16-slides - advanced stress

8/7/2019 16-Slides - Advanced Stress

1/24

Section 16: Neutral Axis andParallel Axis Theorem

16-1


2/24

We will consider the deformation of an ideal, isotropic prismaticbeam

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

Note: we will take these

directions for M0 to be. ,

in the opposite direction to

our convention (Beam 7),and we must remember to

account for this at the end.

16-2 From: Hornsey


3/24

Neutral axis

16-3 From: Hornsey


4/24

6.3 BENDING DEFORMATION OFA TRAI HT MEMBER A neutral surfaceis where longitudinal fibers of the

material will not undergo a change in length.

16-4 From: Wang


5/24

6.3 BENDING DEFORMATION OFA TRAI HT MEMBER Thus, we make the following assumptions:

1. L n i in l xi x wi hin n r l rf

does not experience any change in length2. All cross sectionsof the beam remain lane

and perpendicular to longitudinal axis during

the deformation3. Any deformationof the cross-sectionwithin its

own plane will be neglected

In particular, the zaxis, in plane of x-section andabout which the x-section rotates, is called the

16-5 From: Wang

neu ra ax s


6/24

. By mathematical expression,equilibrium equations of

moment and forces, we get

=-

Equation 6-11max

cM = A y2 dA

The integral represents the moment of inertiaof x-sectional area, computed about the neutral axis.

16-6 From: Wang

.


7/24

. Normal stress at intermediate distance y can bedetermined from

Equation 6-13 MyI

=

is -ve as it acts in the -ve direction (compression) qua ons - an - are o en re erre o as

the flexure formula.

16-7 From: Wang


8/24

* . Beams constructed of two or more differentmaterials are called composite beams

Engineers design beams in this manner to developa 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

beams x-section into one made of a single material,

t en we can app y t e exure ormu a

16-8 From: Wang


9/24

16-9 From: Hornsey


10/24

,

twisting, compression ortension of an object is afunction of its shape

Relationship of appliedforce to distribution ofmass (shape) with

.

16-10 From: Le

Figure from: Browner et al, Skeletal Trauma 2nd Ed,

Saunders, 1998.


11/24

further away materialis spread in an object,greater the stiffness

Stiffness and strengthare proportional toradius4

16-11 From: Justice


12/24

16-12 From: Hornsey


13/24

Moment of Inertia of an Area by Integration econ moments or moments o inertia o

an area with respect to the x and y axes,

== dAxIdAyIx22

Evaluation of the integrals is simplified bychoosing d to be a thin strip parallel to

.

For a rectangular area,h3

31

0

22 bhbdyydAyIx ===

The formula for rectangular areas may also

be applied to strips parallel to the axes,

dxyxdAxdIdxydI yx223

31 ===

16-13 From: Rabiei


14/24

Homework Problem 16.1

inertia of a triangle with respect

to its base.

16-14 From: Rabiei


15/24

Homework Problem 16.2

a) Determine the centroidal polar

moment of inertia of a circular

.

b) Using the result of part a,

determine the moment of inertia

16-15 From: Rabiei

diameter.


16/24

Parallel Axis Theorem

Consider moment of inertia Iof an area A

with respect to the axis AA

=

dAyI

2

The axis BBpasses through the area centroid

and is called a centroidal axis.

( )

++=

+==

dAddAyddAy

dAdydAyI

222

2d+= parallel axis theorem

16-16 From: Rabiei


17/24

Parallel Axis Theorem Moment of inertia ITof a circular area with

respect to a tangent to the circle,

445

4

r

rrrAdT

=

+=+=

Moment of inertia of a triangle with respect to acentroida axis,

2AdII BBAA +=

3361

3212

bh

hbhbhdAABB

===

16-17 From: Rabiei


18/24

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


19/24

yExample:

200

(Dimensions in mm)

10

o

en ro a

Axis

120

125

n =A dA'yA

1

y

y mm6.89=60

20

1= 6020120+

( )2012010200 +1

=000,394

= =

16-19 From: University of Auckland

( )400,4 400,4.

m106.89

3

=


20/24

Example: (Dimensions in mm)y

200

10

z o20

30.4

2

z What is maximum x?

1

89.6

200

zn y+=

20 20

30.4

10

2

3

35.4

89.61 3

bdI

3

1,z =( )( )

3

6.89203

= 46 mm1079.4 =

20 3I 2,z =

3.= 46 mm1019.0 =

23bd 310200

16-20 University of Auckland

3,z12

= .12

+= mm. =


21/24

Example: (Dimensions in mm)y

200

10

z o20

30.4

2

z What is maximum x?

1

89.6

200

zn y+=

20 20

30.4

10

35.42

3

89.61

20

3,z2,z1,zz ++=

46 46


z . .


22/24

Maximum Stress:

y

NAx

40.4 Mxz

89.6

'M

Izx =

MMax

zMax,x yI =

3xz 106.89M = N/m2 orPa


,1026.8


23/24

Homework Problem 16.3SOLUTION:

Determine location of the centroid of

compos e sec on w respec o a

coordinate system with origin at thecentroid of the beam section.

Apply the parallel axis theorem to

determine moments of inertia of beam

section and plate with respect to

beam is increased by attaching a plate

to its upper flange.

composite section centroidal axis.

e erm ne e momen o ner a an

radius of gyration with respect to an

axis which is parallel to the plate and

16-23 From: Rabiei

section.


24/24

.SOLUTION: Com ute the moments of inertia of the

bounding rectangle and half-circle with

respect to the x axis.

e momen o ner a o e s a e area s

obtained by subtracting the moment of

inertia of the half-circle from the moment

Determine the moment of inertia

of the shaded area with respect to

the x axis.

.

16-24 From: Rabiei

16-slides - advanced stress

Documents