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 machineability 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 m1 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 (× 106 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 G10 fiberglass Stagnant Air Styrofoam κ (W m1 K1) 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 kg1 K1 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 kg1 K1) 4184 2500 1000 400 130 comments powerhouse heat capacitor most things! Material Most Plastics Aluminum Copper Steel G10 Fiberglass Wood Normal Glass Invar (Nickel/Iron alloy) Fused Silica Glass α ( × 106 K1) ~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·m3 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 G10 Fiberglass Water Air at STP ρ (kg m3) 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 crosssectional 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 stressstrain 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 G10 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 zdirection • 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 freebody 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 ftlbs
– 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
zaxis F y = m g = λL What’s the deflection?
zaxis 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 Endloaded cantilever beam beam
Fy = F Mext = FL Simplysupported 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 Simplysupported beam with centered weight
Fy = F/2 Fy = F/2 F Mext = FL/2 F Sflex 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 Ithingy Getting
• The “moment of inertia,” or second moment came The into play in every calculation • All deflections inversely proportional to I All
– crosssectional 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 endloaded cantilever 1/8 for uniformly loaded cantilever 1/48 for centerloaded simple beam 5/384 ~ 1/77 for uniformly loaded simple beam – note the cubepower on b: t wice as thick (in the direction of bending) is 8times 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 areasquared 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 yvalue: 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 Ibeam
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 yvalues material • Ibeams 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 Ibeam: I ≈ 20A 2/8 ≈ 2.5A 2 • The Ibeam puts as much material at high yvalue 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 • Ibeam wins handsdown
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 welldefined elastictoplastic transition dA F Shear Stress
wall bolt
γ τ = F/A, where A is bolt’s crosssectional 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·m2) = force over area = F/dA
• dA is now the shear plane (see diagram) – G is the shear modulus (N·m2) – γ is the angular deflection (radians) – aluminum alloys
• 6061T6: 270 (most commonly used in machine shops) • 7075T6: 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× 105 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: endloaded cantilever
Δy is displacement from centerline (halfthickness) • 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 Sbend 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 moreorless 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 3D 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 threelegged stool will never rock
– as opposed to 4legged – 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 diamondshapedpin, so that the device cut/grinding lines is not overconstrained
dowel pin a diamond pin is a homemade modification to a dowel pin: sides are removed so that the pin effectively is a onedim. constraint rather than 2d
Winter 2010 43 dowel pin two dowel pins wrong separation diamond pin
thermal stress, machining error perfect (lucky) fit does not fit but overconstrained
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 delrintipped (plastic) spring plungers to gently push mirror against pads Winter 2010 45 Winter 2010 46 Lecture 3 12 ...
View
Full Document
 Spring '10
 Dr.Adib
 Force, Second moment of area

Click to edit the document details