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116 Cards in this Set
- Front
- Back
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Describe the application of forces on a body. |
In most applications a body is subject to numerous forces however these can not just be simply added. In many cases the vector sum of the forces is 0. So a body is imagined to be made up of infinitesimally small cubes on the surface of which the forces are projected. |
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Define a stress tensor. |
It is a 9 component matrix needed to define a vector. See written notes for general tensors. |
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Define strain. |
The ratio of the change in length. See written notes for equations. See written notes for matrix form. |
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How are normal strains represented by a matrix? How are shear strains represented by a matrix? |
Components along the diagonal are called direct or normal strains. Components off the diagonal are shear strains. The strain tensor must be symmetric. |
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How are normal stresses represented by a matrix? How are shear stresses represented by a matrix? |
Components along the diagonal are called normal stresses and are perpendicular to the plane. Components off the diagonal are shear stresses and are parallel to the surface. The strain tensor must be symmetric. |
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What is pure shear? |
When there are no normal stresses or strains. |
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Define equibiaxial tension. |
The normal strains are all equal. The tensile and the shear strain equal 0. |
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Describe a hydrostatic stress and volumetric strain.. |
By decompressing the stress and strain tensors into their normal and shear components they can be simplified. They are simplified to the hydrostatic stress and volumetric strain (mean normal components). See written notes for equations. |
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Describe deviatoric stresses and strains. |
The invariant nature of the mean normal components means that they can be used to obtain the shear related components from any strain and shear tensor. This means that deviatoric stress and strain tensors can be used with its hydrostatic components to express any stress or strain tensor. See written notes for tensor equations. |
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Define Hooke's law. |
F=kx K is the elastic constant that determines the stiffness of the spring. |
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Describe elasticity. |
There are loads of tensors and equations for this. Including relations: with the Poisson ratio, with the shear modulus, with the strain, with the stress and with elastic strain energy. So definitely see written notes. |
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Describe a tensile test. |
A sample is gripped at both ends and extended using a testing machine. During the test both the force and the extension are recorded. These can be converted into engineering stress and strain by normalising the load values by the sample geometry. See notes for normalising equations. |
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Describe the elastic region. |
Strains are small and increase linearly with stress. In a tensile test only the load is uni-axial and the strain is not 0 perpendicular to the applied load. |
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Describe yielding. |
The onset of the plastic region is determined by the yield point. The stress at which it starts is known as the yield stress. A proof stress is used as a measure of yield strength. |
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Describe how to calculate a proof stress. |
Fit the linear elastic region. Shift it along the strain axis until the desired proof level. Where the fit intercepts the curve is the proof stress. |
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Describe true strain. |
In the plastic regions strains can be very large and can no longer be represented using the small strain approximations. As the strain increases so does the gauge length. See written notes for equations. |
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Describe true stress. |
As a metal deforms the cross sectional area decreases. The ratio of the instant area and the initial area during straining is inversely proportional to the ratio of instant length and initial length of the gauge. For equations see written notes. |
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Describe work/strain hardening. |
It occurs in the plastic region and can be described by several equations. For these expressions see written notes. |
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Describe necking. |
The reduction in cross sectional area that occurs at the ultimate tensile strength. Necking will occur when the stress of the material equals the work hardening rate. For equations see written notes. |
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Describe compression testing. |
During a compression test a sample is deformed uni-axial in compression between parallel platters. For true stress and strain expressions see written notes. |
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Describe plastic deformation as a localised shear event. |
Yield in tension usually takes place as a localised shear event at 45° to the loading direction. The yield strength can be thought of as the materials resistance to shear. See written notes for shear equations. |
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Define the maximum shear stress or Tresca yield criterion. |
For a bi-axial state the maximum shear stress at a point in a material is given by 1/2 of the absolute difference of the principle stresses. In 3D, all of these differences in the principle stresses must be evaluated. For derivation see written notes. |
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Describe Von mises yield criterion. |
The deviatoric stresses are the 'distortion' components of stress. They cause a change in shape not volume. Plastic deformation also is a change in shape and not volume. So a measure of the magnitude of the deviatoric stresses could be used as a yield criterion. For derivation and all expressions see written notes. |
