Briefly answer 1). In an FE simulation with Abaqus, if the unit for the length is...

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Briefly answer 1). In an FE simulation with Abaqus, if the unit for the length is...

Briefly answer

1). In an FE simulation with Abaqus, if the unit for the length is μm and the unit for the force is Netwon, what is the correct unit for the elastic modulus? Why?

2). List the steps involved in the procedure of an ordinary finite element analysis. No need to consider the order of the steps.

3). What are the differences between a first-order tetrahedral element and a second- order tetrahedral element? Write down three differences.

engineering Mechanical-Engineering

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Answer 1

Answer #1

Answer (1) :

As per finite element analysis the correct units to find out about Elastic modulus or Modulus of elasticity (E) .

Definition of "ELASTIC MODULUS" is ratio of stress (sigma) to the strain (Ep

Definition of STRESS is load per unit area and its unit is Newton per metre square (N/M2).

Definition office STRAIN is change in length to the original length , it has no units.

youngs modulus (E) - Stress (o Straru (0) Stress (0) = load (Ⓡ) Area (A) stoalu (E)= change in Length (8) Signal Length (L)

Final the calculation of elastic modulus (E) are :

As per fe Analyses Elastic modeelees(E) E= Stress o strain ) = P/A PL Afl JUL Units (E) = N/m?

Answer (2) :

Steps involved in the procedue of an ordinary finite element analysis.

Step 1 : Selection of an element geometry (Discretizarion) .

Step 2 : Selection of shape functions (Pattern or distribution of unknown variables).

Step 3: Development of finite element equation ( Principles or laws to form governing equation Arara applied boundary conditions by suitable Shape funtions).

Step 4 : Assembly this element equation to form a global eqations.

Step 5 : Solve the unknown equations in step 3 and step 4 .

Step 4 : Results will be obtained.

Answer (3) :

First order tetrahedral element (or) Linear element

(1) linear for first order element have nodes at corners only .Ex Face centred cubic structure.

(2) This linear element mostly used for simple structures and it gives less accurate result heavy structures.

(3) We can calculate the regular shapes of the different simple Structure objects.

Second order tetrahedral element (or) Quadratic element :

(1) In which 2nd order quadratic element will have mid nodes in addition to the cornor nodes.

(2) This second order element has more accurate results for heavy structures and gives more accurate results.

(3) It this module it takes more time to analyse the component because it has more nodes enter takes more time.

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