MECH 260,
Section 102
Introduction to
Mechanics of
Materials
Clarence W. de Silva,
Ph.D., P.Eng.
Professor of Mechanical Engineering
The University of British Columbia
email: desilva@mech.ubc.ca
http:// www.mech.ubc.ca/~ial
C.W. de Silva
Presentation Part 4
Plan
Subject Definition
External Forces/
Moments
Reactions, Internal
Forces/Moments
• Modeling
• Analysis
• Computer Simulation
• Design
• Testing/Diagnosis
• Operation
Stresses
(Normal, Shear)
Strains
(Normal, Shear)
Deflections
Deformations
(Rectilinear, Angular)
Constitutive
(Physical)
Relations
Engineering
Mechanics of
Materials
Statics
MECH 260 Road Map
Course Objectives
Importance
Plan
Review of Statics
Stress
Strain
Mohr’s Circle:
Stress Transformation
Strain Transformation
Design Considerations
Mechanical
Properties of
Materials
Bending
Axial Loading
Torsion
Examples
Applications
Beam Bending:
Shear Stress
Deflection
Statically Indeterminate Beams
Examples
Examples
Design Considerations
Applications
Revision
Plan
Strain:
•
Meaning of Strain
•
Normal Strain
•
Shear Strain
•
Sign Convention
•
Average Strain
•
Thermal Strain
•
Measurement of Strain
Normal Strain
Strain
•
Strain
“Intensity” of deformation at a point on a
section in a body (Compare: Stress
intensity of load
at a point on a section in a body)
• Strains are caused by stresses (Compare:
Deformations are caused by loads)
• Deformations: Change in length of a line segment;
Change in angle between two “perpendicular” line
segments or sections (i.e., change in direction/
orientation of the line segments) at a point (a corner)
•
Stressstrain relationship
“Physical” relationship
governed by material properties of body: “Constitutive
Relation”
•
Note: A rigid body does not deform
does not exhibit
strains (It may experience stresses). Deflections of a rigid
body are “rigid body motions,” not deformations.
Normal Strain
Deformation per unit length of a line segment
Sign Convention: Elongation is positive—called tensile
strain; Contraction is negative—compressive strain.
Note: Sign convention of normal stress (tensile stress +ve;
compressive stress –ve)
Symbol: ε (Greek lowercase epsilon)
Units: m/m or 1 unit of strain or 1 ε (dimensionless).
1 microstrain = 1 μm/m = 1
×
10
6
m/m or
% strain = 100
×
strain
6
1 1 10
−
µε = × ε
e.g., 1 % strain = 0.01 m/m = 0.01 ε = 10,000 με
Local Normal Strain
Average Normal Strain
0
'
lim
x
x
x x
x
∆
∆ ∆
ε
∆
→
−
=
Strains at a given location in a body: Defined by an infinitesimal
(extremely small) element at the location, in the proper directions,
as size
0
Consider elemental line segment OP of length Δx along xaxis
Under stresses, it deforms to O’P’. New length = Δx’ along x
Note: In uniaxial case both OP and O’P’ will fall along same axis
What is shown in the figure is a general case.
The normal strain at O in the x direction is given by:
Local Normal Strain
Average Normal Strain (Cont’d)
avg
L
δ
ε =
Typically, normal strain varies along the body.
Then, an average normal strain may be used to represent the
state of normal strain of the body.
Consider 1D member of length L. Under an axial load it
stretches by δ. Then,
average normal strain in the member:
Note: Average normal strain is defined with respect to a line segment
of “finite” length (not infinitesimal) in an object.
O
O'
P'
P
x
Δx
Δx'
0
Example: Normal Strain
in a Wire Loop
A wooden post of circular crosssection has radius r = 0.7 m.
Reinforcing steel wire is wrapped around it.
Due to humidity, radius of the post increases by Δr = 0.005 m.
Determine: Strain (normal) in the wire
Assume: Initially wire was not under strain.
r
Wire Loops
Wooden Post
(Tree Trunk)
Example: Normal Strain
in a Wire Loop (Cont’d)
Consider only one loop of the wire.
Length of wire loop
L
= 2
πr
Due to swelling of
post
by
r∆
, the new length of the wire loop = 2
π
(
r
+
Δr
)
Change in wire length
2 ( ) 2 2L r r r r∆ π ∆ π π∆= + − =
Normal strain in the wire
2
2
L r r
L r r
∆ π∆ ∆
ε
π
= = =
Not
e
:
We get the same answer even if we consider multiple loops of wire
(check).
Substitute the given data:
0.005
0.007 ε 0.7%
0.7
ε = = =
Example: Variable Normal Strain
Along a Rod
The normal strain along a rod varies according to
x = axial location of rod measured from one end
k and p are +ve constants (depend on material
properties)
Initial unstrained (free) length of rod = L
Determine:
Overall extension (of one end with respect to the other)
of rod due to strain distribution.
Average normal strain in the rod
p
kxε =
Example: Variable Normal Strain
Along a Rod (Cont’d)
x x+δx
x
L
L0
Consider a small segment of length
δx
at the location
x
along the rod.
Extension of this element =
p
x kx xε δ δ× = ×
T
otal extension
L
x
in rod =
summation of
these elemental extension
s
I
n the limit,
this
is given by th
e integral
0
L
p
L
x kx dx=
∫
By integrating, we get
1 1
0
( 1) ( 1)
L
p p
L
k k
x x L
p p
+ +
= =
+ +
Average normal strain =
( 1)
p
L
x k
L
L p
ε = =
+
Example: Cable

Supported
Overhung Platform
Rigid platform of length L = 10 m; Hinged at one end;
Held horizontally using a vertical cable of length b = 5 m (attached to
platform at distance a = 6 m from hinged end.
Due to load on platform, free end moves vertically through
Determine: Cable extension; Average normal strain due to platform
movement.
Assume: Cable was unstrained in the beginning
Example: CableSupported Overhung Platform (Cont’d)
Exact Method:
P
l
atform end C moves to C’; C
able end A moves to A’ due to the platform movement.
A
ngle of rotation
θ
of platform:
0.03
sin 0.003
10
L
y
L
θ= = =
1
sin 0.003 0.003 radθ
−
= =
0.003
AA'2 sin 2 6 sin 0.01801 m
2 2
c a
θ
= = × = × × =
Note
:
In triangle OAA’
angl
es at A and A’
=
(
π

θ
)/2.
A
ngle BAA’ =
2 2 2
π π θ θ
π
−
+ =−
Apply the cosine rule to triangle A’BA:
2 2 2 2 2
2 cos(/2) 2 cos(/2)d b c bc b c bcπ θ θ= + − − = + +
New length of cable =
A’B =
d
2 2 2
5 (0.01801) 2 5 0.01801 cos(0.003/2)d = + − × × ×
d
= 5.01801 m
O
A
C
B
θ
C'
A'
y
L
2
π θ−
2
π
OA = OA ' = a
OC = OC ' = L
AB = b
AA' = c
A'B = d
Example: CableSupported Overhung Platform
(Cont’d)
Cable extension =
d
–
b
= 0.01801 m
Average normal strain
in the cable:
0.01801
0.003602 = 0.3602%
5
d b
b
ε ε
−
= = =
Approximate
Method:
A
ngle of rotation
θ
is small
Angular movement of
end A to its new position A’ which is
equal to
aθ
, is also (approximately) able ex
tension
.
We have
6 0.003 0.018 maθ =× =
This
is
extremely close to the previous answer.
Example: Normal Axial Strain in
Twosegment Rod
Rod has: Slender segment AB = 2 m; Thick segment BC = 1 m
Under axial loading, segment AB attains normal strain
Total elongation of AC due to loading = 0.005 m
Determine: Elongation of AB; Elongation of BC; Strain in BC;
Average strain in the overall rod AC.
0.002
AB
ε = ε
A
B C
P
P
2 m
1 m
Example: Normal Axial Strain in
Twosegment Rod (Cont’d)
Length of AB
2.0 m
AB
L =
E
longation of AB
0.002 2.0 = 0.004 m
AB AB AB
Lδ ε= × = ×
Given:
Total elongation of AC
= 0.005 m
AC
δ
E
longation of BC
0.005  0.004 = 0.001 m
BC
δ =
Length of BC
1.0 m
BC
L =
S
train in BC
0.001
0.001
1.0
BC
BC
BC
L
δ
ε = = = ε
Average strain in AC
0.005
0.0017
3.0
AC
AC
AC
L
δ
ε = = = ε
Shear Strain
Shear Strain
Angle change in a corner of angle π/2.
Sign Convention: Angle reduction is positive (Angle
increase is negative).
Symbol: γ (Greek lowercase gamma)
Units: 1 rad
6
1 rad 1 10 rad
−
µ =×
Local Shear Strain
Average Shear Strain
Shear strain concerns change in angle between line segments (or
change in direction of the line segments)
We must consider at least a 2D (planar) situation to define it.
Consider two orthogonal (perpendicular) and infinitesimal line
segments OP = Δx and OQ = Δy (along xaxis and yaxis).
Under stresses, OP
O’P’ and OQ
O’Q’.
Note: A straight line segment may not deform into a straight line
segment. For an infinitesimal line segment, line joining the two
ends ≈ deformed line segment.
Angle between O’P’ and O’Q’ = θ’
(Local) shear strain at corner O is given by change in angle (in
radians) in the limit:
,0
lim'
2
xy
x y∆ ∆
π
γ θ
→
= −
Local Shear Strain
Average Shear Strain (Cont’d)
γ
avg
L
0
x
y
δ
Q
P
x+Δx
O
O'
Q'
θ'
P'
x
y
y+Δy
Local Shear Strain
Average Shear Strain (Cont’d)
Typically, shear strain varies from point to point in the body
Average shear strain represents state of shear strain in body.
Consider a rectangular body (or part of it) of interest with:
Corner at O and sides falling along x and y axes
Due to shear stresses, suppose one side slides through distance δ
wrt opposite side in x direction.
Length of y direction side = L
Average shear strain in the member:
avg
L
δ
γ =
Note: Average shear strain is defined wrt two perpendicular sides
intersecting at a corner of a rectangular segment of finite
dimensions in an object.
Averaging is over the entire segment under consideration.
Example: Average Shear Strain in a Plastic Plate
Uniform, rectangular plastic plate ABCD; height = 1.6 m; width = 0.8 m
Rigidly fixed at bottom edge AD; Anchored horizontally to a steel post at
corner C using a steel bolt and nut.
Pitch of bolt and nut = 2 mm.
In the beginning, plastic plate is unstrained and held vertically.
Then nut is tightened by 2 turns.
Determine:
(a) Average shear strain in the
plate wrt AB and AD.
(b) Average shear strain in the
plate wrt AD and DC.
A D
C
B
0.8 m
Steel
Post
Steel Bolt
and Nut
Pitch = 2 mm
Plastic
Plate
1.6 m
Example: Average Shear Strain in a Plastic Plate (Cont’d)
Bolt moves horizontally through 2 mm for each full turn of nut
Total horizontal movement of side BC of plate = 2×0.002 m = 0.004 m
Angular shift
avg
γ
of side AB:
0.004
sin 0.0025
1.6
avg
γ = =
(a)
By definition,
avg
γ
=
average shear strain of plate wrt sides AB and AD
(because, it is the angle by which the corner angle BAD has reduced
(+ve according to sign convention)
1
sin 0.0025 0.0025
avg
γ
−
= =
(b) Since the angle ADC has
“increased” by according
sign convention, average shear
strain of plate wrt sides AD and
DC = 0.0025
avg
γ
γ
avg
1.6 m
A D
C’
C
B
B’
0.8 m
Thermal Strain
Thermal Strain
Deformations in a body due to thermal expansions and
contractions caused by temperature changes are represented by
thermal strains.
Coefficient of Thermal Expansion (α): Expansion in a line
segment of unit length due to a temperature increase of 1 degree
in a body.
By definition of normal strain (elongation per unit length), the
thermal strain due to a temperature increase of ΔT degrees is
given by
Note: For homogeneous (uniform, properties do not change
from point to point) and isotropic (nondirectional, properties
do not changed according to the direction) material, α is the
same in every location and in every direction of the body.
Then “shear” thermal strains are not present in the body.
T
Tε α∆=
Example: Thermal Strain of Elevated
Guideway of a Transit System
Guideway: Identical multiple spans of length 40 m placed on support piers.
Expansion slots (in anchoring between guideway ends and support piers)
accommodate change in guideway length (longitudinal strain) due to
temperature changes and load variations.
Temperature Extremes: 20˚C and +40˚C
Coefficient of thermal expansion of a guideway span:
Estimate: Gap between two adjacent guideway span ends (length of an
expansion slot) to accommodate guideway longitudinal thermal strain.
6 o
11.7 10/Cα
−
= ×
Example: Thermal Strain of Elevated
Guideway of a Transit System (Cont’d)
Guideway Span n Guideway Span n+1
Support
Pier n1
Support
Pier n
Support
Pier n+1
40 m
40 m
Maximum temperature variation
ΔT
= 40
–
(

20) = 60˚C
Corresponding thermal strain
6 4
.11.7 10 60 7.02 10 m/m
T
Tε α∆
− −
= = × × = ×
Guideway span length
L
= 40 m
Maximum elongation of a guideway span =
4 3
7.02 10 40 28.08 10 m
T
Lε
− −
× = × × = ×
Hence we will allow for a gap of 30 mm.
Measurement of Strain
Strain Measurement
Strain is measured using strain gauges.
Common are the resistancetype strain gages.
Property Used: Change of electrical resistance in material when
mechanically deformed
Modern Strain Gages: Metallic foil (e.g., using coppernickel
alloy constantan); Semiconductor elements (e.g., silicon with
trace impurity boron).
Relationship for a strain gage element:
δ
ε
R
R
S
s
=
R = strain gauge resistance
δR = change in strain gauge resistance due to strain ε
S
s
= gage factor or sensitivity of the strain gage element
(~ 2 to 6 for most metallic strain gage elements
~ 40 to 200 for semiconductor strain gages)
δR determines strain.
Measured using an electrical circuit (typically dc bridge)
Strain Gauges
Direction of
Sensitivity
Foil
Gri
d
Backing
Film
Solder
Tabs
(For
Leads)
(a)
Single
Element
ThreeElement
Rosettes
(b)
NickelPlated
Copper
Ribbons
Welded
Gold Leads
Doped Silicon
Crystal
(P or N Type)
Phenolic
Glass
Backing
Plate
(a) Metal Foil
Strain Gauge
Semiconductor
Strain Gauge
(b) Strain Gauge Configurations
Rosette Configurations
To Measure: Strains in more
than one direction; principal
strains (which exist in plane
where shear strains are
absent); and shear strain.
They have more than one
direction of sensitivity
Resistance (Wheatstone) Bridge Circuit
v
ref
(Constant Voltage)

+
R
1
A
R
2
R
3
R
4
B
R
L
v
o

+
Load
(High)
Small i
1 3
1 4 2 3
1 2 3 4 1 2 3 4
( )
( ) ( ) ( )( )
ref ref
o ref
Rv R v
R R R R
v v
R R R R R R R R
−
= − =
+ + + +
R
R
R
R
1
2
3
4
=
When the bridge is balanced the output voltage v
o
will be zero.
True for
any R
L
One or more of the resistance elements (R
i
) in the bridge may represent strain gauges.
31
1 2 3 4
o
ref
v RR
v R R R R
= −
+ +
Bridge Equation Under High Load:
Condition for a Balanced Bridge:
Direct Measurement of Bridge Output Voltage
Strain gauge resistance will change due to change in strain. If all four resistors
change we have:
(
)
(
)
(
)
(
)
δ
δ δ δ δ
v
v
R R R R
R R
R R R R
R R
o
ref
=
−
+
−
−
+
2 1 1 2
1 2
2
4 3 3 4
3 4
2
k = bridge constant = 1 in this case
Note: To compensate for temperature changes, temperature coefficients of the
adjacent pairs of resistors in the bridge should be the same
Reference:
De Silva, C.W., Sensors and Actuators—Control System Instrumentation, Taylor &
Francis, CRC Press, Boca Raton, FL, 2007.
4
o
ref
v
R
k
v R
δ
δ
=
Suppose that the initial resistances are all = R and there is only one strain gauge.
The strain gauge element changes its resistance by δR
Then:
Measure the output voltage resulting from the imbalance of resistance elements.
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