UOW ENGG592 Engineering Computing Assignment 2: Spring-Mass and Heat

This assignment solution addresses two key engineering problems: the motion of a damped spring-mass system and transient heat conduction. The spring-mass system analysis involves solving an ordinary differential equation (ODE) to model the displacement of a mass, considering both under-damped and critically damped scenarios. The solution uses numerical methods, specifically the second-order Runge-Kutta Heun method, and includes Matlab code to plot the displacement over time. The heat conduction problem is solved using explicit finite difference and implicit Crank-Nicholson methods. The solution describes the non-dimensional form of the heat equation, boundary conditions, and the discretization process. The assignment provides a comprehensive analysis using numerical techniques to simulate and understand the behavior of both systems.

UNIVERSITY OF WOLLONGONG
FACULTY OF ENGINEERING AND INFORMATION SCIENCES
ENGG592 ENGINEERING COMPUTING
SPRING SESSION -2017
ASSIGNMENT 2
STDUENT NAME
STUDENT REGISTRATION NUMBER
DATE OF SUBMISSION

QUESTION 1
MOTION OF A SPRING MASS SYSTEM (50%)
The motion of a damped spring mass system is described by the following ordinary differential
equation
Where x is displacement from equilibrium position (meter), t is time (second), m is the mass and
equal 20 kg, c is the damping coefficient (N.sec/meter). The damping coefficient, c, takes on two
values of 5 (under damped), 40 (critically damped). The spring constant k=20N/meter. The initial
velocity is zero, and the initial displacement x=1 meter.
Solution
(i) Transform the problem to a system of two first order initial value ODES.
From Newton’s second law,
∑ f =ma
m d2 x
d t2 +c ( dx
dt )+kx =0
Dividing through by mass, m,
d2 x
d t2 + c
m ( dx
dt )+ k
m x=0
The natural frequency of the system and the damping ratio are replaced as shown in
the following equation,
d2 x
d t2 + 2ζ w0 ( dx
dt )+w0
2 x=0
Where,
w0= √ k
m , ζ = c
2 √km
The solution to the differential equation.

λ1,2=−2 ζ ω0 ± √ 4 ζ2 ω0
2 −4 ω0
2
2
Therefore,
λ1,2=−ζ ω0 ± ω0 √ ζ 2−1
The transient solution of the system is given as,
xt =A e
(−ζ ω0 +ω0 √ζ 2−1 ) t +B e
( −ζ ω0 −ω0 √ζ 2−1 ) t
The constants A and B are obtained when the initial conditions are inserted. The final
solution is obtained as,
From the initial conditions,
w0=1 , ζ=0.125
xt =A e
(−0.125+ ω0 √ζ 2−1 ) t + B e
(−ζ ω0−ω0 √ζ 2−1 ) t
(ii) Second order RK Heun method over a period of time
The Heun’s method evaluates the slope at the beginning and at the end of the step.
The generalized idea embodied in the Heun’s method is the Runge-Kutta method
which uses a weighted average of the slope evaluated at multiple in the step.
%% order 647236
tspan=[0:0.1:50];
y0=[0.02;0];
[t,y]=ode45('unforced2',tspan,y0);
plot(t,y(:,1));
grid on
xlabel('time')
ylabel('Displacement')
title('Critically-damped Modelling of MSD system (C=40)')
%% order 647236
tspan=[0:0.1:50];
y0=[0.02;0];
[t,y]=ode45('unforced1',tspan,y0);
plot(t,y(:,1));
grid on
xlabel('time')
ylabel('Displacement')
title('Under-damped Modelling of MSD system (C=5)')
hold on;
%% order 647236
function yp=unforced2(t,y)
c=40;
m=20;
k=20;
yp= [y(2); (-((c/m)*y(2))-((k/m)*y(1)))];

%% order 647236
function yp=unforced1(t,y)
c=5;
m=20;
k=20;
yp= [y(2); (-((c/m)*y(2))-((k/m)*y(1)))];
(iii) Plot the displacement versus time for the two values of the damping coefficient on the
same figure.

QUESTION 2
TRANSIENT HEAT CONDUCTION (50%)
The non-dimensional form for the transient heat conduction in an insulated rod is
∂2 u
d x2 = ∂u
∂ t
The x is the non-dimensional length, t is the non-dimensional time, u is the non-dimensional
temperature. This makes for the following boundary and initial conditions,
Solution
(i) Explicit finite difference method and implicit crank-Nicholson method.
The explicit finite difference method. The relationship between continuous and
discrete problems.
Uniformly spaced in the interval from 0-L such that
xi= ( i−1 ) ∆ x i=1,2, … , N ( N −spatial nodes )
The spacing between the values is computed by,
∆ x= L
N−1
The discrete, t, are uniformly spaced in 0 ≤ t ≤ tmax :
tm= ( m−1 ) ∆ t m=1,2 , … , M

M- is the number of time steps and the ∆ t is the size of a time step.
∆ t= tmax
M −1
The finite difference method involves using discrete approximations like
The implicit crank-Nicholson method
(ii) Matlab implementation

(iii) Non-dimensional temperature versus non-dimensional length