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8/7/2019 16-Slides - Advanced Stress
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Section 16: Neutral Axis andParallel Axis Theorem
16-1
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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
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Neutral axis
16-3 From: Hornsey
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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
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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
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. 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
.
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. 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
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* . 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
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16-9 From: Hornsey
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,
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.
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further away materialis spread in an object,greater the stiffness
Stiffness and strengthare proportional toradius4
16-11 From: Justice
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16-12 From: Hornsey
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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
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Homework Problem 16.1
inertia of a triangle with respect
to its base.
16-14 From: Rabiei
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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.
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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
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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
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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
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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
=
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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. =
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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
16-21 University of Auckland
z . .
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Maximum Stress:
y
NAx
40.4 Mxz
89.6
'M
Izx =
MMax
zMax,x yI =
3xz 106.89M = N/m2 orPa
16-22 University of Auckland
,1026.8
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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.
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.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