Lecture 08 som 05.03.2021
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The document discusses stresses in composite bars made of two or more materials. It defines a composite bar and lists its key features, including equal strain and the total load being the sum of individual loads. It then provides conditions for composite bars, including equal elongation, total load equaling the sum of individual loads, and the modular ratio. Two example problems are worked through demonstrating how to calculate elongation or stresses in a composite bar given various parameters like loads, dimensions, moduli of elasticity and areas of the different materials. The solutions involve setting up and solving equations based on the three stated conditions for composite bars.
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Lecture 08 som 05.03.2021
- 1. BIBIN CHIDAMBARANATHAN STRESSES IN COMPOSITE BARS BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 2. STRESSES IN COMPOSITE BARS ❖ A composite bar may be defined as a bar made up of two or more different materials, joined together, in such a manner that the system extends or contracts as one unit, equally, when subjected to tension or compression. ❖ Features of composite bars: ❖ The extension or contraction of the bar being equal, the strain i.e. deformation per unit length is also equal. ❖ The total external load on the bar is equal to the sum of the loads carried by different materials. BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 3. Conditions for composite bars ❖ Strain constant: Elongation or contraction is equal in each bar 𝛿𝑙1 = 𝛿𝑙2 𝑃1𝐿1 𝐴1𝐸1 = 𝑃2𝐿2 𝐴2𝐸2 ❖ Total load carried by composite bar is equal to sum of load carried by two materials 𝑷 = 𝑷𝟏 + 𝑷𝟐 𝝈𝑨 = 𝝈𝟏𝑨𝟏 + 𝝈𝟐𝑨𝟐 ❖ Modular ratio 𝑬𝟏 𝑬𝟐 = 𝝈𝟏 𝝈𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 4. Problem 01 Determine the total elongation of the composite bar loaded as shown in the figure. Young’s modulus of Al and steel are 80Gpa and 200Gpa respectively. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 =? 