ENGR-1100 Introduction to Engineering Analysis

Section 4Instructor: Professor Suvranu De

Office: JEC 5002

Office Ph: x6096

E-mail: des@rpi.edu

Office hours: Tuesday and Friday 2:00-3:00 pm

Course Coordinator: Mohamed Aboul-Seoud aboulm@rpi.edu 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: akinla@rpi.edu

• Ji Ming Hong (Undergraduate)

hong.j.10@gmail.com

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 from

http://www.rpi.edu/~des/IEA2012Spring.html• Supplementary information: download from

http://www.rpi.edu/~des/IEA2012Spring.html• Connect quick steps: download from

http://www.rpi.edu/~des/IEA2012Spring.html• Matlab tutorial: download from

http://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

MechanicsMechanics

Very large

Very small

StaticsNet force=0

StaticsNet force=0 Dynamics

Net 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/s2

At 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 Symbol

Length meter m

Mass kilogram kg

time second s

Derived unitsQuantity Unit Symbol

Area Square meter m2

Volume Cubic meter m3

Linear velocity Meter per second m/s

Derived units with special nameQuantity Unit Symbol

Plane angle radian rad

Solid angle steradian sr

SI / U.S. customary units conversion

Quantity U.S. customary to SI SI to U.S. customary

Length 1 ft = 0.3048 m 1 m = 3.281 ft

Velocity 1 ft/s = 0.304 m/s 1 m/s = 3.281 ft/s

Mass 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. Magnitude

2. 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

Example

Find 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.”