Lecture 04 som 26.02.2021
0 likes132 views
1) The document discusses deformation of a body due to forces acting on it. It provides equations for stress, strain, and Young's modulus in terms of load, area, change in length, and original length. 2) Three example problems are then presented to demonstrate calculating stress, strain, Young's modulus, and change in length for bars with different geometries under tensile loads. 3) The key equations from Hooke's law and definitions of stress and strain are used to solve for the unknown values in each problem.
Lecture 04 som 26.02.2021
- 1. BIBIN CHIDAMBARANATHAN Deformation of body due to force acting on it BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 2. Deformation of body due to force acting on it Consider a body subjected to a tensile stress We know 𝑠𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝐿𝑜𝑎𝑑 𝐴𝑟𝑒𝑎 𝜎 = 𝑃 𝐴 → 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (1) 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑒 = 𝛿𝐿 𝐿 → 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2) From Hooke’s law, 𝑌𝑜𝑢𝑛𝑔′ 𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 (𝐸) = 𝑆𝑡𝑟𝑒𝑠𝑠 𝑆𝑡𝑟𝑎𝑖𝑛 𝐸 = 𝜎 𝑒 → 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (3) BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 3. Deformation of body due to force acting on it Substitute equation (1) and (2) in (3) 𝐸 = ൗ 𝑃 𝐴 ൗ 𝛿𝐿 𝐿 𝐸 = 𝑃 𝐿 𝐴 𝛿𝐿 𝛿𝑙 = 𝑃 𝐿 𝐴 𝐸 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 (𝜹𝒍) = 𝑷 𝑳 𝑨 𝑬 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 4. Problem 01 A circular cross-sectional bar of diameter 10mm having a length of 1m is subjected by an axial pull of 10KN. Determine stress, strain and young’s modulus of the material. Change in length due to load is 0.01mm. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑑 = 10 𝑚𝑚 𝑙 = 1𝑚 = 1000 𝑚𝑚 𝑃 = 10 𝑘𝑁 = 10 × 103 𝑁 𝛿𝑙 = 0.01 𝑚𝑚 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 =? 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 =? 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 5. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 6. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝐴 = 𝜋 4 × 102 𝑨𝒓𝒆𝒂 𝑨 = 𝟕𝟖. 𝟓𝟒 𝒎𝒎𝟐 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝜎 = 10 × 103 78.54 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝟏𝟐𝟕. 𝟑𝟖 Τ 𝑵 𝒎 𝒎𝟐 𝑑 = 10 𝑚𝑚 𝑃 = 10 𝑘𝑁 = 10 × 103 𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 7. 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑒 = 0.01 1000 𝑺𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝟏 × 𝟏𝟎−𝟓 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 𝐸 = 127.38 1 × 10−5 𝒀𝒐𝒖𝒏𝒈′𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑬 = 𝟏𝟐𝟕. 𝟑𝟖 × 𝟏𝟎𝟓 Τ 𝑵 𝒎 𝒎𝟐 𝑙 = 1𝑚 = 1000 𝑚𝑚 𝛿𝑙 = 0.01 𝑚𝑚 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 127.38 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 8. Problem 02 A rectangular bar of 10mm breath and 12 mm depth is having length of 150 mm. It is subjected to a tensile load of 20 KN. If the young’s modulus is 2 × 105 𝑁/𝑚𝑚2. Determine the change in length of the bar. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑏 = 10 𝑚𝑚 𝑑 = 12 𝑚𝑚 𝑙 = 150 𝑚𝑚 𝑃 = 20 𝑘𝑁 = 20 × 103 𝑁 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 𝛿𝑙 =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 9. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝐴𝑟𝑒𝑎 𝐴 = 𝑙 × 𝑏 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 10. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐴𝑟𝑒𝑎 𝐴 = 𝑙 × 𝑏 𝐴 = 10 × 12 𝑨𝒓𝒆𝒂 𝑨 = 𝟏𝟐𝟎 𝒎𝒎𝟐 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝜎 = 20 × 103 120 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝟏𝟔𝟔. 𝟔𝟔 Τ 𝑵 𝒎 𝒎𝟐 𝑏 = 10 𝑚𝑚 𝑙 = 150 𝑚𝑚 𝑃 = 20 𝑘𝑁 = 20 × 103 𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 11. 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 2 × 105 = 166.66 𝑒 𝑺𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝟖. 𝟑𝟑 × 𝟏𝟎−𝟒 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 8.33 × 10−4 = 𝛿𝑙 150 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒓 𝜹𝒍 = 𝟏. 𝟐𝟓 𝒎𝒎 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 166.66 Τ 𝑁 𝑚 𝑚2 𝑙 = 150 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 12. Problem 03 Determine the diameter of the circular bar if the bar is having a length of 1m subjected to a tensile load of 10KN. The young’s modulus of the material is 2 × 105 𝑁/𝑚𝑚2 and the elongation of the material is 0.001mm. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑙 = 1 𝑚 = 1000 𝑚𝑚 𝑃 = 10 𝑘𝑁 = 10 × 103 𝑁 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝛿𝑙 = 0.001 𝑚𝑚 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑑 =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 13. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑌𝑜𝑢𝑛𝑔′ 𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 14. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑒 = 0.001 1000 𝑺𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝟏 × 𝟏𝟎−𝟔 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 2 × 105 = 𝜎 1 × 10−6 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝟎. 𝟐 Τ 𝑵 𝒎 𝒎𝟐 𝑙 = 1 𝑚 = 1000 𝑚𝑚 𝛿𝑙 = 0.001 𝑚𝑚 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 15. 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 0.2 = 10 × 103 𝐴 𝑨𝒓𝒆𝒂 𝑨 = 𝟓𝟎𝟎𝟎𝟎 𝒎𝒎𝟐 50000 = 𝜋 4 × 𝑑2 𝑑 = 252.31 𝑚𝑚 𝑫𝒊𝒂𝒎𝒆𝒕𝒆𝒓 𝒅 = 𝟐𝟓𝟐. 𝟑𝟏 𝒎𝒎 𝑃 = 10 𝑘𝑁 = 10 × 103 𝑁 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 0.2 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 16. Problem 04 A bar of 30mm diameter is subjected to a pull of 60KN. The measured extension on gauge length of 200mm is 0.1 mm and change in diameter is 0.004mm. Calculate (i) Young’s modulus, (ii) Poisson’s Ratio and (iii) Bulk modulus. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑑 = 30 𝑚𝑚 𝑙 = 200 𝑚𝑚 𝛿𝑙 = 0.1 𝑚𝑚 𝛿𝑑 = 0.004 𝑚𝑚 𝑃 = 60 𝑘𝑁 = 60 × 103 𝑁 𝑌𝑜𝑢𝑛𝑔′ 𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 =? 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′ 𝑠 𝑅𝑎𝑡𝑖𝑜 𝜇 =? 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 17. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑌𝑜𝑢𝑛𝑔′ 𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′𝑠 𝑅𝑎𝑡𝑖𝑜 𝜇 = 𝑒𝑡 𝑒𝑙 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 = 𝛿𝑑 𝑑 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙 = 𝛿𝑙 𝑙 𝑇ℎ𝑒 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑌𝑜𝑢𝑛𝑔’𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 (𝐸), 𝑏𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 (𝐾) 𝑎𝑛𝑑 𝑃𝑜𝑖𝑠𝑠𝑜𝑛’𝑠 𝑟𝑎𝑡𝑖𝑜 (µ) 𝑖𝑠 𝐸 = 3𝐾(1 − 2𝜇) BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 18. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑒 = 0.1 200 𝑺𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝟓 × 𝟏𝟎−𝟒 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝐴 = 𝜋 4 × 302 𝑨𝒓𝒆𝒂 𝑨 = 𝟕𝟎𝟔. 𝟖𝟓𝟖 𝒎𝒎𝟐 𝑑 = 30 𝑚𝑚 𝑙 = 200 𝑚𝑚 𝛿𝑙 = 0.1 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 19. 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝜎 = 60 × 103 706.858 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝟖𝟒. 𝟖𝟖𝟐 Τ 𝑵 𝒎 𝒎𝟐 𝐸 = 84.882 5 × 10−4 𝒀𝒐𝒖𝒏𝒈′𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑬 = 𝟏. 𝟔𝟗𝟕 × 𝟏𝟎𝟓 Τ 𝑵 𝒎 𝒎𝟐 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 𝑃 = 60 𝑘𝑁 = 60 × 103 𝑁 𝐴𝑟𝑒𝑎 𝐴 = 706.858 𝑚𝑚2 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 5 × 10−4 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 20. 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 = 𝛿𝑑 𝑑 𝑒𝑡 = 0.004 30 𝑳𝒂𝒕𝒆𝒓𝒂𝒍 𝒔𝒕𝒓𝒂𝒊𝒏 𝒆𝒕 = 𝟏. 𝟑𝟑 × 𝟏𝟎−𝟒 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙 = 𝛿𝑙 𝑙 𝑒𝑙 = 0.1 200 𝑳𝒐𝒏𝒈𝒊𝒕𝒖𝒅𝒊𝒏𝒂𝒍 𝒔𝒕𝒓𝒂𝒊𝒏 𝒆𝒍 = 𝟓 × 𝟏𝟎−𝟒 𝑑 = 30 𝑚𝑚 𝑙 = 200 𝑚𝑚 𝛿𝑙 = 0.1 𝑚𝑚 𝛿𝑑 = 0.004 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 21. 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′ 𝑠 𝑅𝑎𝑡𝑖𝑜 𝜇 = 𝑒𝑡 𝑒𝑙 𝜇 = 1.33 × 10−4 5 × 10−4 𝑷𝒐𝒊𝒔𝒔𝒊𝒐𝒏′ 𝒔 𝑹𝒂𝒕𝒊𝒐 𝝁 = 𝟎. 𝟐𝟔 𝐸 = 3𝐾 1 − 2𝜇 1.697 × 105 = 3𝐾 1 − 2 × 0.26 𝑩𝒖𝒍𝒌 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑲 = 𝟏. 𝟐𝟐 × 𝟏𝟎𝟓 Τ 𝑵 𝒎 𝒎𝟐 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 = 1.33 × 10−4 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙 = 5 × 10−4 𝐸 = 1.697 × 105 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 22. Problem 05 The following data refer to a mild steel specimen in a laboratory. Diameter of the specimen = 25mm Length of the specimen = 300mm Extension of bar under load of 15KN= 0.045mm Load at yield point = 127.65KN Maximum load = 208.60KN Length of specimen at failure = 375mm Neck diameter = 17.75mm Determine Young’s modulus, yield point stress, ultimate stress, percentage elongation, percentage reduction in area, safe stress with Factor of safety of 2. BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 23. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑑 = 25 𝑚𝑚 𝑙 = 300 𝑚𝑚 𝑃 = 15 𝑘𝑁 = 15 × 103 𝑁 𝑙 = 300 𝑚𝑚 𝛿𝑙 = 0.045 𝑚𝑚 𝑌𝑖𝑒𝑙𝑑 𝑙𝑜𝑎𝑑 𝑃𝑦𝑖𝑒𝑙𝑑 = 127.65 𝐾𝑁 = 127.65 × 103 𝑁 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑙𝑜𝑎𝑑 𝑃𝑚𝑎𝑥 = 208.6 𝐾𝑁 = 208.6 × 103 𝑁 𝐹𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑙𝑓𝑖𝑛𝑎𝑙 = 375 𝑚𝑚 𝑁𝑒𝑐𝑘 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑑1 = 17.75 𝑚𝑚 𝐹𝑂𝑆 = 2 𝑠𝑎𝑓𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 =? 𝑌𝑜𝑢𝑛𝑔’𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 =? 𝑦𝑖𝑒𝑙𝑑 𝑝𝑜𝑖𝑛𝑡 𝑠𝑡𝑟𝑒𝑠𝑠 = ? 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 = ? 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 = ? 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑎𝑟𝑒𝑎 = ? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 24. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑌𝑜𝑢𝑛𝑔′ 𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 𝑌𝑖𝑒𝑙𝑑 𝑝𝑜𝑖𝑛𝑡 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎𝑦𝑖𝑒𝑙𝑑 = 𝑃𝑦𝑖𝑒𝑙𝑑 𝐴 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 = 𝑃𝑚𝑎𝑥 𝐴 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 25. 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 = 𝐹𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ − 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ × 100% 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑎𝑟𝑒𝑎 = 𝑂𝑟𝑖𝑔𝑖𝑛𝑙 𝑎𝑟𝑒𝑎 − 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝑎𝑟𝑒𝑎 𝑂𝑟𝑖𝑔𝑖𝑛𝑙 𝑎𝑟𝑒𝑎 × 100% 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝐴𝑟𝑒𝑎 𝐴1 = 𝜋 4 × 𝑑1 2 𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝐹𝑂𝑆 = 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠𝑎𝑓𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 26. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝐴 = 𝜋 4 × 252 𝑨𝒓𝒆𝒂 𝑨 = 𝟒𝟗𝟎. 𝟖𝟕 𝒎𝒎𝟐 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝑃 𝐴 𝜎 = 15 × 103 490.87 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝟑𝟎. 𝟓𝟓 Τ 𝑵 𝒎 𝒎𝟐 𝑑 = 25 𝑚𝑚 𝑃 = 15 𝑘𝑁 = 15 × 103 𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 27. 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝑙 𝑒 = 0.045 300 𝑺𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝟏. 𝟓 × 𝟏𝟎−𝟒 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 = 𝜎 𝑒 𝐸 = 30.55 1.5 × 10−4 𝒀𝒐𝒖𝒏𝒈′𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑬 = 𝟐. 𝟎𝟑 × 𝟏𝟎𝟓 Τ 𝑵 𝒎 𝒎𝟐 𝜎 = 30.55 Τ 𝑁 𝑚 𝑚2 𝑙 = 300 𝑚𝑚 𝛿𝑙 = 0.045 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 28. 𝑌𝑖𝑒𝑙𝑑 𝑝𝑜𝑖𝑛𝑡 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎𝑦𝑖𝑒𝑙𝑑 = 𝑃𝑦𝑖𝑒𝑙𝑑 𝐴 𝜎𝑦𝑖𝑒𝑙𝑑 = 127.65 × 103 490.87 𝒀𝒊𝒆𝒍𝒅 𝒑𝒐𝒊𝒏𝒕 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈𝒚𝒊𝒆𝒍𝒅 = 𝟐𝟔𝟎. 𝟏𝟕 Τ 𝑵 𝒎 𝒎𝟐 𝑈𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 = 𝑃𝑚𝑎𝑥 𝐴 𝜎𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 = 208.6 × 103 490.87 𝑼𝒍𝒕𝒊𝒎𝒂𝒕𝒆 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈𝒖𝒍𝒕𝒊𝒎𝒂𝒕𝒆 = 𝟒𝟐𝟓. 𝟏𝟕 Τ 𝑵 𝒎 𝒎𝟐 𝐴𝑟𝑒𝑎 𝐴 = 490.87 𝑚𝑚2 𝑃𝑦𝑖𝑒𝑙𝑑 = 127.65 𝐾𝑁 = 127.65 × 103 𝑁 𝑃𝑚𝑎𝑥 = 208.6 𝐾𝑁 = 208.6 × 103 𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 29. 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 = 𝐹𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ − 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ × 100% % 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 = 375 − 300 375 × 100% % 𝒆𝒍𝒐𝒏𝒈𝒂𝒕𝒊𝒐𝒏 = 𝟐𝟓 % 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝐴𝑟𝑒𝑎 𝐴1 = 𝜋 4 × 𝑑1 2 𝐴1 = 𝜋 4 × 17.752 𝑹𝒆𝒅𝒖𝒄𝒆𝒅 𝑨𝒓𝒆𝒂 𝑨𝟏 = 𝟐𝟒𝟕. 𝟑𝟐 𝒎𝒎𝟐 𝑙 = 300 𝑚𝑚 𝐹𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑙𝑓𝑖𝑛𝑎𝑙 = 375 𝑚𝑚 𝑁𝑒𝑐𝑘 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑑1 = 17.75 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 30. 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑎𝑟𝑒𝑎 = 𝑂𝑟𝑖𝑔𝑖𝑛𝑙 𝑎𝑟𝑒𝑎 − 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝑎𝑟𝑒𝑎 𝑂𝑟𝑖𝑔𝑖𝑛𝑙 𝑎𝑟𝑒𝑎 × 100% % 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑎𝑟𝑒𝑎 = 490.6 − 247.32 490.6 × 100% % 𝒓𝒆𝒅𝒖𝒄𝒕𝒊𝒐𝒏 𝒊𝒏 𝒂𝒓𝒆𝒂 = 𝟒𝟗. 𝟓𝟗 % 𝐹𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑠𝑎𝑓𝑒𝑡𝑦 𝐹𝑂𝑆 = 𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠𝑎𝑓𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 2 = 260.17 𝑠𝑎𝑓𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑺𝒂𝒇𝒆 𝒔𝒕𝒓𝒆𝒔𝒔 𝝈𝒔𝒂𝒇𝒆 = 𝟏𝟑𝟎. 𝟎𝟖 Τ 𝑵 𝒎 𝒎𝟐 𝐴𝑟𝑒𝑎 𝐴 = 490.87 𝑚𝑚2 𝑅𝑒𝑑𝑢𝑐𝑒𝑑 𝐴𝑟𝑒𝑎 𝐴1 = 247.32 𝑚𝑚2 𝐹𝑂𝑆 = 2 𝜎𝑦𝑖𝑒𝑙𝑑 = 260.17 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 31. Thank You BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY