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ENGR-1100 Introduction to Engineering Analysis

ENGR-1100 Introduction to Engineering Analysis. Pawel Keblinski Materials Science and Engineering MRC115 Office hours: Tuesday 1-3 phone: (518) 276 6858 email: keblip@rpi.edu. Mechanics -> Static. Mechanics. Mechanics of rigid bodies. Mechanics of deformable bodies. Mechanics of

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ENGR-1100 Introduction to Engineering Analysis

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  1. ENGR-1100 Introduction to Engineering Analysis Pawel Keblinski Materials Science and Engineering MRC115 Office hours: Tuesday 1-3 phone: (518) 276 6858 email: keblip@rpi.edu

  2. Mechanics -> Static Mechanics Mechanics of rigid bodies Mechanics of deformable bodies Mechanics of fluids Kinematics Static Kinetics

  3. Lecture outline • Newton’s law. • Law of gravitation. • Basic concept of physical problem solving. • Fundamentals of vectors

  4. Newton’s Law of Motion • Law #1: Body is in equilibrium • SFi=0 • Body will remain in rest or continue to move with same speed and direction unless unbalanced force is acted on the body or particle.

  5. Law #2: Body not in equilibrium • SFi=0 • Change of motion of a body is proportional to the net force imposed on the body in the direction of the net force • Where: F is the external force acting on the body. • m is the mass of the body. • a is the acceleration of the body in the direction of the force. Newton’s Law of Motion F=ma

  6. Newton’s Law of Motion Law #3: The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense.

  7. 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

  8. 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: Me is the mass of the earth. re is the mean radius of the earth At sea level and latitude 450 g=32.17 ft/s2 = 9.807 m/s2

  9. Example 1 Me.mm r2 F=G M F=1.98*1017 kN r m Determine the gravitational force, in kilonewtons, exerted by the earth on the moon. G= 6.673*10-11 m3/(kg*s2) Me=5.976*1024 kg mm=7.35*1022 kg r=3.844*108 m

  10. Example 2 On earth Me.m r12 Me.m r22 W1=G W2=G Above earth W1/W2=r22/r12=2 D=r2-r1= (√2-1) r1= (√2-1)*6370 km = 2638km At what distance from the surface of the earth, in kilometers, is the weight of the body equal to one-half of its weight on the earth’s surface? W2 W1 D r1 earth The requirement: W2=1/2W1 r2=√2 r1

  11. Class assignment: Exercise set 1-1, 1-2, 1-17. please submit to TA at the end of the lecture 1-1) Calculate the mass m of a body that weighs 600 lb at the surface of the earth. 1-2) Calculate the weight W of a body at the surface of the earth if it has a mass m of 675 kg. 1-17) If a man weight 210 lb at sea level, determine the weight W of the man a) At the top of Mt. Everest (29,028 ft above sea level) b) In a satellite at an altitude of 200 mi. Solution: 1-1) M=18.65 slug 1-2) W=6.62 kN 1-17) a=209 lb b=190.3 lb

  12. Units of measurement • The U.S customary system of units (the British gravitational system) • Base units are foot (ft) for length, pound (lb) for force, and 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) supplementary units. • (3)derived units.

  13. Base units Supplementary units Derived units

  14. SI/U.S. customary units conversion

  15. Dimensional homogeneity • Equation does not depend on the units of measurement. • F=mg is valid whether the force is measured in newtons or lb and the mass in slugs, kilograms or grams, provided g is measured in the same units of length, time and mass as m and F. • The F=9.8m is not dimensional homogeneous since the equation applies only if the length is measured in meters and the time in second.

  16. Method of problem solving Statement of the problem Identification of physical principles Sketch and tabulation Mathematical modeling Solution Checking

  17. Class assignment: Exercise set 1-25 please submit to TA at the end of the lecture Determine the weight W, in U.S. Customary units, of an 85-kg steel bar under standard conditions (sea level at a latitude of 45 degrees). Solution: W=187.4 lb

  18. Scalar and vectors • A scalar quantity is completely described with only a magnitude (a number). • - Examples: mass, density, length, speed, time, temperature. • A vector quantity has both magnitude and direction and obeys the parallelogram law of addition. • - Examples: force, moment, velocity, acceleration.

  19. Vector Terminal point B A Initial point Direction of arrow direction of vector Length of arrow magnitude of vector

  20. F1 R F2 The sum of two vectors – geometrical representation • Two vectors can be vectorially added using the parallelogram law. • Position vector F1 so that its initial point coincides with the terminal point of F2. The vector F1+F2 is represented by the vector R.

  21. y x V = Vxi + Vyj Vectors in rectangular coordinate systems- two dimensional (v1,v2) v (v1,v2) are the terminal point of vector v

  22. y (v1+w1,v2+w2) v2 (w1,w2) w2 (v1,v2) x v1 w1 v+w=(v1+w1,v2+w2) v +w = (v1+ w1)i + (v2+ w2)j The sum of two vectors – analytic representation (two dimensional ) w v

  23. z (a1,a2,a3) a y b (b1,b2,b3) x a+b=(a1 +b1,a2+b2, a3 +b3) a +b = (a1+ b1)i + (a2+ b2)j+ (a3+ b3)k The sum of two vectors – rectangular components (Three dimensional )

  24. z P2(x2 ,y2 ,z2) P1(x1 ,y1 ,z1) w v y x Vectors with initial point not on the origin

  25. 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)

  26. Class assignment: Exercise set 3.1-3a-3c,3e,6a

  27. 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

  28. y x v = v12 + v22 The norm of a vector (the length of a vector) (v1,v2) v v2 v1 From the Theorem of Pythagoras it follows that the length of a vector is :

