The number of elastic constants for a orthotropic material are
Poisson’s ratio is defined as the ratio of
Poisson’s ratio is the ratio of lateral strain to the longitudinal strain.
Which one of the following materials has Poisson’s ratio more than 1?
Poisson’s ratio can not be greater than 0.5.
The maximum values of Poisson’s ratio for an elastic material is
Poisson’s ratio lies between 0 to 0.5.
μ = 0.5 for pure elastic rubber
μ = 0.1 to 0.2 for concrete
μ = 0.25 to 0.33 for steel
The relationship between modulus of elasticity (E) and modulus of rigidity (G) is
What is the relationship between elastic constants E, G and K?
E = 2G(1 + μ) ......(i)
E = 3K(1 - 2μ) ......(ii)
Solving (i) and (ii)
In terms of bulk modulus (K) and modulus of rigidity (G), the Poisson’s ratio can be expressed as
E = 3K(1 - 2μ) ......(i)
E = 2G(1 + μ) .......(ii)
From (i) and (ii)
3K(1 - 2μ) = 2G(1 + μ)
3K - 6μK = 2G + 2μG
3K - 2G = 6μK + 2μG
3K - 2G = 2μ(G + 3K)
What is the relationship between the linear elastic properties Young’s modulus (E), rigidity modulus (G) and bulk modulus (K)?
E = 2G(1 + μ) .......(i)
E = 3K(1 - 2μ) ........(ii)
Solving (i) and (ii), we get
A cylindrical bar of 20 mm diameter and 1 m length is subjected to a tensile test. Its longitudinal strain is 4 times that of its lateral strain. If the modulus of elasticity is 2 x 105 N/mm2, then its modulus of rigidity will be
= 0.8 x 105 N/mm2
E, G, K and μ represent the elastic modulus, shear modulus, bulk modulus and -Poisson's ratio respectively of a linearly elastic, isotropic and homogeneous material. To express the stress strain relations completely for this material, at least
E = 2G(1 + μ) ...(i)
E = 3K (1 - 2μ) ...(ii)
In this cae there are four unknown quantities and only two equation are available. Hence any two of the four unknown quantities must be known for complete stress relationship.