-2- (SME3033)

QUESTION 1 [25 marks]

a) The potential energy of the distributed load p acting on a beam element as shown in Figure

Q1(a) is given by:

[ ] { }∫ ∫ ⎟⎠⎞

⎜⎝⎛==Ω

−qdHplpvdx e

p

1

12ξ

The equivalent nodal forces and moments would be given by:

[ ]∫−1

12ξdH

ple

in which [ ] ⎥⎦⎤

⎢⎣⎡= 4321 22

HlHHlHH ee

and ( )31 32

41 ξξ +−=H , ( )32

2 141 ξξξ +−−=H

( )33 32

41 ξξ −+=H , ( )32

4 141 ξξξ ++−−=H

Show that the equivalent nodal forces and moments due to the distributed load p as shown in

Figure Q1(b) is given by:

T

eeeee plplplplf⎥⎥⎦

⎤

⎢⎢⎣

⎡−=

122122

22

[10 marks]

b) Figure Q1(c) shows a propped cantilever beam loaded with a 1 kN force. The beam has a

rectangular cross section as shown in the figure. By including the weight of the beam in the

analysis, determine:

i) the deflections and rotations at all nodes

ii) the support reactions.

Solve this problem by using two (2) beam elements. For the entire beam the modulus of

elasticity is E = 200 GPa, and the weight density is 10 kN/m3. [15 marks]

Figure Q1(a) Figure Q1(b)

Figure Q1(c)

pl2

e_ pl

2

e __

ple__

2pl e __

2

12 12

-

p

l

1 2

e

QUE

a) S

b) W

c) A

f

o

e

i

i

i

ESTION 2 [

State three m

What is the m

A two-bar p

force of 500

of steel (E =

element conn

) displace

i) normal

ii) reaction

[25 marks]

main characte

major differe

lanar truss s

N is applied

= 200 GPa) a

nectivity, cal

ements of no

stress in elem

n forces at no

ristics of a tw

nce between

structure has

d to the truss

and has a cr

culate the:

de 2,

ment 1,

ode 1.

ELEMENT1 2

y

-3- (SME303

wo-dimensio

n a plane trus

s the geome

at node 2, in

ross-sectiona

T 1st NOD1 2

Figure Q

x

3)

onal plane tru

ss and a plane

try as show

n the directio

al area of 20

E 2nd NOD2 3

Q2

uss structure.

e frame struc

n in Figure

on shown. E

00 mm2. By

DE

. [3

ctures? [2

Q2. A conc

Each member

y using the f

[2

3 marks]

2 marks]

centrated

r is made

following

20 marks]

-4- (SME3033)

QUESTION 3 [25 marks]

The plate shown in Figure Q3 (Young’s modulus = 70 GPa and Poisson’s ratio = 0.3) of thickness

2 mm is loaded with an edge pressure of 2 MN/m2 in the direction normal to the edge. A plane

stress condition is considered to solve for the displacements in the plate and it can be solved by

using two (2) constant strain triangular elements as shown in the figure. By using the following

nodes location and elements connectivity:

NODE X COORDINATE (mm)

Y COORDINATE (mm)

1 231 400 2 0 0 3 500 0 4 731 400

ELEMENT 1ST NODE 2ND NODE 3RD NODE 1 1 2 4 2 2 3 4

the element stiffness matrices (in N/m) are:

( )[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−

−−−

=

21.431.244.469.265.800.587.631.255.162.443.8

62.901.1431.237.369.292.4

7.2231.73.13

1071

sym

k

( )[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−

−−

=

62.901.1431.244.431.237.369.292.469.255.1

7.2231.765.862.43.1300.543.8

21.431.287.6

1072

sym

k

a) Assemble the global stiffness matrix for the whole plate. [8 marks]

b) Calculate the equivalent nodal forces due to the edge pressure. [2 marks]

c) Write the global system of linear equations. [5 marks]

d) Apply the boundary conditions and write the reduced system of linear equations. [5 marks]

e) Calculate the unknown degrees of freedom. [5 marks]

Figure Q3

QUE

a)

b)

c)

ESTION 4 (2

What is the

Explain brie

structure?