  29. a = a12 + a22 + a32 Three dimensional space z (a1,a2,a3) a a3 y a1 a2 x

  30. Example Let u=(2,-2,3), v=(1,-3,4), w=(3,6,-4). Evaluate the following expression: 3u-5v+w Solution 3u-5v+w =3(2,-2,3)-5(1,-3,4)+(3,6,-4) =(6,-6,9)-(5,-15,20)+(3,6,-4)= 3u-5v+w =(4,15,-25) 3u-5v+w = 42+152+(-15)2 =19.9

  31. Class assignment: Exercise set 3.2-3e

  32. A vector F in the positive n direction can be written as follows: F=|F|en=Fen Where enis the unit vector and can be written as follows: FFx Fy Fz en= = i + j + k FFFF Unit vector Any vector can be written as a product of its magnitude and a unit vector in the direction of the given vector.

  33. Class assignment: Find the magnitude of any unit vector

  34. Hanchen Stopped here on Aug 28, 2004

  35. Right-hand system The Cartesian coordinates axes are arranged as a right-hand system.

  36. Multiplication of Cartesian vectors • The scalar (or dot) product. • The vector (or cross) product.

  37. A q B Finding the rectangular scalar component of vector A along the x-axis 0< q <1800 Ax=A•i=A cos(qx) et y A Along any direction n An=A•en=A cos(qn) q en At=A-An The scalar (or dot) product The scalar product of two intersecting vectors is defined as the product of the magnitudes of the vectors and the cosine of the angle between them A•B=B•A=AB cos(q)

  38. Since i, j, k are orthogonal: i•j= j•k= k•j=(1)*(1)*cos(900)=0 i•i= j•j= k•k=(1)*(1)*cos(00)=1 • The scalar product of the two vectors written in Cartesian form are: • A•B = (Ax i + Ay j + Az k) • (Bx i + By j + Bz k) = • Ax Bx (i•i) + Ax By (i•j) + Ax Bz (i•k)+ Ay Bx (j•i) + Ay By (j•j) + Ay Bz (j•k)+ • Az Bx (k•i) + Az By (k•j) + Az Bz (k•k) Therefore: A•B = Ax Bx + Ay By + Az Bz

  39. A•B=AB cos(q) cos(q)= A•B/ AB A = 32+42 = 5 AB= 28.7 B = 22 +22 +52 = 5.74 cos(q)= 26/28.7 q= 25.10 Example Determine the angle q between the following vectors: A=3i +0j +4k and B=2i -2j +5k A•B=3*2+0*(-2)+4*5=26

  40. Class assignment: Exercise set 2-63 z 3 ft 3 ft 4 ft F1=900 lb 3 ft F2=700 lb 6 ft y x • Two forces are applied to an eye bolt as shown in Fig. P2-63. • Determine the x,y, and z scalar components of vector F1. • Express vector F1 in Cartesian vector form. • Determine the angle a between vectors F1 and F2. Fig. P2-63

  41. d1 = x12+ y12+ z12 d1 = (-6)2 +32 +72 z 3 ft 3 ft 4 ft F1=900 lb 3 ft F2=700 lb 6 ft y x Solution =9.7 ft F1x= F1 cos(qx) =900 *{(–6)/9.7}=-557 lb F1y= F1 cos(qy) =900 *{3/9.7}=278.5 lb F1z= F1 cos(qz) =900 *{7/9.7}=649.8 lb b) Express vector F1 in Cartesian vector form. F1 = -557i + 278.5j+ 649.8k lb

  42. d2 = x22+ y22+ z22 z 3 ft d2 = (-6)2 +62 +32 =9 ft 3 ft 4 ft F1=900 lb 3 ft F2=700 lb F1•e2= (-557)*(-2/3)+278.5*2/3+649.8*0.33=771.4lb 6 ft y x cos(a)=771.4/900 a=310 c) Determine the angle a between vectors F1 and F2. cos(a)= F1•e2/ F1 e2 F1 = -557i + 278.5j+ 649.8k lb e2 = -6/9i + 6/9j + 3/9k = -0.67 i + 0.67 j +0.33 k

  43. Orthogonal (perpendicular) vectors (w1 ,w2 ,w3) z w (v1 ,v2 ,v3) v y x wand v are orthogonal if and only ifw·v=0

  44. If u , v, and w are vectors in 2- or 3- space and k is a scalar, then: • u·v= v·u • u·(v+w)= u ·v + u·w • k(u·v)= (ku) ·v = u·(kv) • v·v> 0 if v=0, and v·v= 0 if v=0 Properties of the dot product

  45. F F B B F A Characteristics of a Force • A force is characterizes by the following: • (1) Magnitude, • (2) Direction, • (3) Point of application.

  46. y y qy F qx x F qz 5 m x z 4 m 7 m z Force Representation Two dimensions y y F F 6m q x x 5m Three dimensions

  47. Line of action Push Pull Principle of Transmissibility • The external effect of a force on a rigid body is the same for all points of application of the force along its line of action.

  48. Forces Classification • Contacting or surface forces • For example: push or pull • Body forces • For example: gravitational forces, magnetic forces.

  49. Concurrent forces • A force system is said to be concurrent if the action lines of all forces intersect at a common point

  50. Two Concurrent Forces • Any two concurrent forcesF1andF2acting on a body can be replaced by a single force, called the resultantR. • F1+F2=R • The two forces can be vectorially added using the parallelogram law. F1 R F2

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