University:
Massachusetts Institute of Technology
Course:
3.054 | Cellular Solids: Structure, Properties and Applications
Academic year:

2024
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365

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28
Author:

Bletleduswa3

Lecture 15, Energy Absorption Notes, 3.054 Energy absorption in foams • Impact protection must absorb the kinetic energy of the impact while keeping the peak stress below the threshold that causes injury or damage • Direction of the impact may not be predictable • Impact protection must itself be light e.g. helmet • Capacity to undergo large deformation ( ∼ 0.8, 0.9) at constant σ • Absorb large energies with little increase in peak stress • Foams — roughly isotropic — can absorb energy from any direction - light and cheap • For a given peak stress, foam will always absorb more energy than solid it is made from • Strain rates: Instron typically ˙ ∼ 10−8 to 10−2 /s impact e.g. drop from height of 1 m, if thickness of foam=100mm vimpact = √ 2gh = 2(9.8)(1) = 4.4 m/s; ˙ 4.4m/sec 0.1 m = 44/s p • servo controlled Instrons, drop hammer tests — up to ˙ = 100/s blast: ˙ = 103 − 104 /s — inertial effects impt (we won’t consider this) 1 Energy Absorption Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figure courtesy of Lorna Gibson and Cambridge University Press. 2 Energy absorption mechanisms • Elastomeric foams — elastic buckling of cells — elastic deformation recovered → rebound — also have damping - energy dissipated as heat • Plastic foams, brittle foams • Natural cellular materials — energy dissipated as plastic work or work of fracture — no rebound — may have fiber composite cell walls — dissipate energy by fiber pullout and fracture • Fluid within cells ◦ open cell foams — fluid flow dissipation only important if fluid is viscous, cells are small or rates are high ◦ closed cell foams — compression of cell fluid — energy recovered as unkading 3 Energy absorption diagrams • At stress plateau, energy W increases with little increase in peak stress, σp • As foam densifies, W ∼ constant and σp increases sharply • Ideally, want to be at “shoulder” point • More generally — see Figure • Test series of one type of foam of different ρ∗ /ρs at constant ˙ and temperature, T • Plot W/Es vs. σp /Es for each curve (Es at standard ˙ and T) • Heavy line joins the shoulder points for each curve • Mark ρ∗ /ρs for each foam on that line • Repeat for varying ˙ → join lines for constant ρ∗ /ρs • Build up family of optimum energy absorption curves • Can treat different temperatures, T, in same way 4 Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figures courtesy of Lorna Gibson and Cambridge University Press. 5 Notes: • Elastomeric foams can all be plotted on one curve since E ∗ ∝ Es and σel∗ ∝ Es (normalize W/Es and σp /Es ) • Figure: polyurethane and polyethylene ∗ • polymethacrylimid: σpl ⇒ typical of foams with plastic collapse stress with σys /Es = 1/30 • Can generate energy absorption diagrams from data, or use models for foam properties Modelling energy absorption diagrams Open cell elastomeric foams 6 Elastomeric Foams Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figure courtesy of Lorna Gibson and Cambridge University Press. 7 Flexible Polyurethane Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figures courtesy of Lorna Gibson and Cambridge University Press. 8 Polymethacrylimid Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figures courtesy of Lorna Gibson and Cambridge University Press. 