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English literary canon

Comprehensive canon
The following is a list of canonical authors and canonical works. Names bolded represent highly significant authors or works.

Medieval
Drama
 * The York Cycle
 * Everyman

Poetry
 * Beowulf
 * Brunanburh
 * Cædmon: Cædmon's Hymn
 * Geoffrey Chaucer: The Canterbury Tales, Troilus and Criseyde
 * William Langland: Piers Plowman
 * Cynewulf: The Ascension
 * Deor
 * Discourse of the Soul to its Body
 * Dream of the Rood
 * Pearl Poet: Sir Gawain and the Green Knight, Pearl
 * The Battle of Maldon
 * The Seafarer
 * The Wanderer
 * Widsith

Prose

Renaissance
Drama
 * Frances Beaumont and John Fletcher: Philaster
 * Thomas Dekker: The Shoemaker's Holiday
 * Robert Greene: Friar Bacon and Friar Bungay, James the Fourth
 * John Lyly: Endimion
 * Ben Jonson: The Alchemist, Bartholomew Fair: A Comedy, Volpone, Every Man in His Humour
 * Christopher Marlowe: The Tragical History of Doctor Faustus
 * Philip Massinger: A New Way to Pay Old Debts
 * George Peele: The Old Wives' Tale
 * William Shakespeare
 * Nicholas Udall: Ralph Roister Doister
 * John Webster: The Duchess of Malfi

Poetry
 * John Donne
 * George Herbert: The Temple
 * William Shakespeare
 * Edmund Spenser: The Fairie Queen
 * Alexander Pope

Prose
 * Francis Bacon: New Atlantis
 * Robert Burton: The Anatomy of Melancholy
 * Thomas Deloney: Jack of Newbury
 * George Gascoigne: The Adventures of Master F. J.
 * John Lyly: Euphues: The Anatomy of Wit
 * Thomas Malory: Morte D'Arthur
 * Thomas More: Utopia
 * Thomas Nashe: The Unfortunate Traveller
 * Henry Neville: The Isle of Pines
 * Philip Sidney: Arcadia

English Civil War and Restoration
Drama

Poetry
 * Thomas Carew: The Rapture
 * John Cleveland: The Rebel Scot
 * Abraham Cowley: The Mistress
 * Richard Crashaw: 'Delights of the Muses
 * John Denham: Cooper's Hill
 * John Dryden
 * Robert Herrick: Hesperides
 * Andrew Marvell
 * John Milton: Paradise Lost
 * Francis Quarles: Emblems
 * Thomas Traherne: The Wonder
 * Henry Vaughan: Silex Scintillans
 * Edmund Waller: The Bud, Go, lovely Rose, On a Girdle

Prose
 * Richard Baxter: Saints' Everlasting Rest
 * Aphra Behn: Oroonoko
 * Thomas Browne
 * John Bunyan: The Pilgrim's Progress
 * Thomas Fuller: The Worthies of England
 * Jeremy Taylor: Holy Living, Holy Dying
 * Thomas Urquhart
 * Izaak Walton: The Compleat Angler

Romantic
Drama
 * Henry Fielding: The Modern Husband
 * David Garrick and George Colman: The Clandestine Marriage;
 * John Gay: The Beggar's Opera
 * Oliver Goldsmith: She Stoops to Conquer
 * John O'Keeffe: Wild Oats
 * Percy Shelley: The Cenci, Prometheus Unbound

Poetry
 * William Blake: Songs of Innocence, Songs of Experience, The Marriage of Heaven and Hell, Jerusalem
 * Robert Burns
 * George Gordon Byron: Don Juan, Child Harold's Pilgrimage
 * Samuel Coleridge: The Rime of the Ancient Mariner, Kubla Khan, Cristabel
 * William Collins
 * William Cowper
 * Thomas Gray: Elegy Written in a Country Churchyard
 * Leigh Hunt
 * John Keats: Sleep and Poetry, Lamia, Hyperion, Odes, Fall of Hyperion
 * Samuel Johnson: The Vanity of Human Wishes
 * Charles Lamb
 * Edgar Allan Poe: "The Raven"
 * Alexander Pope: Dunciad, Rape of the Lock
 * Walter Scott
 * Percy Bysshe Shelley: Mont Blanc, The Cloud, Ozymandius, Adonais
 * Christopher Smart: A Song to David
 * James Thomson
 * William Wordsworth: The Prelude, "Lucy" poems, "Matthew" poems, Tintern Abbey

Prose
 * Jane Austen
 * Frances Burney: Evelina
 * Jane Collier:
 * James Fenimore Cooper
 * Daniel Defoe: Robinson Crusoe, Moll Flanders
 * Thomas De Quincey: Confessions of an English Opium-Eater
 * Maria Edgeworth: Belinda
 * Henry Fielding: Joseph Andrews, Tom Jones
 * Charlotte Lennox: The Female Quixote
 * Matthew Lewis: The Monk
 * Henry Mackenzie: The Man of Feeling
 * Ann Radcliffe
 * Samuel Richardson: Pamela, Clarissa
 * Walter Scott: Waverley
 * Mary Shelley: Frankenstein
 * Tobias Smollet: Humphrey Clinker
 * Lawrence Sterne: Tristam Shandy, Sentimental Journey
 * Jonathan Swift: Gulliver's Travels
 * Horace Walpole: The Castle of Otranto

Sarah Scott, Sarah Fielding

Victorian
Drama
 * Gilbert and Sullivan
 * George Bernard Shaw (continuing into 20th Century)
 * Oscar Wilde: The Importance of Being Earnest

Poetry
 * Emily Dickinson
 * Matthew Arnold
 * Elizabeth Barrett Browning
 * Robert Browning
 * Thomas Carlyle
 * Lewis Carroll: Jabberwocky, The Hunting of the Snark
 * Gerard Manley Hopkins
 * A. E. Housman: Shropshire Lad
 * Henry Wadsworth Longfellow

 * Christina Rossetti: Goblin Market
 * Algernon Charles Swinburne
 * Alfred, Lord Tennyson: Ulysses, Idylls of the King, In Memoriam A. H. H., The Lady of Shalott
 * Joanne Slack " A man and his dog"

Prose
 * Mary Elizabeth Braddon: Lady Audley's Secret
 * Charlotte Brontë
 * Emily Brontë
 * Lewis Carroll
 * Wilkie Collins
 * Charles Dickens
 * Benjamin Disraeli: Sybil
 * George Eliot: Middlemarch, The Mill on the Floss, Adam Bede
 * Elizabeth Gaskell: Mary Barton, North and South
 * George Gissing: The Odd Women, New Grub Street
 * Thomas Hardy
 * O. Henry
 * Washington Irving
 * Henry James
 * Herman Melville "Moby Dick"
 * Margaret Oliphant: Miss Marjoribanks
 * Robert Louis Stevenson
 * Bram Stoker: Dracula
 * William Makepeace Thackeray: Vanity Fair
 * Mark Twain
 * H. G. Wells
 * Edith Wharton
 * Oscar Wilde: The Picture of Dorian Gray'
 * Ellen Wood: East Lynne

20th century
Drama
 * Samuel Beckett
 * Harold Pinter
 * J. M. Synge
 * William Butler Yeats

Poetry
 * W. H. Auden
 * Rupert Brooke
 * T. S. Eliot
 * Seamus Heaney
 * Dylan Thomas
 * William Butler Yeats
 * Ezra Pound
 * Sylvia Plath
 * Hart Crane

Prose
 * William Faulkner "Light in August", "The Sound and the Fury", "Go Down Moses", "As I Lay Dying", "Absalom, Absalom!"
 * James Joyce "Ulysses", "A portrait of the artist as a young man", "Finnegans Wake", "Dubliners".
 * Virginia Woolf "The Waves", "Orlando", "To the Lighthouse", "Mrs Dalloway".
 * Saul Bellow
 * Thomas Mann
 * Robert Fitzgerald
 * Agatha Christie
 * Felipe Fernández-Armesto
 * E. M. Forster: A Passage to India
 * Aldous Huxley
 * Rudyard Kipling
 * D. H. Lawrence
 * C. S. Lewis
 * George Orwell - Animal Farm, Nineteen Eighty-Four
 * J. R. R. Tolkien

Developing novels
Novel development

Historic novels
 * Walter Scott: Waverley (1814)
 * Walter Scott: Rob Roy (1817)
 * Walter Scott: Ivanhoe (1819)
 * Fyodor Dostoyevsky: Crime and Punishment (1866)
 * Fyodor Dostoyevsky: The Brothers Karamazov (1880)
 * R. D. Blackmore: Lorna Doone (1869)

Gothic novels
 * Horace Walpole: The Castle of Otranto (1764)
 * Ann Radcliffe: The Mysteries of Udolpho (1794)
 * Matthew Lewis: The Monk (1796)
 * Ann Radcliffe: The Italian (1796)
 * Jane Austen: Northanger Abbey
 * Jonathan Chappell: Jane Slayer (2014)

Sensation novels
 * Wilkie Collins: The Woman in White (1860)
 * Ellen Wood: East Lynne (1861)
 * Mary Elizabeth Braddon: Lady Audley's Secret (1862)
 * Wilkie Collins: Armadale (1866)
 * Wilkie Collins: The Moonstone (1868)

Sentimental novels
 * Henry Mackenzie: The Man of Feeling (1771)

African American Literature
 * Robert Hayden
 * Jay Wright
 * Toni Morrison, Song of Solomon, Beloved, Jazz
 * Jean Toomer
 * Maya Angelou
 * Lucille Clifton
 * Rita Dove
 * Nikki Giovanni
 * Michael S. Harper
 * Yusef Komunyakaa
 * Carl Phillips

Major plays
- William Shakespeare complete works

- Knot of the Heart by David Eldridge

- Equus by Peter Shaffer

- Waiting for Godot by Samuel Beckett

title: English literary canon


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User:Eas4200c.f08.blue.a/Lecture 78

Continuing Torsional Analysis:

Dimensional analysis, from lecture 29, is applied to the differential stress formula from lecture 27 to show that the differential of stress over distance x has the dimensions of force over volume.

$$\left[\frac{\delta \sigma _{ij}}{\delta x_{i}} \right]=\frac{F}{L^{3}}=\frac{force}{volume}$$

In equation 3.14 in Mechanics of aircraft structures by C.T. Sun, it is shown that the of partial derivatives of shear force in the x and y directions is equal to zero.

$$\frac{\partial \tau _{xz}}{\partial x}+\frac{\partial \tau _{yz}}{\partial y}=\frac{\partial \sigma _{xz}}{\partial x}+\frac{\partial \sigma _{yz}}{\partial y}=0$$

From the roadmap described in lecture 16, the Prandtl stress function $$\phi $$ can be found using the following formulas (Sun 3.15):

$$\sigma _{yx}=\frac{\delta \phi }{\delta z}$$ $$\sigma _{zx}=\frac{-\delta \phi }{\delta y}$$

As can be seen, the Prandtl stress function plays the role of a potential function. $$\left(\sigma _{yx},\sigma _{zx} \right)$$ is a component of the "gradient" of the Prandtl function with respect to y and z.

