﻿ 程序代写案例-AMME 2000|学霸联盟

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The University of Sydney
Faculty of Engineering and IT
AMME 2000 Engineering Analysis
This examination paper contains 4 questions
Each question is on a separate page.
All questions carry equal marks.
Candidates should answer any 3 out of 4 questions
There are a total of 5 pages including this cover sheet.
These pages must be handed in with the examination script.
A simple calculator (programmable versions and PDA's not allowed) may be taken into the
exam room.
Take care to write legibly. Write your nal answers in ink, not pencil.
Name:
SID:
Closed Book Duration: 2 hour 00 minutes + 10 minutes reading time
10:00 Tuesday XX
th
June 2XXX
1
Question 1.
Consider a uniform rod with ends at x = 0, 1, and with a temperature distribution T (x , t) that
satises the diusion equation Tt = Txx . Suppose that at t = 0 rod's temperature is a constant T0
for 0 x 1, and that for t > 0 the left end is kept at temperature T0 and the right at 0. Thus
the boundary conditions can be stated as T (0, t) = T0, T (1, t) = 0 for t > 0 and initial conditions
T (x , 0) = T0 , 0 < x < 1.
(a) Sketch the variation of the solution at the initial time and two later time, highlighting the
physical behaviour of the system. [Ans: Sketch the initial condition, the late time steady linear
solution ,and a curve showing the solution at a time 0 << t <<1]. (4 marks)
(b) Derive the steady state late time temperature distribution T s(x , t). [Ans:T s(x , t) = T0(1x)]
(4 marks)
(c) Using separation of variables, solve for the temperature T(x,t) as a sum of a homogeneous
and inhomogeneous solution. Clearly state the initial conditions and boundary conditions
of the homogeneous problem [Ans: ICs: T0x , both BCs:0. Soln: T (x , t) = T0(1 x) +
T0
P1
n=1 Bn sin(nx)e
2nt
where Bn = (2=n) cos(n) ]. (8 marks)
(d) Estimate the temperature at time t=0.1s, x=0.98 using the rst term in the series solution
when T0 = 100 [Ans: 3.49]. (4 marks)
2
Question 2.
The wave equation utt = c
2uxx can be discretised as follows:
un+1i 2ui + un1i
t2
c2 u
n
i+1 2uni + uni1
x2
= 0 (1)
where x is the grid spacing and t the time step size.
(a) Using a Taylor Series expansion, determine the order of accuracy of the approximation uxx =
uni+12uni +uni1
x2
. [Ans: Expand each term as a function of uni in space, simplify. The leading order
term is 2nd order in space.] (6 marks)
(b) Via a von Neumann stability analysis, show that the maximum stable time step size is t =
x=c . [Ans: See notes] (6 marks)
(c) Given the initial conditions:
uni2 = 1.0
uni1 = 1.0
uni = 0.3
uni+1 = 0.0
uni+2 = 0.0
and un1i = 0.25, calculate u
n+1
i using the above scheme given t = x=c . [Ans: u
n+1
i =
0.75] (4 marks)
(d) Discuss the physical behaviour of the solution this equation where the initial conditions are now
u(x , 0) = sinxL and with boundary conditions u(0, t) = u(L, t) = 0.. Sketch the behaviour
of the solution for three representative times. [Ans: The solution is an oscillation in space and
time. In this case it is a mode 1 standing wave between 0 x l . This will not dissipate
in time. The sketch should show a half period sin wave with a positive amplitude (t=0), then
with a negative amplitude after half a full period of oscillation, and then a at dispacement but
with a velocity.] (4 marks)
3
Question 3.
An innitely long cable spanning 0 x 1 is initially at rest and having zero displacement. At
time t = 0s and x = 0m the of the cable is given a disturbance f (t) = t (in metres) for t 1s. For
t > 1s, the forcing term f (t) drops straight back to zero. The governing equation is:
ytt = c
2yxx (2)
where c2 = 400m2=s2
(a) Specify clearly the boundary conditions and initial conditions for this problem. [Ans: ICs: at rest
yt(x , 0) = 0 and level y(x , 0) = 0. The BCs are y(1, t) = 0 and y(0, t) = f (t).] (4 marks)
(b) Solve this problem to determine the variation of displacement y(x , t) with time and space.
[Ans: Use the Laplace Transform to demonstrate that the solution is travelling wave of the
form y(x , t) = f (t x=c)u(t x=c)] (8 marks)
(c) Sketch the solution at three points in time, clearly indicating the shape of the solution.[Ans: See
notes, slide 187. Note that the triangle will have the opposite shape in space as it did in time
(i.e. as the wave passes it will gradually increase in amplitude to size 1 then discontinuously
decay back to zero). (4 marks)
(d) Calculate the displacement at x = 18m for t = 1s [Ans: x=c = 0.9, thus the answer is
f (1 0.9) = f (0.1) = 0.1. The wave has arrived at the point, but has not yet passed it.]
(4 marks)
4
Question 4.
An engineering rm have asked you to compute the stress within a beam xed vertically from a
ceiling as shown in Figure Q4.1. This can be modelled as a one dimensional beam in x with cross-
sectional area A, modulus of elasticity E and total length L. The force per unit length due to gravity
q(x) = Ag . Note that
(a) Discretise the beam using two nite elements. Sketch your discretisation and clearly label the
elements, the nodes, force and the boundary conditions. [Ans: Make a similar sketch to the
one I made in deriving the stiness matrix for the 3 node problem. Label the rst node at the
wall node '0'] (6 marks)
(b) Draw a diagram showing the variation of the shape functions N0, N1 and N2 within each
element, where Ni is the shape function for node i [Ans: Follow the sketch in Section 8.2, or
a version of the one on slide 225. Essential that you show the shape functions going linearly
from 0 to 1 correctly within each element.] (4 marks)
(c) The simplied weak form of the governing equation is:
Z L
0
wxEAuxdx =
Z L
0
wqdx , (3)
where u is the displacement of a single node. Assuming a test function of the form w = v and
discretisation using piecewise constant shape functions Ni , show that the nodal dispacements
ui are given by the solution of the following matrix problem:
EA
l

2 1
1 1

u1
u2

= Agl

1
1=2

(4)
where l is the element length, and qi is the force evaluated at node i [Ans: Straightforward
(6 marks)
(d) Detail how the above matrix problem would change if the element length li is dierent for each
element i . [Ans: Show that the matrix problem becomes:
EA

1=l1 + 1=l2 1=l2
1=l2 1=l2

u1
u2

= Ag

l1=2 + l2=2
l2=2

(5)
] (4 marks)
5
qx,u
L
Figure Q4.1: .
6

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