Centroids, Centers of Gravity
for 3-D Bodies
Locate position vector defining
center of gravity.
G
r
r
- W j
-Wj
Procedure
Divide body into small elements. Equate
force and moment sums to resultants.
- W j = W j W = dW
r (-W j) = r ( W j) rW = rdW
W = dW
THE FRAME
The frame contains at least one multiforce member.
Solution; Step 1
Obtain support forces if possible.
M0 = 0
Fx = 0
Fy = 0
Reactions are statically determinate!
Solution; Steps 2,3
Dismember structure. Apply equilibrium.
Ax
Ay
Bx
C
C
By
C
FRICTION
Friction manifests itself as a force
that resists motion.
Dry (Coulomb) friction acts between
two solid bodies.
Fluid friction acts between two fluids
or a solid and a fluid.
Characteristics
W
Block on plane.
P
P=0
P>0, block motionless
F
N
P>
Method of Sections
The method of sections enables the
calculation of force in one truss member.
It does not require balancing forces at
each joint but requires ensuring
equilibrium over finite portions of the
truss.
Example
Compute force in member BD.
B
Analysis of Structures
Truss. A structure designed to
support load. Composed of straight
two force members connected at pin
joints.
more
Frame. A structure designed to support
load. Composed of members of which at
least one is multi-force.
more
Machin
APPLICATIONS OF
FRICTION
Wedges
Square Threaded Screws
Thrust Bearings
Wheel Friction
Wedges
Wedges are used to raise heavy loads by
small amounts.
What is the minimum value
of force P required to raise
block A?
P
A
Analysis
Free body diagrams.
P
sN2
BELT FRICTION
1. How does T vary around drum?
2. How are T1, T2 related at impending
slip? is angle
T1
of wrap
T2
Assume belt slips right T2>T1.
Belt Element Free Body
Impending slip: F = s N
/
2
F N
/
2
T+ T
/
2
Equilibrium
Fx=0
(T+ T) cos/2 Tcos/2
Mass Moments of Inertia
The inertia characterized by the mass
(m) tends to resist linear motion.
The moment of inertia characterized
by the mass moment of inertia (I)
tends to resist rotational motion.
Mass Rotating About Axis
The applied moment is propo
MOMENTS OF INERTIA
Distributed forces act on volume or
surface elements.
Gravity
Hydrostatic pressure in a fluid
Bending stress in beams
Center of Gravity for Area
Center of gravity
W = t A
W = dW xW = x dW
B
B
y W = y dW
B
First moments Qx = ydA Qy =
The Parallel Axis Theorem
The parallel axis theorem is used with
the method of composite parts.
y
y
x
y
x
x
If Ix is known for centroidal axis how to get
it for any other parallel axis?
Theorem: I = I + Ad
2
Proof:
I WRT lower axis
I WRT centroidal axis
Principle of Virtual Work I
The principle of virtual work is an
alternative formulation of the
principles of equilibrium. It is often
more convenient to use since it
effectively eliminates forces of
constraint.
Work of a Force
Incremental displacement ve
PRODUCTS OF INERTIA
I
Product of inertia: xy = xy dA
y
Ixy may not be
positive!
A
x
y
x
The Rectangle
Compute Ixy.
dA = dxdy
dIxy = xydxdy
y
b
I xy = xdx ydy
0 0
h
dy
h
b
x
122
I xy = b h
4
Geometrical Meaning
Product of inertia
y
I xy = xy dA
y
x
Ixy =
Centroids by Integration
Recall: xA = x dA yA = y dA
B
dy
B
dr
dx
rd
Reduction to Single Integral
Element area:
dA = ydx
Element centroid:
xel = x yel = y/2
yx
yel y
xel
dx
x
b
xA = x dA = xel ydx = xy(x)dx
B
B
a
y(x) y(x)dx
y A = y dA = y el ydx =
B
B
Three-Dimensional
Equilibrium
Equilibrium:
R = F = 0 , MOR = MO = 0
Fx = 0
MOx = 0
Fy = 0
MOy= 0
Fz = 0
MOz = 0
Six equations for 6 unknowns.
