[ESS08] UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING …
Transcript of [ESS08] UNIVERSITY OF BOLTON SCHOOL OF ENGINEERING …
[ESS08]
UNIVERSITY OF BOLTON
SCHOOL OF ENGINEERING
BENG (HONS) IN MECHANICAL ENGINEERING
SEMESTER TWO EXAMINATION 2017/2018
ENGINEERING PRINCIPLES 2
MODULE NO: AME4053
Date: Monday 21st May 2018 Time: 10:00 – 12:00
INSTRUCTIONS TO CANDIDATES: This paper is split into two parts;
Part A and Part B. There are three
questions in each Part.
Answer FOUR questions in total; Part A – Answer TWO questions Part B – Answer TWO questions
All questions carry equal marks.
Marks for parts of questions are
shown in brackets.
This examination paper carries a
total of 100 marks.
CANDIDATES REQUIRE: Formula sheet (attached)
Page 2 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
Part A – Answer TWO questions Q1
a) A small iron ball is dropped from the top of a vertical cliff and takes 2.5 s to
reach the sand beach below. Find:
i) the velocity with which it strikes the sand (5 marks)
ii) the height of the cliff (5 marks)
iii) if the ball penetrates the sand to a depth of 1.5 cm, calculate its average
retardation (5 marks)
b) A car travelling at 50km/h increases its speed to 80km/h in 10 s. If a wheel on
the car has a diameter of 0.4m determine:
i) The initial angular velocity of the wheel (4 marks)
ii) The final angular velocity of the wheel after the 10 seconds (3 marks)
iii) The angular acceleration of the wheel (3 marks)
Total 25 marks
Q2 a) Find the position of the centroid 𝑦 in Figure Q2a
Figure Q2a (10 marks)
Question 2 continues over the page….
PLEASE TURN THE PAGE….
Page 3 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
Question 2 continued….
b) A cantilever has a cross-section as shown in Figure Q2b. Determine, using
the parallel axis theorem, the second moment of area of the section about the
neutral XX.
Figure Q2b
(15 marks)
Total 25 marks
PLEASE TURN THE PAGE….
Page 4 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
Q3 a) Determine the maximum force that can be applied at the mid-span of the
simply supported beam shown in Figure Q3a if the maximum bending stress at
this point is not to exceed 200MPa.
Figure Q3a
(10 marks) b) A hollow drive shaft with inner diameter of 100mm is required to transmit
400kW of power at 2000 rev/min. The angle of twist within a length of 1.5m is not to exceed 1o.
Calculate the minimum external diameter required for the driveshaft. Take G = 80GPa
(15 marks)
Total 25 marks
END OF PART A
PLEASE TURN THE PAGE….
Page 5 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
PART B – Answer TWO questions
Q4) a) Find the first and second derivatives of each of the following;
i) 𝑥3 − 4𝑥2 + 𝑥 + 2 (2 marks)
ii) 6e5𝑥 − 4 cos(2𝑥) (2 marks)
b) Find the derivatives of each of the following;
i) (3𝑥2 − 2𝑥)4 (4 marks)
ii) (4𝑥2 − 3) sin(𝑥) (4 marks)
iii) 𝑥2+5
ln(𝑥) (4 marks)
c) Find and classify the stationary points of the following functions;
i) 𝑥2 − 10𝑥 + 4 (4 marks)
ii) 𝑥3 − 3𝑥2 + 11 (5 marks)
Total 25 marks
PLEASE TURN THE PAGE….
Page 6 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
Q5) a) Consider the following:
𝑓(𝑥) = e𝑥 −2𝑥− 3
It has two roots: one close 𝑥 = 2 and the other close to 𝑥 = −1.
Use the Newton-Raphson method to find these roots accurate to 6
decimal places.
[Recall this occurs when the relative error 𝜀 = |𝑥𝑖−𝑥𝑖−1
𝑥𝑖−1| for an iteration 𝑥𝑖 is
smaller than 10−6. ]
(9 marks)
b) Evaluate the following definite integrals;
(i) ∫ 1
2𝑥3 𝑑𝑥
4
2 (3 marks)
(ii) ∫ ( sin(2𝑥) + 3 cos(6𝑥) )𝑑𝑥𝜋
0 (4 marks)
c) Consider the following integral:
∫ 1
e𝑥 + 𝑥 𝑑𝑥
4
2
.
