STRENGTH OF MATERIALS FOR MECHANICAL ENGINEERS

1. BIBIN CHIDAMBARANATHAN
PROBLEMS
ON
ELASTIC CONSTANTS
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

2. Problem 01
A bar of cross section 8 𝑚𝑚 × 8 𝑚𝑚 is subjected to an axial of 7000 𝑁. The lateral
dimension of the bar is found to be changed to 7.9885 × 7.9885 𝑚𝑚. If the modulus of
rigidity to 0.8 × 105 𝑁/𝑚𝑚2. Determine the µ and E.
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
𝐴𝑟𝑒𝑎 𝐴 = 8 𝑚𝑚 × 8 𝑚𝑚 = 64 𝑚𝑚2 𝑏 = 8 𝑚𝑚 𝑑 = 8𝑚𝑚
𝑃 = 7000 𝑁
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 = 7.9985 𝑚𝑚 × 7.9985 𝑚𝑚
𝐺 = 0.8 × 105 Τ
𝑁 𝑚 𝑚2
𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 =?
𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝐸 =?
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

3. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶
𝐸 = 2𝐺 1 + 𝜇
𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 =
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝑒𝑏 𝑜𝑟 𝑒𝑑
𝑒𝑙
𝑌𝑜𝑢𝑛𝑔′
𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 =
𝑆𝑡𝑟𝑒𝑠𝑠
𝑆𝑡𝑟𝑎𝑖𝑛
𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 =
𝑃
𝐴
𝑒𝑏 =
𝛿𝑏
𝑏
𝑒𝑑 =
𝛿𝑑
𝑑
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

4. 𝑏 = 8 𝑚𝑚 𝑑 = 8𝑚𝑚
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 = 7.9985 𝑚𝑚 × 7.9985 𝑚𝑚
𝛿𝑏 = 8 − 7.9985 = 0.0015 𝑚𝑚
𝛿𝑑 = 8 − 7.9985 = 0.0015 𝑚𝑚
𝑒𝑏 =
𝛿𝑏
𝑏
=
0.0015
8
= 1.875 × 10−4
𝑒𝑑 =
𝛿𝑑
𝑑
=
0.0015
8
= 1.875 × 10−4
𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

5. 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 =
𝑃
𝐴
𝜎 =
7000
64
𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 = 109.37 Τ
𝑁 𝑚 𝑚2
𝐴𝑟𝑒𝑎 𝐴 = 8 𝑚𝑚 × 8 𝑚𝑚 = 64 𝑚𝑚2
𝑃 = 7000 𝑁
𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 =
𝑆𝑡𝑟𝑒𝑠𝑠
𝑆𝑡𝑟𝑎𝑖𝑛
=
𝜎
𝑒𝑙
𝐸 =
109.37
𝑒𝑙
𝑒𝑙 =
109.37
𝐸
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

6. 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 =
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝜇 =
𝑒𝑏 𝑜𝑟 𝑒𝑑
𝑒𝑙
𝜇 =
1.875 × 10−4
109.37
𝐸
𝝁 = 𝟏. 𝟕𝟏𝟒𝟑 × 𝟏𝟎−𝟔 𝑬
𝑒𝑏 = 1.875 × 10−4
𝑒𝑑 = 1.875 × 10−4
𝑒𝑙 =
109.37
𝐸
𝐸 = 2𝐺 1 + 𝜇
𝐸 = 2 × 0.8 × 105 1 + 1.7143 × 10−6 𝐸
𝒀𝒐𝒖𝒏𝒈′𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑬 = 𝟐. 𝟐 × 𝟏𝟎𝟓 Τ
𝑵 𝒎 𝒎𝟐
𝐺 = 0.8 × 105 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

7. 𝜇 = 1.7143 × 10−6 𝐸
𝜇 = 1.7143 × 10−6
× 2.2 × 105
𝑷𝒐𝒊𝒔𝒔𝒊𝒐𝒏′𝒔 𝒓𝒂𝒕𝒊𝒐 𝝁 = 𝟎. 𝟑𝟕𝟕
𝐸 = 2.2 × 105 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

8. Problem 01
Calculate the modulus of rigidity and bulk modulus of a cylindrical bar of diameter
30 𝑚𝑚 and length 1.5 𝑚 if the longitudinal strain in a bar during a tensile stress is four
times the lateral strain. Find the change in volume, when the bar is subjected to a
hydrostatic pressure of 100 𝑁/𝑚𝑚2. Take 𝐸 = 1 × 105 𝑁/𝑚𝑚2
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
𝑑 = 30 𝑚𝑚 𝐿 = 1.5 𝑚 = 1500 𝑚𝑚 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙 = 4 × 𝑒𝑡
ℎ𝑦𝑑𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝜎 = 100 Τ
𝑁 𝑚 𝑚2
𝐸 = 1 × 105 Τ
𝑁 𝑚 𝑚2
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝛿𝑉 =?
𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =?
𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑟𝑖𝑔𝑖𝑑𝑖𝑡𝑦 𝐺 =?
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

9. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶
𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′
𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 =
𝑒𝑡
𝑒𝑙
𝐸 = 2𝐺 1 + 𝜇
𝐸 = 3𝐾 1 − 2𝜇
𝑉𝑜𝑙𝑢𝑚𝑒 𝑉 =
𝜋
4
× 𝑑2 × 𝐿
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 =
𝛿𝑉
𝑉
𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =
𝜎
𝑒𝑣
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

10. 𝑙𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙 = 4 × 𝑒𝑡
𝐸 = 1 × 105 Τ
𝑁 𝑚 𝑚2
𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 =
𝑒𝑡
𝑒𝑙
𝜇 =
𝑒𝑡
4 𝑒𝑡
𝑷𝒐𝒊𝒔𝒔𝒊𝒐𝒏′𝒔 𝒓𝒂𝒕𝒊𝒐 𝝁 = 𝟎. 𝟐𝟓
𝐸 = 2𝐺 1 + 𝜇
1 × 105 = 2𝐺 1 + 0.25
𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝒐𝒇 𝒓𝒊𝒈𝒊𝒅𝒊𝒕𝒚 𝑮 = 𝟒 × 𝟏𝟎𝟒 Τ
𝑵 𝒎 𝒎𝟐
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

11. 𝐸 = 3𝐾 1 − 2𝜇
1 × 105
= 3𝐾 1 − 2 × 0.25
𝑲 = 𝟎. 𝟔𝟔𝟕 × 𝟏𝟎𝟓 Τ
𝑵 𝒎 𝒎𝟐
𝑉𝑜𝑙𝑢𝑚𝑒 𝑉 =
𝜋
4
× 𝑑2 × 𝐿
𝑉 =
𝜋
4
× 302 × 1500
𝑽𝒐𝒍𝒖𝒎𝒆 𝑽 = 𝟏𝟎𝟔. 𝟎𝟐𝟖 × 𝟏𝟎𝟑 𝒎𝒎𝟑
𝜇 = 0.25
𝐸 = 1 × 105 Τ
𝑁 𝑚 𝑚2
𝑑 = 30 𝑚𝑚
𝐿 = 1.5 𝑚 = 1500 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

12. 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 =
𝛿𝑉
𝑉
𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =
𝜎
𝑒𝑣
0.667 × 105 =
100
𝑒𝑣
𝒆𝒗 = 𝟏. 𝟒𝟗𝟗 × 𝟏𝟎−𝟑
1.499 × 10−3
=
𝛿𝑉
106.028 × 103
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒗𝒐𝒍𝒖𝒎𝒆 𝜹𝑽 = 𝟏𝟓𝟗𝟎. 𝟒𝟑 𝒎𝒎𝟑
𝐾 = 0.667 × 105 Τ
𝑁 𝑚 𝑚2
𝜎 = 100 Τ
𝑁 𝑚 𝑚2
𝑉 = 106.028 × 103 𝑚𝑚3
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

13. Problem 01
A bar of 20 𝑚𝑚 diameter is subjected to tensile load of 40 𝐾𝑁. The extension measured
over a gauge length of 200 𝑚𝑚 is 0.12 𝑚𝑚 and contraction in diameter is 0.0036 𝑚𝑚.
Find the Poisson ratio and the elastic constants E, G and K. Also find volumetric strain.
𝑮𝒊𝒗𝒆𝒏 𝒅𝒂𝒕𝒂:
𝑻𝒐 𝒇𝒊𝒏𝒅:
𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝐸𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝐸 =?
𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =?
𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑟𝑖𝑔𝑖𝑑𝑖𝑡𝑦 𝐺 =?
𝑑 = 20 𝑚𝑚 𝑃 = 40 × 103 𝑁 𝐿 = 200 𝑚𝑚 𝛿𝑙 = 0.12 𝑚𝑚
𝛿𝑑 = 0.036 𝑚𝑚
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 =?
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

