Materials and Manufacturing

1Chapter 2 Mechanical Properties

The objective for this chapter is to understand the following topics:

2.1 Introduction to mechanical properties

2.2 Stress-strain relationships

Tensile, compression, bending, shear

2.3 Hardness

Hardness vs. strength

Materials and Manufacturing (AER507), F. Xi

2.4 Effect of Temperature

2.5 Other properties: fatigue, impact and creep.

Textbook: Chapter 3; Reference 2: Chapter 6 and Chapter 8

2.1 Introduction to Mechanical Properties Mechanical properties are concerned about material behavior

when subject to mechanical stress (force), including

- strength, modulus of elasticity, ductility, hardness.

On one hand, design objective is to produce the products that can withstand high force without significant change in geometry and surface, meaning high strength, modulus and hardness.

On the other hand, manufacturing objective is to apply sufficient force so that the material can be cut or deformed to alter its


force so that the material can be cut or deformed to alter its shape. Usually, high strength materials are difficult and expensive to manufacture.

Hence, mechanical properties are an important factor for both design and manufacturing.

22.2 Stress-Strain Relationships Stress (unit area force) = Force / Area

Strain (unit deformation) = Deformation / Length

Stress-strain relationship indicates how much a material will deform under a given force independent of size

F F


Lo

L

Lo L

Stress and Strain Calculation (Engineering) stress-strain (theoretical)

Stress: Ve = F/Ao MPa (psi)

where F - applied force in N (Ib); Ao - original area in mm2 (in2).

1 psi (lb/in2) = 6895 Pa (N/m2)

Strain: e = (L Lo ) / Lo (dimensionless)

where Lo - original length in mm (in); L - length at any point.


Hookes Law: Ve = Ee (in elastic region)

where E - modulus of elasticity (Youngs modulus) in Pa (psi).

inherent material stiffness.

3Stress-Strain Curve

TS

slope = EV = Ee

V = KHn

TS

Y


The Elastic Limit (E.L.) is the limit of elastic deformation, below which the material will not be permanently deformed. Since it is Yield point

Material Strength

not be permanently deformed. Since it is difficult to determine this limit, the yield strength is used instead.

Yield strength (Y or YS) is defined as the stress at which a material deforms from the elastic region to the plastic region.

Y is determined as the stress at which a

Yield point

elasticplastic


Y is determined as the stress at which a 0.2% strain offset from the straight line has occurred. 0.2%

4Material Strength (Ultimate) Tensile Strength (TS) is defined as the maximum

stress. After this point, a localized elongation, known as necking, occurs.

TS = Fmax/Ao

Fracture Strength (FS) is the stress at fracture point.

FS = Fat fracture /Ao

Plastic region


Example 2-1: Tensile Testing

Given:

Lo=125 mm, Ao= 62.5 mm2 (lab measurement)

Test data: (lab record data)Test data: (lab record data)

Load (N) 0 17793 23042 27579 28913 27578 20462

Length (mm) 125 125.23 131.25 140.05 147.01 153.00 160.00

The maximum load is 28913, and the final load data is recorded

immediately prior to fracture.

Problem: It is required to determine the following


(a) Plot the stress strain curve

(b) Y (Yield strength)

(c) E (Youngs modulus)

(d) TS (Tensile strength)

5Tensile Testing Machine

Extensometer 2


Solution to Example 2-1

Plot using Excel. Y = 290 MPa, at 0.0038 after offset 0.2% E = 284.688 / 0.00184 = 154721Mpa

TS 462 608 MP FS? TS = 462.608 MPa, FS?

150200250300350400450500

Stre

ss M

Pa

load length stress Mpa strain0 125 0 0

17793 125.23 284.688 0.00184

23042 131.25 368.672 0.05

27579 140.05 441.264 0.1204


050

100150

0 0.05 0.1 0.15 0.2 0.25 0.3

Strain

S

Ve = F/Ao e = (L Lo ) / Lo

27579 140.05 441.264 0.1204

28913 147.01 462.608 0.17608

27578 153 441.248 0.224

20462 160 327.392 0.28

6Solution to Example 2-1

350

400

Y

0

50

100

150

200

250

300

Stre

ss M

Pa

tan-1 E

Y

290


00 0.001 0.002 0.003 0.004 0.005

Strain

Offset 0.2%

True Stress-Strain Relationships

True stress-strain (for manufacturing)

Stress: V = F/A MPa (psi)

where F - applied force in N (Ib); A - instantaneous area in mm2 (in2).

Strain: dH = dL /L

H = dL / L = ln L/Lowhere L - length at any point in mm (in), i.e. instantaneous length,

L = Lf at fracture.

L

Lo


f

L dL

7True Stress-Strain Relationships

In the elastic region, engineering and true stress-strain are identical, and E is applicable to both cases.

H k L E (i l i i )Hookes Law: V = E H (in elastic region)

In general, they are related as

H = ln (1+e) = ln L/LoProve: since e = (L-Lo) / Lo = L/Lo-1, then 1+ e = L/Lo

V = Ve(1+e) = F/A


Prove: (F/Ao)(1+ (L-Lo) / Lo) = F/A Ao/A 1 = (L Lo)/Lo = L/Lo-1

Since no volume change, then AL = AoLo

Ao/A = L/Lo Ao/A 1 = L/Lo-1

True Stress-Strain Curve - Example 2-1

400450500

500

600

)

050

100150200250300350

0 0.05 0.1 0.15 0.2 0.25 0.3

Strain

Stre

ss M

Pa

0

100

200

300

400

0 0.05 0.1 0.15 0.2 0.25 0.3

true strain %

true

stre

ss (M

pa


Strain true strain %

Ve = F/Ao; e = (L Lo ) / Lo V = F/A= Ve(1+e); H = ln L/LoY? TS? FS? E?

8Ductility

Elongation: (strain at fracture) TS

Ductility Measurement of flexibility and formability.

EL = [(Lf Lo)/Lo] x 100 %

where Lf - length at fracture.

Area reduction:

AR = [(Ao Af)/Ao] x 100 % | [1- Lo/Lf] x 100 % where Af - cross-section area at fracture.

Time (heat treatment)

EL


Metals Ceramics Polymers

EL % 10 60 0 1 500

AR % 20 90 0 -

Plastic Deformation Strain Hardening

In the plastic region after yielding, the stress-strain relationship (for metals) p ( )may be expressed as

V = K H n (flow curve)

where V - true stresstrue strain

Y

Y


H - true strainK - strength coefficient (Mpa, Kpsi)n - strain hardening exponent.

9Plastic Deformation Strain Hardening Strain hardening is to increase elastic region by permanent deformation, with new Y determined by flow curve, e.g. Aluminum alloy:e.g. Aluminum alloy:

TS 350(Mpa) > Y = 240(0.2)0.15=188 > Y 175 (MPa)Table 3.4 Typical values of strength coefficient K and strain hardening exponent n for selected material

K nMaterial Mpa lb/in^2Al, pure, annealed 175 25000 0.2Al alloy, annealed 240 35000 0.15


Al alloy, heat treated 400 60000 0.1Cu, pure, annealed 300 45000 0.5Cu alloy, brass 700 100000 0.35Steel, low C, annealed 500 75000 0.25Steel, high C, annealed 850 125000 0.15Steel alloy, annealed 700 100000 0.15Steel, stainless, austenitic annealed 1200 175000 0.4

True Stress-Strain Curves of Metals

The lower the strength, the longer the elongation.


10

Type of Stress-Strain Relationship

Perfect elastic:

No yielding, only

Elastic and perfect plastic:

Elastic and strain hardening:


fracture, Y=TS.

e.g. brittle materials such as ceramic, cast iron, and polymers

Plastic deformation at the same level, K=Y, n=0.

e.g. sufficiently heat treated metals

Plastic deformation with higher stress K>Y, n>0.

e.g. Most ductile metals.

Tensile Properties of Selected Materials

1000

1200 Brittle

0

200

400

600

800

aled

lloy

lloy

ium led

iron

mic lon

MPa

(E-G

Pa) Y

TSE%x10


Al All

oy an

neale

Al All

o

Stee

l allo

Titan

ium

Nicke

l ann

eale

Cast

iro

Ceram

Polym

er Ny

lo

Brittle materials and perfect plastic materials (polymers) fracture rather than yield.

11

Compression PropertiesIn the compression test, a material specimen is squeezed.

Stress: Ve = F/Ao Strain: e = (h ho ) /ho (negative)

where h - height at any point.

True stress: V = F/A

True strain: H = ln h/ho (negative)

Example: h = 0 9 h = 1

F

hho


Example: h 0.9, ho 1e = 0.9 1 = -0.1H = ln 0.9 = -0.105

Compared to tension: L=1.1, Lo=1e = 0.1, H = ln 1.1 = 0.095

Compression Properties For almost all materials, compression

properties are derived from tensile properties as the true stress-straincurves for both are nearly identicalcurves for both are nearly identical.

The difference is to ignore necking in compression, as materials will not fracture, but barreling.


12

Bending (flexural) Properties

Bending stress: Ve = Mc / Iwhere M bending moment, c distance from the center line, I moment of inertia of the cross section.

Here, M = F/2 x L/2 = FL/4, c = t/2, I = bt3/12

Ve = 1.5 FL/bt2

3 point method


F/2 F/2

M

Bending Properties

The bending test (flexure test) is used to determine the transverse rupture strength (TRS).

F t l tiFor rectangular cross section

TRS = 1.5 FL/bt2 MPa (psi)

where F force, N(lb), L length, mm(in), b and t are the dimensions of the cross section, mm(in).

For circular cross section

TRS = FL/SR3 Mpa (psi)


TRS = FL/SR Mpa (psi)

where I = SR4/4

For brittle materials, TRS |TS

13

Shear PropertiesWhen a material is subject to torsion by twisting, the shear stress occurs, which is defined as

W = F/A MPa (psi)

or W = T/2SR2t (F = T/R, A = 2SRt)

where F force N(lb), A area over which the force is applied mm2 (in2), T applied torque N-mm (lb-in).

LRDJ


Shear PropertiesShear strain is a measure of angular

deflection defined as

J = G/b (radians)

G

b

F

where G - deflection,

b width orthogonal to deflection.

or

J = RD/L

Example: R = 10, L=1, D = 0.01(rad)

RD

L

F


p , , ( )

J = RD/L = 10 x0.01 = 0.1

Note: D - radial directionJ - circumferential

L

14

Shear Properties - main cutting directionIn the elastic region

W = GJ Mpa (psi)

where G shear modulus in MPa (psi)where G shear modulus in MPa (psi).

In the plastic region,

W = KJn

(flow curve is similar to tensile)

For most materials

G = 0 4E or G = E/[2(1+v)]


G = 0.4E or G = E/[2(1+v)]

Shear strength S | 0.7TS

Necking usually does not happen in torsion.

Example: 2024-O: 0.7x27000 = 18900 psi 18000

Mechanical Properties Aluminum

Alloy Temper

Tensile strength TS, psi

Yield strength Y,

psiElongation

%Hardness HB

Shear strength S, psi

Fatigue limit, psi

2014 O 27000 14000 18 45 18000 13000T4,T451 62000 42000 20 105 38000 20000T6,T651 70000 60000 13 135 42000 18000

2017 O 26000 10000 22 45 18000 13000T4,T451 62000 40000 22 105 38000 18000

2024 O 27000 11000 20 47 18000 13000T3 70000 50000 18 120 41000 20000

T36 72000 57000 13 130 42000 18000


T36 72000 57000 13 130 42000 18000T4,T351 68000 47000 20 120 41000 20000

T6 69000 57000 10 125 41000 18000T81,T851 70000 65000 6 128 43000 18000

T86 75000 71000 6 135 45000 180002117 T4 43000 24000 27 70 28000 14000

Anh Nguyen

15

Example 2.2 Mechanical Properties: Tension

Problem: for given D and d of Al, find the required force for extrusion/drawing.

Solution:Chamber

Solution:

Since H = ln L/Lo = ln Ao/A

A = Sd2/4; Ao = SD2/4

Then H = 2 ln (D/d) = 2ln2 = 1.386

In plastic region,

D d

Ram

Shrinking tube

DieF, v


V = K H n = 240(1.386)0.15 = 252 Mpa

Drawing: V = F/A F = V Sd2/4

Extrusion: V = F/Ao F = VSD2/4

Example 2.3 Mechanical Properties:Shear

Problem: for given Al sheet metal of width (w) and thickness (t), determine the cutting force (t =1/4, and w = 6)

Solution:Solution:

Force = S x A = S x w x t = 18000 psi x x 6x12 = 324,000 lbf = 1.44 MN

(1 lbf = 4.4482N) (how to reduce force)


16

2.3 Hardness (friction, grinding)

Hardness is a measure of the material resistance to scratching and wear. It is proportional to TS.

For example, HB of ferrous materials can be approximately related to p , pp yTS as

TS | Kh HB

where Kh = 3.45, TS in MPa;

Kh = 500, TS in psi.

Example: 2024-O:


Estimated TS: 47x500 = 23500 psi

Actual TS: 27000 psi

Hardness Testing MethodsVarious hardness testing methods may be classified into:

1) Size of indentation

Brinell low to high hardness

Vickers (research too), Knoop

2) Depth of impression

Rockwell A K

3) Other


Scleroscope (rebound of a ball) (shore diamond dart dropped from a standard height not accurate, portable),

Mohs (scratch)

17

Hardness Testing Methods

softSD(D D2 d2 )SD(D D d )


H 60 HRHK 150 HRK

Brinell Test

Brinell Hardness number = HB

For harder materials over 500 HB, the cemented carbide ball is d i t d f th t l b llused instead of the steel ball.

Also, high loads (1500 and 3000 Kg) are typically used for harder materials.

It is considered good practice to indicate the load used in the test when reporting HB readings.


18

Rockwell Test

Rockwell = HRA HRK

Apply a minor load (10 Kg) first, then a major load (50 150).

Indentation HR = E - e = t

Different indenters with different loads for different materials.

Commonly used Rockwell scales

Rockwell A carbides, ceramics

B (non) ferrous metals (soft)

C ferrous metals tool steels


C ferrous metals, tool steels

E - softer

2.4 Effect of Temperature

Effect on hardness


Effect on strength

Anh Nguyen

19

2.5 Other Properties - FatigueFatigue failure occurs at the stress low than Y or TS of static loading, after subject to a cyclic loading for a number of cycles. Example: turbine blade failure (cracking).

Dynamic Y or TSStatic Y or TS

Fatigue strength


S-N curve (stress and Number)

strength

Fatigue life

Fatigue Test

1St, maximum stress = 2/3 (TS), the number of cycles to failure is recorded.

2nd, decreasing stress, the number of cycles to failure is recorded.

2/3(TS)

.


Time

20

S-N Behavior

Two distinct types of S-N behavior:

- Fatigue Limit (Endurance Limit): at certain N (number of cycles), the failure stress is no longer decreasingthe failure stress is no longer decreasing.

some ferrous materials, heat treated aluminum alloy, titanium alloys.

usually, fatigue limit | 25-60% TS- Fatigue Strength: defined at N (e.g. 107cycles)

in this case, no fatigue limit, and the failure stress will decrease as


N increases.

- Fatigue Life: defined as the number of cycles at a given stress level.

Impact - Toughness Impact testing is a good measure of material toughness by

applying a shock loading.

Generally materials with high strength and high ductility have highGenerally, materials with high strength and high ductility have high impact resistance. (super alloys, composites)

Charpy test (ft.lb) Izod (ft.lb)


w x t x L

21

Impact Testing

A pendulum is dropped or swung to the specimen.

Diff b t ChTemperature (oF)

Difference between Charpyand Izod test is the support of the specimen.

Measure in energy (force x distance)

Joule (J) = 1 Nm = 0.738 ft-lbf

Impact energy Shear

fracture


Joule (J) 1 Nm 0.738 ft lbf

Also indication of ductile-to-brittle transition.

Temperature (oC)

Properties of Low Alloy Steels for Making Landing Gears

T h

AISI No Treatment Y (psi) TS (psi) EL (%)Area reduction

Hardness HB

Toughess (Izod) (ft.lb)

4340 Normalized 125,000 185,500 12.2 36.3 363 11.7Annealed 68,500 108,000 22 49.9 217 37.7

8740 Normalized 88,000 134,750 16 47.9 269 13


Toughness in line with ductility, but conflict with strength and hardness.

Tough materials may be difficult to cut.

Annealed 60,250 100,750 22.2 46.4 201 29.5ASM databook

22

Creep Creep is the permanent elongation of a component under a static load (force or

heat) maintained for a period of time. Creep is time vs. static force, fatigue is cycle (time) vs. cyclic force. Tm n creep

resistance n Examples of creep failure include gas turbine blades, jet engine components,

rocket motors.rocket motors. Rupture lifetime tf,,

- long for long-life applications such as nuclear power plant components.- short for short-life applications such as turbine blades for military aircraft.

Strain hardening (transition)

Strain hardening & recovery (steady-state)

Necking (tension)


metal

Upon loading (elastic)

(transition)

Rupture lifetime tr

Residual StressResidual stress is caused by inhomogeneous deformation.

Residual stress remains after the material deforms and the force is removed.

Residual stress can be relieved by heat treatment. (paper clip)

elastic unloading

Tensile

Compressive

a b c


elastic

plastic

23

Assignment 1

Problem 1

In example 2-1, i) determine EL and AR; ii) estimate G, S; iii) Is it possible to determine K and n? If yes, how?

Problem 2

In example 2-2, If the diameter is reduced by D/d = 3, check if the material is broken?


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