Lecture 05 som 27.02.2021
•
0 likes•56 views
STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS
Recommended
More Related Content
What's hot
What's hot (20)
Similar to Lecture 05 som 27.02.2021
Similar to Lecture 05 som 27.02.2021 (20)
More from BIBIN CHIDAMBARANATHAN
More from BIBIN CHIDAMBARANATHAN (20)
Recently uploaded
Recently uploaded (20)
Lecture 05 som 27.02.2021
- 1. BIBIN CHIDAMBARANATHAN STRESS ANALYSIS OF BARS OF VARYING SECTIONS BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 2. STRESS ANALYSIS OF BARS OF VARYING SECTIONS ❖ Consider the following non uniform cross sections of a member AB, BC, and CD is subjected with an axial load P as shown in figure. ❖ The length and cross-sectional areas of each section of bar is different and therefore stress induced, strain and change in length too will be different for each section of bar. ❖ Young’s modulus of elasticity of each section might be same or different depending on the material of the each section of bar. ❖ Axial load for each section will be same i.e. P. ❖ Determine the total change in length of the bar of varying sections, by adding change in length of each section of bar. BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 3. STRESS ANALYSIS OF BARS OF VARYING SECTIONS ❖ P = Axial Load applied on the bar ❖ 𝐴1, 𝐴2 𝑎𝑛𝑑 𝐴3 = Area of cross section of section 1, section 2 and section 3 respectively ❖ 𝑙1, 𝑙2 𝑎𝑛𝑑 𝑙3 = Length of section 1, section 2 and section 3 respectively ❖ 𝜎1, 𝜎2 𝑎𝑛𝑑 𝜎3 = Stress induced for the section 1, section 2 and section 3 respectively ❖ 𝑒1, 𝑒2 𝑎𝑛𝑑 𝑒3= Strain developed for the section 1, section 2 and section 3 respectively ❖ E = Young’s Modulus of the bar BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 4. 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝐵, 𝜎1 = 𝐿𝑜𝑎𝑑 𝐴𝑟𝑒𝑎 = 𝑃 𝐴1 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝐵, (𝛿𝑙1) = 𝑃 𝑙1 𝐴1𝐸 Similarly 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐵𝐶, (𝜎2) = 𝑃 𝐴2 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐵𝐶, (𝛿𝑙2) = 𝑃 𝑙2 𝐴2𝐸 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐶𝐷, (𝜎3) = 𝑃 𝐴3 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐶𝐷, (𝛿𝑙3) = 𝑃 𝑙3 𝐴3𝐸 Similarly BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 5. 