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## Week 1

by: Eleanor Notetaker

86

0

9

# Week 1 coe 3001 I

Eleanor Notetaker
Georgia Tech
Deformable Bodies
Dr. Kennedy

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First, a review of ideas from COE 2001 (the prerequisite for this class). Next, we define some basic terms such as stress and strain and how to calculate them.
COURSE
Deformable Bodies
PROF.
Dr. Kennedy
TYPE
Class Notes
PAGES
9
WORDS
KARMA
25 ?

## Popular in Engineering and Tech

This 9 page Class Notes was uploaded by Eleanor Notetaker on Saturday August 22, 2015. The Class Notes belongs to coe 3001 I at Georgia Institute of Technology taught by Dr. Kennedy in Summer 2015. Since its upload, it has received 86 views. For similar materials see Deformable Bodies in Engineering and Tech at Georgia Institute of Technology.

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Date Created: 08/22/15
COE 3001 Mechanics of Deformable Bodies Week 1 Introduction Statics COE 2001 Objective Find forces that keep a system in equilibrium 0 Rigid bodies no deformation Deformable bodies COE 3001 0 Study the stresses and strains in structures and their displacements Objectives 0 Find stresses and strain 0 Find structural displacements Stress and strain in an axially loaded bar P p P P tensio compressio n n 0 Bar loaded along its length p ground Truss Real life example The Wheels of a plane are b constantly under compression 01 i Free Body Diagram FBD m P P lt gt gt L 6 lt gt lt gtlt gt 0 Bar deforms under load 3 3 0 SEaXial deformation Let s say we want to find the stress of a rod in equilibrium at some point in the rod not either of the edges h P P gt COE 3001 Mechanics of Deformable Bodies Week 1 COE 3001 Mechanics of Deformable Bodies Week 1 The first thing we will want to do is break up the bar where we would like to measure the pI39CS SUI39C NXEinterna force lt gt n I gt P N X N X Of b P What is NX We were previously told that the bar was in equilibrium That means that no ALL parts of the bar must therefore be in equilibrium Therefore when the bar is broken up there must be an internal force that is equal to the pressure given NXP Why does the chart show NX pointing right in the left diagram and left in the right diagram NX is essentially a force we created to explain what is happening in the broken up diagram Therefore when both pieces are put back together to reform the original rod any added forces will need to cancel out Therefore NX on both parts need to be equal and opposite 0 Enlarge cut 0 stress E stress ForceArea0 units Nm2 Pa or psi 0 Assuming that the force is constant over the surface the internal force equals area stress Sign convention O gt 0 tensiontensile stress 0 lt0 compressive stress Normal Strain P P a lt N gt p M p 4 p 5 L 52 L 54 6 E I E axial stress E E unitless is constant no matter how many times bar is broken up COE 3001 Mechanics of Deformable Bodies Week 1 strain typically written as or microstrain 1 is considered high amount of strain Mechanical Properties of Materials 0 Depends on the material 0 Measure experimentally 0 Tensile test K O39y S 7 4 m P l P PAo I I 85 L 39 The slope of the graph is equal to the material39s Young39s modulus Here we can use Hooke39s Law 0E When the curve ends stops becoming linear the first point it has hit Sp Proportionality limit The curve begins to move linearly until it reaches quotthe point of no returnquot or cry The material then starts yielding and when stress begins to increase again it is known as strain hardening When the material reaches the maximum stress possible the ultimate stress it begins necking until it reaches the final point on the curve and rupturesfails V Elastic response Inelastic response I o O L l V W1th the elast1c 5 response the stress stops before 0y letting the material return back to its normal state Meanwhile with the inelastic response the material is pushed past this point so it the energy cannot follow the same path to equilibrium Young39s modulus E has units of Paksi same as stress 0 E 70 X 109 Pa for Aluminum 70 GPa 0 E 200 X 109 Pa for standard steel 200 GPa Durability ability of a material to undergo large deformation before failure Brittle material does not deform significantly before failure Ex Aluminum alloy 0 yield point unclear define the yield point as my 2 offset E002 0 slope follows approximate slope at origin Poisson39s ratio 0 Materials mnerct laterally when subject to axial load 39I lt HP E J COE 3001 Mechanics of Deformable Bodies Week 1 O O 52 DO 62 0 lateral construction proportional to the axial strain E 5 L axial strain E39 50 D0 E lateral or transverse strain Poisson39s ratio For steel or aluminum 8 I D a 1 3 E 30000 ksi Find P 140 kib shortening of the tube D 3 lateral strain 839 d1 45 in Adz andAd2 lt 5 d2 6 in 83 d 2 2 A211 32 1 2123722 P 02 72 113216139 The stress is negative due to the fact that the force is acting down on the object causing it to be a compressive force So it is just as accurate to say the stress is quot1132 ksiquot or quot1132 ksi in compressionquot oEa gta 377 X106 377u Lateral straining I BIT gt8 ao1132H 5 5 a D gtia 39 gt6D1a d1509x103ez 0 l 50 2 23 39 gt6D2a39d2679 x103EZ Shortening a g gt 6 2 La L COE 3001 Mechanics of Deformable Bodies Week 1 Q N gt o UE gt O E 0 L E EA EA Hooke39s law from physics EA EA Pk5gt PT6 gtk Shear Stress and Strain 0 Normatresss t i irection that is perpendiclar to urface 0 Since the pin connection is in equilibrium we can quotcutquot the piece at any point and both parts should be in equilibrium 0 Since the top part needs an internal force pushing the object the left an equal and opposite internal force must be pushing the bottom part to the right 0 average sheer stress is obtained by diViding sheer force by area V T X T sheer stress Shear stresses on perpendicular planes 0 For equlibrium Z Forces in Xdir leC T3bCZO 0 outof lane dimension E c Z Moments TibCa T23Cb 0 T1 T2 COE 3001 Mechanics of Deformable Bodies Week 1 The forces on the left and bottom of the box cancel out due to the fact that they are acting on the original so are multiplied by O Shear stress 0 sheer stress y also called the engineering shear stress 39I 0 length remains the same 0 area remains the same Sign convention Pos1 1ve s eer stress Neg pos1 am You H e sign of sheer stress by determining the forces39 direction respective the center of the shape Hooke39s Law for shear 0 For material below the proportional limit T GV G E shear modulus G has units of forcearea E 0 For many types of materials 2 1 0 ShCCl bu am CAGLIIHLC A Z X 2 E Hooke39s law for strain T GV T F G abG COE 3001 Mechanics of Deformable Bodies Week 1 d hF tany h d htany hy abG tany y 17 Allowable stresses and factors of safety Strength of a structure 0 ability to carry loads Without damage Design Given a set of loads determine the geometrysize of structure so that stressesdisplacement do not exceed some values stressdisplacement allowables actual strength Factorso sa e E f f ty required strength

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