A uniform

normally fro

the rod is co

portion of th

at a temper

W/m2°C. U

assumed to

the rod is as

i) deter

ii) sketc

iii) amou

25 marks)

reason of us

efly, how doe

rod of 20 m

om a furnace

overed by ins

he rod, which

rature T∞ = 2

se three (3)

experience c

ssumed as 80

rmine temper

ch the temper

unt of heat lo

ing thin fins?

es heat transf

mm diameter

e wall that is

sulation of th

h has a lengt

25°C and pr

one-dimens

convection w

0% of its diam

rature To at th

rature profile

oss from the

-5- (SME303

?

fer in a thin f

r and therma

at a tempera

hickness Lins

th of Lo = 20

rovides a co

sional eleme

with the sam

meter,

he end of the

e along the e

entire rod to

Figure Q

3)

fin differ fro

al conductiv

ature of Tw =

= 100 mm,

00 mm, is ex

onvective he

nts to mode

e ambient ai

e insulated se

entire length

o the ambient

Q4

m that throu

vity of k = 6

100°C. For

as shown in

xposed to an

at transfer c

l the rod. If

ir conditions

ection,

of the rod,

t air.

[2

ugh a plane w

[3

60 W/m°C p

safety reason

Figure Q4.

ambient air,

coefficient o

f the tip of th

and the thic

[2

2 marks]

wall

3 marks]

protrudes

n, part of

The bare

which is

f h = 50

he rod is

ckness of

20 marks]

-6- (SME3033)

FORMULAE

( )

( )

( )

( )324

33

322

31

141

3241

141

3241

ξξξ

ξξ

ξξξ

ξξ

++−−=

−+=

+−−=

+−=

H

H

H

H

( ) 44332211 22qHlqHqHlqHv ee +++=ξ

( ) ( )[ ]43212 136136 qlqqlqlEIM ee

e

++−−+= ξξξξ

[ ]43213 226 qlqqlqlEIV eee

+−+=

________________________________________________________________________

[ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

−−=

4

3

2

1

qqqq

mlmllE

e

σ

[ ]2 2

3

2 2

12 6 12 66 4 6 212 6 12 6

6 2 6 4

e e

e e e e e

e ee

e e e e

l ll l l lEIk

l lll l l l

−⎡ ⎤⎢ ⎥−⎢ ⎥=⎢ ⎥− − −⎢ ⎥−⎣ ⎦

{ }2 2

2 12 2 12

Te e e e epl pl pl plf

⎡ ⎤= −⎢ ⎥⎣ ⎦

[ ] [ ]

2 2

2 2

2 2

2 2

trusse

l lm l lmlm m lm mAEk k

l l lm l lmlm m lm m

⎡ ⎤− −⎢ ⎥− −⎢ ⎥= =⎢ ⎥− −⎢ ⎥− −⎢ ⎥⎣ ⎦

-7- (SME3033)

[ ] [ ] [ ][ ]BDBAtk Tee

e = 634221

533211

qNqNqNvqNqNqNu

++=++=

[ ]JAe det21

=

ηξηξ −−=== 1321 NNN

[ ] [ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

122131132332

211332

123123

000000

det1

yxyxyxxxx

yyy

JB

32313

32313

yyyyxxxx

++=++=

ηξηξ

[ ] ⎥⎦

⎤⎢⎣

⎡=

2323

1313

yxyx

J

[ ] 13232313det yxyxJ −=

[ ]( )

1000101

121

2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

=ν

νν

νED

{ } ( ) ( ) ( ) ( )[ ]Tyxyxee TTTTltT2

21−=

[ ] ( )( )

( )( )

( )

000101

21121 ⎥

⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

−+=

ννν

νν

ννED

{ } [ ][ ]{ } [ ]Txyyxe qBD τσσσ ==

________________________________________________________________________

[ ] 1 11 1

eT

e

kkl

−⎡ ⎤= ⎢ ⎥−⎣ ⎦

[ ] 2 11 23

eT

hlht⎡ ⎤

≈ ⎢ ⎥⎣ ⎦

{ } 11

ehT lrt∞

∞

⎧ ⎫≈ ⎨ ⎬

⎩ ⎭

[ ] [ ]( ){ } { }e e e eT Tk h T r∞+ =

( )e avg sH h T T A∞= −

-8- (SME3033)