𝑇𝑜𝑡𝑎𝑙 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = 𝛿𝑙1 + 𝛿𝑙2 + 𝛿𝑙3 𝛿𝑙 = 𝑃 𝑙1 𝐴1𝐸 + 𝑃 𝑙2 𝐴2𝐸 + 𝑃 𝑙3 𝐴3𝐸 𝛿𝑙 = 𝑃 𝐸 𝑙1 𝐴1 + 𝑙2 𝐴2 + 𝑙3 𝐴3 Let us consider that Young’s modulus of elasticity of each section is different, then the total change in length of the bar. 𝛿𝑙 = 𝑃 𝑙1 𝐴1𝐸1 + 𝑙2 𝐴2𝐸2 + 𝑙3 𝐴3𝐸3 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 6. Problem 06 A straight bar of 450 mm long is 20 mm in diameter for the first 250 mm length and 10mm diameter for the remaining length. If the bar is subjected to an axial pull of 10KN find the extension of the bar. Take 𝐸 = 2 × 105 𝑁/𝑚𝑚2. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑙 = 450 𝑚𝑚 𝑃 = 10 𝑘𝑁 = 10 × 103 𝑁 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 𝛿𝑙 =? 250 mm 200 mm 𝜑 20 𝑚𝑚 𝜑 10 𝑚𝑚 𝑃 𝑃 𝑑1 = 20 𝑚𝑚 𝑑2 = 10 𝑚𝑚 𝑙1 = 250 𝑚𝑚 𝑙2 = 200 𝑚𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 7. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝐴𝑟𝑒𝑎 of section 1 𝐴1 = 𝜋 4 × 𝑑1 2 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 (𝛿𝑙) = 𝑃 𝐸 𝑙1 𝐴1 + 𝑙2 𝐴2 𝐴𝑟𝑒𝑎 of section 2 𝐴2 = 𝜋 4 × 𝑑2 2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 8. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐴𝑟𝑒𝑎 of section 1 𝐴1 = 𝜋 4 × 𝑑1 2 𝐴𝑟𝑒𝑎 of section 2 𝐴2 = 𝜋 4 × 𝑑2 2 𝑑1 = 20 𝑚𝑚 𝑑2 = 10 𝑚𝑚 𝐴1 = 𝜋 4 × 202 𝑨𝒓𝒆𝒂 𝐨𝐟 𝐬𝐞𝐜𝐭𝐢𝐨𝐧 𝟏 𝑨𝟏 = 𝟑𝟏𝟒. 𝟏𝟔 𝒎𝒎𝟐 𝐴2 = 𝜋 4 × 102 𝑨𝒓𝒆𝒂 𝐨𝐟 𝐬𝐞𝐜𝐭𝐢𝐨𝐧 𝟐 𝑨𝟐 = 𝟕𝟖. 𝟓𝟒 𝒎𝒎𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 9. 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 (𝛿𝑙) = 𝑃 𝐸 𝑙1 𝐴1 + 𝑙2 𝐴2 𝑬𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒓 𝜹𝒍 = 𝟎. 𝟏𝟔𝟔 𝒎𝒎 𝛿𝑙 = 10 × 103 2 × 105 250 314.16 + 200 78.54 𝑃 = 10 𝑘𝑁 = 10 × 103 𝑁 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝑙1 = 250 𝑚𝑚 𝑙2 = 200 𝑚𝑚 𝐴1 = 314.16 𝑚𝑚2 𝐴2 = 78.54 𝑚𝑚2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 10. Problem 07 An axial pull of 35KN is acting on a bar consisting of three lengths: 20cm with 2cm diameter, 25cm with 3cm diameter and 22cm with 5cm diameter. If the young’s modulus = 2 × 105 𝑁/𝑚𝑚2. Determine (i) stress in each section, (ii) Total extension of the bar. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝑃 = 35 𝑘𝑁 = 35 × 103 𝑁 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝑇𝑜𝑡𝑎𝑙 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 𝛿𝑙 =? 𝑑1 = 20 𝑚𝑚 𝑑2 = 30 𝑚𝑚 𝑙1 = 20 𝑐𝑚 = 200 𝑚𝑚 𝑙2 = 25 𝑐𝑚 = 250 𝑚𝑚 𝑙3 = 22 𝑐𝑚 = 220 𝑚𝑚 𝑑3 = 50 𝑚𝑚 𝑆𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 =? BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 11. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝐴𝑟𝑒𝑎 of section1 𝐴1 = 𝜋 4 × 𝑑1 2 𝐴𝑟𝑒𝑎 of section 2 𝐴2 = 𝜋 4 × 𝑑2 2 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 𝜎1 = 𝑃 𝐴1 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 (𝜎2) = 𝑃 𝐴2 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 (𝜎3) = 𝑃 𝐴3 𝐴𝑟𝑒𝑎 of section 3 𝐴3 = 𝜋 4 × 𝑑3 2 𝑇𝑜𝑡𝑎𝑙 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 (𝛿𝑙) = 𝑃 𝐸 𝑙1 𝐴1 + 𝑙2 𝐴2 + 𝑙3 𝐴3 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 12. 𝐴𝑟𝑒𝑎 of section1 𝐴1 = 𝜋 4 × 𝑑1 2 𝐴𝑟𝑒𝑎 of section 2 𝐴2 = 𝜋 4 × 𝑑2 2 𝐴𝑟𝑒𝑎 of section 3 𝐴3 = 𝜋 4 × 𝑑3 2 𝑑1 = 20 𝑚𝑚 𝑑2 = 30 𝑚𝑚 𝑑3 = 50 𝑚𝑚 𝐴1 = 𝜋 4 × 202 𝑨𝒓𝒆𝒂 𝐨𝐟 𝐬𝐞𝐜𝐭𝐢𝐨𝐧𝟏 𝑨𝟏 = 𝟑𝟏𝟒. 𝟏𝟔 𝐦𝐦𝟐 𝐴2 = 𝜋 4 × 302 𝑨𝒓𝒆𝒂 𝐨𝐟 𝐬𝐞𝐜𝐭𝐢𝐨𝐧 𝟐 𝑨𝟐 = 𝟕𝟎𝟔. 𝟖𝟔 𝐦𝐦𝟐 𝐴3 = 𝜋 4 × 502 𝑨𝒓𝒆𝒂 𝐨𝐟 𝐬𝐞𝐜𝐭𝐢𝐨𝐧 𝟑 𝑨𝟑 = 𝟏𝟗𝟔𝟑. 𝟓 𝐦𝐦𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 13. 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1 𝜎1 = 𝑃 𝐴1 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2 (𝜎2) = 𝑃 𝐴2 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3 (𝜎3) = 𝑃 𝐴3 𝐴1 = 314.16 mm2 𝑃 = 35 𝑘𝑁 = 35 × 103 𝑁 𝐴2 = 706.86 mm2 𝐴3 = 1963.5 mm2 𝜎1 = 35 × 103 314.16 𝑻𝒆𝒏𝒔𝒊𝒍𝒆 𝒔𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟏 𝝈𝟏 = 𝟏𝟏𝟏. 𝟒𝟔 𝑵/𝒎𝒎𝟐 𝜎2 = 35 × 103 706.86 𝑻𝒆𝒏𝒔𝒊𝒍𝒆 𝒔𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟐 𝝈𝟐 = 𝟒𝟗. 𝟓𝟑 𝑵/𝒎𝒎𝟐 𝜎3 = 35 × 103 1963.5 𝑻𝒆𝒏𝒔𝒊𝒍𝒆 𝒔𝒕𝒓𝒆𝒔𝒔 𝒊𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟑 𝝈𝟑 = 𝟏𝟕. 