Approximate the value of this integral with 4 strips using:
(i) the trapezoidal rule; and (4 marks)
(ii) Simpson’s rule. (5 marks)
Give your answers to 4 decimal places.
Total 25 marks
Page 7 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
PLEASE TURN THE PAGE….
Q6) (a) Find the particular solution of the following differential equations:
(i) {
𝑦′ = 2𝑥
𝑦
𝑦(0) = 3
(5 marks)
(ii) { 𝑦′ + 2𝑥 = e3𝑥
𝑦(0) = 1
(5 marks)
(iii) {
𝑦′ + 2𝑥𝑦 = 𝑥
𝑦(0) = 2. (5 marks)
(b) Find the particular solution of the differential equation:
{
𝑦′′ − 4𝑦′ + 13𝑦 = 0
𝑦(0) = 0
𝑦′(0) = 3
(10 marks)
Total 25 marks
END OF QUESTIONS
FORMULAE SHEETS OVER THE PAGE….
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Formula Sheet. Stress and Strain:
E
L
u
A
F
G
strainshear
A
F
lat
long
lat
long
volumetric
E
V
V
KV
V
etcE
etcEE
EEE
zz
xy
y
zyx
x
.................
.......
12
213
EG
EK
Static Equilibrium:
∑𝐹𝑥 = 0 ; ∑𝐹𝑦 = 0 ; ∑ 𝐹𝑧 = 0 ; ∑𝑀𝑥 = 0 ; ∑𝑀𝑦 = 0 ; ∑𝑀𝑧 = 0
PLEASE TURN THE PAGE….
Page 9 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
Thin Pressure Vessels:
hoop
pd
t
2
longitudinal
pd
t
4
longitudinal
pd
tEl
41 2
diametral
p
tEd
41 2
For Cylindrical Shells: Vpd
tEV
45 4
For Spherical Shells: Vpd
tEV
3
41
2nd Moments of Area
Rectangle I = 12
bd3
Circle I = 64
4πd Polar J =
32
4d
Parallel Axis Theorem
Ixx = IGG + Ah2
Bending
R
E
yI
M
Torsion
G
rJ
T
Motion
v = u + at 2 = 1 + t
v2 = u2 + 2as 221
22
s = tvu
2 t
2
21
s = ut + ½ at2 21
2
1tt
Speed = Distance Acceleration = Velocity
Time Time
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s = r
V = r
a = r
Torque and Angular
TP
mkI
IT
2
Energy and Momentum
Potential Energy = mgh
Kinetic Energy
Linear = ½ mv2
Angular =½ I2
Momentum
Linear = mv
Angular = I
Vibrations
Linear Stiffness
Fk
Circular frequency m
kn
Frequency n
nn
Tf
1
2
maF
T
Tf
xa
trxrv
trx
2
1
sin
cos
2
22
PLEASE TURN THE PAGE….
