Section 16: Neutral Axis and Parallel Axis Theorem 16-1
Section 16: Neutral Axis and Parallel Axis Theorem
16-1
Geometry of deformationGeometry 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 curved axisbeam s curved axis
Note: we will take these directions for M0 to be positive However they arepositive. 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 A STRAIGHT MEMBERA STRAIGHT MEMBER• A neutral surface is where longitudinal fibers of the
material will not undergo a change in length.g g g
16-4 From: Wang
6.3 BENDING DEFORMATION OF A STRAIGHT MEMBERA STRAIGHT MEMBER• Thus, we make the following assumptions:
1. Longitudinal axis x (within neutral surface)1. Longitudinal axis x (within neutral surface) does not experience any change in length
2. All cross sections of the beam remain planepand 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
t l i16-5 From: Wang
neutral axis
6 4 THE FLEXURE FORMULA6.4 THE FLEXURE FORMULA• By mathematical expression,
equilibrium equations of moment and forces, we get
∫A y dA = 0Equation 6-10 ∫A y dA 0Equation 6 10
σEquation 6-11
σmax
cM = ∫A y2 dA
• The integral represents the moment of inertia of x-sectional area, computed about the neutral axis. We symbolize its value as I
16-6 From: Wang
We symbolize its value as I.
6 4 THE FLEXURE FORMULA6.4 THE FLEXURE FORMULA• Normal stress at intermediate distance y can be
determined from
Equation 6-13MyI
σ = −
• σ is -ve as it acts in the -ve direction (compression)E ti 6 12 d 6 13 ft f d t• Equations 6-12 and 6-13 are often referred to as the flexure formula.
16-7 From: Wang
*6 6 COMPOSITE BEAMS6.6 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 l th fl f lthen we can apply the flexure formula
16-8 From: Wang
16-9 From: Hornsey
Moments of InertiaMoments of Inertia
• Resistance to bending,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 axisrespect to an axis.
16-10 From: LeFigure from: Browner et al, Skeletal Trauma 2nd Ed,
Saunders, 1998.
Implant ShapeImplant Shape
• Moment of Inertia: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 IntegrationS d f i i f• Second moments or moments of inertia of an area with respect to the x and y axes,
∫∫ == dAxIdAyI yx22 ∫∫ yx
• Evaluation of the integrals is simplified by choosing dΑ to be a thin strip parallel to one of the coordinate axes.one of the coordinate axes.
• For a rectangular area,322
h3
31
0
22 bhbdyydAyI x === ∫∫
• The formula for rectangular areas may also be applied to strips parallel to the axes,
dxyxdAxdIdxydI yx223
31 ===
16-13 From: Rabiei
Homework Problem 16.1
Determine the moment ofDetermine 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 integrationarea by direct integration.
b) Using the result of part a, determine the moment of inertia of a circular area with respect to a
16-15 From: Rabiei
of a circular area with respect to a diameter.
Parallel Axis Theorem• Consider moment of inertia I of an area A
with respect to the axis AA’
∫= dAyI 2
• The axis BB’ passes through the area centroid and is called a centroidal axis.
22 ( )
∫∫∫
∫∫+′+′=
+′==
dAddAyddAy
dAdydAyI22
22
2
2AdII += parallel axis theorem
16-16 From: Rabiei
Parallel Axis Theorem• Moment of inertia IT of a circular area with
respect to a tangent to the circle,
( ) 22412 ( )4
45
224412
r
rrrAdIIT
π
ππ
=
+=+=
• Moment of inertia of a triangle with respect to a id l icentroidal axis,
( )211312
2AdII BBAA += ′′
( )3
361
231
213
1212
bh
hbhbhAdII AABB
=
−=−= ′′
16-17 From: Rabiei
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
yExample:200
(Dimensions in mm)
z C t id l
10
zo
Centroidal Axis
120125
n ∫ ⋅=A
dA'yA1y
y mm6.89=
60
20
1y = ( )( )[ 12510200 × ( )( )]6020120 ×+( )2012010200
y×+×
=
( )[ ]000,144000,2501y +=000,394
= mm55.89=
( )( )[ 12510200 × ( )( )]6020120+
16-19 From: University of Auckland
( )[ ]000,144000,250400,4
y +400,4
mm55.89m106.89 3−×=
Example: (Dimensions in mm)y200
10 • What is I ?
z o20
30.4
2AII
What is Iz?• What is maximum σx?
1
89.6
200
zn yAII +=
20 2030.4
10
23
35.4
89.61 3bdI
3
1,z = ( )( )3
6.8920 3
= 46 mm1079.4 ×=
bd3 ( )( )43020 3
20 3bdI
3
2,z = ( )( )3
4.3020 3
= 46 mm1019.0 ×=
23
yAbdI + ( )( ) ( )( )23
4351020010200 461028316-20 University of Auckland
3,z yA12
I += ( )( ) ( )( )24.351020012
×+= 46 mm1028.3 ×=
Example: (Dimensions in mm)y200
10 • What is I ?
z o20
30.4
2AII
What is Iz?• What is maximum σx?
1
89.6
200
zn yAII +=
20 2030.4
10
35.423
89.61IIII
203,z2,z1,zz IIII ++=
46 mm10268I ×=⇒ 46 m10268 −×=
16-21 University of Auckland
z mm1026.8I ×=⇒ m1026.8 ×=
Maximum Stress:
y
NA x40.4 Mxz
89.6
'yMxz 'yIz
xzx ⋅−=σ
xzMMax
z
xzMax,x y
I⋅−=σ
( ) ( )36
xzMaxx 106.89M −×−⋅−=⇒ σ (N/m2 or Pa)
16-22 University of Auckland
( ) ( )6Max,x 1026.8 −×( )
Homework Problem 16.3SOLUTION:
• Determine location of the centroid of it ti ith t tcomposite section with respect to a
coordinate system with origin at the centroid of the beam section.
The strength of a W14x38 rolled steel
• Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to The strength of a W14x38 rolled steel
beam is increased by attaching a plate to its upper flange.
D t i th t f i ti d
composite section centroidal axis.
Determine the moment of inertia and radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the
16-23 From: Rabiei
passes through the centroid of the section.
Homework Problem 16 4Homework Problem 16.4SOLUTION:
• Compute the moments of inertia of theCompute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis.
Th t f i ti f th h d d i• The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle
Determine the moment of inertia of the shaded area with respect to the x axis.
of inertia of the rectangle.
16-24 From: Rabiei