Lecture 07 som 04.03.2021
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STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
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Lecture 07 som 04.03.2021
- 1. BIBIN CHIDAMBARANATHAN PRINCIPLE OF SUPERPOSITION BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 2. PRINCIPLE OF SUPERPOSITION ❖ When there are numbers of loads are acting together on an elastic material, the resultant strain will be the sum of individual strains caused by each load acting separately. A B C D 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝑳𝟏 𝑳𝟐 𝑳𝟑 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 3. A B 𝑷𝟏 𝑷𝟐 𝑳𝟏 𝑷𝟑 𝑷𝟒 A B C D 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝑳𝟏 𝑳𝟐 𝑳𝟑 𝐷𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵 𝛿𝑙𝐴𝐵 = 𝑃𝐴𝐵𝐿𝐴𝐵 𝐴𝐴𝐵𝐸𝐴𝐵 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 4. B C 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝑳𝟐 A B C D 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝑳𝟏 𝑳𝟐 𝑳𝟑 𝐷𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐵𝐶 𝛿𝑙𝐵𝐶 = 𝑃𝐵𝐶𝐿𝐵𝐶 𝐴𝐵𝐶 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 5. A B C D 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝑳𝟏 𝑳𝟐 𝑳𝟑 C D 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝑳𝟑 𝐷𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐶𝐷 𝛿𝑙𝐶𝐷 = 𝑃𝐶𝐷𝐿𝐶𝐷 𝐴𝐶𝐷 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 6. 𝑻𝒐𝒕𝒂𝒍 𝐝𝐞𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧 𝜹𝒍 = 𝑷𝑨𝑩𝑳𝑨𝑩 𝑨𝑨𝑩𝑬𝑨𝑩 + 𝑷𝑩𝑪𝑳𝑩𝑪 𝑨𝑩𝑪𝑬𝑩𝑪 + 𝑷𝑪𝑫𝑳𝑪𝑫 𝑨𝑪𝑫𝑬𝑪𝑫 𝑻𝒐𝒕𝒂𝒍 𝐝𝐞𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧 𝜹𝒍 = 𝟏 𝑬 𝑷𝑨𝑩𝑳𝑨𝑩 𝑨𝑨𝑩 + 𝑷𝑩𝑪𝑳𝑩𝑪 𝑨𝑩𝑪 + 𝑷𝑪𝑫𝑳𝑪𝑫 𝑨𝑪𝑫 𝑇𝑜𝑡𝑎𝑙 deformation 𝛿𝑙 = 𝛿𝑙𝐴𝐵 + 𝛿𝑙BC + 𝛿𝑙𝐶𝐷 𝛿𝑙𝐴𝐵 = 𝑃𝐴𝐵𝐿𝐴𝐵 𝐴𝐴𝐵𝐸𝐴𝐵 𝛿𝑙𝐵𝐶 = 𝑃𝐵𝐶𝐿𝐵𝐶 𝐴𝐵𝐶 𝛿𝑙𝐶𝐷 = 𝑃𝐶𝐷𝐿𝐶𝐷 𝐴𝐶𝐷 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 7. Problem 01 A brass bar having cross sectional area of 1000 𝑚𝑚2 is subjected to axial forces as shown below. Find the total elongation of the bar. Take 𝐸 = 1.05 × 105 𝑁/𝑚𝑚2. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 (𝛿𝑙) =? 𝟓𝟎 𝒌𝑵 𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵 𝟏𝟎 𝒌𝑵 𝟔𝟎𝟎 𝒎𝒎 𝟏 𝒎 𝟏. 