Recall that the gradient of a scalar function f(x,y,z) is simply the partial differential of f in each direction (x, y, and z) or (i, j, and k).

$$\vec{\bigtriangledown} f(x,y,z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k} $$

In lecture 27, the class found that the sum of the partial derivative with respect to y of stress in the yx direction and the partial derivative with respect to z of the stress in the zx direction results in zero. Now we can replace the stresses with the Prandtl stress function equivalents.

$$\frac{\partial \sigma _{yx}}{\partial y}+\frac{\partial \sigma _{xz}}{\partial z}=0$$

$$\frac{\partial }{\partial y}\left(\frac{\partial \phi }{\partial z} \right)+\frac{\partial }{\partial z}\left(\frac{-\partial \phi }{\partial y} \right)=0$$

$$\frac{\partial ^{2}\phi }{\partial y\partial z}-\frac{\partial ^{2}\phi }{\partial z\partial y}=0$$

Since $$\phi$$ is continuous and smooth then the second mixed derivative is interchangable.

Now the Prandtl stress function $$\phi (x,y)$$ can be introduced in terms of the shear stress as in equation 3.15 (Sun, 70).

$$\tau _{xz}=\frac{\partial \phi }{\partial y}$$ $$\tau _{yz}=-\frac{\partial \phi }{\partial x}$$

It follows that:

$$\gamma _{xz}=\frac{\partial w}{\partial x}+\frac{\partial u}{\partial z}=\frac{\partial w}{\partial x}-\theta y$$

$$\gamma _{yz}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=\frac{\partial w}{\partial y}+\theta x$$

$$\frac{\partial \gamma _{yz}}{\partial x}-\frac{\partial \gamma _{xz}}{\partial y}=2\theta $$

$$\gamma _{yz}=\frac{1}{G}\tau _{yz}$$

$$\gamma _{xz}=\frac{1}{G}\tau _{xz}$$

$$\frac{\partial \tau _{yz}}{\partial x}-\frac{\partial \tau _{xz}}{\partial y}=2G\theta $$

$$\frac{\partial ^{2}\phi }{\partial x^{2}}+\frac{\partial ^{2}\phi }{\partial y^{2}}=-2G\theta $$

$$\frac{\partial ^{2}\phi }{\partial y^{2}}-\frac{\partial ^{2}\phi }{\partial z^{2}}=-2G\theta=\bigtriangledown ^{2}\phi $$

This equation for the Prandtl Equation is equivalent to step F of the Torsional analysis Roadmap.

To find a function for the Prandtl Function, a two dimensional case is first looked at:

In the above image, a cross section of a non-uniform bar is shown with a force (t) pu.

This image represents the stress vector (t) on an infinitesimal surface of a non-uniform beam.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:Eas4200c.f08.blue.a/Lecture 78


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User:Eas4200c.f08.gator.edwards/Plate Buckling

First, we start by creating a 3-D perspective plot of the buckling shape of a simply supported rectangular plate.In the formula for u_z used to create the plot, c = 1, a = 1.5 m, b = 1.0 m, and therefore $$\frac{a}{b}=1.5$$

For Case 1, we set m = 1 and n = 2

For Case 2, we set m = 2 and n = 1

Now we found the period T of the function using:

$$ \sin \left( \frac{m \pi (x + T)}{a} \right) = \sin \left( \frac{m \pi x}{a} \right) $$

therefore,

$$ \left( \frac{m \pi (x + T)}{a} \right) = \left( \frac{m \pi x}{a} \right) + 2\pi $$

canceling out the x terms, we have

$$\frac{m\pi}{a}T=2\pi$$

$$\Rightarrow T=\frac{2a}{m} $$

Looking at this, we can determine that m is indeed the number of half-wavelengths.

Reproducing the k_c vs. a/b graph found on the MIT lecture notes page:

CRITICAL STRESS

Now we will create another perspective plot similar to before, except now the plate will be under clamped conditions instead of the previous simply supported conditions.


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User:Eml4500.f08.team.dwyer.jd/HW5 Nov 3 (Meeting 29)

The Principle of Virtual Work Continued
The initial conditions for the above elastic bar are described as follows:
 * At t=0, prescribe $$u(x,t=0)=\bar{u}(x)$$
 * $$\bar{u}(x)$$ is a known function of displacement
 * Also at t=0, $$\frac{\partial u}{\partial x}(x,t=o)=\dot{u}(x,t=0)=\bar{v}(x)$$
 * $$\bar{v}(x)$$ is a known function of velocity

Equation (1)  $$\frac{\partial }{\partial x}[A(x)E(x)\frac{\partial u}{\partial x}]+f(x,t)=m(x)\ddot{u}$$

Knowing that the Discrete Equation of Motion is as follows:  $$-\mathbf{k}*\mathbf{d}+\mathbf{F}=\mathbf{M}*\mathbf{\ddot{d}}$$

Noticing that there is a one to one correspondence between the two equations, for example $$\mathbf{M}*\mathbf{\ddot{d}}$$ is similar to $$m(x)\ddot{u}$$, the above Discrete EOM can be rewritten as:

Equation (2)  $$\mathbf{M}*\mathbf{\ddot{d}}+\mathbf{k}*\mathbf{d}=\mathbf{F}$$

Equation (2) is a multiple degree of freedom equation (MDOF).

Single Degree of Freedom Example

A single degree of freedom (SDOF) situation is shown above. The equation of motion is: Equation (3): $$m\ddot{d}+kd=F$$(3)

 * 2 initial conditions are needed since it is a 2nd order derivative.

Deriving Equation (2) from Equation (1)

 * To derive Equation (2) from Equation (1) the following integral is used.

Equation (4)

$$ \int_{x=0}^{x=L} w(x)\left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)dx=0 $$

Let $$ \left({\frac{\partial}{\partial x}\left[(EA)\frac{\partial u}{\partial x}\right]+f-m\ddot{u}}\right)=g(x) $$

Equation 4 then becomes $$ \int_{x=0}^{x=L} w(x)g(x)dx=0 $$ for all $$ w\left(x\right) $$

Since Equation 4 holds for all $$ w\left(x\right) $$, select $$ w\left(x\right)=g\left(x\right) $$,

then Equation 4 becomes $$ \int_{x=0}^{x=L}g^2dx=0 $$ ⇨ $$ g\left(x\right)=0 $$

Rectangular Truss Zero Eigenvalues Plot Revisited
The purpose of this exercise was to plot the displacements corresponding to the zero eigenvalues of the stiffness matrix for the rectangular truss presented below.

In the above diagram a = b = 1 m, and for case (a) E = 2 and A = 3.


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Medical physics

Medical physics is the application of physics to medicine. Physics in mainly used in the process of imaging, such as X-Rays, CT and MRI; but it may also be used in the treatment of disease, such as radiotherapy. Many of the applications of physics in medicine use radiation or radioactivity.

X-rays
X-rays are a form of electromagnetic radiation. They are formed by colliding fast-moving high energy electrons into a dense metal. As the the electrons interact with the metal they slow down, losing energy which is radiated as x-rays. The importance of x-rays in medicine is that the rays are stopped by bone much more than soft body tissues. So by placing a body part between the x-ray source and some photographic film an image of the patient's bones is formed. This is very useful for diagnosing fractures or other bone problems, but it is difficult to see detail in the soft tissues. To image the blood vessels or other soft structures, contrast media may be injected. These are substances which absorb the x-rays, so appear on the image. For example, a vein containing contrast media will appear on an x-ray, so by injecting the contrast into a patient, their blood vessels can be studied.

Computed Tomography (CT)
CT imaging uses x-rays to obtain 3D images of a patient. X-ray images are obtained at angles all around the patient and then reconstructed to produce slices through the body.

Magnetic Resonance Imaging (MRI)
To understand how MRI works requires a degree in particle physics. Hydrogen atoms have only one electron, so produce their own small amount of magnetic field. Normally these fields are pointing in all different directions, but when they are placed near a much stronger magnet (the large tube used in MRI), all the fields align with the external field. When the small fields are hit by a radio-frequency pulse, they change direction and it takes some time for the fields to go back into allignment with the large magnet. The time it takes for the re-alignment to occur depends on what substances the hydrogen is chemically bonded to. So fat with have a different timing to muscle (which is mostly water). Images can then be formed based on how long it takes the atoms in each region to realign.

title: Medical physics


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User:Eml4500.f08.bottle.brockmiller/hw2

HW2

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User:Jooooohn

Welcome to Jooooohn
Points of Interest

Art History, Theory & Criticism

Philosophy

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User:Eml4500.f08.wiki1.schaet/hw6

Using this general formula for stiffness matrix that works with non-constant E and A.

$$\mathbf{k}^{(i)}=\frac{1}{6L} \left( {2E_1A_1}+{(E_1A_2+E_2A_1)}+ {2E_2A_2}\right) \begin{bmatrix} 1 & -1\\ -1 & 1\end{bmatrix}$$

The matlab function of the axial forces and the function that calls it were modified to accomodate teh variances in E and A.

Boundary conditions:

Element 1

$$E_1^{(1)}=2$$ $$E_2^{(1)}=4$$ $$A_1^{(1)}=0.5$$ $$A_2^{(1)}=1.5$$

Element 2

$$E_1^{(2)}=3$$ $$E_2^{(2)}=7$$ $$A_1^{(2)}=1$$ $$A_2^{(2)}=3$$

Averages from before:

$$E_{avg}^{(1)}=3$$ $$A_{avg}^{(1)}=1$$

$$E_{avg}^{(2)}=5$$ $$A_{avg}^{(2)}=2$$

The plot of the undeformed beam, the deformed beam using the new equation, and the deformed beam using just the averages of E and A is as follows. The undeformed beam is the blue dotted line, the deformed beam using the averages is the solid red line, and the deformed beam using the varied E and A values is the solid green line. The matlab code that plots this image is as follows:  Two-Bar Plot

The matlab code used to attain the deformed beam using the varied E and A is as follows:  2 Bar Truss with varied E and A

The altered m-file, DeltaPlaneTrussElement.m, is as follows:  DeltaPlaneTrussElement.m Note: The deformed beam using the average values of E and A can be found in a previous hw.

Three and four bar eigenvalue problem revisited
The images for this section are as follows:

Three Bar Rectangular Truss
A for all elements is 1 E for all elements is 1 L for all elements is 1.

The following matlab code computed the results of this problem:

The results of this say that the 'Matrix is singular to working precision'. Essentially, this shows that the system is too unstable to compute results. If any load was applied at any node, the system would show NaN for the results. This is why a fourth bar is required.

AssumingThe eignevectors are found as follows: The plots of these eignevectors are as follows: image
 * code

Four Bar Rectangular Truss
The four bar rectangular truss system is as follows:
 * $$A^{(1)}=A^{(2)}=A^{(3)}=3$$
 * $$E^{(1)}=E^{(2)}=E^{(3)}=2$$
 * $$L^{(1)}=L^{(2)}=L^{(3)}=1$$ and $$L^{4}=\sqrt{2}$$

The following matlab code computed the results of this problem:
 * code

The following are the results:
 * code

The eignevectors are found as follows: The plots of these eignevectors are as follows: image
 * code

title: User:Eml4500.f08.wiki1.schaet/hw6


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User:Eml4500.f08.bottle.ranto/HW6

Electric Pylon Analysis
Our team was asked to conduct an analysis of a 91 bar, 60 m high electric pylon. The properties of the pylon are:

Material = 300 M Steel Height = 60 m Young's Modulus = 200 GPa Area = 4 cm2 Density = 7.8 g/cm3

MATLAB was implemented in solving for the displacements, stresses in each bar, the reaction forces, the eigenvectors, and periods of oscillation.

