ENGR-1100 Introduction to Engineering Analysis
Section 4Instructor: Professor Suvranu De
Office: JEC 5002Office Ph: x6096
E-mail: [email protected] hours: Tuesday and Friday 2:00-3:00 pmCourse Coordinator: Mohamed Aboul-Seoud
[email protected] x2317
Teaching assistants: TA
• Ademola Akinlalu (Graduate)– Office hours: M 3P-5PM;T 10AM -12Noon; W:
12Noon-2PM; R 3PM-5PM; F 10AM-12 Noon– Office: JEC 1022– E-mail: [email protected]
• Ji Ming Hong (Undergraduate)[email protected]
Some Important Points• Studio course (combined lesson & problem session)• ALL CLASSES ARE
CLOSED-LAPTOP unless otherwise stated.
• Bring your relevant textbook, calculators, pencil, engineering computation paper to class EVERYDAY
• Importance of laptop/MATLAB to solve complicated problems
• Important tools: syllabus, 2 textbooks (listed in syllabus), laptop, pencil and, engineering computation paper
Course websites• Course official web site: http://www.rpi.edu/dept/core-eng/WWW/IEA• My website for the course:http://www.rpi.edu/~des/IEA2012Spring.html• McGraw-Hill Connect website (for Home works) for
this section:http://connect.mcgraw-hill.com/class/2012-iea-4• McGraw-Hill Connect help:http://create.mcgraw-hill.com/wordpress-mu/success-academy/
Course handouts• Course syllabus: download fromhttp://www.rpi.edu/~des/IEA2012Spring.html• Supplementary information: download fromhttp://www.rpi.edu/~des/IEA2012Spring.html• Connect quick steps: download fromhttp://www.rpi.edu/~des/IEA2012Spring.html• Matlab tutorial: download fromhttp://www.rpi.edu/~des/IEA2012Spring.html
Course format• Mini lectures• Daily Class Activities (CA) 5% (drop 4 lowest
grades). NO makeup for CA. • Daily Homeworks (HW). 15% (drop 2 lowest
grades). HWs due next day of class 12 noon. NO LATE SUBMISSIONS!
• Three mid term exams (2/15, 3/21, 4/18) in SAGE 3510: 2@20% + 1@15%, total 55%– Exam times: Wednesday 8 – 9:50 am – Make-up exams (2/22 , 3/28, 4/25) in TBD 5:00-6:50pm – Grade challenges must be within a week (6-8pm Exam 1:
2/20, 2/21; Exam 2: 3/26, 3/27; Exam 3: 4/23, 4/24)– No make-ups for missed make-ups!
• 1 final exam (time: TBA) : 25%
Course objectives
Formulation and solution of static equilibrium problems for particles and rigid bodies.
A bit of linear algebra: solution of sets of linear equations as they arise in mechanics and matrix operations.
Use your laptop (running Matlab) for the manipulation of vector quantities and solution of systems of equations (only as an aid to completely solve “realistic” problems)
Lecture outline
• Newton’s laws• Units of measurement• Vectors
Mechanics
Mechanics
Mechanics ofrigid bodies
Mechanics of deformable bodies
Mechanics of fluids
Mechanics is the branch of science that deals with the state of rest or motion of bodies under the action of forces
In this class we will exclusively deal with the mechanics of rigid bodies.Few basic principles but exceedingly wide applications
Mechanics
Very large
Very small
StaticsNet force=0
DynamicsNet force 0
In this class we will deal with the statics of rigid bodies.
Physical Problem
Physical Model
Mathematical model
(set of equations)
Does answer make sense?
Physical idealizations: particles, rigid body, concentrated forces, etc.
Physical laws: Newton’s lawsApplied to each interacting body(free body diagram)
Solution of equations: Using pen+paper/own code/ canned software like Matlab
Happy YES!
No!
