Transcript Structural Engineering - Indian Institute of Technology

17-07-2015
ME101: Engineering
Mechanics (3-1-0-8)
Jan-May 2014
Division I/III (Room No.: L1)
Instructor
Dr. Arunasis Chakraborty
Room M-106
Department of Civil Engineering, IITG
[email protected]
Phone: 258 2430
Web: www.iitg.ernet.in/arunasis
shilloi.iitg.ernet.in/~arunasis
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Syllabus for Tutorial 5:
 Potential energy
 Equilibrium & Stability
 Center of Gravity
 Moment of Inertia
 First and second moment of area
 Radius of gyration
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Potential Energy and Stability:
 Till now equilibrium configurations of mechanical
systems composed of rigid members was considered
for analysis using method of virtual work.
 Extending the method of virtual work to account for
mechanical systems which include elastic elements in
the form of springs (non-rigid elements).
 Introducing the concept of Potential Energy, which will
be used for determining the stability of equilibrium.
 Work done on an elastic member is stored in the
member in the form of Elastic Potential Energy Ve.
 This energy is potentially available to do work on
some other body during the relief of its
compression or extension.
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Elastic Potential Energy (Ve):
Consider a linear and elastic spring
compressed by a force F  F = kx
k = spring constant or stiffness of the spring
Work done on the spring by F during a
movement dx: dU = F dx
Elastic potential energy of the spring for
compression x = total work done on the spring
 Potential Energy of the spring = area of the
triangle in the diagram of F versus x from 0 to x
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Elastic Potential Energy (Ve):
During increase in compression from x1 to x2:
Work done on the springs = change in Ve
 Area of trapezoid from x1 to x2
During a virtual displacement dx of the spring:
virtual work done on the spring = virtual change
in elastic potential energy
During the decrease in compression of the spring as it is relaxed from x=x2 to x=x1,
the change (final minus initial) in the potential energy of the spring is negative
 If dx is negative, dVe is negative
When the spring is being stretched, the force acts in the direction of
the displacement  Positive Work on the spring  Increase in the
Potential Energy
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Elastic Potential Energy (Ve):
Work done by the linear spring on the body to which the spring
is attached (during displacement of the spring) is the negative
of the change in the elastic potential energy of the spring (due
to equilibrium).
Torsional Spring: resists the rotation
K = Torsional Stiffness (torque per radian of twist)
θ = angle of twist in radians
Resisting torque, M = Kθ
The Potential Energy:
This is similar to the expression for the linear extension spring
Units of Elastic Potential Energy  Joules (J)
(same as those of Work)
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Gravitational Potential Energy (Vg):
Treatment of Gravitational force till now:
For an upward displacement dh of the body,
work done by the weight (W=mg) is:dU = - mgdh
For downward displacement (with h measured
positive downward): dU = mgdh
Alternatively, work done by gravity can be expressed
in terms of a change in potential energy of the body.
The Gravitational Potential Energy of a body is defined as the
work done on the body by a force equal and opposite to the
weight in bringing the body to the position under consideration
from some arbitrary datum plane where the potential energy is
defined to be zero. Vg is negative of the work done by the
weight.
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Gravitational Potential Energy (Vg):
Note: Vg =0 at h=0
 At height h above the datum
plane, Vg of the body = mgh
 If the body is at distance h