𝑃 = 35 𝑘𝑁 = 35 × 103 𝑁 𝐿𝑆 = 800 𝑚𝑚 𝐴𝑆1 = 100 𝑚𝑚2 𝐴𝑆2 = 100 𝑚𝑚2 𝐴𝑆 = 𝐴𝑆1 + 𝐴𝑆2 = 100 + 100 = 200 𝑚𝑚2 𝐴𝐴𝑙 = 200 𝑚𝑚2 𝐸𝑠 = 200 𝐺𝑃𝑎 = 200 × 103 Τ 𝑁 𝑚 𝑚2 𝐸𝐴𝑙 = 80 𝐺𝑃𝑎 = 80 × 103 Τ 𝑁 𝑚 𝑚2 𝐿𝐴𝑙 = 800 𝑚𝑚 𝟑𝟓 𝒌𝑵 𝟐𝟎𝟎 𝒎𝒎 𝟐 𝟏𝟎𝟎 𝒎𝒎 𝟐 𝟏𝟎𝟎 𝒎𝒎 𝟐 𝟖𝟎𝟎 𝒎𝒎 𝒔𝒕𝒆𝒆𝒍 𝒔𝒕𝒆𝒆𝒍 𝑨𝒍𝒖𝒎𝒊𝒏𝒊𝒖𝒎 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 5. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝛿𝑙𝑠 = 𝛿𝑙𝐴𝑙 𝛿𝑙𝑠 = 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝜎𝑠 𝐿𝑠 𝐸𝑠 𝛿𝑙𝐴𝑙 = 𝑃𝐴𝑙 𝐿𝐴𝑙 𝐴𝐴𝑙 𝐸𝐴𝑙 = 𝜎𝐴𝑙 𝐿𝐴𝑙 𝐸𝐴𝑙 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝛿𝑙𝑠 = 𝛿𝑙𝐴𝑙 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝑃𝐴𝑙 𝐿𝐴𝑙 𝐴𝐴𝑙 𝐸𝐴𝑙 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝐴𝑙 𝐿𝐴𝑙 𝐸𝐴𝑙 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: 𝟑𝟓 𝒌𝑵 𝟐𝟎𝟎 𝒎𝒎 𝟐 𝟏𝟎𝟎 𝒎𝒎 𝟐 𝟏𝟎𝟎 𝒎𝒎 𝟐 𝟖𝟎𝟎 𝒎𝒎 𝒔𝒕𝒆𝒆𝒍 𝒔𝒕𝒆𝒆𝒍 𝑨𝒍𝒖𝒎𝒊𝒏𝒊𝒖𝒎 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 6. 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝑃𝑠 + 𝑃𝐴𝑙 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝐴𝑙 × 𝐴𝐴𝑙 𝟑𝟓 𝒌𝑵 𝟐𝟎𝟎 𝒎𝒎 𝟐 𝟏𝟎𝟎 𝒎𝒎 𝟐 𝟏𝟎𝟎 𝒎𝒎 𝟐 𝟖𝟎𝟎 𝒎𝒎 𝒔𝒕𝒆𝒆𝒍 𝒔𝒕𝒆𝒆𝒍 𝑨𝒍𝒖𝒎𝒊𝒏𝒊𝒖𝒎 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 7. 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝛿𝑙𝑠 = 𝛿𝑙𝐴𝑙 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝑃𝐴𝑙 𝐿𝐴𝑙 𝐴𝐴𝑙 𝐸𝐴𝑙 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝐴𝑙 𝐿𝐴𝑙 𝐸𝐴𝑙 𝜎𝑠 × 800 200 × 103 = 𝜎𝐴𝑙 × 800 80 × 103 𝜎𝑠 = 2.56 × 𝜎𝐴𝑙 𝐿𝑆 = 800 𝑚𝑚 𝐸𝑠 = 200 × 103 Τ 𝑁 𝑚 𝑚2 𝐸𝐴𝑙 = 80 × 103 Τ 𝑁 𝑚 𝑚2 𝐿𝐴𝑙 = 800 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 8. 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝑃𝑠 + 𝑃𝐴𝑙 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝐴𝑙 × 𝐴𝐴𝑙 35 × 103 = 2.56 𝜎𝐴𝑙 × 200 + 𝜎𝐴𝑙 × 200 35 × 103 = 7000 𝜎𝐴𝑙 𝝈𝑨𝒍 = 𝟓𝟎 Τ 𝑵 𝒎 𝒎𝟐 𝝈𝑺 = 𝟏𝟐. 𝟓 Τ 𝑵 𝒎 𝒎𝟐 𝜎𝑠 = 2.56 × 𝜎𝐴𝑙 𝜎𝑠 = 2.56 × 𝜎𝐴𝑙 𝑃 = 35 𝑘𝑁 = 35 × 103 𝑁 𝐴𝑆 = 200 𝑚𝑚2 𝐴𝐴𝑙 = 200 𝑚𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 9. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝛿𝑙𝑠 = 𝜎𝑠 𝐿𝑠 𝐸𝑠 𝛿𝑙 = 𝜎𝑠 𝐿𝑠 𝐸𝑠 𝛿𝑙 = 12.5 × 800 200 × 103 𝑻𝒐𝒕𝒂𝒍 𝒆𝒍𝒐𝒏𝒈𝒂𝒕𝒊𝒐𝒏 𝜹𝒍 = 𝟎. 𝟓 𝒎𝒎 𝐿𝑆 = 800 𝑚𝑚 𝐸𝑠 = 200 × 103 Τ 𝑁 𝑚 𝑚2 𝜎𝑆 = 12.5 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 10. Problem 01 Determine the magnitude of stress developed in a composite bar. 