ch01-401.pdf

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ررقمل يلاتلا كنللا يف داوم رابتخا :

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و ـلا اذهل ررقملا باتكلا نع ينغت لا عبطلاب يه course

مسقلا لبِق نم و

باتك وه :

Materials Science and Engineering : An Introduction William D. Callister, Jr.

7th Edition, March 2006, ISBN: 978-0-471-73696-7, Wiley و وه باتكلا مدختسملا اساسأ

سيردتل ررقم ملع داوم لكشب ماع يف بلغأ تاعماج ملاعلا كلذ بجو هيونتلا نأ ـلا lecture notes ةقفرملا

ام يه

لاإ notes طقف و لصلأا وه باتكلا

، و عم كلذ ناكمإب بلاطلا ةدايزل تامولعملا نأ

اونيعتسي اضيأ

عجارملاب ةيلاتلا

عم ةاعارم اهنأ طقف ةدايزل تامولعملا

و لصلأا يف عجرملا يساسلأا قباسلا

: 1. The Science and Engineering of materials

Donald R. Askeland Chapman and Hall

2. Peter A. Thornton & Vito J. Clangelo Fundamentals of engineering materials Prentice-Hall

تاعاسلا يساردلُا لصفلا اذهل ةيبتكملا

يتلأا :

مويلا ةعاسلا نم

ةعاسلا يلإ

ءاثلاثلا 12:00

12:30

ءاعبرلاا 12:00

12:30

سيمخلا 12:00

12:30

يف ضعب عيباسلأا نكمي نأ ريغتت ديعاوم تاعاسلا ةيبتكملا ةجيتن فورظل

ةئراط وجرأف رذعلا كاذنيح و فوس نلعي يف كلت ءانثلأا نع ديعاوملا

ةديدجلا يلع باب ةفرغ بتكملا

عم صلاخ يتاينمت قيفوتلاب

نإ ءاش الله ,

Final Exam 50 Marks MidTerm Exam 30 Marks Lab Work

(8+6+3+3=15)

20 Marks

2

Total 100 Marks

uqu.edu.sa/tabenawy

TENSION TEST CHAPTER 1:

HARDNESS TEST & HARDENABILITY CHAPTER 2:

Tension Test to be discussed + Report IMPACT TEST

CHAPTER 3:

FATIGUE TEST CHAPTER 4:

Hardness and Impact Tests to be discussed + Reports MidTerm Exam & Fatigue Test Report

CREEP TEST CHAPTER 5:

Material Testing course

Tension Test

4

CHAPTER 1

Types of loading:

1. Axial loading (Normal): applied all along the axis line.

Tensile and compressive stress

2. Transverse loading: perpendicular to the longitudinal axis line.

Direct shear stress

3. Torsional loading: twisting action of material.

Torsional shear stress

Types of Loading

& Stresses

TENSION TEST

6

Tensile Testing Concept

The extension of the specimen length, Dl=(l-l

₀

) is measured against the applied loads, P . This is usually done using calibrated load cell and X-Y plotter or through a PC computer system as shown in the figure

In tension test, known loads

are applied to specimen of

standard dimensions. The

specimen has initial gauge

length of l

₀

and cross

section, A

₀

7

The P - D l curve can be used to calculate the terms of stress, s , and strain, e , instead of load, P, and specimen length extension D l.

The output data of the load, P, and the extension of the specimen length, Dl, are given in relationship as shown in the given figure.

Stress-Strain Curve

8

Accordingly, the stress, s, and strain,

e

^,

are calculated as the following:

Strain , e , is defined as the change in specimen length due to load, Dl=(l-l

₀

) divided by the initial length, l

₀

Stress , s, is defined as the load, P, divided by the initial area of the tested specimen, A

₀

The load (P)-extension (Dl) curve given previously, is recalculated according to the given

s

and

e

equations in order to determine another curve, which is called stress- strain curve (s- e)

• Where the stress is increased in linear relationship with strain

Elastic Zone :

• If the tested metal specimen is stressed, the specimen is

strained to the

corresponding strain and when the stress is removed the specimen return to its original dominions

• within elastic zone. As the volume of the

specimen, V , is constant, then:

V = A

₀

l

₀

= A

₁

l

₁

= A

₂

l

₂

= A

₃

l

₃

10

• where the stress is increased in exponential relationship

Plastic Deformation Zone:

• If the tested specimen is stressed, the specimen is strained to the corresponding strain and when the stress is removed the specimen dose not return to its original dominions, instead resulting permanent deformation is achieved which is called plastic deformation.

