JNTU
JNTU B.Tech II ELEMENTS OF MECHANICAL ENGINEERING Previous Question Paper (August/September - 2022 - R13)
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Main
A bar of elastic material is subjected to directed stress in a longitudinal direction, and its strains in the two directions at right angles are reduced to one-half and one third respectively to those which normally occur in an ordinary tension member. If E=200kN/mm2 and 1/m = 0.25, what are the value of elastic constants?
Answer
Given that the material is subjected to a longitudinal stress, let the longitudinal strain be εx = σ/E. In an ordinary tension member, the lateral strains are εy = εz = -μ(σ/E), where μ = 1/m = 0.25. The problem states that the actual lateral strains are reduced to one-half and one-third of these values. Thus, the actual strains are εx = σ/E, εy = -0.5(0.25)(σ/E) = -0.125(σ/E), and εz = -0.333(0.25)(σ/E) = -0.0833(σ/E).
The generalized Hooke's Law relates stress and strain as: εx = (1/E)[σx - ν(σy + σz)], εy = (1/E)[σy - ν(σx + σz)], and εz = (1/E)[σz - ν(σx + σy)]. Substituting the known values of εy and εz and solving the system of equations for σy and σz in terms of σx (where σx = σ), we find the effective elastic constants for this anisotropic behavior.
The elastic constants are calculated as follows:
- Young's Modulus (E): Given as 200 kN/mm2.
- Poisson's Ratio (ν): Given as 1/m = 0.25.
- Modulus of Rigidity (G): Calculated using the relation G = E / (2(1 + ν)) = 200 / (2(1 + 0.25)) = 200 / 2.5 = 80 kN/mm2.
- Bulk Modulus (K): Calculated using the relation K = E / (3(1 - 2ν)) = 200 / (3(1 - 2(0.25))) = 200 / (3(0.5)) = 200 / 1.5 = 133.33 kN/mm2.
Derive the relation among all moduli E, G and K.
Answer
The relationship between the modulus of elasticity (E), modulus of rigidity (G), and bulk modulus (K) is derived using Poisson's ratio (μ). The fundamental relations are:
To eliminate μ, solve the first equation for μ: μ = (E / 2G) - 1. Substitute this into the second equation:
Rearranging this leads to the relation:
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