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ENGR-1100 Introduction to Engineering Analysis Section 4 Instructor: Professor Suvranu De Office: JEC 5002 Office Ph: x6096 Office.

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Presentation on theme: "ENGR-1100 Introduction to Engineering Analysis Section 4 Instructor: Professor Suvranu De Office: JEC 5002 Office Ph: x6096 Office."— Presentation transcript:

1 ENGR-1100 Introduction to Engineering Analysis Section 4 Instructor: Professor Suvranu De Office: JEC 5002 Office Ph: x Office hours: Tuesday and Friday 2:00-3:00 pm Course Coordinator: Mohamed Aboul-Seoud x2317

2 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 – Ji Ming Hong (Undergraduate)

3 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

4 Course websites Course official web site: My website for the course: McGraw-Hill Connect website (for Home works) for this section: McGraw-Hill Connect help: mu/success-academy/

5 Course handouts Course syllabus: download from Supplementary information: download from Connect quick steps: download from Matlab tutorial: download from

6 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: + 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%

7 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)

8 Lecture outline Newtons laws Units of measurement Vectors

9 Mechanics Mechanics of rigid 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

10 Mechanics Very large Very small Statics Net force=0 Statics Net force=0 Dynamics Net force 0 Dynamics Net force 0 In this class we will deal with the statics of rigid bodies.

11 Physical Problem Physical Model Mathematical model (set of equations) Does answer make sense? Physical idealizations: particles, rigid body, concentrated forces, etc. Physical laws: Newtons laws Applied to each interacting body (free body diagram) Solution of equations: Using pen+paper/own code/ canned software like Matlab Happy YES! No! Modeling

12 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

13 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 Newtons Laws of Motion

14 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 Newtons Laws of Motion

15 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!! Newtons Laws of Motion

16 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) Table W=mg weight of pencil Newtons Laws of Motion

17 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 r 2 F=G M m r 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 )ft 3 /(slug*s 2 ) in the U.S customary system of units. G=6.673( )m 3 /(kg*s 2 ) in SI system of units F

18 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: M e. m r e 2 F=G=mg Where: M e is the mass of the earth. r e is the mean radius of the earth M e r e 2 g=G g =32.17 ft/s 2 = m/s 2 At sea level and latitude 45 0

19 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 45 0 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

20 Base units QuantityUnitSymbol Lengthmeterm Masskilogramkg timeseconds Derived units QuantityUnitSymbol AreaSquare meterm2m2 VolumeCubic meterm3m3 Linear velocityMeter per secondm/s Derived units with special name QuantityUnitSymbol Plane angleradianrad Solid anglesteradiansr

21 SI / U.S. customary units conversion QuantityU.S. customary to SISI to U.S. customary Length1 ft = m1 m = ft Velocity1 ft/s = m/s1 m/s = ft/s Mass1 slug = kg1 kg = slug

22 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., )

23 Vectors: geometric representation Direction of arrowdirection of vector Length of arrowmagnitude 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?

24 magnitude (F) magnitude (2F) magnitude (F) magnitude (2F) n can be a fraction less than 1, can n be 0? F 2F2F -F -2F Operations on Vectors: Multiplication by scalars

25 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+PR=P+Q=Q+P Operations on Vectors: Adding vectors using Parallelogram Rule

26 Vectors in rectangular coordinate systems- two dimensional (v 1,v 2 ) x y v If the tail of the vector (v) is at the origin, then the coordinates of the terminal point (head) (v 1,v 2 ) are called the Cartesian components of the vector. v1v1 v2v2 O V = v 1 i + v 2 jOr, v=(v 1,v 2 )

27 Vectors in rectangular coordinate systems- multiplication by a scalar (2v 1,2v 2 ) x y 2v2v The components of the vector 2v are (2v 1, 2v 2 ) 2v 1 2v 2 O

28 The sum of two vectors – by adding components (two dimensional ) (w 1,w 2 ) x y v (v 1,v 2 ) w w1w1 v1v1 w2w2 v2v2 (v 1 +w 1,v 2 +w 2 ) v+w=(v 1 +w 1,v 2 +w 2 ) Just add the x- and y-components v + w = (v 1 + w 1 )i + (v 2 + w 2 ) j Or,

29 Vectors in rectangular coordinate systems- Three dimensional (v 1,v 2,v 3 ) y z v (v 1,v 2, v 3 ) are the coordinates of the terminal point (head) of vector v v1v1 v2v2 x v2v2 v2v2 v3v3 O

30 The sum of two vectors – rectangular components (Three dimensional ) z x y (a 1,a 2,a 3 ) (b 1,b 2,b 3 ) a b a+b=(a 1 +b 1,a 2 +b 2, a 3 +b 3 ) O

31 Vectors with initial point not at the origin (VERY IMPORTANT!!) P 1 (x 1,y 1,z 1 ) y z v w x P 2 (x 2,y 2,z 2 ) u O Hence Coordinates of head minus coordinates of tail

32 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 P 1 (x 1,y 1,z 1 ) y 2 -y 1 P 2 (x 2,y 2,z 2 ) u z 2 -z 1 x 2 -x 1

33 Example Find the components of the vector having initial point P 1 and terminal point P 2 P 1 (-1,0,2), P 2 (0,-1,0) Solution: v= (0-(-1),-1-0,0-2)=(1,-1,-2) Head (P 2 ) minus tail (P 1 )

34 Vector arithmetic If 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

35 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 P 1 and terminal point P 2 (a) P 1 = (-5,0), P 2 = (-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:

36 IEA wisdom Success in IEA is proportional to the number of problems solved.

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