Section 16: Neutral Axis and Parallel Axis Theorem 16-1 Geometry - - PowerPoint PPT Presentation

Section 16: Neutral Axis and Parallel Axis Theorem

16-1

Geometry of deformation 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 curved axis beam s curved axis Note: we will take these directions for M0 to be positive However they are

• positive. However, they are

in the opposite direction to

• ur 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 MEMBER A 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 MEMBER A 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 plane

p and perpendicular to longitudinal axis during the deformation

• 3. Any deformation of the cross-section within its
• wn 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 i

16-5 From: Wang

neutral axis

6 4 THE FLEXURE FORMULA 6.4 THE FLEXURE FORMULA

• By mathematical expression,

equilibrium equations of moment and forces, we get ∫A y dA = 0 Equation 6-10 ∫A y dA 0 Equation 6 10 σ Equation 6-11 σmax c M = ∫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 FORMULA 6.4 THE FLEXURE FORMULA

• Normal stress at intermediate distance y can be

determined from Equation 6-13 My I 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 BEAMS 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 l th fl f l then we can apply the flexure formula

16-8 From: Wang

16-9 From: Hornsey

Moments of Inertia Moments 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 axis respect to an axis.

16-10 From: Le Figure from: Browner et al, Skeletal Trauma 2nd Ed, Saunders, 1998.

Implant Shape Implant 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 Integration

S d f i i f

• Second moments or moments of inertia of

an area with respect to the x and y axes,

∫ ∫

= = dA x I dA y I

y x 2 2

∫ ∫

y x

• Evaluation of the integrals is simplified by

choosing dΑ to be a thin strip parallel to

• ne of the coordinate axes.
• ne of the coordinate axes.
• For a rectangular area,

3 2 2 h 3 3 1 2 2

bh bdy y dA y I x = = =

∫ ∫

• The formula for rectangular areas may also

be applied to strips parallel to the axes, dx y x dA x dI dx y dI

y x 2 2 3 3 1

= = =

16-13 From: Rabiei

Homework Problem 16.1

Determine the moment of 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 area by direct integration. b) Using the result of part a, determine the moment of inertia

• f a circular area with respect to a

16-15 From: Rabiei

• f 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’

∫

= dA y I

2

• The axis BB’ passes through the area centroid

and is called a centroidal axis.

2 2

( )

∫ ∫ ∫ ∫ ∫

+ ′ + ′ = + ′ = = dA d dA y d dA y dA d y dA y I

2 2 2 2

2

2

Ad I I + = 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,

( ) 2

2 4 1 2

( )

4 4 5 2 2 4 4 1 2

r r r r Ad I IT π π π = + = + =

• Moment of inertia of a triangle with respect to a

id l i centroidal axis,

( )

2 1 1 3 1 2 2

Ad I I

B B A A

+ =

′ ′

( )

3 36 1 2 3 1 2 1 3 12 1 2

bh h bh bh Ad I I

A A B B

= − = − =

′ ′

16-17 From: Rabiei

Moments of Inertia of Composite Areas

• The moment of inertia of a composite area A about a given axis is
• btained by adding the moments of inertia of the component areas

A1, A2, A3, ... , with respect to the same axis.

16-18 From: Rabiei

y Example: 200 (Dimensions in mm) z

C t id l

10 z

• Centroidal

Axis

120 125 n

∫

⋅ =

A

dA ' y A 1 y

y mm 6 . 89 =

60 20

1 y =

( )( ) [

125 10 200×

( )( )]

60 20 120 × +

( )

20 120 10 200 y × + × =

( )[ ]

000 , 144 000 , 250 1 y + = 000 , 394 = mm 55 . 89 =

( )( ) [

125 10 200×

( )( )]

60 20 120 +

16-19 From: University of Auckland

( )[ ]

000 , 144 000 , 250 400 , 4 y + 400 , 4 mm 55 . 89 m 10 6 . 89

3 −

× =

Example: (Dimensions in mm) y

200 10

• What is I ?

z

• 20

30.4

2

A I I

What is Iz?

• What is maximum σx?

1 89.6 200

z n

y A I I + =

20 20 30.4 10

2 3

35.4 89.6

1

3 bd I

3 1 , z

=

( )( )

3 6 . 89 20

3

=

4 6 mm

10 79 . 4 × = bd3

( )( )

4 30 20

3

20

3 bd I

3 2 , z

=

( )( )

3 4 . 30 20

3

=

4 6 mm

10 19 . × =

2 3

y A bd I +

( )( ) ( )( )

2 3

4 35 10 200 10 200

4 6

10 28 3

16-20 University of Auckland 3 , z

y A 12 I + =

( )( ) ( )( )

2

4 . 35 10 200 12 × + =

4 6 mm

10 28 . 3 × =

Example: (Dimensions in mm) y

200 10

• What is I ?

z

• 20

30.4

2

A I I

What is Iz?

• What is maximum σx?

1 89.6 200

z n

y A I I + =

20 20 30.4 10 35.4

2 3

89.6

1

I I I I

20

3 , z 2 , z 1 , z z

I I I I + + =

4 6 mm

10 26 8 I × = ⇒

4 6 m

10 26 8

−

× =

16-21 University of Auckland z

mm 10 26 . 8 I × = ⇒ m 10 26 . 8 × =

Maximum Stress: y NA x

40.4

Mxz

89.6

' y Mxz ' y Iz

xz x

⋅ − = σ

xz

M

Max z xz Max , x

y I ⋅ − = σ

( ) ( )

3 6 xz Max x

10 6 . 89 M

−

× − ⋅ − = ⇒ σ

(N/m2 or Pa)

16-22 University of Auckland

( ) ( )

6 Max , x

10 26 . 8

−

×

( )

Homework Problem 16.3

SOLUTION:

• Determine location of the centroid of

it ti ith t t composite 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 4 Homework Problem 16.4

SOLUTION:

• Compute the moments of inertia of the

Compute 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
• btained by subtracting the moment of

inertia of the half-circle from the moment

• f inertia of the rectangle

Determine the moment of inertia

• f the shaded area with respect to

the x axis.

• f inertia of the rectangle.

16-24 From: Rabiei