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ENGR-1100 Introduction to Engineering Analysis Section 4. Instructor: 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.
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ENGR-1100 Introduction to Engineering AnalysisSection 4 Instructor: 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 • AdemolaAkinlalu (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-LAPTOPunless 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 is the branch of science that deals with the state of rest or motion of bodies under the action of forces Mechanics Mechanics of rigid bodies Mechanics of deformable bodies Mechanics of fluids In this class we will exclusively deal with the mechanics of rigid bodies. Few basic principles but exceedingly wide applications
Very large Mechanics Statics Net force=0 Dynamics Net force 0 Very small In this class we will deal with the statics of rigid bodies.
Modeling Physical Problem Physical idealizations: particles, rigid body, concentrated forces, etc. Physical Model Physical laws: Newton’s laws Applied to each interacting body (free body diagram) Mathematical model (set of equations) Solution of equations: Using pen+paper/own code/ canned software like Matlab Does answer make sense? No! YES! Happy
Physical Idealizations 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.
Newton’s Laws of Motion • 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 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 Vector equation
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!!
Newton’s Laws of Motion Force (F) Force (F) Pencil W=mg weight of pencil Table R (force acting on the pencil) R (force acting on the table) • Need to isolate the bodies and consider the forces actingonthem (FreeBody Diagram). • Be careful about which force in the pair we are talking about!
M.m r2 M F=G r m Law of gravitation 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: 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
At the surface of the earth: Me.m re2 F=G =mg Me re2 g=G Mass and weight 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. Where: Meis the mass of the earth. reis the mean radius of the earth At sea level and latitude 450 g =32.17 ft/s2 = 9.807 m/s2
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 units Derived units Derived units with special name
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., )
Vectors: geometric representation A vector is geometrically represented as a line segment with an arrow indicating direction Line of action F Length represents magnitude (F) Head Tail Direction of arrow direction of vector Length of arrow magnitude of vector Question: What is a vector having the same magnitude and line of action, but opposite sense?
Operations on Vectors: Multiplication by scalars magnitude (2F) magnitude (F) 2F F magnitude (2F) magnitude (F) -2F -F n can be a fraction less than 1, can n be 0?
Operations on Vectors: Adding vectors using Parallelogram Rule 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 theParallelogram law P P + R Q Q R=P+Q=Q+P • To obtain the resultant, add two vectors using parallelogram law • Addition of vectors is commutative (order does not matter)
y x V = v1i + v2j Vectors in rectangular coordinate systems- two dimensional (v1,v2) v v2 v1 O 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. v=(v1,v2) Or,
y x Vectors in rectangular coordinate systems- multiplication by a scalar (2v1,2v2) 2v 2v2 2v1 O The components of the vector 2v are (2v1, 2v2)
y (v1+w1,v2+w2) v2 (w1,w2) w2 (v1,v2) x v1 w1 v + w = (v1+ w1)i + (v2+ w2)j The sum of two vectors – by adding components (two dimensional ) w v Just add the x- and y-components v+w=(v1+w1,v2+w2) Or,
Vectors in rectangular coordinate systems- Three dimensional z (v1,v2,v3) v v2 v2 v3 y O v1 v2 x (v1,v2,v3) are the coordinates of the terminal point (head) of vector v
z (a1,a2,a3) a y b (b1,b2,b3) x The sum of two vectors – rectangular components (Three dimensional ) O a+b=(a1 +b1,a2+b2, a3 +b3)
z y Vectors with initial point not at the origin (VERY IMPORTANT!!) u P2(x2 ,y2 ,z2) P1(x1 ,y1 ,z1) w v O x Coordinates of headminus coordinates of tail Hence
z y Vectors with initial point not at the origin (VERY IMPORTANT!!) z2-z1 u P2(x2 ,y2 ,z2) P1(x1 ,y1 ,z1) y2-y1 x2-x1 O x The components are the projections of the vector along the x-, y- and z-axes
Example Solution: v= (0-(-1),-1-0,0-2)=(1,-1,-2) Find the components of the vector having initial point P1 and terminal point P2 P1(-1,0,2), P2(0,-1,0) Head (P2) minus tail (P1)
Vector arithmetic If u,v,w are vectors in 2- or 3-space and k and l are scalar, then the following relationship holds: • u+v=v+u • u+0=0+u=u • k(lu)=(kl)u • (k+l)u=ku+lu • (u+v)+w=u+(v+w) • u+(-u)=0 • k(u+v)= ku+ kv • 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) 2. Let u = (-3,1,2), v = (4,0,-8) and w = (6,-1,-4). Find the x, y and z components of: (a) 6u + 2v (b) -3(v – 8w)
IEA wisdom “Success in IEA is proportional to the number of problems solved.”