𝟖𝟑 𝑵/𝒎𝒎𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 14. 𝑇𝑜𝑡𝑎𝑙 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 (𝛿𝑙) = 𝑃 𝐸 𝑙1 𝐴1 + 𝑙2 𝐴2 + 𝑙3 𝐴3 𝑃 = 35 𝑘𝑁 = 35 × 103 𝑁 𝐸 = 2 × 105 Τ 𝑁 𝑚 𝑚2 𝑙1 = 20 𝑐𝑚 = 200 𝑚𝑚 𝑙2 = 25 𝑐𝑚 = 250 𝑚𝑚 𝑙3 = 22 𝑐𝑚 = 220 𝑚𝑚 𝛿𝑙 = 35 × 103 2 × 105 200 314.16 + 250 706.86 + 220 1963.5 𝐴1 = 314.16 mm2 𝐴2 = 706.86 mm2 𝐴3 = 1963.5 mm2 𝑻𝒐𝒕𝒂𝒍 𝑬𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒓 𝜹𝒍 = 𝟎. 𝟏𝟖𝟑𝟔 𝒎𝒎 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 15. Problem 08 A member formed by connecting a steel bar to an aluminium bar shown in the figure. Assuming that the bars are presented from buckling sideways. Calculate the magnitude of the force ’P’ that will cause total length of the member to decrease 0.25mm. The values of elastic modulus for steel and Al are 2.1 × 105 𝑁/𝑚𝑚2 𝑎𝑛𝑑 7 × 104 𝑁/𝑚𝑚2. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑻𝒐 𝒇𝒊𝒏𝒅: 𝛿𝑙 = 0.25 𝑚𝑚 𝐸1 = 2.1 × 105 Τ 𝑁 𝑚 𝑚2 M𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑐𝑒 𝑃 =? 𝐴1 = 5 𝑐𝑚 × 5 𝑐𝑚 𝑙1 = 30 𝑐𝑚 = 300 𝑚𝑚 𝐸2 = 7 × 104 Τ 𝑁 𝑚 𝑚2 𝑙2 = 38 𝑐𝑚 = 380 𝑚𝑚 𝐴2 = 10 𝑐𝑚 × 10 𝑐𝑚 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 16. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝑇𝑜𝑡𝑎𝑙 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 (𝛿𝑙) = 𝑃 𝑙1 𝐴1𝐸1 + 𝑙2 𝐴2𝐸2 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝐴1 = 5 𝑐𝑚 × 5 𝑐𝑚 𝐴2 = 10 𝑐𝑚 × 10 𝑐𝑚 𝐴1 = 50 𝑚𝑚 × 50 𝑚𝑚 𝑨𝟏 = 𝟐𝟓𝟎𝟎 𝒎𝒎𝟐 𝐴2 = 100 𝑚𝑚 × 100 𝑚𝑚 𝑨𝟐 = 𝟏𝟎𝟎𝟎𝟎 𝒎𝒎𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 17. 𝑇𝑜𝑡𝑎𝑙 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 (𝛿𝑙) = 𝑃 𝑙1 𝐴1𝐸1 + 𝑙2 𝐴2𝐸2 𝐴1 = 2500 𝑚𝑚2 𝐴2 = 10000 𝑚𝑚2 𝐸1 = 2.1 × 105 Τ 𝑁 𝑚 𝑚2 𝑙1 = 30 𝑐𝑚 = 300 𝑚𝑚 𝐸2 = 7 × 104 Τ 𝑁 𝑚 𝑚2 𝑙2 = 38 𝑐𝑚 = 380 𝑚𝑚 0.25 = 𝑃 300 2500 × 2.1 × 105 + 380 10000 × 7 × 104 𝛿𝑙 = 0.25 𝑚𝑚 𝑃 = 224.37 × 103 𝑁 𝐌𝒂𝒈𝒏𝒊𝒕𝒖𝒅𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒇𝒐𝒓𝒄𝒆 𝑷 = 𝟐𝟐𝟒. 