1
材料导论第一章绪论课程安排第一讲绪论，力学性质第二讲聚合物材料第三讲复合材料第四讲陶瓷材料第五讲金属学基础课中考试40％
第六讲金属材料第七讲材料与环境第八讲电性能第九讲磁性能，光学性质课末考试60％
What are materials?
Materials are the matter of the
universe,These substances have
properties that make them useful in
structures,machines,devices,products,
and systems.
材料是宇宙间可用于制造有用物品的物质。
钢铁：67%
高分子：12%
合金铝：4％
纺织品：12％
钢铁：62%
高分子：18%
合金铝：6％
纺织品：12％
1986年奔驰汽车
1996年奔驰汽车材料成熟曲线铝铜碳钢通用塑料不锈钢超级合金特殊金属传统工程塑料高性能工程塑料工程塑料合金光导纤维树脂基复合材料金属基复合材料结构陶瓷
MATERIALS SCIENCE &
ENGINEERING
involves the generation and
application of knowledge relating
the composition,structure,and the
processing of materials to their
properties and uses.
材料组成、结构、加工与材料性质、使用之间关系的发现与应用
2
WHY STUDY MATERIALS
SCIENCE AND ENGINEERING
Materials scientists and engineers are
specialists who are totally involved in the
investigation and design of materials.
原材料制取预加工生产制品服务期丢弃/回收工程化材料环
Many times,a materials problem is one
of selecting the right material from the
many thousands that arc available.
WHY STUDY MATERIALS
SCIENCE AND ENGINEERING
Does the material possess the
necessary mechanical,electrical,and
thermal properties?
Can the material be formed to the desired
shape?
Will the properties of the material alter with
time during service?
Will the material be adversely affected by
the environmental conditions and resist
corrosion and other forms of attack?
Will the material be acceptable on aesthetic
grounds?
Will the material give sufficient degree of
reliability and quality?
Can the product be made at an
acceptable cost?
Can the product be
recycled?
Processing
→ Structure
→ Properties
→ Performance
3
Structure
Subatomic structure
Atomic level structure
Microscopic
Macroscopic
Properties
Mechanical
Electrical
Thermal
Magnetic
Optical
Deteriorative
Physical
材料金属材料高分子材料粉末金属粉末金属有色金属有色金属钢铁塑料橡胶粘合剂涂料纤维复合材料陶瓷材料玻璃结晶陶瓷碳材料半导体
METALS
good conductors of electricity and heat
not transparcnt to visible light;
polished metal surface has a lustrous appearance
quite strong,yet deformable
The elements inherently metallic in nature Ceramics
compounds between metallic and nonmetallic
elements
typically insulative to the passage of electricity and
heat
more resistant to high temperatures and harsh
environments
hard but very brittle
4
Ceramic compounds indicated by a combination of
metallic elements with nonmetallic elements
POLYMERS
?Electrical insulation
?Thermal insulation
?Chemical resistance
?Magnetic inertness
?Extremely light weight
?Toughness
?Transparency
?Colorability
The elements associated with
commercial polymers COMPOSITES
A composite material is a combination of two
or more chemically different materials,
having a distinct interface between them,
which act together to produce a desired
(tailored) set of properties,
A composite is designed to combine the best
characteristic of each constituent material.
EXAMPLES OF COMPOSITES
SEMICONDUCTORS
Semiconductors have electrical properties that are
intermediate between the electrical conductors and
insulators.
Furthermore,the electrical characteristics of these
materials are extremely sensitive to the presence of
minute concentrations of impurity atoms.
5
The elements involved in
semiconductors
Biomaterials arc employed in components
implanted into the human body for
replacement of diseased or damaged body
parts,
BIOMATERIALS
ADVANCED MATERIALS
Materials that are utilized in high-
technology applications are sometimes
termed advanced materials.
Lasers,integrated circuits,magnetic information
storage,liquid crystal displays (LCDs),fiber
optics,spacecraft,aircraft,and military rocketry
and the thermal protection system for the space
shuttle orbiter.
Examples of High-tech
The properties of materials can be engineered by
controlling the sizes of these building blocks in
the 1-100 nm size range and their assembly,
Nanostructured Materials
The building blocks of these materials,be it
metal,ceramic or polymers,are nanometer size
particles.
6
材料导论第二章材料的性质材料性质阻燃性
晶相结构相结构微观结构耐腐蚀性组成化学性质颜色光学性质磁学性质电学性质热学性质熔点物理性质动态力学性质硬度蠕变性质韧性压缩性质拉伸性质力学性质材料性质阻燃性
晶相结构相结构微观结构耐腐蚀性组成化学性质颜色光学性质磁学性质电学性质热学性质熔点物理性质动态力学性质硬度蠕变性质韧性压缩性质拉伸性质力学性质材料性质阻燃性
晶相结构相结构微观结构耐腐蚀性组成化学性质颜色光学性质磁学性质电学性质热学性质熔点物理性质动态力学性质硬度蠕变性质韧性压缩性质拉伸性质力学性质材料性质
7
2.1力学性质
2.1.1 拉伸性质测试拉伸性质的样品屈服点极限强度断裂点弹性区塑性区典型高分子材料的应力－应变曲线
A：低碳钢；B：中碳钢；C：熟石膏；D：碳化钨；
E：灰铸铁（压缩）；F：灰铸铁（拉伸）
偏屈服强度的确定
8
断裂伸长率＝
截面积收缩率＝
100
0
0
×
l
ll
f
100
0
0
×
A
AA
f
材料的延展性工程应力：
工程应变：
真应力：
真应变：
0
A
F
=σ
0
0
l
ll?
=ε
A
F
t
=σ
∫ ==
A
A
l
l
l
dl
0
0
lnln
真应力应变曲线与工程应力应变曲线对比例2－1设计一铝棒以承受200kN (0.2MN)的力。
为确保安全，棒上最大应力不能超过170MPa。
棒的长度至少为3.8m，受力时弹性形变不能超过
6mm。所用铝材的弹性模量为69GPa。
223
20
11801018.1
/170
2.0
mmm
mMN
MNF
A =×===
σ
解：先利用工程应力的定义计算棒的截面积：
截面积可以为任何形状,为方便加工，设计一圆棒，其直径为d:
最大容许弹性形变为6mm，而170MPa应力所对应的应变约为0.0025，由工程应变的定义可确定棒的最大长度：
mmdmm
d
A 8.381180
4
2
2
0
===即
π
mmml
ll
l
l
ll
4.224000025.0
6
0
000
0
====
=
=即ε
所以，为同时满足最大应力最小伸长两项条件，
棒的截面积至少为1820mm
2
，即直径至少48mm。
但规定的最小长度为3.8m。加长棒的长度，截面积必须随之变大。3.8m长的棒的最小应变为：
这一应变相应于110MPa的应力，小于最大应力
170MPa。则最小截面积为：
00158.0
3800
6
0
==
=
l
l
ε
223
20
18201082.1
/110
2.0
mmm
mMN
MNF
A =×===
σ
9
2.1力学性质
2.1.2剪切强度与挠曲强度剪切强度剪应力：
0
A
F
=τ
剪切应变：
h
s
δ
γ =
剪切模量：G=τ/γ
弯曲强度
2
2
3
wh
FL
挠曲强度＝
挠曲模量＝
δ
3
3
4wh
FL
例2-2：一玻璃纤维增强复合材料的挠曲强度为315MPa，
挠曲模量为124GPa。一样品宽12mm，厚9.5mm,长
200mm，置于相距125mm两圆辊之间。计算使样品断裂所需的力以及样品断裂时的挠曲。假设无塑性形变。
解：将样品尺寸代入挠曲强度公式：
NF
F
F
wh
FL
1819
173.0
315
173.0
5.9122
1253
2
3
315
22
==
=
××
×
==
又由挠曲模量公式：
mm
wh
FL
7.0
5.9124
1819125
4
10124
3
3
3
2
3
=
××
×
==×
δ
δδ
10
2.1力学性质
2.1.3 韧性用应力－应变曲线表征韧性冲击强度材料的脆－韧转变
2.1力学性质
2.1.4 疲劳强度交变应力
11
轴的直径与所受交变应力的关系：
3
18.10
d
lF
=σ
σ－N曲线耐久比
----疲劳极限与拉伸强度之比铁合金：~0.5
其它材料：0.25 ~ 0.45
例2-3：一工具钢的轴长度必须为2.44m，必须承受0.05MN的交变应力1年。轴的转速为每分钟一周。设计符合上述要求的轴。
解：一年时间轴所经历的交变应力周数为：
N＝（1周/min）×60×24×365 ＝5.256×10
5
周/年由σ-N 图，轴上经受的应力必须小于
500MPa。利用公式：
3
18.10
d
lF
=σ
mmmd
d
MNm
MPa
136135.0
05.044.218.10
500
3
==
××
=
直径为136mm的轴已能满足上述要求。但如果我们对设计作进一步改进，可得到永不破坏的材料。
仍由σ-N 图，疲劳极限为410MPa，则最小直径应为：
mmmd
d
MNm
MPa
145145.0
05.044.218.10
410
3
==
××
=
12
2.1力学性质
2.1.5 蠕变蠕变曲线铁铬镍合金应力
－
断裂曲线
Larson-Miller参数
L.M.＝（T/1000）（A＋Blnt）
例2-4：设计一个铸铁链用于砖窑。链必须在
600oC下受力22kN连续运转5年。
解：铸铁的Larson-Miller参数为：
1000
)ln78.036(
..
tT
ML
+
=
链必须运转5年，即43800h。
7.38
1000
)]43800ln(78.036)[(273600(
.,=
++
=ML
由前图，所施应力不能大于20MPa。链受力为22kN，则截面积应为：
22
2
11000011.0
/20
022.0
/ mmm
mMN
MN
FA ==== σ
则半边链的截面积为550mm
2
，即直径为26.5mm。
13
2.1力学性质
2.1.6 抗扭强度
P
A
γ
θ
x
y
r
rT
C
C’
z
扭矩Torque T = Pr
θ
T
Torque – twist diagram
τ
max
=
torsional yield
strength
r
J
Tr
=
max
τ
J – polar moment inertia
例2-5 How much torque (T) is needed to produce
a shear stress of 79.6 MPa on the surface of a 40-
mm-diameter shaft with a polar moment inertia
(J) equal to 25.12 × 10
-8
m
4
SOLUTION
P
T
r
J
Tr
T == τPr
m1kNmN8.999
0.02m
m1012.25MPa6.79
48
=
××
=
=
r
J
T
τ
14
2.1力学性质
2.1.7 硬度莫氏硬度一级：滑石二级：石膏三级：方解石四级：萤石五级：磷灰石六级：正长石七级：石英八级：黄玉九级：刚玉十级：金刚石布氏硬度维氏硬度洛氏硬度
15
硬度与拉伸强度的关联钢的拉伸强度（MPa）大约为布式硬度的3.4倍，误差在10%左右。
例2-6 Given the formula for determining the
Brinell hardness number (HB),where P is the
applied load (kg),D is the diameter of the steel
ball (mm),and d is the diameter of the
indentation (mm),
))(2/(
22
dDDD
P
HB
=
π
What is the Brinell hardness number if a 10-mm
ball with a load of 3000kg produces an
indentation with a diameter of 2.88 mm?
作业：
P41 2-5,2-7,2-10,2-11,2-12
材料导论第一章绪论课程安排第一讲绪论，力学性质第二讲聚合物材料第三讲复合材料第四讲陶瓷材料第五讲金属学基础课中考试40％
第六讲金属材料第七讲材料与环境第八讲电性能第九讲磁性能，光学性质课末考试60％
What are materials?
Materials are the matter of the
universe,These substances have
properties that make them useful in
structures,machines,devices,products,
and systems.
材料是宇宙间可用于制造有用物品的物质。
钢铁：67%
高分子：12%
合金铝：4％
纺织品：12％
钢铁：62%
高分子：18%
合金铝：6％
纺织品：12％
1986年奔驰汽车
1996年奔驰汽车材料成熟曲线铝铜碳钢通用塑料不锈钢超级合金特殊金属传统工程塑料高性能工程塑料工程塑料合金光导纤维树脂基复合材料金属基复合材料结构陶瓷
MATERIALS SCIENCE &
ENGINEERING
involves the generation and
application of knowledge relating
the composition,structure,and the
processing of materials to their
properties and uses.
材料组成、结构、加工与材料性质、使用之间关系的发现与应用
2
WHY STUDY MATERIALS
SCIENCE AND ENGINEERING
Materials scientists and engineers are
specialists who are totally involved in the
investigation and design of materials.
原材料制取预加工生产制品服务期丢弃/回收工程化材料环
Many times,a materials problem is one
of selecting the right material from the
many thousands that arc available.
WHY STUDY MATERIALS
SCIENCE AND ENGINEERING
Does the material possess the
necessary mechanical,electrical,and
thermal properties?
Can the material be formed to the desired
shape?
Will the properties of the material alter with
time during service?
Will the material be adversely affected by
the environmental conditions and resist
corrosion and other forms of attack?
Will the material be acceptable on aesthetic
grounds?
Will the material give sufficient degree of
reliability and quality?
Can the product be made at an
acceptable cost?
Can the product be
recycled?
Processing
→ Structure
→ Properties
→ Performance
3
Structure
Subatomic structure
Atomic level structure
Microscopic
Macroscopic
Properties
Mechanical
Electrical
Thermal
Magnetic
Optical
Deteriorative
Physical
材料金属材料高分子材料粉末金属粉末金属有色金属有色金属钢铁塑料橡胶粘合剂涂料纤维复合材料陶瓷材料玻璃结晶陶瓷碳材料半导体
METALS
good conductors of electricity and heat
not transparcnt to visible light;
polished metal surface has a lustrous appearance
quite strong,yet deformable
The elements inherently metallic in nature Ceramics
compounds between metallic and nonmetallic
elements
typically insulative to the passage of electricity and
heat
more resistant to high temperatures and harsh
environments
hard but very brittle
4
Ceramic compounds indicated by a combination of
metallic elements with nonmetallic elements
POLYMERS
?Electrical insulation
?Thermal insulation
?Chemical resistance
?Magnetic inertness
?Extremely light weight
?Toughness
?Transparency
?Colorability
The elements associated with
commercial polymers COMPOSITES
A composite material is a combination of two
or more chemically different materials,
having a distinct interface between them,
which act together to produce a desired
(tailored) set of properties,
A composite is designed to combine the best
characteristic of each constituent material.
EXAMPLES OF COMPOSITES
SEMICONDUCTORS
Semiconductors have electrical properties that are
intermediate between the electrical conductors and
insulators.
Furthermore,the electrical characteristics of these
materials are extremely sensitive to the presence of
minute concentrations of impurity atoms.
5
The elements involved in
semiconductors
Biomaterials arc employed in components
implanted into the human body for
replacement of diseased or damaged body
parts,
BIOMATERIALS
ADVANCED MATERIALS
Materials that are utilized in high-
technology applications are sometimes
termed advanced materials.
Lasers,integrated circuits,magnetic information
storage,liquid crystal displays (LCDs),fiber
optics,spacecraft,aircraft,and military rocketry
and the thermal protection system for the space
shuttle orbiter.
Examples of High-tech
The properties of materials can be engineered by
controlling the sizes of these building blocks in
the 1-100 nm size range and their assembly,
Nanostructured Materials
The building blocks of these materials,be it
metal,ceramic or polymers,are nanometer size
particles.
6
材料导论第二章材料的性质材料性质阻燃性
晶相结构相结构微观结构耐腐蚀性组成化学性质颜色光学性质磁学性质电学性质热学性质熔点物理性质动态力学性质硬度蠕变性质韧性压缩性质拉伸性质力学性质材料性质阻燃性
晶相结构相结构微观结构耐腐蚀性组成化学性质颜色光学性质磁学性质电学性质热学性质熔点物理性质动态力学性质硬度蠕变性质韧性压缩性质拉伸性质力学性质材料性质阻燃性
晶相结构相结构微观结构耐腐蚀性组成化学性质颜色光学性质磁学性质电学性质热学性质熔点物理性质动态力学性质硬度蠕变性质韧性压缩性质拉伸性质力学性质材料性质
7
2.1力学性质
2.1.1 拉伸性质测试拉伸性质的样品屈服点极限强度断裂点弹性区塑性区典型高分子材料的应力－应变曲线
A：低碳钢；B：中碳钢；C：熟石膏；D：碳化钨；
E：灰铸铁（压缩）；F：灰铸铁（拉伸）
偏屈服强度的确定
8
断裂伸长率＝
截面积收缩率＝
100
0
0
×
l
ll
f
100
0
0
×
A
AA
f
材料的延展性工程应力：
工程应变：
真应力：
真应变：
0
A
F
=σ
0
0
l
ll?
