STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS

1. BIBIN CHIDAMBARANATHAN
PRINCIPLE
OF
SUPERPOSITION
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

2. PRINCIPLE OF SUPERPOSITION
❖ When there are numbers of loads are acting together on an elastic material, the
resultant strain will be the sum of individual strains caused by each load acting
separately.
A B
C D
𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒
𝑳𝟏
𝑳𝟐 𝑳𝟑
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

3. A B
𝑷𝟏 𝑷𝟐
𝑳𝟏
𝑷𝟑
𝑷𝟒
A B
C D
𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒
𝑳𝟏
𝑳𝟐 𝑳𝟑
𝐷𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐴𝐵 𝛿𝑙𝐴𝐵 =
𝑃𝐴𝐵𝐿𝐴𝐵
𝐴𝐴𝐵𝐸𝐴𝐵
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

4. B C
𝑷𝟏
𝑷𝟐 𝑷𝟑
𝑷𝟒
𝑳𝟐
A B
C D
𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒
𝑳𝟏
𝑳𝟐 𝑳𝟑
𝐷𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐵𝐶 𝛿𝑙𝐵𝐶 =
𝑃𝐵𝐶𝐿𝐵𝐶
𝐴𝐵𝐶
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

5. A B
C D
𝑷𝟏 𝑷𝟐 𝑷𝟑 𝑷𝟒
𝑳𝟏
𝑳𝟐 𝑳𝟑
C D
𝑷𝟏
𝑷𝟐
𝑷𝟑 𝑷𝟒
𝑳𝟑
𝐷𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝐶𝐷 𝛿𝑙𝐶𝐷 =
𝑃𝐶𝐷𝐿𝐶𝐷
𝐴𝐶𝐷
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

6. 𝑻𝒐𝒕𝒂𝒍 𝐝𝐞𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧 𝜹𝒍 =
𝑷𝑨𝑩𝑳𝑨𝑩
𝑨𝑨𝑩𝑬𝑨𝑩
+
𝑷𝑩𝑪𝑳𝑩𝑪
𝑨𝑩𝑪𝑬𝑩𝑪
+
𝑷𝑪𝑫𝑳𝑪𝑫
𝑨𝑪𝑫𝑬𝑪𝑫
𝑻𝒐𝒕𝒂𝒍 𝐝𝐞𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧 𝜹𝒍 =
𝟏
𝑬
𝑷𝑨𝑩𝑳𝑨𝑩
𝑨𝑨𝑩
+
𝑷𝑩𝑪𝑳𝑩𝑪
𝑨𝑩𝑪
+
𝑷𝑪𝑫𝑳𝑪𝑫
𝑨𝑪𝑫
𝑇𝑜𝑡𝑎𝑙 deformation 𝛿𝑙 = 𝛿𝑙𝐴𝐵 + 𝛿𝑙BC + 𝛿𝑙𝐶𝐷 𝛿𝑙𝐴𝐵 =
𝑃𝐴𝐵𝐿𝐴𝐵
𝐴𝐴𝐵𝐸𝐴𝐵
𝛿𝑙𝐵𝐶 =
𝑃𝐵𝐶𝐿𝐵𝐶
𝐴𝐵𝐶
𝛿𝑙𝐶𝐷 =
𝑃𝐶𝐷𝐿𝐶𝐷
𝐴𝐶𝐷
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

7. Problem 01
A brass bar having cross sectional area of 1000 𝑚𝑚2 is subjected to axial forces as shown
below. Find the total elongation of the bar. Take 𝐸 = 1.05 × 105 𝑁/𝑚𝑚2.
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 (𝛿𝑙) =?
𝟓𝟎 𝒌𝑵
𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵
𝟏𝟎 𝒌𝑵
𝟔𝟎𝟎 𝒎𝒎 𝟏 𝒎 𝟏. 𝟐 𝒎
A B C D
𝐴 = 1000 𝑚𝑚2
𝐿1 = 600 𝑚𝑚
𝐿2 = 1𝑚 = 1000 𝑚𝑚
𝐿3 = 1.2 𝑚 = 1200 𝑚𝑚
𝐸 = 1.05 × 105 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

8. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂:
𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 =
𝑃1𝐿1
𝐴1𝐸1
+
𝑃2𝐿2
𝐴2𝐸2
+
𝑃3𝐿3
𝐴3𝐸3
𝛿𝑙 =
𝑃1𝐿1 + 𝑃2𝐿2 + 𝑃3𝐿3
𝐴𝐸
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

9. 𝟓𝟎 𝒌𝑵
𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵
𝟏𝟎 𝒌𝑵
𝟔𝟎𝟎 𝒎𝒎 𝟏 𝒎 𝟏. 𝟐 𝒎
A B C D
1 3
2
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐴 = 50 𝑘𝑁 𝑇
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐵 = 80 − 20 − 10 = 50 𝑘𝑁 𝑇
∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟏, 𝑷𝟏 = 𝟓𝟎 𝒌𝑵 𝑻
A B
𝟓𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵
𝟔𝟎𝟎 𝒎𝒎
𝟏𝟎 𝒌𝑵
𝟖𝟎 𝒌𝑵
1
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

10. 𝟓𝟎 𝒌𝑵
𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵
𝟏𝟎 𝒌𝑵
𝟔𝟎𝟎 𝒎𝒎 𝟏 𝒎 𝟏. 𝟐 𝒎
A B C D
1 3
2
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐵 = 50 − 80 = −30 𝑘𝑁 = 30 𝑘𝑁 𝐶
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐶 = −20 − 10 = −30 𝑘𝑁 = −30 𝑘𝑁 𝐶
∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟐, 𝑷𝟐 = 𝟑𝟎 𝒌𝑵 𝑪
B C
𝟓𝟎 𝒌𝑵
𝟖𝟎 𝒌𝑵 𝟏𝟎 𝒌𝑵
𝟐𝟎 𝒌𝑵
𝟏𝟎𝟎𝟎𝒎𝒎
2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

11. 𝟓𝟎 𝒌𝑵
𝟖𝟎 𝒌𝑵 𝟐𝟎 𝒌𝑵
𝟏𝟎 𝒌𝑵
𝟔𝟎𝟎 𝒎𝒎 𝟏 𝒎 𝟏. 𝟐 𝒎
A B C D
1 3
2
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝑐 = 50 + 20 − 80 = −10 𝑘𝑁 = 10 𝑘𝑁 𝐶
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐷 = −10 = −10 𝑘𝑁 = −10 𝑘𝑁 𝐶
∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟑, 𝑷𝟑 = 𝟏𝟎 𝒌𝑵 𝑪
C D
𝟓𝟎 𝒌𝑵
𝟖𝟎 𝒌𝑵
𝟐𝟎 𝒌𝑵 𝟏𝟎 𝒌𝑵
𝟏𝟐𝟎𝟎 𝒎𝒎
3
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

12. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 =
𝑃1𝐿1 + 𝑃2𝐿2 + 𝑃3𝐿3
𝐴𝐸
𝛿𝑙 =
50 × 103 × 600 + −30 × 103 × 1000 + −10 × 103 × 1200
1000 × 1.05 × 105
𝛿𝑙 = −0.114 𝑚𝑚
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝜹𝒍 = 𝟎. 𝟏𝟏𝟒 𝒎𝒎
𝐴 = 1000 𝑚𝑚2
𝐿1 = 600 𝑚𝑚
𝐿2 = 1𝑚 = 1000 𝑚𝑚
𝐿3 = 1.2 𝑚 = 1200 𝑚𝑚
𝐸 = 1.05 × 105 Τ
𝑁 𝑚 𝑚2
𝑃1 = 50 𝑘𝑁 𝑇
𝑃2 = 30 𝑘𝑁 𝐶
𝑃3 = 10 𝑘𝑁 𝐶
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

