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A bar made of material whose Young's modulus is equal to E and Poisson's ratio to mu is subjected to the hydrostatic pressure p. Find: (a) the fractional decrement of its volume, (b) the relationship between the compressibility beta and the elastic constants E and mu. Show that Poisson's ratio mu cannot exceed 1//2. |
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Answer» Solution :Consider a cube of unit length before pressure is applied. The pressure acts on each FACE. The pressure on the OPPOSITE faces constitute a TENSILE stress producing longitudinal compression and lateral extension. The COMPRESSIONS is `p/E` and the lateral extension is `mup/E` The net result is a compression `p/E(1-2mu)` in each side. Hence `(DeltaV)/(V)=-(3p)/(E)(1-2mu)` because from symmetry `(DeltaV)/(V)=3(Deltal)/(l)` (b) Let us consider a cube under an equal compressive stress `sigma`, acting on all its faces. Then, VOLUME strain `=-(DeltaV)/(V)=(sigma)/(k)`, (1) where k is the bulk modulus of elasticity. So `sigma/k=(3sigma)/(E)(1-2mu)` or, `E=3k(1-2mu)=3/beta(1-2mu)` (as `k=1/beta`) `mule1/2` if E and `beta` are both to remain positive. |
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