14. 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 ∶
𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙 =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐿𝑒𝑛𝑔𝑡ℎ
=
𝛿𝑙
𝑙
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
=
𝛿𝑑
𝑑
𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′
𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 =
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝑒𝑡
𝑒𝑙
.
𝐴𝑟𝑒𝑎 𝐴 =
𝜋
4
× 𝑑2
𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 =
𝑃
𝐴
𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 =
𝑆𝑡𝑟𝑒𝑠𝑠
𝑆𝑡𝑟𝑎𝑖𝑛
=
𝜎
𝑒𝑙
𝐸 = 3𝐾 1 − 2𝜇
𝐸 = 2𝐺 1 + 𝜇
𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =
𝜎
𝑒𝑣
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

15. 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙 =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐿𝑒𝑛𝑔𝑡ℎ
=
𝛿𝑙
𝑙
𝑒𝑙 =
0.12
200
𝑳𝒐𝒏𝒈𝒊𝒕𝒖𝒅𝒊𝒏𝒂𝒍 𝒔𝒕𝒓𝒂𝒊𝒏 𝒆𝒍 = 𝟔 × 𝟏𝟎−𝟒
𝐿 = 200 𝑚𝑚
𝛿𝑙 = 0.12 𝑚𝑚
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
=
𝛿𝑑
𝑑
𝑒𝑡 =
0.036
20
𝑳𝒐𝒏𝒈𝒊𝒕𝒖𝒅𝒊𝒏𝒂𝒍 𝒔𝒕𝒓𝒂𝒊𝒏 𝒆𝒕 = 𝟏. 𝟖 × 𝟏𝟎−𝟒
𝑑 = 20 𝑚𝑚
𝛿𝑑 = 0.036 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

16. 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′
𝑠 𝑟𝑎𝑡𝑖𝑜 𝜇 =
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛
=
𝑒𝑡
𝑒𝑙
.
𝜇 =
1.8 × 10−4
6 × 10−4
𝝁 = 𝟎. 𝟑
𝑒𝑙 = 6 × 10−4
𝑒𝑡 = 1.8 × 10−4
𝐴𝑟𝑒𝑎 𝐴 =
𝜋
4
× 𝑑2
𝐴 =
𝜋
4
× 202
𝑨𝒓𝒆𝒂 𝑨 = 𝟏𝟎𝟎𝝅 𝒎𝒎𝟐
𝑑 = 20 𝑚𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

17. 𝑆𝑡𝑟𝑒𝑠𝑠 𝜎 =
𝑃
𝐴
𝜎 =
40 × 103
100𝜋
𝑺𝒕𝒓𝒆𝒔𝒔 𝝈 = 𝟏𝟐𝟕. 𝟑𝟐 Τ
𝑵 𝒎 𝒎𝟐
𝑃 = 40 × 103
𝑁
𝐴 = 100𝜋 𝑚𝑚2
𝑌𝑜𝑢𝑛𝑔′𝑠 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐸 =
𝑆𝑡𝑟𝑒𝑠𝑠
𝑆𝑡𝑟𝑎𝑖𝑛
=
𝜎
𝑒𝑙
𝐸 =
127.32
6 × 10−4
𝒀𝒐𝒖𝒏𝒈′𝒔 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑬 = 𝟐. 𝟏 × 𝟏𝟎𝟓 Τ
𝑵 𝒎 𝒎𝟐
𝑒𝑙 = 6 × 10−4
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

18. 𝐸 = 3𝐾 1 − 2𝜇
2.1 × 105 = 3𝐾 1 − 2 × 0.3
𝐸 = 2.1 × 105 Τ
𝑁 𝑚 𝑚2
𝜇 = 0.3
𝑩𝒖𝒍𝒌 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑲 = 𝟏. 𝟕𝟓 × 𝟏𝟎𝟓 Τ
𝑵 𝒎 𝒎𝟐
𝐸 = 2𝐺 1 + 𝜇
2.1 × 105
= 2𝐺 1 + 0.3
𝑹𝒊𝒈𝒊𝒅𝒊𝒕𝒚 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑮 = 𝟎. 𝟖𝟕 × 𝟏𝟎𝟓 Τ
𝑵 𝒎 𝒎𝟐
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

19. 𝐵𝑢𝑙𝑘 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝐾 =
𝜎
𝑒𝑣
1.75 × 105 =
127.32
𝑒𝑣
𝑽𝒐𝒍𝒖𝒎𝒆𝒕𝒓𝒊𝒄 𝒔𝒕𝒓𝒂𝒊𝒏 𝒆𝒗 = 𝟕. 𝟐 × 𝟏𝟎−𝟒
𝜎 = 127.32 Τ
𝑁 𝑚 𝑚2
𝐾 = 1.75 × 105 Τ
𝑁 𝑚 𝑚2
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

