Lecture 06 som 03.03.2021
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Lecture 06 som 03.03.2021
- 1. BIBIN CHIDAMBARANATHAN ELONGATION OF UNIFORMLY TAPERING CIRCULAR ROD BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 2. Prerequisite A D B A’ D’ B’ E E’ C C’ 𝐷1 𝐷2 𝐷𝑥 𝑥 L 𝐵𝐶 = 𝐷1 − 𝐷2 2 𝐷1 = 𝐵𝐶 + 𝐶𝐶′ + B′ C’ 𝐷1 − 𝐷2 = 𝐵𝐶 + B′ C’ 𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡, 𝐵𝐶 = 𝐵′𝐶′ 𝐷1 − 𝐷2 = 2𝐵𝐶 𝐷1 = 𝐵𝐶 + 𝐷2 + B′C’ BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 3. A D B A’ D ’ B’ E E’ C C’ 𝐷1 𝐷2 𝐷𝑥 𝑥 L 𝐷𝐸 𝑥 = 𝐷1 − 𝐷2 2 𝐿 𝐷𝐸 = 𝐷1 − 𝐷2 2 𝐿 𝑥 → 𝑒𝑞𝑛 (1) 𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑡ℎ𝑒 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑜𝑓 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒𝑠, ∆𝐴𝐷𝐸 𝑎𝑛𝑑 ∆𝐴𝐵𝐶 𝑎𝑟𝑒 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝐷𝐸 𝑥 = 𝐵𝐶 𝐿 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 4. Prerequisite A D B A’ D’ B’ E E’ C C’ 𝐷1 𝐷2 𝐷𝑥 𝑥 L 𝐷1 = 𝐷𝐸 + 𝐷x + D′ E’ 𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡, 𝐷𝐸 = 𝐷′𝐸′ 𝐷1 = 𝐷𝐸 + 𝐸𝐸′ + D′E’ 𝐷x = 𝐷1 − 𝐷𝐸 − D′E’ 𝐷x = 𝐷1 − 𝐷𝐸 − 𝐷𝐸 𝐷x = 𝐷1 − 2𝐷𝐸 → 𝑒𝑞𝑛 (2) BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 5. 𝐷𝑥 = 𝐷1 − 2𝐷𝐸 𝐷𝑥 = 𝐷1 − 2 𝐷1 − 𝐷2 2 𝐿 𝑥 𝑫𝒙 = 𝑫𝟏 − 𝑫𝟏 − 𝑫𝟐 𝑳 𝒙 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝐷𝐸 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2) BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 6. ELONGATION OF UNIFORMLY TAPERING CIRCULAR ROD ❖ Let us consider the uniformly tapering circular rod ❖ length of the uniformly tapering circular rod is 𝐿 ❖ larger diameter of the rod is 𝐷1 at one end ❖ circular rod will be uniformly tapered and hence other end diameter of the circular rod will be smaller L 0 𝑥 𝑑𝑥 P P 𝐷1 𝐷2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 7. ❖ let us assume that diameter of other end is 𝐷2. ❖ Uniformly tapering circular rod is subjected with an axial tensile load 𝑃 ❖ Let us consider one infinitesimal smaller element of length 𝑑𝑥 and its diameter will be at a distance 𝑥 from its larger diameter end. L 0 𝑥 𝑑𝑥 P P 𝐷1 𝐷2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 8. Let us consider that diameter of infinitesimal smaller element is 𝐷𝑥 Diameter ൯ 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑟𝑜𝑑 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑥 (𝐷𝑥 = 𝐷1 − 𝐷1 − 𝐷2 𝐿 𝑥 𝐷𝑥 = 𝐷1 − 𝑘𝑥 𝑊ℎ𝑒𝑟𝑒, 𝑘 = 𝐷1 − 𝐷2 𝐿 Area of cross section Area of cross section of circular bar at a distance x from its larger diameter end is 𝐴𝑥 ൯ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 (𝐴𝑥 = 𝜋 4 × 𝐷𝑥 2 𝐴𝑥 = 𝜋 4 × 𝐷1 − 𝑘𝑥 2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 9. Stress Let us consider that stress induced in circular bar at a distance x from its larger diameter end is 𝜎𝑥 ൯ 𝑆𝑡𝑟𝑒𝑠𝑠 (𝜎𝑥 = 𝐿𝑜𝑎𝑑 𝐴𝑟𝑒𝑎 𝜎𝑥 = 𝑃 𝐴𝑥 𝜎𝑥 = 𝑃 𝜋 4 × 𝐷1 − 𝑘𝑥 2 𝜎𝑥 = 4𝑃 𝜋 𝐷1 − 𝑘𝑥 2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 10. Strain Let us consider that strain induced in circular bar at a distance 𝑥 from its larger diameter end is 𝑒𝑥 ൯ 𝑆𝑡𝑟𝑎𝑖𝑛 (𝑒𝑥 = 𝑠𝑡𝑟𝑒𝑠𝑠 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑒𝑥 = 𝜎𝑥 𝐸 𝑒𝑥 = 4𝑃 𝜋 𝐷1 − 𝑘𝑥 2 𝐸 𝑒𝑥 = 4𝑃 𝜋𝐸 𝐷1 − 𝑘𝑥 2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 11. Change in length of infinitesimal smaller element Change in length of infinitesimal smaller element will be determined by recalling the concept of strain. 𝛿𝑙𝑥 = 𝑒𝑥 𝑑𝑥 𝛿𝑙𝑥 = 4𝑃 𝜋𝐸 𝐷1 − 𝑘𝑥 2 𝑑𝑥 Total change in length The total change in length of the uniformly tapering circular rod is obtained by integrating the above equation from 0 to L. 𝛿𝑙 = න 0 𝐿 4𝑃 𝜋𝐸 𝐷1 − 𝑘𝑥 2 𝑑𝑥 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 12. 𝛿𝑙 = 4𝑃 πE න 0 𝐿 1 𝐷1 − 𝑘𝑥 2 𝑑𝑥 𝛿𝑙 = 4𝑃 πE න 0 𝐿 𝐷1 − 𝑘𝑥 −2 𝑑𝑥 𝛿𝑙 = 4𝑃 πE 𝐷1 − 𝑘𝑥 −2+1 −2 + 1 −𝑘 0 𝐿 𝛿𝑙 = 4𝑃 πE 𝐷1 − 𝑘𝑥 −1 𝑘 0 𝐿 𝛿𝑙 = 4𝑃 πEk 1 𝐷1 − 𝑘𝑥 0 𝐿 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 13. 𝛿𝑙 = 4𝑃 πEk 1 𝐷1 − 𝑘𝐿 − 1 𝐷1 − 0 𝛿𝑙 = 4𝑃 πEk 1 𝐷1 − 𝑘𝐿 − 1 𝐷1 Substituting 𝑘 = 𝐷1−𝐷2 𝐿 in the above equation 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝛿𝑙 = 4𝑃 πE 𝐷1 − 𝐷2 𝐿 1 𝐷1 − 𝐷1 − 𝐷2 𝐿 𝐿 − 1 𝐷1 𝛿𝑙 = 4𝑃𝐿 πE 𝐷1 − 𝐷2 1 𝐷1 − 𝐷1 + 𝐷2 − 1 𝐷1 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 14. 𝛿𝑙 = 4𝑃𝐿 π E 𝐷1 − 𝐷2 1 𝐷2 − 1 𝐷1 𝛿𝑙 = 4𝑃𝐿 π E 𝐷1 − 𝐷2 𝐷1 − 𝐷2 𝐷1 𝐷2 𝜹𝒍 = 𝟒𝑷𝑳 𝛑 𝐄 𝑫𝟏 𝑫𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 15. Problem 01 Find the modulus of elasticity of the rod, which tapers uniformly from 30mm to 15mm diameter in a length of 350mm. The rod is subjected to an axial load of 5.5KN and extension of the rod is 0.025mm. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝐷1 𝐷2 L 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝐸 =? 𝐷1 = 30 𝑚𝑚 𝐷2 = 15 𝑚𝑚 𝐿 = 350 𝑚𝑚 𝑃 = 5.5 𝑘𝑁 = 5.5 × 103 𝑁 𝛿𝑙 = 0.025 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 16. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑑 𝛿𝑙 = 4𝑃𝐿 𝜋 𝐸 𝐷1 𝐷2 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑑 𝛿𝑙 = 4𝑃𝐿 𝜋 𝐸 𝐷1 𝐷2 0.025 = 4 × 5.5 × 103 × 350 𝜋 × 𝐸 × 30 × 15 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝒐𝒇 𝒆𝒍𝒂𝒔𝒕𝒊𝒄𝒊𝒕𝒚 𝑬 = 𝟐. 𝟏𝟕𝟗 × 𝟏𝟎𝟓 Τ 𝑵 𝒎 𝒎𝟐 𝐷1 = 30 𝑚𝑚 𝐷2 = 15 𝑚𝑚 𝐿 = 350 𝑚𝑚 𝑃 = 5.5 𝑘𝑁 = 5.5 × 103 𝑁 𝛿𝑙 = 0.025 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 17. BIBIN CHIDAMBARANATHAN ELONGATION OF UNIFORMLY TAPERING RECTANGULAR BAR BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 18. ELONGATION OF UNIFORMLY TAPERING RECTANGULAR BAR ❖ A rectangular bar is basically taper uniformly from one end to another end throughout the length and therefore its one end will be of larger width and other end will be of smaller width. ❖ However, thickness of the rectangular rod will be constant throughout the length of bar. 𝑎 𝑏 L 𝑥 𝑑𝑥 0 P P 𝑡 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 19. ❖ Let us consider the uniformly tapering rectangular bar. ❖ length of the uniformly tapering rectangular bar is 𝐿. ❖ Width of larger end of the bar is ‘𝑎’ ❖ Rectangular bar will be uniformly tapered and hence width of the rectangular bar at the other end will be smaller. 𝑎 𝑏 L 𝑥 𝑑𝑥 0 P P BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 20. ❖ let us assume that width of another end is ‘ 𝑏 ’. Let us assume that thickness of the rectangular bar is ′𝑡′. ❖ Let us consider that uniformly tapering rectangular bar is subjected with an axial tensile load 𝑃. ❖ Let us consider one infinitesimal smaller element of length 𝑑𝑥 and its width will be at a distance 𝑥 from its larger end. 𝑎 𝑏 L 𝑥 𝑑𝑥 0 P P BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 21. Width Let us consider that width of infinitesimal smaller element is 𝐵𝑥 ൯ 𝑊𝑖𝑑𝑡ℎ 𝑜𝑓 𝑏𝑎𝑟 𝑎𝑡 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑥 (𝐵𝑥 = 𝑎 − 𝑎 − 𝑏 𝐿 𝑥 𝐵𝑥 = 𝑎 − 𝑘𝑥 𝑊ℎ𝑒𝑟𝑒, 𝑘 = 𝑎 − 𝑏 𝐿 Area of cross section Area of cross section of bar at a distance 𝑥 from its larger end is 𝐴𝑥 ൯ 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 (𝐴𝑥 = 𝑊𝑖𝑑𝑡ℎ × 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝐴𝑥 = 𝐵𝑥 × 𝑡 𝐴𝑥 = 𝑎 − 𝑘𝑥 × 𝑡 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 22. Stress Let us consider that stress induced in bar at a distance 𝑥 from its larger end is 𝜎𝑥 ൯ 𝑆𝑡𝑟𝑒𝑠𝑠 (𝜎𝑥 = 𝐿𝑜𝑎𝑑 𝐴𝑟𝑒𝑎 = 𝑃 𝐴𝑥 𝜎𝑥 = 𝑃 𝑎 − 𝑘𝑥 × 𝑡 Strain Let us consider that strain induced in bar at a distance x from its larger diameter end is 𝑒𝑥 ൯ 𝑆𝑡𝑟𝑎𝑖𝑛 (𝑒𝑥 = 𝑠𝑡𝑟𝑒𝑠𝑠 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 = 𝜎𝑥 𝐸 𝑒𝑥 = 𝑃 𝑎 − 𝑘𝑥 × 𝑡 𝐸 𝑒𝑥 = 𝑃 𝐸𝑡 𝑎 − 𝑘𝑥 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 23. Change in length of infinitesimal smaller element Change in length of infinitesimal smaller element will be determined by recalling the concept of strain. 𝛿𝑙𝑥 = 𝑒𝑥 𝑑𝑥 𝛿𝑙𝑥 = 𝑃 𝐸𝑡 𝑎 − 𝑘𝑥 𝑑𝑥 Total change in length The total change in length of the uniformly tapering rectangular bar is obtained by integrating the above equation from 0 to L. 𝛿𝑙 = න 0 𝐿 𝑃 𝐸𝑡 𝑎 − 𝑘𝑥 𝑑𝑥 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 24. 𝛿𝑙 = 𝑃 Et න 0 𝐿 1 𝑎 − 𝑘𝑥 𝑑𝑥 𝛿𝑙 = 𝑃 Et 𝑙𝑛 𝑎 − 𝑘𝑥 0 𝐿 × − 1 𝑘 𝛿𝑙 = −𝑃 kEt 𝑙𝑛 𝑎 − 𝑘𝐿 − 𝑙 𝑛 𝑎 𝛿𝑙 = −𝑃 kEt 𝑙𝑛 𝑎 − 𝑘𝐿 − 𝑙 𝑛(𝑎 − 0) 𝛿𝑙 = 𝑃 kEt 𝑙𝑛 𝑎 − 𝑙𝑛 𝑎 − 𝑘𝐿 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 25. Substituting 𝑘 = 𝑎−𝑏 𝐿 in the above equation 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝛿𝑙 = 𝑃 𝑎 − 𝑏 𝐿 Et 𝑙𝑛 𝑎 − 𝑙𝑛 𝑎 − 𝑎 − 𝑏 𝐿 𝐿 𝛿𝑙 = 𝑃𝐿 𝑎 − 𝑏 Et ൯ 𝑙𝑛 𝑎 − 𝑙𝑛 𝑎 − (𝑎 − 𝑏 𝛿𝑙 = 𝑃𝐿 𝑎 − 𝑏 Et 𝑙𝑛 𝑎 − 𝑙𝑛 𝑎 − 𝑎 + 𝑏 𝛿𝑙 = 𝑃𝐿 𝑎 − 𝑏 Et 𝑙𝑛 𝑎 − 𝑙𝑛 𝑏 𝜹𝒍 = 𝑷𝑳 𝒂 − 𝒃 𝐄𝐭 𝒍𝒏 (𝒂/𝒃) BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 26. Problem 01 A rectangular steel bar of length 400mm and thickness 10mm having 𝐸 = 2 × 105 𝑁/ 𝑚𝑚2. The extension due to load is 0. 21mm. The bar tapers uniformly in width from 100mm to 50mm. Determine the axial load on the bar. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑎 𝑏 L 𝐿 = 400 𝑚𝑚 𝑡 = 10 𝑚𝑚 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝛿𝑙 = 0.21 𝑚𝑚 𝑎 = 100 𝑚𝑚 𝑏 = 50 𝑚𝑚 𝐴𝑥𝑖𝑎𝑙 𝑙𝑜𝑎𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑎𝑟 𝑃 =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 27. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑑 𝛿𝑙 = 𝑃𝐿 𝐸𝑡 𝑎 − 𝑏 ) Τ ln(𝑎 𝑏 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑑 𝛿𝑙 = 𝑃𝐿 𝐸𝑡 𝑎 − 𝑏 ) Τ ln(𝑎 𝑏 0.21 = 𝑃 × 400 2 × 105 × 10 × 100 − 50 ) Τ ln(100 50 𝑨𝒙𝒊𝒂𝒍 𝒍𝒐𝒂𝒅 𝒐𝒏 𝒕𝒉𝒆 𝒃𝒂𝒓 𝑷 = 𝟏𝟕𝟒. 𝟒 𝒌𝑵 𝐿 = 400 𝑚𝑚 𝑡 = 10 𝑚𝑚 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝛿𝑙 = 0.21 𝑚𝑚 𝑎 = 100 𝑚𝑚 𝑏 = 50 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 28. Thank You BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY