This file for education only.

1. Chapter 8
The mechanical properties of solid

2. 8.1 Elasticity of material
When some force hit to the object and
can affect to the objects change in shape and
size. Elasticity of materials has two types:
elastic deformation and plastic deformation
+ elasticity deformation: If the material
receives tensile stress there will be minor leaks,
But if the pressure is pressed down
(compression stress) the material will shrink
and when the impact is extinguished, the
material will return to its original state
shows the change of the ceramic unit when affected.
Chapter 8
The mechanical properties of solid
1

3. + plastic deformation: a permanent deformation or change in shape of a solid body
without fracture under the action of a sustained force
2

4. 8.2 Stress and strain of solid
If the force is exerted, 𝑙0 cut surface 𝐴0 with force F and then make the If the force exerted by the
object is cross-sectional area with a force F and then the object is exerted by a certain amount of
stress.
𝜎 =
𝐹
𝐴 𝑂
8.1
𝜎 : stress there are units Pa. / 1 Pa=1𝑁/𝑚2
and 1𝑝𝑠𝑖 = 6,89 × 103
Pa
𝐹: is the instantaneous load applied perpendicular to the specimen cross section, in units of (N)
𝐴 𝑂: is the specimen cross-sectional area prior load application
When the affected material changes shape, the material will be subjected to an oscillation (𝑙) by
a ratio that changes to the original length (𝑙0) called strain.
𝜀 =
𝑙−𝑙0
𝑙0
=
∆𝑙
𝑙0
8.2
𝜀: strain of solid
∆𝑙: is the deformation elongation or change in length at some instant, as referenced to the original length.
𝑙: is the length 3

5. From the value of strain, we can obtain the eigenvalues of elongation from the equation
% elongation = 𝜀 × 100% 8.3
Example1 : Round pipe with diameter 1,25 cm and can receive material 2500 Kg. Calculate the pressure at Mpa unit.
𝜎 =
𝐹
𝐴0
=
𝑚𝑔
𝜋𝑟2
=
2500 × 9.8
𝜋
1,25 × 10−2
2
2 = 2 × 108 𝑁/𝑚2 = 200𝑀𝑃𝑎
4

6. Example2: 0,5 in wide aluminum block. By 8 in. but if using tensile force, can measured in length 2,65 in.
Calculate the strain and the percentage of the elasticity.
Solution
𝜀 =
𝑙−𝑙0
𝑙0
=
2,65−2
2
= 0,325
% elongation = 𝜀 × 100%
= 0,325 × 100% = 32,5%
5

7. The elastic modulus will have the stress value, and the durability will be linearly described by Hook's law.
𝝈 = 𝑬𝜺 8.4
The SI unit for the modulus of elasticity is gigapascal, GPa, where 1 GPa 109 N/m2
103 MPa. When E is the modulus of elasticity or Young's modulus, The values ​​of E Relation of the
bond strength between material atoms and temperatures, If the bond is very strong the material will
have a high E value and when the temperature increases the value of E will decrease.
The relationship between stress and strain 6
숙제: 사진을 설명
This picture shows the relationship between
stress and strain, which is a linear line,
when we keep putting strain on an object to
the yield point, also known as the yield
strength, but when giving putting on
strength up to object will continue to deform
to the point where it falls before the fracture
to break. We call it the ultimate strength.

8. In the event that the soft material undergoes a permanent change after the release of pressure and it will
be some time before this break point is necessary for the shape of the material, such as the bending steel plate
to be made to create a bumper.
7
Able to be manufactured into the bumper of the car

9. shows the 0,2% offset yield strength
calculated at the strain point = 0,002
8
Yield stress
Intersection
사진을 설명
The graph shows that stress and strain are very similar to lines, but there is no point see.
Therefore the yield stress is set at 0,2 percentage offset or stress value is 0,0002. then the long
lines and of the line and the line showing the intersecting points are is the maximum stress
boundary value.