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Integration
∫ 𝑥𝑛 𝑑𝑥 = 𝑥𝑛+1
𝑛 + 1+ 𝐶 (𝑛 ≠ −1)
∫ (𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 =(𝑎𝑥 + 𝑏)𝑛+1
𝑎(𝑛 + 1)+ 𝑐 (𝑛 ≠ −1)
∫1
𝑥𝑑𝑥 = 𝑙𝑛|𝑥| + 𝐶
∫1
𝑎𝑥 + 𝑏𝑑𝑥 =
1
𝑎ln|𝑎𝑥 + 𝑏| + 𝑐
∫ ⅇ𝑥 𝑑𝑥 = ⅇ𝑥 + 𝑐
∫ ⅇ𝑚𝑥 𝑑𝑥 =1
𝑚ⅇ𝑚𝑥 + 𝑐
∫ cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐
∫ cos 𝑛𝑥 𝑑𝑥 =1
𝑛sin 𝑛𝑥 + 𝑐
∫ sin 𝑥 𝑑𝑥 = −cos 𝑥 + 𝑐
∫ sin 𝑛𝑥 𝑑𝑥 = −1
𝑛cos 𝑛𝑥 + 𝑐
∫ sec2 𝑥 𝑑𝑥 = tan 𝑥 + 𝑐
∫ sec2 𝑛𝑥 𝑑𝑥 =1
𝑛tan 𝑛𝑥 + 𝑐
∫1
√1 − 𝑥2𝑑𝑥 = sin−1 𝑥 + 𝑐
∫1
√𝑎2 − 𝑥2𝑑𝑥 = sin−1 (
𝑥
𝑎) + 𝑐
∫1
1 + 𝑥2𝑑𝑥 = tan−1 𝑥 + 𝑐
∫1
𝑎2 + 𝑥2𝑑𝑥 =
1
𝑎𝑡𝑎𝑛−1 (
𝑥
𝑎) + 𝑐
Page 12 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
∫ 𝑢 𝑑𝑣
𝑑𝑥 𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣
𝑑𝑢
𝑑𝑥 𝑑𝑥 (Integration by parts)
Recall that in the following 𝒚′ =𝒅𝒚
𝒅𝒙, 𝒇′ =
𝒅𝒇
𝒅𝒙 and 𝒈′ =
𝒅𝒈
𝒅𝒙:
Product rule: 𝒚 = 𝒇(𝒙) ∙ 𝒈(𝒙) ⇒ 𝒚′ = 𝒇′(𝒙) ∙ 𝒈(𝒙) + 𝒇(𝒙) ∙ 𝒈′(𝒙)
Quotient rule: 𝒚 =𝒇(𝒙)
𝒈(𝒙) ⇒ 𝒚′ =
𝒇′(𝒙)∙𝒈(𝒙)−𝒇(𝒙)∙𝒈′(𝒙)
𝒈(𝒙)𝟐
Chain rule: 𝒚 = 𝒇( 𝒈(𝒙) ) ⇒ 𝒚′ = 𝒇′( 𝒈(𝒙) ) ∙ 𝒈′(𝒙)
Page 13 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
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Differential Equations
Auxiliary equations for differential equations of the form
𝑎𝑑2𝑦
𝑑𝑥2+ 𝑏
𝑑𝑦
𝑑𝑥+ 𝑐𝑦 = 0
1) Real and Different roots 𝛼 and 𝛽
𝑦 = 𝐴ⅇ𝛼𝑥 + 𝐵ⅇ𝛽𝑥
2) Repeated (Real and Equal) root 𝛼
𝑦 = ⅇ𝛼𝑥(𝐴 + 𝐵𝑥)
3) Complex roots (𝑝 ± 𝑖𝑞)
𝑦 = ⅇ𝑝𝑥(𝐴 cos 𝑞𝑥 + 𝐵 sin 𝑞𝑥)
Numerical Methods
Approximating integrals
∫ 𝑓(𝑥) 𝑑𝑥 ≈ ℎ
2[𝑦0 + (𝑦1 + 𝑦2 + 𝑦3 + ⋯ + 𝑦𝑛−1) + 𝑦𝑛 ] (Trapezoidal rule)
𝑏
𝑥=𝑎
≈ ℎ
3[𝑦0 + (4𝑦1 + 2𝑦2 + 4𝑦3 + ⋯ + 4𝑦𝑛−1) + 𝑦𝑛 ] (Simpson′s rule)
where ℎ =𝑏−𝑎
𝑛 is the step size and 𝑛 indicates the number of strips.
(Note: for Simpson’s rule, 𝑛 must be even).
Root finding
For a function 𝑓(𝑥), the solutions to 𝑓(𝑥) = 0 can be found using the following
iterative scheme:
𝑥𝑖+1 = 𝑥𝑖 −𝑓(𝑥𝑖)
𝑓′(𝑥𝑖)
This is the Newton-Raphson method.
Page 14 of 14 School Of Engineering BEng (Hons) in Mechanical Engineering Semester 2 Examination 2017/2018 Engineering Principles 2 Module No. AME4053
END OF PAPER