13. Problem 01
A member ABCD is subjected to point loads 𝑃1, 𝑃2, 𝑃3 and 𝑃4 as shown in the figure.
Calculate the force 𝑃2 necessary for equilibrium, if 𝑃1 = 45 𝑘𝑁, 𝑃3 = 450 𝑘𝑁 and 𝑃4 =
130 𝑘𝑁. Determine the total elongation of the member assuming 𝐸 = 2.1 × 105 𝑁/𝑚𝑚2.
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑒𝑚𝑏𝑒𝑟(𝛿𝑙) =?
A
B
𝟏𝟐𝟎𝟎 𝒎𝒎
C
𝟔𝟎𝟎 𝒎𝒎
D
𝟗𝟎𝟎 𝒎𝒎
𝟔𝟐𝟓 𝒎𝒎𝟐
𝟐𝟓𝟎𝟎 𝒎𝒎𝟐
1250 𝒎𝒎𝟐
𝑷𝟏
𝑷𝟐 𝑷𝟑
𝑷𝟒
𝐴𝐴𝐵 = 625 𝑚𝑚2
𝐴𝐵𝐶 = 2500 𝑚𝑚2
𝐴𝐶𝐷 = 1250 𝑚𝑚2
𝐿𝐴𝐵 = 1200 𝑚𝑚
𝐿𝐵𝐶 = 600 𝑚𝑚
𝐿𝐶𝐷 = 900 𝑚𝑚 𝑃1 = 45 𝑘𝑁 𝑃3 = 450 𝑘𝑁 𝑃4 = 130 𝑘𝑁
𝐹𝑜𝑟𝑐𝑒 (𝑃2) =?
𝐸 = 2.1 × 105 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

14. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂:
𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 =
𝑃𝐴𝐵𝐿𝐴𝐵
𝐴𝐴𝐵𝐸𝐴𝐵
+
𝑃𝐵𝐶𝐿𝐵𝐶
𝐴𝐵𝐶𝐸𝐵𝐶
+
𝑃𝐶𝐷𝐿𝐶𝐷
𝐴𝐶𝐷𝐸𝐶𝐷
𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 =
1
𝐸
𝑃𝐴𝐵𝐿𝐴𝐵
𝐴𝐴𝐵
+
𝑃𝐵𝐶𝐿𝐵𝐶
𝐴𝐵𝐶
+
𝑃𝐶𝐷𝐿𝐶𝐷
𝐴𝐶𝐷
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

15. A
B
𝟏𝟐𝟎𝟎 𝒎𝒎
C
𝟔𝟎𝟎 𝒎𝒎
D
𝟗𝟎𝟎 𝒎𝒎
𝟔𝟐𝟓 𝒎𝒎𝟐
𝟐𝟓𝟎𝟎 𝒎𝒎𝟐
1250 𝒎𝒎𝟐
𝑷𝟏
𝑷𝟐 𝑷𝟑
𝑷𝟒
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑙𝑒𝑓𝑡 = 𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑡𝑜𝑤𝑎𝑟𝑑𝑠 𝑟𝑖𝑔ℎ𝑡
𝑃1 + 𝑃3 = 𝑃2 + 𝑃4
45 + 450 = 𝑃2 + 130
𝑷𝟐 = 𝟑𝟔𝟓 𝒌𝑵
𝑃1 = 45 𝑘𝑁
𝑃3 = 450 𝑘𝑁
𝑃4 = 130 𝑘𝑁
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

16. A
B
𝟏𝟐𝟎𝟎 𝒎𝒎
C
𝟔𝟎𝟎 𝒎𝒎
D
𝟗𝟎𝟎 𝒎𝒎
𝟔𝟐𝟓 𝒎𝒎𝟐
𝟐𝟓𝟎𝟎 𝒎𝒎𝟐
1250 𝒎𝒎𝟐
𝑷𝟏
𝑷𝟐 𝑷𝟑
𝑷𝟒
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 1
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐴 = 45 𝑘𝑁 𝑇
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐵 = 365 − 450 + 130 = 45 𝑘𝑁 𝑇
∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟏, 𝑷𝑨𝑩 = 𝟒𝟓 𝒌𝑵 𝑻
A B
𝟒𝟓 𝒌𝑵
𝟒𝟓𝟎 𝒌𝑵
𝟏𝟐𝟎𝟎 𝒎𝒎
𝟏𝟑𝟎 𝒌𝑵
𝟑𝟔𝟓 𝒌𝑵
𝟔𝟐𝟓 𝒎𝒎𝟐
𝑃1 = 45 𝑘𝑁
𝑃3 = 450 𝑘𝑁
𝑃4 = 130 𝑘𝑁
𝑃2 = 365 𝑘𝑁
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

