Lecture 13 som 11.03.2021
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The document derives expressions for Young's modulus in terms of other elastic constants like modulus of rigidity and bulk modulus. It considers a solid cube subjected to shear forces, which causes diagonal strains. Using relationships between shear strain, normal stress, longitudinal strain, and lateral strain, it arrives at the expression E = 2G(1+μ), where E is Young's modulus, G is modulus of rigidity, and μ is Poisson's ratio. It then considers a cube under mutually perpendicular tensile stresses and derives another expression for E in terms of bulk modulus.
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Lecture 13 som 11.03.2021
- 1. BIBIN CHIDAMBARANATHAN RELATIONSHIP BETWEEN ELASTIC CONSTANTS BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 2. EXPRESSION FOR YOUNGS MODULUS IN TERMS OF MODULUS OF RIGIDITY LMST is a solid cube subjected to a shearing force F. Let 𝜏 be the shear stress produced in the faces MS and LT due to this shearing force. The complementary shear stress consequently produced in the faces ML and ST is also 𝜏. Due to the shearing force, the cube is distorted to 𝐿′𝑀′𝑆′𝑇′ and as such, the edge M moves to M’ and S to S’ and diagonal LS to LS’. 𝑭 𝑳 𝑻 𝑺′ 𝑴 𝑴′ 𝝉 𝝉 𝝉 𝝉 𝝋 𝟒𝟓° 𝟒𝟓° 𝝋 𝑺 𝑵 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 3. 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐿𝑆′ − 𝐿𝑆 𝐿𝑆 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐿𝑆′ − 𝐿𝑁 𝐿𝑆 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑁𝑆′ 𝐿𝑆 → 1 𝐿𝑆 = 𝐿𝑁 𝐹𝑟𝑜𝑚 ∆𝑙𝑒 𝑁𝑆𝑆′ 𝐶𝑜𝑠 45 = 𝑁𝑆′ 𝑆𝑆′ 𝑭 𝑳 𝑻 𝑺′ 𝑴 𝑴′ 𝝉 𝝉 𝝉 𝝉 𝝋 𝟒𝟓° 𝟒𝟓° 𝝋 𝑺 𝑵 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 4. 𝑁𝑆′ = 𝑆𝑆′ 𝐶𝑜𝑠 45 𝑁𝑆′ = 𝑆𝑆′ 2 → 2 𝐹𝑟𝑜𝑚 𝑅𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑑 ∆𝑙𝑒𝐿𝑆𝑇 𝐿𝑆2 = 𝐿𝑇2 + 𝑆𝑇2 𝐿𝑆 = 𝐿𝑇2 + 𝑆𝑇2 𝐿𝑆 = 𝑆𝑇2 + 𝑆𝑇2 𝐿𝑆 = 2𝑆𝑇2 𝐿𝑆 = 2 𝑆𝑇 → 3 𝑭 𝑳 𝑻 𝑺′ 𝑴 𝑴′ 𝝉 𝝉 𝝉 𝝉 𝝋 𝟒𝟓° 𝟒𝟓° 𝝋 𝑺 𝑵 𝐹𝑜𝑟 𝑐𝑢𝑏𝑒 𝑎𝑙𝑙 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 5. Substituting equations (2) and (3) in equation (1) 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑆 Τ 𝑆′ 2 2 𝑆𝑇 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑆𝑆′ 2 𝑆𝑇 → 4 𝑅𝑖𝑔𝑖𝑑𝑖𝑡𝑦 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝐺 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 𝐺 = 𝜏 𝜑 𝑁𝑆′ = 𝑆𝑆′ 2 → 2 𝐿𝑆 = 2 𝑆𝑇 → 3 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑁𝑆′ 𝐿𝑆 → 1 𝜑 = 𝜏 𝐺 → 5 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 6. 