Code
The following MATLAB code was developed and run:

Results
The results array for the axial stress on each bar is: Stresses =

2.1562e+006 -1.3071e+006 1.3468e+006 1.2911e+006 -7.0304e+005 -4.7596e+006 -2.465e+006 1.4236e+006 3.8538e+006 -5.7655e+006 -37882      1755.6  3.8655e+006 2.9395e+006 -3.6988e+006 -5.7661e+006 3.5896e+005 5.242e+006 1.9797e+006 -3.7749e+006 -6.5159e+006 5.5784e+005 6.4388e+005 5.9656e+006 -8.1704e+005 -3.9616e+006 2.32e+006 -8.1307e+005 -5.7556e+006 -28474 -9.9353e+005 -15214 5.9216e+006 55813 -4.8717e+006 1.3915e+006 -3023.3 -5.7588e+006 -4.854e+006 1.3904e+006 -1.8831e+005 2.258e+006 -6.32e+005 -4.0102e+006 1.0133e-007 1.4647e-007 64147 2.7001e+005 2.7149e+005 -2.0778e+006 -4.4271e+006 -5.8777e+006 -7.3549e+006 -8.662e+006 -8.6957e+006 2.5807e-008 -6.1883e-008 -5.397e-008 9.7638e-008 -1.7704e+005 2.2069e+005 -1.9193e+005 1.9887e+005 2.1464e+006 -2.083e+006 2.2895e+006 -2.2643e+006 2.1035e+006 -2.1464e+006 1.4138e+006 -1.3772e+006 1.5835e+006 -3.6928e+006 -97393      -23690        31208  9.0479e+006 9.3927e-008 9.0317e+006 -2.3815e-007 9.0511e+006 -1.3175e+005 -2.9645e+005 1.7966e+006 4.4876e+006 6.6233e+006 7.7755e+006 -1.0741e-008 -2.4017e-008 -1.9214e-008 -1.0741e-008

The highest stresses occur at: MaxComp = -8.6957e+006 CompBar = 55

MaxTens = 9.0511e+006 TensBar = 81

The following is a plot of the deformed (solid black line) vs. undeformed (dashed blue line) pylon structure:

Notice that the bars in which the maximum stresses occur are labeled with arrows.

Effects of Eigenvectors
The lowest eigenvalues and their corresponding oscillation periods are:

lam =  132.16       2468.4       2940.3

T = 0.54654      0.12647      0.11587

Plots of the effects of the three lowest eigenvalues are shown below. The blue dashed line represents the undeformed structure while the solid black line represents the deformed structure.

Statically Determinant
The final step in this problem was to determine whether the structure can be solved using statics. Since there are only two fixed points and four bars for which reactions must be computed, it can be determined that this problem is statically indeterminate.

title: User:Eml4500.f08.bottle.ranto/HW6


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User:Eas4200c.f08.gator.edwards/Airfoil

The goal of this week's addition to the MATLAB program is to determine the buckling strength of the airfoil skin. Only the sections of skin between the stringers were considered, that is sections BF and EH. The load conditions remained My = -125 kN.cm and Mz = 50 kN.cm.

Next the provided Excel file was used to find ultimate stresses for varying aspect ratios. These were plotted with the maximum stress found in the Matlab program. The idea is that since the airfoil skin is not clamped or simply supported but somewhere in between so the maximum normal stress should lie between the two curves taken from the Excel file.

It is clear that the maximum stress is outside the boundaries of the curves. This means that the skin alone cannot stand up to the bending moments applied and will buckle unless it is reinforced by stringers.

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User:Eml4500.f08.wiki1.ambrosio/hw6

Electric Pylon
The highest axial stresses are 1376.8 lbs in element 1 and -1370.1 lbs in element 4.

The figure below shows the undeformed and deformed pylon. The dashed blue line represents the undeformed pylon while the solid red line shows the deformed pylon with a 20,000 weighting factor applied.

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User:Eml4500.f08.lulz.abcd/HW6

HW: What if EA is constant?
If EA is constant, prove the below stiffness matrix to be true:

$$\mathbf{k}= \mathbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}$$

We are given this general form of the stiffness matrix:

$$\mathbf{k}^{(i)}=\int^{x_{i+1}}_{x_i}\mathbf{B}^T(x)(EA)(x)\mathbf{B}(x)dx$$

We also can solve $$\mathbf{B}$$ and $$\mathbf{B}^T$$ into terms of $$L^{(i)}=x_{i+1}-x_i$$:

$$\mathbf{B}(x) = \begin{bmatrix} N'_i(x) & N'_{i+}(x)\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{x_i-x_{i+1}} & \frac{1}{x_{i+1}-x_i}\\ \end{bmatrix} = \begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}}\\ \end{bmatrix}$$

$$\mathbf{B}^T(x) = \begin{bmatrix} \frac{1}{-L^{(i)}}\\ \frac{1}{L^{(i)}}\\ \end{bmatrix}$$

From here it is simple substitution and algebra to solve the proof since (EA)(x) can be removed from the integral as a constant.

$$\mathbf{k}^{(i)}=(EA)\int^{x_{i+1}}_{x_i}\begin{bmatrix} \frac{1}{-L^{(i)}} & \frac{1}{L^{(i)}}\\ \end{bmatrix}\begin{bmatrix} \frac{1}{-L^{(i)}}\\ \frac{1}{L^{(i)}}\\ \end{bmatrix}dx=(EA)\int^{x_{i+1}}_{x_i}\begin{bmatrix} \frac{1}{{L^{(i)}}^2} & -\frac{1}{{L^{(i)}}^2}\\ -\frac{1}{{L^{(i)}}^2} & \frac{1}{{L^{(i)}}^2}\\ \end{bmatrix}dx=(EA)\begin{bmatrix} \frac{x_{i+1}-x_i}{{L^{(i)}}^2} & -\frac{x_{i+1}-x_i}{{L^{(i)}}^2}\\ -\frac{x_{i+1}-x_i}{{L^{(i)}}^2} & \frac{x_{i+1}-x_i}{{L^{(i)}}^2}\\ \end{bmatrix}=(EA)\begin{bmatrix} \frac{L^{(i)}}{{L^{(i)}}^2} & -\frac{L^{(i)}}{{L^{(i)}}^2}\\ -\frac{L^{(i)}}{{L^{(i)}}^2} & \frac{L^{(i)}}{{L^{(i)}}^2}\\ \end{bmatrix}$$

Dividing out $$L^{(i)}$$ from this and reducing the fractions yields the target equation:

$$\mathbf{k}= \mathbf{k}^{(i)} = \frac{EA}{L^{(i)}}\begin{bmatrix}1 & -1 \\ -1 & 1 \end{bmatrix}$$

HW: Find an expression for $$\mathbf{k}^{(i)}(\tilde{x})$$
We are given the general formula for the stifness matrix and want to find a more useful way to express it in the $$\tilde{x}$$ coordinate system:

$$\mathbf{k}^{(i)}=\int^{x_{i+1}}_{x_i}\mathbf{B}^T(x)(EA)(x)\mathbf{B}(x)dx$$

Remembering how to convert the coordinate systems we can convert the general formula to the $$\tilde{x}$$ coordinate system:

$$\tilde{x}:=x-x_i$$

$$d\tilde{x}=dx$$

$$\mathbf{k}^{(i)}=\int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0}\mathbf{B}^T(\tilde{x})(EA)(\tilde{x})\mathbf{B}(\tilde{x})d\tilde{x}$$

Next we consider the known solutions for A and E from the notes:

$$A(\tilde{x}) = N_1^{(i)}(\tilde{x})A_1 + N_2^{(i)}(\tilde{x})A_2$$

$$E(\tilde{x}) = N_1^{(i)}(\tilde{x})E_1 + N_2^{(i)}(\tilde{x})E_2$$

Substituting these yields:

$$\mathbf{k}^{(i)}=\int^{\tilde{x}=L^{(i)}}_{\tilde{x}=0}\mathbf{B}^T(\tilde{x})[(N_1^{(i)}(\tilde{x})E_1+N_2^{(i)}(\tilde{x})E_2)(N_1^{(i)}(\tilde{x})A_1+N_2^{(i)}(\tilde{x})A_2)]\mathbf{B}(\tilde{x})d\tilde{x}$$

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User talk:Eas4200c.f08.radsam/HW6

lot of
Use clear instead of a lot of  .Crochet.david 20:35, 21 November 2008 (UTC)

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User:Jtneill/Electronic marking

 * Single-spacing is easier to read on screen.
 * Set up a shortcut (e.g., Alt-T) to a tick symbol; this can easily be placed to indicate particularly good sentences (double, triple ticks etc. can indicate extremeley good).

title: User:Jtneill/Electronic marking


text:
User:Uf.team.aero/HW7

= Shear Loading on thin walled open section =

Consider the following open section consisting of 4 stringers:

The steps involved to solve for the shear flow:
 * 1) Find the location of the centroid $$\left(y_c,z_c \right)$$
 * 2) Find $$I_y,I_z,I_{yz}$$
 * 3) Find $$k_y,k_z,k_{yz}$$
 * 4) Follow the path "s" to find the shear flow

= Shear Loading Resistance of Thin-walled Sections =

It is necessary for the analysis of aircraft structures to understand the response of an arbitrary thin-walled section to an applied shear load. As such, the following outline will be used in analysis.

S) Single-cell Sections
 * S.1) Without Stringers
 * S.2) With Stringers

M) Multi-cell Sections
 * M.1) Without Stringers
 * M.2) With Stringers

The analysis is as follows:

S.1) Single-cell Sections Without Stringers
It is known from previous work that shear flow in a thin-wall can be assumed constant. The resultant of the shear flow can be found as the line integral of the shear flow projected onto the axis of interest. The figure below shows an arbitrary shape containing some shear flow.

To calculate the resultant of the shear flow along the z-axis, consider the shear flows from A to B and B to A.

 * $$ R^{Z} = R_{AB}^{Z} + R_{BA}^{Z} = -q\bar{A'B'} + q \bar{B'A'} = 0 $$

As a result, the single-cell section without stringers is shown to be incapable of resisting a shear loading in the z-direction, VZ. Since the Z-axis is chosen arbitrarily, any direction can chosen. Thus, in general, a single-cell section without stringers is incapable of resisting any shear loading whatsoever.

S.2) Single-cell Sections With Stringers
There exist single-cell sections with stringers, such that the shear flow is no longer constant about the perimeter of the section. It is assumed that the thin walls play very little role in shear loading resistance, such that the shear flow can be assumed piece-wise constant between stringers. This is shown in the figure below.

