ME 108 Statics - Gaziantep Üniversitesierklig/me108/0_lecture1.pdfUnits used in Engineering Statics...
Transcript of ME 108 Statics - Gaziantep Üniversitesierklig/me108/0_lecture1.pdfUnits used in Engineering Statics...
ME 108 Statics
Ahmet Erkliğ [email protected]
Office No: 315 4th floor of Mechanical Engineering Departmen
What is Engineering Mechanics? Engineering Mechanics is a discipline that studies the response of
solids, structures, fluids and devices made of these materials to
mechanical, thermal, and other types of load events. It emphasizes a
basic science approach that draws heavily on physics, mathematics
and computer science.
Mechanics is crucial to addressing modern problems in:
• aerospace engineering • mechanical engineering
• civil engineering • nuclear engineering
• materials science • geological engineering
• agricultural engineering • biomedical engineering
• electrical engineering (electronics packaging)
• chemical engineering (polymeric material modeling)
Particles vs Objects
• Particles:
–Point mass
–No geometry
–Rotation is not important
• Objects
–Contain mass
–Have geometry
–Rotation is important
The basic principle we will use is Newton’s 2nd law
of motion
We will spend considerable time becoming
proficient in applying this law to engineering
problems.
What is Statics?
Statics is the study of particle and rigid body structure
equilibrium. Static means unchanging position in time.
amF
The Basic Concept
If an object is at rest (or moving with a constant velocity), we can say quite a lot about the forces acting on it – and within it!
What You Will Learn in Statics?
• How to determine the resultant of multiple forces and distributed forces.
• How to determine the moment of forces about points and axes.
• How to draw “free body diagrams” and identify all of the forces acting on a body.
• How to use “equilibrium equations” to determine unknown or unspecified forces and moments acting on (and within) bodies, including structures and machines.
• How to identify and quantify friction forces. • How to determine centroids, centers of mass/gravity, and
moments of inertia for shapes and objects.
And what will you do with these competencies???
• Use information about forces and moments to analyze and design parts, assemblies, mechanisms, and structures that are appropriate to their intended (and sometimes unintended) functions.
• Use information about centroids, centers of mass/gravity, and moments of inertia in support of the above.
Why is Statics Important?
• ME 108 is the fundamental building block for all of engineering courses.
• Good habits developed in Statics will help you in subsequent courses.
• Performance in Statics is factored into your “Math/Science GPA” for admission to a department
ME 208 dynamics ME 223, 307-308
advanced dynamics
robotics
satellite mechanics
vibrations
advanced mech of
materials
stress analysis
structural engineering
machine design
finite element analysis
others
others
ME 108 Statics ME 223
mech of materials 1
Tips for Success in Statics (i.e., what is expected!)
• Do all assigned reading – before class.
• Solve extra problems
• Don’t spend more than 20 minutes/problem. Get help!
• Attend lecture regularly.
• Ask questions!!!
Courtesy During Lectures
• Arrive for lecture on time.
• No sleeping, reading newspapers, conversations, etc. during lecture. I expect your attention to be focused on statics for 50 minutes.
• Please wait until class is dismissed before leaving.
Text Book
Text Book: Engineering Mechanics Vol-1;Statics
By J. L. Meriam and L. G. Kraige
Auxiliary Book: Vector Mechanics for Engineers; Statics
By Ferdinand P. Beer and E. R. Johnston
Engineering Mechanics – Statics
By R.C. Hibbeler
Many books related with the Statics in the Library
Grading
2 midterm 2 x 20%
Homeworks+Quiz 20%
final 40%
100%
Exams
1st midterm 26-07-2011
2nd midterm 09-08-2011
final ??-08-2011
Course Overview
• Basic principles
• Force systems
• Equilibrium
• Structural analysis and machines
• Centroids, distributed load systems
• Beams
• Area and mass moments of inertia
Fundamentals, units, calculations
• Units
– Length – need to know position and geometry of objects
– Time – need to determine succession of events
– Mass – related to amount of stuff in a body, found using gravitational attraction
– Weight – force due to gravity acting on a mass, W=mg, where g=9.8m/s2
Fundamentals, units, calculations
• Basic Quantities
– Force – push or pull on a body, can be direct (contact) or indirect (no contact)
–Moment – turning effect caused by a force applied at some distance away from the axis of rotation
Fundamentals, units, calculations
Engineering Concepts
– Idealizations – all real problems are simplified
to some degree
– Particle – mass acting is if it were concentrated
at a singe point
– Rigid Body – particle collection in a shape that
doesn’t change with applied force
– Concentrated Force – force acting as if it were
at a single point
1st law: A particle remains at rest, or continues to move in a
straight line with uniform velocity, if there is no unbalanced
force acting on it.
2nd law: The acceleration of a particle is proportional to the
resultant force acting on the particle, and is in the direction of
this force.
3rd law: The forces of action and reaction between interacting
bodies are equal in magnitude, opposite in direction, and co-
linear.
Newton’s laws of motion
case staticfor 0
amF
amF
Newton’s law of gravitational attraction
Consider two particles of mass m1 and m2. Newton stated:
F G
m1m2
r 2
F = force of attraction between the two particles.
G = universal gravitational constant, experimentally found
to be = 66.73 (10-12) m3/(kg-s2).
r = distance between particles.
weight of a particle
Consider a particle with mass m1=m, and let the earth’s mass be
denoted by m2. Letting W=F where we call W the weight, we
can write
W=mg where g = acceleration due to gravity = Gm2/r2
Observation: We commonly think of g as a constant. However, in
our application of F=Gm1m2/r2, the earth is not a particle, it has
nonuniform density, and r depends on position. Nonetheless,
commonly accepted values for g are:
g = 32.2 ft/s2 = 386.4 in/s2
= 9.81 m/s2
Units used in Engineering Statics – SI Units Dimension Unit SI Symbol Length (l) meter m Mass (m) kilogram kg Time (t) second s Derived Unit Basic Units Symbol Area (A) m2 - Volume (V) m3 - Density (ρ) kg/m3 - Velocity (v) m/s - Acceleration (a) m/s2 - Force (F) kgm/s2 N Energy (E, U, K, etc.) kgm2/s2 J
Be consistent when using units!
Units System length mass force time
English foot, ft slug==lb-sec 2/ft pound, lb second, s
metric (SI) meter, m kilogram, kg Newton==kg-m/s2, N second, s
Remark: the relationship among these units is governed by
f=ma. By selecting 3 fundamental units, a 4th “derived” unit can
be obtained. The derived units are:
English system:
metric (SI):
1 slug f
a
1 lb
1 ft/s2
or...1 lb (1 slug)(1 ft/s2 )
1 Newton ma 1 kg 1 m/s2
Suggestion: Avoid
slugs and simply
work with
fundamental units.
Unit Conversions
If p=expression, to change the units of p, multiply the
r.h.s. by dimensionless factors of unity until the desired units
are obtained.
e.g., Convert the velocity v = 10 m/s to English units.
Since 1 ft 0.3048 m, 1 ft
0.3048 m
0.3048 m
0.3048 m1
v 10 m
s1 10
m
s
1 ft
0.3048 m1
32.808 ft/s
Unit Conversions Length Mass
1 km = 0.62137 mi. 1 kg = 2.2 lbs
1 mile = 5,280 ft = 1.6093 km 1 lb = 453.59 g = 16 oz.
1 meter = 1.0936 yds 1 metric ton = 1000kg
1 inch = 2.54 cm (exact) 1 slug = 32 lb = 14.594 kg
Volume Pressure
1 mL = 1 cm3 1 atm = 760 torr (mm Hg)
1 L = 1 dm3 = 1.0567 qts. 1 atm = 14.70 lb/in2
1 gal = 4 qts = 8 pts = 3.7854 L 1 bar = 100,000 Pa
1 in3 = 16.4 cm3 1atm = 101325 Pa
1 qt = 0.946 L
Force Energy
1N = 0.2248lbf 1 J = 1 kg-m2/s2
1N = 2.248x10-4 kip 1 cal. = 4.184 J
1 kip = 1000lbf
1N = 7.233 poundal (lbm ft/s2)
Useful to reduce numbers to manageable size,
especially SI units.
factor prefix symbol
109
giga
G
106 mega M
103 kilo k
10-3 milli m
10-6 micro
10-9 nano n
Prefixes
Numerical calculations Dimensional Homogeneity:
In addition to being numerically correct, an equation must
also be dimensionally correct. Always carry units along with
your calculations.
Accuracy:
You cannot create accuracy with your calculations.
e.g., 12.34 * 12.3 = 151.782
In lengthy calculations, retain one or two extra significant
digits*. Your final results should be reported with the same
accuracy as your data (in this case ... 152.)
* Occasional exceptions to this.
example 1
• convert 175 lb/ft3 to SI units:
• convert 6 ft/hr to SI units:
• convert 1.13 kN-m to English units:
A:
a) 27.5 kN/m3
b) 0.508 mm/s
c) 833 ft-lb
example 2
Two steel spheres with 100 mm diameter are placed
beside one another at the earth’s surface. Compute the
gravitational force W (weight) and the force of mutual
attraction F in Newtons.
The unit weight of steel is
= 0.284 lb/in3
A: W = 40.3 N, F = 1.13 (10-7) N
Problem Solving Strategy 1. Interpret: Read carefully and determine what is
given and what is to be found/ delivered. Ask, if not clear. If necessary, make assumptions and indicate them.
2. Plan: Think about major steps (or a road map) that you will take to solve a given problem. Think of alternative/creative solutions and choose the best one.
3. Execute: Carry out your steps. Use appropriate diagrams and equations. Estimate your answers. Avoid simple calculation mistakes. Reflect on / revise your work.
FORCE SYSTEMS - Force
Definitions:
A scalar is a quantity that is completely characterized by a
single number. Mass, volume, and length are examples of
scalars.
A vector is a quantity that has both magnitude (or size) and
direction. Velocity, force (e.g., weight) and position are
examples of vectors.
Common notation for vector quantities:
F or F
Vector Notation To report a vector, its magnitude and direction need to be
stated. We will discuss a few convenient methods for doing
this including:
• Simple statement of magnitude and direction relative to
some arbitrary reference direction (effective for 2-D).
e.g.,
• Statement in terms of rectangular Cartesian components
(effective in both 2-D and 3-D ... more on this shortly)
Basic Vector Operations
There is more than one way of adding vectors:
• parallelogram addition
• head-to-tail: note that this can be done in any order
Addition:
Multiplication: • multiplication of a vector by a scalar changes its
magnitude but NOT its direction.
Subtraction:
• equivalent to multiplying the vector to be subtracted
by -1 and adding it to the vector from which it is
being subtracted.
Summary:
A
B
R = A+B
R = A+B
R = B+A
A
A
A
B
B
B
Scalar multiplication: s(A + B) = sA + sB, s is a scalar
Vector addition is commutative: A + B = B + A
Resolution of a vector into components
Imagine two directions, a and b in 2-dimensional space.
Provided a and b are not parallel, any vector can be
resolved into components in the a and b directions.
a
b
R
a
b
R A
B
R = A + B
Vectors A and B are often called the projections of R in
the a and b directions, respectively.
more ...
more on resolution of vectors ...
When the a and b directions are
orthogonal (i.e., a and b intersect
at a right angle) it is usually
straightforward to determine the
vector components.
When the a and b directions
are non-orthogonal, it is usually
more work to determine the
vector components. The law of
sines and law of cosines may be
useful.
a
b
R A
B
a
b
R A
B
more ...
Rectangular Components:
A force F is said to have been resolved into two rectangular
components if its components are directed along the
coordinate axes. Introducing the unit vectors i and j along
the x and y axes,
x
y
q
Fx = Fx i
Fy = Fy j
F
i
j
F = Fx i + Fy j
Fx = F cos q Fy = F sin q
tan q = Fy
Fx
F = Fx + Fy 2 2
When three or more coplanar forces act on a particle, the
rectangular components of their resultant R can be obtained
by adding algebraically the corresponding components of the
given forces.
Rx = SRx Ry = SRy
The magnitude and direction of R can be determined from
tan q = Ry
Rx R = Rx + Ry
2 2
Addition of Several Vectors
• Step 1: resolve each force into its components
• Step 2: add all the x components together and add all the y components together. These two totals become the resultant vector.
• Step 3: find the magnitude and angle of the resultant vector.
Example 1
Combine the two forces P and T, which act on the
fixed structure at B, into a single equivalent force R.
P = 500 N
T = 200 N
3m
T
P
A
B
C D
5m
75o
Law of cosines:
NR
R
cabbac
5.396
)4.48cos()500)(200(2500200
)cos(2
222
222
Law of sines:
4.48sin
5.396
sin
200
q
2.22q
NR 5.396
22.2
Example 2
A barge is pulled by 2 tugboats. The resultant of the forces
exerted by the tugboats is a 5000 N force directed along
the center axis of the barge. Determine tension in each rope
if α =45 degrees
Barge
A
B
30
α
Example 3
• Given: Three concurrent forces acting on a bracket.
• Find: The magnitude and angle of the resultant force.
Example 4
A ring supports three forces. What is the resultant
force applied to the ring?
more ... A: R = 998 N @ 134.9°
Add force vectors to get the resultant.
Because vector addition is commutative, the order of addition is
not important.
Example 5
25 F = 110 N
Wind
A sail boat tacks into the wind such that the force
perpendicular to the sail is 110 N. Resolve this force
into two components, one parallel to the keel and one
perpendicular to the keel.
A: F= 46.5 N
F = 99.7 N