be taken to ensure that the specimens did not have any

notching or cracks from manufacturing or any surface defects that would adversely affect

the tensile tests.

2. Before loading the specimens in the Tensometer, the computer system connected to

the machine should be set up by inputting the necessary information of gauge length and

width of the specimen. The computer system will then be prepared to record data and

output necessary load-deflection graphs.

3. The specimens then be loaded into the Tensometer, and a tensile test will be

performed. The data which will be recorded electronically in text files and the load-

deflection curve will be shown on the computer screen as a visual representation.

Evaluation:

Selection of material is performed by matching its mechanical properties to the required

service conditions. Once the required properties are known, the material selection can be

made. Obtaining the physical properties through an experiment reflective of the

application is one of the key steps that must be done before a selection can be made,

since properties from idealized tests do not necessarily reflect the application

environment of the material.

Tensile tests help characterize the strength, stiffness, and plasticity of a polymer. By

obtaining the stress-strain curve of a polymer the yield strength, necking point, nonlinear

elastic deformation, linear elastic deformation, plastic deformation, and tensile strength

can be identified.

The tested specimens were of an amorphous structure, also known as “glassy” polymers.

Plastic deformation in amorphous polymers occurs inhomogeneously in two main forms:

crazing, and shear banding.

Crazing consists of narrow zones of highly deformed polymer containing voids. The zones

are oriented perpendicular to the stress axis. In the crazed zone, the molecular chains

get aligned along the stress axis, but are “interspersed” with voids. The typical craze

thickness in glassy polymers is in the order of 5µm or less. Since the lateral surface

contraction associated with the craze is negligible by comparison, the density of the craze

decreases. Generally, crazing occurs in brittle glassy polymers such as the specimens

tested.

Shear bands form at 45° to the largest principle stress. The polymer molecular chains get

oriented within the shear bands without any accompanying change in volume. Shear

yielding can take two forms: diffuse shear yielding and localized shear band formation. In

localized shear, the shear is concentrated in thin planar regions, and the process involves

a “cooperative” movement of molecular chains, resulting in a band formation of about 45°

to the stress axis. Shear yielding is usually the dominant mode of deformation in ductile

amorphous thermoplastics. Since Acrylic and Lexan are brittle amorphous

thermoplastics, shear yielding was not the dominant form of deformation.

Reference

http://www.sv.vt.edu/classes/MSE2094_NoteBook/97ClassProj/exper/bailey/www/bailey

.html (Accessed 1/12/14)

http://nptel.ac.in/courses/Webcoursecontents/IIScBANG/Material%20Science/pdf/Lectur

e_Notes/MLN_08.pdf (Accessed: 3/12/14)

http://biofinagroup.com/polymer-degradation-how-it-works/ (Accessed: 8/12/14)

http://encyclopedia2.thefreedictionary.com/Aging+of+Polymers (Accessed: 8/12/14)

https://www.ndeed.org/EducationResources/CommunityCollege/NDTIntro/cc_intro001.h

tm (Accessed: 8/12/14)

http://liquidpenetrant.com/penetrant-inspection-process/ (Accessed: 8/12/14)

http://www.ndt.net/article/0698/hayes/hayes.htm (Accessed: 8/12/14)

http://www.wermac.org/others/ndt_dyepenetrant.html (Accessed: 8/12/14)