In the above figure,

 * $$ q_{ij} = constant $$

where subscript ij represents the shear flow in the wall along the path from stinger i to stringer j. Furthermore,

 * $$ q_{12} \neq q_{23} \neq q_{31} $$

It is proposed that for a balance of forces, the resultant from all shear flows must balance the applied shear loading condition. For example, if some shear load in the z-direction is placed on the section, then the resultant of shear flow is given as

 * $$ R_Z = V_Z \neq 0 $$

To analyze a section with stringers, the previous method and equations used for an open section must be modified to account for a closed section. Specifically, the available equations are applicable to an open section, such that a path can be drawn beginning at a free surface (where the shear flow is known to be zero. As such, the method of superposition is used to calculate the total shear flow in a closed section with stringers.

First a constant shear flow, q0 is assumed to be present throughout the perimeter of the section as if no stringers are present as shown in the figure below.

Next, the original section is assumed to be cut at a given point in the wall between two stringers. Thus, the resulting section is an open section with a free edge. Since the shear flow at a free edge is zero, and since the shear flow is assumed to be constant between two stringers, the wall section containing the cut is defined to have zero shear flow. Then, the method and equations for analyzing an open section can be implemented to find the piece-wise constant shear flows between the stringers, namely q12', q23', q31'. For the given cut location, it is seen that $$ q_{31}' = 0 $$. This is shown in the figure below.

The result of the superposition is given in equation form as

 * $$ q_{ij} = q_0 + q_{ij}' $$

-Notes from 8 Dec-

 * $$0= {q }^{(1)} + {q }^{(2)}+{q }^{(3)}+{q }^{(4)}=-{q }^{(3)} \neq 0$$

or

 * $$ 0 = \sum_{e=1}^{4} q^{(e)}$$

Note, from 38_2

 * $$ q^{(e)}= n_{z}Q_{z}^{(e)}+n_{y}Q_{y}^{(e)}$$

where

 * $$ n_{z}=-(k_y V_y-k_{yz}V_z)$$

and (for HW7 answer)

 * $$ n_{y}=-(k_z V_z-k_{yz}V_y)$$

so

 * $$ \Sigma q^{(e)}=n_z \sum_{1 }^{4 } Q_z^{(e)}+n_y \sum_1^4Q_y^{(e)}$$

on 41_2,we spoke of ways to determine Q for an area. Refer to the following figure:

Where,

 * $$Q_y ( \hat{z})= \int zdA$$

 * $$M_y=-EI_{yz} \frac{ \partial ^2 v_o}{ \partial z^2} - EI_y \frac{\partial ^2 w_o}{\partial x^2 } $$

and

 * $$M_z=-EI_{z} \frac{ \partial ^2 v_o}{ \partial x^2} - EI_{yz} \frac{\partial ^2 w_o}{\partial x^2 } $$

Where

 * $$\frac{ \partial ^2 v_o}{ \partial x^2}$$

can be noted as $$\chi_y$$

and

 * $$\frac{\partial ^2 w_o}{\partial x^2 }$$

can be noted as $$ \chi_z$$

The equations are rearranged in the following manner to find q for the cell:

 * $$ \left [ \begin{array}{rr} M_y\\ M_z \end{array} \right ] = \left [ \begin{array}{cc} I_y & I_{yz} \\ I_{yz} & I_z \end{array} \right ] \left [ \begin{array}{cc} -E X_z \\ -E X_y \end{array} \right ] $$

 * $$\epsilon _{xx} = -y \chi_y - z \chi_z=  \left [ \begin{array}{cc} z & y \end{array} \right ] \begin{Bmatrix}

- \chi _z \\ - \chi _y \end{Bmatrix}$$

 * $$\sigma _{xx} = E \epsilon _{xx} = E \left [ \begin{array}{cc} z & y \end{array} \right ] \begin{Bmatrix}

- \chi _z \\ - \chi _y \end{Bmatrix} =E \left [ \begin{array}{cc} z & y \end{array} \right ]\frac{1}{E} \check{I}^{-1}\begin{Bmatrix} - M _y \\ - M _z \end{Bmatrix}$$

where

 * $$\check{I} = \begin{bmatrix}

I_y & I_{yz}\\ I_{yz} & I_z \end{bmatrix}$$

and

 * $$\check{I}^{-1} = \frac{1}{D}\begin{bmatrix}

I_z & -I_{yz}\\ -I_{yz} & I_y \end{bmatrix}$$

and to find q:

 * $$q=-\int_{a}^{}{\frac{\partial \sigma _{xx}}{\partial x}}dA$$

where

 * $${\frac{\partial \sigma _{xx}}{\partial x}}=\begin{bmatrix}

z & y \end{bmatrix}I^{-1}\begin{bmatrix} \frac{dM_y}{dx}\\ \\ \frac{dM_z}{dx} \end{bmatrix}$$

M.1) Multicell Sections Without Stringers
NOTES FOR NOV. 21

The figure below depicts an example of a multicell section without stringers

The resultant of the shear flow for a figure of this case can be found by

 * $$R^z= \sum_{i=1}^n R_i^z$$

where n is the number of cells and $$R_i^z$$ is the resultant of the shear flow in the i-th cell.

It has been previously shown that any cell without strings is incapable of resisting shear in the Z direction and therefore

 * $$R_i^z = 0$$

for all i. It then directly follows that

 * $$R^z = 0$$

for multicell sections without stringers.

M.2) Multicell Sections With Stringers
Below is an example of a multicell section with stringers.

To solve for the resultant shear flow in such a system two states need to be analyzed then combined with the principle of superposition. State P1 below shows the closed cell constant shear flow, whereas state P2 below shows the open cell piecewise constant shear flow.

States P1 and P2 shown above are components of the total resultant shear flow, which can be formulated as

 * $$R^z = R^{z1}+R^{z2}$$

where $$R^z$$ is the resultant force of the section depicted at the beginning of the section, $$R^{z1}$$ is the resultant of the state P2, and $$R^{z2}$$ is the resultant of the state P2. Furthermore, from the analysis of single celled section we can expect that

 * $$R^{z1} = \sum{i=1}^n R_i^{z1} = 0 \qquad$$ and $$\qquad R^{z2} = R^{z2} V_2 \ne 0$$

Stringer 3
$$\displaystyle q^{(3)} = -\int_{A_3} \frac{d \sigma_{XX}}{dx} dA_3$$ The 3 designates that this is the shear flow contributed by stringer 3.

Recall:

$$\displaystyle V_y = \frac{M_z}{dx}$$

$$\displaystyle V_z = \frac{M_y}{dx}$$

p.34-3: $$ \displaystyle q_{(3)} = -(k_y V_y - k_{yz} V_z) Q_z^{(3)} - (k_z V_z - k_{yz} V_y) Q_y^{(3)}$$ → Eq. 1 $$ \displaystyle => Q_z^{(3)} = \int_{A_s} y dA$$ and $$ \displaystyle Q_y^{(3)} = \int_{A_s} z dA $$

Stringer 2
$$\displaystyle \tilde q_{23} = \tilde q_{12} - \tilde q_{24} + q^{(2)}$$

Because of the way the cell was cut: $$\displaystyle \tilde q_{12} = 0$$ $$\displaystyle \tilde q_{24} = 0$$

Therefore $$\displaystyle \tilde q_{23} = q^{(2)}$$

$$ \displaystyle q_{(2)} = -(k_y V_y - k_{yz} V_z) Q_z^{(2)} - (k_z V_z - k_{yz} V_y) Q_y^{(2)}$$ → Eq. 1 $$ \displaystyle => Q_z^{(2)} = \int_{A_s} y dA = y_2 A_2$$ and $$ \displaystyle Q_y^{(2)} = \int_{A_s} z dA = z_2 A_2$$

Stringer 4
$$\displaystyle \tilde q_{43} = \tilde q_{24} - \tilde q_{41} + q^{(4)}$$

Because of the way the cell was cut: $$\displaystyle \tilde q_{24} = 0$$ $$\displaystyle \tilde q_{41} = 0$$

Therefore $$\displaystyle \tilde q_{43} = q^{(4)}$$

$$ \displaystyle q_{(4)} = -(k_y V_y - k_{yz} V_z) Q_z^{(4)} - (k_z V_z - k_{yz} V_y) Q_y^{(4)}$$ → Eq. 1 $$ \displaystyle => Q_z^{(4)} = \int_{A_s} y dA = y_4 A_4$$ and $$ \displaystyle Q_y^{(4)} = \int_{A_s} z dA = z_4 A_4$$

Superposition (again)
$$\displaystyle q_{ij} = \tilde q_{ij} + q_{k}$$

$$\displaystyle q_{12} = \tilde q_{12} + q_{1}$$

$$\displaystyle q_{31} = \tilde q_{31} + q_{1}$$

$$\displaystyle q_{24} = \tilde q_{24} + q_{2}$$

$$\displaystyle q_{43} = \tilde q_{43} + q_{4}$$

$$\displaystyle q_{41} = \tilde q_{41} + q_{1}$$

=> 3 Unknowns $$\displaystyle q_1, q_2, and q_3$$: need 3 equations
 * 1) Moment Equation: Take moment of $$\displaystyle V_y, V_z, and {q_{12} ... q_{41}}$$ about any convenient point. (Usually where lines of action of $$\displaystyle V_y and V_z$$ intersect)
 * 2) $$\displaystyle \theta_1 = \theta_2$$
 * 3) $$\displaystyle \theta_2 = \theta_3$$

Solving Problem P2 continued
There are two methods to solve the shear flows of the section with hypothetical cuts (Problem P2):

 * 1) Complete method: free body diagram
 * 2) Consequence of complete method: conservation of shear flows

To expedite the solution of simple problems, we will further examine the conservation of shear flow approach. It is seen that the flow into or out of a stringer, both from the wall sections and the stringer itself, must be conserved. That is, the shear flow arriving at the stringer location must equal that leaving the stringer location. For example, consider the figure below.

To solve for $$ q_{j6}' $$, consider the summation of inflows and outflows.

 * $$ q_{j6}' = q_{2j}' - q_{j5}' - q_{j8}' + q^{(j)} $$

Thus, the outward shear flow from stringer j to stringer 6 is equal to the inward shear flows from 2 to j and the contribution from stringer j itself less the outward shear flow towards stringers 5 and 8. In this manner, shear flow about stringer j is conserved.

Likewise,

 * $$ -q_{2j}' = -q_{j6}' - q_{j5}' - q_{j8}' + q^{(j)} $$

It is seen that the shear flow in from stringer 2 to j and the shear flow from stringer j itself are supplying the shear outflows to stringers 5, 6 and 8. Again, this makes sense from a conservation perspective.

Isolating a Stringer
What if a closed, multicell section is cut in such a manner that a single stringer is isolated? What are the results? Is this allowed?

To answer this question, consider a cut of the interior walls such that stringer 3 is completely isolated. In this case, $$ q_{23}' = q_{31}' = q_{34}' = 0 $$. The resulting shear flow conservation of stringer 3 would produce

 * $$ q_{31}' = q_{23}' - q_{34}' + q^{(3)} $$

Based upon the previous statement that the shear flows in the cut walls are identically zero, the above statement would yield

 * $$ q^{(3)} = -\int_{A_3}^{}{\frac{d \sigma_{xx}}{dx}dA_3} = 0 $$.

