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)
ENGINEERING MECHANICS-«STATICS EC3221
FALL 2011
EXAM 2
October 14, 2011
Name St) I off} 0 m5
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Recitation instructor
Recitation Day
Problem #1
Problem #2
Problem #3
Problem #4
TOTAL PROBLEM ONE
A ca
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 = Fxi + Fyj
j
i
x
More Vectors
Given the magnitude (F)
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
Forces in Space
Force F is defined by magnitude F, two
angles and , and point of
application (origin).
y
F
x
z
Rectangular Components
Given the size F and angles , obtain
components Fx, Fy, Fz of force F.
Step One.
F=Fh+Fy
y
F
x
z
Fh = F sin
Fy = F cos
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
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
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 =
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
1
1
1
U 12 = dU = F dr = Fcos dr
2
1
Examples
F constant
M constant
2
2
U 12 = dU = F dr = F ( r 2 r1 )
1
2
ECS 221: Engineering Mechanics
Trusses Method of Joints
Lecture 23
Truss Analysis - Basics
Very Efficient Structural Solution for
spanning long distances
All joints are pins, thus no moments
All members carry load along own axis i.e.
tension or compres
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
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
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
Equilibrium of Two and
Three Force Bodies
2-force body: rigid body
subject to two forces only.
F2
F1
more
Theorem. A 2-force body is in equilibrium if
and only if both (non-zero) forces have the
same magnitude, line of action and opposite
sense.
F2
F1
Equilibrium of Rigid
Bodies
A necessary and sufficient
condition for equilibrium of a
rigid body is for the resultant
force and the resultant moment
about a point to vanish.
more
Equilibrium:
R = F = 0 , MOR = MO = 0
Fx = 0 Fy = 0 Fz = 0
MOx = 0 MOy = 0
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
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
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