University FEA Assignment: Contact Analysis and Elliptical Constraints

This document presents a detailed solution to a Finite Element Analysis (FEA) assignment. The first part of the solution focuses on contact analysis using ABAQUS, simulating a hollow cylinder subjected to compressive forces from rigid indenters. The solution explores two cases: one where the cylinder material fails based on von Mises stress and another where failure is determined by maximum tensile stress, providing calculated deformation values at failure and failure locations. The second part of the solution addresses an elliptical constraint problem, deriving the constraint equation and solving it using both the penalty method and Lagrange multipliers method. The solution includes the mathematical derivations, the constraint equations, and the Mathematica code used to solve the problem. The document provides a comprehensive understanding of the problem, the methods used to solve it, and the results obtained.

Question 1. A simulation was performed on the ring with outer radius of 100 mm and inner radius of 30
mm.
Material property –
Young’s modulus is 2000MPa and Poisson’s ratio of 0
For 1st case - The cylinder is made out of a material whose maximum von Mises stress is 400MPa
If the von-Mises stress is checked then it crosses value of 400 MPa at step time of 6E-3 seconds. Total
applied deformation over 0.01 seconds is 20 mm hence:
Allowed deformation = (0.006/0.01) x 20 = 12 mm
The location of failure (max stress) is found to be the bottom contact region near the cylindrical
indenter.
Von-Mises’ stress plot for 1st case

For 2nd Case - The cylinder is made out of a brittle material whose maximum tensile stress is 400MPa
For this case, max principal stress is checked since brittle material fails by maximum principal stress
theory.
Max deformation at failure = (0.0095/0.01) x 20 = 19 mm
As per the stress plot, the failure location is visible at the center of the ring inside the hole.
Figure – Maximum principal stress plot for brittle material
Question 2.
Contraint equation g ( u1 ,u2 )=0 for the given ellipse

Figure – Elliptical mechanism
Given condition - F=E=l0= A=a=2b=1
Center is assumed to be = (0,0)
Ellipse equation
x2
a2 + y2
b2 =1
Now if the center of the ellipse goes to a new location (da+a,0) then the equation is written as:
( x−da−a)2
a2 + y2
b2 =1
Where x = da +u1 and y = u2
Hence the equation becomes:
(u1 −a)2
a2 + u2
2
b2 =1
It is given that a = 2b = 1 hence the equation is now:
g=0.25u1
2−0.5 u1 +u2
2=0
This equation is solved using 2 methods:
1. Penalty method
u1=0.9647
u2=0.4994
2. Langrange Multiplier
u1=0.9693
u2=0.4998

The code utilized is given below:
Clear[u1, u2, PE, s, lambda]
L0 = A = F = EE = a = 1;
(*Penalty Method*)
b = 0.5;
(*Calculating length along the ellipse*)
LNEW = Sqrt[(L0 + u1)^2 + u2^2];
(*Calculating strain by length change*)
STRAIN = (LNEW - L0) / L0;
betapenal = 1000;
SOLU1 = FullSimplify[1 / 2 * STRAIN^2 * L0 - 1 * u1 + betapenal / 2 * (0.25 * u1^2 - 0.5
* u1 + u2^2)^2] ;
Eq1 = D[SOLU1, u1];
Eq2 = D[SOLU1, u2];
s = FindRoot[{Eq1 ⩵ 0, Eq2 ⩵ 0}, {u1, 1}, {u2, 0.5}]
(*Lagrange Multipliers Method*)
SOLU2 = FullSimplify[1 / 2 * STRAIN^2 * L0 - 1 * u1 + lambda * (0.25 * u1^2 - 0.5 * u1
+ u2^2)] ;
Eq3 = D[SOLU2, lambda];
s2 = FindRoot[{Eq1 ⩵ 0, Eq3 ⩵ 0}, {u1, 1}, {u2, 0.5}]
{u1 → 0.964705, u2 → 0.499435}
{u1 → 0.969274, u2 → 0.499764}