below the datum plane, Vg of the
body = -mgh
Virtual change in the gravitational potential energy: dVg = + mgdh
dh is the upward virtual displacement of the mass centre of the body
If mass centre has downward virtual displacement  dVg = - mgdh
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Gravitational Potential Energy (Vg):
Datum Plane for zero potential energy is
arbitrary
• because only the change in potential
energy matters
• the change remains the same irrespective
of location of datum plane
Vg is also independent of the path followed
in arriving at a particular level h.
• The body of mass m has same potential
energy change in going from datum 1 to
datum 2 (3 possible paths) because dh
remains the same for all three paths.
Units of Vg are the same as those for the work and
elastic potential energy: Joules (J)
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Energy Equation:
Work done by the linear spring on the body to which the spring is
attached (during displacement of the spring) is the negative of the
change in the elastic potential energy of the spring.
Work done by the gravitational force or weight mg is the negative of
the change in gravitational potential energy
Virtual Work eqn to a system with springs and with changes in the
vertical position of its members  replace the work of the springs and
the work of the weights by negative of the respective potential energy
changes
Total Virtual Work dU = work done by all active forces (dU’) other than
spring forces and weight forces + the work done by the spring forces and
weight forces, i.e., -(dVe + dVg)
V = Ve + Vg  Total Potential Energy of the system
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Active Force Diagrams:
Use of two equations
dU=0
Restating the Principle of Virtual Work for a mechanical system with
elastic members and members that undergo changes in position:
Virtual work done by all external active forces (other than the gravitational
and spring forces accounted for in the potential energy terms) on a
mechanical system in equilibrium = the corresponding change in the total
elastic and gravitational potential energy of the system for any and all
virtual displacements consistent with the constraints
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Stability of Equilibrium:
Consider a system where movement is
accompanied by changes in gravitational
and elastic potential energy:
If work done by all active forces other than
spring forces and weight is zero
 dU’ = 0 (No work is done on
the system by the non-potential forces)
Equilibrium configuration of a mechanical system
is one for which the total potential energy V of the
system has a stationary value
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Stability of Equilibrium:
For a SDOF system (where potential
energy and its derivatives are
continuous functions of a single
variable, x that describes the
configuration), it is equivalent to state
that:
 For systems with multiple DOF, partial derivative of V wrt
each coordinate must be zero for equilibrium.
A mechanical system is in equilibrium
when the derivative of its total potential
energy is zero
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Stability of Equilibrium (SDOF):
Three Conditions under which this eqn applies.
when total potential energy is:
 A Minimum (Stable Equilibrium)
 A Maximum (Unstable Equilibrium)
 A Constant (Neutral Equilibrium)
Total Potential Energy:
Minimum
Maximum
Constant
• A small displacement away from the STABLE position results in an increase in
potential energy and a tendency to return to the position of lower energy.
• A small displacement away from the UNSTABLE position results in a decrease in
potential energy and a tendency to move farther away from the equilibrium
position to a position of still lower energy.
• For a NEUTRAL position, a small displacement one way or the other
