chapter 11 - materials engineering (1)

Exploring EngineeringExploring Engineering

Chapter 11Chapter 11Materials EngineeringMaterials Engineering

Topics to be CoveredTopics to be Covered Stress

Strain

Elastic (Young’s) Modulus

Toughness

Yield Strength

Material Properties

Exploring Engineering

Material Use Throughout TimeMaterial Use Throughout Time


Stresses and BugsStresses and Bugs


Fracture Along an Atomic PlaneFracture Along an Atomic Plane

a) Normal state b) Fracture

The force required to separate the atomic planes is about 100,000 MPa or 100 GPa (giga pascals).

But typical metals have an fracture strength of only about 500 MPa!


The Grain Structure of CopperThe Grain Structure of CopperThis is because metals are full of impurities and “grain boundaries.”

The strongest materials today are made of single crystals. The largest single crystal is believed to be at the center of the earth where an enormous single crystal of iron 1220 km in diameter exists.


Tensile Strength of Atomic Tensile Strength of Atomic Chains and Nano-wiresChains and Nano-wires

Copper atomic chains, nano-wires, and shells


Stress and StrainStress and StrainStress (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,

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.


Tensile and shear ModulusTensile and shear ModulusFor 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.


““Toughness” is the Area Toughness” is the Area Under the Stress-Strain curveUnder the Stress-Strain curve

Area = Toughness


Simplified Elastic - Plastic Simplified Elastic - Plastic Deformation ModelDeformation Model

30%

55 MPa

Elasticdeformation

Plasticdeformation


A More Accurate GraphA More Accurate Graph


More Accurate GraphsMore Accurate Graphs


Animal Muscle Stress-StrainAnimal Muscle Stress-Strain


Silicone ElastomersSilicone Elastomers

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


Biological Tissue Stress-StrainBiological Tissue Stress-StrainMaterials 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.


Bone Stress-StrainBone Stress-Strain

Compressive Stress-Strain Curves


Composite MaterialsComposite Materials


Product Design Product Design Influences the Choice of Influences the Choice of

MaterialsMaterials

Spring Bum

per


Not all crashes are head on – Not all crashes are head on – the the

material may yield locally in low material may yield locally in low speed impactsspeed impacts

1,000. kgat 2..5 mph

Bumper,

Immoveablepole


Chapter 11 – ProblemChapter 11 – Problem

0.0050 m

.10 m

Before

0.10 m

.10 m

?

After

0.10 m


Problem StatementProblem StatementA 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.


SolutionSolutionNeed: Compressive strain on saucer made of polymer = ___ .

Know: Stress-strain relationship. Initial thickness of the saucer = 0.0050 m. Contact area from diameter 0.10 m. Mass = 0.15 kg and E= 2.0 × 102

How: Hooke’s Law, ε = σ/E. 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 (negative because it is a compressive stress)


More SolutionMore SolutionThe 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 applies.


Chapter 11 – ProblemChapter 11 – Problem

Portion of shield affected by the micrometeoriteis a cylinder with the samediameter as the micrometeorite

Thickness, T

Micrometeorite

Shield

Diameter = 1 micron


A Simplified Stress-Strain CurveA Simplified Stress-Strain Curve

1.0 S

tress

(M

pa)

-100

200

Strain, (%)

-200

-1.0

YieldStress Plastic

deformation


Problem StatementProblem StatementConsider 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).


Solution Set-upSolution Set-upNeed: 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.


SolutionSolutionSolve: 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.Impact area of micrometeorite = π(D2/4) = π(1.0×10-6)2/4 = 7.85×10-13 m2. Mass 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 T gives T = 9.0 mm.


Chapter 11 - ProblemChapter 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. Assume the properties of the polymer are symmetric in tension and in compression.


Problem Set-upProblem Set-upNeed: 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.


SolutionSolution

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.


Finite Element Analysis (FEA)Finite Element Analysis (FEA)Stress analysisStress analysis


FEA Fluid MechanicsFEA Fluid Mechanics


FEA Heat TransferFEA Heat Transfer


Finite Element Analysis (FEA)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 flawsReverse engineering Determination of manufacturing processes and sequences


SummarySummaryMaterials engineers develop and specify materials. They 1) Define material requirements; 2) Consider metals and polymers; 3) Understand the internal microstructure of materials that can be crystalline and/or amorphous; 4) Use a stress-strain diagram to express materials properties in terms of the five engineering variables: stress, strain, elastic limit, yield strength, and toughness; and 5) Use the results determined for those properties to carry out materials selection.


chapter 11 - materials engineering (1)

Documents