Transcript Computer Aided Design

Normal Strain and Stress
Normal Strain and Stress, Stress strain
diagram, Hooke’s Law
1
Strain
• When a body is subjected to load, it
will deform and can be detected
through the changes in length and the
changes of angles between them.
• The deformation is measured through
experiment and it is called as strain.
• The important of strain: it will be
related to stress in the later chapter
2
Normal Strain
Normal strain is detected by the changes in length.
l 'l
 
l
l

l
Note

(epsilon)
l’: length after deformed
l: original length.
:
• dimensionless
• very small (normally is mm (=10-6 m))
• 480(10)-6 m/m = 480 mm/m = 480 “micros” = 0.0480
%
3
Example 1
When load P is applied, the RIGID lever arm rotates by 0.05o.
Calculate the normal strain of wire BD
Foundation: L/L
Knowledge required: geometrical equation
Rigid: no deformation on the lever
4
Geometry: The mathematics
Sine and Cosine Rule
L1  L 2 2  L32  2( L 2)( L3) cos(  )
5
Example 1
When force P is applied to the
rigid lever arm
LBD after deformed is DB’
Cosine rule can be applied
here
LBD'  LAB'  LAD  2( LAB' )( LAD ) cos(   0.05)
2
2
Strain:
 BD 
LDB '  LDB
LDB
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Example 1
When force P is applied to the
rigid lever arm
LBD after deformed is DB’
Cosine rule can be applied here
LBD'  LAB'  LAD  2( LAB' )( LAD ) cos(   0.05)
2
2
LBD'  300.3491mm
Strain:
 BD 
LDB '  LDB
LDB
 BD  0.00116mm/ mm
7
Example 2
The force applied to the handle of
the rigid lever the arm to rotate
clockwise through an angle of 3o
about pin A. Determine the average
normal strain developed in the
wire. Originally, the wire is
unstretched.
Discuss the approach?
8
Solution
LB’D = 0.6155 m
= 0.0258 m/m
9
Simple Tensile Test
•
Strength of a material can only be
determined by experiment
•
The test used by engineers is the
tension or compression test
•
This test is used primarily to
determine the relationship between
the average normal stress and
average normal strain in common
engineering materials, such as
metals, ceramics, polymers and
composites
10
Conventional Stress–Strain Diagram
• Nominal or engineering stress is obtained by
dividing the applied load P by the specimen’s original
cross-sectional area.

P
A0
• Nominal or engineering strain is obtained by
dividing the change in the specimen’s gauge length
by the specimen’s original gauge length.


L0
Conventional Stress–Strain Diagram
Conventional Stress–Strain Diagram
Elastic Behaviour
•
•
•
•
A straight line
Stress is proportional to strain, i.e., linearly elastic
Upper stress limit, or proportional limit; σpl
If load is removed upon reaching elastic limit, specimen will return to its original shape
Yielding
•
•
•
•
•
Material deforms permanently; yielding; plastic deformation
Yield stress, σY
Once yield point reached, specimen continues to elongate (strain) without any increase in
load
Note figure not drawn to scale, otherwise induced strains is 10-40 times larger than in
elastic limit
Material is referred to as being perfectly plastic
Conventional Stress–Strain Diagram
Strain Hardening.
•
•
•
Ultimate stress, σu
While specimen is elongating, its x-sectional area will decrease
Decrease in area is fairly uniform over entire gauge length
Necking
•
•
At ultimate stress, cross-sectional area begins to decrease in a
localized region of the specimen.
Specimen breaks at the fracture stress.
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Stress–Strain Behavior of
Ductile and Brittle Materials
Ductile Materials
• Material that can
subjected to large strains
before it ruptures is
called a ductile material.
Brittle Materials
• Materials that exhibit
little or no yielding
before failure are
referred to as brittle
materials.
Stress–Strain Behavior of Ductile and Brittle Materials
Yield Strength
• 0.02% strain for ductile material
Strain hardening
• When ductile material is loaded into the
plastic region and then unloaded, elastic
strain is recovered.
• The plastic strain remains and material is
subjected to a permanent set.
Hooke’s Law
• Hooke’s Law defines the linear relationship
between stress and strain within the elastic
region.
  E
σ = stress
E = modulus of elasticity or Young’s modulus
ε = strain
• E can be used only if a material has linear–
elastic behaviour.
E can be derived from stress
and strain graph.
What is it?
Strain Energy
•
•
When material is deformed by external
loading, it will store energy internally
throughout its volume.
Energy is related to the strains called
strain energy.
Modulus of Resilience
•
When stress reaches the proportional
limit, the strain-energy density is the
modulus of resilience, ur:
2

1 pl
1
ur   pl  pl 
2
2 E
Example
•
The stress–strain diagram for an aluminum alloy
that is used for making aircraft parts is shown.
When material is stressed to 600 MPa, find the
permanent strain that remains in the specimen
when load is released. Also, compute the
modulus of resilience both before and after the
load application.
•
Approach to the problem:
– Parallel to elastic line
– Both slope is equal
– Distance CD can be calculated based on the
slope
– Permanent strain: 0.023 – distance CD
Solution
• When the specimen is subjected to the load, the strain is
approximately 0.023 mm/mm.
• The slope of line OA is the modulus of elasticity,
• From triangle CBD,
450
E
 75.0 GPa
0.006
 
 
BD 600106
E

 75.0 109  CD  0.008mm/mm
CD
CD
Solution:
This strain represents the amount of recovered elastic
strain.
The permanent strain is
 OC  0.023 0.008  0.0150mm/mm (Ans)
Computing the modulus of resilience,
ur initial  1  pl  pl  1 4500.006  1.35 MJ/m3
(Ans)
2
2
ur  final  1  pl  pl  1 6000.008  2.40 MJ/m3 (Ans)
2
2
Note that the SI system of units is measured in joules, where 1 J = 1 Nm
Modulus of Toughness
• Modulus of toughness, ut,
represents the entire area
under the stress–strain
diagram.
• It indicates the strain-energy
density of the material just
before it fractures.
Example
The bar DA is rigid and is originally
held in the horizontal position
when the weight W is supported
from C. If the weight causes B to
be displaced downward 0.625mm,
determine the strain in wires DE
and BC. Also if the wires are made
of A-36 steel and have a crosssectional area of 1.25 mm2,
determine the weight W.
Discuss the approach????
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1) Calculate the displacement of D.
D
1.5

B
0.9
1.5
)
0.9
 D  1.0417m m
 D  0.625(
2) Based on displacement on D, calculate the strain and normal stress
D 
D
LD

1.0417
 1.157(10) 3 m m/ m m
900
* strain in mm/mm,
stress and E in MPa, F in
N and length in mm
 D  E  200(10)31.157(10)3  231.4MPa
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3) Based on normal stress at wire DE, calculate the T of wire D
TED
 D  231.4MPa 
A
TED   D A  289.3N
4) Calculate W, based on FBD of bar DA
M
A
0
 TDE (1.5)  W (0.9)  0
W  482.2 N
5) Calculate normal stress of wire CB and strain of wire CB
 BC 
TBC 482 .2

 385 .7 MPa
A
1.25
Strain can not be calculated as normal stress goes beyond yield stress (Sy = 250
MPa), elastic property is no more applied. Therefore it requires the stress and
strain curve to predict the strain
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Poisson’s Ratio
• n (nu), states that in the elastic range, the ratio of
these strains is a constant since the deformations
are proportional.
 lat
v
 long
Poisson’s ratio is dimensionless.
Typical values are 1/3 or 1/4.
• Negative sign since longitudinal elongation
(positive strain) causes lateral contraction
(negative strain), and vice versa.
Example
A bar made of A-36 steel has the dimensions shown. If an axial force of
P is applied to the bar, determine the change in its length and the
change in the dimensions of its cross section after applying the load.
The material behaves elastically.
Discuss the approach
Approach:
Property A-36: E , n
1.  = P/A
2. z =  / E
3. Lz = L * z
4. x = y = -n z
5. Lx = L * x
Ly = L * y
Solution
1) The normal stress in the bar :
 
 
P
80 103
z  
 16.0 106 Pa
A 0.10.05
4) The contraction strains in both the
x and y directions are
 x   y  vst  z  0.3280106   25.6 mm/m
2) From the table for A-36 steel,
Est = 200 GPa
 z 16.0106 
6


z 


80
10
mm/mm
9
Est 20010 
3) The axial elongation of the bar
is therefore
 z   z Lz  80106 1.5  120mm (Ans)
5) The changes in the dimensions
of the cross section are
 x   x Lx  25.6106 0.1  2.56mm (Ans)
 y   y Ly  25.6106 0.  05  1.28mm (Ans)