Since the area A3 is nonzero and the normal stress is nonzero, this cannot be true. As such, the method of cutting a multicell section cannot be performed in such a way as to isolate any one stringer.

A continuation from stringer isolation
When looking at a closed cell, stringer that is in equilibrium, may help in simplification for the following stringers:

For stringer 1, we may write the following equation for equilibrium:

 * $$\tilde{q_{12}} = \tilde{q}_{31} + \tilde{q}_{41} + q_1$$

where $$\tilde{q}_{31}=0$$

After simplification, the equation becomes:

 * $$\tilde{q_{12}} = \tilde{q}_{41} + q_1$$

For stringer 2, accordingly we may write the following equation for equilibrium:

 * $$\tilde{q_{24}} = \tilde{q}_{12} + \tilde{q}_{23} + q_2$$

where $$\tilde{q}_{23}=0$$

After simplification, the equation becomes:

 * $$\tilde{q_{24}} = \tilde{q}_{12} + q_2$$

= Plotting Buckling Shape under Shear Stress =

First, we can express $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ in terms of $$C_{11}$$.

For $$\vartheta = 1.5$$ we can accomplish the following:

 * Find $$\lambda$$ for $$\vartheta = 1.5$$ using equation (30)

 * Evaluate for $$k_{sys}$$ using equation (26)

 * Calculate $$[k_{ij}]$$

 * $$\begin{bmatrix} k_{12} & k_{23} & \cdots & k_{25} \\ \vdots & \ddots & & \vdots \\ \vdots & & \ddots & \vdots \\k_{32} & \cdots & \cdots & k_{55}\end{bmatrix} \begin{bmatrix} C_{22} \\ \vdots \\ \vdots \\ C_{33}\end{bmatrix} = \begin{bmatrix} -\frac{u_z}{a}C_{11} \\ 0 \\ 0 \\ 0\end{bmatrix}$$

Then we will need to solve $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ intersect with $$C_{11}$$. If $$K$$ matrix is expressed as $$\underline{\overline{K}}$$ then its inverse is $$\underline{\overline{K}}^{-1}$$. Therefore, the calculated values for $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix}$$ can be related as:

 * $$\begin{Bmatrix} C_{22}, C_{13}, C_{31}, C_{33}\end{Bmatrix} = \underline{\overline{K}}^{-1}\ \begin{bmatrix} -\frac{u_z}{a}C_{11} \\ 0 \\ 0 \\ 0\end{bmatrix}$$

Looking at this result,

 * $$u_z = C_{11}sin(\frac{\pi x}{a})sin(\frac{\pi y}{b}) + C_{22}sin(\frac{2\pi x}{a})sin(\frac{2\pi y}{b}) + C_{13}sin(\frac{\pi x}{a}) sin(\frac{3\pi y}{b}) + C_{31}sin(\frac{3\pi x}{a})sin(\frac{\pi y}{b}) + C_{33}sin(\frac{3\pi x}{a})sin(\frac{3\pi y}{b})$$

= NACA Airfoil Continued =

This weeks homework incorporated finding the stresses in the airfoil due to shear loads imparted on the wing by drag and lift forces. This would be calculated for both single cell and multicell sections. For the first single cell section, we started with finding just the shear flow and determined from our code the following data:

 * {| style="width:100%" border="0"

Single Cell Airfoil
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

Length of FB = 0.25343 m

q_HF = 626.4307 N/m

q_FB = -16006.5727 N/m

q_BE = 626.4307 N/m

q_EH = 32480.5985 N/m

 * }

By dividing our obtained value of shear flow in each member by the thickness of the panel, 2 mm, would could compute the value of shear stress in the panel. Doing this for the panel EH, we determined $$\tau = 1.624e7\ Pa$$. This value is close to the experimental clamped critical stress for large aspect ratios. This is something we would expect as an aircraft wing has a relatively large aspect ratio.

The image to the right shows the buckling shape when adding the five mode; m = 1,2,1,3,3 and n = 1,2,3,1,3.

= Comparison of WebCT & Mediawiki =

As a team, we are in agreement that Mediawiki and WebCT each have benefits and drawbacks. The overriding comparisons are in the areas of collaboration, layout and privacy.

As a means of collaboration, Mediawiki excels. It is established for the expressed purpose of collaboration. It benefits from common internet accessibility, the history of edits and a common document. When all group members are able to see the same document real time and edit as necessary, it makes things much smoother. The ability to track which user made which edits is very useful for gauging contributions and correcting mistakes. WebCT uses discussion boards for collaboration, which severely limits the scope of projects capable. Since no common document is established, there is no means to understand edits rather than constantly exchanging a master document via email or messaging. Ultimately, for this class, the scope of collaboration did not extend as far as was perhaps intended. In general, the assignment was divided based on lecture days and each person made their own posts with little editing. In particular, collaboration across groups was very limited if it even existed at all.

The layout of WebCT lends itself to a more logical repository of common files, such as syllabus, homework assignments, solutions, grades, etc. The file structure is very straightforward for WebCT. However, Mediawiki doesn't have as good a hierarchical structure. This can be confusing and circular at times.

Because WebCT requires an individual login and provides the ability to privately upload individual assignments, it outperforms Mediawiki in terms of privacy. WebCT is a much more individualized platform as compared to Mediawiki. Also, the universal accessibility of Mediawiki reduces the problems of limited bandwidth often encountered during high traffic periods by WebCT.

Regardless of the platform, we feel that submitting homework and notes as was done in this class is not the most effective use of time. It would have been far easier to write out homework assignments and solutions by hand rather than in an electronic format. It was often frustrating to type out the Latex code for equations, taking far longer than just writing it by hand. Furthermore, drawing figures by means of Visio or Inkscape was also troublesome and frustrating. The return on investment for this time was very small. It served more to frustrate and waste time than to reinforce learning. Mediawiki certainly represents a great resource, but may be better used for presenting a final report or project rather than day-to-day homework assignments.

= Contributing Team Members =

 * Jared Lee --Eas4200c.f08.aero.lee 21:57, 21 November 2008 (UTC)
 * Ray Strods --Eas4200c.f08.aero.strods 20:32, 24 November 2008 (UTC)
 * Oliver Oyama --Eas4200c.fo8.aero.oyama 20:29, 8 December 2008 (UTC)
 * Thomas McGilvray--Eas4200c.f08.aero.mcgilvray 23:24, 8 December 2008 (UTC)
 * William Kurth --Eas4200c.f08.aero.kurth 15:44, 9 December 2008 (UTC)
 * Gonzalo Barcia --Eas4200c.f08.aero.barcia 16:44, 9 December 2008 (UTC)
 * Melisa Gaar -- Eas4200c.f08.aero.gaar 21:02, 9 December 2008 (UTC)

title: User:Uf.team.aero/HW7


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User:Eas4200c.f08.WIKI.E/extra2

This is equation one which will be combined with two later equations to give us the generalized 3-D case:

$$t_y=\sigma_{yy}n_y+\sigma_{yz}n_z\;$$

Note: $$\left[t_y \right]=\frac{F}{L^2}$$

$$\vec{t}\;$$:traction vector (distribution surface force)

$$\left[t_y \right]=\left[\sigma \right]$$

This is equation 2:

$$t_z=\sigma _{yz}\cdot n_y+\sigma _{zz}n_z$$

This is equations one and two combined into matrix format:

$$\begin{Bmatrix} t_y\\ t_z \end{Bmatrix}=\begin{bmatrix} \sigma_{yy} &\sigma_{yz} \\ \sigma_{zy}& \sigma_{zz} \end{bmatrix}\begin{Bmatrix} n_y\\ n_z \end{Bmatrix}$$

Generalized to 3-D case

$$\begin{Bmatrix} t_1\\ t_2\\ t_3 \end{Bmatrix}=\begin{bmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} &\sigma_{22} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{bmatrix} \begin{Bmatrix} n_1\\ n_2\\ n_3 \end{Bmatrix}$$

$$t_i=\sum_{j=1}^{3}{\sigma_{ij}n_j}$$ $$i=1,2,3$$

$$\left\{ti \right\}_{3x1}=\left[\sigma_{ij} \right]_{3x3}\left\{n_j \right\}_{3x1}$$

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User:EAS4200C.F08.WIKI.A/My Contribution: 6.1

Here we will find the sum of the forces in the y-direction.
$$\displaystyle \begin{Vmatrix} \bar{n} \end{Vmatrix}= 1$$ $$\displaystyle \sum{F_y} = -\sigma_{yy}\cdot (dz\cdot 1) - \sigma(dy\cdot1) + t_y(ds\cdot1) = 0 = \begin{cases}dz =ds\cos \theta \\ dy = ds\sin\theta \end{cases}$$

$$\displaystyle 0 = -\sigma_{yy}n_y - \sigma_yzn_z + t_y ds$$

$$\displaystyle n_y = \cos\theta, n_z = \sin\theta $$ is the unit normal vector unit depth along x-direction.

This is equation one which will be combined with two later equations to give us the generalized 3-D case:

$$t_y=\sigma_{yy}n_y+\sigma_{yz}n_z\;$$

Note: $$\left[t_y \right]=\frac{F}{L^2}$$

$$\vec{t}\;$$:traction vector (distribution surface force)

$$\left[t_y \right]=\left[\sigma \right]$$

This is equation 2:

Taking the sum of the forces in the z-direction
$$t_z=\sigma _{yz}\cdot n_y+\sigma _{zz}n_z$$

This is equations one and two combined into matrix format:  $$\begin{Bmatrix} t_y\\ t_z \end{Bmatrix}=\begin{bmatrix} \sigma_{yy} &\sigma_{yz} \\ \sigma_{zy}& \sigma_{zz} \end{bmatrix}\begin{Bmatrix} n_y\\ n_z \end{Bmatrix}$$

Generalized to 3-D case

$$\begin{Bmatrix} t_1\\ t_2\\ t_3 \end{Bmatrix}=\begin{bmatrix} \sigma_{11} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} &\sigma_{22} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33} \end{bmatrix} \begin{Bmatrix} n_1\\ n_2\\ n_3 \end{Bmatrix}$$

 $$t_i=\sum_{j=1}^{3}{\sigma_{ij}n_j}$$ $$i=1,2,3$$

 $$\left\{ti \right\}_{3x1}=\left[\sigma_{ij} \right]_{3x3}\left\{n_j \right\}_{3x1}$$


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User:Eas4200c.fo8.aero.lee/HW6

 See my comments below. Please don't remove these comment boxes; just add your comment to these comment boxes in case you fix the problems.

Warning: User Eas4200c.f08.lee was not registered; please register to create this user account, since otherwise the page may be deleted.

Also, don't put the figures on the right of the collapsible tables, since these figures would block the link to expand/collapse these tables; see your figures to the right of your matlab codes in collapsible tables below. Instead, you may want to put these figures to the left of the collapsible tables. Another way is to shorten the width of the collapsible tables.

