material science for chemical engineers

1. MATERIAL SCIENCE
ENG. KAREEM. H. MOKHTAR

2. STRESS AND STRAIN
• Stress =
• Strain =
•
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
= Y (young’s modulus)
F
A
eL = DL
L o
Big
Small Weak material
Strong material

3. STRESS

4. STRESS
• 1- Compressive stress = F
A o
2
m
N

5. STRESS
original area
before loading
Area, A
Ft
Ft
s =
Ft
A o
2
f
2
m
N
or
in
lb
=
2- Tensile stress (s)

6. STRESS
• 3- Shear stress (t)
Area, A
Ft
Ft
Fs
F
F
Fs
t =
Fs
Ao

7. STRAIN
e = d
Lo
d/2
dL/2
Lo
wo
• Lateral strain:
eL = L
wo
d
• Tensile strain:

8. STRAIN
• Shear strain
g = Dx/H
H

9. YOUNG’S MODULUS
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝐿
𝐿𝑜
D

10. EXAMPLE 1
Givens
Weight = 1000N
Area of the beam is 2cm x 2cm
Lo = 1.75 m
Find delta L
Material Fe  youngs modulus
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝐿
𝐿𝑜
=
1000
2 𝑥 2 𝑥 10−4
?
1.75
D

11. EXAMPLE 2
F
Givens
Weight= 50 Kg
Radius of the heel= 0.5 cm
30% of the woman's weight acts on the
heel
Solution
• stress = F
A o
=
50 ∗ 0.3 ∗ 9.8
𝑝𝑖
4
∗ 0.012
= 1871662 𝑁/𝑚2

12. EXAMPLE 4
Givens
D= 1mm
Lo= 4m
m= 500 kg
Find the elongation
young’s modulus (AL) = 7 * 10^10 N/m2
Solution
young’s modulus=
𝑆𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝐿
𝐿𝑜
=
500∗9.8
𝑝𝑖
4
∗ 10−6
?
4
= 6.9* 1010

13. MAX STRESS (BREAKING STRESS)
• It is the max stress a material can handle before it breaks

14. EXAMPLE 3
• Would the wire breaks or not?
• F/ A = 500 * 9.8 / pi*0.0005^2
• What is the maximum weight this wire could handle
• StressMAX = mmax* g / A
Mmax= 17.6 kg

15. SHEAR MODULUS (S)
•
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝑑𝑒𝑙𝑡𝑎 (𝑥)
𝐻
H

16. SHEAR STRESS
• It is used in cutting materials
• The area used is the area that is going to be cut

17. EXAMPLE 6
• H = 5 cm
• W= 20 cm
• L= 2cm
• F= 1000N
• Fe shear modulus= 7.7 * 10^10
W
H
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝑑𝑒𝑙𝑡𝑎 (𝑥)
𝐻
𝑑𝑒𝑙𝑡𝑎 (𝑥) = 1.67 * 10-7 m

18. EXAMPLE 8
Givens
The diameter of the drilling bit= 4.2 cm
Thickness of the sheet = 5mm
Length of the sheet = 5 m
Width of the sheet = 3 m
Shear stress of steel = 4 x 10^8
Solution
Shear stress= F/A
F= 4 *10^8 * pi * 0.042*0.005 = 263893.8 N

19. BULK MODULUS
𝐵𝑢𝑙𝑘 𝑠𝑡𝑟𝑒𝑠𝑠
𝐵𝑢𝑙𝑘 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
− 𝑑𝑒𝑙𝑡𝑎 (𝑣)
𝑉𝑜
Compressibility =
1
𝐵𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠
The higher the bulk modulus
the more force you need to
Change the volume of the material

20. EXAMPLE 6
• A Bowling ball made of steel sunk in an ocean, find the change in it’s volume if
you know that the ocean is 10000 m in depth and the original volume of the ball
is 1m^3
𝐵𝑢𝑙𝑘 𝑠𝑡𝑟𝑒𝑠𝑠
𝐵𝑢𝑙𝑘 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝑑𝑒𝑙𝑡𝑎 (𝑣)
𝑉𝑜
= Bulk modulus