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APPLIED MECHANICS

APPLIED MECHANICS. By : Amit Phogat (lecturer in M.E.). ENGINEERING MECHANICS. CHAPTER TOPIC 1 INTRODUCTION AND REVIEW OF MATHEMATICS 2 FORCES AND RESULTANTS

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APPLIED MECHANICS

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  1. APPLIED MECHANICS By : Amit Phogat (lecturer in M.E.)

  2. ENGINEERING MECHANICS CHAPTERTOPIC 1 INTRODUCTION AND REVIEW OF MATHEMATICS 2 FORCES AND RESULTANTS 3 MOMENTS AND COUPLES 4 EQUILIBRIUM 5 FRICTION 6     KINEMATICS OF LINEAR & ROTATIONAL MOTION 7 KINETICS OF LINEAR MOTION 8 KINETICS OF ROTATIONAL MOTION Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 1 of 19 EE31102005

  3. CHAPTER 1 :Introduction And Review of Mathematics 1.1 Introduction Basic mechanics involves the study of two principal areas – statics and dynamics. Statics is the study of forces on objects or bodies which are at rest or moving at a constant velocity, and the forces are in balance, or in staticequilibrium. A ball at rest may have several forces acting on it, such as gravitational force (weight) and a force opposing that gravity (reaction). The ball is at rest or static, has forces in balance or EQUILIBRIUM Dynamics is the study of forces on moving bodies, and the forces are in dynamic equilibrium. Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 2 of 19 EE31102005

  4. The concept of applied mechanics is useful when it comes to analyzing stress, designing of machine structures and hydraulics, etc. Physical QuantityUnitSymbol 1. Length meter m 2. Mass kilogram kg 3. Time second s There are only seven basic units in the SI system but only three are frequently used in statics and dynamics: Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 3 of 19 EE31102005

  5. For large or small figures, multiples or submultiples are used. For example: Multiples Submultiples 1 kilogram is 1 kg or 103 g. 1 millimeter is 1 mm or 10-3 m. 1 megagram is 1 Mg or 106 g. 1 micrometer is 1 m or 10-6m. 1 gigagram is 1 Gg or 109g. 1 nanometer is 1 nm or 10-9m. Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 4 of 19 EE31102005

  6. The following SI derived units are frequently used in this course: Force – The unit of force is the newton (N) .1 newton is the force applied to a 1 kg mass to give it an acceleration of 1 m/s2 (i.e 1 N = 1 kg.m/s2).Or : Force = mass x acceleration = kg x m/s2 = kg m/s2Hence a 1 kg mass has a force or weight due to gravity, equal to (1 kg x 9.81 m/s2) = 9.81 N, where g = 9.81 m/s2 . Moment – It is the product of a force and its perpendicular distance, and the unit is newton-meter or N-m. Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 5 of 19 EE31102005

  7. 1.2 Mathematics RequiredThe followings are the mathematics skills that are important for this module : • Quadratic equations • Simultaneous equations • Trigonometry functions of a right-angle triangle • Sine and cosine rules Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 6 of 19 EE31102005

  8. 1.2.1 Quadratic Equations The equation has the standard form as follows ax2 + bx + c = 0 (1.1) The standard solution to this equation is x = – b  ( b2 – 4ac) (1.2) 2a Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 7 of 19 EE31102005

  9. Example 1 Solve for x in the equation 5x2 + 12x – 2 = 0.Comparing equation 1.1 above, and substitute a=5, b=12 and c= –2 into equation 1.2, the solution is : X = – 12  [(12)2 – 4(5)( – 2)] 2(5) = – 12  13.56 10 = +0.156 or – 2.56 Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 8 of 19 EE31102005

  10. 1.2.2 Simultaneous Equation The equation has two unknowns x and y in the form of ax + by + c = 0 (1.3) px + qy + r = 0 (1.4)Example 2Find the values of x and y satisfying the given equations: 4x + 3y + 10 = 0 (1)20x + 30y + 5 = 0 (2)There are 2 methods of solving these equations Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 9 of 19 EE31102005

  11. Method of Substitution We can start by expressing x in terms of y, or y in terms of x. Let us choose to express x in terms of y, thus from (1) x = -3y -10 (3) 4Substituting (3) into (2) , yielding 20( -3y -10) + 30y + 5 = 0 4 -15y -50 + 30y + 5 = 015y – 45 = 0y = 45 = 315To find x, substitute the value of y into (3)x = (-3 x 3 - 10) = -19 4 4 Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 10 of 19 EE31102005

  12. Method of EliminationThis method looks for a way to eliminate one of the unknowns. This can be done by making the constant factor of that unknown or variable the same in both equations by multiplying or dividing one equation by a selected constant: Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 11 of 19 EE31102005

  13. 4x + 3y + 10 = 0 (1)20x + 30y + 5 = 0 (2) Divide (2) by 5 4x + 6y + 1 = 0 (3) Subtract (3) by (1) 3y - 9 = 0 y = 3 Substitute the value of y into (1) or (2) 4x + 3(3) + 10 = 0 4x = - 9 - 10 x = - 19 4 Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 12 of 19 EE31102005

  14. 1.2.3 Trigonometry Functions Of A Right-Angle Triangle  sine  = opposite side = o = cosine  (1.5) hypotenuse h cosine  = adjacent side = a = sine  (1.6) hypotenuse h tangent  = opposite side = o (1.7) adjacent side a tangent  = sin cos h o  a  Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 13 of 19 EE31102005

  15. For triangles that are not right-angle, the following two laws are important in vector algebra introduced in chapter two later: 1.2.4 Sine And Cosine Rules Cosine Rule a2 = b2 + c2 – 2bc cos  (1.8) b2 = a2 + c2 – 2ac cos  c2 = a2 + b2 – 2ab cos  Sine Rule a = b = c sin  sin  sin  Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 14 of 19 EE31102005

  16. If the cosine rule is applied to a right-angle triangle where  = 90 0 , and applying to equation (1.8), c a 90 0 b i.e. a2 = b2 + c2 – 2bc cos 90 0 since cos 90 0 = 0 a2 = b2 + c2 (Pythagoras Theorem) Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 15 of 19 EE31102005

  17. Example 3Find the length of the unknown side a and the angle . Cosine rule : a2 = b2+c2-2bccos 200 6m 4m  i.e. a2 = 62+42-2x6x4cos200 = 36 +16-6x4xcos200 = 6.9 a = 2.63m Sine rule : 2.63 = 6 sin 200 sin  a sine = 6 x sin 200 2.63 = 51.30 Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 16 of 19 EE31102005

  18. But we know this to be in the second quadrant,Hence  = 180 – 51.4 = 128.6 0 Check : 62 = 2.632 - 42 - 2x2.63x4 cos  cos  = 2.632 + 42 – 62 2x2.63x4 = - 0.634  = 128.6 0 Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 17 of 19 EE31102005

  19. 1.2.5 Geometry Some of the basic rules are shown below: Sum of supplementary angles = 180 0  +  = 180 0   A straight line intersecting two parallel lines  = ,  =   = ,  =      Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 18 of 19 EE31102005

  20. Similar triangles ABC and ADE, by proportion AB = BC = AC AD DE AE Hence if AB = 6, AD = 3 and BC = 4, Then, 6 = 4 3 DE DE = (3 x 4) 3 = 2 B D A C E End of Chapter 1 Engineering Mechanics – Chapter 1: Introduction & Review of Mathematics Slide 19 of 19 EE31102005

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