2. EXPRESSION FOR YOUNGS MODULUS IN TERMS OF MODULUS OF RIGIDITY
LMST is a solid cube subjected to a
shearing force F. Let 𝜏 be the shear stress
produced in the faces MS and LT due to
this shearing force. The complementary
shear stress consequently produced in the
faces ML and ST is also 𝜏. Due to the
shearing force, the cube is distorted to
𝐿′𝑀′𝑆′𝑇′ and as such, the edge M moves to
M’ and S to S’ and diagonal LS to LS’.
𝑭
𝑳 𝑻
𝑺′
𝑴 𝑴′ 𝝉
𝝉
𝝉
𝝉
𝝋
𝟒𝟓°
𝟒𝟓°
𝝋
𝑺
𝑵
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
7. Substituting equation (5) in equation (7)
𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =
Τ
𝜏 𝐺
2
𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =
𝜏
2𝐺
→ 8
𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =
𝜑
2
→ 7
The load on all sides is equal. In a cube the area of cross section is same on all sides.
𝑁𝑜𝑟𝑚𝑎𝑙 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠
𝜎𝑁 = 𝜏 → 9
𝜑 =
𝜏
𝐺
→ 5
Substituting equation (9) in equation (8)
𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =
𝜎𝑁
2𝐺
→ 10
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
10. EXPRESSION FOR YOUNGS MODULUS IN TERMS OF BULK MODULUS
Figure shows a cube ABCDEFGH which is
subjected to three mutually perpendicular tensile
stresses of equal intensity.
Let
𝑨 𝑩
𝑪
𝑫
𝑮
𝑭
𝑬
𝝈
𝝈
𝝈
𝝈
𝑯
𝐿 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝐶𝑢𝑏𝑒
𝛿𝑙 = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑏𝑒
𝐸 = 𝑌𝑜𝑢𝑛𝑔′𝑠 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑢𝑏𝑒
𝜎 = 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑒𝑠𝑠 𝑎𝑐𝑡𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑓𝑎𝑐𝑒𝑠
𝜇 = 𝑃𝑜𝑖𝑠𝑠𝑜𝑛′
𝑠 𝑟𝑎𝑡𝑖𝑜
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
11. Now let us consider the strain of one of the sided of the cube (say AB) under the
action of three perpendicular stresses. This side will suffer the following three
strains.
1. 𝑆𝑡𝑟𝑎𝑖𝑛 𝑜𝑓 𝐴𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠 𝑜𝑛 𝐴𝐸𝐻𝐷 𝑎𝑛𝑑 𝐵𝐹𝐺𝐶. 𝑇ℎ𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝑖𝑠 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑎𝑛𝑑 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 Τ
𝜎 𝐸
2. 𝑆𝑡𝑟𝑎𝑖𝑛 𝑜𝑓 𝐴𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠 𝑜𝑛 𝐴𝐸𝐹𝐵 𝑎𝑛𝑑 𝐷𝐻𝐺𝐶. 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑎𝑛𝑑 𝑖𝑠
𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 − 𝜇 ×
𝜎
𝐸
3. 𝑆𝑡𝑎𝑟𝑖𝑛 𝑜𝑓 𝐴𝐵 𝑑𝑢𝑒 𝑡𝑜 𝑠𝑡𝑟𝑒𝑠𝑠𝑒𝑠 𝑜𝑛 𝐴𝐵𝐶𝐷 𝑎𝑛𝑑 𝐸𝐹𝐺𝐻. 𝑇ℎ𝑖𝑠 𝑖𝑠 𝑎𝑙𝑠𝑜 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑎𝑛𝑑
𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 − 𝜇 ×
𝜎
𝐸
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY
12. Hence the total strain of AB is given by
𝛿𝑙
𝐿
=
𝜎
𝐸
− 𝜇
𝜎
𝐸
− 𝜇
𝜎
𝐸
𝛿𝑙
𝐿
=
𝜎
𝐸
1 − 2𝜇 → (1)
Now original volume of cube (𝑉) = 𝐿3 → (2)
If 𝛿𝑙 is the change in length, 𝛿𝑉 is the change in volume
𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (2) 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝐿
𝑑𝑉 = 3𝐿2𝛿𝑙 → (3)
𝑨 𝑩
𝑪
𝑫
𝑮
𝑭
𝑬
𝝈
𝝈
𝝈
𝝈
𝑯
BIBIN.C / ASSOCIATE PROFESSOR / MECHANICAL ENGINEERING / RMK COLLEGE OF ENGINEERING AND TECHNOLOGY