Discuss and define Isotropy, Anisotropy and Orthotropy - Stress Analysis Worksheet Assessment Answers

Stress Analysis Worksheet Assignment Question 1 (i) Discuss and define Isotropy, Anisotropy and Orthotropy with regards to engineering materials. (ii) Regarding Finite Element Analysis: a. List the three main types of elements and explain how to choose elements? b. Common sources of errors in FEA? (iii) After conducting a photoelastic investigation of a sample, you remove the load and notice that a fringe pattern still exists. What could this mean? Question 2 A shaft has a maximum allowable shear stress of 25 MPa and must carry a torque of 150 N.m. It is required to have a change in diameter from 35 mm to 42 mm. How would you design this section? Show calculations to justify you answer.

(i) Isotropy : Material property is uniform in all direction, for example metals material with an infinite number of symmetry planes (i.e. every plane is a plane of symmetry) isisotropic, and requir es only 2 elastic constants Anisotropy : anisotropic material do not have planes of symmetry. It requires 21 elastic constants for full characterization. An orthotropic material has minimum two orthogonal planes of symmetry. In these two planes the material properties do not vary within planes. It require 9 independent variables to form the matrices. (ii) Three main type of elements. They are a) Rod b) Beam c) Plate Common sources of error are : Round off error , and discretization error. Discretization error is of two type

1. a) Error arising due to size of element
2. b) Error which do not vanish as size of element reduces ,and accuracy of applying the boundary conditions

(iii) That means material has crossed the elastic range Answer 2 Applied torque = 150 Nm Area moment of inertia = 7260 mm³ for 35 mm shaft Shear stress = 150*0.035/(7260*10^-3) M Pa= 0.72 Mpa, so solid rod is very strong we can use tube with Id=33 mm Area moment of inertia = 881 mm³ Shear stress = 150*0.035/(881*10^-3) M Pa= 5.96 Mpa, we still have factor of safty >4 With O.D of 45 mm, Id can be taken as 44 mm thus wall thickness is reduced from 1 mm to 0.5 mm Area moment of inertia = 767 mm³ Shear stress = 150*0.045/( 767*10^-3) M Pa= 8.8 Mpa, we still have factor of safety around 3 Reduction in weight of shaft = (35*1 - 45*0.5)*100/35 = 35.7% Answer 3 Answer 4