10. 8.2.1 True stress and true stain
- During the test, the material pulls out and at the same time, the cutting surface area of
the material will shrink. As shown in Figure. But search engine prices need to be
drawn to the default content as
𝜎𝑡 =
𝐹
𝐴𝑖
8.5
𝐴𝑖: instantaneous cross-sectional area
𝜀𝑡 = න
𝑙0
𝑙𝑖
𝑑𝑙
𝑙
= 𝑙𝑛
𝑙𝑖
𝑙0
𝑙𝑖: instantaneous length
- If there is no V (volume) change at deformation, it is 𝑙0 𝐴0 = 𝑙𝑖 𝐴𝑖 𝑜𝑟
𝑙𝑖
𝑙0
=
𝐴0
𝐴𝑖
will get
𝜀𝑡 = 𝑙𝑛
𝐴0
𝐴𝑖
8.7
9
8.6

11. Elastic Properties
𝑣 = −
𝜀 𝑥
𝜀 𝑧
=
𝜀 𝑦
𝜀 𝑧
Poisson’s ratio~ the ratio of the lateral to axial strains
For metals & alloys, v =0.25~0.35
In the case of isotropic materials, E, G, and v relations:
E = 2𝑔 1 + 𝑣 8.9
Many of the common materials have a relationship of E = 2,5~3G
8.8
10

12. 8.2.2 Shear stress and shear strain
1) Shear stress 𝛕
- Shear stress (𝜏) is defined as the resisting force offered by the body per unit area of cross-
section, when a tangential force is applied on the body.
- When ever there is an balanced force is acting on a body on two different sections, there may
be the chance of shear stress
𝜏 =
𝐹𝑠
𝐴
8.10
11

13. 2) shear strain, 𝛾
Shear strain is the ratio of deformation to original dimensions. In the case of shear strain, it is
the amount of deformation perpendicular to a given line rather than parallel to it.
𝛾 =
𝛼
ℎ
= tgθ 8.11
If the shear occurs within the boundary, we can write the relation between the stress and the tolerance.
𝜏 = 𝐺𝛾 8.12
12

14. Young's Moroccan values and material values
Material type
modulus of young
× 106
𝑝𝑠𝑖 (𝐺𝑃𝑎)
modulus of shear
× 106
𝑝𝑠𝑖 (𝐺𝑃𝑎)
Aluminium allloy 10,5 (72,4) 4,0 (27,5)
Cu 16,0 (110) 6,0 (41,4)
Stainless steel 28,0 (193) 9,5 (65,6)
Ti 17,0 (117) 6,5 (44,8)
8.3 Hardness in material
Material hardness is the property of the material which enables it to resist plastic deformation, usually by
penetration or by indentation. The term of hardness is also referred to stiffness or temper, or to resistance to
bending, scratching, abrasion, or cutting
13

15. Types of Hard Tests are
divided into four main
categories:
(1) Brinell
(2) Vickers microhardness
(3) Knoop microhardness
(4) Rockwell
14
Show different hardness testing techniques

16. Hardness level Mineral Absolute hardness
1 Talc 1
2 Gypsum 2
3 Calcite 9
4 Fluorite 21
5 Apatite 48
6 Feldspar 72
7 Quartz 100
8 Topaz 200
9 Corundum 400
10 Diamond 1500
Table 8.3 Display of mosh scale hardness Mohs scale of various minerals
15

17. 8.4 Ductile of Material
toughness is the ability of a material to absorb energy and plastically deform without
fracturing. One definition of material toughness is the amount of energy per unit volume
that a material can absorb before rupturing. 사진을 설명
Toughness of Soft and Brittle Materials
16
- fail suddenly
by cracking
or spintering
- Much
weaker in
tension than
in
compressio
- no
significant
plastic
deformation
before
fracture
- able to
deform
significantl
y in to the
inelastic
range
- sustains
significant
plastic
deformation
prior to
fracture
calculate the toughness
𝑻𝒐𝒖𝒈𝒉𝒏𝒆𝒔𝒔 =
𝝈𝜺
𝟐
=
𝝈 𝟐
𝟐𝑬
8.13

18. shows (a) cracks, (b) center, and (c) the relationship between pressure and distance to the edge.
17

19. Stress of toughness that is resistant to breakage be obtained from
𝒌 𝒕 = 𝒀𝝈 𝝅𝒂
𝑘 𝑡: stress concentration factor
𝜎: stress
𝑌: geometry constant =1
𝑎: the length of edge crack half of center crack
The coefficient of stress intensity factor that makes the laminate is called the fracture
toughness 𝑘𝐼𝑐 of the material can be obtained from the equation:
𝑘𝐼 = 𝑌𝜎 𝜋𝑎
𝑘𝐼𝑐 = 𝑌𝜎𝑓 𝜋𝑎
𝜎𝑓: breakage pressure
𝑘𝐼𝑐: plain strain fracture toughness
18
8.14설명

20. Table 8.4 𝑘1𝑐 and durability of some materials
Materials
𝑘1𝑐 Yield strength
MPa 𝑚 𝑘𝑠𝑖 𝑖𝑛 MPa ksi
Aluminum alloys
2024-T851
7075-T651
7178-T651
26,4
24,2
23,1
24
22
21
455
495
570
66
72
83
Titanium alloys
Ti-6Al-4V 55,0 50 1035 150
Alloy steels
4340(low –alloy stell)
17-7pH (precipitation hardening)
350 maraging steel
60,4
76,9
55,0
55
70
50
1515
1435
1550
220
208
225
19

21. Example 8.5 A metal plan to be used in a structure shall be able to withstand (tensile stress) 207MPa, if the aluminum
alloy 2024-T851 is used in this work. How much is the blame for what will happen within 2a? Thus, the work will
remain stable without breaking Y = 1.
Solution:
k1c = Yσf πa
mm,
,
,
k
a
f
C
185
207
426
143
11
22
1
=





=








=

The largest internal cracks are
2𝑎 = 2 × 5,18 = 10,36 𝑚𝑚
20

22. 8.5 Deformation Mechanisms
metallurgy and materials science, deformation mechanisms refer to the various
mechanisms at the grain scale that are responsible for accommodating large plastic strains
in rocks, metals and other materials.
Most likely the permanent deformation, which is often caused by the material being
affected, causes such radical changes that even if we remove it, the material will never be
able to return to its original state.
(a) Surface scroll bar (0001) (b) SEM photograph showing sliding bar on surface of copper wire
a b
21

23. 8.6 Fracture in materials
A fracture is the separation of an object or material into two or more pieces under the action of stress. Divided into two
categories: ductile fracture and brittle fracture
2) brittle fracture
Brittle Fracture is the sudden, very rapid cracking of equipment
under stress where the material exhibited little or no evidence of
ductility or plastic degradation before the fracture occurs.
1) ductile fracture
In ductile fracture, there is absorption of massive
amounts of energy and slow propagation before the
fracture occurs.
22

24. 8.7 Fatigue in metal
Metal fatigue, weakened condition induced in metal parts of machines, vehicles, or
structures by repeated stresses or loadings, ultimately resulting in fracture under a stress
much weaker than that necessary to cause fracture in a single application.
23Displays breakage due to defective axle load

25. The graph showing stress received from pressure versus time: (a) receives the pull / pull
force (B) Pull / pull force and (c) Received force from the spectrum
24

26. 8.8 Summary
- Stress base is a measure of mechanical action or impact force on the material also on the cross-sectional area, stress
indicates the amount of the deformation.
- Some metals can do simple tested for stress and strain with 4 types: 1. testing under tensile strength, 2. testing under
compression,3. testing under torque and 4. testing under shear. Tensile testing is a fundamental method of stressed
materials that undergo a contraction, stress and strain are interlinked, the change constant is the contraction modulus
for pulling and compressing force. Poisson’s ratio is the ratio of crossing resistance to long-term strain of a material
in the case of isotropic materials.
- Hardness is a measure of happiness in resistance to permanent deformation, because somewhere, a popular hardness
test is not a small indenter. To press down on the surface of the material and calculate the hardness number or the
depth of the indentation.
25