Post on 09-Jan-2016
description
Dr. Ali Keshavarz
General Principles
Equilibrium of Bodies
At rest or moving with constant
velocity
Mechanics of Materials
Physical science concerned with the state of rest or
motion of bodies that are subject to forces
Solid Mechanics - Lec. 2 1
Dr. Ali Keshavarz
Newtons Laws of Motion First Law.
A particle originally at rest or moving in a straight line with constant velocity, will remain in this state provided that the particle is not subjected to an unbalanced force.
Second Law. A particle acted upon by an unbalanced force F experiences an
acceleration, a that has the same direction as the force and a magnitude that is proportional to the force.
Third Law. The mutual forces of action and reaction between two particles
are equal, opposite, and collinear.
mF a
Solid Mechanics - Lec. 2 2
Dr. Ali Keshavarz
Newtons Law of Gravitational Attraction
1 22
m mF Gr
1 2
12 3 2
: Force of gravitation between two particles and : Mass of each particle
: Distance between two particles: Universal constant of gravitation
66.73 10 m / kg s
Fm mrGG
Solid Mechanics - Lec. 2 3
Dr. Ali Keshavarz
Significant Figures Accuracy of a number is specified by the number of significant figures it contains. A
significant figure is any digit, including a zero, provided it is not used to specify the location of the decimal point for the number, i.e., 0.5 has only one significant digit.
Example 57 098 and 44.893 (Both numbers have 5 significant digits) When numbers begin or end with zeros, it gets little confusing. Consider 400. In this kind of situations, express the number in engineering notation
(exponent is used in multiples of 3). Then, 400 = 0.4x103. Only 1 Significant digit. 2500 for example can be written as 2.5x103 with 2 significant digits or can be written
as 2.50x103 with 3 significant digits. This is done to specify more accuracy. 0.00546 can be written as 5.46x10-3 with 3 significant digits. Rounding Off Numbers
Rule: Use of 3 significant digits is usually enough in final answers. In intermediate calculations keep a higher number of significant digits.
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Dr. Ali Keshavarz
Force Vectors Scalar. A quantity characterized by a positive or negative number such as mass,
volume, length. Vector. A quantity that has
A magnitude (how big is your vectors length compared to a given reference) A direction (on a line you can have 2 direction choices) A line of action
Examples: Position, velocity, force When specifying direction: Always know your reference, i.e., CW from an axis.
Solid Mechanics - Lec. 2 5
Dr. Ali Keshavarz
Vector Math Multiplication and division of a vector with a scalar
Changes the magnitude only (aA or A/a or -A) Vector addition (Commutative Law: A + B = B + A)
Parallelogram law or triangular construction Vector subtraction: (A B = A + (-B))
Same as addition concept Just multiply the vector being subtracted and add two
Solid Mechanics - Lec. 2 6
Dr. Ali Keshavarz
Vector Resolution Resolution of a vector
Resolve into 2 components on 2 known line of actions 2 known line of actions are not necessarily perpendicular WHY? We may need to resolve due to geometry
Solid Mechanics - Lec. 2 7
Dr. Ali Keshavarz
Trigonometric Laws and Force Notation Sine Law and Cosine
Law Trigonometric
relations Geometric relations Force Notation
Scalar and Vector500 N and F or F
F
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Dr. Ali Keshavarz
Vector Addition of 3 or More Forces Application of successive parallelogram law will
lead to the result. This can get involved and can be error prone in
terms of geometry and trigonometry.
