AA530 – (Solution)

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1. [20 points] (Linear elastic isotropic material) Derive G = E / 2 (1+v) for linear elastic isotropic materials. E, G (or  according to the Bower textbook), and v are Young’s modulus, shear modulus, and Poisson’s ratio, respectively.

2. [30 points] (Constitutive relationship) Derive constitutive relationship under the plane strain and stress conditions (i.e., derive expressions in Sec. 3.2.3. in the Bower book from the full 3D expressions).

3. [20 points] (Constitutive relationship) Given a state of strain at a point in an isotropic material,

é 0.001 ê
e=ê 0.001
êë -0.002 0.001 0
0.005 -0.002 ùú
0.005 ú
0 úû
Determine stress tensor. Assume E = 200 GPa and v = 0.3.

4. [30 points] (Coordinate transformation) An equi-triangular strain rosette is mounted on the surface of a body as shown below.

1

The strain gauge readings are:

𝜀0° = 0.005, 𝜀60° = 0.002, 𝜀120° = −0.001

The material properties are E = 200 GPa and v = 0.3. Find the stress tensor along the xy axes at point O.

2

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