𝟑𝟕 𝒌𝑵 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 18. Problem 09 The bar shown in the figure subjected to a tensile load of 160 kN. If the stress in the middle section is limited to 150 𝑁/𝑚𝑚2 , determine the diameter of the middle portion. Find the length of the middle portion if the total elongation of the bar is to be 0.2mm. Youngs modulus of the materials is 2.5 × 105 𝑁/𝑚𝑚2. 𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂: 𝑃 = 160 𝑘𝑁 = 160 × 103 𝑁 𝐸 = 2.5 × 105 Τ 𝑁 𝑚 𝑚2 𝑙 = 40 𝑐𝑚 = 400 𝑚𝑚 𝑑1 = 𝑑3 = 6 𝑐𝑚 = 60 𝑚𝑚 𝜎2 = 150 𝑁/𝑚𝑚2 𝛿𝑙 = 0.2 𝑚𝑚 𝑻𝒐 𝒇𝒊𝒏𝒅: Diameter of the middle portion 𝑑2 =? Length of the middle portion 𝑙2 =? 𝑙 = 𝑙1 + 𝑙2 + 𝑙3 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 19. 𝑆𝑡𝑟𝑒𝑠𝑠 at section 2 𝜎2 = 𝑃 𝐴2 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶ 𝑇𝑜𝑡𝑎𝑙 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 (𝛿𝑙) = 𝑃 𝐸 𝑙1 𝐴1 + 𝑙2 𝐴2 + 𝑙3 𝐴3 𝐴𝑟𝑒𝑎 of section 1 𝐴1 = 𝜋 4 × 𝑑1 2 𝐴𝑟𝑒𝑎 of section 2 𝐴2 = 𝜋 4 × 𝑑2 2 𝐴𝑟𝑒𝑎 of section 3 𝐴3 = 𝜋 4 × 𝑑3 2 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 20. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏: 𝑆𝑡𝑟𝑒𝑠𝑠 at section 2 𝜎2 = 𝑃 𝐴2 150 = 160 × 103 𝐴2 𝑃 = 160 𝑘𝑁 = 160 × 103 𝑁 𝜎2 = 150 𝑁/𝑚𝑚2 𝑨𝒓𝒆𝒂 𝒐𝒇 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟐 𝑨𝟐 = 𝟏𝟎𝟔𝟔. 𝟔𝟕 𝐦𝐦𝟐 𝐴𝑟𝑒𝑎 of section 2 𝐴2 = 𝜋 4 × 𝑑2 2 1.06 × 103 = 𝜋 4 × 𝑑2 2 𝑫𝒊𝒂𝒎𝒆𝒕𝒆𝒓 𝒐𝒇 𝒎𝒊𝒅𝒅𝒍𝒆 𝒑𝒐𝒓𝒕𝒊𝒐𝒏 𝒅𝟐 = 𝟑𝟔. 𝟖𝟓 𝒎𝒎 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 21. 𝐴𝑟𝑒𝑎 of section1 𝐴1 = 𝜋 4 × 𝑑1 2 𝐴1 = 𝜋 4 × 602 𝑨𝒓𝒆𝒂 𝐨𝐟 𝐬𝐞𝐜𝐭𝐢𝐨𝐧𝟏 𝑨𝟏 = 𝟐𝟖𝟐𝟕. 𝟒𝟑 𝐦𝐦𝟐 𝑨𝒓𝒆𝒂 𝐨𝐟 𝐬𝐞𝐜𝐭𝐢𝐨𝐧 𝟑 𝑨𝟑 = 𝟐𝟖𝟐𝟕. 𝟒𝟑 𝐦𝐦𝟐 𝑙1 + 𝑙3 = 𝑙 − 𝑙2 𝑙 = 𝑙1 + 𝑙2 + 𝑙3 𝑙 = 40 𝑐𝑚 = 400 𝑚𝑚 𝑙1 + 𝑙3 = 400 − 𝑙2 𝑑1 = 𝑑3 = 6 𝑐𝑚 = 60 𝑚𝑚 𝑨𝟏 = 𝑨𝟑 = 𝟐𝟖𝟐𝟕. 𝟒𝟑 𝐦𝐦𝟐 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 22. 𝑇𝑜𝑡𝑎𝑙 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 (𝛿𝑙) = 𝑃 𝐸 𝑙1 𝐴1 + 𝑙2 𝐴2 + 𝑙3 𝐴3 𝛿𝑙 = 𝑃 𝐸 𝑙1 𝐴1 + 𝑙2 𝐴2 + 𝑙3 𝐴1 𝛿𝑙 = 𝑃 𝐸 𝑙1 + 𝑙3 𝐴1 + 𝑙2 𝐴2 + 𝐴1 = 𝐴3 = 2827.43 mm2 𝑙1 + 𝑙3 = 400 − 𝑙2 𝑃 = 160 𝑘𝑁 = 160 × 103 𝑁 𝐸 = 2.5 × 105 Τ 𝑁 𝑚 𝑚2 𝛿𝑙 = 0.2 𝑚𝑚 𝐴2 = 1066.67 mm2 0.2 = 160 × 103 2.5 × 105 400 − 𝑙2 2827.43 + 𝑙2 1066.67 + 𝑙2 = 207.14 𝑚𝑚 Length of the middle portion (𝒍𝟐) = 𝟐𝟎𝟕. 𝟏𝟒 𝒎𝒎 BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
- 23. Thank You BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY