MOD I1
Modelling and Simulation
Lesson 1
Week 43 2009
Lecturer:
Poul Vaggemose
Today’s Highlights
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Resistance in wire
Poisson’s ratio
Young’s modulus
Thermal expansion
Sensor circuit ¼, ½ and full bridge
Find the right Strain Gauge
Hooke’s law
Wheatstone bridge exercise
Resistance in wire
Area of wire = A = pi
Area
of wire = A = pi*r*r
r r
Radius of wire = r
Resistivity () = R*A / L
R A = 
R*A
LL
R =  L / A
A = pi*r*r
R =  L / pi*r*r
Thermal
Common Name
Conductivity
(W/cm K)
Stainless Steel
type 304
0.15
Density
(g/cm3)
7.9
Coeff. of linear
Electrical
Modulus of
expansion
Resistivity
elasticity
(Ԩ-1)
(μΩ cm)
(GPa)
17.3
72
195
Tensile strength
(MPa)
550
Approx. melting
point
(°C)
1425
Calculate resistance in wire
Example 1
Pi = 3,14
Radius r = 0,00005 m
Resistivity  = 72 μ ohm-cm = 0,72 μ ohm-m
L
Length
th L = 3
3,82
82 cm
R=?
R = * L / pi*r*r
R = 0,72 micro ohm-m * 3,82 cm / 3,14*0,00005m*0,00005m
R = 0,72 micro ohm-m*0.0382m / 3,14*0,00005m*0,00005m
R = 3,50
3 50 ohm
==========
Calculate resistance in wire
Example 2 (thicker wire)
Pi = 3,14
Radius r = 0,00006 m
Resistivity  = 72 μ ohm-cm = 0,72 μ ohm-m
L
Length
th L = 3
3,82
82 cm
R=?
R = * L / pi*r*r
R = 0,72 micro ohm-m * 3,82 cm / 3,14*0,00006m*0,00006m
R = 0,72 micro ohm-m*0.0382m / 3,14*0,00006m*0,00006m
R=2
2,43
43 ohm
h
==========
Calculate resistance in wire
Example 3 (thinner wire)
Pi = 3,14
Radius r = 0,00004 m
Resistivity  = 72 micro ohm-cm = 0,72 micro ohm-m
L
Length
th L = 3
3,82
82 cm
R=?
R = * L / pi*r*r
R = 0,72 micro ohm-m * 3,82 cm / 3,14*0,00004m*0,00004m
R = 0,72 micro ohm-m*0.0382m / 3,14*0,00004m*0,00004m
R=5
5,47
47 ohm
h
==========
Poisson’s ratio
If a bar is loaded with a tensile force then it is extends by
Delta l / l = ε (l) longitudinal strain; its thickness is also reduced
by Delta b / b = ε (t) transverse contraction. ε(t) is positive, ε (l)
is negative.
Hi t W
Hint:
Works
k lik
like a pulled
ll d rubber
bb b
band.
d
ε (l) = longitudinal strain
ε ((t)) = transverse contraction
| ε (t) / ε (l) | = √ (Poisson’s ration)
The Poisson’s ratio coefficient for steel is
√ = 0,26
Young’s modulus
With “linearly elastic” materials there is a linear rise in the σ/ε curve in
the region of elastic deformation
deformation. The slope in this section of the
diagram characterized the material’s rigidity. This is expressed as the
ratio between σ and ε is termed the modulus of elasticity or Young’s
modulus having the symbol E.
σ = nominal stress
ε = strain
E = elasticity
E = σ/ε
The Elasticity or Young’s modulus for steel is
E = 210 000 N/mm2
Thermal expansion
Every object alters its dimensions when its temperature changes.
Heating produces expansion
expansion, cooling produces contraction
‫ =ט‬thermal expansion
α = longitudinal expansion coefficient or linear thermal expansion
coefficient, states the relative change in length in μ m/m per K.
α = (l2 – l1 / l1 )* 1/ delta ‫ט‬
α = delta l / (l1 * delta ‫ = )ט‬1/K
The Thermal expansion for Invar steel (64Fe; 36 Ni) is
‫ = ט‬1,5 (at 100 degree Celsius)
‫ = ט‬9,4 (at 500 degree Celsius)
‫ = ט‬13,1 (at 800 degree Celsius)
Sensor circuit
The Wheatstone ¼ bridge:
Vs
Suitable for use under
environment of less
ambient Temperature
changes; no temperature
compensation
R1
A
R4
B
R2
Sensor circuit
The Wheatstone ½ bridge:
Bending stress
Temperature compensation;
thermal effect of lead wires
cancelled;
compressive/ tensile strain
cancelled.
Uniaxial stress
(uniform tension/compression)
Temperature compensation;
thermal effect of leadwires
cancelled.
Vs
R1
A
B
R2
Sensor circuit
The Wheatstone 1/1 bridge:
Temperature
compensation;
thermal effect
of lead wires
cancelled;
compressive/
p
tensile strain
cancelled.
Vs
A
B
SG series - Exercise
The Strain Gauge Exercise
Use the “Strain Gages and Accessories” book from HBM
to find the right strain gage series
series.
SG-Data:
Not a manufacture of transducer
No biaxial stress state
Full bridge SG
No lacking space
Homogeneous field of strain
Temperature from -70 degree to 200 degree
SG series = ?
Strain Gage
SG series = Y series
SG serie stock type - Exercise
The Strain Gauge Exercise
Use the “Strain Gages and Accessories” book from HBM
to find the right strain gage stock type
type.
SG-Data:
Steel SG
Nominal resistance = 350 ohm
Dimension 3 mm pr inch (Measuring grid “a”)
Max. Perm. Effective bridge excitation voltage = 7 volt
Solder terminals = LS7
SG/Y serie stock type = ?
Strain Gage
Y series = 1-LY11-3/350
Hooke’s Law - Exercise
Hooke’s law : material stress = strain * modulus of elasticity.
a
SG
F
h
F = m*g = Force on steel bar
M=F*a = Bending moment
S= b*h*h/6 =
σ = M / S = m*g*a / (b*h*h/6)
σ = ε*E = Material stress
a = 100 mm
b = 20 mm
h = 1 mm
m = 0,5 Kg
g = 9,8 m/s*s
E = 210 000 N/mm*mm
N/mm mm
σ =?
ε =?
b
σ = 147
ε = 0,0007
Sensor circuit
http://www.rdpelectronics.com/ex/hiw-sglc.htm
A strain gauge is a long length of conductor arranged in a zigzag pattern on a
membrane.
When it is stretched, its resistance increases. Strain gauges are mounted in
the same direction as the strain and often in fours to form a full 'Wheatstone
Bridge'.
The diagram above represents what might happen if a strip of metal were
fitted with four gauges.
An downward
A
d
db
bend
d stretches
t t h th
the gauges on th
the ttop and
d compresses th
those on
the bottom.
Wheatstone bridge - Exercise
The Wheatstone bridge
a3
SG2 SG3
ε+ F
h
SG1
SG4
F = m*g
M = F*a
S = b*h*h/6
σ = M / S = m*g*a
* * / (b*h*h/6)
ε=σ/E
Delta U / U = K/4* (-ε1+ ε2+ ε3 – ε4)
ε-
b
Wheatstone bridge - Exercise
σ = M / S = m*g*a / (b*h*h/6)
ε=σ/E
Delta U / U = K/4* (-ε1+ ε2+ ε3 – ε4)
a1 = 129 mm
a2
2 = 131 mm
a3 = 99 mm
a4 = 101 mm
K=2
b = 20 mm
h = 1 mm
m = 0,5 Kg
U = 7 volt
g = 9,8 m/s*s
E = 210 000 N/mm*mm
Delta U = ?
Delta U = 0,0112 volt