3-D Supports
1. Reactions with known line of action.
Displacement constrained in one direction
only. One un
Engineering Mechanics
Statics
ECS221
A. J. Levy
Course Structure
M,W lecture + F clinic or exam
Graded homework (13%)
Four class exams (54%)
Final exam (33%)
Course evaluation
Introduction
Primitive concepts
Space (coordinate system, length; scalar)
Tim
Rectangular Components; Unit
Vectors
Rectangular Components
y
y
F
=
x
Fy
Fx
x
Unit Vectors
A vector directed along the positive x or y
axis with unit magnitude.
Fx = Fxi , Fy = Fyj
y
F = Fx + Fy
F = F x i + F yj
j
i
x
More Vectors
Given the magnitude (
Statics of Particles
Law of sines
a sin = b sin
Law of cosines
c2 = a2 + b2 - 2ab cos
c
a
b
Representation of a Force
A force (P) is defined by a point of
application (A), a magnitude (P), and a
direction (30o from horizontal).
5N
A
30o
Force Resultants
Equilibrium of a Particle
Particles (mass points)
Equilibrium Continued
A particle is in equilibrium if the resultant
force acting on it is the zero vector. R = F
=0
Planar equilibrium.
R = F = 0 (Fxi + Fyj) = 0i + 0j
Fx = 0
Fy = 0
The Free Body Diagram
Forces in Space
Force F is defined by magnitude F,
two angles and , and point of
application (origin).
y
F
z
x
Rectangular Components
Given the size F and angles , obtain
components Fx, Fy, Fz of force F.
Step One.
F=Fh+Fy
y
F
z
x
Fh = F sin
Fy = F cos
Rigid Bodies; Equivalent
Force Systems
Transmissibility Principle
F
=
Force F is a sliding vector.
F
Vector (Cross) Product
Multiply 2 vectors P Q resulting in a
vector.
Define magnitude of product.
Define direction of product.
V = P Q
Magnitude: V =
Couples
A system of 2 forces of equal magnitude
parallel lines of action and opposite sense.
The Couple
F
-F
Moment of a Couple
MO = rA F + rB (-F) = (rA- rB) F = r F
-F
rB
rA
O
r
F
Moment of a couple:
M = r F
More
M = r F , M = rF sin = Fd
Magnitude:
Reduction of a System of
Forces
Problem. Find the simplest
representation of a system of
forces such that its effect on a rigid
body is unaltered.
Recall
Step 1.
Step 2.
F
O
r
M
F
O
When moving a force off its line of action
we must add a couple with mo
MOMENTS OF INERTIA
Distributed forces act on volume or
surface elements.
Gravity
Hydrostatic pressure in a fluid
Bending stress in beams
Center of Gravity for Area
Center of gravity W = tA
W dW xW x dW
B
B
yW y dW
B
First moments Qx = ydA Qy = xdA
Bendi
STATICALLY INDETERMINATE
PROBLEMS II
Summary:
Rigid body equilibrium
F
x
= 0,
F
y
= 0,
M
P
=0
Kinematics of infinitesimal rigid displacements (u=u0+ r)
ux = u0x -y, u y = u 0y + x plane Cartesian compon ents
Lumped support models e.g. linear and torsional
Rigid Bodies; Equivalent Force
Systems
Transmissibility Principle
F
=
Force F is a sliding vector.
F
Vector (Cross) Product
Multiply 2 vectors PQ resulting in a
vector.
Define magnitude of product.
Define direction of product.
V = PQ
Magnitude: V = PQ
Principle of Virtual Work II
Recall: work done by force F in
incremental displacement dr is
dU = Fdr = F dr cos
For a finite displacement:
2
2
2
U12 dU F dr Fcos dr
1
1
1
2
1
Examples
2
2
dU F dr F r 2 r1
F constant U12
1
1
M constant
U dU Md M
2
12
2
APPLICATIONS OF FRICTION
Wedges
Square Threaded Screws
Thrust Bearings
Wheel Friction
Wedges
Wedges are used to raise heavy loads by small
amounts.
What is the minimum value
of force P required to raise
block A?
P
A
Analysis
Free body diagrams.
P
sN2