𝟐 𝒎 A B C D 𝐴 = 1000 𝑚𝑚2 𝐿1 = 600 𝑚𝑚 𝐿2 = 1𝑚 = 1000 𝑚𝑚 𝐿3 = 1.2 𝑚 = 1200 𝑚𝑚 𝐸 = 1.05 × 105 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 8. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝑃1𝐿1 𝐴1𝐸1 + 𝑃2𝐿2 𝐴2𝐸2 + 𝑃3𝐿3 𝐴3𝐸3 𝛿𝑙 = 𝑃1𝐿1 + 𝑃2𝐿2 + 𝑃3𝐿3 𝐴𝐸 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 9. 𝟓𝟎 𝒌𝑵 𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵 𝟏𝟎 𝒌𝑵 𝟔𝟎𝟎 𝒎𝒎 𝟏 𝒎 𝟏. 𝟐 𝒎 A B C D 1 3 2 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐴 = 50 𝑘𝑁 𝑇 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐵 = 80 − 20 − 10 = 50 𝑘𝑁 𝑇 ∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟏, 𝑷𝟏 = 𝟓𝟎 𝒌𝑵 𝑻 A B 𝟓𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵 𝟔𝟎𝟎 𝒎𝒎 𝟏𝟎 𝒌𝑵 𝟖𝟎 𝒌𝑵 1 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 10. 𝟓𝟎 𝒌𝑵 𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵 𝟏𝟎 𝒌𝑵 𝟔𝟎𝟎 𝒎𝒎 𝟏 𝒎 𝟏. 𝟐 𝒎 A B C D 1 3 2 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐵 = 50 − 80 = −30 𝑘𝑁 = 30 𝑘𝑁 𝐶 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐶 = −20 − 10 = −30 𝑘𝑁 = −30 𝑘𝑁 𝐶 ∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟐, 𝑷𝟐 = 𝟑𝟎 𝒌𝑵 𝑪 B C 𝟓𝟎 𝒌𝑵 𝟖𝟎 𝒌𝑵 𝟏𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵 𝟏𝟎𝟎𝟎𝒎𝒎 2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 11. 𝟓𝟎 𝒌𝑵 𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵 𝟏𝟎 𝒌𝑵 𝟔𝟎𝟎 𝒎𝒎 𝟏 𝒎 𝟏. 𝟐 𝒎 A B C D 1 3 2 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝑐 = 50 + 20 − 80 = −10 𝑘𝑁 = 10 𝑘𝑁 𝐶 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐷 = −10 = −10 𝑘𝑁 = −10 𝑘𝑁 𝐶 ∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟑, 𝑷𝟑 = 𝟏𝟎 𝒌𝑵 𝑪 C D 𝟓𝟎 𝒌𝑵 𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵 𝟏𝟎 𝒌𝑵 𝟏𝟐𝟎𝟎 𝒎𝒎 3 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 12. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝑃1𝐿1 + 𝑃2𝐿2 + 𝑃3𝐿3 𝐴𝐸 𝛿𝑙 = 50 × 103 × 600 + −30 × 103 × 1000 + −10 × 103 × 1200 1000 × 1.05 × 105 𝛿𝑙 = −0.114 𝑚𝑚 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝜹𝒍 = 𝟎. 𝟏𝟏𝟒 𝒎𝒎 𝐴 = 1000 𝑚𝑚2 𝐿1 = 600 𝑚𝑚 𝐿2 = 1𝑚 = 1000 𝑚𝑚 𝐿3 = 1.2 𝑚 = 1200 𝑚𝑚 𝐸 = 1.05 × 105 Τ 𝑁 𝑚 𝑚2 𝑃1 = 50 𝑘𝑁 𝑇 𝑃2 = 30 𝑘𝑁 𝐶 𝑃3 = 10 𝑘𝑁 𝐶 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 13. Problem 01 A member ABCD is subjected to point loads 𝑃1, 𝑃2, 𝑃3 and 𝑃4 as shown in the figure. Calculate the force 𝑃2 necessary for equilibrium, if 𝑃1 = 45 𝑘𝑁, 𝑃3 = 450 𝑘𝑁 and 𝑃4 = 130 𝑘𝑁. Determine the total elongation of the member assuming 𝐸 = 2.1 × 105 𝑁/𝑚𝑚2. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑚𝑏𝑒𝑟(𝛿𝑙) =? A B 𝟏𝟐𝟎𝟎 𝒎𝒎 C 𝟔𝟎𝟎 𝒎𝒎 D 𝟗𝟎𝟎 𝒎𝒎 𝟔𝟐𝟓 𝒎𝒎𝟐 𝟐𝟓𝟎𝟎 𝒎𝒎𝟐 1250 𝒎𝒎𝟐 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝐴𝐴𝐵 = 625 𝑚𝑚2 𝐴𝐵𝐶 = 2500 𝑚𝑚2 𝐴𝐶𝐷 = 1250 𝑚𝑚2 𝐿𝐴𝐵 = 1200 𝑚𝑚 𝐿𝐵𝐶 = 600 𝑚𝑚 𝐿𝐶𝐷 = 900 𝑚𝑚 𝑃1 = 45 𝑘𝑁 𝑃3 = 450 𝑘𝑁 𝑃4 = 130 𝑘𝑁 𝐹𝑜𝑟𝑐𝑒 (𝑃2) =? 𝐸 = 2.1 × 105 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 14. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝑃𝐴𝐵𝐿𝐴𝐵 𝐴𝐴𝐵𝐸𝐴𝐵 + 𝑃𝐵𝐶𝐿𝐵𝐶 𝐴𝐵𝐶𝐸𝐵𝐶 + 𝑃𝐶𝐷𝐿𝐶𝐷 𝐴𝐶𝐷𝐸𝐶𝐷 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 1 𝐸 𝑃𝐴𝐵𝐿𝐴𝐵 𝐴𝐴𝐵 + 𝑃𝐵𝐶𝐿𝐵𝐶 𝐴𝐵𝐶 + 𝑃𝐶𝐷𝐿𝐶𝐷 𝐴𝐶𝐷 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 15. A B 𝟏𝟐𝟎𝟎 𝒎𝒎 C 𝟔𝟎𝟎 𝒎𝒎 D 𝟗𝟎𝟎 𝒎𝒎 𝟔𝟐𝟓 𝒎𝒎𝟐 𝟐𝟓𝟎𝟎 𝒎𝒎𝟐 1250 𝒎𝒎𝟐 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑙𝑒𝑓𝑡 = 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑟𝑖𝑔ℎ𝑡 𝑃1 + 𝑃3 = 𝑃2 + 𝑃4 45 + 450 = 𝑃2 + 130 𝑷𝟐 = 𝟑𝟔𝟓 𝒌𝑵 𝑃1 = 45 𝑘𝑁 𝑃3 = 450 𝑘𝑁 𝑃4 = 130 𝑘𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 16. A B 𝟏𝟐𝟎𝟎 𝒎𝒎 C 𝟔𝟎𝟎 𝒎𝒎 D 𝟗𝟎𝟎 𝒎𝒎 𝟔𝟐𝟓 𝒎𝒎𝟐 𝟐𝟓𝟎𝟎 𝒎𝒎𝟐 1250 𝒎𝒎𝟐 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐴 = 45 𝑘𝑁 𝑇 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐵 = 365 − 450 + 130 = 45 𝑘𝑁 𝑇 ∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟏, 𝑷𝑨𝑩 = 𝟒𝟓 𝒌𝑵 𝑻 A B 𝟒𝟓 𝒌𝑵 𝟒𝟓𝟎 𝒌𝑵 𝟏𝟐𝟎𝟎 𝒎𝒎 𝟏𝟑𝟎 𝒌𝑵 𝟑𝟔𝟓 𝒌𝑵 𝟔𝟐𝟓 𝒎𝒎𝟐 𝑃1 = 45 𝑘𝑁 𝑃3 = 450 𝑘𝑁 𝑃4 = 130 𝑘𝑁 𝑃2 = 365 𝑘𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 17. A B 𝟏𝟐𝟎𝟎 𝒎𝒎 C 𝟔𝟎𝟎 𝒎𝒎 D 𝟗𝟎𝟎 𝒎𝒎 𝟔𝟐𝟓 𝒎𝒎𝟐 𝟐𝟓𝟎𝟎 𝒎𝒎𝟐 1250 𝒎𝒎𝟐 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐵 = 45 − 365 = −320 𝑘𝑁 = −320 𝑘𝑁 𝐶 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐶 = −450 + 130 = −320 𝑘𝑁 = −320 𝑘𝑁 𝐶 ∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟐, 𝑷𝑩𝑪 = 𝟑𝟐𝟎 𝒌𝑵 𝑪 B C 𝟒𝟓 𝒌𝑵 𝟑𝟔𝟓 𝒌𝑵 𝟏𝟑𝟎 𝒌𝑵 𝟒𝟓𝟎 𝒌𝑵 𝟔𝟎𝟎 𝒎𝒎 𝟐𝟓𝟎𝟎 𝒎𝒎𝟐 𝑃1 = 45 𝑘𝑁 𝑃3 = 450 𝑘𝑁 𝑃4 = 130 𝑘𝑁 𝑃2 = 365 𝑘𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 