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material science 2 .pptx
- 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.
- 4. STRESS • 1- Compressivestress = 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- Shearstress (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.
- 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= 50Kg 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 (BREAKINGSTRESS) • It is the max stress a material can handle before it breaks
- 14. EXAMPLE 6 • Wouldthe 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 STRAININ 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) • 𝑠ℎ𝑒𝑎𝑟𝑠𝑡𝑟𝑒𝑠𝑠 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐹 𝐴 𝑑𝑒𝑙𝑡𝑎 (𝑥) 𝐻 H
- 19. SHEAR STRESS • Itis 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 2.4m Max stress = 𝐹 𝐴 4 x 108 = 𝐹 2.4 ∗0.002 F must be greater than 1920000 N
- 22. EXAMPLE 12 Givens The diameterof 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 𝐵𝑢𝑙𝑘𝑠𝑡𝑟𝑎𝑖𝑛 = 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 • ABowling 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) WILLIT 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) THESTRAIN 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.