Eml4500.f08 13:27, 23 November 2008 (UTC)

=Plate Buckling= A plate under uniform stress in one dimension (see Figure 1) undergoes deformation according to the following equation:
 * $$u_z(x,y)= \sum_{m=1}^{ \infty }{}\sum_{n=1}^{\infty}\psi_{mn} (x,y) $$ where $$\psi$$ is expressed as
 * $$\psi_z(x,y)=csin(\frac{m\pi x}{a})sin(\frac{n\pi x}{b})$$

and $$u$$ is the out of plane deformation. (Can be thought of as deformation in z direction.) Please note this is for a simply supported plate.

The resulting shape when looking at the plate normal to the xy plane is called the buckling mode shape.

The critical loading, $$P_x$$ is given by the following equation:
 * $$(P_x)=k_c\frac{\pi ^{2}D}{b}$$

 * $$k_c(m,\frac{a}{b})=(\frac{mb}{a}+\frac{a}{mb})^2$$

and

 * $$D=\frac{Eh^3}{12(1-\upsilon ^2)}$$

Where h is the thickness of the plate in the z direction.

The critical buckling stress is:
 * $$(\sigma_{xx})_{cr}=\frac{(P_x)_{cr}}{bh}=k_c \frac{\pi^2D}{b^2h}$$

Components related to the minimum critical value are denoted with a star, e.g. $$\sigma^\star_{cr}$$ The minimum critical buckling stress is dependent on $$k^\star_c$$, which in turn is dependent on $$m^\star$$ by the following equation:
 * $$k_c^\star(a/b)=\min_{m = 1, 2, \cdots}\ k_c(m, a/b)=\min_{m = 1, 2, \cdots}\left( \frac {m b}{a} +\frac{a}{m b}\right)^2 =

\left( \frac{m^\star b}{a}+\frac{a}{m^\star b}\right)^2$$ where
 * $$ m^\star(a/b)=\underset{m=1,2, \cdots}{\rm argmin}\ k_c(m, a/b) =\underset{m=1,2, \cdots}{\rm argmin}\left(\frac{m b}{a}

+ \frac {a}{m b}\right)^2$$ The transition from m to m+1 waves occurs when:
 * $$\frac{a}{b}=\sqrt{m(m+1)}$$

For a square plate with all 4 edges clamped, the buckling mode shape approximates to:
 * $$ \psi(x,y)=\frac{c}{4}\left(1 - \cos \frac{2 \pi x}{a}\right)\left(1 - \cos \frac{2 \pi y}{b} \right)$$

This gives a “bubble” like shape, as seen in the matlab figure below. It makes logical sense since the slope of the buckled shape must approach zero (dz/dx and dz/dy) at all clamped edges.

Shear Load, Simply Supported
Load is denoted as $$N_{xy}$$, which is force per unit length, or stress times h, or plate thickness.

Stored strain energy is equated to the work done by shear load U=W.

The Nxy term is in the equation for strain energy, and the critical Nxy is expressed as:
 * $$\displaystyle

(N_{xy})_{cr} =-\frac{ab D}{32}\frac{\displaystyle\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}C_{mn}^2\left(\frac{m^2 \pi^2}{a^2}+       \frac{n^2 \pi^2}{b^2}\right)^2}{\displaystyle\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\sum_{p=1}^{\infty}\sum_{q=1}^{\infty} C_{mn}C_{pq}\frac{mnpq}{(m^2 - p^2)(n^2 - q^2)}} $$ After further analysis and derivation, crit. buckling stress is:
 * $$(\sigma_{xy})^{\star,5}_{cr}=\pm k^{\star, 5}_c \frac{\pi^2D}{b^2h}, k^{\star,5}_c = \frac{\pi^2}{32}\frac{1}{\lambda\vartheta }$$

=A note on dimensional analysis=

A bracket notation (ex. [m]) is used to indicate the dimensions associated with a variable. Here are some examples:

 * $$[f]=\frac{F}{L}$$

this means the dimensions of "f" are force per length.

Since

 * $$[A]=L^2$$

then

 * $$\left[\frac{f}{A} \right]= \frac{F}{L^{3}}$$

meaning the dimensions of "f" per unit area are force per volume, or length cubed. Note, L is only a unit of measure, not a specific dimension.

The following describes a body force:

 * $$\frac{d\sigma }{dx}+\frac{f(z)}{A}=0\Rightarrow [\sigma ]=\frac{F}{L^2}\Rightarrow \left[\frac{d \sigma}{dx} \right]= \frac{[d \sigma ]}{[dx]}=\frac{F/L^2}{L} = \frac{F}{L^3}$$

To show that axial strain is nondimensional:

 * $$[\epsilon ] = \frac{du}{dx} = \frac{L}{L} = 1...nondimensional...$$

Shear strain is also nondimensional:

 * $$[\gamma ]=\frac{[\epsilon _{yy}]}{[\epsilon _{xx}]}=1...nondimensional...$$

Also note
 * $$\left[\frac{d \sigma_{ij}}{dx_i} \right]=\frac{F}{L^3}=\frac{Force}{Volume}$$

Intro
We introduce a function, $$\phi$$, which plays the role of a potential function. We call this the Prandtl stress function. $$\sigma_{yx}$$ and $$\sigma_{zx}$$ are components of the gradient of $$\phi$$ with respect to y and z.

In wikipedia, a gradient is defined as is a vector field which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change. In notation, it is defined as
 * $$\bigtriangledown f(x,y,z)= \frac{\partial f}{\partial x}i + \frac{\partial f}{\partial y}j + \frac{\partial f}{\partial z}k$$

where f is the potential.

For our case,
 * $$\sigma_{yx}=\frac{\partial \phi}{\partial z}$$ and
 * $$\sigma_{zx}=-\frac{\partial \phi}{\partial y}$$

We can plug these values into the following equation:
 * $$\frac{\partial \sigma_{yx}}{\partial y}+ \frac{\partial \sigma_{zx}}{\partial z}=0$$

to obtain
 * $$\frac{\partial }{\partial y}\frac{\partial \phi}{\partial z}+\frac{\partial }{\partial z} \frac{(-\partial \phi)}{\partial y}=\frac{\partial^2 \phi}{\partial y \partial z}- \frac{\partial^2 \phi}{\partial z \partial y}=0$$

Phi is continuous and smooth, making the 2nd mixed derivative interchangeable, i.e.
 * $$\frac{\partial^2 \phi}{\partial y \partial z}= \frac{\partial^2 \phi}{\partial z \partial y}$$

These equations can be compared with Mechanics of Aircraft Structures(Sun) Ch. 3 eqs 3.15-3.19. Using the author's notation:
 * $$\frac{\partial \tau_{yz}}{\partial x}-\frac{\partial \tau_{xz}}{\partial y}=2G\theta$$

and
 * $$\frac{\partial^2 \phi}{\partial x^2}+ \frac{\partial^2 \phi}{\partial y^2}=-2G\theta$$

The remainder of section 3.2 brings us to our roadmap equation for J (Roadmap eq 7) and notes that once the Prandtl stress function is solved, GJ, the torsional rigidity (the bar's ability to resist deformation due to torsion) is also found. Torsional analysis of structures fall into two cases - that of structures with a thin walled cross section (also referred to as closed cross section) and those of structures with a solid cross section.

Case 1: Thin Walled Cross Section
A prime example of this case is that of the NACA airfoils we have been studying throughout the semester. For thin walled cross section the boundary conditions for Prandtl stress function, $$ \phi $$, are that $$\phi$$ is a constant on inner and outer surfaces of the cross section. The figure below depicts the inner and outer surfaces, $$ S_1 $$ and $$ S_0 $$, respectively.

The boundary conditions for this case are $$\phi=C_0$$ on $$S_0$$ and $$\phi=C_1$$ on $$S_1$$.

Case 2: Solid Cross Section
This case consists of any bar or beam that has a solid cross section such as the one depicted below.

The constant value of $$\phi$$ at the boundary solid cross sections is arbitrary and so it can be set to zero on that surface. Thus the boundary in this case is $$\phi = 0$$ on $$S_0$$

Uniform Bar With Solid Circular Cross Section
As an example, we will analyze the following bar with a circular cross section as shown below.

For this case, $$\phi$$ can be defined as

 * $$\phi(y,z) = C\left(\frac{y^2}{a^2}+\frac{z^2}{b^2}-1\right)$$

It might also be noted that for the case of a circular cross section a=b. As mentioned before for the case of a solid cross section $$\phi=0$$ on $$S_0$$. Recalling the relation:

 * $$\nabla^2\phi=-2G\Theta$$

an expression for C can be obtained by solving for the laplacian of $$\phi$$.

 * $$\nabla^2\phi = \frac{4C}{a^2} = -2G\Theta$$
 * $$\Rightarrow C = -\frac{1}{2}a^2G\Theta$$

To analyze the Prandtl stress function, $$ \phi $$, as a function of the radius of a circular cross section of diameter $$ 2\alpha $$, consider the definition of the radius as

 * $$ r = \left(y^2 + z^2 \right)^{1/2} $$

Then, a plot of $$ \phi(r) $$ vs. r would yield the figure below

Now, recall from previous lectures the following equations

 * $$ T = 2\int_{A}^{}{\phi dA} = 2C\left(\frac{J}{a^2}-A\right) $$

 * $$ J = \int_{A}^{}{r^2 dA} = \frac{1}{2} \pi a^4 $$

 * $$ \tau = \frac{Tr}{J} $$

Combine these equations with the following

 * $$ \sigma_{yx} = \frac{\partial \phi}{\partial z} = -G \theta z $$

 * $$ \sigma_{zx} = -\frac{\partial \phi}{\partial y} = G \theta y $$

to obtain

 * $$ u_x\left(y,z\right) = 0 $$

This implies that there is no axial displacement at any point along a bar with uniform, solid, circular cross-section. This was assumed to be true during ad hoc derivations of twist; however, we see now that the Theory of Elasticity proves it to be so. A rigorous proof is presented below.

 Error: See my comments in Team VQCrew (with annotation). Eml4500.f08 14:04, 23 November 2008 (UTC)

= Traction Force Equation: Derivation =

Geometric Manipulation Using an Infinitesimal Element
Goal: Using the traction force equation to derive the equation:

$$\displaystyle[t_i]_{3x1}=[\sigma_{ij}]_{3x3}[n_i]_{3x1} $$

Consider the following figure below of an infinitesimal element to be able to be able to derive geometrically, the traction force from the stress tensor and normal vector at the surface of the element.

From the figure the vector $$\boldsymbol\vec{n}$$ has the vector equation: $$\displaystyle \boldsymbol \vec{n}=n_y \boldsymbol \vec{j}+n_z \boldsymbol \vec{k}$$

From this equation, $$\boldsymbol\vec{n}$$ is the unit normal vector and is given as $$\displaystyle ||\boldsymbol \vec{n}||=1$$

Looking at the figure, the elemental geometry helps in the establishing the following relationship between dy and dz:

$$\displaystyle dy=ds*sin(\theta)$$ $$\displaystyle dz=ds*cos(\theta)$$

Additionally, the normal vector equation can be used to rewrite the vector in terms of angle $$\theta$$ and conclude also that:

$$\displaystyle n_y=cos(\theta)$$ $$\displaystyle n_z=sin(\theta)$$

Looking at Figure and continuing through this derivation, the next logical choice would be to take the sum the forces in the y and z directions using the equilibrium equations. It is important to note that the depth into the page is taken as the x-direction and is neglected for this case since the depth is considered to be unity, or 1.

As given in class, the derivation for the equilibrium in the y direction of the sum of forces is given as:

$$\sum {F}_y=0=-\sigma_{yy}*(dz*1)-\sigma_{yz}*(dy*1) +t_y*(d_s*1)$$

Substituting values for dy and dz into the equilibrium equation above, the following equation is produced:

$$\sum {F}_y=0=-\sigma_{yy}ds n_y-\sigma_{yz}ds n_z +t_yd_s$$

Further simplification of the equilibrium equation above and solving for $$\displaystyle t_y : $$

$$\displaystyle t_y= \sigma_{yy}n_y+\sigma_{yz}n_z(1)$$

As part of HW 6, the equilibrium equation for the sum of forces in the z-direction are given below:

Using equations (1) and (2) for tractional force components, we can arrange the equations in the form of a matrix, reflected below to produce equation 3:

$$\begin{Bmatrix} t_y\\ t_z\\ \end{Bmatrix} = \begin{bmatrix} \sigma_{yy} & \sigma_{yz}\\ \sigma_{zy} & \sigma_{zz}\\ \end{bmatrix} \begin{Bmatrix} n_y\\ n_z\\ \end{Bmatrix}(3) $$ Similarly for a general 3D case, a similar matrix may be constructed using indicial notation, shown below:

$$\begin{Bmatrix} t_1\\ t_2\\ t_3\\ \end{Bmatrix} = \begin{bmatrix} \sigma_{11} & \sigma_{12}& \sigma_{13}\\ \sigma_{21} & \sigma_{22}& \sigma_{23}\\ \sigma_{31} & \sigma_{32}& \sigma_{33}\\ \end{bmatrix} \begin{Bmatrix} n_1\\ n_2\\ n_3\\ \end{Bmatrix}(4) $$

An advantage of using this indicial notation allows us to rewrite the entire matrix equation in the form of a single equation that is more compact. This compact notation is shown below:

$$t_i=\sum_{j=1}^3{\sigma_{ij}n_j},i=1,2,3$$

In this equation looking at the stress value, where the subscript i, and j designates the row, and column of the matrix respectively.

Generalizing this formula, produces our end goal and solves for the desired equation: $$\displaystyle [t_i]_{3x1}=[\sigma_{ij}]_{3x3}[n_i]_{3x1} $$

Roadmap Incorporation
Looking at the roadmap on lecture page 16-2, we recall the following three equations:

$$\displaystyle T=2\int\int_{A}^{}{}\phi dA(5)$$ $$\displaystyle T = GJ\theta~(6) $$ $$\displaystyle J=\frac{-4}{\nabla^{2}\phi }\int \int _{A}\phi dA(7)$$

These three equations can be combined, and the equation (8) shown below can be derived by solving the three equations shown above for the Lagrangian of phi.

$$\frac{\partial^2{\phi}}{\partial{y}^2}+\frac{\partial^2{\phi}}{\partial{z}^2}=-2G\theta(8)$$

This becomes important because $$\phi$$ is taken to be constant on the lateral surface of the bar.

= Flexural Shear in Thin-walled Sections =

Motivation
It is known that for a generic loading condition of a beam, there exists not only a bending moment but also a transverse shear load. Consider the beam in the figure below. Some generic loading condition is applied to the straight beam on the left. A cut can be made in the beam at a distance x from the wall. The bending moment and shear loads are drawn for each side of the exposed section, where M(x) represents the moment as a function of distance x and V(x) represents the shear load as a function of distance x.

= Unsymmetrical Thin-walled Cross Section =

General Equation for unsymmetrical cross sections $$\displaystyle \int_{A_s} \frac{\sigma_{xx}}{dx} dA = -q_s $$ Eq. 5.1

Symmetric Cross Section (about y- axis)
$$ \displaystyle \sigma_{xx} = \frac{M_y z}{I_y}$$

$$ \displaystyle q(s) = \frac{-V_z Q_y}{I_y}$$ Eq. 5.2

$$ \displaystyle Q_y = \int_{A_s} z dA = z_c A_s$$

Unsymmetric Cross Section
$$ \displaystyle \sigma_{xx} = (k_y M_z - k_{yz} M_y)y + (k_z M_y - k_{yz} M_z)z$$ → Eq. 5.3

This is true for the following k values from p.26-2 of notes: $$ \displaystyle k_y = \frac{I_y}{D}$$ $$k_{yz} = \frac{I_{yz}}{D}$$ $$k_z = \frac{I_z}{D}$$

The expression for $$\displaystyle \sigma_{xx}$$ can also be put into matrix form:

$$\displaystyle \sigma_{xx} = \begin{bmatrix} z & y\\ \end{bmatrix} \begin{bmatrix} k_z & -k_{yz}\\ -k_{yz} & k_y\\ \end{bmatrix} \begin{Bmatrix} M_y\\ M_z\\ \end{Bmatrix} $$

$$ \displaystyle q(s) = -(k_y V_y - k_{yz} V_z) Q_z - (k_z V_z - k_{yz} V_y) Q_y$$ → Eq. 5.5

$$ \displaystyle => Q_z = \int_{A_s} y dA$$ and $$ \displaystyle Q_y = \int_{A_s} z dA $$

= Matlab Analysis =

The main purpose of this weeks homework problem is to examine the bending stress in the skin of the NACA airfoil. In order to determine this, we will use exact approximations as well as approximations for simply supported and clamped boundary conditions since the airfoil's spars and stringers cause the skin to fall somewhere in between the two.

The first part involves using the previous homework's inertia matrix, centroid, and applied moments to determine the normal bending stress in the skin between the two spars as a function of the y-coordinate using the equations:

 * $$\sigma_{xx}=E\epsilon_{xx}=\frac{I_yM_z-I_{yz}M_y}{I_yI_z-(I_{yz})^2}y+\frac{I_zM_y-I_{yz}M_z}{I_yI_z-(I_{yz})^2}z$$

\begin{bmatrix} I_{22} \\ I_{33} \\ I_{23} \\ \end{bmatrix}=

\begin{bmatrix} 1.298\ x\ 10^{-5}\ m^4 \\ 6.346\ x\ 10^{-7}\ m^4 \\ -9.494\ x\ 10^{-8}\ m^4 \\ \end{bmatrix}$$

 * $$Centroid(y,z)=(0.210\ m,\ 0.210\ m)$$

 The figure does not look right: The normal stress $$\displaystyle \sigma_{xx}$$ is expected to rise and fall with the coordinate $$\displaystyle z$$ of the airfoil skin, i.e., the function would look curved similar to the curve of the airfoil skin, instead of a straight line. See, e.g., Team VQCrew (with annotations) Eml4500.f08 11:36, 26 November 2008 (UTC)

From this we find that the average stresses are $$\sigma_{xx,FB}=26.5\ MPa$$ and $$\sigma_{xx,EH}=32.1\ MPa$$.

Next we wish to explore what shapes the skin would undergo due to these bending stress. For an aspect ratio of 1.5, we can see in Figures XXX and XXX that depending on m and n, different modes of buckling occur. The variables m and n determine the number of half wave lengths in the x and y direction respectively, which is readily apparent in the figures.

We are interested in finding the period T of this function that determines the shape of the buckled plate to prove the m is indeed the number of half waves. This involves finding the period of $$ \sin ( \frac{m \pi x}{a} )$$  by considering   $$\sin ( \frac{m \pi (x + T)}{a} ) = \sin ( \frac{m \pi x}{a} )$$. We know that:

 * $$\sin( \theta ) = \sin( \theta + 2\pi k)=\sin(\frac{m \pi x}{a}+\frac{m \pi T}{a})$$

 * $$Period=T=\frac{2\pi}{\frac{m\pi}{a}}=\frac{2a}{m}$$

For constant a=1, we see that the period is 2/m. For m=1, we see the period is 2 and therefore there is one half of the wave at a=1. For m=2, the period is 1 and two half waves at a=1, so on and so forth.

 I moved this figure to the left, so the figure does not block the link to show the collapsible table containing your matlab code, and added a few blank lines, so this collapsible table does not get pushed too far to the right. Eml4500.f08 11:36, 26 November 2008 (UTC)

In the determining of critical stresses, a variable of large importance is k_c. This number is a function purely of m, the number of half wave lengths, and the aspect ratio of the plate. The figure to the right shows how k_c varies with these variables. This factor is used in critical stress calculations for simply supported plates.

 I moved this figure to the left, so the figure does not block the link to show the collapsible table containing your matlab code, and added a few blank lines, so this collapsible table does not get pushed too far to the right. Eml4500.f08 11:36, 26 November 2008 (UTC)

As we discussed earlier, there are stringers and spars connect to the skin so it is not just simply supported. The other approximation for bending of a flat plate is clamped boundary condition. Here the two ends are not allowed to rotate, making the slope of plate zero at the boundary. The figure to the right shows how this sort of boundary condition affects the buckling mode of the plate.

The final section is interested in modeling the skin section in between the two spars as flat plates and determining the critical stresses in this plate. We will compare the value reached earlier for average stress to each of the two boundary conditions. In this figure, we vary the aspect ratio of plate while holding the width constant to that of the actual skin section. The two cases shown are for the clamped and simply supported boundaries mentioned earlier. We can see that at small aspect ratios, our average stress for the top plate, which the figure was based upon, falls in between the two curves. At higher aspect ratio's however (which is what we would expect for an aircraft wing), the stress is higher the than the critical stress, implying that the skin does not work well to resist bending, which is what we expected.

