move notes out of year subfolders

uni/mmme/2053_mechanics_of_solids/.n2w.yml

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+itags: [ mmme2053 ]

uni/mmme/2053_mechanics_of_solids/asymmetrical_bending.md

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+---
+author: Akbar Rahman
+date: \today
+title: MMME2053 // Asymmetrical Beam Bending
+tags: [ mmme2053, beam_bending, asymmetrical_beam_bending ]
+uuid: 7afb5f13-4d55-4e00-927a-5d622520d844
+---
+
+## Procedure for Solving for Bending Stress and Neutral Axis Position in Asymmetrical Bending Problems
+
+1. [Determine the principal axes](#principal-axes-and-principal-2nd-moments-of-area) of the section (about which $I_{xy}= 0$)
+2. [Resolve bending moments onto these axes](#resolving-onto-principal-axes)
+3. [Determine angle of neutral axis](#position-of-the-neutral-axis)
+4. Evaluate bending stress at any position in the section, such as extremes away from neutral axis, which give maximum bending stress
+
+## Worked Example
+
+- [Worked Example 1 (PDF)](./worked_examples/MMME2053 AB WE1 Slides.pdf)
+- [Worked Example 2 (PDF)](./worked_examples/MMME2053 AB WE2 Slides.pdf)
+
+# Product Moments of Area
+
+To analyse asymmetrically loaded sections, we need the second moments of area $I_{yy}$, and $I_{xx}$
+but we also need $I_{xy}$, the product moment of area:
+
+$$I_{xy} = \int_A xy \mathrm{d}A$$
+
+# Parallel Axis Theorem
+
+The parallel axis theorem allows calculation of 2nd moments of area and product moments of area with
+respect to $x'$ and $y'$ axes:
+
+![](./images/vimscrot-2023-02-05T16:58:38,302909671+00:00.png)
+
+\begin{align*}
+I_{x'x'} &= \int_A y'^2 \mathrm{d}A \\
+&= \int_A (y+b)^2 \mathrm{d}A \\
+&= \int_A (y^2 + b^2 + 2by) \mathrm{d}A \\
+\\
+I_{x'x'} &= I_{xx} + Ab^2
+\end{align*}
+
+Similarly you can get
+
+\begin{align*}
+I_{y'y'} &= I_{yy} + Aa^2 \\
+I_{x'y'} &= I_{xy} + abA
+\end{align*}
+
+# Principal Axes and Principal 2nd Moments of Area
+
+Once the second moments of area and product moments are found, they can be used to plot a Mohr's circle where:
+
+- Point A is plotted at $(I_{xx},  I_{xy})$
+- Point B is plotted at $(I_{yy}, -I_{xy})$
+- Points P and Q show the positions of the principal 2nd moments of area, $I_p$, and $I_q$.
+- $\theta$ is the angular position $e$ of the principal axes with respect to the $x$-$y$ axes
+
+  The principal axes are the axes where the product moment of area is 0.
+
+![](./images/vimscrot-2023-02-05T16:59:07,932521138+00:00.png)
+
+$$C = \frac{I_{xx} + I_{yy}}{2}$$
+
+$$R = \sqrt{\left(\frac{I_{xx}-I_{yy}}{2}\right)^2 + I_{xy}^2}$$
+
+$$I_p = C + R$$
+
+$$I_q = C - R$$
+
+$$\sin{2\theta} = \frac{I_{xy}}{R}$$
+
+# Analyse Bending of a Beam with Asymmetric Section by Resolving Bending Moments onto Principal Axes
+
+![](./images/vimscrot-2023-02-05T16:59:59,303885015+00:00.png)
+
+If a bending moment, $M_x$ is applied about the x-axis only, then the stress in the flanges will
+cause bending to takeplace about both x and y axes.
+This is a consequence of $I_{xy} \neq 0$.
+
+To avoid this moment coupling effect, it is usually convenient to solve bending problems by
+considering bending about the principal axes, for which $I_{xy} = 0$.
+
+## Resolving onto Principal Axes
+
+![](./images/vimscrot-2023-02-05T17:00:19,100442577+00:00.png)
+
+If bending moments $M_x$ and $M_y$ are applied about the x and y axes respectively, these can be
+resolved onto the principal axes, P and Q:
+
+![](./images/vimscrot-2023-02-05T17:02:36,461283527+00:00.png)
+
+\begin{align*}
+\cos\theta &= \frac{M_{P_x}}{M_x} \rightarrow M_{P_x} = M_x\cos\theta\\ 
+\sin\theta &= \frac{-M_{Q_x}}{M_x} \rightarrow M_{Q_x} = -M_x\sin\theta\\
+\end{align*}
+
+![](./images/vimscrot-2023-02-05T17:02:45,578419970+00:00.png)
+
+Similarly we get:
+
+\begin{align*}
+M_{P_y} = M_y\sin\theta\\
+M_{Q_y} = M_y\cos\theta
+\end{align*}
+
+Therefore:
+
+\begin{align*}
+M_P = M_{P_x} + M_{P_y} = M_x\cos\theta + M_y\sin\theta \\
+M_Q = M_{Q_x} + M_{Q_y} = -M_x\sin\theta + M_y\cos\theta
+\end{align*}
+
+## Bending Stress at Position (P, Q)
+
+![](./images/vimscrot-2023-02-05T17:03:08,322998726+00:00.png)
+
+$$\sigma = \frac{M_PQ}{I_P} - \frac{M_QP}{I_Q}$$
+
+Note the -ve sign, as a positive stress results in a -ve moment about the y-axis.
+
+## Position of the Neutral Axis
+
+![](./images/vimscrot-2023-02-05T17:03:20,351506450+00:00.png)
+
+The neutral axis is where $\sigma = 0$:
+
+\begin{align*}
+\frac{M_PQ}{I_P} &= \frac{M_QP}{I_Q} \\
+\frac Q P &= \frac{M_QI_P}{M_PI_Q} \\
+\\
+\alpha &= \arctan\frac{Q}{P}
+\end{align*}
+
+The maximum stress is located in cross section point which is furthest from the neutral axis.

