STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS

1. BIBIN CHIDAMBARANATHAN
ELONGATION OF BAR
DUE TO ITS SELF
WEIGHT
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

2. ELONGATION OF BAR DUE TO ITS SELF WEIGHT
❖ Consider a bar AB hanging freely under its own
weight
Let
❖ L = Length of the bar
❖ A = Cross sectional Area of the bar
❖ E = Young’s modulus of the bar material
❖ ω = Specific weight of the bar material
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

3. 𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑎 𝑠𝑚𝑎𝑙𝑙 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑥 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 𝑎𝑡 𝑎 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑥 𝑓𝑟𝑜𝑚 𝐵.
weight of the bar of length 𝑥, (𝑃) = 𝜔. 𝐴. 𝑥
Elongation of the small section of the bar due to weight of the bar for a small section
of length 𝑥.
𝛿𝑙𝑥 = 𝑒. 𝑑𝑥
𝛿𝑙𝑥 =
𝜎
𝐸
. 𝑑𝑥
𝛿𝑙𝑥 =
𝑃
𝐴 𝐸
. 𝑑𝑥
𝛿𝑙𝑥 =
𝜔 𝐴 𝑥
𝐴 𝐸
. 𝑑𝑥
𝛿𝑙𝑥 =
𝜔 𝑥
𝐸
. 𝑑𝑥
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

4. Total elongation of bar may be found out by integrating the above equation
between the limits 0 and 𝐿
𝑇𝑜𝑡𝑎𝑙 𝑒𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝛿𝑙 = න
0
𝐿
𝜔 𝑥
𝐸
. 𝑑𝑥
𝛿𝑙 =
𝜔
𝐸
න
0
𝐿
𝑥. 𝑑𝑥
𝛿𝑙 =
𝜔
𝐸
𝑥2
2 0
𝐿
𝛿𝑙 =
𝜔
𝐸
𝐿2
2
− 0
𝛿𝑙 =
𝜔 𝐿2
2𝐸
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

5. 𝛿𝑙 =
𝑊
𝐴 𝐿
𝐿2
2𝐸
𝑇𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑊 = 𝜔 𝐴 𝐿
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 𝜔 =
𝑊
𝐴 𝐿
Elongation of bar due to weight of the bar (𝜹𝒍) =
𝑾𝑳
𝟐𝑨𝑬
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

6. Problem 01
A copper alloy wire of 1.5 𝑚𝑚 diameter and 30 𝑚 long is hanging freely from a tower.
What will be its elongation due to self-weight? Take specific weight of copper and its
modulus of elasticity as 89.4 𝑘𝑁/𝑚3
and 90 𝐺𝑃𝑎 respectively.
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
Elongation due to self weight (𝛿𝑙)=?
𝑑 = 1.5 𝑚𝑚 𝐿 = 30 𝑚 = 30 × 103
𝑚𝑚 𝜔 = 89.2 𝑘 Τ
𝑁 𝑚3 = 89.2 × 10−6 Τ
𝑁 𝑚 𝑚3
𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝐸 = 90 𝐺𝑃𝑎 = 90 × 103 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

7. 𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝛿𝑙 =
𝜔 𝐿2
2𝐸
𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝛿𝑙 =
𝜔 𝐿2
2𝐸
𝛿𝑙 =
89.2 × 10−6 × 30 × 103
2 × 90 × 103
𝑬𝒍𝒐𝒏𝒈𝒂𝒕𝒊𝒐𝒏 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒆𝒍𝒇 𝒘𝒆𝒊𝒈𝒉𝒕 𝜹𝒍 = 𝟎. 𝟒𝟓 𝒎𝒎
𝑑 = 1.5 𝑚𝑚
𝐿 = 30 𝑚 = 30 × 103 𝑚𝑚
𝜔 = 89.2 × 10−6 Τ
𝑁 𝑚 𝑚3
𝐸 = 90 × 103 Τ
𝑁 𝑚 𝑚2
𝛿𝑙 = 0.45 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

8. Problem 01
An alloy wire of 2 𝑚𝑚2 cross sectional area and 12 𝑁 weight hangs freely under its own
weight. Find the maximum length of the wire, if its extension is not to exceed 0.6 𝑚𝑚.
Take E for wire material as 150 𝐺𝑃𝑎.
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
Maximum length of the wire (L)=?
𝐴 = 2 𝑚𝑚2
𝑊 = 12 𝑁 𝛿𝑙 = 0.6 𝑚𝑚
𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝐸 = 150 𝐺𝑃𝑎 = 150 × 103 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

9. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝛿𝑙 =
𝑊𝐿
2𝐴𝐸
0.6 =
12 × 𝐿
2 × 2 × 150 × 103
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒘𝒊𝒓𝒆 𝑳 = 𝟑𝟎𝟎𝟎𝟎 𝒎𝒎 = 𝟑𝟎 𝒎
𝐸𝑙𝑜𝑛𝑔𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝛿𝑙 =
𝑊𝐿
2𝐴𝐸
𝐴 = 2 𝑚𝑚2
𝑊 = 12 𝑁
𝛿𝑙 = 0.6 𝑚𝑚
𝐸 = 150 × 103 Τ
𝑁 𝑚 𝑚2
𝐿 = 30000 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

10. Problem 01
A steel wire ABC 16 𝑚 long having cross sectional area of 4 𝑚𝑚2 weighs 20 𝑁 as shown
in fig. If the modulus of elasticity for the wire material is 200𝐺𝑃𝑎, find the deflection at C
and B.
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
Deflection at C (𝛿𝑙𝑐 ) and B (𝛿𝑙𝐵 )=?
𝐿 = 16 𝑚 = 16 × 103 𝑚𝑚 𝐴 = 4 𝑚𝑚2
𝑊 = 20 𝑁
𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝐸 = 200 𝐺𝑃𝑎 = 200 × 103 Τ
𝑁 𝑚 𝑚2
𝑩
𝑨
𝑪
𝟖
𝒎
𝟖
𝒎
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

11. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐶 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑟𝑒 𝐴𝐶 𝛿𝑙𝑐 =
𝑊𝐿
2𝐴𝐸
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐵 𝛿𝑙𝐵 = 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝐴𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 +
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑟𝑒 𝐵𝐶
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐵 𝛿𝑙𝐵 = 𝛿𝑙1 + 𝛿𝑙2
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑟𝑒 𝐴𝐵 𝛿𝑙1 =
WAB × 𝐿𝐴𝐵
2𝐴𝐸
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑟𝑒 𝐵𝐶 𝛿𝑙2 =
PBC × 𝐿𝐴𝐵
𝐴𝐸
𝑩
𝑨
𝑪
𝟖
𝒎
𝟖
𝒎
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

12. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐶 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑟𝑒 𝐴𝐶 𝛿𝑙𝑐 =
𝑊𝐿
2𝐴𝐸
𝑩
𝑨
𝑪
𝟖
𝒎
𝟖
𝒎
𝛿𝑙𝑐 =
20 × 16 × 103
2 × 4 × 200 × 103
𝑫𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒘𝒊𝒓𝒆 𝒂𝒕 𝑪 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒆𝒍𝒇 𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒘𝒊𝒓𝒆 𝑨𝑪 𝜹𝒍𝒄 = 𝟎. 𝟐𝒎𝒎
𝛿𝑙𝑐 = 0.2 𝑚𝑚
𝐿 = 16 × 103
𝑚𝑚
𝐴 = 4 𝑚𝑚2
𝑊 = 20 𝑁
𝐸 = 200 × 103 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

13. 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑒𝑙𝑓 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑟𝑒 𝐴𝐵 𝛿𝑙1 =
WAB × 𝐿𝐴𝐵
2𝐴𝐸
𝛿𝑙1 =
10 × 8 × 103
2 × 4 × 200 × 103
𝛿𝑙1 = 0.05 𝑚𝑚
𝑫𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒘𝒊𝒓𝒆 𝒂𝒕 𝑩 𝒅𝒖𝒆 𝒕𝒐 𝒔𝒆𝒍𝒇 𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒘𝒊𝒓𝒆 𝑨𝑩 𝜹𝒍𝟏 = 𝟎. 𝟎𝟓 𝒎𝒎
𝑩
𝑨
𝑪
𝟖
𝒎
𝟖
𝒎
𝐿𝐴𝐵 = 8 × 103
𝑚𝑚
𝐴 = 4 𝑚𝑚2
𝑊 = 20 𝑁
𝐸 = 200 × 103 Τ
𝑁 𝑚 𝑚2
𝑊𝐴𝐵 =
W
2
= 10 𝑁
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

14. 𝑩
𝑨
𝑪
𝟖
𝒎
𝟖
𝒎
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑟𝑒 𝐵𝐶 𝛿𝑙2 =
P × 𝐿𝐴𝐵
𝐴𝐸
𝛿𝑙2 =
10 × 8 × 103
4 × 200 × 103
𝛿𝑙2 = 0.1 𝑚𝑚
𝑫𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒘𝒊𝒓𝒆 𝒂𝒕 𝑩 𝒅𝒖𝒆 𝒕𝒐 𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒘𝒊𝒓𝒆 𝑩𝑪 𝜹𝒍𝟐 = 𝟎. 𝟏 𝒎𝒎
𝐿AB = 8 × 103
𝑚𝑚
𝐴 = 4 𝑚𝑚2
𝑊 = 20 𝑁
𝐸 = 200 × 103 Τ
𝑁 𝑚 𝑚2
P =
W
2
= 10 𝑁
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

15. 𝑩
𝑨
𝑪
𝟖
𝒎
𝟖
𝒎
𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑖𝑟𝑒 𝑎𝑡 𝐵 𝛿𝑙𝐵 = 𝛿𝑙1 + 𝛿𝑙2
𝛿𝑙𝐵 = 0.05 + 0.1
𝑫𝒆𝒇𝒍𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒘𝒊𝒓𝒆 𝒂𝒕 𝑩 𝜹𝒍𝑩 = 𝟎. 𝟏𝟓 𝒎𝒎
𝛿𝑙1 = 0.05 𝑚𝑚
𝛿𝑙2 = 0.1 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

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