Add notes on metals

mechanical/mmme1029_materials.md

@@ -902,3 +902,166 @@ is any multiple, $n$, of $\lambda$. Or:
 $$n\lambda = 2d\sin\theta$$
 
 This is Bragg's Law.
+
+# Metals
+
+## Defects on the Atomic Scale
+
+Defects on the atomic scale have a significant effect on yield stress, ultimate tensile stress, and
+ultimate fracture stress.
+
+The yield stress of a real metal(-alloy) is much lower than the theoretical yield stress for
+the perfect metal(-alloy) crystal.
+This difference is because of the defects in the metal, particularly dislocations, as the
+dislocations allow the ions to slide past each other at much lower yield stresses.
+
+The 5 types of defects are:
+
+- Grain boundaries
+- Vacancies (missing ion)
+- Dislocations (missing row of ions)
+- Impurity ions
+- Crystalline includison
+
+![](./images/vimscrot-2021-12-22T13:02:28,180694109+00:00.png)
+
+### Dislocation Movement vs. Simple Sliding
+
+The layers of ions in a crystalline metal could simply over each other:
+
+![](./images/vimscrot-2021-12-22T13:16:39,506214227+00:00.png)
+
+However, the stress required for simple sliding is much higher than the stress required to move a
+dislocation.
+This is because dislocation motion is successive sliding of the partial plane of ions under applied
+shear stress (black arrow).
+The vacancy in the slip plane (yellow arrow) moves in steps in sequence from left to right.
+
+![](./images/vimscrot-2021-12-22T13:19:51,513367988+00:00.png)
+
+If there are no dislocations then plastic deformation is delayed to a higher applied stress,
+meaning the yield stress of the metal would be much higher.
+
+Dislocations move more easily on specific planes and in specific directions called the
+slip planes and slip directions which make up what is known as the
+[slip system](#slip-systems-in-metals).
+
+There are a very large amount of dislocations in metals and alloys.
+Dislocation density is expressed as total length of dislocations per unit volume.
+
+## Single Crystal Metals
+
+![](./images/vimscrot-2021-12-22T12:58:16,773351925+00:00.png)
+
+Normally when a molten metal is cooled to a solid, then lots of tiny crystals (grains) grow in
+different directions until they impinge.
+The grain boundaries are a source of mechanical weakness.
+
+A single crystal metal is one for which the casting is cooled to form just one giant crystal:
+
+1. The molten metal is cast into a mould
+2. At the very base of the mould, the temperature is dropped and the alloy crystallises into many
+   little crystals
+3. The crystals grow upwars through the liquid and meet a spiral tube and are constricted
+4. This tube only allows one crystal to grow through the spiral and then into the main mould
+
+## Polycrystalline Metals
+
+Most normal metals you see everyday are polycrystalline.
+
+![](./images/vimscrot-2021-12-22T12:58:38,918714742+00:00.png)
+
+![Acid etched surface of a polycrystalline metal](./images/vimscrot-2021-12-22T12:59:11,940527867+00:00.png)
+
+## Elastic and Plastic Strain in Metals
+
+When you apply a tensile stress to a mteal, this will produce a shear stress in any part of the
+metallic lattice that is not parallel or perpendicular to the applied stress.
+Under the action of shear stress, the metallic lattice will tend to experience a combination of
+elastic strain and plastic strain:
+
+![](./images/vimscrot-2021-12-22T13:46:45,608572706+00:00.png)
+
+## Raising the Yield Stress of a Metal
+
+There are 4 main ways to raise the yield stress of a metal:
+
+- Make a solid-solution---by metal alloying or atomic addition
+- Precipitate crystalline inclusions---by metal alloying or atomic additions and then heat treatment
+- Work-harden --- by processing and/or cold-working
+- Decrease the grain size --- by processing and/or heat-treatment
+
+### Make a Solid-Solution
+
+Adding an alloying element, B, to the host, A, forms a solid-solution as the ions or atoms of B
+dissolve in A.
+
+The impurity particles of B are a different size from the particles of A, distorting the metal
+lattice.
+The larger the difference in radii of the particles, the bigger the distortion.
+
+![Substitutional addition replaces ions in the host](./images/vimscrot-2021-12-22T14:21:25,321894455+00:00.png)
+
+![Interstitial addition adds particles between the ions in the host](./images/vimscrot-2021-12-22T14:21:32,009562988+00:00.png)
+
+The particles of B tend to diffuse to dislocations and immobilise them.
+This is why alloying increases the yield stress.
+
+Impurity particles generate lattice strain in the structure too:
+
+- Smaller particles introduce a compressive strain in the surrounding lattice
+- Larger particles introduce a tensile strain in the surrounding lattice
+
+![How Ni content in Cu affects Yield and Ultimate Tensile Stress](./images/vimscrot-2021-12-22T14:25:58,567632767+00:00.png)
+
+### Precipitating Crystalline Inclusions
+
+When adding an element, B, to a host, A, exceeds the solubility, the result is the formation of a
+solid-solution with a fixed ratio of B to A, but also precipitated crystals of a different ratio of
+B to A.
+
+![](./images/vimscrot-2021-12-22T14:30:14,141230179+00:00.png)
+
+Crystalline inclusions are really difficult to shear, especially if they are small, numerous, and
+have high Vickers' hardness.
+This slows down dislocation movement, increasing yield stress.
+
+### Work-Hardening and Cold Working
+
+We can use room temperature deformation to increase the number of dislocations present in a metal.
+As the % cold-work (%CW) is increased, the number of dislocations present also increases:
+
+$$\% CW = \frac{A_0 - A_d}{A_0} \times 100\%$$
+
+where $A_0$ is the initial cross sectional area and $A_d$ is the final cross sectional area.
+
+A carefully prepared sample has a dislocation density, $\rho_d$ of around $10^3$ mm mm$3$,
+whereas for a heavily deformed sample it is around $10^{10}$.
+
+A high density of dislocations means they are more likely to get entangled with each other,
+making it harder for dislocations to move.
+Therefore as $\rho_d$ increases, yield stress does too.
+
+### Decreasing the Grain Size
+
+- Most metals are polycrystalline with many grains.
+- Different grains will have a different crystal orientation.
+- Grains impede dislocation motion
+
+As you decrease grain size, you get more grain boundaries which basically creates more barriers
+to prevent slip.
+
+This is because a dislocation would have to change orientation across a grain boundary and "ionic
+disorder in the grain boundary results in discontinuity of slip" (A.B Seddon University of
+Nottingham 2020) (I think that's repeating it but it said it on the slideshow sooo...).
+
+So for any given metal, the fine grained is harder and has greater yield stress than the coarse
+grained version of it.
+
+#### Hall Petch Equation
+
+$$\sigma_{yield} = \sigma_0 + k_yd^{-0.5}$$
+
+where $d$ is the grain size and $\sigma_0$ and $k_y$ are material constants.
+
+Therefore a plot of $\sigma_{yield}$ against $d^{-0.5}$ would results in a straight line.