𝑭 𝑳 𝑻 𝑺′ 𝑴 𝑴′ 𝝉 𝝉 𝝉 𝝉 𝝋 𝟒𝟓° 𝟒𝟓° 𝝋 𝑺 𝑵 𝐹𝑟𝑜𝑚 ∆𝑙𝑒 𝑇𝑆𝑆′ 𝑡𝑎𝑛 𝜑 = 𝑆𝑆′ 𝑆𝑇 𝜑 𝑖𝑠 𝑣𝑒𝑟𝑦 𝑠𝑚𝑎𝑙𝑙, 𝑠𝑜 𝑡𝑎𝑛 𝜑 = 𝜑 𝜑 = 𝑆𝑆′ 𝑆𝑇 → 6 Substituting equation (6) in equation (4) 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑆𝑆′ 2 𝑆𝑇 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝜑 2 → 7 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝑆𝑆′ 2 𝑆𝑇 → 4 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 7. Substituting equation (5) in equation (7) 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = Τ 𝜏 𝐺 2 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝜏 2𝐺 → 8 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝜑 2 → 7 The load on all sides is equal. In a cube the area of cross section is same on all sides. 𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝜎𝑁 = 𝜏 → 9 𝜑 = 𝜏 𝐺 → 5 Substituting equation (9) in equation (8) 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝜎𝑁 2𝐺 → 10 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 8. 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙 = 𝜎 𝐸 𝑒𝑙 = 𝜎𝑁 𝐸 → 11 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 𝑜𝑟 1 𝑚 = 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 = 𝑒𝑡 𝑒𝑙 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 = 𝑒𝑙 × 𝜇 → 12 Substituting equation (11) in equation (12) 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 = 𝜎𝑁 𝐸 × 𝜇 → 13 𝑇𝑜𝑡𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝑒𝑙 + 𝑒𝑡 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 9. Add equation (11) and equation (13) 𝑇𝑜𝑡𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝜎𝑁 𝐸 + 𝜎𝑁 𝐸 × 𝜇 𝑇𝑜𝑡𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝜎𝑁 𝐸 1 + 𝜇 → 14 Equate equation (10) and equation (14) 𝜎𝑁 𝐸 1 + 𝜇 = 𝜎𝑁 2𝐺 1 𝐸 1 + 𝜇 = 1 2𝐺 𝐄 = 2𝐆 1 + 𝛍 𝑒𝑙 = 𝜎𝑁 𝐸 → 11 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 = 𝜎𝑁 𝐸 × 𝜇 → 13 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝜎𝑁 2𝐺 → 10 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 10. EXPRESSION FOR YOUNGS MODULUS IN TERMS OF BULK MODULUS Figure shows a cube ABCDEFGH which is subjected to three mutually perpendicular tensile stresses of equal intensity. Let 𝑨 𝑩 𝑪 𝑫 𝑮 𝑭 𝑬 𝝈 𝝈 𝝈 𝝈 𝑯 𝐿 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝐶𝑢𝑏𝑒 𝛿𝑙 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑏𝑒 𝐸 = 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑏𝑒 𝜎 = 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑓𝑎𝑐𝑒𝑠 𝜇 = 𝑃𝑜𝑖𝑠𝑠𝑜𝑛′ 𝑠 𝑟𝑎𝑡𝑖𝑜 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 11. Now let us consider the strain of one of the sided of the cube (say AB) under the action of three perpendicular stresses. This side will suffer the following three strains. 1. 𝑆𝑡𝑟𝑎𝑖𝑛 𝑜𝑓 𝐴𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠 𝑜𝑛 𝐴𝐸𝐻𝐷 𝑎𝑛𝑑 𝐵𝐹𝐺𝐶. 𝑇ℎ𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝑖𝑠 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑎𝑛𝑑 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 Τ 𝜎 𝐸 2. 𝑆𝑡𝑟𝑎𝑖𝑛 𝑜𝑓 𝐴𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠 𝑜𝑛 𝐴𝐸𝐹𝐵 𝑎𝑛𝑑 𝐷𝐻𝐺𝐶. 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑎𝑛𝑑 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 − 𝜇 × 𝜎 𝐸 3. 𝑆𝑡𝑎𝑟𝑖𝑛 𝑜𝑓 𝐴𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠 𝑜𝑛 𝐴𝐵𝐶𝐷 𝑎𝑛𝑑 𝐸𝐹𝐺𝐻. 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑎𝑛𝑑 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 − 𝜇 × 𝜎 𝐸 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 12. Hence the total strain of AB is given by 𝛿𝑙 𝐿 = 𝜎 𝐸 − 𝜇 𝜎 𝐸 − 𝜇 𝜎 𝐸 𝛿𝑙 𝐿 = 𝜎 𝐸 1 − 2𝜇 → (1) Now original volume of cube (𝑉) = 𝐿3 → (2) If 𝛿𝑙 is the change in length, 𝛿𝑉 is the change in volume 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2) 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝐿 𝑑𝑉 = 3𝐿2𝛿𝑙 → (3) 𝑨 𝑩 𝑪 𝑫 𝑮 𝑭 𝑬 𝝈 𝝈 𝝈 𝝈 𝑯 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 13. 𝐷𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (3) 𝑏𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2) 𝑑𝑉 𝑉 = 3𝐿2𝛿𝑙 𝐿3 𝑑𝑉 𝑉 = 3 𝛿𝑙 𝐿 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝛿𝑙 𝐿 𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (1) 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑤𝑒 𝑔𝑒𝑡 𝑑𝑉 𝑉 = 3𝜎 𝐸 1 − 2𝜇 → (4) V = 𝐿3 → (2) 𝑑𝑉 = 3𝐿2𝛿𝑙 → (3) 𝛿𝑙 𝐿 = 𝜎 𝐸 1 − 2𝜇 → (1) BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 14. 𝐾 = 𝜎 𝑑𝑉 𝑉 𝐾 = 𝜎 3𝜎 𝐸 1 − 2𝜇 𝐾 = 𝐸 3 1 − 2𝜇 𝑬 = 𝟑𝑲 𝟏 − 𝟐𝝁 𝐵𝑢𝑙𝑘 𝑚𝑜𝑑𝑢𝑙𝑢𝑠, 𝐾 = 𝜎 𝑒𝑣 𝑑𝑉 𝑉 = 3𝜎 𝐸 1 − 2𝜇 → (4) 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (4) 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑤𝑒 𝑔𝑒𝑡 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 15. 𝐸 = 3𝐾 1 − 2𝜇 𝐹𝑟𝑜𝑚 𝑡ℎ𝑖𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑡ℎ𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑓𝑜𝑟 𝑝𝑜𝑖𝑠𝑠𝑜𝑛′ 𝑠 𝑟𝑎𝑡𝑜 𝜇 𝑖𝑠 𝑜𝑏𝑡𝑎𝑖𝑛𝑒𝑑 𝑎𝑠 𝝁 = 𝟑𝑲 − 𝑬 𝟔𝑲 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 16. RELATIONSHIP BETWEEN E, G AND K 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2) 𝐸 = 3𝐾 1 − 2𝜇 = 3𝐾 1 − 2 𝑚 → (1) 𝐸 = 2G 1 + μ = 2𝐶 1 + 1 𝑚 → (2) 𝐸 = 2𝐶 1 + 1 𝑚 1 + 1 𝑚 = 𝐸 2𝐶 1 𝑚 = 𝐸 − 2𝐶 2𝐶 𝑚 = 2𝐶 𝐸 − 2𝐶 → (3) BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 17. 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑚 𝑖𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (1), 𝑤𝑒 𝑔𝑒𝑡, 𝐸 = 3𝐾 1 − 2 2𝐶 𝐸 − 2𝐶 𝐸 = 3𝐾 ቆ1 − 𝐸 − 2𝐶 𝐶 𝐸 3𝐾 = 1 − 𝐸 − 2𝐶 𝐶 𝐸 3𝐾 = 𝐶 − 𝐸 + 2𝐶 𝐶 𝐸 3𝐾 = −𝐸 + 3𝐶 𝐶 𝐸 3𝐾 = − 𝐸 𝐶 + 3𝐶 𝐶 𝐸 = 3𝐾 1 − 2𝜇 = 3𝐾 1 − 2 𝑚 → (1) 𝑚 = 2𝐶 𝐸 − 2𝐶 → (3) BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 18. 𝐸 3𝐾 + 𝐸 𝐶 = 3 𝐸 1 3𝐾 + 1 𝐶 = 3 𝐸 𝐶 + 3𝐾 3𝐾𝐶 = 3 𝐸 = 9𝐾𝐶 𝐶 + 3𝐾 𝑬 = 𝟗𝑲𝑪 𝟑𝑲 + 𝑪 = 𝟗𝑲𝑮 𝟑𝑲 + 𝑮 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 19. Thank You BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY