Exploring Engineering

Chapter 11

Materials Engineering

Topics to be Covered

Stress

Strain

Elastic (Young’s) Modulus

Toughness

Yield Strength

Material Properties

Material Use Throughout Time

Stresses and Bugs

Fracture 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 where 1 Pa = 1N/m2).

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

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.

Tensile Strength of Atomic Chains and Nano-wires

Copper atomic chains, nano-wires, and shells

Stress 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 = 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 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.

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

Area = Toughness

Simplified Elastic - Plastic Deformation Model

30%

55 MPa

Elasticdeformation

Plasticdeformation

More Accurate Graph

More More Accurate Graphs

Animal Muscle Stress-Strain

Silicone Elastomers

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

Biological 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-Strain

Compressive Stress-Strain Curves

Composite Materials

Product Design Influences the Choice of Materials

Spring

Bum

per

Not all crashes are head on – the material may yield locally in low speed

impacts

1,000. kgat 2..5 mph

Bumper,

Immoveablepole

Chapter 11 – Problem

0.0050 m

.10 m

Before

0.10 m

.10 m

?

After

0.10 m

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.

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.)

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!

Chapter 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 Curve

1.0

Str

ess,

,

MP

a

-100

200

Strain, %

-200

-1.0

Yieldstress

Yie

ldst

rain

Plasticdeformation

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).

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.

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 or 1.85MPa. 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.

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.

Problem 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.

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.

Finite Element Analysis (FEA)Stress analysis

FEA Fluid Mechanics

FEA Heat Transfer

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

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