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