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Dr. Ali Keshavarz
Adding Coplanar Forces: Cartesian Vector Notation
x yF F F i j, : unit vectors
magnitude = 1 (unity)direction: +/- sign
i j
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Dr. Ali Keshavarz
Resultant of Coplanar Forces
1 1 1
2 2 2
3 3 3
yx
yx
yx
F FF F
F F
F i jF i jF i j
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Dr. Ali Keshavarz
The Resultant
1 2 3R F F F FR Rx RyF F F i j
1 2 3
1 2 3
Rx x x x
Ry y y y
F F F FF F F F
Rx x
Ry y
F FF F
2 2
1tan
R Rx Ry
Ry
Rx
F F F
FF
Solid Mechanics - Lec. 2 12
Dr. Ali Keshavarz
Unit Vector is used to specify direction
Rectangular 3D Components of A Vector
x y z
x y zA A A
A A A AA i j k
or
0
A AAAA
Au A u
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Dr. Ali Keshavarz
Revisit the Unit Vector
or
0
A AAAA
Au A u
cos cos cos
yx zA
A
AA AA A A A
Au i j k
u i j k2 2 2cos cos cos 1
cos cos cosA
x y z
AA A AA A A
A ui j k
i j k
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Dr. Ali Keshavarz
Addition and Subtraction of Cartesian Vectors
x y z
x y z
A A AB B B
A i j kB i j k
( ) ( ) ( )x x y y z xA B A B A B
R A Bi j k
Concurrent Force Systems:
R x y zF F F F F i j kSolid Mechanics - Lec. 2 15
Dr. Ali Keshavarz
Position Vectors
x y z r i j kSolid Mechanics - Lec. 2 16
Dr. Ali Keshavarz
Dot Product How do we find angle
between two lines? Dot Product = Scalar
Product (result is scalar) Commutative Law:
AB = BA Multiply by a Scalar:
a(AB) = (aA)B = A(aB) =(AB)a
Distributive Law: A(B+D) = (AB) + (AD)
cos(0 180 )
AB
A B
10
i i j j k ki j i k k j
x y z
x y z
A A AB B B
A i j kB i j k
x x y y z zA B A B A B A BSolid Mechanics - Lec. 2 17
Dr. Ali Keshavarz
Applications of Dot Product
The angle formed between 2 vectors
Components of a vector parallel/perpendicular to a line
1cos 0 180AB
A B
||
||
coscos ( )
A AA
A u A u u
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Dr. Ali Keshavarz
Force System ResultantsMoment of a Force
Moment: A measure of the tendency of the
force to cause a body to rotate about a point or
axis
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Dr. Ali Keshavarz
Moment of a Force
Scalar Formulation Magnitude:
Direction: Right-Hand Rule
OM Fd
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Dr. Ali Keshavarz
Cross Product The result is a vector The order of multiplication
does matter Magnitude:
Direction:
sinC AB
( sin ) CAB C AB u
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Dr. Ali Keshavarz
Cross ProductCartesian Vector Formulation
i i = j j = kk = 0i j k jk i k i = j
j i k k j i ik = j
x y z
x y z
A A AB B B
A i j kB i j k
( ) ( ) ( )x y z y z z y x z z x x y y xx y z
A A A A B A B A B A B A B A BB B B
i j k
AB i j k
MINUS SIGNImportant!
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Dr. Ali Keshavarz
Moment of a ForceVector Formulation The moment of force F about point O The moment of force F about the
moment axis passing through O and perpendicular to the plane containing O and F
r is the position vector from O to anypoint lying on the line of action of F
Magnitude:
Direction: Right-hand rule
O M rF
sin ( sin )OM rF F r Fd
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Dr. Ali Keshavarz
Moment of a ForceCartesian Vector Formulation
x y z
x y z
r r rF F F
r i j kF i j k
( )( ) ( )
( ) ( ) ( )
( ) ( ) ( )O yO x O z
O x y z
x y z
y z z y x z z x x y y x
MM M
O x O y O z
r r rF F F
r F r F r F r F r F r F
M M M
i j kM rF
i j k
i j k
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Dr. Ali Keshavarz
Resultant Moment ofA System of Forces
( )OR M rF
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Dr. Ali Keshavarz
Equilibrium Conditions
O
F 0M 0
FOR A RIGID BODY
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Dr. Ali Keshavarz
Springs
Linear spring (ideal)
= stiffness of the spring (N/m or lb/ft)
Undeformed length of the spring,
F ks
0l
k
if 0, Pushing Force, if 0, Pulling Force, ss
FF
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Dr. Ali Keshavarz
Cables and Pulleys
Assumptions: They have NO weight They DONT stretch They ONLY can
support tension Assuming NO friction
in the pulley: For equilibrium, applied
force is balanced at the other end of the pulley.
Solid Mechanics - Lec. 2 28
Dr. Ali Keshavarz
Equilibrium in 2D: Free-Body Diagrams Due to symmetry of geometry and loading
conditions, many engineering problems can be simplified to 2 dimensions.
All known (given) and unknown external forcesacting on a body must be identified.
FBD: An isolated sketch of a body from its surroundings showing all the forces and couple moments that the surroundings exert on the body.
Solid Mechanics - Lec. 2 29
Dr. Ali Keshavarz
FBD Procedure Draw the outline shape
Make sure the body is totally isolated or cut free from its constraints and connections and sketch an outlined shape
Show all forces and couple moments Identify all external forces and couple moments that act on the
body Applied loadings Reactions occurring at the supports and contact with other bodies Weight of a body
Identify each loading and give directions Label given forces and couple moments with their magnitudes
and directions. Label with letters unknowns resolved into x, y components, i.e., Ax, Ay, etc., when necessary.
