Material science

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
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
= 1.87 ∗ 106
𝑁/𝑚2

12. EXAMPLE 4
Givens
D= 1mm
Lo= 4m
m= 500 kg
Find the elongation
young’s modulus (AL) = 6.9* 1010 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 6
• Would the wire breaks or not?
• F/ A = 500 * 9.8 / pi*0.00052
• If the stress is > max stress  the wire will break
• If the stress is < max stress  it will not break
• What is the maximum weight this wire could handle
• StressMAX = mmax* g / A
Mmax= 6 kg

15. STRESS AND STRAIN IN DIFFERENT FORCE
DIRECTIONS
Tension Compression Shear
Stress= F/A Stress= F/A Stress= F/A
Strain= delta (L)/Lo Strain= delta (L)/Lo Strain= delta (x)/H

16. EXAMPLE 3
• F = ?
• Strain = 1%
• L= 1cm
• Steel youngs modulus = 21x1010 N/m2
• Solution
• 21 MPa

17. EXAMPLE 8
• Lo = 40m
• Dia= 1cm
• Delta(L) = 2m
• Weight = 900N climber
• Y = ?
• Solution
• Y=229,183,118 N/m2

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

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

20. EXAMPLE 9
• H = 5 cm
• W= 20 cm
• L= 2cm
• F= 1000N
• Carbon steel shear modulus= 7.7 * 1010
W
H
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝑑𝑒𝑙𝑡𝑎 (𝑥)
𝐻
𝑑𝑒𝑙𝑡𝑎 (𝑥) = 1.67 * 10-7 m

21. EXAMPLE 10
5 m
2.4 m
Max stress =
𝐹
𝐴
4 x 108 =
𝐹
2.4 ∗0.002
F must be greater than 1920000 N

22. EXAMPLE 12
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

23. 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

24. EXAMPLE 13
• 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 1m3
𝐵𝑢𝑙𝑘 𝑠𝑡𝑟𝑒𝑠𝑠
𝐵𝑢𝑙𝑘 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝑑𝑒𝑙𝑡𝑎 (𝑣)
𝑉𝑜
= Bulk modulus

25. EXAMPLE 14
• Compressibility =
− 𝑑𝑒𝑙𝑡𝑎 (𝑣)
𝑉𝑜
𝐹
𝐴
• Substitute
• Compressibility = 3 x 10-10 m3/ N

26. EXAMPLE 15
Shear modulus (S) =
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠
𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝐹
𝐴
𝑑𝑒𝑙𝑡𝑎 (𝑥)
𝐻

27. SOLVE THIS QUESTION
• A carbon steel cylinder that has a young’s modulus of 200 GPa, bulk modulus of
140 GPA, And max shear stress of 256000 N/m2
• The radius of the upper and lower faces are 5 cm and the height is 20 cm
• A) If the specimen is subjected to a force of 100 KN. How long will it be stretched?
• B) Will it be cut using a force of 20000. N
• C) The strain at which this specimen starts its plastic region is at 6 x 10-5 , after
releasing the applied force will it return back or not? And why (by calculations)

28. SOLUTION : A) IF THE SPECIMEN IS SUBJECTED TO A
FORCE OF 100 KN. HOW LONG WILL IT BE STRETCHED?
• Y = stress/strain
• Stress = F/A = 100,000/pi * 0.052 = 12732395.4 N/m2
• Strain = delta (l) / 0.2
• Using Y= 200 Gpa
• Delta (l) = 0.000013 m

29. SOLUTION: B) WILL IT BE CUT USING A FORCE OF 20000 N
• Shear stress = F/A = 20000/pi*0.052
• And given that max shear stress is 256000 N/m2
• As the shear stress is

30. ANSWER: C) THE STRAIN AT WHICH THIS SPECIMEN STARTS ITS
PLASTIC REGION IS AT 6 X 10-5 , AFTER RELEASING THE APPLIED FORCE
WILL IT RETURN BACK OR NOT? AND WHY (BY CALCULATIONS)
• From a) the change in length is found to be 0.000013 m
• The strain = delta (l)/Lo = 0.000065 = 6.5*10-5
• As the strain is higher than the strain at which the specimen enters the plastic
region, it will not return back to its original shape.