𝐸𝑠 = 2 × 105 𝑁/𝑚𝑚2 and 𝐸𝑏 = 1 × 105 𝑁/𝑚𝑚2. Area of steel and brass are 800 𝑚𝑚2 and 500 𝑚𝑚2 respectively. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝐸𝑠 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝐸𝑏 = 1 × 105 Τ 𝑁 𝑚 𝑚2 𝐴𝑆 = 800 𝑚𝑚2 𝐴𝑏 = 500 𝑚𝑚2 𝐿𝑆 = 250 𝑚𝑚 𝐿𝑏 = 200 𝑚𝑚 𝑃 = 25 𝑘𝑁 = 25 × 103 𝑁 𝑆𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑡𝑒𝑒𝑙 𝜎𝑠 𝑎𝑛𝑑 𝑏𝑟𝑎𝑠𝑠 𝜎𝑏 =? 𝟐𝟓 𝒌𝑵 𝟐𝟎𝟎 𝒎𝒎 𝟓𝟎 𝒎𝒎 𝑩𝒓𝒂𝒔𝒔 𝑩𝒓𝒂𝒔𝒔 𝒔𝒕𝒆𝒆𝒍 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 11. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝛿𝑙𝑠 = 𝛿𝑙𝑏 𝛿𝑙𝑠 = 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝜎𝑠 𝐿𝑠 𝐸𝑠 𝛿𝑙𝑏 = 𝑃𝑏 𝐿𝑏 𝐴𝑏 𝐸𝑏 = 𝜎𝑏 𝐿𝑏 𝐸𝑏 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝛿𝑙𝑠 = 𝛿𝑙𝑏 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝑃𝑏 𝐿𝑏 𝐴𝑏 𝐸𝑏 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝑏 𝐿𝑏 𝐸𝑏 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: 𝟐𝟓 𝒌𝑵 𝟐𝟎𝟎 𝒎𝒎 𝟓𝟎 𝒎𝒎 𝑩𝒓𝒂𝒔𝒔 𝑩𝒓𝒂𝒔𝒔 𝒔𝒕𝒆𝒆𝒍 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 12. 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝑃𝑠 + 𝑃𝑏 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝑏 × 𝐴𝑏 𝟐𝟓 𝒌𝑵 𝟐𝟎𝟎 𝒎𝒎 𝟓𝟎 𝒎𝒎 𝑩𝒓𝒂𝒔𝒔 𝑩𝒓𝒂𝒔𝒔 𝒔𝒕𝒆𝒆𝒍 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 13. 𝜎𝑠 × 200 1 × 105 = 𝜎𝑏 × 250 2 × 105 𝝈𝒃 = 𝟎. 𝟔𝟐𝟓 𝝈𝒔 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝑏 𝐿𝑏 𝐸𝑏 𝐴𝑆 = 800 𝑚𝑚2 𝐴𝑏 = 500 𝑚𝑚2 𝐿𝑆 = 250 𝑚𝑚 𝐿𝑏 = 200 𝑚𝑚 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝑏 × 𝐴𝑏 25 × 103 = 𝜎𝑠 × 800 + 0.625 × 𝜎𝑠 × 1000 𝜎𝑠 = 17.54 Τ 𝑁 𝑚 𝑚2 𝑺𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒔𝒕𝒆𝒆𝒍 𝝈𝒔 = 𝟏𝟕. 𝟓𝟒 Τ 𝑵 𝒎 𝒎𝟐 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 14. 𝜎𝑏 = 0.625 𝜎𝑠 𝜎𝑠 = 17.54 Τ 𝑁 𝑚 𝑚2 𝜎𝑏 = 10.96 Τ 𝑁 𝑚 𝑚2 𝜎𝑏 = 0.625 × 17.54 𝑺𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒃𝒓𝒂𝒔𝒔 𝝈𝒃 = 𝟏𝟎. 𝟗𝟔 Τ 𝑵 𝒎 𝒎𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 15. Problem 01 A solid copper rod 36 mm diameter is rigidly forced at both ends of a steel tube of 45 mm inside diameter and 50 mm outer diameter. The compound section is then subjected to an axial pull of 98 kN. Determine the stress induced in rod and tube and total elongation of composite section in a length of 1 m. E for copper= 1.1 × 105 𝑁/ 𝑚𝑚2 and for steel is 2 × 105 𝑁/𝑚𝑚2. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑆𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 steel 𝑡𝑢𝑏𝑒 𝜎𝑠 𝑎𝑛𝑑 copper 𝑟𝑜𝑑 𝜎𝑐 =? 𝟏 𝒎 𝟗𝟖 𝒌𝑵 𝟗𝟖 𝒌𝑵 𝑑𝐶 = 36 𝑚𝑚 𝐷𝑠 = 50 𝑚𝑚 𝑑𝑆 = 45 𝑚𝑚 𝑃 = 98 𝑘𝑁 = 98 × 103 𝑁 𝐿 = 𝐿𝐶 = 𝐿𝑠 = 1 𝑚 = 1000 𝑚𝑚 𝐸𝑠 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝐸𝐶 = 1.1 × 105 Τ 𝑁 𝑚 𝑚2 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 (𝛿𝑙) =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 16. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝛿𝑙𝑠 = 𝛿𝑙𝑐 𝛿𝑙𝑠 = 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝜎𝑠 𝐿𝑠 𝐸𝑠 𝛿𝑙𝑐 = 𝑃𝑐 𝐿𝑐 𝐴𝑐 𝐸𝑐 = 𝜎𝑐 𝐿𝑐 𝐸𝑐 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝛿𝑙𝑠 = 𝛿𝑙𝑐 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝑃𝑐 𝐿𝑐 𝐴𝑐 𝐸𝑐 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝑐 𝐿𝑐 𝐸𝑐 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: 𝟏 𝒎 𝟗𝟖 𝒌𝑵 𝟗𝟖 𝒌𝑵 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 17. 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝑃𝑠 + 𝑃𝑐 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝑐 × 𝐴𝑐 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑡𝑢𝑏𝑒 (𝐴𝑆) = 𝜋 4 × 𝐷𝑆 2 − 𝑑𝑆 2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 𝑟𝑜𝑑 (𝐴𝐶) = 𝜋 4 × 𝑑𝑐 2 𝟏 𝒎 𝟗𝟖 𝒌𝑵 𝟗𝟖 𝒌𝑵 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 18. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝐶 𝐿𝐶 𝐸𝐶 𝜎𝑠 × 1000 2 × 105 = 𝜎𝐶 × 1000 1.1 × 105 𝝈𝒔 = 𝟏. 𝟖𝟏𝟖 𝝈𝒄 𝐿 = 𝐿𝐶 = 𝐿𝑠 = 1 𝑚 = 1000 𝑚𝑚 𝐸𝑠 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝐸𝐶 = 1.1 × 105 Τ 𝑁 𝑚 𝑚2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑡𝑢𝑏𝑒 (𝐴𝑆) = 𝜋 4 × 𝐷𝑆 2 − 𝑑𝑆 2 𝐴𝑆 = 𝜋 4 × 502 − 452 𝑨𝑺 = 𝟑𝟕𝟑. 𝟎𝟔 𝒎𝒎𝟐 𝐷𝑠 = 50 𝑚𝑚 𝑑𝑆 = 45 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 19. 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 𝑟𝑜𝑑 (𝐴𝐶) = 𝜋 4 × 𝑑𝑐 2 𝐴𝐶 = 𝜋 4 × 362 𝑨𝑪 = 𝟏𝟎𝟏𝟕. 𝟖𝟏𝟔 𝒎𝒎𝟐 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝐶 × 𝐴𝐶 98 × 103 = 1.818 𝜎𝑐 × 373.064 + 𝜎𝐶 × 1017.876 𝜎𝐶 = 57.7 Τ 𝑁 𝑚 𝑚2 𝑑𝐶 = 36 𝑚𝑚 𝑃 = 98 𝑘𝑁 = 98 × 103 𝑁 𝐴𝑆 = 373.06 𝑚𝑚2 𝜎𝑠 = 1.818 𝜎𝑐 𝑺𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒄𝒐𝒑𝒑𝒆𝒓 𝒓𝒐𝒅 (𝝈𝑪) = 𝟓𝟕. 𝟕 Τ 𝑵 𝒎 𝒎𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 20. 𝜎𝑠 = 1.818 𝜎𝑐 𝜎𝑠 = 1.818 × 57.7 𝑺𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒔𝒕𝒆𝒆𝒍 𝒕𝒖𝒃𝒆 (𝝈𝒔) = 𝟏𝟎𝟓. 𝟎𝟒𝟐 𝑵/𝒎𝒎𝟐 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝛿𝑙𝑠 = 𝛿𝑙𝐶 𝛿𝑙𝑠 = 𝜎𝑠 𝐿𝑠 𝐸𝑠 𝛿𝑙𝑠 = 105.042 × 1000 2 × 105 𝛿𝑙𝑠 = 0.5252 𝑚𝑚 𝜎𝐶 = 57.7 Τ 𝑁 𝑚 𝑚2 𝑻𝒐𝒕𝒂𝒍 𝒆𝒍𝒐𝒏𝒈𝒂𝒕𝒊𝒐𝒏 𝜹𝒍 = 𝟎. 𝟓𝟐𝟓𝟐 𝒎𝒎 𝐿 = 𝐿𝐶 = 𝐿𝑠 = 1 𝑚 = 1000 𝑚𝑚 𝐸𝑠 = 2 × 105 Τ 𝑁 𝑚 𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 21. Problem 01 Reinforced short concrete column 250 × 250 𝑚𝑚2 in section is reinforced with 8 steel bars. The total area of the steel bar is 2500 𝑚𝑚2 . The column carries a load of 400 kN. If the modulus of elasticity of the steel is 15 times of concrete. Find the stress in concrete and steel. Also find area of the steel required so that column can support 500 kN load. The maximum stress in concrete is 4.5𝑁/𝑚𝑚2 . 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑆𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑡𝑒𝑒𝑙 𝜎𝑠 𝑎𝑛𝑑 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝜎𝐶 =? 𝑇𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎 𝐴 = 250 × 250 𝑚𝑚2 𝐴 = 62500 𝑚𝑚2 𝐴𝑟𝑒𝑎 𝑜𝑓 8 𝑠𝑡𝑒𝑒𝑙 𝑏𝑎𝑟 𝐴𝑆 = 2500 𝑚𝑚2 𝑃 = 400 𝑘𝑁 = 400 × 103 𝑁 𝐸𝑠 = 15 𝐸𝐶 𝐸𝑠 𝐸𝐶 = 15 𝐸𝑠 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝜎𝐶 = 4.5 Τ 𝑁 𝑚 𝑚2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑒𝑒𝑙 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑠𝑜 𝑡ℎ𝑎𝑡 𝑐𝑜𝑙𝑢𝑚𝑛 𝑐𝑎𝑛 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 500 𝑘𝑁 𝑙𝑜𝑎𝑑 𝐿𝑠 = 𝐿𝐶 𝑃 = 500 𝑘𝑁 = 500 × 103 𝑁 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 22. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝛿𝑙𝑠 = 𝛿𝑙𝑐 𝛿𝑙𝑠 = 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝜎𝑠 𝐿𝑠 𝐸𝑠 𝛿𝑙𝑐 = 𝑃𝑐 𝐿𝑐 𝐴𝑐 𝐸𝑐 = 𝜎𝑐 𝐿𝑐 𝐸𝑐 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝛿𝑙𝑠 = 𝛿𝑙𝑐 𝑃𝑠 𝐿𝑠 𝐴𝑠 𝐸𝑠 = 𝑃𝑐 𝐿𝑐 𝐴𝑐 𝐸𝑐 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝑐 𝐿𝑐 𝐸𝑐 𝑭𝒐𝒓𝒎𝒖𝒍𝒂: BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 23. 