• Within this zone the volume constancy relation is also valid, then:

V= A

₀

l

₀

= A

₁

l

₁

= A

₂

l

₂

= A

₃

l

₃

= A

₄

l

₄

= A

₅

l

₅

• where the stress is increased in instability relationship with strain.

Necking Zone (Fracturing Zone):

• The deformation along the stress-strain curve from point 6 to point 8 is not homogenous dissimilar with the deformation before point 6, therefore, the volume constancy relation is not valid within this zone

• Within the plastic Fracturing Zone, the specimen tries to localize the deformation in particular zone, which produce neck shape.

the reduction in the cross-section area will not be able to withstanding the true stress and fracturing will be occurred at point 8.

• The applied load and the corresponding calculated stress regarding to original specimen cross-section area, A₀, would be reduce. Finally

12

13

Useful Parameters Relationships

1- Elastic limit (s

_el

):

Highest stress imposed on the material such that there is no permanent deformation remaining just before point 3.

The corresponding elastic strain is (e_el) also shown in the figure.

2- Yield Stress (s

_y

):

Yield stress is defined as stress

required to produce small specific amount of plastic deformation. It will be considered that the value of s_y is the stress calculated at point 3` in the figure.

However, in some cases where it is difficult to recognize the yielding zone, offset strain method of 0.002 (0.2%) is used to define

s

_y. The corresponding yield strain (

e

) is also shown in the figure.

Engineering Stresses:

14

3- Engineering Ultimate Tensile Stress (s

_u

):

The ultimate tensile stress is the measure of the maximum stress that the material can withstand under the condition of homogenous or stability deformation. The value of

s

_u ^is

the stress calculated at point 6 in the shown figure and its corresponding strain is

e

_u as shown in the figure.

4- Engineering

Fracturing stress (s

_F

):

The fracturing stress is the measure of the stress that the

material is fracturing. The value of s_F is the stress calculated at point 8 in the figure and its corresponding maximum strain is

e

_F as shown in the figure.

5-Modulus of Elasticity (E):

It is defined as the slope of the initial straight-line portion of the stress-strain curve. It can be determined graphically as the slope between point 0 and point 3. In some cases, it will be assumed that

s

_el

 s

_y ^and

e

_el 

e

_y, then the modulus of elasticity, E, is defined as the following:

It is important to note that in the previous equation, it is assumed that the end of elastic limit stress approximately equals yield stress

s

_y, and the ratio of modulus of elasticity given in the equation is only valid in elastic zone.

16

In fact modulus of elasticity is a measure of an important metallic property called Stiffness )ةبلاصلا(. It is clear that higher E resulting to lower elastic strain corresponding to the application of the given stress which mean higher stiffness.

This can be observed in the shown figure for different metallic materials namely:

steel; copper and aluminum.

The figure indicates lower elastic strain for steel compared to copper and aluminum for the same applied stress. Also it is clear from the figure that

E

^of

steel is much higher than the one of copper and aluminum. Accordingly, stiffness of steel > stiffness of copper > stiffness of aluminum

The importance of stiffness can, for instance, be observed in the design of aircraft wings and reinforcement rods in which the design is controlled by the amount of elastic deformation.

6-Modulus of Resilience (M/R) U r :

Resilience )ةيعوجرلا( is the ability of the material to observe energy when it is elastically deformed. It is commonly measured by the term modulus of resilience, M/R, which is the energy per unit volume required to stress the material from the condition of zero stress to its yield stress,

s

_y.