17. A
B
𝟏𝟐𝟎𝟎 𝒎𝒎
C
𝟔𝟎𝟎 𝒎𝒎
D
𝟗𝟎𝟎 𝒎𝒎
𝟔𝟐𝟓 𝒎𝒎𝟐
𝟐𝟓𝟎𝟎 𝒎𝒎𝟐
1250 𝒎𝒎𝟐
𝑷𝟏
𝑷𝟐 𝑷𝟑
𝑷𝟒
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 2
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐵 = 45 − 365 = −320 𝑘𝑁 = −320 𝑘𝑁 𝐶
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐶 = −450 + 130 = −320 𝑘𝑁 = −320 𝑘𝑁 𝐶
∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟐, 𝑷𝑩𝑪 = 𝟑𝟐𝟎 𝒌𝑵 𝑪
B C
𝟒𝟓 𝒌𝑵
𝟑𝟔𝟓 𝒌𝑵 𝟏𝟑𝟎 𝒌𝑵
𝟒𝟓𝟎 𝒌𝑵
𝟔𝟎𝟎 𝒎𝒎
𝟐𝟓𝟎𝟎 𝒎𝒎𝟐
𝑃1 = 45 𝑘𝑁
𝑃3 = 450 𝑘𝑁
𝑃4 = 130 𝑘𝑁
𝑃2 = 365 𝑘𝑁
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

18. A
B
𝟏𝟐𝟎𝟎 𝒎𝒎
C
𝟔𝟎𝟎 𝒎𝒎
D
𝟗𝟎𝟎 𝒎𝒎
𝟔𝟐𝟓 𝒎𝒎𝟐
𝟐𝟓𝟎𝟎 𝒎𝒎𝟐
1250 𝒎𝒎𝟐
𝑷𝟏
𝑷𝟐 𝑷𝟑
𝑷𝟒
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 3
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝑐 = 45 − 365 + 450 = 130 𝑘𝑁 = 130 𝑘𝑁 𝑇
𝐹𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑠𝑖𝑑𝑒 𝐷 = 130 𝑘𝑁 = 130 𝑘𝑁 𝑇
∴ 𝑭𝒐𝒓𝒄𝒆 𝒂𝒄𝒕𝒊𝒏𝒈 𝒐𝒏 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝟑, 𝑷𝑪𝑫 = 𝟏𝟑𝟎 𝒌𝑵 (𝑻)
C D
𝟒𝟓 𝒌𝑵
𝟑𝟔𝟓 𝒌𝑵
𝟒𝟓𝟎 𝒌𝑵 𝟏𝟑𝟎 𝒌𝑵
𝟗𝟎𝟎 𝒎𝒎
1250 𝒎𝒎𝟐
𝑃1 = 45 𝑘𝑁
𝑃3 = 450 𝑘𝑁
𝑃4 = 130 𝑘𝑁
𝑃2 = 365 𝑘𝑁
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

19. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 =
1
𝐸
𝑃𝐴𝐵𝐿𝐴𝐵
𝐴𝐴𝐵
+
𝑃𝐵𝐶𝐿𝐵𝐶
𝐴𝐵𝐶
+
𝑃𝐶𝐷𝐿𝐶𝐷
𝐴𝐶𝐷
𝛿𝑙 =
1
2.1 × 105
45 × 103 × 1200
625
+
−320 × 103 × 600
2500
+
130 × 103 × 1200
1250
𝛿𝑙 = 0.4914 𝑚𝑚
𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 𝜹𝒍 = 𝟎. 𝟒𝟗𝟏𝟒 𝒎𝒎
𝐴𝐴𝐵 = 625 𝑚𝑚2
𝐴𝐵𝐶 = 2500 𝑚𝑚2
𝐴𝐶𝐷 = 1250 𝑚𝑚2
𝐿𝐴𝐵 = 1200 𝑚𝑚
𝐿𝐵𝐶 = 600 𝑚𝑚
𝐿𝐶𝐷 = 900 𝑚𝑚
𝑃𝐴𝐵 = 45 𝑘𝑁 𝑇
𝑃𝐵𝐶 = 320 𝑘𝑁 𝐶
𝑃CD = 130 𝑘𝑁 𝑇
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

20. Problem 01
A tensile load of 40 kN is acting on a rod of diameter 40 mm and of length 4 m. A bore of
diameter 20 mm is made centrally on the rod. To what length the rod should be bored so
that the total extension will increase 30% under the same tensile load. Take 𝐸 = 2.1 ×
105 𝑁/𝑚𝑚2.
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
𝑙𝑒𝑛𝑔𝑡ℎ 𝑡ℎ𝑒 𝑟𝑜𝑑 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑏𝑜𝑟𝑒𝑑 (𝑥) =?
𝑃 = 40 𝑘𝑁 = 40 × 103 𝑁 𝐷 = 40 𝑚𝑚 𝐿 = 4 𝑚 = 4000 𝑚𝑚 𝐸 = 2 × 105 Τ
𝑁 𝑚 𝑚2
𝑑𝐵 = 20 𝑚𝑚 𝛿𝑙𝑆𝐵 = 1.3 𝛿𝑙 𝐷S = 40 𝑚𝑚
𝟒𝟎 𝒌𝑵
𝟒 𝒎
𝟒𝟎 𝒌𝑵
𝟒𝟎 𝒌𝑵
𝒙 𝒎
𝟒𝟎 𝒌𝑵
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

21. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂:
𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝐴 =
𝜋
4
× 𝐷2
𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝛿𝑙 =
𝑃𝐿
𝐴𝐸
𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝑠 =
𝜋
4
× 𝐷𝑆
2
𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝐵 =
𝜋
4
× (𝐷𝑆
2
−𝑑𝐵
2
)
𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑢𝑛𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝛿𝑙𝑆 =
𝑃𝐿𝑠
𝐴𝑆𝐸
𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝛿𝑙𝐵 =
𝑃𝐿𝐵
𝐴𝐵𝐸
𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑓𝑡𝑒𝑟 𝑏𝑜𝑟𝑒 𝑖𝑠 𝑚𝑎𝑑𝑒 𝛿𝑙𝑆𝐵 = 𝛿𝑙𝑆 + 𝛿𝑙𝐵
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

22. 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝐴 =
𝜋
4
× 𝐷2
𝐴 =
𝜋
4
× 402
𝑪𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏𝒂𝒍 𝒂𝒓𝒆𝒂 𝑨 = 𝟏𝟐𝟓𝟔. 𝟔𝟒 𝒎𝒎𝟐
𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝛿𝑙 =
𝑃𝐿
𝐴𝐸
𝛿𝑙 =
40 × 103
× 4000
1256.64 × 2 × 105
𝑻𝒐𝒕𝒂𝒍 𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 (𝜹𝒍) = 𝟎. 𝟔𝟑𝟔𝟔 𝒎𝒎
𝟒𝟎 𝒌𝑵
𝟒 𝒎
𝟒𝟎 𝒌𝑵
𝑃 = 40 𝑘𝑁 = 40 × 103 𝑁
𝐷 = 40 𝑚𝑚
𝐿 = 4 𝑚 = 4000 𝑚𝑚
𝐸 = 2 × 105 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

