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MIT 16 01 - Lecture Notes

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Page 1Lecture M2So what Engineering Science methods do we use to do Structural Engineering?We will use: Solid Mechanics (mainly Unified) and Structural MechanicsSolid Mechanics is a branch of Newtonian Mechanics dealing with behavior of solids.Mechanics: forces, masses and motion ---Review 8:01 in U lecturesNewton's Law:Fddtmv=() (F = ma)In Unified we will deal with structures in static equilibrium, ΣF = 0, but in general inaerospace engineering we often have to deal with structural dynamics (coupling ofstructural response and dynamics, particularly vibrations) and aeroelasticity (coupling ofaerodynamic loading with structural response, also particular with regard to vibrations)What is a solid?Page 2During this term we will deal with:"THREE (GREAT) PRINCIPLES OF SOLID MECHANICS"Consider a body acted on by a force, restrained by three springs(forces in spring = kδ) Hooke's Law3 Principles:1. Equilibrium: Forces must balance2. Compatibility of Displacements the springs remain connected -displacements at P must match.3. Constitutive Relations (Force - deflection)How much force is needed to cause a certain deflection (or vice versa).e.g. spring Fk=δWe will come back to all of these f ormally in the coming weeks.Page 3"All About Equilibrium"The concept of equilibrium is the first of the 3 great principles of solid mechanics we willconsider.For a structure, we will need to consider the general case, but let's build up to this byconsidering the simplest case:First, for formality, we will often "idealize" a relatively complicated force system assomething simpler. This relies on the concept of "equipollence".Equipollent Forces (equipollence means "equally powerful"),Definition: two forces are equipollent if they have the same total force and total momentabout the same (arbitrary) point.Example: Two force systems, A and B, viewed with respect to the origin Force F1 at a distance d from the origin Force F1 at origin and moment M A BFF FFMFd MFdABAB= ⇔ == ⇔ =1111NOT Equilibrium (∑ = 0) but equipollence (∑ = FM,)Generalize in 3-DMust have ∑∑F and Mii (external)The same for both cases.Page 4Examples of equipollence: Center of Gravity for a solid bodyaerodynamic forces acting on a wing.The actual distribution of pressure is given on the left. On the right we have reduced it toan equipollent set of forces. Either a lift force LW acting through the center of pressure,or a lift force, LW and a moment, MW acting about the aerodynamic center.Now we can move on to look at equilibrium itself, starting with the simplest case:Page 5Equilibrium of a ParticleA particle is a body whose mass is concentrated at a point.Forces acting on a particle produce acceleration.Newton's 2nd Law gives:∑ =()iiFMaIn many cases in structures we have:a = 0 (statics) - note airplane in steady flight a = 0, v = constantSo this reduces to (static) equilibrium∑ =()Fi0Which represents 3 equations - vector addition∑ =()ixiF 0 Sum of forces in x = 0∑ =()iyiF 0Sum of forces in y = 0∑ =()iziF 0Sum of forces in z = 0Page 6Equilibrium of a System of ParticlesLook at a system of particles (we'll consider 3) with forces Pi() acting on each particle i.Particles are connected by internal forces - exact nature of which does not matter at thistime.Isolate particle (1) - go back to equilibrium a particleSo for particle (1), draw a "free body diagram". (We'll discuss this more in laterlectures.). Replace internal forces with equipollent forces, RijRij() represents the reaction force in the connection between particles i and j.Draw reactions initially in tension (i..e pulling away from particle)Page 7Consider equilibrium of a connection between two particlesNewton's Law of action-reaction states:RR12 21() ()= −i.e. Point in opposite directions, but equal magnitude.or more generally:RRijji() ( )= −Returning to the equilibrium of particle (1) in our system, we can now apply equilibriumPRRma112 13110++= =(1)Similarly for particles 2 and 3PRR ma221 23220++ = =(2)PRR ma331 32330++ = = (3)Combining these expressions gives (1), (2) + (3) and substituting for RRijji() ( )= −gives:PP P R R R R R R12312 12 23 23 13 130+++ − + − + − =Hence:PP P1230++=Page 8Which gives us the key result that internal forces do not appear in overall equilibrium ofthe forces acting on the system.Is this enough to define the overall equilibrium of the system of particles?We have dealt with the requirement of no linear acceleration.So there is a second set of equilibrium requirements.Moment EquilibriumFor there to be no rotation, the moments about a (any) point must be zero.(Why "any"? Pure moment magnitude does not change as point changes.)Consider each particle and the moments acting on it. Take moments about some"convenient" point:Particle 1Page 9Moment due to a force: rF×Consider moment of forces acting on (1) about origin, OrPR R1112 130()()() ()×+ + =()M1Similarly for | and }(2) rPR R2221 230()()() ( )×++ =()M2(3) rPR R3331 320()()() ( )×++ =()M3Again, use Newton's Law of action-reaction for the internal forcesRRijji() ( )= −And we must further note that these pairs act along the same line of action, i.e., exert nonet momentMathematically:rR r R1133 310() ( ) ( ) ( )×+×=Combining M1, M2, M3∑×=()rPnn0(+ inertial terms)or∑=Mn0i.e. only the external forces affect the overall equilibrium of the system of


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MIT 16 01 - Lecture Notes

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