=ε
A
F
t
=σ
∫ ==
A
A
l
l
l
dl
0
0
lnln
真应力应变曲线与工程应力应变曲线对比例2－1设计一铝棒以承受200kN (0.2MN)的力。
为确保安全，棒上最大应力不能超过170MPa。
棒的长度至少为3.8m，受力时弹性形变不能超过
6mm。所用铝材的弹性模量为69GPa。
223
20
11801018.1
/170
2.0
mmm
mMN
MNF
A =×===
σ
解：先利用工程应力的定义计算棒的截面积：
截面积可以为任何形状,为方便加工，设计一圆棒，其直径为d:
最大容许弹性形变为6mm，而170MPa应力所对应的应变约为0.0025，由工程应变的定义可确定棒的最大长度：
mmdmm
d
A 8.381180
4
2
2
0
===即
π
mmml
ll
l
l
ll
4.224000025.0
6
0
000
0
====
=
=即ε
所以，为同时满足最大应力最小伸长两项条件，
棒的截面积至少为1820mm
2
，即直径至少48mm。
但规定的最小长度为3.8m。加长棒的长度，截面积必须随之变大。3.8m长的棒的最小应变为：
这一应变相应于110MPa的应力，小于最大应力
170MPa。则最小截面积为：
00158.0
3800
6
0
==
=
l
l
ε
223
20
18201082.1
/110
2.0
mmm
mMN
MNF
A =×===
σ
9
2.1力学性质
2.1.2剪切强度与挠曲强度剪切强度剪应力：
0
A
F
=τ
剪切应变：
h
s
δ
γ =
剪切模量：G=τ/γ
弯曲强度
2
2
3
wh
FL
挠曲强度＝
挠曲模量＝
δ
3
3
4wh
FL
例2-2：一玻璃纤维增强复合材料的挠曲强度为315MPa，
挠曲模量为124GPa。一样品宽12mm，厚9.5mm,长
200mm，置于相距125mm两圆辊之间。计算使样品断裂所需的力以及样品断裂时的挠曲。假设无塑性形变。
解：将样品尺寸代入挠曲强度公式：
NF
F
F
wh
FL
1819
173.0
315
173.0
5.9122
1253
2
3
315
22
==
=
××
×
==
又由挠曲模量公式：
mm
wh
FL
7.0
5.9124
1819125
4
10124
3
3
3
2
3
=
××
×
==×
δ
δδ
10
2.1力学性质
2.1.3 韧性用应力－应变曲线表征韧性冲击强度材料的脆－韧转变
2.1力学性质
2.1.4 疲劳强度交变应力
11
轴的直径与所受交变应力的关系：
3
18.10
d
lF
=σ
σ－N曲线耐久比
----疲劳极限与拉伸强度之比铁合金：~0.5
其它材料：0.25 ~ 0.45
例2-3：一工具钢的轴长度必须为2.44m，必须承受0.05MN的交变应力1年。轴的转速为每分钟一周。设计符合上述要求的轴。
解：一年时间轴所经历的交变应力周数为：
N＝（1周/min）×60×24×365 ＝5.256×10
5
周/年由σ-N 图，轴上经受的应力必须小于
500MPa。利用公式：
3
18.10
d
lF
=σ
mmmd
d
MNm
MPa
136135.0
05.044.218.10
500
3
==
××
=
直径为136mm的轴已能满足上述要求。但如果我们对设计作进一步改进，可得到永不破坏的材料。
仍由σ-N 图，疲劳极限为410MPa，则最小直径应为：
mmmd
d
MNm
MPa
145145.0
05.044.218.10
410
3
==
××
=
12
2.1力学性质
2.1.5 蠕变蠕变曲线铁铬镍合金应力
－
断裂曲线
Larson-Miller参数
L.M.＝（T/1000）（A＋Blnt）
例2-4：设计一个铸铁链用于砖窑。链必须在
600oC下受力22kN连续运转5年。
解：铸铁的Larson-Miller参数为：
1000
)ln78.036(
..
tT
ML
+
=
链必须运转5年，即43800h。
7.38
1000
)]43800ln(78.036)[(273600(
.,=
++
=ML
由前图，所施应力不能大于20MPa。链受力为22kN，则截面积应为：
22
2
11000011.0
/20
022.0
/ mmm
mMN
MN
FA ==== σ
则半边链的截面积为550mm
2
，即直径为26.5mm。
13
2.1力学性质
2.1.6 抗扭强度
P
A
γ
θ
x
y
r
rT
C
C’
z
扭矩Torque T = Pr
θ
T
Torque – twist diagram
τ
max
=
torsional yield
strength
r
J
Tr
=
max
τ
J – polar moment inertia
例2-5 How much torque (T) is needed to produce
a shear stress of 79.6 MPa on the surface of a 40-
mm-diameter shaft with a polar moment inertia
(J) equal to 25.12 × 10
-8
m
4
SOLUTION
P
T
r
J
Tr
T == τPr
m1kNmN8.999
0.02m
m1012.25MPa6.79
48
=
××
=
=
r
J
T
τ
14
2.1力学性质
2.1.7 硬度莫氏硬度一级：滑石二级：石膏三级：方解石四级：萤石五级：磷灰石六级：正长石七级：石英八级：黄玉九级：刚玉十级：金刚石布氏硬度维氏硬度洛氏硬度
15
硬度与拉伸强度的关联钢的拉伸强度（MPa）大约为布式硬度的3.4倍，误差在10%左右。
例2-6 Given the formula for determining the
Brinell hardness number (HB),where P is the
applied load (kg),D is the diameter of the steel
ball (mm),and d is the diameter of the
indentation (mm),
))(2/(
22
dDDD
P
HB
=
π
What is the Brinell hardness number if a 10-mm
ball with a load of 3000kg produces an
indentation with a diameter of 2.88 mm?
作业：
P41 2-5,2-7,2-10,2-11,2-12