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Describe the slab model for longitudinal modulus. |
2 materials bonded perfectly together used as a model to describe a composite materials mechanical behaviour. Leads to the Voigt equation which shows that the modulus of the composite along the loading direction is a linear function of the volume fraction of the constituent phases and proportional to the Young's modulus of both phases. For derivation see written notes. |
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Describe the slab model for trasverse modulus. |
2 materials bonded perfectly together used as a model to describe a composite materials mechanical behaviour. Leads to the Reuss model. It shows that in the transverse direction the modulus of a composite varies non-linearly with reinforcement volume fraction and Young modulus of the 2 phases. For derivation see written notes |
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What are the mechanical rules of mixtures? |
The Voigt and Reuss model make up the rules of mixtures. Can also be used for compositions where the reinforcement phase is in fibrous form. See written notes for equations. |
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What is different about the rules of mixtures for short fibre composites? |
The assumptions of equal stress and strain are no longer valid. The exact value of the composite moduli must lie between the bounds of the longitudinal and transverse moduli. |
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Define stress. Define strain. |
The intensity of force loading a material. The intensity of the shape change. |
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Define stiffness. |
The resistance of a material to elastic deformation. The stiffer a material the less it will deform elastically for a given stress. |
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Define Young's modulus. |
It is a measure of elastic behaviour. Elastic behaviour is a result of stretching and contracting of atomic bonds. |
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Define brittle behaviour. Define ductile behaviour. Define plastic behaviour. |
A completely brittle material is one which deforms purely elastically to failure. A ductile material will deform permanently at a critical stress before reaching a maximum stress and failing. Ductile materials undergo plastic deformation once a critical stress/ yield stress is reached. |
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Define ductility. |
The ability to be deformed permanently without fracture. Defined as the plastic strain to failure. Deformation absorbs energy resulting in high fracture resistance. |
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Define an edge dislocation. |
An edge dislocation can be thought of as an extra half plane of atoms that terminates within a crystal. |
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Describe the motion of an edge dislocation. |
A dislocation can move easily in the presence of a shear stress because it only requires local rearrangement of the bonds around the dislocation core. The motion of the dislocation through the crystal is like a displacement wave which causes a small net shear displacement when it exits the crystal. Macroscopic plastic deformation in crystals is thus propagated in small increments by the motion of millions of dislocations. |
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How is a dislocation defined? |
A dislocation is defined in terms of its burgers vector and the glide plane on which it moves. |
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Describe how to find the burger vector. |
To find the Burgers vector a circuit is constructed around the dislocation. When repeated in a perfect crystal the line closing the circuit is the vector. |
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What is the relationship between the dislocation line, the shear direction and the burgers vector for an edge dislocation? |
The dislocation moves in the same direction as the shear. The Burgers vector and the shear direction are perpendicular to the dislocation line. |
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Describe a screw dislocation. Which direction is the burgers vector in? |
Screw dislocations are generated in shear by a tearing action. This causes a helical, or screw thread, like displacement. The Burgers vector is parallel to the dislocation line. |
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Describe cross slip. |
A dislocation can glide only in the plane which contains both its line and its Burgers vector. An edge dislocation can't cross slip as the Burger vector lies in a specific glide plane. A screw dislocation can move onto another slip plane as the Burgers vector lies parallel to the dislocation line. |
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What factor effects the strain rate? |
The strain rate is related to the average dislocation velocity. See written notes for mathematical derivation. |
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What is the shear force on a dislocation? |
For expressions and derivation see written notes. |
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Define Peierls Nabarro Stress. |
It is the lattices resistance to dislocation glide. It is the force needed to push a dislocation through a perfect lattice in a pure crystal. It is a direct consequence of the periodic structure of the crystal lattice and depends on the nature of the inter atomic bonding. |
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Why does plastic deformation occurs easily in pure metals? |
Plastic deformation occurs easily in pure metals because the delocalised non-directional metallic bond and high packing density allows dislocations to move easily. |
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Why does plastic deformation not occur easily in covalent crystals? |
Dislocation motion is very difficult due to the high strength and highly directional nature of the bonds. Also more open structure. |
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Why does plastic deformation not occur easily in ionic crystals? |
