Answer Happy

We consider the formation of isotropic elastic bodies. Calculate the deformation of a (cyrindrical) rod under the uniform gravity. Note that gravity for the part of rod is given by pg (g: gravitational acceleration) with p: the mass per unit volume. "stretch deformation" In this problem, the rod stands on the ground under gravity. If we define the ď₁ axis in the verti cally upward (opposite direction of gravity), the cross section with x₁ (the length from bottom) re ceives a compressive stress by pgs(L-₁), where s: the area of cross section and L: the length of t he rod.
。 Stretch deformativ Eq Q Ezz Eij →x, a rods (cylinder), one side fixed to the wall. cross-sectin aren: S, force F As we define di-axis in the direction of the rod, Stress has P₁. = I and Pij F =0(1+1,871) S E₁₁ λ + эм злом) Ритэм Рис E33 ه S = Eq@. F A Pu 2μ(3X12μ) P11 0 (jj) > 0 1 + 3/4> 0 =>d+M >o F: tensim => elongation F: pressure =) contractin where yo P₁₁ = YE₁₁ Pu Ell S F лм M(3X +34) P₁₁ P₁₁ - & M(3λ +2μ) M+μ = Young's mochilus