Course E - Mechanical Behaviours of Materials Flashcards
what is tensile testing, explain the different regions and the graph that is formed
- specimen is stretched
- force and corresponding extension of sample is recorded
- done until failure
1) as material extends, force rises and is linear with extension, in this region all deformation is elastic
2) yield point, from linear/elastic —> non-linear/plastic deformation
3) Ultimate tensile strength, highest point on graph, max force a material can withstand
4) failure
what can we say about unloading from a tensile test
- if the only extension that occurred was in the elastic region then it is all recovered
- any plastic deformation is not recovered but the elastic deformation that had occurred is
what is the limitation with tensile testing
- it is very dependent on the size/shape of the material
define stress, give the equation for (true) stress
stress, σ, is the normal force, F, acting on a surface per unit area, A,
σ = F/A
units = Pa
what is an issue that occurs with stress, what alternative expression can we use for stress
- when under stress, the sample can deform and hence the cross-sectional area that the force acts on can change
- hence we tend to just use engineering stress where the initial cross sectional area is taken
σ(eng) = F / Ao
define strain/ give the expression for (true) strain
- strain, ε, is defined as the relative change in the linear dimension of the material in the direction of the force
δε = δl/l
integrate between limits (li and lo) gives
ε(true) = ln(li/lo)
what is engineering strain, give the expression for it
ε(eng) = li-lo/lo = li/lo -1
define shear stress
shear stress, τ, is the force per unit area parallel to a surface
τ = F/Ao
define shear strain, γ
- to do with the change in shape that occurs under a shear stress
- generally under a shear stress a sample will ‘squish’/ deform
- focusing on one vertex we can note the distance by which the opposite side has shifted, this is Δyo, the side length is xo
γ = Δyo/xo = tan(φ)
define Poisson’s ratio, give the equation for it
- we know that under deformation, the shape of a material changes
- i.e. it elongated parallel to the force
- for a normal stress on z we can say
εx = εy = - v εz
where
v = Poisson’s ratio
- generally (0.2 - 0.5)
what is a stress-strain curve, why do we use them, what does the gradient of the linear part give
- similar to force-extension graph but now it is independent of the material geometry
- it follows the exact same shape as a force- ext. graph but has slightly different features
- yield point –> yield stress σy
- ultimate tensile strength –> ultimate tensile stress, σ(UTS)
- strain to failiure εf = strain at failiure
- gradient of linear part = Young’s Modulus, E
define elastic strain, give the equations linking stress and strain for normal and shear stresses
“In an elastic deformation, any strain in response to stress will be immediately and completely recoverable on removal of the stress”
σ = Eε
τ = Gγ
E = Young’s modulus
G = shear modulus
how can we relate E and G using poisson’s ratio, v
E = 2G(1+v)
how can we calculate the elastic strain energy/work done in elastic strain PER UNIT VOLUME for a material
consider work done as
FδL = σAδL = σALδε = VEε δε
W = ∫(0 to εmax) VEε dε = 1/2 Vεmax^2
so
Wv = 1/2 Eεmax^2 (normal stress)
Wv = 1/2 Gγ^2 (shear stress)
what is the link between the work done per unit volume in elastic deformation and the stress-strain graph
- the work done PER UNIT VOLUME is the area under a stress-strain graph
how can we model bonds/ roughly explain the shape of their potential against bond length graph
- when the atoms are far apart, their binding energy –> 0
- as the atoms approach each other PE decreases, potential well, attractive forces
- Umin = PEmin = equilibrium separation of atoms at 0K
- if atoms are close than the bond length at Umin, ro, then their energy rises/repulsive forces
The shape of the potential can be approximated using the Leonard-Jones potential
U(LJ) = Umin[(ro/r)^12 - 2(ro/r)^6]
DON’T LEARN THIS
give a summary of how we can approximate Young’s moduli of a simple cubic crystal using the Leonard Jones potential
for small displacements from ro we assume its roughly linear so
F = dU/dr
(explains F = kx)
we consider stretching a crystal on [100]
If the crystal is simple cubic then:
- each bond has an area ro^2 perp. to normal stress
σ = F/ro^2 = 1/ro^2 dU/dr
E = dσ/dε = dσ/dr dr/dε = (1/ro^2 d^2U/dr^2 |r = ro) (ro)
E = (1/ro) (d^2U/dr^2 | r = ro)
why do we use composites
- we can combine ceramics and polymers for better properties
- generally ceramics/polymers are stronger as fibres
- these fibres are embedded in a matrix material
- orientation within a plane can be random or aligned
- plies of aligned fibres can be stacked
- greatest reinforcing effect when fibres are aligned parallel but this makes material V anisotropic
how can we do analysis on composites/ what model should we use
we can use a slab model:
- we consider ‘slabs’ of the matrix material and fibre material
- the volumes of the slabs correspond to their true volume fractions
how can we determine/estimate the axial modulus of a composite using a slab model
- consider an axial stress on the slabs, this is where the stress acts on the side faces where both slabs are visible
- Voigt Model
- we assume extension by each slab is equal
σc = σfVf + σmVm = σfVf + σm(1-Vf)
σ = Eε
Ecεc = EfεfVf + Emεm(1-Vf)
we assume εc = εf = εm
so
Ec = EfVf + Em(1-Vm)
- this is a reasonable estimate - it is just a weighted average
how can we determine/estimate the transverse modulus of a composite using a slab model
We use Reuss model:
- we assume equal stress
- this time we consider stress acting on the two faces ‘top and bottom’ where only one of the slabs is visible
εc = εfVf + εm(1-Vf)
ε = σ/E
σc/Ec = σfVf/Ef + σm(1-Vf)/Em
σc = σf = σm
so
Ec = EfEm / (EmVf + Ef(1-Vf))
- this is a poor model, lower bound estimate as it does not consider shielding of areas around fibres by fibres
how is temperature related to kinetic energy
3/2 KT = <1/2mv^2> = avr. Ek
microscopically/ on an atomic level, how can we explain thermal expansion
- due to the asymmetry of the Leonard-Jones potential, we can see by taking the mean ro value at different temperatures that ro increases with temp
- this is done by considering higher Y lines on the U(LJ) graph
macroscopically give the equation for what happens in thermal expansion
εt = α ΔT
α = coefficient of thermal expansion