uni/mmme/2053_mechanics_of_solids/fatigue.md

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+---
+author: Akbar Rahman
+date: \today
+title: MMME2053 // Fatigue
+tags: [ mmme2053, fatigue, materials, uni, engineering ]
+uuid: 23852418-9fbb-44b8-a697-3c8b566e5143
+---
+
+
+# Stages of Fatigue
+
+## 1: Crack Initiation
+
+- happens on a micro-structural level
+- causes the start fatigue cracks
+- persistent slip bands form at the surface
+
+  - they are the result of dislocations moving along crystallographic planes
+  - leads to slip band intrusions and extrusions on the surface
+  - act as stress concentrations, **leading to crack initiation**A
+
+![](./images/vimscrot-2022-11-03T14:11:47,770744805+00:00.png)
+
+- crystallographic slip is controlled by shear stresses rather than normal stresses
+- therefore cracks tend to initially grow in a plane of maximum shear stress range
+- this leads to short cracks, usually on the order of a few grains
+
+![The loading in this figure is horizontal tension](./images/vimscrot-2022-11-03T14:13:13,910050859+00:00.png)
+
+## 2: Crack Propagation
+
+- the fatigue cracks tend to join together with more cycles
+- they grow along planes of maximum tensile stress
+
+
+
+![](./images/vimscrot-2022-11-03T14:15:30,939765129+00:00.png)
+
+## 3: Final Fracture
+
+- occurs when crack reaches critical length
+- results in either
+
+  - ductile tearing (plastic collapse)
+  - cleavage (brittle fracture)
+
+# Total Life Approach (Estimating Lifetime of a Part)
+
+- based on lab tests
+
+  - carried out under controlled loading conditions
+  - either stress or strain controlled loading conditions
+  - performed on idealised specimens
+  - specimens usually have finely polished defects (minimises surface roughness effects, affecting
+    stage 1 crack initiation)
+
+- tests give number of loading cycles to the initiation of a measurable crack as a function of
+  applied stress or strain parameters
+- measurability is dictated by the accuracy of the crack detection method used
+- this is typically between 0.75 mm to 1.00 mm
+
+- the challenge of fatigue design is to then relate the tests to actual component lives under
+  real loading conditions
+- traditionally, most fatigue testing was based stress controlled conditions with mean stress,
+  $S_m = 0$, which is known as a fully reversed load
+- the data was presented in the form of S-N curves (either semi-log or log-log plots) of alternating
+  stress, $S_a$, against cycles to failure, $N$ (failure defined as fracture)
+
+![](./images/vimscrot-2022-11-03T14:31:30,946882039+00:00.png)
+
+- figure \ref{fig:typical-s-n} contains schematic representations of two typical S-N curves
+- part _(a)_ shows a continuously sloping curve
+- part _(b)_ shows a discontinuity ("knee") in the curve---this is only found in a few materials
+  (notably low strength steels) between $10^6$ and $10^7$ cycles under non-corrosive conditions
+
+![\label{fig:typical-s-n}](./images/vimscrot-2022-11-03T14:34:36,090286120+00:00.png)
+
+- the curves are normally drawn through the median life value
+- therefore represent 50 percent expected failure
+
+- fatigue strength, $S_e$, is a hypothetical value of stress range at failure for exactly $N$ cycles
+  as obtained from an S-N curve
+- fatigue limit (or endurance limit) is the limiting value of the median fatigue strength as $N$
+  becomes very large ($>10^8$)
+
+# Effect of Mean Stress
+
+- mean stress has a significant effect on fatigue behaviour in cyclically loaded components
+- in figure \ref{fig:effect-of-mean-stress} you can see tensile mean stresses reduce fatigue life
+- compressive stresses increase fatigue life
+
+![\label{fig:effect-of-mean-stress}](./images/vimscrot-2022-11-03T14:31:30,946882039+00:00.png)
+
+- effect of mean stress commonly represented as a plot of $S_a$ against $S_m$ for a given fatigue
+  life
+- attempts have been made to generalise the relationship, as shown in figure \ref{fig:s_a-s_m}
+
+![\label{fig:s_a-s_m}](./images/vimscrot-2022-11-03T14:45:42,986596633+00:00.png)
