03_materials - Properties/Mechanics of Materials UCSD:...

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Unformatted text preview: Properties/Mechanics of Materials UCSD: Physics 121; 2010 Why we need to know about materials • Stuff is made of stuff – what should your part be made of? – what does it have to do? – how thick should you make it • The properties we usually care about are: – – – – – – – – – – stiffness electrical conductivity thermal conductivity heat capacity coefficient of thermal expansion density hardness, damage potential machine-ability surface condition suitability for coating, plating, etc. Materials Properties Mechanics Winter 2010 2 Electrical Resistivity Electrical Resistivity • Expressed as ρ in Ω·m Expressed – resistance = ρ · L /A • where L is length and A is area UCSD: Physics 121; 2010 Thermal Conductivity Conductivity • Expressed as κ in W m-1 K -1 Expressed – power transmitted = κ ·A ·ΔT/t, UCSD: Physics 121; 2010 – conductivity is 1/ρ Material Silver Gold Copper Aluminum Stainless Steel ρ (× 10-6 Ω·m) 0.0147 0.0219 0.0382 0.047 0.06– 0.12 comments $$ $$$$ cheapest good conductor • where A is area, t is thickness, and Δ T is the temperature across the material Material Silver Copper Gold Aluminum Stainless Steel Glass, Concrete,Wood Many Plastics G-10 fiberglass Stagnant Air Styrofoam κ (W m-1 K-1) 422 391 295 205 10– 25 0.5–3 ~0.4 0.29 0.024 0.01– 0.03 comments room T metals feel cold great for pulling away heat why cookware uses S.S. buildings room T plastics feel warm strongest insulator choice but usually moving… can be better than air! 4 Winter 2010 3 Winter 2010 Lecture 3 1 Properties/Mechanics of Materials Specific Heat (heat capacity) • Expressed as c p in J kg-1 K-1 kg – energy stored = cp·m·ΔT • where m is mass and Δ T is the temperature change UCSD: Physics 121; 2010 Coefficient of Thermal Expansion • Expressed as α = δ L /L per degree K Expressed – length contraction = α ·Δ T·L , UCSD: Physics 121; 2010 • where ΔT is the temperature change, and L is length of material Material water alcohol (and most liquids) wood, air, aluminum, plastic brass, copper, steel platinum cp (J kg-1 K-1) 4184 2500 1000 400 130 comments powerhouse heat capacitor most things! Material Most Plastics Aluminum Copper Steel G-10 Fiberglass Wood Normal Glass Invar (Nickel/Iron alloy) Fused Silica Glass α ( × 10-6 K-1) ~100 24 20 15 9 5 3–5 1.5 0.6 comments best structural choice Winter 2010 5 Winter 2010 6 Density • Expressed as ρ = m/V in kg·m-3 Expressed in UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Stress and Strain • Everything is a spring! – nothing is infinitely rigid Material Platinum Gold Lead Copper, Brass, Steels Aluminum Alloys Glass G-10 Fiberglass Water Air at STP ρ (kg m-3) 21452 19320 11349 7500– 9200 2700– 2900 2600 1800 1000 1.3 comments tell this to Indiana Jones • You know Hooke’s Law: You Hooke Law: F = k·δ L – where k is the spring constant (N/m), δ L is length change – for a given material, k should be proportional to A/L – say k = E·A/L, where E is some elastic constant of the material glass and aluminum v. similar • Now divide by cross-sectional area F/A = σ = k ·δ L/A = E·ε σ = E·ε – where ε is δ L/L: the fractional change in length • This is the stress-strain law for materials – σ is the s tress, and has units of pressure – ε is the s train, and is u nitless 7 Winter 2010 8 Winter 2010 Lecture 3 2 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Stress and Strain, Illustrated • A bar of material, with a force F bar applied, will change its size by: δL/L = ε = σ /E = F/AE Elastic Modulus • Basically like a spring constant – for a hunk of material, k = E ( A/L), but E is the only part of this that is intrinsic to the material: the rest is geometry L F • Units are N/m2, or a pressure (Pascals) Material E (GPa) 350 190– 210 100– 120 70 50– 80 16 6– 15 2– 3 10 • Strain is a very useful number, F being dimensionless • Example: Standing on an aluminum rod: – – – – – – E = 70× 109 N·m−2 (Pa) say area is 1 cm2 = 0.0001 m2 say length is 1 m weight is 700 N σ = 7 × 106 N/m2 ε = 10− 4 → δL = 100 µm A δL σ = F/A ε = δ L/L Tungsten Steel Brass, Bronze, Copper Aluminum Glass G-10 fiberglass Wood most plastics σ = E·ε – compression is width of human hair Winter 2010 9 Winter 2010 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Bending Beams tension: stretched In the Moment • Since each mass/volume element is still, the net each force is zero – Each unit pulls on its neighbor with same force its neighbor pulls on it, and on down the line – Thus there is no net moment (torque) on a mass element, and thus on the whole beam • otherwise it would rotate: angular momentum would change neutral “ plane” compression • A bent beam has a stretched outer surface, a compressed inner s tretched surface, and a neutral surface somewhere between • If the neutral length is L, and neutral radius is R, then the strain If and at some distance, y , from the neutral surface is (R + y)/R − 1 at – ε = y/R – because arclength for same Δθ is proportional to radius – note L = RΔθ – But something is exerting the bending influence Bending Moments dV • So stress at y is σ = Ey/R So Ey/R Winter 2010 11 Winter 2010 And we call this “ something” the moment (balanced) 12 Lecture 3 3 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 What’s it take to bend it? • At each infinitesimal cross section in rod with coordinates (x, y) and area dA = dxdy: and dxdy – – – – – dF = σ dA = ( Ey/R) dA where y measures the distance from the neutral surface the moment (torque) at the cross section is just d M = y·dF so dM = Ey2dA/R integrating over cross section: Energy in the bent beam • We know the force on each volume element: – dF = σ ·dA = E·ε·dA = (Ey/R)dA • We know that the length changes by δL = εdz = σ·dz/E dz z-direction • So energy is: – dW = dF·δL = d F·ε·dz = E·ε·dA × ε·dz = E( y/R) 2dxdydz • Integrate this throughout volume – where we have defined the “moment of inertia” as • So W = M(L/R) ≈ Mθ ∝ θ2 So – where θ is the angle through which the beam is bent Winter 2010 13 Winter 2010 14 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Calculating beam deflection • We start by making a free-body diagram so that all forces and torques are balanced – otherwise the beam would fly/rotate off in some direction Tallying the forces/moments • Ax = 0; Ay = 21,000 lbs • Mext = (4)(4000) + (8)(3000) + (14)(2000) + (4)(4000) (8)(3000) (14)(2000) (11)(6)(2000) = 200,000 ft-lbs – In this case, the wall exerts forces and moments on the beam (though Ax=0) – This example has three point masses and one distributed load Winter 2010 15 – last term is integral: – where λ is the force per unit length (2000 lbs/ft) Winter 2010 16 Lecture 3 4 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 F y = m g = λL A Simpler Example z-axis F y = m g = λL What’s the deflection? z-axis Mext = λ<z>Δz = λ ( L /2)L = ½ λ L 2 Mext = λ<z>Δz = λ ( L /2)L = ½ λ L 2 force per unit length = λ; total f orce = m g = λL force per unit length = λ; total f orce = m g = λL • A cantilever beam under its own weight (or a uniform weight) c antilever – Fy and M ext have been defined above to establish force/moment balance – At any point, distance z along the beam, we can sum the moments about this point and find: • At any point, z, along the beam, the unsupported moment is given by: At along unsupported g iven • From before, we saw that moment and radius of curvature for the beam are related: – M = EI/R • – validating that we have no net moment about any point, and thus the beam will not spin up on its own! Winter 2010 17 And the radius of a curve is the reciprocal of the second derivative: r adius – d2y/dz2 = 1/R = M/EI – so for this beam, d2y/dz2 = M/EI = Winter 2010 18 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Calculating the curve • If we want to know the deflection, y, as a function of distance, z, along the beam, and have the second distance, derivative… • Integrate the second derivative twice: Bending Curve, Illustrated • Plastic ruler follows expected cantilever curve! – where C and D are constants of integration – at z =0, we define y=0, and note the slope is zero, so C and D are likewise zero – so, the beam follows: – with maximum deflection at end: Winter 2010 19 Winter 2010 20 Lecture 3 5 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 End-loaded cantilever beam beam Fy = F Mext = FL Simply-supported beam under own weight Fy = mg/2 = λL/2 Fy = mg/2 = λL/2 F force per unit length = λ; total f orce = m g = λL • Playing the same game as before (integrate moment from z to L): from – which integrates to: • This support cannot exert a moment – at z =0, y =0 → D = 0; at z=L/2, slope = 0 → C = −L3/12 – and at z=0, y =0 and slope=0 → C = D = 0, yielding: Winter 2010 21 Winter 2010 22 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Simply-supported beam with centered weight Fy = F/2 Fy = F/2 F Mext = FL/2 F S-flex beam “walls” are held vertical; beam flexes in “S” shape total M(z) = 2Mext − F z − F(L − z) = 0 for all z F Mext = FL/2 • Working only from 0 < z < L/2 (symmetric): Working • Playing the same game as before (integrate moment from z to L): from – which integrates to: – integrating twice, setting y (0) = 0, y ’ ( L /2) = 0: – and the max deflection (at z=L/2): – and at z=0, y =0 and slope=0 → C = D = 0, yielding: as it should be Winter 2010 23 Winter 2010 24 Lecture 3 6 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Cantilevered beam formulae Simply Supported beam formulae Winter 2010 25 Winter 2010 26 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Lessons to be learned • All deflections inversely proportional to E All – the stiffer the spring, the less it bends Getting a feel for the I-thingy Getting • The “moment of inertia,” or second moment came The into play in every calculation • All deflections inversely proportional to I All – cross-sectional geometry counts • All deflections proportional to applied force/weight – in linear regime: Hooke’ s l aw • All deflections proportional to length cubed – pay the price for going long! – beware that if beam under own weight, m g ∝ L also (so L 4) • Calculating this for a variety of simple cross sections: • Rectangular beam: b a • Numerical prefactors of maximum deflection for same Numerical of load/length were: – – – – 1/3 for end-loaded cantilever 1/8 for uniformly loaded cantilever 1/48 for center-loaded simple beam 5/384 ~ 1/77 for uniformly loaded simple beam – note the cube-power on b: t wice as thick (in the direction of bending) is 8-times better! – For fixed area, win by fraction b/a • Thus support at both ends helps: cantilevers suffer Winter 2010 27 Winter 2010 28 Lecture 3 7 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Moments Later Later • Circular beam – work in polar coordinates, with y = r sinθ And more moments • Circular tube, continued – if R 2 = R, R1 = R −t, for small t: I ≈ (A 2/4π)(R/t) – for same area, thinner wall stronger (until crumples/dents compromised integrity) a radius, R • Rectangular Tube – wall thickness = t – note that the area-squared fraction (1/4π) is very close to that for a square beam (1/12 when a = b) – so for the same area, a circular cross section performs almost as well as a square b • Circular tube – and if t is small compared to a & b: and for a square g eom.: inner radius R1, outer radius R2 or, outer radius R, thickness t – note that for a = b (square), side walls only contribute 1/4 of the total moment of inertia: best to have more mass at larger y-value: this is what makes the integral bigger! Winter 2010 29 Winter 2010 30 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 The final moment • The I-beam a Lessons on moments • Thickness in the direction of bending helps to the third power – always orient a 2×4 with the “ 4 ” side in the bending direction – we will ignore the minor contribution from the “web” connecting the two flanges b – note this is just the rectangular tube result without the side wall. If you want to put a web member in, it will add an extra b3t/12, roughly – in terms of area = 2at: • For their weight/area, tubes do better by putting material at high y-values material • I-beams maximize the moment for the same reason • For square geometries, equal material