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material science 2 .pptx

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This document discusses material science concepts including stress, strain, Young's modulus, shear modulus, and bulk modulus. It provides definitions and formulas for these terms and includes examples of calculating stress, strain, elongation, shear stress, and changes in volume using given material properties and applied forces. Several practice problems are worked through applying these concepts to situations like stretching a cylinder, cutting a sheet, and determining changes to a bowling ball underwater.

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MATERIAL SCIENCE ENG. KAREEM. H. MOKHTAR
STRESS AND STRAIN Stress = Strain = ＄ ＄ = Y (youngs modulus) F A eL = DL L o Big Small Weak material Strong material
STRESS
STRESS 1- Compressive stress = F A o 2 m N
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)
STRESS 3- Shear stress (t) Area, A Ft Ft Fs F F Fs t = Fs Ao
STRAIN e = d Lo d/2 dL/2 Lo wo Lateral strain: eL = L wo d Tensile strain:
STRAIN Shear strain g = Dx/H H
YOUNGS MODULUS ＄ ＄ = 瑞 D
EXAMPLE 1 Givens Weight = 1000N Area of the beam is 2cm x 2cm Lo = 1.75 m Find delta L Material Fe youngs modulus Y = ＄ ＄ = 瑞 = 1000 2 2 104 ? 1.75 D
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 4 0.012 = 1.87 106 /2
EXAMPLE 4 Givens D= 1mm Lo= 4m m= 500 kg Find the elongation youngs modulus (AL) = 6.9* 1010 N/m2 Solution youngs modulus= ＄ ＄ = 瑞 = 5009.8 4 106 ? 4 = 6.9* 1010
MAX STRESS (BREAKING STRESS) It is the max stress a material can handle before it breaks
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
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
EXAMPLE 3 F = ? Strain = 1% L= 1cm Steel youngs modulus = 21x1010 N/m2 Solution 21 MPa
EXAMPLE 8 Lo = 40m Dia= 1cm Delta(L) = 2m Weight = 900N climber Y = ? Solution Y=229,183,118 N/m2
SHEAR MODULUS (S) ＄ ＄ = ＄ () H
SHEAR STRESS It is used in cutting materials The area used is the area that is going to be cut
EXAMPLE 9 H = 5 cm W= 20 cm L= 2cm F= 1000N Carbon steel shear modulus= 7.7 * 1010 W H ＄ ＄ = ＄ () ＄ () = 1.67 * 10-7 m
EXAMPLE 10 5 m 2.4 m Max stress = 4 x 108 = 2.4 0.002 F must be greater than 1920000 N
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
BULK MODULUS 巨 ＄ 巨 ＄ = ＄ () Compressibility = 1 巨 The higher the bulk modulus the more force you need to Change the volume of the material
EXAMPLE 13 A Bowling ball made of steel sunk in an ocean, find the change in its volume if you know that the ocean is 10000 m in depth and the original volume of the ball is 1m3 巨 ＄ 巨 ＄ = ＄ () = Bulk modulus
EXAMPLE 14 Compressibility = ＄ () Substitute Compressibility = 3 x 10-10 m3/ N
EXAMPLE 15 Shear modulus (S) = ＄ ＄ = ＄ ()
SOLVE THIS QUESTION A carbon steel cylinder that has a youngs 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)
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
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
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.
material science 2 .pptx

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material science 2 .pptx

  • 2. STRESS AND STRAIN Stress = Strain = ＄ ＄ = Y (youngs modulus) F A eL = DL L o Big Small Weak material Strong material
  • 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:
  • 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 Y = ＄ ＄ = 瑞 = 1000 2 2 104 ? 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 4 0.012 = 1.87 106 /2
  • 12. EXAMPLE 4 Givens D= 1mm Lo= 4m m= 500 kg Find the elongation youngs modulus (AL) = 6.9* 1010 N/m2 Solution youngs modulus= ＄ ＄ = 瑞 = 5009.8 4 106 ? 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) ＄ ＄ = ＄ () 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 ＄ ＄ = ＄ () ＄ () = 1.67 * 10-7 m
  • 21. EXAMPLE 10 5 m 2.4 m Max stress = 4 x 108 = 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 巨 ＄ 巨 ＄ = ＄ () Compressibility = 1 巨 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 its volume if you know that the ocean is 10000 m in depth and the original volume of the ball is 1m3 巨 ＄ 巨 ＄ = ＄ () = Bulk modulus
  • 25. EXAMPLE 14 Compressibility = ＄ () Substitute Compressibility = 3 x 10-10 m3/ N
  • 26. EXAMPLE 15 Shear modulus (S) = ＄ ＄ = ＄ ()
  • 27. SOLVE THIS QUESTION A carbon steel cylinder that has a youngs 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.

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