+
+- modified Goodman line assumes linear relationship, where gradient and intercept are defined by
+  fatigue life, $S_e$, and material UTS, $S_u$, respectively
+- Gerber parabola employs same intercepts but relationship is a parabola
+- Soderberg line assumes linear relationship but the x intercept (mean axis end point) is taken
+  as yield stress, $S_y$
+
+- these curves can be extended into the compressive mean stress region to give increasing allowable
+  alternating stress with increasing compressive mean stress
+- this is normally taken to be horizontal for design purposes and conservatism
+
+
+# Effect of Stress Concentrations
+
+- fatigue failure is most commonly associated with notch-type features
+- stress concentrations associated with notch-type features typically leads to local plastic strain
+  and eventually fatigue cracking
+- the estimation of stress concentration factors (SCFs) are typically expressed in terms of an
+  elastic SCF, $K_t$:
+
+  $$K_t = \frac{\sigma^{\text{el}}_{\text{max}}}{\sigma_{\text{nom}}}$$
+
+- the fatigue strength of a notched component can be predicted with the fatigue notch factor, $K_f$,
+  which is defined as the ratio of the fatigue strengths:
+
+  $$K_f= \frac{S_a^{\text{smooth}}}{S_a^{\text{notch}}}$$
+
+  > i thought $S_a$ is alternating stress and $S_e$ is fatigue strength but the
+  > [uni slides](./lecture_slides/fatigue_and_failure_1.pdf) (slide 18) say otherwise :sob:
+  > TODO: find out what it's meant to be for sure
+
+  - however $K_f$ is found to vary with both alternating stress level and mean stress level and thus
+    number of cycles
+
+- figure \ref{fig:effect-of-notch} shows the effect of a notch, with $K_t = 3.4$, on the fatigue
+  behaviour of wrought aluminium alloy
+
+
+ ![\label{fig:effect-of-notch}](./images/vimscrot-2022-11-03T15:22:29,993209954+00:00.png)
+
+# S-N Design Procedure for Fatigue
+
+- constant life diagrams plotted as $S_a$ against $S_m$ (also known as Goodman diagrams) 
+  (figure \ref{fig:goodman-diagram}) can be used in design to give safe estimates of fatigue life
+  and loads
+
+![\ref{fig:goodman-diagram}](./images/vimscrot-2022-11-03T15:40:37,514148113+00:00.png)
+
+- the fatigue strength for zero mean stress is is reduced by the fatigue notch factor, $K_f$
+- $K_t$ is used if $K_f$ is not known
+- for static loading of a ductile component with a stress concentration, failure still occurs
+  when mean stress, $S_m$, is equal to UTS
+- failure at intermediate values of mean stress is assumed to be given by the dotted line
+
+- in order to avoid yield of whole cross-section of component, maximum nominal stress must be less
+  than the yield stress, $S_y$:
+
+  $$S_m + S_a < S_y$$
+
+## Safety Factor, $F$
+
+- determined from the position of the point relative to the safe/fail boundary:
+
+  $$\frac1F = \frac{S_aK_f}{S_e} + \frac{S_m}{S_u}$$
+
+  <details>
+  <summary>
+
+  Derivation
+
+  <summary>
+
+  $$F = \frac{OB}{OA}$$
+
+  from similar triangles we get
+
+  $$\frac{S_a}{\frac{S_u}{F} - S_m} = \frac{S_e}{K_fS_u}$$
+
+  </details>
+
+
+# Failure Examples
+
+## Bicycle Crank Arm
+
+![](./images/vimscrot-2022-11-03T14:37:49,949154012+00:00.png)
+
+![](./images/vimscrot-2022-11-03T14:37:57,733079705+00:00.png)
+
+## D.H.-106 Comet Failure
+
+- 1st production jet liner (debut in 1952)
+- several crashed in 1954 led to an inquiry
+- water tank testing and examination of a recovered fuselage showed that failure originated at a
+  square corner window
+- future designs used oval windows
+
+
+![](./images/vimscrot-2022-11-03T14:39:35,250528225+00:00.png)
+
+![](./images/vimscrot-2022-11-03T14:39:41,710507511+00:00.png)
+
+# Glossary (of Symbols)
+
+- notch stress concentration factor, $K_f$
+- stress concentration factor, $K_t$
+- alternating stress, $S_a$
+- fatigue strength, $S_e$ --- hypothetical value of stress range at failure for exactly $N$ cycles
+- mean stress, $S_m$
+- ultimate tensile stress, $S_u$
+- yield strength, $S_y$