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝑃𝑠 + 𝑃𝑐 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝑐 × 𝐴𝑐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 24. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝐶 𝐿𝐶 𝐸𝐶 𝝈𝒔 = 𝟏𝟓 𝝈𝑪 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 (𝐴𝐶) = 𝐴 − 𝐴𝑠 𝐴𝐶 = 62500 − 2500 𝑨𝑪 = 𝟔𝟎𝟎𝟎𝟎 𝒎𝒎𝟐 𝜎𝑠 = 𝐸𝑠 𝜎𝐶 𝐿𝐶 𝐸𝐶 𝐿𝑠 𝐸𝑠 𝐸𝐶 = 15 𝐿𝑠 = 𝐿𝐶 𝐴 = 62500 𝑚𝑚2 𝐴𝑆 = 2500 𝑚𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 25. 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝐶 × 𝐴𝐶 400 × 103 = 15 𝜎𝑐 × 2500 + 𝜎𝐶 × 60000 𝜎𝐶 = 4.1 𝑁/𝑚𝑚2 𝜎𝑠 = 15 𝜎𝐶 𝐴𝐶 = 60000 𝑚𝑚2 𝐴𝑆 = 2500 𝑚𝑚2 𝑃 = 400 𝑘𝑁 = 400 × 103 𝑁 𝜎𝑠 = 15 𝜎𝑐 𝜎𝑠 = 15 × 4.1 𝜎𝑠 = 61.54 Τ 𝑁 𝑚 𝑚2 𝑺𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒔𝒕𝒆𝒆𝒍 𝝈𝒔 = 𝟔𝟏. 𝟓𝟒 Τ 𝑵 𝒎 𝒎𝟐 𝑺𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒄𝒐𝒏𝒄𝒓𝒆𝒕𝒆 (𝝈𝑪) = 𝟒. 𝟏 𝑵/𝒎𝒎𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 26. 𝒄𝒂𝒔𝒆 𝒊𝒊 𝑨𝒓𝒆𝒂 𝒐𝒇 𝒔𝒕𝒆𝒆𝒍 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 𝒕𝒐 𝒔𝒖𝒑𝒑𝒐𝒓𝒕 𝟓𝟎𝟎 𝒌𝑵 𝒍𝒐𝒂𝒅 𝜎𝐶 = 4.5 Τ 𝑁 𝑚 𝑚2 𝑃 = 500 𝑘𝑁 = 500 × 103 𝑁 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 1: 𝑆𝑡𝑟𝑎𝑖𝑛 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜎𝑠 𝐿𝑠 𝐸𝑠 = 𝜎𝐶 𝐿𝐶 𝐸𝐶 𝜎𝑠 = 𝐸𝑠 𝜎𝐶 𝐿𝐶 𝐸𝐶 𝐿𝑠 𝜎𝑠 = 15 𝜎𝐶 𝜎𝑠 = 15 × 4.5 𝜎𝑠 = 67.5 Τ 𝑁 𝑚 𝑚2 𝐿𝑠 = 𝐿𝐶 𝐸𝑠 𝐸𝐶 = 15 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 27. 𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 2: 𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 = 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑙𝑜𝑎𝑑𝑠 𝑃 = 𝜎𝑠 × 𝐴𝑠 + 𝜎𝐶 × 𝐴𝐶 In this case, the total area remains unchanged. If the steel area is changed then concrete area will also change. 𝐴𝐶 = 𝐴 − 𝐴𝑠 500 × 103 = 67.5 × 𝐴𝑠 + 4.5 × 𝐴 − 𝐴𝑠 500 × 103 = 67.5 × 𝐴𝑠 + 4.5 × 62500 − 𝐴𝑠 𝑨𝒓𝒆𝒂 𝒐𝒇 𝒔𝒕𝒆𝒆𝒍 (𝑨𝑺) = 𝟑𝟒𝟕𝟐. 𝟐𝟐 𝒎𝒎𝟐 𝐴𝑆 = 3472.22 𝑚𝑚2 𝑃 = 500 𝑘𝑁 = 500 × 103 𝑁 𝐴 = 62500 𝑚𝑚2 𝜎𝑠 = 67.5 Τ 𝑁 𝑚 𝑚2 𝜎𝐶 𝑚𝑎𝑥 = 4.5 Τ 𝑁 𝑚 𝑚2 𝐴 = 𝐴𝐶 + 𝐴𝑠 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