# Restriction on Material Parameters Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : Restriction on Material Parameters Civil Engineering (CE) Notes | EduRev

The document Restriction on Material Parameters Civil Engineering (CE) Notes | EduRev is a part of the Civil Engineering (CE) Course Advanced Solid Mechanics - Notes, Videos, MCQs & PPTs.
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Restriction on material parameters
Next, we would like to limit the range of values that these material parameters can take so that the response predicted by using these constitutive relations confirm with the observations.
Let us now see why Young’s and shear modulus cannot be ∞. From the definition of the Young’s modulus we find that the axial displacement due to an applied axial load has to be zero if E = ∞. Similarly, if G = ∞ the change in angle due to an applied shear stress is zero. These would happen only if the body is rigid since, no strain develops despite stress being applied. However, the focus of the study here is deformable bodies. Hence, we obtain the condition that E < ∞, G < ∞.

However, the bulk modulus can be ∞. From the definition of bulk modulus, it is clear that if K = ∞ for some material, then the volumetric strain developed due to applied hydrostatic pressure, for these materials has to be zero. This means that the volume of the body made of this material does not change, the material is incompressible. Some materials like rubber, polymers are known to be nearly incompressible. Moreover, it is also known that the volume of these materials do not change in any deformation. This means that these materials are capable of undergoing only isochoric deformations. Constitutive relations for such incompressible materials are obtained in section 6.7.1.

Since, we expect that tensile stress produce elongation and compressive stresses produce shortening, the three modulus - Young’s, shear and bulk - should be positive. Since, E is positive, for G to be positive and finite, equation (6.87) requires (1 + ν) > 0. Thus,

ν > −1.                                          (6.92)

Note that, this is a strict inequality because if (1 + ν) = 0, G = ∞, (since 0 < E < ∞) which is not permissible.
Similarly, for K to be positive, it transpires from equation (6.91) that (1 − 2ν) ≥ 0, since E is positive. Hence,

ν ≤ 0.5.                                         (6.93)

Here we allow for equality because K can be ∞.
Combining these both restrictions (6.92) and (6.93) on the Poisson’s ratio,

− 1 < ν ≤ 0.5.                                   (6.94)

Thus, Poisson’s ratio can be negative, and it has been measured to be negative for certain foams. What this means is that as the body is stretched along a particular direction, the cross sectional area over which the load is distributed can increase. However, for most materials, especially metals, this cross sectional area decreases and therefore the Poisson’s ratio is positive.
Finally, we show that the three modulus values cannot be zero. From the definition of these modulus, they being zero means that, any amount of strain can develop even when no stress is applied. This means that there can be displacement without the force, when the modulus value is zero. Since, there has to be force for displacement, this means that the modulus cannot be zero.
In table 6.1 the restrictions on various parameters are summarized. The point to note is that one of the Lam`e constants, λ has no restrictions. From equation (6.80), it can be seen that if 0 ≤ ν ≤ 0.5, then from the restrictions on E and ν it can be said that λ ≥ 0. However, λ < 0 for certain foams whose Poisson’s ratio is negative. Therefore λ has no restrictions.

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## Advanced Solid Mechanics - Notes, Videos, MCQs & PPTs

42 videos|61 docs

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