uni/mmme/2053_mechanics_of_solids/fracture.md

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+---
+author: Akbar Rahman
+date: \today
+title: MMME2053 // Fracture
+tags: [ uni, mmme2053, fracture, materials, engineering ]
+uuid: 17315e63-3870-428b-b65d-a5d249768c05
+---
+
+# Fracture
+
+- Consider the stress concentration factor (SCF) for an elliptical hole in a large, linear-elastic
+  plate subjected to a remote, uniaxial stress
+
+  ![](./images/vimscrot-2022-11-03T16:16:29,022777996+00:00.png)
+
+- It can be shown that SCF can be expressed as:
+
+  $$K_t = \frac{\sigma_\text{max}^\text{el}}{\sigma_\text{nom}} = 1 + 2\frac{a}{b}$$
+
+- Therefore as $b \rightarrow 0$, the hole degenerates to a crack and $\frac ab \rightarrow \infty$
+  $\therefore K_t \rightarrow \infty$, provided the material behaviour remains linear elastic
+
+# Basis of the Energy Approach to Fracture Mechanics
+
+There are three modes of loading cases: $K_\text{I}$, $K_\text{II}$, $K_\text{III}$.
+
+- Generally, for geometries with finite boundaries, $K_\text{I}$ is used:
+
+  $$K_\text{I} = Y\sigma\sqrt{a\pi}$$
+
+  where $Y$ is a function of the crack and $a$ is never mentioned in
+  [the slides](./lecture_slides/fatigue_and_failure_2.pdf) (slide 6)
+
+  > TODO: find out what $Y$ and $a$ are
+
+- $K_\text{I}$ is the Mode-1 stress-intensity factor which defined the magnitude of the elastic stress
+  field in the vicinity of the crack tip
+- $K_\text{II}$ and $K_\text{III}$ are similar
+- The energy release release rate under mixed loading is given by
+
+  $$K_\text{total} = K_\text{I} + K_\text{II} + K_\text{III}$$
+
+![](./images/stress-intensity-factors.png)
+
+## Typical Fracture Toughness Values
+
+Material                      | $\sigma_y$ / Nm$^{-2}$ | $K_\text{Ic}$ / Nm$^{-1.5}$
+----------------------------- | ---------------------- | ---------------------------
+Mild steel                    | 220                    | 140 to 200
+Pressure vessel steel (HY130) | 1700                   | 170
+Aluminium Alloys              | 100 to 600             | 45 to 23
+Cast Iron                     | 200 to 1000            | 20 to 6
+
+# Fatigue Crack Growth
+
+- for a wide range of conditions, there is a logarithmic linear between crack growth rate and
+  intensity factor range during cyclic loading of cracked components
+- it allows crack growth to be modelled and estimated based on
+
+  - knowledge of crack and component geometry
+  - loading conditions
+  - empirical crack growth data
+
+Considering a load cycle as shown in figure \ref{fig:p-vs-t} which gives rise to a load acting on
+a cracked body
+
+![A graph of P vs t \label{fig:p-vs-t}](./images/P_vs_t.png)
+
+- The load range and crack geometry gives rise to a cyclic variation in stress intensity factor:
+
+  $$\Delta K = K_\text{max} - K_\text{min}$$
+
+- Paris showed that subsequent crack growth can be modelled by following equation
+
+  $$\frac{\mathrm{d}a}{\mathrm{d}N} = C\Delta K^m$$
+
+  where $C$ and $m$ are empirically determined material constants.
+
+
+- Fatigue crack growth data is often as $\log \frac{\mathrm{d}a}{\mathrm{d}N}$ against $\log{\Delta K}$
+
+  ![](./images/fatigue_and_failure_2_-019.png)
+
+- Below $K_\text{th}$, no observable crack growth occurs
+- Region II shows a near linear relationship---this is the region which fail by fatigue failure spend
+  most of their life
+- In region III rapid crack grown occurs and little life is involved
+
+- the fatigue crack growth life of the component can be obtained by integrating the Paris equation
+  between the limits of the initial crack size and final crack size, given that you know the
+  stress intensity factor
+
+
+## Typical Values for $\Delta K_\text{th}$, $m$, and $\Delta K$
+
+Material        | $\Delta K_\text{th}$ | $m$ | $\Delta K$ for $\frac{\mathrm{d}a}{\mathrm{d}N} = 10^{-6}$
+--------------- | -------------------- | --- | -------------
+Mild Steel      | 4 to 7               | 3.3 | 6.2
+Stainless Steel | 4 to 6               | 3.1 | 6.3
+Aluminium       | 1 to 2               | 2.9 | 2.9
+Copper          | 1 to 3               | 3.9 | 4.3
+Brass           | 2 to 4               | 4.0 | 4.3 to 66.3
+Nickel          | 4 to 8               | 4.0 | 8.8

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