= Contributing team members =

 * Gonzalo Barcia--Eas4200c.f08.aero.barcia 03:05, 21 November 2008 (UTC)
 * William Kurth--Eas4200c.f08.aero.kurth 14:56, 21 November 2008 (UTC)
 * Melisa Gaar--Eas4200c.f08.aero.gaar 19:21, 21 November 2008 (UTC)
 * Thomas McGilvray--Eas4200c.f08.aero.mcgilvray 19:38, 21 November 2008 (UTC)
 * Ray Strods--Eas4300c.f08.aero.strods 19:19, 21 November 2008 (UTC)
 * Jared Lee--Eas4200c.f08.aero.lee 19:43, 21 November 2008 (UTC)
 * Oliver Oyama--Eas4300c.f08.aero.oyama 19:54, 21 November 2008 (UTC)

title: User:Eas4200c.fo8.aero.lee/HW6


title: File:HW6diag8.jpg


text: File:Travis 1.jpg

title: File:Travis 1.jpg


title: File:HW6diag9.jpg


text: File:Travis 2.jpg

title: File:Travis 2.jpg


text: File:Travis 3.jpg

title: File:Travis 3.jpg


text: File:Travis 4.jpg

title: File:Travis 4.jpg


text:
User:Eml4500.f08.lulz.strack/hw6

ElectricPylon Display.m
Command Window Output >> ElectricPylonSystem

d =

0           0            0            0  -0.00019023   0.00017465   -0.0001277 -1.4074e-006 -0.00016006  -0.0002967  -0.00012671   0.00032602  -0.00012798    -0.000128  -0.00012792  -0.00054327   0.00044932   0.00030901   0.00046381  -0.00061057     0.001017   0.00064583    0.0010325  -0.00019427    0.0010502  -0.00097027    0.0017167    0.0010338    0.0017162   0.00023259    0.0016833  -0.00064959     0.001683   -0.0013293    0.0023452   0.00064079    0.0021863   -0.0010297    0.0047122    0.0023902    0.0039576   -0.0024165    0.0063605    0.0057328    0.0063605    0.0037877    0.0063605    0.0029803    0.0063614     0.002384    0.0063657    0.0016463    0.0063711   0.00086194    0.0063182  -0.00024268    0.0062289   -0.0013182    0.0061363   -0.0023672    0.0060413   -0.0033231    0.0058839   -0.0051278    0.0055152    -0.012587    0.0067826    0.0042648     0.006996   -0.0063359     0.007012    0.0034574    0.0074006   -0.0043591    0.0071505    0.0029772    0.0071478    0.0020508    0.0071421    0.0012278     0.007182   0.00032026    0.0072816  -0.00073249    0.0074073   -0.0018633    0.0075643   -0.0033872    0.0090005    0.0029772     0.010455   -0.0033872

reactions =

149.64     -501.65      -149.64       1501.7

results =

1.0781e-005 2.1562e+006       862.48 -6.5356e-006 -1.3071e+006     -522.85 6.7339e-006 1.3468e+006       538.71 6.4556e-006 1.2911e+006       516.45 -3.5152e-006 -7.0304e+005     -281.22 -2.3798e-005 -4.7596e+006     -1903.8 -1.2325e-005 -2.465e+006      -986.01 7.1182e-006 1.4236e+006       569.46 1.9269e-005 3.8538e+006       1541.5 -2.8828e-005 -5.7655e+006     -2306.2 -1.8941e-007      -37882      -15.153 8.778e-009      1755.6      0.70224 1.9328e-005 3.8655e+006       1546.2 1.4698e-005 2.9395e+006       1175.8 -1.8494e-005 -3.6988e+006     -1479.5 -2.883e-005 -5.7661e+006     -2306.4 1.7948e-006 3.5896e+005       143.58 2.621e-005  5.242e+006       2096.8 9.8986e-006 1.9797e+006       791.89 -1.8874e-005 -3.7749e+006       -1510 -3.258e-005 -6.5159e+006     -2606.4 2.7892e-006 5.5784e+005       223.14 3.2194e-006 6.4388e+005       257.55 2.9828e-005 5.9656e+006       2386.3 -4.0852e-006 -8.1704e+005     -326.82 -1.9808e-005 -3.9616e+006     -1584.6 1.16e-005   2.32e+006       928.02 -4.0653e-006 -8.1307e+005     -325.23 -2.8778e-005 -5.7556e+006     -2302.2 -1.4237e-007      -28474       -11.39 -4.9677e-006 -9.9353e+005     -397.41 -7.6072e-008      -15214      -6.0857 2.9608e-005 5.9216e+006       2368.6 2.7907e-007       55813       22.325 -2.4358e-005 -4.8717e+006     -1948.7 6.9573e-006 1.3915e+006       556.59 -1.5116e-008     -3023.3      -1.2093 -2.8794e-005 -5.7588e+006     -2303.5 -2.427e-005 -4.854e+006      -1941.6 6.9522e-006 1.3904e+006       556.17 -9.4154e-007 -1.8831e+005     -75.323 1.129e-005  2.258e+006       903.22 -3.16e-006  -6.32e+005       -252.8 -2.0051e-005 -4.0102e+006     -1604.1 -1.0133e-019 -2.0266e-008 -8.1063e-012 7.3233e-019 1.4647e-007  5.8586e-011 3.2073e-007       64147       25.659 1.3501e-006 2.7001e+005       108.01 1.3575e-006 2.7149e+005        108.6 -1.0389e-005 -2.0778e+006     -831.13 -2.2136e-005 -4.4271e+006     -1770.9 -2.9388e-005 -5.8777e+006     -2351.1 -3.6774e-005 -7.3549e+006     -2941.9 -4.331e-005 -8.662e+006      -3464.8 -4.3478e-005 -8.6957e+006     -3478.3 0           0            0  3.0942e-019  6.1883e-008  2.4753e-011 -2.6985e-019 -5.397e-008 -2.1588e-011 3.0512e-019 6.1024e-008  2.4409e-011 -8.8522e-007 -1.7704e+005     -70.818 1.1034e-006 2.2069e+005       88.276 -9.5965e-007 -1.9193e+005     -76.772 9.9436e-007 1.9887e+005       79.549 1.0732e-005 2.1464e+006       858.56 -1.0415e-005 -2.083e+006      -833.19 1.1448e-005 2.2895e+006       915.81 -1.1321e-005 -2.2643e+006     -905.71 1.0517e-005 2.1035e+006       841.38 -1.0732e-005 -2.1464e+006     -858.56 7.0689e-006 1.4138e+006       565.51 -6.8859e-006 -1.3772e+006     -550.87 7.9177e-006 1.5835e+006       633.42 -1.8464e-005 -3.6928e+006     -1477.1 -4.8696e-007      -97393      -38.957 -1.1845e-007      -23690      -9.4762 1.5604e-007       31208       12.483 4.5239e-005 9.0479e+006       3619.2 4.6964e-019 9.3927e-008  3.7571e-011 4.5158e-005 9.0317e+006       3612.7 -7.9385e-019 -1.5877e-007 -6.3508e-011 4.5256e-005 9.0511e+006       3620.5 -6.5877e-007 -1.3175e+005     -52.702 -1.4822e-006 -2.9645e+005     -118.58 8.9829e-006 1.7966e+006       718.63 2.2438e-005 4.4876e+006         1795 3.3117e-005 6.6233e+006       2649.3 3.8878e-005 7.7755e+006       3110.2 -5.3704e-020 -1.0741e-008 -4.2963e-012 0           0            0 -4.8034e-020 -9.6069e-009 -3.8428e-012 0           0            0

max_compressive_stress =

-8.6957e+006

element_number =

55

max_tensile_stress =

9.0511e+006

element_number =

81

min_eigenvalues =

132.16      2468.4       2940.3

min_eigen_num =

82   85    86

title: User:Eml4500.f08.lulz.strack/hw6


text:
Wikiversity talk:Supporting Wikiversity participants

New user experiences
In order to answer this question at the page I encourage people to write about their experiences and start with my own. (Even if this might be not conform to some of the many policies)

I came here from a link of wikipedia, just after following the link to wikibooks. Bevor coming here, I did not have an account at Wikipedia and did not dare to edit articles. So I did neither plan to edit Wikiversity pages. Instead I wanted to find out what Wikiversity is and try learning resourcdes about Microbiology. So I followed the first links on the welcome page (Learning_projects and then Learning models, …). Soon, I was easily distracted by controversial and sometimes redundant discussions about what wikiversity is and what it is not, different views on learning, and policies. I did not really understand these topics at this point. I took me a while to come back to the welcome page and the Introduction guide, which ended again with rather specific pages (Colloquium,Portal:Education). Eventually, I followed links to Welcome,_newcomers, which answered better what Wikiversity is about.

I argue that we do not need to encourage users to make edits when welcoming them. I suggest to guide them to specific learning activities and aid them in starting editing at the point when they encounter this task in the learning activity. Starting with a specific activity is a wiki-way and much easier than explaining participation at welcome. As you see this written by me, I got excited about wikiversity and I got involved. This happend with following a course (School:WikiService), and with starting contribute to authoring a learning activity while learning the topic (Activity creation,Wikipedia_service-learning_courses). All three articles were just stubs, however, I could learn about wikiversity by starting creating. --tomaschwutz 22:10, 21 November 2008 (UTC)
 * "I suggest to guide them to specific learning activities" <-- At one time the main page did this, then learning activities were hidden away under several levels of portals designed by participants who rejected the idea that Wikiversity is for learn-by-doing projects. "we do not need to encourage users to make edits when welcoming them" <-- I do not agree. 99.99999% of Wikiversity content does not yet exist. It should be clear to new Wikiversity participants that Wikiversity is a place where learners are invited to edit and help create content, not just participate in existing learning projects. --JWSchmidt 14:45, 22 November 2008 (UTC)

 * I'd take the comments above about why the main page looks as it does with a grain of salt :-). It's true however that the best way to participate in WV is to jump in and edit... in some ways the learning process is greatly enhanced by adding your notes and/or viewpoint on the topic you're interested in, because that's how other participants will learn about your interests and in the best case you'll find people who share them and can study with you. Personally I find one of the best ways to see what's really happening is to look at the recent changes, where you'll be able to see what other people have been working on today. If something interests you, you can either watch the progress for a while or just jump right in. --SB_Johnny talk 15:41, 22 November 2008 (UTC)


text:
User:EML4500.f08.delta 6.ramirez/112108

Continued examination of dimensions for various variables.

$$ [\varepsilon ] = \frac{[d_4]}{[dx]} = \frac{L}{L} = 1 $$

Further,

$$ [\sigma] = [E] = \frac{F}{L^2} $$

$$ [A] = L^2 $$

$$ [I] = \int x^2dx = L^4 $$

$$ [\frac{EA}{L}] = [\tilde{k}_{11}]= \frac{L^2\frac{F}{L^2}}{L} = \frac{F}{L} $$

$$ [\tilde{k}_{11}\tilde{d}_1] = [\tilde{k}_{11}][\tilde{d}_1] = \frac{F}{L}L = F $$

$$ [\tilde{k}_{23}\tilde{d}_3] = [\tilde{k}_{23}][\tilde{d}_3] = \frac{[6][E][I]}{[L^2]} = \frac{1 \frac{F}{L^2} L^4}{L^2} 1 = F $$

Note: $$ [\tilde{d}_3] $$ = 1

Goal: Derive element force displacement relationship in global coordinate system from element force displacement relationship in local coordinates.

In elaboration, obtaining

$$ k^{(e)}_{6x6} d^{(e)}_{6x1} = f^{(e)}_{6x1} $$

and

from

$$ \tilde{k}^{(e)}_{6x6} \tilde{d}^{(e)}_{6x1} = \tilde{f}^{(e)}_{6x1} $$

In order to move on $$ \tilde{T}^{(e)}_{6x6} $$ must be analyzed, which has a block diagonal structure.

Note: It is easy to expand the $$ \tilde{T}^{(e)}_{6x6} $$ matrix to account for the frame problem since the rotational axis $$ z = \tilde{z} $$

Examining the relationship between $$ \tilde{d}_i $$ and $$ d_i $$ gives,

Note: $$ \tilde{d}_3, \tilde{d}_6, d_3, d_6 $$ represent rotational degrees of freedom.

Furthermore, we will now derive $$ \hat{k}^{(e)} $$ from PVW, focusing only on bending effect.

$$ \frac{\partial^2}{\partial x^2}((EI) \frac{\partial^2v}{\partial x^2}) - f_t(x) = m(x) \ddot{v} $$

Note: EI is a function of x, $$ f_t(x) $$ represents the distributive transverse load, and v represents the transverse displacement.

title: User:EML4500.f08.delta 6.ramirez/112108