SOLUTIONS MANUAL
,Table of Contents
Chapter Title Page
2 Properties of Engineering Materials 2
3 Structure of Metals and Alloys 4
7 Casting of Metallic Materials 6
8 Fundamentals of Metal Forming 8
9 Bulk Forming of Metallic Materials 13
10 Sheet Metal Forming Processes 18
11 High Velocity Forming and High Energy Rate Forming 22
13 Processing of Polymeric Materials 24
14 Composite Materials and their Fabrication Processes 25
15 Fundamentals of Traditional Machining Processes 26
16 Machine Tools for Traditional Machining 32
17 Fundamentals Nontraditional Machining 35
18 Numerical Control of Machine Tools 37
19 Industrial Robots and Hexapods 41
20 Surface Technology 42
21 Joining Processes 46
24 Quality Control 50
1
, Chapter 2
Properties of Engineering Materials
1- What is the stress in a wire whose diameter is 1 mm, which supports a load of
100 Newton
Solution:
load 100 400
1000 2 127 MPa
area 1 2
4 1000
127 N / mm 2
2- Suppose that the wire in the above problem is cupper and is 30 m long. How
much will it stretch? Repeat for a steel and aluminum wires
Solution:
Cu wire:
127000000
E 10x1010
Total elongation for copper wire ( l cu )
l
Since
l
127000000
l cu xl x30 0.038 m =38 mm
10x1010
Steel wire:
127000000
E 21x1010
Total elongation for copper wire ( l st )
l
Since
l
127000000
l st xl x30 0.018 m =18 mm
21x1010
Aluminum wire:
127000000
E 7 x1010
Total elongation for copper wire ( l al )
2
, l
Since
l
127000000
l al xl x30 0.054 m =54 mm
7 x1010
Comment: Steel is stiffer than both cupper and aluminum. It is exactly 3 times more stiff than
aluminum
3- Estimate the HB hardness of an alloy steel of ultimate tensile strength TS=1500
MPa using equation 2.9. Use the hardness conversion table to determine the
HRC hardness.
Solution:
TS (MPa) = 3.5 HB
l500
HB 428
3.5
Using the conversion table 2.3, the corresponding HRC of that alloy steel = 46
Referring to Figure 2.5 the hardness values are realized.
4- A constructional component made of Al-alloy can be represented as a bar of
diameter 20 mm and length 400 mm is subjected to pure tension; calculate:
a- Extension of the bar if the imposed load= 8 kN
b-The load at which the bar suffers permanent deformation
c- Maximum load the bar can withstand fracture
Assume the following mechanical properties: E=70 GPa, YS=496 MPa, TS=560 MPa
Solution:
x 20 2
a- Ao 314 mm 2
4
80000
Im posed tensile strength 255 N/mm 2
314
255 MPa
It is thus less than the YS, which means that the deformation will be purly elastic.
b- YS 0.2 496 N/mm 2
Fy 0.2 xA o 496 x 314 156 kN
c- Fmax TSxA o 560 x 314 176 kN
3
,Table of Contents
Chapter Title Page
2 Properties of Engineering Materials 2
3 Structure of Metals and Alloys 4
7 Casting of Metallic Materials 6
8 Fundamentals of Metal Forming 8
9 Bulk Forming of Metallic Materials 13
10 Sheet Metal Forming Processes 18
11 High Velocity Forming and High Energy Rate Forming 22
13 Processing of Polymeric Materials 24
14 Composite Materials and their Fabrication Processes 25
15 Fundamentals of Traditional Machining Processes 26
16 Machine Tools for Traditional Machining 32
17 Fundamentals Nontraditional Machining 35
18 Numerical Control of Machine Tools 37
19 Industrial Robots and Hexapods 41
20 Surface Technology 42
21 Joining Processes 46
24 Quality Control 50
1
, Chapter 2
Properties of Engineering Materials
1- What is the stress in a wire whose diameter is 1 mm, which supports a load of
100 Newton
Solution:
load 100 400
1000 2 127 MPa
area 1 2
4 1000
127 N / mm 2
2- Suppose that the wire in the above problem is cupper and is 30 m long. How
much will it stretch? Repeat for a steel and aluminum wires
Solution:
Cu wire:
127000000
E 10x1010
Total elongation for copper wire ( l cu )
l
Since
l
127000000
l cu xl x30 0.038 m =38 mm
10x1010
Steel wire:
127000000
E 21x1010
Total elongation for copper wire ( l st )
l
Since
l
127000000
l st xl x30 0.018 m =18 mm
21x1010
Aluminum wire:
127000000
E 7 x1010
Total elongation for copper wire ( l al )
2
, l
Since
l
127000000
l al xl x30 0.054 m =54 mm
7 x1010
Comment: Steel is stiffer than both cupper and aluminum. It is exactly 3 times more stiff than
aluminum
3- Estimate the HB hardness of an alloy steel of ultimate tensile strength TS=1500
MPa using equation 2.9. Use the hardness conversion table to determine the
HRC hardness.
Solution:
TS (MPa) = 3.5 HB
l500
HB 428
3.5
Using the conversion table 2.3, the corresponding HRC of that alloy steel = 46
Referring to Figure 2.5 the hardness values are realized.
4- A constructional component made of Al-alloy can be represented as a bar of
diameter 20 mm and length 400 mm is subjected to pure tension; calculate:
a- Extension of the bar if the imposed load= 8 kN
b-The load at which the bar suffers permanent deformation
c- Maximum load the bar can withstand fracture
Assume the following mechanical properties: E=70 GPa, YS=496 MPa, TS=560 MPa
Solution:
x 20 2
a- Ao 314 mm 2
4
80000
Im posed tensile strength 255 N/mm 2
314
255 MPa
It is thus less than the YS, which means that the deformation will be purly elastic.
b- YS 0.2 496 N/mm 2
Fy 0.2 xA o 496 x 314 156 kN
c- Fmax TSxA o 560 x 314 176 kN
3
