Section 16: Neutral Axis and
Parallel Axis Theorem
16-1
Geometry of deformation
•
•
We will consider the deformation of an ideal, isotropic prismatic beam
– the cross section is symmetric about y-axis
All parts of the beam that were originally aligned with the longitudinal axis
bend into circular arcs
– plane sections of the beam remain plane and perpendicular to the
beam’s
beam
s curved axis
Note: we will take these
directions for M0 to be
positive However
positive.
However, they are
in the opposite direction to
our convention (Beam 7),
and we must remember to
account for this at the end.
16-2
From: Hornsey
Neutral axis
16-3
From: Hornsey
•
6.3 BENDING DEFORMATION OF
STRAIGHT
MEMBER
A neutralAsurface
is where longitudinal
fibers of the
material will not undergo
g a change
g in length.
g
16-4
From: Wang
6.3 BENDING DEFORMATION OF
MEMBER
• Thus, A
weSTRAIGHT
make the following
assumptions:
1. Longitudinal axis x (within neutral surface)
does not experience any change in length
2. All cross sections of the beam remain plane
p
and perpendicular to longitudinal axis during
the deformation
3. Any deformation of the cross-section within its
own plane will be neglected
• In particular, the z axis, in plane of x-section and
about which the x-section rotates, is called the
neutral
t l axis
i
16-5
From: Wang
6 4 THE FLEXURE FORMULA
6.4
• By mathematical expression,
equilibrium equations of
moment and forces, we get
Equation 6-10
6 10
Equation 6-11
∫A y dA = 0
M=
σmax
c
∫A
y2 dA
• The integral represents the moment of inertia of xsectional area, computed about the neutral axis.
We symbolize its value as I.
I
16-6
From: Wang
6 4 THE FLEXURE FORMULA
6.4
• Normal stress at intermediate distance y can be
determined from
My
Equation 6-13 σ = −
I
• σ is -ve as it acts in the -ve direction (compression)
• Equations
E
ti
6
6-12
12 and
d6
6-13
13 are often
ft referred
f
d to
t as
the flexure formula.
16-7
From: Wang
*6
6.6
6 COMPOSITE BEAMS
• Beams constructed of two or more different
materials are called composite beams
• Engineers design beams in this manner to develop
a more efficient means for carrying applied loads
• Flexure formula cannot be applied directly to
determine normal stress in a composite beam
• Thus a method will be developed to “transform” a
beam’s x-section into one made of a single material,
th we can apply
then
l th
the fl
flexure fformula
l
16-8
From: Wang
16-9
From: Hornsey
Moments of Inertia
• Resistance to bending,
twisting, compression or
tension of an object is a
function of its shape
• Relationship of applied
force to distribution of
mass (shape) with
respect to an axis
axis.
16-10
Figure from: Browner et al, Skeletal Trauma 2nd Ed,
From: Le
Saunders, 1998.
Implant Shape
• Moment of Inertia:
further away material
is spread in an object,
greater the stiffness
• Stiffness and strength
are proportional to
radius4
16-11
From: Justice
16-12
From: Hornsey
Moment of Inertia of an Area by Integration
• S
Secondd moments or moments off inertia
i
i off
an area with respect to the x and y axes,
I x = ∫ y 2 dA
I y = ∫ x 2 dA
• Evaluation of the integrals is simplified by
choosing dΑ to be a thin strip parallel to
one of the coordinate axes.
• For a rectangular area,
h
I x = ∫ y dA = ∫ y 2bdy = 13 bh 3
2
0
• The formula for rectangular areas may also
be applied to strips parallel to the axes,
dI x = 13 y 3 dx
16-13
From: Rabiei
dI y = x 2 dA = x 2 y dx
Homework Problem 16.1
Determine the moment of
inertia of a triangle with respect
to its base.
16-14
From: Rabiei
Homework Problem 16.2
a) Determine the centroidal polar
moment of inertia of a circular
area by direct integration.
integration
b) Using the result of part a,
determine the moment of inertia
of a circular area with respect to a
diameter.
16-15
From: Rabiei
Parallel Axis Theorem
• Consider moment of inertia I of an area A
with respect to the axis AA’
I = ∫ y 2 dA
• The axis BB’ passes through the area centroid
and is called a centroidal axis.
I = ∫ y 2 dA = ∫ ( y ′ + d )2 dA
= ∫ y ′ 2 dA + 2d ∫ y ′dA + d 2 ∫ dA
I = I + Ad 2
16-16
From: Rabiei
parallel axis theorem
Parallel Axis Theorem
• Moment of inertia IT of a circular area with
respect to a tangent to the circle,
( )
I T = I + Ad 2 = 14 π r 4 + π r 2 r 2
= 54 π r 4
• Moment of inertia of a triangle with respect to a
centroidal
id l axis,
i
I AA′ = I BB′ + Ad 2
I BB′ = I AA′ − Ad
1 bh 3
= 36
16-17
From: Rabiei
2
1 bh 3
= 12
( )
2
1
1
− 2 bh 3 h
Moments of Inertia of Composite Areas
• The moment of inertia of a composite area A about a given axis is
obtained by adding the moments of inertia of the component areas
A1, A2, A3, ... , with respect to the same axis.
16-18
From: Rabiei
Example:
(Dimensions in mm)
y
200
10
z
120
y = 89.6 mm
60
125
o
Centroidal
C
t id l
Axis
1
y = ∫ y'⋅dA
A A
n
20
1
[(200 × 10 )(125 ) + (120 × 20 )(60 )]
y=
(200 × 10 + 120 × 20 )
1
394,000
[250,000 + 144,000] =
y=
= 89.55 mm
(4,400 )
4,400
−3
=
89
.
6
×
10
m
16-19
From: University of Auckland
y
Example:
(Dimensions in mm)
200
10
• What is Iz?
• What is maximum σx?
z
120
30.4
o
In = Iz + A y
89.6
2
200
20
2
I z ,3
10
3
20
30.4
35.4
bd3 (20 )(89.6 )3
Iz ,1 =
=
= 4.79 × 10 6 mm 4
3
89.6
1
3
bd3 (20 )(30.4 )3
I z ,2 =
=
= 0.19 × 10 6 mm 4
3
3
20
3
2
bd3
(
200 )(10 )
2
=
+ Ay =
(
)(
)
+ 200 × 10 35.4 = 3.28 × 10 6 mm 4
12
12
16-20
University of Auckland
y
Example:
(Dimensions in mm)
200
10
• What is Iz?
• What is maximum σx?
z
120
30.4
o
In = Iz + A y
89.6
2
200
20
2
1
10
3
20
30.4
35.4
89.6
Iz = Iz ,1 + Iz ,2 + Iz ,3
20
⇒ Iz = 8.26 × 10 6 mm 4 = 8.26 × 10 −6 m4
16-21
University of Auckland
Maximum Stress:
y
40.4
Mxz
x
NA
89.6
σx = −
⇒ σ x ,Max
16-22
Mxz
⋅ y'
Iz
Mxz
σ x ,Max = −
⋅ yMax
Iz
Mxz
−3
=−
⋅
−
89
.
6
×
10
8.26 × 10 −6
(
(
)
University of Auckland
)
((N/m2 or Pa))
Homework Problem 16.3
SOLUTION:
• Determine location of the centroid of
composite
it section
ti with
ith respectt to
t a
coordinate system with origin at the
centroid of the beam section.
The strength of a W14x38 rolled steel
beam is increased by attaching a plate
to its upper flange.
• Apply the parallel axis theorem to
determine moments of inertia of beam
section and plate with respect to
composite section centroidal axis.
Determine
D
t
i the
th momentt off inertia
i ti andd
radius of gyration with respect to an
axis which is parallel to the plate and
passes through the centroid of the
section.
16-23
From: Rabiei
Homework Problem 16
16.4
4
SOLUTION:
• Compute the moments of inertia of the
bounding rectangle and half-circle with
respect to the x axis.
• Th
The momentt off inertia
i ti off the
th shaded
h d d area is
i
obtained by subtracting the moment of
inertia of the half-circle from the moment
of inertia of the rectangle.
rectangle
Determine the moment of inertia
of the shaded area with respect to
the x axis.
16-24
From: Rabiei