Mathematically, U_r can be calculated from the figure as the area under elastic line of the

s-e

curve as the following:

18

7-Modulus of Toughness (M/T) U

_t

:

Mathematically, M/T can be calculated as shown in the given figure as the area under stress- strain curve.

Toughness )ةناتملا( can simply be defined as the ability of the material to observe energy when it is deformed up to fracturing.

It is commonly measured by the term modulus of toughness, M/T, which is the energy per unit volume required to stress the material from the condition of zero stress up to fracturing stress.

It is convenient to mathematically calculate M/T by the following equation and by approximate calculating the area under the stress-strain curve as shown in the given figure.

8-Poisson’s ratio

When a tensile stress is imposed on a metal specimen, an elastic elongation and accompanying strain e_z result in the direction of the applied stress.

There will be constrictions in the lateral (x and y) directions perpendicular to the applied stress.

Then, the compressive strains e_x and e_y may be determined. If the applied stress is uniaxial (only in the z direction), and the material is isotropic, then e_x = e_y. A parameter termed Poisson’s ratio is defined as the ratio of the lateral and axial strains, or

The negative sign is included in the expression so that u will always be positive, since e_x and e_y and will always be of opposite sign. Theoretically, Poisson’s ratio for isotropic materials should be 0.25; furthermore, the maximum value for (or that value for which there is no net volume change) is 0.50. For many metals and other alloys, values of Poisson’s ratio range between 0.25 and 0.35.

Room-Temperature Elastic Moduli, and Poisson’s Ratio for Various Metal Alloys

22

Effect of Temperature On Tensile Properties

Generally, increasing temperature leads to change the shape of the stress-strain curve as shown in the figure.

The properties of s

_y

, s

_u

and El % (elongation percentage) are also

given in the shown figure as the function of temperature.

24

Brittle and ductile Metallic Materials

The given figure indicates stress-strain curve given for different metallic materials. It is possible to recognize the behavior of the different metallic materials as the following

Curve 1:

represents ductile metal

behavior, where it is

possible to distinguish

between the different

values of the yield,

ultimate and fracture

stresses, i.e. s

_y

 s

_u

 s

_f

Curve 2:

represents semi-ductile

metal behavior, where it is

possible to distinguish the

different values of the

yield and ultimate

stresses. However, it is

impossible to distinguish

between the values of the

ultimate and fracture

stresses, i.e. s

_y

 s

_u

= s

_f

26

Curve 3:

represents brittle metal

behavior, where it is

impossible to distinguish

between the different

values of the yield,

ultimate and fracture

stresses, as shown in the

figure, i.e. s

_y

= s

_u

= s

_f

True Stress And Strain

From the given figure, the decline in the stress necessary to continue deformation past the maximum, point M, seems to indicate that the metal is becoming weaker. This is not at all the case; as a matter of fact, it is increasing in strength.

However, the cross-sectional area is decreasing rapidly within the neck region, where deformation is occurring. This results in reduction in the load-bearing capacity of the specimen. The stress is computed on the basis of the original cross-sectional area before any deformation, and does not take into account this reduction in area at the neck.

28

It is more convenient to use a true stress–

true strain scheme. True stress s_T is defined as the load F divided by the instantaneous cross-sectional area over which deformation is occurring:

It is also more convenient to represent strain as true strain

e

_T defined as:

If no volume change occurs during deformation, that is, if

True and engineering stress and strain are related according to

These equations are valid only to the onset of necking; beyond this point true stress and strain should be computed from actual load, cross-sectional area, and gauge length measurements

A

_i

l

_i =

A

_o

l

_o

and Then

True stress–True strain behaviors

A comparison of typical tensile engineering stress–strain and true stress–strain behaviors.

Necking begins at point M on the engineering curve, which corresponds to M

`

ⁱⁿ the true curve. The “corrected” true stress-strain curve takes into account the complex stress state within the neck region.

For some metals and alloys the region of the true stress–strain curve from the onset of plastic deformation to the point at which necking begins may be approximated by

K and n are constants; these values will vary from alloy to alloy, and will also depend on the condition of the material. The parameter n is often termed the strain hardening exponent and has a value less than unity. Values of n and K for several alloys are contained in the following table.

30

31

Exercise 1

32 0

200 400 600 800 1000

0 5 10 15

Stress [MPa]

Strain [%]

The results of a tensile test of a 8 mm diameter and 80 gauge length stainless steel test bar is given in the following chart:

Question 1

The final measured gauge length after failure of 93.93 mm

and the final measured diameter of 5.4 mm at the

fractured surface.

33

1. Define the yield stress, s_y, ultimate tensile stress, s_u, and fracture stress, s_f and the corresponding strains e_y, e_u and e_f as well as the values of E, M/R and M/T

0 200 400 600 800 1000

0 0.5 1

Stress [MPa]

Strain [%]

sel= 420 MPa eel= 0.002

sy= 820 MPa ey= 0.00625

850 900 950 1000 1050

2 7 12 17

Stress [MPa]

Strain [%]

su= 1030 MPa eu= 0.13

sf= 855 MPa ef= 0.175

E= 420/0.002= 210000 MPa M/R= 0.5X420X0.002X1000= 420 kJ/m³

3

34 0

200 400 600 800 1000 1200

0 2 4 6 8 10 12 14 16 18

Stress [MPa]

Strain [%]

TRUE Eng.

2. Define the s

_T

- e

_T

between yield and maximum

stresses, i.e s

_T

= K e

ⁿ

relationship

35 900

950 1000 1050 1100 1150 1200

4 6 8 10 12

Stress [MPa]

Strain [%]

True

eT sT

1 0.06 1080 2 0.10 1135

𝒏 =

𝒍𝒐𝒈 𝝈_𝑻𝟏 𝝈_𝑻𝟐 𝒍𝒐𝒈 𝝐_𝑻𝟏 𝝐_𝑻𝟐

=

𝒏 = 0.1

𝑲 = 𝝈_𝑻𝟏

𝝐_𝑻𝟏^𝒏 + 𝝈_𝑻𝟐 𝝐_𝑻𝟐^𝒏 𝟐 𝑲 = 1430

𝝈_𝑻 = 𝟏𝟒𝟑𝟎 𝝐_𝑻 ^𝟎.𝟏

36

3. Estimate the ductility of the stainless steel tested material

𝑬𝒍% = 𝒍

_𝑭

− 𝑳

_𝟎

𝒍

_𝟎

𝐱 𝟏𝟎𝟎

𝑹𝑨% = 𝑨

_𝟎

− 𝑨

_𝑭

𝑨

_𝟎

𝐱 𝟏𝟎𝟎

El%= 𝟗𝟑. 𝟗𝟑 − 𝟖𝟎

𝟖𝟎 × 𝟏𝟎𝟎 = 17.4%

RA%= 𝟖 ^𝟐 − 𝟓. 𝟒 ^𝟐

𝟖 ^𝟐 × 𝟏𝟎𝟎 = 54.4%

37

A tensile test bar of aluminium has a diameter of 12.8 mm and gauge length of 200 mm. The following

information has been collected after tensile test:

Question 2

If the modulus of elasticity of aluminium, E, is 70000 MPa.

Calculate the following:

38

 

y 6

y o 2 2 6 2

o

P 7600 7600 10 MN

59 06MPa

A d 12 8 10 m

4 4

. .



s    





 

 

6

u o 2 2 6 2

o

P 13730 13730 10 MN

106 7 MPa

A d 12 8 10 m

4 4

max

.

.



s    





 

a) Yield Stress s

_y

, UTS ( s

_u

)

o

y F

F

d 12 8mm 200mm

P 7600 N P 13730 N P 9810 N 245 5mm d 4 3mm

o max F

. , l

, ,

l . , .

 

  

 

39

b) Fracture Stress s

_F

, Elongation% at fracture

 

6 F

F o 2 2 6 2

o

P 9810 9810 10 MN

76 24 MPa

A d 12 8 10 m

4 4

. .



s    





 

F o

o

l l

l

. .

Elongation% 100 245 5 200

100 22 75%

200

   

   

o

y F

F

d 12 8mm 200mm

P 7600 N P 13730 N P 9810 N 245 5mm d 4 3mm

o max F

. , l

, ,

l . , .

 

  

 

40

c) Modulus of resilience

o

y F

F

d 12 8mm 200mm

P 7600 N P 13730 N P 9810 N 245 5mm d 4 3mm

o max F

. , l

, ,

l . , .

 

  

 

41

d) Strain at yielding point

y 59 06 MPa . E = 70000 MPa

s 

42

Tensile test has been carried out on standard copper specimen and the following results were found:

Question 3

It is required to calculate:

a) Yield Stresss_y (in MPa), of the copper specimen.

b) If a copper round solid-rod of length of 1.0 m made of the same tested copper material, is stretched under an axial load. Calculate the increase in the rod length before being plastically deformed.

c) If a cylinder made of the same tested copper material is subjected to a load of 7 kN. Check whether the copper cylinder is deformed in the elastic deformation zone or not. Consider the outer diameter of the cylinder is 10 mm and thickness of the cylinder wall is 2.0 mm.

43

a) Yield Stress s_y (in MPa), of the copper specimen.

. .

. .





 s

  s

s    

s  

2 y

2 3 y

3

2

3

y 3 2

y 2

M R 1

2 E

M N m 1 100 10

2 MN

m 112500 m

M N m MN

2 100 10 112500

m m

150 MN 150 MPa

m

44

σ = 150 MPay

b) If a copper round solid-rod of length of 1.0 m made of the same tested copper material, is stretched under an axial load. Calculate the increase in the rod length before being plastically deformed.

45

σ = 150 MPay

c) If a cylinder made of the same tested copper material is subjected to a load of 7 kN. Check whether the copper cylinder is deformed in the elastic deformation zone or not. Consider the outer diameter of the cylinder is 10 mm and thickness of the cylinder wall is 2.0 mm.

   

copper cylinder is deformed in the elastic deformation





 

            





0

-3

2 2

2 2 6 2

0 i

y

Load P

σ = Cross section Area A

P 7 10 MN

σ =

d d 10 6 10 m

4 4

σ = 139.3 MPa σ = 150 MPa

46

A cylindrical specimen of steel having a diameter of 15.2 mm and length of 250 mm is deformed elastically in tension with a force of 48900 N.

Suppose that the modulus of elasticity is 207 GPa and Poisson’s ratio (𝜈) is 0.3, determine the following:

a) The amount by which this specimen will elongate in the direction of the applied stress ( ∆𝑙 ).

b) The change in diameter of the specimen ( ∆𝑑 ).

Will the diameter increase or decrease?

Question 4

𝜎 =

_𝜋⁴⁸⁹⁰⁰

4 15.2 ²

= 269.5 MPa 𝐸 = 269.5

𝜀 ∴ 𝜀 = 269.5

207 × 10

³

= 1.3 × 10

⁻³

𝜀 = ∆𝑙

𝑙

₀

∴ ∆𝑙 = 250 × 1.3 × 10

⁻³

= 0.325 𝑚𝑚

𝜇 = − 𝜖_𝑑 𝜖 =

− ∆𝑑 𝑑₀

∆𝑙 𝑙₀

= 0.3 ∴ ∆𝑑 = − 0.3 × 1.3 × 10⁻³ × 15.2

= − 6.0 × 10⁻³ ∴ ∆𝑑 = − 6.0 × 10⁻³

47

Example Problem 6.1

Example Problem 6.2

Example Problem 6.3

Example Problem 6.4

Example Problem 6.5

48

Load requirements for compressing specimens of various diameters made of a material with a yield stress of 1380 MPa

Diameters: A = 28.4 mm, B = 25.4 mm, C = 20.3 mm, D = 12.7 mm.

Length-to-diameter ratio (L/D) = 3

Compression Test

Shear Loading 49

Direct Shear Test