area, and a thickness 1/20 of width (where appropriate), we get: – – – – – square solid: I ≈ A 2/12 ≈ 0.083A 2 circular solid: I ≈ A 2/4π ≈ 0 .080A2 square tube: I ≈ 20A 2/24 ≈ 0.83A 2 circular tube: I ≈ 10A 2/4π ≈ 0.80A 2 I-beam: I ≈ 20A 2/8 ≈ 2.5A 2 • The I-beam puts as much material at high y-value as The as it can, where it maximally contributes to the beam stiffness – the web just serves to hold these flanges apart Winter 2010 31 10× better than solid form func. of assumed 1/20 ratio • I-beam wins hands-down Winter 2010 32 Lecture 3 8 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Beyond Elasticity • Materials remain elastic for a while – returning to exact previous shape Breaking Stuff • Once out of the elastic region, permanent damage results – thus one wants to stay below the yield stress – yield strain = yield stress / elastic modulus Material Tungsten* Steel Brass, Bronze, Copper Aluminum Glass* Wood most plastics* Yield Stress (MPa) 1400 280– 1600 60– 500 270– 500 70 30– 60 40– 80 Yield Strain 0.004 0.0015–0.0075 0.0005–0.0045 0.004–0.007 0.001 0.0025–0.005 0.01–0.04 34 • But ultimately plastic (permanent) deformation sets in – and without a great deal of e xtra effort * ultimate stress quoted (see next slide for reason) Winter 2010 33 Winter 2010 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Notes on Yield Stress • The entries in red in the previous table represent The red ultimate stress rather than yield stress – these are materials that are brittle, experiencing no plastic deformation, or plastics, which do not have a well-defined elastic-to-plastic transition dA F Shear Stress wall bolt γ τ = F/A, where A is bolt’s cross-sectional area hanging mass huge force, F • There is much variability depending on alloys – the yield stress for steels are • • • • • stainless: 280– 700 machine: 340– 700 high strength: 340–1000 tool: 520 spring: 400– 1600 (want these to be elastic as long as possible) • τ = Gγ – τ is the shear stress (N·m-2) = force over area = F/dA • dA is now the shear plane (see diagram) – G is the shear modulus (N·m-2) – γ is the angular deflection (radians) – aluminum alloys • 6061-T6: 270 (most commonly used in machine shops) • 7075-T6: 480 • The shear modulus is related to E, the elastic modulus The – E/G = 2(1+ν ) – ν is called Poisson’ s ratio, and is typically around 0.27– 0.33 35 Winter 2010 36 Winter 2010 Lecture 3 9 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Practical applications of stress/strain • Infrared spectrograph bending (flexure) – dewar whose inner shield is an aluminum tube 1/8 inch (3.2 mm) thick, 5 inch (127 mm) radius, and 1.5 m long – weight is 1 00 N ewtons – loaded with optics throughout, so assume (extra) weight i s 20 kg → 200 Newtons – If gravity loads sideways (when telescope is near horizon), what is maximum deflection, and what is maximum angle? – calculate I ≈ ( A 2/4π)(R/t) = 2× 10-5 m4 – E = 70×109 – ymax = mgL3/8EI = 90 µm deflection – y’max = m gL2/6EI = 80 µR angle Applications, continued • A stainless steel flexure to permit parallel displacement d • Now the effect of these can be assessed in connection with the optical performance – each flexing member has length L = 13 mm, width a = 25 mm, and bending thickness b = 2.5 mm, separated by d = 150 mm – how much range of motion do we have? – stress greatest on skin (max tension/compression) – Max strain is ε = σ y/E = 280 MPa / 200 GPa = 0.0014 – strain is y/R, so b/2R = 0.0014 → R = b /0.0028 = 0.9 m – θ = L/R = 0.013/0.9 = 0.014 radians (about a degree) – so max displacement is about d·θ = 2.1 mm – energy in bent member is E IL/R2 = 0.1 J per member → 0 .2 J total – W = F·d → F = (0.2 J)/(0.002 m) = 100 N (~ 20 lb) Winter 2010 38 Winter 2010 37 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Flexure Design • Sometimes you need a design capable of flexing a flexing certain amount without breaking, but want the thing to be as stiff as possible under this deflection – strategy: • work out deflection formula; • decide where maximum stress is (where moment, and therefore curvature, is greatest); • work