Modeling
Continuum: For most engineering applications assume matter to be a continuous distribution rather than a conglomeration of particles.Rigid body: A continuum that does not undergo any deformation.Particle: No dimensions, only has mass. Important simplifying assumption for situation where mass is more important than exactly how it is distributed.Point force: A body transmits force to another through a finite area of contact. But it is sometimes easier to assume that a finite force is transmitted through an infinitesimal area.
Physical Idealizations
Law I: (Principle of equilibrium of forces) A particle remains at rest or continues to move in a straight line with uniform velocity (this is what we mean by being “in equilibrium”) if there is no unbalanced force acting on it.
• Inertial reference frame • Necessary condition for equilibrium• Foundation of Statics
0......321
nFFFFF Vector equation
Newton’s Laws of Motion
Law II: (Nonequilibrium of forces) The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force.
• is the resultant force acting on a particle of mass ‘m’.
• Foundation of Dynamics• Necessary condition for equilibrium
corresponding to
nFFFFF
......321 amF
0
a
Vector equation
F a
Newton’s Laws of Motion
Law III: (Principle of action and reaction) If one body exerts a force on a second body, then the second body exerts a force on the first body that is (1) equal in magnitude, (2) opposite in direction and (3) collinear (same line of action).
• EXTREMELY IMPORTANT to keep this in mind when working out problems!!
reactionaction FF
Newton’s Laws of Motion
• Need to isolate the bodies and consider the forces acting on them (Free Body Diagram).
• Be careful about which force in the pair we are talking about!
Pencil
R (force acting on the pencil)R (force acting on the table)
Force (F)Force (F)
Table
W=mg weight of pencil
Newton’s Laws of Motion
Two bodies of mass M and m are mutually attracted to each other with equal and opposite forces F and –F of magnitude F given by the formula:
Law of gravitation
M.m
r2F=G
M
mr
where r is the distance between the center of mass of the two bodies; and G is the Universal Gravitational Constant.
G=3.439(10-8)ft3/(slug*s2) in the U.S customary system of units.
G=6.673(10-11)m3/(kg*s2) in SI system of units
F
The mass m of a body is an absolute quantity.
The weight W of a body is the gravitational attraction exerted on the body by the earth or by another massive body such as another planet.
Mass and weight
At the surface of the earth:Me
.m
re2
F=G =mgWhere: Me is the mass of the earth.
re is the mean radius of the earth Me
re2
g=G g =32.17 ft/s2 = 9.807 m/s2At sea level and latitude 450
Units of measurement• The U.S customary system of units (the
British gravitational system)• Base units are foot (ft) for length, the pound (lb) for
force, and the second (s) for time• Pound is defined as the weight at sea level and altitude
of 450 of a platinum standard
• The international system of units (SI) • Three class of units
• (1) base units• (2) derived units • (3) derived units with special name
Base unitsQuantity Unit SymbolLength meter mMass kilogram kgtime second s
Derived unitsQuantity Unit SymbolArea Square meter m2
Volume Cubic meter m3
Linear velocity Meter per second m/s
Derived units with special nameQuantity Unit SymbolPlane angle radian radSolid angle steradian sr
SI / U.S. customary units conversion
Quantity U.S. customary to SI SI to U.S. customaryLength 1 ft = 0.3048 m 1 m = 3.281 ftVelocity 1 ft/s = 0.304 m/s 1 m/s = 3.281 ft/sMass 1 slug = 14.59 kg 1 kg = 0.06854 slug
Scalar and vectors• A scalar quantity is completely described
by a magnitude (a number).-Examples: mass, density, length, speed, time,
temperature. • A vector quantity has
1. Magnitude2. Direction (expressed by the line of action + sense)3. Obey parallelogram law of addition
-Examples: force, moment, velocity, acceleration.We will represent vectors by bold face symbols (e.g., F)
in the lecture. But, when you write, you can use the symbol with an arrow on top (e.g., ) F
Vectors: geometric representation
Direction of arrow direction of vector
Length of arrow magnitude of vector
Line of action
Head
Tail
Length represents magnitude (F)
F
A vector is geometrically represented as a line segment with an arrow indicating direction
Question: What is a vector having the same magnitude and line of action, but opposite sense?
magnitude (F)
magnitude (2F)
magnitude (F)
magnitude (2F)
n , n is a scalar (negative or positive, integer or fraction)Fn can be a fraction less than 1, can n be 0?