18. A B 𝟏𝟐𝟎𝟎 𝒎𝒎 C 𝟔𝟎𝟎 𝒎𝒎 D 𝟗𝟎𝟎 𝒎𝒎 𝟔𝟐𝟓 𝒎𝒎𝟐 𝟐𝟓𝟎𝟎 𝒎𝒎𝟐 1250 𝒎𝒎𝟐 𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝑐 = 45 − 365 + 450 = 130 𝑘𝑁 = 130 𝑘𝑁 𝑇 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐷 = 130 𝑘𝑁 = 130 𝑘𝑁 𝑇 ∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟑, 𝑷𝑪𝑫 = 𝟏𝟑𝟎 𝒌𝑵 (𝑻) C D 𝟒𝟓 𝒌𝑵 𝟑𝟔𝟓 𝒌𝑵 𝟒𝟓𝟎 𝒌𝑵 𝟏𝟑𝟎 𝒌𝑵 𝟗𝟎𝟎 𝒎𝒎 1250 𝒎𝒎𝟐 𝑃1 = 45 𝑘𝑁 𝑃3 = 450 𝑘𝑁 𝑃4 = 130 𝑘𝑁 𝑃2 = 365 𝑘𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 19. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 1 𝐸 𝑃𝐴𝐵𝐿𝐴𝐵 𝐴𝐴𝐵 + 𝑃𝐵𝐶𝐿𝐵𝐶 𝐴𝐵𝐶 + 𝑃𝐶𝐷𝐿𝐶𝐷 𝐴𝐶𝐷 𝛿𝑙 = 1 2.1 × 105 45 × 103 × 1200 625 + −320 × 103 × 600 2500 + 130 × 103 × 1200 1250 𝛿𝑙 = 0.4914 𝑚𝑚 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝜹𝒍 = 𝟎. 𝟒𝟗𝟏𝟒 𝒎𝒎 𝐴𝐴𝐵 = 625 𝑚𝑚2 𝐴𝐵𝐶 = 2500 𝑚𝑚2 𝐴𝐶𝐷 = 1250 𝑚𝑚2 𝐿𝐴𝐵 = 1200 𝑚𝑚 𝐿𝐵𝐶 = 600 𝑚𝑚 𝐿𝐶𝐷 = 900 𝑚𝑚 𝑃𝐴𝐵 = 45 𝑘𝑁 𝑇 𝑃𝐵𝐶 = 320 𝑘𝑁 𝐶 𝑃CD = 130 𝑘𝑁 𝑇 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 20. Problem 01 A tensile load of 40 kN is acting on a rod of diameter 40 mm and of length 4 m. A bore of diameter 20 mm is made centrally on the rod. To what length the rod should be bored so that the total extension will increase 30% under the same tensile load. Take 𝐸 = 2.1 × 105 𝑁/𝑚𝑚2. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑙𝑒𝑛𝑔𝑡ℎ 𝑡ℎ𝑒 𝑟𝑜𝑑 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑏𝑜𝑟𝑒𝑑 (𝑥) =? 𝑃 = 40 𝑘𝑁 = 40 × 103 𝑁 𝐷 = 40 𝑚𝑚 𝐿 = 4 𝑚 = 4000 𝑚𝑚 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝑑𝐵 = 20 𝑚𝑚 𝛿𝑙𝑆𝐵 = 1.3 𝛿𝑙 𝐷S = 40 𝑚𝑚 𝟒𝟎 𝒌𝑵 𝟒 𝒎 𝟒𝟎 𝒌𝑵 𝟒𝟎 𝒌𝑵 𝒙 𝒎 𝟒𝟎 𝒌𝑵 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 21. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝐷2 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝛿𝑙 = 𝑃𝐿 𝐴𝐸 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝑠 = 𝜋 4 × 𝐷𝑆 2 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝐵 = 𝜋 4 × (𝐷𝑆 2 −𝑑𝐵 2 ) 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑢𝑛𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝛿𝑙𝑆 = 𝑃𝐿𝑠 𝐴𝑆𝐸 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝛿𝑙𝐵 = 𝑃𝐿𝐵 𝐴𝐵𝐸 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑓𝑡𝑒𝑟 𝑏𝑜𝑟𝑒 𝑖𝑠 𝑚𝑎𝑑𝑒 𝛿𝑙𝑆𝐵 = 𝛿𝑙𝑆 + 𝛿𝑙𝐵 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 22. 