20. BIBIN CHIDAMBARANATHAN
VOLUMETRIC STRAIN
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

21. VOLUMETRIC STRAIN OF A RECTANGULAR BAR SUBJECTED
TO AN AXIAL LOAD IN THE DIRECTION OF ITS LENGTH
❖ Consider rectangular bar of length l, width b, and depth d which is subjected to an
axial load P in the direction of its length as shown in fig.
𝒅
𝑷
𝑷
𝑳
𝒃
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

22. Let
𝛿𝑙 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ
𝛿𝑏 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑤𝑖𝑑𝑡ℎ
𝛿𝑑 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑒𝑝𝑡ℎ
𝐹𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 = 𝐿 + 𝛿𝑙
𝐹𝑖𝑛𝑎𝑙 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 = 𝑏 + 𝛿𝑏
𝐹𝑖𝑛𝑎𝑙 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 = 𝑑 + 𝛿𝑑
𝒅
𝑷
𝑷
𝑳
𝒃
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

23. 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 𝑉 = 𝐿 𝑏 𝑑
𝐹𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 = 𝐿 + 𝛿𝑙 𝑏 + 𝛿𝑏 𝑑 + 𝛿𝑑
= 𝐿 𝑏 + 𝐿 𝛿𝑏 + 𝑏 𝛿𝑙 + 𝛿𝑙 𝛿𝑏 𝑑 + 𝛿𝑑
= 𝐿 𝑏 𝑑 + 𝐿 𝛿𝑏 𝑑 + 𝑏 𝛿𝑙 𝑑 + 𝛿𝑙 𝛿𝑏 𝑑 + 𝐿 𝑏 𝛿𝑑 + 𝐿 𝛿𝑏 𝛿𝑑 + 𝑏 𝛿𝑙 𝛿𝑑 + 𝛿𝑙 𝛿𝑏 𝛿𝑑
𝐼𝑔𝑛𝑜𝑟𝑖𝑛𝑔 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑜𝑓 𝑠𝑚𝑎𝑙𝑙 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑖𝑒𝑠
𝐹𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 = 𝐿 𝑏 𝑑 + 𝑏 𝑑 𝛿𝑙 + 𝑙 𝑏 𝛿𝑑 + 𝑙 𝑑 𝛿𝑏
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝛿𝑉 = 𝐹𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 − 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝛿𝑉 = 𝐿 𝑏 𝑑 + 𝑏 𝑑 𝛿𝑙 + 𝑙 𝑏 𝛿𝑑 + 𝑙 𝑑 𝛿𝑏 − 𝐿 𝑏 𝑑
𝛿𝑉 = 𝑏 𝑑 𝛿𝑙 + 𝑙 𝑏 𝛿𝑑 + 𝑙 𝑑 𝛿𝑏
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

24. 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
=
𝛿𝑉
𝑉
𝑒𝑣 =
𝑏 𝑑 𝛿𝑙 + 𝑙 𝑏 𝛿𝑑 + 𝑙 𝑑 𝛿𝑏
𝐿 𝑏 𝑑
𝑒𝑣 =
𝑏 𝑑 𝛿𝑙
𝐿 𝑏 𝑑
+
𝑙 𝑏 𝛿𝑑
𝐿 𝑏 𝑑
+
𝑙 𝑑 𝛿𝑏
𝐿 𝑏 𝑑
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 =
𝛿𝑙
𝐿
+
𝛿𝑑
𝑑
+
𝛿𝑏
𝑏
𝐵𝑢𝑡,
𝛿𝑙
𝐿
= 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙
𝛿𝑑
𝑑
𝑜𝑟
𝛿𝑏
𝑏
= 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

25. 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 = 𝑒𝑙 + 𝑒𝑡 + 𝑒𝑡
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 = 𝑒𝑙 + 2𝑒𝑡
𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡 = −𝜇 × 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙
𝑒𝑡 = −𝜇 × 𝑒𝑙
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 = 𝑒𝑙 + 2 −𝜇 × 𝑒𝑙
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 = 𝑒𝑙 1 − 2 𝜇
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 = 𝑒𝑙 1 −
2
𝑚
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

26. VOLUMETRIC STRAIN OF A CYLINDRICAL ROD
❖ Consider a cylindrical rod which is subjected to an axial load P.
Let
❖ d – diameter of the rod
❖ L – Length of the rod
❖ Due to tensile load P, there will be an increase in length of the rod, but the diameter
of the rod will decrease as shown in fig.
𝑷
𝑷
𝑳 𝜹𝒍
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

27. 𝐹𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ = 𝐿 + 𝛿𝑙
𝐹𝑖𝑛𝑎𝑙 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 𝑑 + 𝛿𝑑
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑜𝑑 𝑉 =
𝜋
4
× 𝑑2 𝐿
𝐹𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 =
𝜋
4
× 𝑑 − 𝛿𝑑 2 𝐿 + 𝛿𝐿
=
𝜋
4
× 𝑑2 − 2 𝑑 𝛿𝑑 + 𝛿𝑑2 𝐿 + 𝛿𝐿
=
𝜋
4
× 𝑑2𝐿 − 2 𝑑 𝛿𝑑 𝐿 + 𝛿𝑑2 𝐿 + 𝑑2𝛿𝐿 − 2 𝑑 𝛿𝑑 𝛿𝐿 + 𝛿𝑑2𝛿𝐿
𝐼𝑔𝑛𝑜𝑟𝑖𝑛𝑔 𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑜𝑓 𝑡𝑤𝑜 𝑠𝑚𝑎𝑙𝑙 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑖𝑒𝑠 𝑎𝑛𝑑 ℎ𝑖𝑔ℎ𝑒𝑟 𝑝𝑜𝑤𝑒𝑟𝑠
𝐹𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 =
𝜋
4
× 𝑑2𝐿 − 2 𝑑 𝛿𝑑 𝐿 + 𝑑2𝛿𝐿
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

28. 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝛿𝑉 = 𝐹𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 − 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
𝛿𝑉 =
𝜋
4
× 𝑑2𝐿 − 2 𝑑 𝛿𝑑 𝐿 + 𝑑2𝛿𝐿 −
𝜋
4
× 𝑑2 𝐿
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝛿𝑉 =
𝜋
4
× 𝑑2𝛿𝐿 − 2 𝑑 𝛿𝑑 𝐿
𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣 =
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
=
𝛿𝑉
𝑉
𝑒𝑣 =
𝜋
4
× 𝑑2𝛿𝐿 − 2 𝑑 𝛿𝑑 𝐿
𝜋
4
× 𝑑2 𝐿
𝑒𝑣 =
𝑑2𝛿𝐿 − 2 𝑑 𝛿𝑑 𝐿
𝑑2 𝐿
𝑉 =
𝜋
4
× 𝑑2 𝐿
𝑉f =
𝜋
4
× 𝑑2𝐿 − 2 𝑑 𝛿𝑑 𝐿 + 𝑑2𝛿𝐿
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

29. 𝑒𝑣 =
𝑑2𝛿𝐿
𝑑2 𝐿
−
2 𝑑 𝛿𝑑 𝐿
𝑑2 𝐿
𝑒𝑣 =
𝛿𝐿
𝐿
−
2 𝛿𝑑
𝑑
𝐵𝑢𝑡,
𝛿𝑙
𝐿
= 𝐿𝑜𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑙
𝛿𝑑
𝑑
= 𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑡
𝒆𝒗 = 𝒆𝒍 − 𝟐𝒆𝒅
𝑽𝒐𝒍𝒖𝒎𝒆𝒕𝒓𝒊𝒄 𝒔𝒕𝒓𝒂𝒊𝒏 = 𝒔𝒕𝒓𝒂𝒊𝒏 𝒊𝒏 𝒍𝒆𝒏𝒈𝒕𝒉 − 𝟐 × 𝒔𝒕𝒓𝒂𝒊𝒏 𝒊𝒏 𝒅𝒊𝒂𝒎𝒆𝒕𝒆𝒓
𝑒𝑣 =
𝑑2𝛿𝐿 − 2 𝑑 𝛿𝑑 𝐿
𝑑2 𝐿
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

30. VOLUMETRIC STRAIN OF RECTANGULAR BAR SUBJECTED TO
THREE FORCES WHICH ARE MUTUALLY PERPENDICULAR
❖ Consider a rectangular bar of dimensions x, y and z subjected to three direct tensile
stresses along 3 mutually perpendicular axes as shown in fig.
𝒙
𝒙
𝒛
𝒛
𝒚
𝒚
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑟 𝑉 = 𝑥 𝑦 𝑧
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

31. Taking log on both sides, we have
log 𝑉 = log 𝑥 + log 𝑦 + 𝑙𝑜𝑔 𝑧
1
𝑉
𝑑𝑉 =
1
𝑥
𝑑𝑥 +
1
𝑦
𝑑𝑦 +
1
𝑧
𝑑𝑧
𝑑𝑉
𝑉
=
𝑑𝑥
𝑥
+
𝑑𝑦
𝑦
+
𝑑𝑧
𝑧
→ 1
But,
𝛿𝑉
𝑉
=
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒
= 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 𝑒𝑣
𝑉 = 𝑥 𝑦 𝑧
log 𝑉 = log (𝑥 𝑦 𝑧)
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

32. 𝛿𝑥
𝑥
=
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
= 𝑠𝑡𝑟𝑎𝑖𝑛 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑒𝑥
𝛿𝑦
𝑦
=
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑖𝑛 𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑖𝑛 𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
= 𝑠𝑡𝑟𝑎𝑖𝑛 𝑖𝑛 𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑒𝑦
𝛿𝑧
𝑧
=
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑖𝑛 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑖𝑛 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
= 𝑠𝑡𝑟𝑎𝑖𝑛 𝑖𝑛 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑒𝑧
Substitute these values in equation (1)
𝑒𝑣 = 𝑒𝑥 + 𝑒𝑦 + 𝑒𝑧
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

33. Now, Let
𝜎𝑥 = 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑥 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝜎𝑦 = 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑦 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝜎𝑧 = 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑧 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝜇 = 𝑃𝑜𝑖𝑠𝑠𝑖𝑜𝑛′𝑠 𝑟𝑎𝑡𝑖𝑜
𝐸 = 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦
Now,
𝜎𝑥 will produce a tensile strain equal to
𝜎𝑥
𝐸
in the direction of 𝑥 and compressive
strain 𝜇
𝜎𝑥
𝐸
in the direction of 𝑦 and 𝑧.
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

34. ❖ 𝜎𝑦 will produce a tensile strain equal to
𝜎𝑦
𝐸
in the direction of 𝑦 and compressive
strain 𝜇
𝜎𝑦
𝐸
in the direction of 𝑥 and 𝑧.
❖ 𝜎𝑧 will produce a tensile strain equal to
𝜎𝑧
𝐸
in the direction of 𝑧 and compressive
strain 𝜇
𝜎𝑧
𝐸
in the direction of 𝑥 and 𝑦.
❖ Hence 𝜎𝑦 and 𝜎𝑧 will produce compressive strains equal to 𝜇
𝜎𝑦
𝐸
and 𝜇
𝜎𝑧
𝐸
in the
direction of 𝑥.
Net tensile strain along the 𝑥 direction is given by
𝑒𝑥 =
𝜎𝑥
𝐸
− 𝜇
𝜎𝑦
𝐸
− 𝜇
𝜎𝑧
𝐸
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

35. Similarly,
𝑒𝑦 =
𝜎𝑦
𝐸
− 𝜇
𝜎𝑧
𝐸
+
𝜎𝑥
𝐸
𝑒𝑧 =
𝜎𝑧
𝐸
− 𝜇
𝜎𝑥
𝐸
+
𝜎𝑦
𝐸
Adding all strains, we get
𝑒𝑥 + 𝑒𝑦 + 𝑒𝑧 =
𝜎𝑥
𝐸
− 𝜇
𝜎𝑦
𝐸
+
𝜎𝑧
𝐸
+
𝜎𝑦
𝐸
− 𝜇
𝜎𝑧
𝐸
+
𝜎𝑥
𝐸
+
𝜎𝑧
𝐸
− 𝜇
𝜎𝑥
𝐸
+
𝜎𝑦
𝐸
𝑒𝑥 =
𝜎𝑥
𝐸
− 𝜇
𝜎𝑦
𝐸
+
𝜎𝑧
𝐸
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

36. 𝛿𝑉
𝑉
= 𝑒𝑣 =
1
𝐸
1 − 2𝜇 𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧
If the value of
𝛿𝑉
𝑉
is positive, it represents increase in volume whereas negative
value of
𝛿𝑉
𝑉
represents decrease in volume.
𝑒𝑥 + 𝑒𝑦 + 𝑒𝑧 =
1
𝐸
𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧 −
2𝜇
𝐸
𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧
𝑒𝑣 =
1
𝐸
−
2𝜇
𝐸
𝜎𝑥 + 𝜎𝑦 + 𝜎𝑧
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY

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