Solid Mechanics - Lec. 2 30
Dr. Ali Keshavarz
Weight and Center of Gravity
Forces due to gravitational field acting on a body can be considered as a system of parallel forces.
Using our previous analysis, we can find a single resultant force: Weight of the body.
Location of Weights point of application is called the center of gravity.
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Dr. Ali Keshavarz
Support Reactions in 2D
There are common engineering joints/supports. Consider attachments of a simple horizontal beam.
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Dr. Ali Keshavarz
Common 2D Supports
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Dr. Ali Keshavarz
Examples: Support ReactionsCable
Rocker
SmoothSurface
Fixed
SmoothPin
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Dr. Ali Keshavarz
Equations of Equilibrium: 2D
O
F 0M 0
000
x
y
O
FFM
MO is the sum of couple moments and moments of all forces about an axis
perpendicular to x-y plane and passing through the arbitrary point O. O can be
on the body or off the body.
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Dr. Ali Keshavarz Solid Mechanics - Lec. 2 36
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 37
Dr. Ali Keshavarz
Procedure for Analysis Free-Body Diagram
Establish x-y axes in any suitable orientation Draw an outlined shape of the body Show all the forces and couple moments acting on the body Label all loadings and specify their directions with respect to x, y axes. For an unknown load (force/moment), directions can be assumed along
the known line of action Indicate dimensions necessary for calculating moments
Equations of Equilibrium Apply the moment equation of equilibrium about a point O that lies at
the intersection of the lines of actions of two unknown forces. This way, you can eliminate the moment effect of those forces
Choose the most convenient x-y axes so that the resolutions of forces along x and y would be easier
If a negative result found, it indicates the chosen direction must be reversed
Solid Mechanics - Lec. 2 38
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 39
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 40
Dr. Ali Keshavarz
Equilibrium in 3D: Support Reactions
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Dr. Ali Keshavarz
Examples: 3D Supports
SmoothPin
ThrustBearing
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Dr. Ali Keshavarz
Equations of Equilibrium: 3D
O
F 0M 0
000
x
y
z
FFF
000
x
y
z
MMM
x y z
O x y z
F F FM M M
F i j k 0M i j k 0
You can use these equations to solve for 6 unknowns!
Solid Mechanics - Lec. 2 43
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 44
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 45
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 46
Dr. Ali Keshavarz
Structural Analysis
Simple Trusses
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Dr. Ali Keshavarz
Planar Trusses: Roof
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Dr. Ali Keshavarz
Planar Trusses: Bridge
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Dr. Ali Keshavarz
Method of Joints
To design a truss, we must know the force in each member.
If FBD of the entire truss is used, the forces in the members are internal forces and they cannot be determined.
Considering the FBD of a joint solves this problem.
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Dr. Ali Keshavarz
Method of Joints
Analysis must start from a joint with at least one known force and at most 2 unknowns. 0
0x
y
FF
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Dr. Ali Keshavarz
Determining Directions
Always assume the unknown member forces to be tension.Positive result Tension (T)Negative result Compression (C)Once an unknown force is found, use the
correct direction (T or C) in subsequent step. Determine the correct direction by
inspection when possible.
Solid Mechanics - Lec. 2 52
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 53
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 54
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 55
Dr. Ali Keshavarz
Method of Sections
If a body is in equilibrium, any part of that body must also be in equilibrium.
Using the same idea we can cut or section the members of an entire truss.
This can be desirable to determine force of a specific member.
Solid Mechanics - Lec. 2 56
Dr. Ali Keshavarz
Method of Sections
Since we have 3 equations, try to cut through a maximum of 3 members with unknown forces.
0, 0, 0x y OF F M
Solid Mechanics - Lec. 2 57
Dr. Ali Keshavarz
Determining Directions
Always assume the unknown member forces at the cut section to be tension.Positive result Tension (T)Negative result Compression (C)
Determine the correct direction by inspection when possible
Solid Mechanics - Lec. 2 58
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 59
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 60
Dr. Ali Keshavarz
APPLICATIONS
To design the structure for supporting a water tank, we will need to know the weights of the tank and water as well as the locations where the resultant forces representing these distributed loads are acting.
How can we determine these weights and their locations?
Solid Mechanics - Lec. 2 61
Dr. Ali Keshavarz
APPLICATIONS (continued)
One concern about a sport utility vehicle (SUVs) is that it might tip over while taking a sharp turn.
One of the important factors in determining its stability is the SUVs center of mass.
Should it be higher or lower for making a SUV more stable?
How do you determine its location?Solid Mechanics - Lec. 2 62
Dr. Ali Keshavarz
Center of Gravity and Centroid for a Body
xdW ydW zdW
x y zdW dW dW
W V dW dV Specific WeightWeight per unit
volume
x dV y dV z dV
x y zdV dV dV
Solid Mechanics - Lec. 2 63
Dr. Ali Keshavarz
Center of Massg
DensityMass per unit
volume
x dV y dV z dV
x y zdV dV dV
dm dV
xdm ydm zdmx y z
dm dm dm
Solid Mechanics - Lec. 2 64
Dr. Ali Keshavarz
CentroidCentroid
Geometric centerHomogenous
V V V
V V V
xdV ydV zdVx y z
dV dV dV
A A A
A A A
xdA ydA zdAx y z
dA dA dA
L L L
L L L
xdL ydL zdLx y z
dL dL dL
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Dr. Ali Keshavarz
Symmetry
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Dr. Ali Keshavarz
STEPS FOR DETERMING AREA CENTROID
1. Choose an appropriate differential element dA at a general point (x,y). Hint: Generally, if y is easily expressed in terms of x(e.g., y = x2 + 1), use a vertical rectangular element. If the converse is true, then use a horizontal rectangular element.
2. Express dA in terms of the differentiating element dx (or dy).
3. Determine coordinates of the centroid of the rectangular element in terms of the general point (x,y).
4. Express all the variables and integral limits in the formula using either x or y depending on whether the differential element is in terms of dx or dy, respectively, and integrate.
Note: Similar steps are used for determining CG, CM, etc.. These steps will become clearer by doing a few examples.
)~,~( yx
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Dr. Ali Keshavarz
EXAMPLE
Solution
1. Since y is given in terms of x, choose dA as a vertical rectangular strip.
x,y
x , y~~2. dA = y dx = (9 x2) dx
3. x = x and y = y / 2~~
Given: The area as shown.
Find: The centroid location (x , y)
Plan: Follow the steps.
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Dr. Ali Keshavarz
EXAMPLE (continued)
4. x = ( A x dA ) / ( A dA )~
0
0
0 x ( 9 x2) d x [ 9 (x2)/2 (x4) / 4] 30 ( 9 x2) d x [ 9 x (x3) / 3 ] 3
= ( 9 ( 9 ) / 2 81 / 4 ) / ( 9 ( 3 ) ( 27 / 3 ) )
= 1.13 ft
3= =
3
33.60 ftA y dA 0 ( 9 x2) ( 9 x2) dx
A dA 0 ( 9 x2) d x
3
=y = = ~
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Dr. Ali Keshavarz
More Example
Given: The area as shown.
Find: The x of the centroid.
Plan: Follow the steps.
Solution
1. Choose dA as a horizontal rectangular strip.(x1,,y) (x2,y)
2. dA = ( x2 x1) dy
= ((2 y) y2) dy
3. x = ( x1 + x2) / 2
= 0.5 (( 2 y) + y2 )
~
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Dr. Ali Keshavarz
More Example (continued)
4. x = ( A x dA ) / ( A dA )~
A dA = 0 ( 2 y y2) dy[ 2 y y2 / 2 y3 / 3] 1 = 1.167 m2
1
0
A x dA = 0 0.5 ( 2 y + y2 ) ( 2 y y2 ) dy= 0.5 0 ( 4 4 y + y2 y4 ) dy= 0.5 [ 4 y 4 y2 / 2 + y3 / 3 y5 / 5 ] 1
= 1.067 m30
1
1
~
x = 1.067 / 1.167 = 0.914 m
Solid Mechanics - Lec. 2 71
Dr. Ali Keshavarz
Composite Bodies
xW yW zWx y zW W W
CompositeMade of connected
simpler shapes
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Dr. Ali Keshavarz
APPLICATIONS
The I-beam is commonly used in building structures.
When doing a stress analysis on an I - beam, the location of the centroid is very important.
How can we easily determine the location of the centroid for a given beam shape?
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Dr. Ali Keshavarz
APPLICATIONS (continued)
Cars, trucks, bikes, etc., are assembled using many individual components.
When designing for stability on the road, it is important to know the location of the bikes center of gravity (CG).
If we know the weight and CG of individual components, how can we determine the location of the CG of the assembled unit?
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Dr. Ali Keshavarz
CONCEPT OF A COMPOSITE BODY
Many industrial objects can be considered as composite bodies made up of a series of connected simpler shaped parts or holes, like a rectangle, triangle, and semicircle.
Knowing the location of the centroid, C, or center of gravity, G, of the simpler shaped parts, we can easily determine the location of the C or G for the more complex composite body.
a
be
d
ab
ed
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Dr. Ali Keshavarz
This can be done by considering each part as a particle and following the procedure as described.
This is a simple, effective, and practical method of determining the location of the centroid or center of gravity.
CONCEPT OF A COMPOSITE BODY(continued)
a
be
d
ab
ed
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Dr. Ali Keshavarz
STEPS FOR ANALYSIS
1. Divide the body into pieces that are known shapes. Holes are considered as pieces with negative weight or size.
2. Make a table with the first column for segment number, the second column for weight, mass, or size (depending on the problem), the next set of columns for the moment arms, and, finally, several columns for recording results of simple intermediate calculations.
3. Fix the coordinate axes, determine the coordinates of the center of gravity of centroid of each piece, and then fill-in the table.
4. Sum the columns to get x, y, and z. Use formulas like
x = ( xi Ai ) / ( Ai ) or x = ( xi Wi ) / ( Wi )This approach will become clear by doing examples!
Solid Mechanics - Lec. 2 77
Dr. Ali Keshavarz
EXAMPLE
Given: The part shown.
Find: The centroid of the part.
Plan: Follow the steps for analysis.
Solution:
1. This body can be divided into the following pieces: rectangle (a) + triangle (b) + quarter circular (c) semicircular area (d)
a
bc
d
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Dr. Ali Keshavarz
EXAMPLE (continued)
Steps 2 & 3: Make up and fill the table using parts a, b, c, and d.
39.8376.528.0
274.59
- 2/3
5431.5 9 0
1.51
4(3) / (3 )4(1) / (3 )
37
4(3) / (3 )0
184.5
9 / 4 / 2
RectangleTriangleQ. CircleSemi-Circle
A y( in3)
A x( in3)
y(in)
x(in)
Area A(in2)
Segment
abc
d
Solid Mechanics - Lec. 2 79
Dr. Ali Keshavarz
x = ( x A) / ( A ) = 76.5 in3/ 28.0 in2 = 2.73 iny = ( y A) / ( A ) = 39.83 in3 / 28.0 in2 = 1.42 in
4. Now use the table data and these formulas to find the coordinates of the centroid.
EXAMPLE (continued)
C
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Dr. Ali Keshavarz
More Example
Given: Two blocks of different materials are assembled as shown.
The densities of the materials are:
A = 150 lb / ft3 andB = 400 lb / ft3.Find: The center of gravity of this
assembly.
Plan: Follow the steps for analysis.Solution
1. In this problem, the blocks A and B can be considered as two segments.
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Dr. Ali Keshavarz
More Example (continued)
Weight = w = (Volume in ft3)wA = 150 (0.5) (6) (6) (2) / (12)3 = 3.125 lb
wB = 400 (6) (6) (2) / (12)3 = 16.67 lb
56.2553.1229.1719.79
6.2550.00
3.12550.00
12.516.67
23
13
41
3.12516.67
AB
wz (lbin)
w y (lbin)
w x (lbin)
z (in)y (in)x (in)w (lb)Segment
Solid Mechanics - Lec. 2 82
Dr. Ali Keshavarz
~x = ( x w) / ( w ) = 29.17/19.79 = 1.47 iny = ( y w) / ( w ) = 53.12/ 19.79 = 2.68 inz = ( z w) / ( w ) = 56.25 / 19.79 = 2.84 in ~
~
More Example (continued)
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Dr. Ali Keshavarz
Moment of Inertia for Areas
kz dF dA kzdA
2dM dFz kz dA 2M k z dA
C
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Dr. Ali Keshavarz
Moment of Inertia
2
2
x A
y A
I y dA
I x dA
2o x yAJ r dA I I
UNITSm4, mm4, ft4, in4
2 2 2r x y Polar Moment of Inertia
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Dr. Ali Keshavarz
Parallel Axis Theorem2
22
( )
2
x yA
y yA A A
I y d dA
y dA d y dA d dA
2
2
x x y
y y x
I I Ad
I I Ad
2O CJ J Ad
Moment of Inertia for Composite Areas:If moment of inertia of simple shapes thatmake up the final shape is known or can bedetermined about a specified axis, then themoment of inertia of the composite shape isthe algebraic sum of the moments of its parts.
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Dr. Ali Keshavarz Solid Mechanics - Lec. 2 87
Dr. Ali Keshavarz Solid Mechanics - Lec. 2 88