F2F
-F -2F
Operations on Vectors: Multiplication by scalars
Task: Add two vectors ( P and Q ) to obtain a “resultant” vector (R) that has the same effect as the original vectors Vectors are added using the Parallelogram law
• To obtain the resultant, add two vectors using parallelogram law
• Addition of vectors is commutative (order does not matter)
P
Q
R
P
Q
R=P+Q=Q+P
Operations on Vectors: Adding vectors using Parallelogram Rule
Vectors in rectangular coordinate systems- two dimensional
(v1,v2)
x
y
v
If the tail of the vector (v) is at the origin, then the coordinates of the terminal point (head) (v1,v2) are called the Cartesian components of the vector.
v1
v2
O
V = v1 i + v2 j Or, v=(v1,v2)
Vectors in rectangular coordinate systems- multiplication by a
scalar
(2v1,2v2)
x
y
2v
The components of the vector 2v are (2v1, 2v2)
2v1
2v2
O
The sum of two vectors – by adding components (two dimensional )
(w1,w2)
x
y
v(v1,v2)
w
w1v1
w2
v2
(v1+w1,v2+w2)
v+w=(v1+w1,v2+w2)
Just add the x- and y-components
v + w = (v1 + w1 )i + (v2 + w2 ) jOr,
Vectors in rectangular coordinate systems- Three dimensional
(v1,v2,v3)
y
z
v
(v1,v2,v3) are the coordinates of the terminal point (head) of vector v
v1
v2
x
v2
v2
v3O
The sum of two vectors – rectangular components (Three dimensional )
z
x
y
(a1,a2,a3)
(b1,b2,b3)
a
b
a+b=(a1 +b1,a2+b2, a3 +b3)
O
Vectors with initial point not at the origin (VERY IMPORTANT!!)
P1(x1 ,y1 ,z1)
y
z
v
w
x
P2(x2 ,y2 ,z2)u
1 1 1 1
2 2 2 2
( , , )
( , , )
OP x y z
OP x y z
w
v
O
w + u v
u = v - w
1 2 2 1 2 1 2 1( , , )PP x x y y z z u =
Hence
Coordinates of head minus coordinates of tail
Vectors with initial point not at the origin (VERY IMPORTANT!!)
y
z
x
O
The components are the projections of the vector along the x-, y- and z-axes
P1(x1 ,y1 ,z1) y2-y1
P2(x2 ,y2 ,z2)uz2-z1
x2-x1
ExampleFind the components of the vector having initial point P1 and terminal point P2
P1(-1,0,2), P2(0,-1,0)
Solution:
v= (0-(-1),-1-0,0-2)=(1,-1,-2)
Head (P2) minus tail (P1)
Vector arithmeticIf u,v,w are vectors in 2- or 3-space and k and l are scalar, then the following relationship holds:
(a) u+v=v+u
(b) u+0=0+u=u
(c) k(lu)=(kl)u
(d) (k+l)u=ku+lu
(e) (u+v)+w=u+(v+w)
(f) u+(-u)=0
(g) k(u+v)= ku+ kv
(h) 1u=u
Class assignment: (on a separate piece of paper with your name and RIN on top) please submit to TA at the end of the lecture
1. Find the component of the vector having initial point P1 and terminal point P2
(a) P1 = (-5,0), P2 = (-3,1)
(a) 6u + 2v
(b) -3(v – 8w)
2. Let u = (-3,1,2), v = (4,0,-8) and w = (6,-1,-4). Find the x, y and z components of:
IEA wisdom “Success in IEA is proportional to the number of problems solved.”
