[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)

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….


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



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



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



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



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….


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



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


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



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


∫ 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)



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.


END OF PAPER

[ess08] university of bolton school of engineering …

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