out formula for maximum stress; • combine to get stress as function of displacement • invert to get geometry of beam as function of tolerable stress Flexure Design, cont. • Note that the ratio F/I appears in both the ymax and σ max formulae Note F/I (can therefore eliminate) where h is beam thickness • If I can tolerate some fraction of the yield stress can σmax = σy /Φ, where Φ is the safety factor (often 2) – example: end-loaded cantilever Δy is displacement from centerline (half-thickness) • so now we have the necessary (maximum) beam thickness that can tolerate a displacement y max without exceeding the safety can without factor, Φ factor, • You will need to go through a similar procedure to work out the thickness of a flexure that follows the S-bend type (prevalent in the Lab 2) L ab Winter 2010 39 Winter 2010 40 Lecture 3 10 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Notes on Bent Member Flexure Design Bent Kinematic Design Kinematic Design • Physicists care where things are – position and orientation of optics, detectors, etc. can really matter • When the flex members have moments at both ends, they curve into more-or-less an arc of constant radius, accomplishing angle θ • R = EI/M , and θ = L /R = ML/EI, where L is the length of the EI and ML flexing beam (not the whole assembly) beam • σmax = E εmax = EΔy /R = hθE /2L , so h = (σ y/Φ E ) ×(2L /θ) so – where h = 2Δ y and R = L/θ • Much of the effort in the machine shop boils down to holding things where they need to be – and often allowing controlled adjustment around the nominal position • Any rigid object has 6 degrees of freedom – three translational motions in 3-D space – three “ Euler” angles of rotation • take the earth: need to know two coordinates in sky to which polar axis points, plus one rotation angle (time dependent) around this axis to nail its orientation • Kinematic design seeks to provide minimal/critical Kinematic design constraint Winter 2010 41 Winter 2010 42 UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Basic Principles Principles • A three-legged stool will never rock – as opposed to 4-legged – each leg removes one degree of freedom, leaving 3 • can move in two dimensions on planar floor, and can rotate about vertical axis part with holes Diamond Pin Idea part with holes part with holes • A pin & hole constrain two translational degrees of freedom • A second pin constrains rotation – though best if it’s a diamond-shaped-pin, so that the device cut/grinding lines is not over-constrained dowel pin a diamond pin is a home-made modification to a dowel pin: sides are removed so that the pin effectively is a one-dim. constraint rather than 2-d Winter 2010 43 dowel pin two dowel pins wrong separation diamond pin thermal stress, machining error perfect (lucky) fit does not fit but over-constrained Winter 2010 constrains only rotation diamond pin must be ground on grinder from dowel pin: cannot buy 44 Lecture 3 11 Properties/Mechanics of Materials UCSD: Physics 121; 2010 UCSD: Physics 121; 2010 Kinematic Summary Kinematic Summary • Combining these techniques, a part that must be located precisely will: – sit on three legs or pads – be constrained within the plane by a dowel pin and a diamond pin References and Assignment • For more on mechanics: – Mechanics of Materials, by Gere and T imoshenko • For a boatload of stress/strain/defflection examples For examples worked out: – Roark’s Formulas for Stress and Strain • Reflective optics will often sit on three pads – when making the baseplate, can leave three bumps in appropriate places • only have to be 0.010 high or so • Reading from text: – Section 1.5; 1.5.1 & 1.5.5 – Section 1.6; 1.6.1, 1.6.5, 1.6.6 – use delrin-tipped (plastic) spring plungers to gently push mirror against pads Winter 2010 45 Winter 2010 46 Lecture 3 12 ...
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This note was uploaded on 11/09/2010 for the course AEROSPACE AE 1202 taught by Professor Dr.adib during the Spring '10 term at Sharif University of Technology.

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