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

November 12, 2018
##### Author : Julia Miles

Solution Code: 1DCJ

## Question:

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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.

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## Solution:

(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) is isotropic, 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

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%

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