Assignment #2 - Dual Material Shaft FEM Analysis
September 19th, 2025
Prepared by:
Dawson Mork (23541659)
Seamus MacDougall (42214478)
Prepared for:
Hadi Mohammadi, Ph.D., P.Eng.
Instructor, ENGR 492
School of Engineering
University of British Columbia, Okanagan Campus
hadi.mohammadi@ubc.ca
Introduction
The finite element method is a network of discrete approximations that model a larger stress analysis
problem. This method is helpful when an analytical or exact solution is too cumbersome, or is simply
unsolvable.
Methodology
The first step in understanding this problem is to determine the degrees of freedom in the system that is
being analyzed. For this assignment we can assume that this is a spring element, where the element can
be stretched and compressed but no bending occurs. Based on this we can apply hookes law:
𝐹 = 𝐾∆𝑥
F= Force
K= spring constant
x= displacement
For the purposes of this assignment we can relate how solids elongate to a spring constant.
𝑃𝐿
δ = 𝐸𝐴
Rearranging the equation to relate to Hooke's law we get
𝑃= δ*
𝐸𝐴
𝐿
P = Force (F)
δ = displacement (x) = U
(EA/L) = spring constant/material properties (K)
After relating the forces acting on the two material cylinders to a spring system we can analyze each
spring and node in matrices. The Matrix form:
For Element 1:
For Element 2:
Once in matrix form the next step is to enlarge each matrix to be able to combine them:
Element 1: ​
​
​
​
​
​
Element 2:
We can then combine the matrices:
Now we can determine boundary conditions:
𝐹1 = − 𝑝 = 𝑅
𝐹2 = 0
𝐹3 = 𝑃
Substituting the boundary conditions in:
We can now solve for the unknown displacements:
Ⅰ: 𝐾1 * 𝑈1 − 𝐾1 * 𝑈2 = 𝑅
𝐾1 * (𝑈1 − 𝑈2 ) = 𝑅 [1]
Ⅱ: (𝐾1 + 𝐾2) * 𝑈2 − 𝐾2 * 𝑈3 = 0 [2]
− 𝐾2 * 𝑈2 + 𝐾2 * 𝑈3 = 𝑃
𝐾2(𝑈3 − 𝑈2) = 𝑃 [3]
Results
𝐾1 =
𝐾2 =
𝐸𝐴*𝐴
𝐿𝐴
𝐸𝑆*𝐴
𝐿𝑆
9
−4
=
(70*10 )*(1.13097*10 )
0.2
=
(200*10 )*(1.13097*10 )
0.3
9
= 39583950 [𝑁/𝑚]
−4
= 75398000 [𝑁/𝑚]
By solving a system of linear equations ([2] & [3]) and our spring rates K1 and K2, we find that the
displacement U2 and U3 are 2.5263*10^-4 [m] and 3.8526*10^-4 [m], respectively.
With the interface and free-end displacements known, we can now use equation [1] to solve for the
reaction force. We expect the reaction force to be equal (or very close) to the applied force “P” as no other
external forces act on the system.
−4
𝑅 = 𝐾1 * (𝑈1 − 𝑈2 ) = 39583950 * (0 − 2. 5263 * 10 ) =
− 10000. 093 [𝑁]
Discussion
By applying the 2-spring element system methods learned in class, we were able to build our local
stiffness matrices, then combine them into a global stiffness matrix which we then solved in parts to find
the displacements and the reaction force at the base.
With only one applied reaction force, and one fixed end, U1 and F2 are both zero which simplifies the
process. Other external forces could be applied but would have to be known in order to solve for the
displacements.
Because of the singular applied force, we expected the reaction force to be equal and opposite. Our results
clearly show this and are only 0.093 N off of the expected value of 10000 N which supports our theory
and acts as a nice little proof to verify our calculations.
Conclusion
In summary, our results behaved as expected and reasonable values were obtained for the displacements
and the reaction force.
Summary:
Interface displacement = U2 =................2.5263*10^-4 [m]
Free end displacement = U2 + U3 =.......1.1379*10^-3 [m] or 1.1379 [mm]
Reaction force =......................................-10000.093 [N]