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝐷2 𝐴 = 𝜋 4 × 402 𝑪𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒂𝒓𝒆𝒂 𝑨 = 𝟏𝟐𝟓𝟔. 𝟔𝟒 𝒎𝒎𝟐 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝛿𝑙 = 𝑃𝐿 𝐴𝐸 𝛿𝑙 = 40 × 103 × 4000 1256.64 × 2 × 105 𝑻𝒐𝒕𝒂𝒍 𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 (𝜹𝒍) = 𝟎. 𝟔𝟑𝟔𝟔 𝒎𝒎 𝟒𝟎 𝒌𝑵 𝟒 𝒎 𝟒𝟎 𝒌𝑵 𝑃 = 40 𝑘𝑁 = 40 × 103 𝑁 𝐷 = 40 𝑚𝑚 𝐿 = 4 𝑚 = 4000 𝑚𝑚 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 23. 𝟒𝟎 𝒌𝑵 𝒙 𝒎 𝟒𝟎 𝒌𝑵 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑓𝑡𝑒𝑟 𝑏𝑜𝑟𝑒 𝛿𝑙𝑆𝐵 = 1.3 𝛿𝑙 𝛿𝑙𝑆𝐵 = 1.3 × 0.6366 𝑻𝒐𝒕𝒂𝒍 𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒂𝒇𝒕𝒆𝒓 𝒃𝒐𝒓𝒆 𝜹𝒍𝑺𝑩 = 𝟎. 𝟖𝟐𝟕 𝒎𝒎 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐿𝐵 = 𝑥 𝑚 𝐿𝐵 = 1000𝑥 𝑚𝑚 𝛿𝑙 = 0.6366 𝑚𝑚 𝑳𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒔𝒐𝒍𝒊𝒅 𝒑𝒐𝒓𝒕𝒊𝒐𝒏 𝑳𝑺 = 𝟒𝟎𝟎𝟎 − 𝟏𝟎𝟎𝟎𝒙 𝐿 = 4 𝑚 = 4000 𝑚𝑚 𝟒 𝒎 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 24. 𝟒𝟎 𝒌𝑵 𝒙 𝒎 𝟒𝟎 𝒌𝑵 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝑠 = 𝜋 4 × 𝐷𝑆 2 𝐴𝑆 = 𝜋 4 × 402 𝑨𝑺 = 𝟒𝟎𝟎𝝅 𝒎𝒎𝟐 𝐷𝑆 = 40 𝑚𝑚 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝐵 = 𝜋 4 × (𝐷𝑆 2 −𝑑𝐵 2 ) ቁ 𝐴𝐵 = 𝜋 4 × (40 2 − 202 𝑨𝑩 = 𝟑𝟎𝟎𝝅 𝒎𝒎𝟐 𝑑𝐵 = 20 𝑚𝑚 𝐷S = 40 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 25. 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑢𝑛𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝛿𝑙𝑆 = 𝑃𝐿𝑠 𝐴𝑆𝐸 𝛿𝑙𝑆 = 40 × 103 400𝜋 × 2 × 105 × 4000 − 1000𝑥 𝜹𝒍𝑺 = 𝟒 − 𝒙 𝟐𝝅 𝒎𝒎 𝑃 = 40 𝑘𝑁 = 40 × 103 𝑁 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝐴𝑆 = 400𝜋 𝑚𝑚2 𝐿𝑆 = 4000 − 1000𝑥 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝛿𝑙𝐵 = 𝑃𝐿𝐵 𝐴𝐵𝐸 𝛿𝑙𝐵 = 40 × 103 300𝜋 × 2 × 105 × 1000𝑥 𝜹𝒍𝑩 = 𝟒𝒙 𝟔𝝅 𝒎𝒎 𝐿𝐵 = 1000𝑥 𝑚𝑚 𝐴𝐵 = 300𝜋 𝑚𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 26. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑓𝑡𝑒𝑟 𝑏𝑜𝑟𝑒 𝑖𝑠 𝑚𝑎𝑑𝑒 𝛿𝑙𝑆𝐵 = 𝛿𝑙𝑆 + 𝛿𝑙𝐵 0.827 = 4 − 𝑥 2𝜋 + 4𝑥 6𝜋 𝒍𝒆𝒏𝒈𝒕𝒉 𝒕𝒉𝒆 𝒓𝒐𝒅 𝒔𝒉𝒐𝒖𝒍𝒅 𝒃𝒆 𝒃𝒐𝒓𝒆𝒅 (𝒙) = 𝟑. 𝟔 𝒎 𝛿𝑙𝑆 = 4 − 𝑥 2𝜋 𝑚𝑚 𝛿𝑙𝐵 = 4𝑥 6𝜋 𝑚𝑚 𝛿𝑙𝑆𝐵 = 0.827 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 27. Thank You BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY