# Introduction to Mechanics of Materials

Mechanics

Oct 29, 2013 (4 years and 6 months ago)

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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
e-mail: 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
Course Objectives
Importance
Plan
Review of Statics
Stress
Strain
Mohr’s Circle:
Stress Transformation
Strain Transformation
Design Considerations
Mechanical
Properties of
Materials
Bending
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

at a point on a section in a body)
• Strains are caused by stresses (Compare:
• 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)

Stress-strain 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 x-axis
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 1-D 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 cross-section 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: Cable-Supported 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

= =

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: Cable-Supported 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

, 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
Two-segment Rod
Rod has: Slender segment AB = 2 m; Thick segment BC = 1 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
Two-segment 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)

6

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 2-D (planar) situation to define it.
Consider two orthogonal (perpendicular) and infinitesimal line
segments OP = Δx and OQ = Δy (along x-axis and y-axis).
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

,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
(b) Average shear strain in the
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 (non-directional, 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 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 n-1
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

− −
× = × × = ×

Hence we will allow for a gap of 30 mm.

Measurement of Strain
Strain Measurement
Strain is measured using strain gauges.
Common are the resistance-type strain gages.
Property Used: Change of electrical resistance in material when
mechanically deformed
Modern Strain Gages: Metallic foil (e.g., using copper-nickel
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
(a)
Single
Element
Three-Element
Rosettes
(b)
Nickel-Plated
Copper
Ribbons
Welded
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

-
+
(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
= −
+ +

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.