Von-Mises’ stress plot

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}