results in no change in potential energy and no tendency to move either
way.
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Stability of Equilibrium (SDOF):
Conditions for equilibrium and stability of a system with a SDOF x:
Equilibrium: dV = 0
Stable Equilibrium:
Unstable Equilibrium:
Neutral Equilibrium:
dx
dV
d 2V
= 0, 2  0
dx
dx
dV
d 2V
= 0, 2  0
dx
dx
2
dV d V d 3V
=
= 3 = ... = 0
dx dx2
dx
If d2V/dx2 = 0 at the equilibrium position, examine the sign of a
higher derivative to ascertain the type of equilibrium.
• If the order of the lowest remaining non-zero derivative is even,
the equilibrium will be stable or unstable according to whether
the sign of this derivative is positive or negative.
• If the order of the derivative is odd, the equilibrium is classified
as unstable.
Stability criteria for MDOF systems require more advanced
treatment.
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Stability of Equilibrium (SDOF):
V:
Minimum
Maximum
Constant
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Example:
A 10 kg cylinder is suspended by
the spring. Plot the potential
energy of the system and show
that it is minimum at the
equilibrium position.
Solution: No external active forces. Choosing datum
plane for zero potential energy at the position where the
spring is un-extended.
For any position x: Elastic potential energy: Ve = ½ kx2
Gravitational Potential Energy: Vg = - mgx (-ve of work
done) Total Potential Energy: V = Ve + Vg = ½ kx2 - mgx
Equilibrium occurs where dV/dx = 0
kx – mg = 0  x = mg/k
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Solution:
d2V/dx2 = + k  Stable Equilibrium
since positive
Substituting values of m and k
V = ½ (2000)x2 – 10(9.81)x
x = 10(9.81)/2000 = 0.049 m = 49 mm
Plot V vs x graph for various values of
x
Minimum value of V occurs at x = 49
mm where dV/dx = 0 and d2V/dx2 is
positive
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Example:
Both ends of the uniform bar of mass m
slide freely along the guides. Examine the
stability conditions for the position of
equilibrium. The spring of stiffness k is undeformed when x=0
Solution: The system consists of the
spring and the bar There are no external
active forces. The Figure shows active force
diagram. Choosing the x-axis as the datum
plane for zero gravitational potential
energy.
In the displaced position x:
Ve = ½ kx2 = ½ k (b2sin2θ)
Vg = mg (b/2)cosθ (-ve of work done)
V = Ve + Vg = ½ k (b2sin2θ) + mg (b/2)cosθ
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Solution:
Equilibrium occurs for dV/dθ = 0
kb2sinθcosθ - ½ mgbsinθ = 0
(kb2cosθ - ½ mgb)sinθ = 0  Two
Solutions
Two equilibrium positions at:
sinθ = 0 (and θ = 0) &
cosθ = mg/(2kb)
Determine the stability for each of the two
equilibrium position by examining the sign of
d2V/dθ2
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Solution:
d2V/dθ2 = kb2(cos2θ – sin2θ) – ½ mgb cosθ
= kb2(2cos2θ – 1) – ½ mgb cosθ
Solution I: sinθ = 0 and θ = 0
d2V/dθ2 = kb2(1 – mg/2kb)
If mg/2kb < 1, i.e., mg/2b < k  d2V/dθ2 is positive (Stable)
If mg/2kb > 1, i.e., mg/2b > k  d2V/dθ2 is negative (Unstable)
 If the spring is sufficiently stiff, the bar will return to its original
vertical position, even though there is no force in the spring at that
position.
Solution II: cosθ = mg/(2kb) and θ = cos-1(mg/2kb)
d2V/dθ2 = kb2 [(mg/2kb)2 – 1]
 this will be always be negative (Unstable) because cosθ, i.e.,
mg/(2kb), must be less than unity
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Quick Review:
Equilibrium configuration of a mechanical
system is one for which the total potential
energy V of the system has a stationary
value
Equilibrium: dV = 0
Stable Equilibrium:
Unstable Equilibrium:
Neutral Equilibrium:
dx
dV
d 2V
= 0, 2  0
dx
dx
dV
d 2V
= 0, 2  0
dx
dx
2
dV d V d 3V
=
= 3 = ... = 0
dx dx2
dx
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Example:
Determine the force P to maintain
equilibrium
2
Ve =
1 

 
K  2b sin    = 2 Kb 2 sin 2
2 
2
 2 


Vg = 2m g - b cos 
2




dU = Pd  4b sin  = 2 Pb cos d
2
2

dU = dVe + dVg

 


 2 Pb cos d = d  2kb2 sin 2  + d  - 2m gbcos 
2
2
2


 1

P = kb sin + m g t an
2 2
2
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Example:
A homogenous block of mass m
rests on top surface of the
cylinder. Show that this is a
condition of unstable equilibrium
if h > 2R
Solution: Choosing the base of the
cylinder as the datum plane for zero
gravitational potential energy.
V = Ve + Vg = 0 + mgy (-ve of work done)
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Solution:
At equilibrium:
θ = 0 is the equilibrium position that satisfies this equation
Determine the stability of the system at θ = 0 by examining the sign of d2V/dθ2
if h/2 > R, i.e., h > 2R because then d2V/dθ2 will be negative.
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Example:
Determine the maximum value of
length for which equilibrium at vertical
position is stable.
Solution:
1
l
K 2 + m g cos 
2
2
dV
l
= K - m g sin  
d
2
d 2V
l
=
K
m
g
cos 
2
d
2
d 2V
0 @  =0
d 2
l
K - mg  0
2
Note: Sol. for stable
2K
l
configuration
mg
V = Ve + Vg =
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Center of Mass & Centroid:
Concentrated Forces: If dimension of the contact area is
negligible compared to other dimensions of the body
then the contact forces may be treated as Concentrated
Forces
Important issues are – Magnitude of
the equivalent force and its point of
action
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Center of Mass & Centroid:
Distributed Forces: If forces are applied over a region whose
dimension is not negligible compared with other pertinent
dimensions  proper distribution of contact forces must be
accounted for to know intensity of force at any location.
Line Distribution
(Ex: UDL on beams)
Body Distribution
(Ex: Self weight)
Area Distribution
Ex: Water Pressure
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Center of Mass & Centroid:
A body of mass m in equilibrium
under the action of tension in the cord
and resultant W of the gravitational
forces acting on all particles of the
body → The resultant is collinear with
the cord
Suspend the body from different points
on the body
• Dotted lines show lines of action of the resultant force in each case.
• These lines of action will be concurrent at a single point G
As long as dimensions of the body are smaller compared with those of the
earth → we assume uniform and parallel force field due to the
gravitational attraction of the earth.
The unique Point G is called the Center of
Gravity of the body (CG)
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Determination of CG:
Apply Principle of Moments
Moment of resultant gravitational force W
about any axis equals sum of the moments
about the same axis of the gravitational
forces dW acting on all particles treated as
infinitesimal elements.
Weight of the body W = ∫dW
Moment of weight of an element (dW) @
x-axis = ydW
Sum of moments for all elements of body
= ∫ydW
From Principle of Moments: ∫ydW = ӯ W
xdW

x=
W
ydW

y=
W
zdW

z=
W
Numerator of these expressions represents the sum of the
moments i.e. Product of W and corresponding coordinate of G
represents the moment of the sum  Moment Principle
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Determination of CG:
x=
 xdW
W
y=
 ydW
W
z=
 zdW
W
Substituting W = mg and dW = gdm

xdm

x=
m
ydm

y=
m
zdm

z=
m
In vector notations:
Position vector for elemental mass: r = xi + yj + zk
Position vector for mass center G: r = xi + yj + zk
→
rdm

r=
The above equations are the
components of this single vector equation
Density ρ of a body = mass per unit volume
 Mass of a differential element of volume dV  dm = ρdV
 ρ may not be constant throughout the body
m
xdV

x=
 dV
ydV

y=
 dV
zdV

z=
 dV
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Centre of Mass:
Following equations independent of g
 xdm
x=
m
 ydm
y=
m
 zdm
z=
m
 rdm
r=
m
xdV

x=
 dV
ydV

y=
 dV
zdV

z=
 dV
 They define a unique point, which is a function of distribution of
mass
 This point is Center of Mass (CM)
 CM coincides with CG as long as gravity field is treated as uniform
and parallel
 CG or CM may lie outside the body
CM always lie on a line or a plane of symmetry in a homogeneous body
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Centroid of Line, Area & Volume:
xdL

x=
L
ydL

y=
L
zdL

z=
L
x=
 xdA
A
y=
 ydA
A
z=
 zdA
A
x=
 xdV
V
y=
 ydV
V
dV
Lines: Slender rod, Wire
Cross-sectional area = A
ρ and A are constant over L
dm = ρAdL; Centroid = CM
Areas: Body with small but
constant thickness t
Cross-sectional area = A
ρ and A are constant over A
dm = ρtdA; Centroid = CM
z=
 zdV
V
V
Volumes: Body with
volume V
ρ constant over V
dm = ρdV
Centroid = CM
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Centroid of Line, Area & Volume:
Guidelines for Choice of Elements for Integration
Order of Element Selected for Integration
 A first order differential element should be selected in preference to a higher
order element  only one integration should cover the entire figure
A = ∫dA = ∫ ldy
A = ∫ ∫ dx dy
V = ∫ dV = ∫ πr2 dy
V = ∫ ∫ ∫ dxdydz
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Centroid of Line, Area & Volume:
Guidelines for Choice of Elements for Integration
Continuity – Choose an element that can be integrated in one
continuous operation to cover the entire figure
 The function representing the body should be continuous
 Only one integral will cover the entire figure
Continuity in the expression
for the width of the strip
Discontinuity in the expression
for the height of the strip at
x = x1
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Centroid of Line, Area & Volume:
Guidelines for Choice of Elements for Integration
Discarding Higher Order Terms
Higher order terms may be dropped
Note: Vertical strip of area under the curve is given by the first order term
dA = ydx
The second order triangular area 0.5dxdy may be discarded
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Centroid of Line, Area & Volume:
Guidelines for Choice of Elements for Integration
Choice of Coordinates: Coordinate system should best match the
boundaries of the figure
 easiest coordinate system that satisfies boundary conditions should
be chosen
Boundaries of this area (not circular) Boundaries of this circular sector are
can be easily described in rectangular best suited to polar coordinates
coordinates
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Centroid of Line, Area & Volume:
Guidelines for Choice of Elements for Integration
Centroidal Coordinate of Differential Elements: While expressing
moment of differential elements, take coordinates of the centroid of
the differential element as lever arm (not the coordinate describing
the boundary of the area)
Modified
Equations
x=
 xc dA
A
y=
 yc dA
A
z=
 z c dA
A
x=
 xc dV
V
y=
 yc dV
V
z=
 zc dV
V
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Centroid of Line, Area & Volume:
Guidelines for Choice of Elements for Integration
1.
2.
3.
4.
5.
Order of Element Selected for Integration
Continuity
Discarding Higher Order Terms
Choice of Coordinates
Centroidal Coordinate of Differential Elements
xdL

x=
ydL

y=
zdL

z=
xc dA

x=
y c dA

y=
z c dA

z=
xc dV

x=
yc dV

y=
z c dV

z=
L
A
V
L
A
V
L
A
V
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Examples (Centroid of Line):
Locate the centroid of the circular arc
Solution: Polar coordinate system is better
Since the figure is symmetric: centroid lies on the x axis
Differential element of arc has length dL = rdӨ
Total length of arc: L = 2αr
x-coordinate of the centroid of differential element: x=rcosӨ
xdL

x=
L
ydL

y=
L
zdL

z=
L
For a semi-circular
arc: 2α = π 
centroid lies at 2r/π
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Center of Mass and Centroids:
Integration vs App. Summation: Irregular Area
In some cases, the boundaries of an area or volume might not
be expressible mathematically or in terms of simple geometrical
shapes  App. Summation may be used instead of integration
Divide the area into several strips
Area of each strip = hΔx
Moment of this area about x- and y-axis
= (hΔx)yc and (hΔx)xc
 Sum of moments for all strips
divided by the total area will give
corresponding coordinate of the centroid
x=
 Axc
A
y=
 Ay c
A
Accuracy may be improved by reducing
the width of the strip
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Example (Centroid of an Area):
Locate the centroid of the
triangle along h from the base
Solution:
dA = xdy ; x/(h-y) = b/h
Total Area, A = ½(bh)
yc = y
x=
 xc dA
A
y=
 yc dA
A
z=
 z c dA
A
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Examples:
Determine the CG of the area
confined by two lines y=x & x2=ay
y
a
Solution:
a

x2 
dA =  y2 - y1 dx =  x - dx
a

y2
y1
x
x
dx
a
1 3
1 3
1  3 x6 
a4
I x =  y2 dx -  y1 dx =   x - 3 dx =
3
3
3 0
a 
24
2
4


x
a
I y =  x 2 dA =  x 2  x - dx =
a 
20

0
a
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Example (Centroid of an Area):
Locate the centroid of the shaded area
Solution: Divide the area into four elementary
shapes: Total Area = A1 + A2 - A3 - A4
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Area Moments of Inertia:
Previously considered distributed forces which were proportional
to the area or volume over which they act.
- The resultant was obtained by summing or integrating over the
areas or volumes.
- The moment of the resultant about any axis was determined by
computing the first moments of the areas or volumes about
that axis.
Will now consider forces which are proportional to the area or
volume over which they act but also vary linearly with distance
from a given axis.
- the magnitude of the resultant depends on the first moment of
the force distribution with respect to the axis.
- The point of application of the resultant depends on the second
moment of the distribution with respect to the axis.
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Area Moments of Inertia:

• Consider distributed forces F whose magnitudes are
proportional to the elemental areas A on which they
act and also vary linearly with the distance of A
from a given axis.
• Example: Consider a beam subjected to pure bending.
Internal forces vary linearly with distance from the
neutral axis which passes through the section centroid.

F = kyA
R = k  y dA = 0  y dA = Qx = first moment
2
M = k  y 2 dA
 y dA = second moment
First Moment of the whole section about the x-axis =
y A = 0 since the centroid of the section lies on the x-axis.
Second Moment or the Moment of Inertia of the beam
section about x-axis is denoted by Ix and has units of
(length)4 (never –ve)
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Area Moments of Inertia by Integration:
Second moments or moments of
inertia of an area with respect to
the x and y axes
I x =  y 2 dA
I y =  x 2 dA
Evaluation of the integrals is
simplified by choosing dA to be a
thin strip parallel to one of the
coordinate axes.
Therefore, for a rectangular area
h
I x =  y 2 dA =  y 2bdy = 13 bh3
0
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Area Moments of Inertia by Integration:
Second moments or moments of
inertia of an area with respect to
the x and y axes
I x =  y 2 dA
I y =  x 2 dA
The formula for rectangular areas may also
be applied to strips parallel to the axes
dI x = 13 y 3dx
dI y = x 2 dA = x 2 y dx
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Polar Moment of Inertia:
The polar moment of inertia is an
important parameter in problems
involving torsion of cylindrical
shafts and rotations of slabs.
J 0 = I z =  r 2 dA
The polar moment of inertia is related to the
rectangular moments of inertia


J 0 = I z =  r 2 dA =  x 2 + y 2 dA =  x 2 dA +  y 2 dA
= Iy + Ix
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Radius of Gyration of an Area:
Consider area A with moment of inertia Ix.
Imagine that the area is concentrated in a thin
strip parallel to the x axis with equivalent Ix.
I
I x = k x2 A
kx = x
A
kx = radius of gyration with respect
to the x axis
Similarly,
Iy
2
I y = ky A
ky =
A
JO
J O = I z = k O2 A = k z2 A k O = k z =
kO2 = k z2 = k x2 + k y2
A
Radius of Gyration, k is a measure of distribution
of area from a reference axis
Radius of Gyration is different from centroidal
distances
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Example:
Determine the moment of
inertia of a triangle with
respect to its base.
Solution:
 A differential strip parallel to the x axis is chosen for dA
dI x = y 2dA
dA = l dy
 For similar triangles,
l h- y
=
b
h
l =b
h- y
h
dA = b
bh3
I x=
12
 Integrating dIx from y = 0 to y = h,
h- y
dy
h


h- y
bh 2
I x =  y dA =  y b
dy =  hy - y 3 dy
h
h0
0
2
h
2
h
b  y3 y 4 
= h - 
h 3
4
0
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Example:
a) Determine the centroidal polar moment of
inertia of a circular area by direct
integration.
b) Using the result of part a, determine the
moment of inertia of a circular area with
respect to a diameter.
Solution:
An annular differential area element
is chosen,
dJ O = u 2dA
dA = 2 u du
r
From symmetry, Ix = Iy
2
r 4 = 2I x
r
J O =  dJ O =  u 2 u du  = 2  u 3du
2
0
JO =

JO = I x + I y = 2I x
0

2
I diameter = I x =

4
r4
r4
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