The bonding is less directional than covalent bonds but slip is very restricted. On most planes slip causes atoms of the same charge to become neighbours. There are however a few slip systems where slip can occur without this happening. This leads to the brittle character of ceramic materials which generally fracture by crack propagation at a lower stress. than their very high yield stress |
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Why do dislocations move in slip systems. |
Dislocations move preferentially on specific slip systems this is because they have the lowest Peierls Nabarro Stress |
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How many slip systems does an FCC structure have? |
There are 4 {111} planes forming a tetrahedron. There are 3 <110> directions in each plane Thus there are 12 slip systems. High symmetry of slip in a cubic structure. |
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How many slip systems does an HCP structure have? |
There is 1 {0001} plane, all slip planes are parallel. There are 3 <1120> directions in each plane. Thus there are only 3 easy slip systems. |
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Describe the critical resolved shear stress. |
A crystal will start to slip when the shear stress on the first mobile dislocation exceeds a critical stress and it starts to move. The critical resolved shear stress (CRSS) is when the shear stess in a slip plane {hkl} in the direction of slip exceeds the resistance to dislocation glide. See notes for derivation. |
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Describe the yield stress in a polycrystalline material. |
Neighbouring grains each yield on the slip system with the highest resolved shear stress. In a poly-crystal the yield point is the stress at which dislocation activity can initiate and propagate throughout the material. |
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Describe dislocation multiplication. |
Plastic deformation requires large increase in the number of mobile dislocations. These are generated from dislocation sources within the grains and by the multiplication of existing mobile dislocations. |
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Describe the stress fields around dislocations. |
For a screw dislocation it is symmetric and involves a pure shear parallel to the dislocation. For an edge dislocation the distortion is only in a plane normal to the dislocation and involves mainly tensile and compressive components in the direction of the burgers vector. For derivations of stress field see written notes. |
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Describe the interactions between dislocations. |
Dislocations on the same glide plane with the same sign will repel each other. Dislocations on the same glide plane with opposite sign will attract and annihilate. For mathematical expressions see written notes. |
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Define line tension. |
The strain field of a dislocation gives rise to an increase in strain energy in the crystal. Increasing the length of a dislocation will requires energy. Therefore a dislocation must exert a line tension. See written notes for equations. |
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Describe how to calculate the force required to bend a dislocation. |
Derivation in written notes. |
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Describe how dislocation interactions with obstacles increases the yield strength. |
The bending of the dislocation line between obstacles increases the line length and requires work. The force at which the dislocation can 'cut' or 'escape' from the obstacles will affect the bend angle which depends on the obstacle strength and if a line bends more it will interact with more obstacles. |
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Describe how solid solution strengthening increases the yield strength. |
Foreign atoms in solid solutions cause an elastic distortion of the crystal lattice. The lattice distortion interacts with the strain field around the dislocation. The increase in strength is related to the misfit atoms strain and concentration of solute. |
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Describe the interaction of strong solute interactions. |
In interstitial solid solution stronger solute dislocation interactions can occur that cause a yield point effect. At the dislocation core there is more space and the solute atoms can relax their lattice strain. Separating a carbon atom from the dislocation creates more lattice strain which provides a pinning force on the dislocation this increases the yield stress. At a critical stress dislocation break away from their carbon atmosphere and rapidly multiply. The rapid generation of mobile dislocations propagates as a slip band which leads to a yield drop until the band propagates across the entire gauge width. |
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Describe how precipitation strengthening increases the yield strength in the case of small soft particles. |
If the shear strength of the particle is exceeded they can be cut by a dislocation. The increase in strength is related to the area of particles intersecting the glide plane and the particles shear strength. The obstacle strength increases with size. |
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Describe how precipitation strengthening increases the yield strength in the case of strong particles. |
Dislocations are forced to bow around strong particles and do this via the arowan looping mechanism. See notes for mathematical explanation of mechanism. |
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How does particle size and spacing effect the a particles effect on the yield strength? |
A particles resistance to shear increases with its size and the particle spacing increases with size reducing the arowan stress. Small particles shearing dominates. Large particles looping dominates. |
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Describe the effect of the grain size on the yield strength. |
Grain boundaries act as barriers to dislocations. Dislocations can't cross grain boundaries as the slip plane is not continuous. For the whole material to yield plasticity must propagate from grain to grain but dislocations can not cross a grain boundary. Dislocations pule up on grain boundaries creating stress concentrations until the stress is high enough to stimulate a new dislocation in the next grain and so allow plasticity to propagate. The number of dislocations in pile is proportional to the slip line length.
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Describe the effect of strain/ work hardening on a crystal. |
The dislocation density increases with plastic strain at low temperatures (cold working). This leads to an increase in the yield stress due to the increase rate of dislocation interactions with strain. |
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Describe the 1st stage of strain hardening. |
Easy Glide. One slip system operating. Stress fields of dislocations moving on parallel glides interact. Strain field of dislocations moving on parallel slide planes overlap increasing resistance to glide. The flow stress increases with increasing dislocation density as the average separation of the dislocations decreases. |
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Describe the 2nd stage of strain hardening. |
Multiple slip. As the flow stress increases it becomes possible for other less favourable slip systems to operate. Dislocations moving on different slip systems interact much more strongly. This can cause dislocations to become locked. Dense tangles of dislocations develop (forests) these rapidly increase the yield strength. For mathematically expressions see written notes. |
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Describe the 3rd stage of strain hardening. |
Saturation of strain hardening. Eventually a reduction in strain hardening rate occurs due to a saturation of the dislocation density due to recovery. Cross slip and thermally activated climb allows dislocations to move onto new glide planes and lower the net lattice strain in the crystal. Dislocations of opposite sign try to annihilate. Dislocations of the same signs re-arrange into walls to reduce their net strain energy. This results in the formation of cellular dislocation structures or sub grains. |
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Describe the effect of strain/ work hardening on a poly-crystalline material. |
Same as for a crystal except their is no easy glide stage. Ad there are many crystal orientations some grains will always be orientated suitably for multiple slip from the initiation of yielding. |
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Describe the effect of severe strain hardening. |
Processes like cold rolling result in very large strains. This can lead to a big increase in strength but with an accompanying loss of tensile ductility due to a reduction in the strain hardening capacity |
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Define fracture resistance/ toughness. |
Is related to the ability of a material to absorb energy during fracture. A crack will grow if the elastic energy stored in the material due to the applied stress is released at a greater rate than the rate of energy required for the crack to grow. |
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Describe the appearance of a surface that has failed by ductile failure. |
Rough 'dimpled' surface. |
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Describe the appearance of a surface that has failed by brittle failure. |
Flat.
Smooth. Some propagation marks. |
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Describe the 2 types of brittle failure. |
Trans-granular- Across the grains. Inter-granular- Follows the grain boundaries. In crystalline materials trans-granular fracture is associated with specific crystal planes (cleavage planes). In amorphous materials there is no preferred cleavage plane. |
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Describe stress concentration on crack tips. |
Lines of force can not connect across cracks. Force is diverted to neighbouring material. A high load must be carried locally in a small volume. Max stress at the tip of an elliptical crack is given by the expression in the written notes. |
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Describe the Griffin theory of fast fracture. |
If the crack is unstable it can grow by its self under the applied stress. For diagrams, equations and derivation see written notes. |
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Describe the brittle mechanisms of crack growth. |
Automatically sharp cracks. Stress concentration at crack tip becomes extremely high as no mechanism for blunting. Exceeds theoretical strength of material. |
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Describe the ductile mechanisms of crack growth. |
Crack grows by ductile tearing. Blunts crack reducing stress concentration. Micro-voids nucleate at particles in plastic zone ahead of crack tip. |
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Why are composites tough? |
Fibres are very thin and contain few defects. Any defects they do contain are limited in size by the fibre diameter. Large increase in fracture surface. Friction sliding of fibres from sockets. |
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Define fatigue. |
Failure that occurs in structures subjected to fluctuating/cyclic stress. Mainly a problem in metals but can occur in polymers and ceramics by other mechanisms. Failure can occur even through the maximum stresses are lower than the materials yield point. |
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Describe cyclic or dynamic stress. |
Cyclic stresses are commonly caused by vibration, fluctuating loads or reciprocating motion. Both the stress range or amplitude and the mean stress affect the fatigue rate. See written notes for equations. |
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What are the 3 stages of fatigue? When does failure occur? |
1.Initiate of crack. 2.Propagation of crack. 3.Failure. Failure occurs when the remaining material can not carry the peak load. It occurs by ductile rapture or fast fracture. |
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Describe low cycle fatigue. |
Number of cycles to fatigue< 10^4. Bulk specimen deforms plastically. Typical of a repeated service overload. Stress amplitude> Yield stress. |
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Describe high cycle fatigue. |
Number of cycles to fatigue>10^4. Bulk specimen deforms elastically. Dangerous as failure can occur below the yield stress at the design load. Stress amplitude< Yield stress. |
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Describe an S-N curve. |
Data is plotted as stress amplitude (y) against number of cycles to failure (x). Some materials show a very flat curve which effectively gives rise to a fatigue limit. A tensile mean stress reduces fatigue life as tensile stress is required for crack growth. |
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Describe initiation in a fatigue system. |
For low cycle fatigue whole component deforms plastically. For high cycle fatigue sample is deforming elastically but on a local scale at stress concentrations local plastic deformation is possible. The high cycle fatigue life of any un-cracked component is dominated by the time to initiate a crack. Initiation of a fatigue failure is very sensitive to stress concentrations. |
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Describe crack nucleation in a 'persistent slip band'. |
Cracks develop in shear plane are arrested by grain boundaries. One crack breaks through and becomes large enough to develop its own plastic zone at the crack tip. The crack then propagates on the plane of maximum tensile stress. |
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Describe crack propagation under a cyclic stress. |
Crack propagation occurs by a progressive crack blunting and sharpening process. Tensile stress causes opening of the crack which blunts the tip which creates new surface. Compressive stress then causes closing of the tip. New surface can not be destroyed and is folded by extension of the crack. |
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Describe the two levels of markings formed on the fracture surface of a material that has failed due to crack propagation under cyclic stress. |
Bench marks- Microscopic lines due to the occasional large jumps in the stress amplitude producing sudden large crack advances. Striations- Micro-scale lines corresponding to the crack tip advance each stress cycle. |
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Define Paris' crack growth rate law. |
Crack growth data is also used to predict life following initiation. For derivation of the law see written notes. |
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What are the 2 main methods of combating fatigue. |
1. Avoid stress concentrations. 2. Induce surface compressive residual stress |
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Define creep. |
It is a time dependant permanent straining of a material under a constant applied load which can occur over a very long time. Usually only important at high temperature relative to melting point. |
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Describe the 3 stages of creep. |
1.Primary creep: Strain increases at a diminishing rate. rate of hardening greater that the rate of softening. 2.Secondary creep: Minimum creep rate. Rate of hardening is nearly equal to rate of softening. 3.Tertiary creep: Acceleration to failure by creep rupture. For graphs and expressions see written notes. |
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Which of the 3 stages of creep determines the creep life? |
Creep life is determined by the creep rupture time. This is dominated by the 2nd stage which is the longest. |
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How does increasing the temperature or stress effect the creep life? |
Increasing the temperature or applied stress decreases the creep life. The creep rate is very sensitive to both. |
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Describe dislocation creep. |
T>0.4 Tm. Occurs at high stress and moderate temperatures in metals. Caused by the climb of dislocations around obstacles. Climb undoes the pinning effect of obstacles on dislocations resulting in a constant creep rate. There is high sensitivity to the applied stress. |
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Describe diffusional creep. |
Occurs in poly-crystalline metals and ceramics. Cable creep: Atoms diffuse along grain boundaries. Naberro-herring creep: atoms diffuse through the crystal lattice. See written notes for equations. |
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Describe grain boundary sliding. |
Grain boundaries are more open structures that the crystal lattice. At high temperatures grain boundaries can slide. This can cause voids to form at areas of stress concentration. |
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Describe 4 ways of preventing creep. |
Use materials with a high melting point. Restrict dislocation creep. Design a material with a high density of stable second phase particles to restrict dislocation glide. Stop grain boundary sliding. Use an aligned grain structure or a single crystal. Reduce operating temperature. Insulate and actively cool. Thermal barrier coatings. |
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Describe the stiffness of polymers. |
Polymers have very low stiffness despite being contently bonded. In most polymers stiffness is dominated by secondary bonding and chain flexing. Specialised fibres such as Kevlar can be stiff. |
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Describe elastomers. |
They obey the definition of elasticity. They do not deform linearly. Their stiffness is very low as they deform by bond rotation and molecular uncoiling between cross-linked points. Large extensions are due to their being a stress large enough to unravel the chains. Other extensions arise due to the stress extending the sections between cross-links. |
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Describe a rubber. |
Rubbers are made of long flexible molecules. The natural state of such a molecule is to be curled up. When a stress is applied the molecule can uncoil due to the bond rotations along the backbone of the molecule. Rubbers are lightly cross-linked to prevent long molecules slipping relative to each other. Rubber can thus deform elastically to very large strains at low stresses and store large amounts of energy compared to other materials. |
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Define vulcanisation. |
The process of adding sulphur to form cross-links in a rubber. |
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Describe high performance polymers. |
High levels of crystallinity can be induced by drawing the polymers during processing, which aligns the molecules giving a very strong fibre texture. This can lead to a large increase in strength and stiffness and is exploited in the manufacture of fibres such as nylon and Kevlar. Strong fibre alignment is obtained by 'melt spinning' (extruding the molten polymer solution through small holes). |
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Describe how the mechanical behaviour of polymers is effected by temperature and strain rate. |
Increasing strain rate makes them more elastic and stronger. Increasing temperature makes them less elastic and weaker. This is caused by the effect temperature has on chain mobility and the fact that chain motion is time dependant. |
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Define the glass transition temperature. |
It is the temperature above which the c-c bonds of the molecules backbone become able to rotate due to thermal activation which allows motion of chain segments. |
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Describe a polymer below its glass transition temperature. |
Glassy region. Behaves as a brittle glass. Only elastic extension prior to failure. Stretching of bonds. High modulus and strength. Low ductility and toughness. |
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Describe a polymer above its glass transition temperature. |
Rubbery region. Viscoelastic behaviour. Chain segment rotation and slippage. Low modulus and strength. High ductility and toughness. |
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Describe a polymer at its glass transition temperature. |
Transition region. Intermediate temperatures (leathery). Neither glass nor rubber. Viscoelastic behaviour. Limited ductility. |
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Describe the structural factors effecting the glass transition temperatures. |
The glass transition temperature increases with; rigidity of molecule backbone, bulky side groups, stronger secondary boning and crystallinity. |
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How does molecular weight effect the polymer with respect to its glass transition temperature? |
Below transition temperature little effect. Above transition temperature it extends the rubbery plateau by increasing chain entanglements. There is a minimum molecular weight required before a linear polymer can have useful mechanical properties above its transition temperature. Cross linking extends rubber region. |
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Describe the yield behaviour and fracture of a polymer below its glass transition temperature. |
Low ductility and so fails by brittle failure. Unstable growth of largest flow. Fracture follows plane with maximum tensile stress. |
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Describe the yield behaviour and fracture of a polymer above its glass transition temperature. |
Ductile linear polymers exhibit a yield drop and large tensile ductility. Cold drawing is the development of a stable neck in ductile linear polymers once the yield stress is exceeded. Alignment of molecules occurs in neck which greatly increases strength. This stabilises the neck which spreads along the gauge length. Failure occurs by ductile rapture. |
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Describe the creep behaviour of polymers. |
Due to chain segment mobility polymers above their transition temperature are very prone to creep. This is a problem as it leads to dimensional instability. Creep can occur by both viscoelastic and viscous flow. |
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Define viscoelasticity. |
Ideal elastic behaviour. Obeys Hooke's law. |
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Define viscous behaviour. |
Stress is proportional to the strain rate and not dependant on strain. See written notes for equations. |
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Describe the viscoelastic behaviour of polymers. |
Above the transition temperature the behaviour of most [polymers can be thought of as being between that of elastic solids and viscous liquids. At lower temperatures and high rates of strain they display a more elastic like behaviour. At higher temperatures and low rates of strain they behave in a more viscous manner. In an ideal viscoelastic material the strain lags behind the stress but when the stress is removed it is fully recoverable with time (an-elastic). Most polymers also exhibit some viscous flow under load. Polymers will thus combine elements of all these behaviours depending on the temperature and strain rate. |
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What are the models of viscoelstic behaviour? |
Maxwell model. Voigt Model. For maths and graphs see written notes. |