9 (a) Linear elastic region < 0 1 σp2 W = 2 E∗ 1 σp2 2 W 1 = Es 2 Es (ρ∗ /ρs )2 (b) Stress plateau 0 < < D dW = σel∗ d W = 0.05 (ρ∗ /ρs )2 ( − 0 ) Es ◦ family of vertical lines on figure ◦ plateau ends at densification strain D ◦ then W/Es vs. σp /Es becomes horizontal (c) At end of stress plateau ∼ D ◦ maximum energy absorbed just before reach D (shoulder point) Wmax − = 0.05 (ρ∗ /ρs )2 (1 − 1.4 ρ∗ /ρs ) (assuming 0 << D and neglecting 0 ) Es ◦ optimum choice of foam is one with shoulder point that lies at σp = σD ◦ envelope of shoulder points, for optimum foams, at: ∗ σp = σD = 0.05 Es (ρ /ρs ) 2 ∗ ρ /ρs = 20 σ 1/2 p Es 10 Open-cell Elastomeric Foams: Modelling (VISCOUS DISSIPATION) Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figure courtesy of Lorna Gibson and Cambridge University Press. 11 Substituting into equation for Wmax /Es : 20 σ 1/2 i σp h Wmax p = 1 − 1.4 Es Es Es σ 1/2 i σp h Wmax p = 1 − 6.26 Es Es Es • Line of slope 1 at low stresses, falling to 7/8 at higher σ (d) Densification – when foam fully densified and compressed to a solid, then energy absorption curve joins that for the fully dense elastomer W 1 σp2 = Es 2 Es Note: • Model curves have same shape as expts. • Model shows W/Es depends on σp /Es and ρ∗ /ρs only — one diagram for all elastomer foams • For a given W/Es , σp /Es for the foam less than that of the fully dense solid, by a factor of 10−3 to 10−1 12 Closed-cell Elastomeric Foams: Modelling Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figure courtesy of Lorna Gibson and Cambridge University Press. 13 Modeling: open-cell foams that yield • Analysis similar to elastomeric foams, with σpl replacing σel • Note that some closed cell foams that yield, face contribution to E ∗ σpl negligible • Neglect fluid contribution ∗ σpl = 0.3 σys (ρ∗ /ρs )3/2 1 σp 2 W 1 = (a) Linear elastic regime: same as elastomeric foam: Es 2 Es (ρ∗ /ρs )2 W σys ρ∗ 3/2 = 0.3 ( − 0 ) (b) Stress plateau: Es Es ρs Wmax σys ρ∗ 3/2 ≈ 0.3 (c) End of stress plateau: (1 − 1.4 ρ∗ /ρs ) Es Es ρ s • optimum choice of foam — absorbs maximum energy without σp rising sharply at D 14 Plastic Foams: Modelling Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figures courtesy of Lorna Gibson and Cambridge University Press. 15 • Curve of optimum energy absorption (heavy line on figure) is envelope that touches W − σp curve at shoulder points ρ∗ 3.3 σp 2/3 = • For σp , ρs σys σ 2/3 o σD n Wmax D = • Substituting in Wmax /Es equation: 1 − 3.1 Es Es σys • Model curves explain general features of experimental curves • Modeling — curves less general than for elastomers — this cure for a particular value of σys /Es = 1/100 (typical value for polymers) 16 Design and selection of foams for impact protection • Typically know object to be protected and some details about it mass, m contact area, A max drop height, h (or energy to be absorbed, u) max allowable acceleration, a (e.g. head injury - 100g) peak stress allowable, σp • Variables: foam material, density, thickness Example 1 Given: mass, m=0.5 kg contact area, A=0.01 m2 drop height, h=1 m max deceleration, a=10g foam: flexible polyurethane Es = 50 MPa Find: optimum foam density optimum foam thickness 17 Example 1: Find Foam Density and Thickness Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figures courtesy of Lorna Gibson and Cambridge University Press. 18 • Energy to be absorbed, u=mgh=(0.5 kg)(10 m/s 2 )(1m)=5 J • Maximum allowable force on package = F=ma=(0.5 kg)(10g)=50 N • Peak stress, σp =F/A=50 N / 0.01 m2 =5 kN/m2 • Normalized peak stress, σp /Es =5 kPa / 50 MPa = 10−4 • Draw vertical line on energy absorption diagram at σp /Es = 10−4 √ • Need to know ˙ ≈ v/t velocity v = 2gh = 4.5 m/s • Iterative approach — choose arbitrary thickness, t • e.g. t1 = 1 m ˙ = 4.5/s W/Es = 5.25 × 10−5 (u=WAt1 ) W = 2620 J/m3 t2 = WA/u = 0.19 m ˙2 = 24/s W/Es = 6.6 × 10−5 W = 3300 J/m3 t1 = 0.001 m ˙ = 4.5 × 103 /s W/Es = 7.4 × 10−5 W = 3700 J/m3 t2 = 0.14 m ˙2 = 32/s W/Es = 6.7 × 10−5 W = 3350 J/m3 19 Third iteration: t3 =0.15 m (both W) Optimum density (Fig) ρ∗ /ρs ∼ 0.01 Note: t converges quickly even from very different initial guesses for t Example 2 Given m A t h a = = = = = 2.5 kg 0.025 m2 20 mm 1m 100 g Find foam material foam density Calculate W, σp , ˙ mgh (2.5 kg)(10 m/s2 )(1 m) = = 5 × 10−4 J/m3 W = 2 At 0.025 m (0.02 m) ma (2.5 kg)(100)(10 m/s2 ) Fmax = = 105 N/m2 σp = = 2 A 0.025 m A q √ 2(10 m/s2 )(1 m) 4.5 m/s v 2gh ˙ = = = = 225 /s = 0.02 m 0.02 m t t Select arbitrary value of Es =100 MPa Plot W/Es = 5 × 10−4 σp /Es = 10−3 point A 20 Example 2: Find Foam Material and Density Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figures courtesy of Lorna Gibson and Cambridge University Press. 21 • Construct a line of slope 1 through this point (broken line) • Moving along this line simply changes Es • Select the point where the broken line intersects the appropriate ˙ ∼ 102 /s (point B) • Read off values of W/Es = 1.8 × 10−3 σp /Es = 37 × 10−3 • Resulting value of Es = 28 MPa ⇒ low modulus, flexible polyurethane • Replotting on more detailed figure: ρ∗ /ρs = 0.1 • If point A above all energy contours and lone of slope 1 does not intersect them, specification cannot be achieved, A or t has to increase • If point A below all contours, then A and t larger than need to be — can be reduced 22 Case study: design of car head rest • Head rest should absorb kinetic energy of head while keeping force less than that which would cause injury • Example in book: mass of head = 25 kg max. deceleration = a = 50 g = 500 m/s2 area of contact, A = 0.01 m2 thickness of padding t = 0.17 m max. allowable force F = ma = 1250 N max. allowable stress σp = F/A = 125 kN/m2 2 energy to be absorbed/vol, W = 1/2Atmv = 735 v2 J/m3 peak strain rate ˙=v/t [s−1 ] current material — flexible polyester foam ρ∗ /ρs = 0.06 from plot: for σp = 125 kN/m2 W 103 J/m3 q= 5 ×q maximum collision velocity = v = W 735 = 5×103 735 23 = 2.6 m/s = 5.8 mph Car Head Rest Design Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figures courtesy of Lorna Gibson and Cambridge University Press. 24 Alternative design #1 • Consider energy absorption diagram for elastomeric foams • Add sales for polyester (using Es =15 MPa) • For σp = 125 kN/m2 could use polyester foam ρ∗ /ρs =0.2 then W/Es = 2.6 × 10−3 and v = 7.3 m/s =16 mph Alternative design #2 • Use different material e.g. low density open cell polyethylene Es = 200 MPa • σp /Es = 0.125 200 = 6.3 × 10−4 • At ˙ = v/t ≈ 100/s (estimated) W/Es = 3.2 × 10−4 (from fig) W = (3.2 × 10−4 )(200M P a) = 6.4 × 104 J/m4 q q 4 W v = 735 = 6.4×10 = 9.3 m/s = 21 mph 735 • Reading from figure 1: ρ∗ /ρs = 0.03 25 Case study: foams for bicycle helmets US: 600-700 bicycle deaths/year > 90% not wearing a helmet ∼ 50,000 cyclists injured (2009) (US Nat. Hwy Traffic Safety Admin Bicycle Helmet Safety Inst.) • Helmets consist of solid outer shell and foam liner (e.g. expanded PS) • Liner thickness typically 20 mm • Wish to absorb as much energy as possible while keeping peak acceleration less than that to cause head injury • Foam liner ◦ Redistributes load over larger area, reducing stress on head ◦ Peak stress on head limited by plateau stress of foam (as long as don’t reach densification) ◦ Max. tolerable acceleration = 300 g (if for a few milliseconds) ◦ Mass of head ≈ 3 kg Fmax = ma = (3 kg)(300)(10 m/s2 )=9 kN ◦ As foam crushes, it distributes load over area ∼ A ∼ 0.01m2 (may be high) σp = 9kN = 0.9 MPa 0.01m2 26 Figure ⇒ EPS ρ∗ = 0.05 Mg/m3 absorbs W = 0.8 MJ / m3 • Diagram allows easy identification of possible candidate materials • More complete analysis can then be done • Energy absorbed U=0.8 × 106 J/m3 × 0.01 m2 × 0.02 m = 160 J (u=WAt) s r 2U 2(160)kg m2 2 = =10 m/s ≈ 22 mph • 1/2 mv = U ; vmax = m 3 kg s2 27 Case Study: Foams for Bicycle Helmets Gibson, L. J., and M. F. Ashby. Cellular Solids: Structure and Properties. 2nd ed. Cambridge University Press, © 1997. Figures courtesy of Lorna Gibson and Cambridge University Press. 28

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