Exploring Engineering

1. Exploring Engineering Chapter 11 Materials Engineering

2. Topics to be Covered • Stress • Strain • Elastic (Young’s) Modulus • Toughness • Yield Strength • Material Properties

3. Material Use Throughout Time

4. Stresses and Bugs

5. b) Fracture a) Normal state Fracture Along an Atomic Plane The force required to separate the atomic planes is about 100,000 MPa or 100 GPa (giga pascals where 1 Pa = 1N/m2). But typical metals have an fracture strength of only about 500 MPa!

6. The Grain Structure of Copper This is because metals are full of impurities and “grain boundaries.” The strongest materials today are made of single crystals. The largest terrestial single crystal is believed to be at the center of the Earth where an enormous single crystal of iron 1220 km in diameter may exist.

7. Tensile Strength of Atomic Chains and Nano-wires Copper atomic chains, nano-wires, and shells

8. Stress and Strain • Stress (σ) is defined as the force applied per unit area, σ = F/A, and has units of N/m2 or pascals (Pa). • The stress will cause a deformation called strain ε = ΔL/L where ΔL is the change in length and L is the original length. Strain is dimensionless, and is often given in % for convenience.

9. Tensile and shear Modulus • For a perfectly elastic material, the sample will obey Hooke’s law: E = σ/ε where E is the tensile elastic (or Young’s) modulus. • The shear modulus G = τ/γ(where τ is the shear stress and γ is the shear strain. • Distortion of the sample will occur as the cross-sectional area of the specimen deceases as its length increases. This is called Poisson’s ratio (ν), and relates the tensile modulus to the shear modulus as: • G = E/(2(1 + ν)) For most materials ν is between 0 and 0.5.

10.   “Toughness” is the Area Under the Stress-Strain curve Area = Toughness

11. 55 MPa  Plastic deformation Elastic deformation  30% Simplified Elastic - Plastic Deformation Model

12. More Accurate Graph

13. More More Accurate Graphs

14. Animal Muscle Stress-Strain

15. Silicone Elastomers Elastomers can have high elongation, but low-to-moderate tensile strengths

16. Biological Tissue Stress-Strain Materials that simulate the mechanical and acoustical properties of biological tissues can be used in experiments to assess blunt forces trauma. Applications include understanding automobile crash injuries, nonlethal projectiles, and the performance of blast-resistant structures.

17. Bone Stress-Strain Compressive Stress-Strain Curves

18. Composite Materials

19. Spring Bumper Product Design Influences the Choice of Materials

20. Bumper, Immoveable pole 1,000. kg at 2..5 mph Not all crashes are head on – the material may yield locally in low speed impacts

21. 0.10 m 0.10 m .10 m .10 m ? 0.0050 m After Before Chapter 11 – Problem

22. Problem Statement • A flat saucer made of a polymer (E=2.0 × 102) has an initial thickness of 0.0050 m. • A ceramic coffee cup of diameter 0.10 m and mass 0.15 kg is placed on a plate made of the same polymer as indicated above. What is the final thickness of the plate beneath the cup, assuming that the force of the cup acts directly downward, and is not spread horizontally by the saucer? Comment on your answer’s plausibility.

23. Solution • Need: Compressive strain on saucer made of polymer = ___ . • Know: Stress-strain relationship. Initial thickness of the saucer = 0.0050 m. Contact area calculate from diameter 0.10 m. Mass = 0.15 kg and E= 2.0 × 102 Note: Sometime E is not given in the problem and you have to find it from the graph. • How: 1) Hooke’s Law, ε = σ/E. 2) Given that the applied stress on the saucer acts over its footprint, σ = Mg/ACup. Solve: Area of contact of cup = πd2/4 = 7.85 × 10-3 m2; hence σ = force/area = -0.15 × 9.81/ 7.85 × 10-3 [kg][m/s2][1/m2] = -188 Pa (- since compressive.)

24. More Solution • The resulting strain is then ε = σ/E = -188/(2.0 × 102 × 106) [Pa][1/MPa][Mpa/Pa] = - 9.4 × 10-7 (dimensionless). • Since the original saucer thickness is 0.0050 m, its compression is εT = - 9.4 × 10-7 × 0.0050 = - 4.7 × 10-9 m or 4.7 nm. • The resulting thickness of the silicone ‘saucer’ under the cup is effectively unchanged (0.0050 - 4.7 × 10-9 = 0.0050 m). • Note that 4.7 nm is in the nano range and a continuum stress-strain model no longer even applies!

25. Shield Portion of shield affected by the micrometeorite is a cylinder with the same diameter as the micrometeorite Diameter = 1 micron Micrometeorite Thickness, T Chapter 11 – Problem

26. Yield stress 200 Plastic deformation Stress, ,MPa Yield strain 1.0 -1.0 Strain,  % -100 -200 A Simplified Stress-Strain Curve

27. Problem Statement • Consider a “micrometeorite” to be a piece of mineral that is approximately a sphere of diameter 1.×10-6 m and density 2.00 × 103 kg/m3. It travels through outer space at a speed of about 5.0 × 103 m/s relative to a spacecraft. Your job as an engineer is to provide a micrometeorite shield for the spacecraft. Assume that if the micrometeorite strikes the shield, it affects only a volume of the shield 1.0 × 10-6 m in diameter and extending through the entire thickness of the shield. • Using the stress-strain diagram for steel presented in this chapter, Figure 11, determine the minimum thickness a steel micrometeorite shield would have to be to protect the spacecraft from destruction (even though the shield itself might be dented, cracked or even destroyed in the process).

28. Solution Set-up • Need: Thickness of shield, T = ____ mm. • Know: KE of micrometeorite and volume of steel affected; also the steel with properties shown in Figure 11 is symmetric with respect to tension and compression. Its compressive yield is - 2.0 × 102 MPa at ε = - 0.15% and its fracture occurs at ε = - 1.0%. • Micrometeorite density 2.00 × 103 kg/m3 and a relative speed = 5.0 × 103 m/s. • How: Compare the shield’s toughness with the KE/volume material. If contact area is A, (½)(mV2/AT) = toughness gives T, where toughness is area to failure under σ, ε diagram.

29. Solution • Solve: Steel toughness = ½ × (-2.0 × 102) × (-0.0015) + (-2.0 × 102) × (-0.01 – (-0.0015)) [MPa] = 0.15 (elastic) + 1.7 (plastic) [MN/m2] = 1.85 MN/m2 = 1.85 × 106 N/m2 or1.85MPa. • Impact area of micrometeorite = π(D2/4) = π × (1.0 × 10-6)2/4 = 7.85 × 10-13 m2. • Mass3 of micrometeorite = ρ × (πD3/6) = 2.00 × 103 × π × (1.0 × 10-6)3/6 [kg/m3][m3] = 1.05 × 10-15 kg. • KE released = ½ mv2 = ½ × 1.05 × 10-15 × (5.0 × 103)2 [kg][m/s]2 = 1.31 × 10-8 Nm. • Toughness = (½)(mV2/AT) or 1.85 × 106 N/m2 = 1.31 × 10-8/(7.85 × 10-13 × T) [Nm][1/m3] and solving for the thickness gives T = 9.0 mm. • This implies a pretty hefty mass of steel to be orbited. Also much faster micrometeorites can be imagined than 500 m/s relative to the space craft.

30. Chapter 11 - Problem • Using the stress/strain properties for a polymer (Figure 13 and Table 1), determine whether a sheet of this polymer 0.10 m thick could serve as a micrometeorite shield, if this time the shield must survive a micrometeorite strike without being permanently dented or damaged. (This assumption would have to be carefully checked since this polymer is very deformable) • Assume the properties of the polymer are symmetric in tension and in compression.

31. Problem Set-up • Need: Polymer shield will survive undamaged ___ Yes/No? • Know: KE of micrometeorite = 1.31 × 10-8 J; impact area = 7.85 × 10-13 m and σ = 55 MPa and ε = 0.30 at yield. T = 0.10 m. • How: Compare the shield’s toughness with the KE/volume material.

32. Solution • Solve: Toughness at yield for polymer = ½ × 55 × 106 × 0.30 [N/m2] = 8.3 × 106 J/m3. • KE deposited/volume material = 1.31×10-8 /(7.85×10-13×0.10) [J][1/m3] = 1.67 × 105 J/m3 < 8.3 × 106 J/m3. • Hence 0.1 m of this polymer will survive micrometeorite unscathed.

33. Finite Element Analysis (FEA)Stress analysis

34. FEA Fluid Mechanics

35. FEA Heat Transfer

36. Finite Element Analysis (FEA) The use of FEA is widely accepted today, in almost any engineering discipline. The following is a list of possible applications • Static and dynamic analysis • Analysis of motion, fit, interference and function • Analysis of weight and centre of gravity of components and assemblies • Product life cycles • Troubleshooting of design flaws • Reverse engineering • Determination of manufacturing processes and sequences

37. Summary • Materials are best mechanically characterized by performing a stress – strain analysis • the data can be expressed in Hooke’s Law as E = σ/ε in which E is the elastic modulus, σ is stress and ε is the resulting strain. • For elastic materials E is a constant • The area under a σ – ε curve is called “toughness” and is a measure of how much energy can be absorbed and the material remain intact