23. 𝟒𝟎 𝒌𝑵
𝒙 𝒎
𝟒𝟎 𝒌𝑵
𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑓𝑡𝑒𝑟 𝑏𝑜𝑟𝑒 𝛿𝑙𝑆𝐵 = 1.3 𝛿𝑙
𝛿𝑙𝑆𝐵 = 1.3 × 0.6366
𝑻𝒐𝒕𝒂𝒍 𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝒂𝒇𝒕𝒆𝒓 𝒃𝒐𝒓𝒆 𝜹𝒍𝑺𝑩 = 𝟎. 𝟖𝟐𝟕 𝒎𝒎
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐿𝐵 = 𝑥 𝑚
𝐿𝐵 = 1000𝑥 𝑚𝑚
𝛿𝑙 = 0.6366 𝑚𝑚
𝑳𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒔𝒐𝒍𝒊𝒅 𝒑𝒐𝒓𝒕𝒊𝒐𝒏 𝑳𝑺 = 𝟒𝟎𝟎𝟎 − 𝟏𝟎𝟎𝟎𝒙
𝐿 = 4 𝑚 = 4000 𝑚𝑚
𝟒 𝒎
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

24. 𝟒𝟎 𝒌𝑵
𝒙 𝒎
𝟒𝟎 𝒌𝑵
𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑢𝑛𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝑠 =
𝜋
4
× 𝐷𝑆
2
𝐴𝑆 =
𝜋
4
× 402
𝑨𝑺 = 𝟒𝟎𝟎𝝅 𝒎𝒎𝟐
𝐷𝑆 = 40 𝑚𝑚
𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝐴𝐵 =
𝜋
4
× (𝐷𝑆
2
−𝑑𝐵
2
)
ቁ
𝐴𝐵 =
𝜋
4
× (40
2
− 202
𝑨𝑩 = 𝟑𝟎𝟎𝝅 𝒎𝒎𝟐
𝑑𝐵 = 20 𝑚𝑚
𝐷S = 40 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

25. 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑢𝑛𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝛿𝑙𝑆 =
𝑃𝐿𝑠
𝐴𝑆𝐸
𝛿𝑙𝑆 =
40 × 103
400𝜋 × 2 × 105
× 4000 − 1000𝑥
𝜹𝒍𝑺 =
𝟒 − 𝒙
𝟐𝝅
𝒎𝒎
𝑃 = 40 𝑘𝑁 = 40 × 103
𝑁
𝐸 = 2 × 105 Τ
𝑁 𝑚 𝑚2
𝐴𝑆 = 400𝜋 𝑚𝑚2
𝐿𝑆 = 4000 − 1000𝑥
𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑜𝑓 𝑏𝑜𝑟𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝛿𝑙𝐵 =
𝑃𝐿𝐵
𝐴𝐵𝐸
𝛿𝑙𝐵 =
40 × 103
300𝜋 × 2 × 105
× 1000𝑥
𝜹𝒍𝑩 =
𝟒𝒙
𝟔𝝅
𝒎𝒎
𝐿𝐵 = 1000𝑥 𝑚𝑚
𝐴𝐵 = 300𝜋 𝑚𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

26. 𝑇𝑜𝑡𝑎𝑙 𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑎𝑓𝑡𝑒𝑟 𝑏𝑜𝑟𝑒 𝑖𝑠 𝑚𝑎𝑑𝑒 𝛿𝑙𝑆𝐵 = 𝛿𝑙𝑆 + 𝛿𝑙𝐵
0.827 =
4 − 𝑥
2𝜋
+
4𝑥
6𝜋
𝒍𝒆𝒏𝒈𝒕𝒉 𝒕𝒉𝒆 𝒓𝒐𝒅 𝒔𝒉𝒐𝒖𝒍𝒅 𝒃𝒆 𝒃𝒐𝒓𝒆𝒅 (𝒙) = 𝟑. 𝟔 𝒎
𝛿𝑙𝑆 =
4 − 𝑥
2𝜋
𝑚𝑚
𝛿𝑙𝐵 =
4𝑥
6𝜋
𝑚𝑚
𝛿𝑙𝑆𝐵 = 0.827 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

27. Thank You
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY