Lecture 09 som 05.03.2021
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Thermal stress and strain occur when a material is heated or cooled but prevented from expanding or contracting freely. Thermal stress is caused by a temperature change creating thermal strain, which induces stress if expansion/contraction is constrained. The document provides formulas to calculate thermal strain, stress, and examples calculating these values for rods with given temperature changes, materials properties, and constraints. It also addresses cases where the rod ends yield partially versus being fully constrained.
Lecture 09 som 05.03.2021
- 1. BIBIN CHIDAMBARANATHAN THERMAL STRESS & THERMAL STRAINS BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 2. Thermal Stress & Thermal Strains ❖ Stress induced in a body due to change in the temperature is known as thermal stress and the corresponding strain is called thermal strain. ❖ Thermal stress induces in a body when the temperature of the body is raised or lowered and the body is not allowed to expand or contract freely. ❖ No stress will be induced in a body when it is allowed to expand or contract freely. BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 3. ❖ Considered a body (for example a rod) which is heated to a specified temperature. Let ❖ L = Original length of the rod, ❖ T = Rise in temperature, ❖ E = Young’s modulus, ❖ α = Coefficient of linear expansion and ❖ 𝛿𝑙 = Extension produced in the rod BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 4. ❖ Due to the increase in the temperature, there is an extension produced in the rod. ❖ When the rod is allowed to expand freely, the extension produced in the rod is given by 𝑬𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒑𝒓𝒐𝒅𝒖𝒄𝒆𝒅 𝜹𝒍 = 𝜶 𝑻 𝑳 ❖ AB is the original length of the rod, and ❖ BB’ is the extension produced in the rod. BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 5. ❖ Suppose a compressive load P is applied at the end BB’, Due to this compressive load, there is a decrease in the length of the rod from (L + dL) to L. ❖ Compressive stress and strain is produced in the rod and is given by 𝑪𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒗𝒆 𝒔𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝜶 𝑻 𝑳 𝑳 + 𝜶 𝑻 𝑳 = 𝜶 𝑻 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝒔𝒕𝒓𝒂𝒊𝒏 × 𝑬 = 𝜶 𝑻 𝑬 ❖ Thrust or load on the rod is given by 𝑳𝒐𝒂𝒅 𝒐𝒓 𝑻𝒉𝒓𝒖𝒔𝒕 𝑷 = 𝒔𝒕𝒓𝒆𝒔𝒔 × 𝑨 = 𝜶 𝑻 𝑬 × 𝑨 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 6. ❖ When the ends of the rod is rigidly fixed with the supports and the rod is heated to a certain temperature, here the extension in the rod is prevented because we have rigidly fixed it’s both the ends. ❖ Stress and strain is setup in the rod. The stresses and strain setup in the rod is known as thermal stresses and strains. Thermal strain is given by ❖ Thermal stress-induced is given by 𝑻𝒉𝒆𝒓𝒎𝒂𝒍 𝒔𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝑬𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒑𝒓𝒆𝒗𝒆𝒏𝒕𝒆𝒅 𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝜶 𝑻 𝑳 𝑳 = 𝜶 𝑻 𝑻𝒉𝒆𝒓𝒎𝒂𝒍 𝒔𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝜶 𝑻𝑬 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 7. ❖ 𝐼𝑓 𝑡ℎ𝑒 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑠 𝑎𝑡 𝑡ℎ𝑒 𝑒𝑛𝑑𝑠 𝑦𝑖𝑒𝑙𝑑 𝑏𝑦 𝑎𝑛 𝑎𝑚𝑜𝑢𝑛𝑡 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑡ℎ𝑒 𝛿, 𝑡ℎ𝑒𝑛 𝑇ℎ𝑒 𝑎𝑐𝑡𝑢𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 = 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑟𝑖𝑠𝑒 𝑖𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 − 𝛿 𝐴𝑐𝑡𝑢𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 = 𝐴𝑐𝑡𝑢𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ = 𝛼 𝑇 𝐿 − 𝛿 𝛿 𝐴𝑐𝑡𝑢𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝛼 𝑇 𝐿 − 𝛿 𝛿 × 𝐸 𝑇ℎ𝑒 𝑎𝑐𝑡𝑢𝑎𝑙 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 = 𝛼 𝑇 𝐿 − 𝛿 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 8. Problem 01 A rod is 2 m long at a temperature of 10°C. Find the expansion of the rod, when the temperature is raised to 80°C. If this expansion is prevented, find the stress induced and strain in the material of the rod. Take 𝐸 = 1 × 105 𝑁/𝑚𝑚2 and 𝛼 = 0.000012/°𝐶. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝐿 = 2 𝑚 = 2000 𝑚𝑚 𝑇𝑖 = 10°𝐶 𝑇𝑓 = 80°𝐶 𝑇 = 𝑇𝑓 − 𝑇𝑖 = 80 − 10 = 70°𝐶 𝐸 = 1 × 105 Τ 𝑁 𝑚 𝑚2 𝛼 = Τ 0.000012 ° 𝐶 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑑 𝛿𝑙 =? 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 =? 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 9. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝛼 𝑇𝐸 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝐿 = 𝛼 𝑇 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑑 𝛿𝑙 = 𝛼 𝑇 𝐿 Expansion is prevented BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 10. 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝛼 𝑇𝐸 𝜎 = 0.000012 × 70 × 1 × 105 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝟖𝟒 Τ 𝑵 𝒎 𝒎𝟐 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝐿 = 𝛼 𝑇 𝑒 = 0.000012 × 70 𝑺𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝟖. 𝟒 × 𝟏𝟎−𝟒 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝑇 = 𝑇𝑓 − 𝑇𝑖 = 80 − 10 = 70°𝐶 𝐸 = 1 × 105 Τ 𝑁 𝑚 𝑚2 𝛼 = Τ 0.000012 ° 𝐶 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 11. 𝐸𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑑 𝛿𝑙 = 𝛼 𝑇 𝐿 𝛿𝑙 = 0.000012 × 70 × 2000 𝑬𝒙𝒑𝒂𝒏𝒔𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒓𝒐𝒅 𝜹𝒍 = 𝟏. 𝟔𝟖 𝒎𝒎 𝐿 = 2 𝑚 = 2000 𝑚𝑚 𝑇 = 𝑇𝑓 − 𝑇𝑖 = 80 − 10 = 70°𝐶 𝛼 = Τ 0.000012 ° 𝐶 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 12. Problem 01 A steel rod 3cm diameter and 5m long. If it is connected to two grips and rod is maintained at a temperature of 95°C. Determine stress, strain and pull exerted when the temperature falls to 30°C, 𝐸 = 2 × 105 𝑁/𝑚𝑚2 and 𝛼 = 1.2 × 10−6/°𝐶. (i) Ends are not yield; (ii) Ends are yield by 0.12cm. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑃𝑢𝑙𝑙 𝑒𝑥𝑒𝑟𝑡𝑒𝑑 𝑃 =? 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 =? 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 =? 𝑑 = 3 cm = 30 𝑚𝑚 𝐿 = 5 𝑚 = 5000 𝑚𝑚 𝑇𝑖 = 95°𝐶 𝑇𝑓 = 30°𝐶 𝑇 = 𝑇𝑖~𝑇𝑓 = 95 − 30 = 65°𝐶 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝛼 = 1.2 × Τ 10−6 ° 𝐶 𝛿 = 0.12 𝑐𝑚 = 1.2 𝑚𝑚 (i) Ends are not yield 𝐶𝑎𝑠𝑒 𝑖𝑖 ∶ 𝐸𝑛𝑑 𝑎𝑟𝑒 𝑦𝑖𝑒𝑙𝑑 𝑏𝑦 0.12 𝑐𝑚 𝑌𝑖𝑒𝑙𝑑 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎𝑦 =? 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 =? 𝑃𝑢𝑙𝑙 𝑒𝑥𝑒𝑟𝑡𝑒𝑑 𝑃 =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 13. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝛼 𝑇𝐸 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝐿 = 𝛼 𝑇 (ii) Ends are yield by 0.12cm. (i) Ends are not yield 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝑌𝑖𝑒𝑙𝑑 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎𝑦 = 𝛼 𝑇𝐿 − 𝛿 𝐿 𝐸 𝑃𝑢𝑙𝑙 𝑜𝑟 𝑙𝑜𝑎𝑑 𝑃 = 𝜎𝑦 × 𝐴 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛼 𝑇𝐿 − 𝛿 𝐿 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 14. 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 𝛼 𝑇𝐸 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝟏𝟓𝟔 Τ 𝑵 𝒎 𝒎𝟐 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝜎 = 1.2 × 10−6 × 65 × 2 × 105 𝐴𝑟𝑒𝑎 𝐴 = 𝜋 4 × 𝑑2 𝐴 = 𝜋 4 × 302 𝑨𝒓𝒆𝒂 𝑨 = 𝟕𝟎𝟔. 𝟖𝟓 𝒎𝒎𝟐 𝑇 = 𝑇𝑖~𝑇𝑓 = 95 − 30 = 65°𝐶 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝛼 = 1.2 × Τ 10−6 ° 𝐶 𝑑 = 30 𝑚𝑚 (i) Ends are not yield BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 15. 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛿𝑙 𝐿 = 𝛼 𝑇 𝑺𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝟎. 𝟕𝟖 × 𝟏𝟎−𝟑 𝑃𝑢𝑙𝑙 𝑜𝑟 𝑙𝑜𝑎𝑑 𝑃 = 𝜎 × 𝐴 𝑃 = 156 × 706.85 𝑷𝒖𝒍𝒍 𝒐𝒓 𝒍𝒐𝒂𝒅 𝑷 = 𝟏𝟏𝟎. 𝟐𝟏𝟒 × 𝟏𝟎𝟑 𝑵 𝜎 = 156 Τ 𝑁 𝑚 𝑚2 𝐴 = 706.85 𝑚𝑚2 𝑒 = 1.2 × 10−6 × 65 𝑇 = 𝑇𝑖~𝑇𝑓 = 95 − 30 = 65°𝐶 𝛼 = 1.2 × Τ 10−6 ° 𝐶 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 16. 𝑪𝒂𝒔𝒆 𝒊𝒊 ∶ 𝑬𝒏𝒅 𝒂𝒓𝒆 𝒚𝒊𝒆𝒍𝒅 𝒃𝒚 𝟎. 𝟏𝟐 𝒄𝒎 𝑌𝑖𝑒𝑙𝑑 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎𝑦 = 𝛼 𝑇𝐿 − 𝛿 𝐿 𝐸 𝜎𝑦 = 1.2 × 10−6 × 65 × 5000 − 1.2 5000 × 2 × 105 𝒀𝒊𝒆𝒍𝒅 𝑺𝒕𝒓𝒆𝒔𝒔 𝝈𝒚 = 𝟏𝟎𝟖 Τ 𝑵 𝒎 𝒎𝟐 𝑃𝑢𝑙𝑙 𝑜𝑟 𝑙𝑜𝑎𝑑 𝑃 = 𝜎𝑦 × 𝐴 𝑃 = 108 × 706.85 𝑷𝒖𝒍𝒍 𝒐𝒓 𝒍𝒐𝒂𝒅 𝑷 = 𝟕𝟔. 𝟑𝟒 × 𝟏𝟎𝟑𝑵 𝐿 = 5 𝑚 = 5000 𝑚𝑚 𝑇 = 𝑇𝑖~𝑇𝑓 = 95 − 30 = 65°𝐶 𝛼 = 1.2 × Τ 10−6 ° 𝐶 𝛿 = 0.12 𝑐𝑚 = 1.2 𝑚𝑚 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝐴 = 706.85 𝑚𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 17. 𝐿 = 5 𝑚 = 5000 𝑚𝑚 𝑇 = 𝑇𝑖~𝑇𝑓 = 95 − 30 = 65°𝐶 𝛼 = 1.2 × Τ 10−6 ° 𝐶 𝛿 = 0.12 𝑐𝑚 = 1.2 𝑚𝑚 𝑆𝑡𝑟𝑎𝑖𝑛 𝑒 = 𝛼 𝑇𝐿 − 𝛿 𝐿 𝑒 = 1.2 × 10−6 × 65 × 5000 − 1.2 5000 𝑺𝒕𝒓𝒂𝒊𝒏 𝒆 = 𝟓. 𝟒 × 𝟏𝟎−𝟒 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 18. Thank You BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY