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Course Overview

Course Overview. Statics (Freshman Fall) Dynamics (Freshman Spring) Strength of Materials (Sophomore Fall) Mechanism Kinematics and Dynamics (Sophomore Spring ) Aircraft structures (Sophomore Spring and Junior Fall) Vibration(Senior). Engineering Mechanics.

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Course Overview

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  1. Course Overview Statics (Freshman Fall) Dynamics (Freshman Spring) Strength of Materials (Sophomore Fall) Mechanism Kinematics and Dynamics (Sophomore Spring ) Aircraft structures (Sophomore Spring and Junior Fall) Vibration(Senior) Engineering Mechanics Statics: force distribution on a system Dynamics: x(t)= f(F(t)) displacement as a function of time and applied force Strength of Materials: δ = f(F) deflection of deformable bodies subject to static applied force Vibration: x(t) = f(F(t)) displacement on particles and rigid bodies as a function of time and frequency

  2. Engineering Mechanics: Statics Chapter 1 General Principles • Chapter Objectives • Basic quantities and idealizations of mechanics • Newton’s Laws of Motion and Gravitation • Principles for applying the SI system of units • General guide for solving problems Chapter Outline • Mechanics • Fundamental Concepts • Units of Measurement • The International System of Units • Numerical Calculations • General Procedure for Analysis

  3. 1.1 Mechanics • Mechanics can be divided into 3 branches: - Rigid-body Mechanics - Deformable-body Mechanics - Fluid Mechanics • Rigid-body Mechanics deals with - Statics – Equilibrium of bodies, at rest, and Moving with constant velocity - Dynamics – Accelerated motion of bodies

  4. 1.2 Fundamentals Concepts Basic Quantities • Length - locate the position of a point in space • Mass - measure of a quantity of matter • Time - succession of events • Force - a “push” or “pull” exerted by one body on another Idealizations • Particle - has a mass and size can be neglected • Rigid Body - a combination of a large number of particles • Concentrated Force - the effect of a loading

  5. 1.2 Newton’s Laws of Motion • First Law - “A particle originally at rest, or moving in a straight line with constant velocity, will remain in this state provided that the particle is not subjected to an unbalanced force.” • Second Law - “A particle acted upon by an unbalanced forceFexperiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force.” • Third Law - “The mutual forces of action and reaction between two particles are equal and, opposite and collinear.”

  6. Chapter 2 Force Vector • Scalars and Vectors • Vector Operations • Vector Addition of Forces • Addition of a System of Coplanar Forces • Cartesian Vectors • Addition and Subtraction of Cartesian Vectors • Position Vectors • Force Vector Directed along a Line • Dot Product Chapter Outline

  7. 2.1 Scalar and Vector • Vector • – A quantity that has magnitude and direction, e.g. position, force and moment, presented as A and its magnitude (positive quantity) as • Scalar – A quantity characterized by a positive or negative number, indicated by letters in italic such as A, e.g. Mass, volume and length • Vector Subtraction R’ = A – B = A + ( - B ) • Finding a Resultant Force • Parallelogram law is carried out to find the resultant force FR = ( F1 + F2 )

  8. 2.4 Addition of a System of Coplanar Forces Scalar Notation • Components of forces expressed as algebraic scalars • Cartesian Vector Notation • Cartesian unit vectorsiandjare used to designate the x and y directions with unit vectorsiandj • Coplanar Force Resultants Coplanar Force Resultants Scalar notation

  9. Example 2.6 The link is subjected to two forces F1 and F2. Determine the magnitude and orientation of the resultant force. Cartesian Vector Notation F1 = { 600cos30°i+ 600sin30°j } N F2 = { -400sin45°i + 400cos45°j } N Thus, FR = F1 + F2 = (600cos30º - 400sin45º)i + (600sin30º + 400cos45º)j = {236.8i + 582.8j}N The magnitude and direction of FRare determined in the same manner as before.

  10. 2.5 Cartesian Vectors • Right-Handed Coordinate System A rectangular or Cartesian coordinate system is said to be right-handed provided: • Rectangular Components of a Vector • A vector A may have one, two or three rectangular components along the x, y and z axes, depending on orientation • By two successive applications of the parallelogram law • A = A’ + Az • A’ = Ax + Ay • Combing the equations, A can be expressed as • A = Ax + Ay + Az

  11. 2.5 Direction Cosines of a Cartesian Vector • Orientation of A is defined as the coordinate direction angles α, β and γ measured between A and the positive x, y and z axes, 0°≤ α, β and γ ≤ 180 ° • The direction cosines of A is

  12. a + b + g = 2 2 2 cos cos cos 1 a + + = 2 2 2 o o cos cos 60 cos 45 1 ( ) ( ) 2 2 a = - - = cos 1 0 . 5 0 . 707 0 . 5 ± ( ) - a = = 1 o cos 0 . 5 60 Example 2.8 Since two angles are specified, the third angle is found by Express the force F as Cartesian vector. or By inspection, β= 60º since Fx is in the +x direction Given F = 200N Checking:

  13. 2.7 Position Vector: Displacement and Force Position Displacement Vector • r = xi + yj + zk F can be formulated as a Cartesian vector

  14. Example 2.13 The man pulls on the cord with a force of 350N. Represent this force acting on the support A as a Cartesian vector and determine its direction. Solution:

  15. 2.9 Dot Product Cartesian Vector Formulation - Dot product of Cartesian unit vectors i·i = 1 j·j = 1 k·k = 1 i·j = 0 i·k = 1 j·k = 1 Laws of Operation 1. Commutative law A·B = B·A or ATB=BTA 2. Multiplication by a scalar a(A·B) = (aA)·B = A·(aB) = (A·B)a 3. Distribution law A·(B + D) = (A·B) + (A·D) 4. Cartesian Vector Formulation - Dot product of 2 vectors A and B 5. Applications - The angle formed between two vectors - The components of a vector parallel and perpendicular to a line

  16. Example 2.17 The frame is subjected to a horizontal force F = {300j} N. Determine the components of this force parallel and perpendicular to the member AB. Since Thus

  17. Solution Since result is a positive scalar, FAB has the same sense of direction as uB. Perpendicular component Magnitude can be determined from F┴ or from Pythagorean Theorem

  18. Chapter 3 Equilibrium of a Particle • Concept of the free-body diagram for a particle • Solve particle equilibrium problems using the equations of equilibrium Chapter Outline • Condition for the Equilibrium of a Particle • The Free-Body Diagram • Coplanar Systems • Three-Dimensional Force Systems Chapter Objectives

  19. 3.2 The Free-Body Diagram • Spring • Linear elastic spring: with spring constant or stiffness k.F = ks • Cables and Pulley • Cables (or cords) are assumed negligible weight and cannot stretch • Tension always acts in the direction of the cable • Tension force must have a constant magnitude for equilibrium • For any angle , the cable is subjected to a constant tension T Procedure for Drawing a FBD 1. Draw outlined shape 2. Show all the forces Active forces: particle in motion and Reactive forces: constraints that prevent motion 3. Identify each of the forces

  20. Example 3.1 The sphere has a mass of 6kg and is supported. Draw a free-body diagram of the sphere, the cord CE and the knot at C. FBD at Sphere Two forces acting, weight 6kg (9.81m/s2) = 58.9N and the force on cord CE. Cord CE Two forces acting: sphere and knot For equilibrium, FCE = FEC FBD at Knot 3 forces acting: cord CBA, cord CE and spring CD

  21. 3.4 Three-Dimensional Systems Example 3.7 Determine the force developed in each cable used to support the 40kN crate. FBD at Point A To expose all three unknown forces in the cables. Equations of Equilibrium Expressing each forces in Cartesian vectors, FB = FB(rB / rB) = -0.318FBi – 0.424FBj + 0.848FBk FC = FC(rC / rC) = -0.318FCi – 0.424FCj + 0.848FCk FD = FDi and W = -40k

  22. Solution For equilibrium, ∑F = 0; FB + FC + FD + W = 0 – 0.318FBi – 0.424FBj + 0.848FBk – 0.318FCi – 0.424FCj + 0.848FCk + FDi - 40k = 0 ∑Fx = 0; – 0.318FB – 0.318FC+ FD = 0 ∑Fy= 0; – 0.424FB – 0.424FC= 0 ∑Fz= 0; 0.848FB + 0.848FC– 40 = 0 Solving, FB = FC = 23.6kN FD = 15.0kN

  23. Chapter 4 Force System Resultants Chapter Objectives • Concept of moment of a force in two and three dimensions • Method for finding the moment of a force about a specified axis. • Define the moment of a couple. • Determine the resultants of non-concurrent force systems • Reduce a simple distributed loading to a resultant force having a specified location Chapter Outline • Moment of a Force – Scalar Formation • Cross Product • Moment of Force – Vector Formulation • Principle of Moments • Moment of a Force about a Specified Axis • Moment of a Couple • Simplification of a Force and Couple System • Further Simplification of a Force and Couple System • Reduction of a Simple Distributed Loading

  24. 4.1 Moment of a Force – Scalar Formation • Moment of a force about a point or axis – a measure of the tendency of the force to cause a body to rotate about the point or axis • Torque – tendency of rotation caused by Fx or simple moment (Mo)z • Magnitude • For magnitude of MO, MO = Fd (Nm) • where d = perpendicular distance from O to its line of action of force • Direction • Direction using “right hand rule” Resultant Moment MRo = ∑Fd

  25. 4.2 Cross Product C = A x B = (ABsinθ)uC • Cross product of two vectors A and B yields C, which is written as Laws of Operations 1. Commutative law is not valid A x B ≠ B x A Rather, A x B = - B x A 2. Multiplication by a Scalar a( A x B ) = (aA) x B = A x (aB) = ( A x B )a 3. Distributive Law A x ( B + D ) = ( A x B ) + ( A x D ) Proper order of the cross product must be maintained since they are not commutative

  26. 4.2 Cross Product Cartesian Vector Formulation • Use C= AB sinθ on a pair of Cartesian unit vectors • A more compact determinant in the form as • Moment of force F about point O can be expressed using cross product • MO = r x F • For magnitude of cross product, • MO = rFsinθ • Treat ras a sliding vector. Since d = r sinθ, • MO = rFsinθ = F (rsinθ) = Fd

  27. 4.3 Moment of Force - Vector Formulation For force expressed in Cartesian form, with the determinant expended,

  28. Example 4.4 The resultant moment about O is Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector.

  29. 4.4 Principles of Moments • Since F = F1 + F2, MO = r x F = r x (F1 + F2) = r x F1 + r x F2 4.5 Moment of a Force about a Specified Axis • Vector Analysis • For magnitude of MA, • MA = MOcosθ = MO·ua • where ua = unit vector • In determinant form,

  30. Example 4.8 Determine the moment produced by the force Fwhich tends to rotate the rod about the AB axis.

  31. Solution:

  32. 4.6 Moment of a Couple • Couple • two parallel forces of the same magnitude but opposite direction separated by perpendicular distance d • Resultant force = 0 Scalar Formulation • Magnitude of couple moment M = Fd • M acts perpendicular to plane containing the forces Vector Formulation • For couple moment, M = r x F • Equivalent Couples • 2 couples are equivalent if they produce the same moment • Forces of equal couples lie on the same plane or plane parallel to one another

  33. Example 4.12 Determine the couple moment acting on the pipe. Segment AB is directed 30° below the x–y plane. Take moment about point O, M = rA x (-250k) + rB x (250k) = (0.8j) X (-250k) + (0.66cos30ºi + 0.8j – 0.6sin30ºk) x (250k) = {-130j}N.cm Take moment about point A M = rAB x (250k) = (0.6cos30°i – 0.6sin30°k) x (250k) = {-130j}N.cm Take moment about point A or B, M = Fd = 250N(0.5196m) = 129.9N.cm M acts in the –j direction M = {-130j}N.cm

  34. 4.7 Simplification of a Force and Couple System • Equivalent resultant force acting at point O and a resultant couple moment is expressed as • If force system lies in the x–y plane and couple moments are perpendicular to this plane,

  35. 4.8 Simplification of a Force and Couple System Concurrent Force System • A concurrent force system is where lines of action of all the forces intersect at a common point O • Coplanar Force System • Lines of action of all the forces lie in the same plane • Resultant force of this system also lies in this plane

  36. 4.8 Simplification of a Force and Couple System • Consists of forces that are all parallel to the z axis

  37. 4.9 Distributed Loading • Magnitude of Resultant Force • Magnitude of dF is determined from differential area dA under the loading curve. • For length L, • Large surface area of a body may be subjected to distributed loadings, often defined as pressure measured in Pascal (Pa): 1 Pa = 1N/m2 • Location of Resultant Force • dF produces a moment of xdF =x w(x) dx about O • Solving for

  38. Example 4.21 Determine the magnitude and location of the equivalent resultant force acting on the shaft. Solution For the colored differential area element, For resultant force For location of line of action,

  39. Chapter 5 Equilibrium of a Rigid BodyObjectives • Develop the equations of equilibrium for a rigid body • Concept of the free-body diagram for a rigid body • Solve rigid-body equilibrium problems using the equations of equilibrium Outline • Conditions for Rigid Equilibrium • Free-Body Diagrams • Equations of Equilibrium • Two and Three-Force Members • Free Body Diagrams • Equations of Equilibrium • Constraints and Statical Determinacy

  40. 5.1 Conditions for Rigid-Body Equilibrium • The equilibrium of a body is expressed as • Consider summing moments about some other point, such as point A, we require

  41. 5.2 Free Body Diagrams Support Reactions • If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction. • If rotation is prevented, a couple moment is exerted on the body.

  42. 5.2 Free Body Diagrams (1) cable (2) θ θ F θ θ θ weightless link F F (3) θ roller θ F (4) roller or pin in confined smooth slot F F θ θ θ (5) θ θ rocker F

  43. 5.2 Free Body Diagrams (6) smooth contacting surface θ θ F (7) member pin connected F F θ θ θ (8) Fy F smooth pin or hinge φ Fx θ (9) Member fixed connected to collar on smooth rod M F Fy (10) F φ fixed support Fx M

  44. 5.2 Free Body Diagram Weight and Center of Gravity • Each particle has a specified weight • System can be represented by a single resultant force, known as weight W of the body • Location of the force application is known as the center of gravity

  45. Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg. Solution • Free-Body Diagram • Support at A is a fixed wall • Two forces acting on the beam at A denoted as Ax, Ay, with moment MA • Unknown magnitudes of these vectors • For uniform beam, • Weight, W = 100(9.81) = 981N • acting through beam’s center of gravity, 3m from A

  46. 5.4 Two- and Three-Force Members Two-Force Members • When forces are applied at only two points on a member, the member is called a two-force member • Only force magnitude must be determined Three-Force Members When subjected to three forces, the forces are concurrent or parallel

  47. 5.5 3D Free-Body Diagrams (1) F cable (2) smooth surface support F (3) roller F (4) Fz Fy Fx ball and socked Mz (5) Fz Fx Mx single journal bearing

  48. 5.5 Free-Body Diagrams Mz Mz (6) Fz Fz Fx My single journal bearing with square shaft Mx Mx (7) Fy single thrust bearing Mz (8) Fz My single smooth pin Fx Fy Mz (9) Fz Fy My Fx single hinge Mx Mz (10) Fz Fy Fx My fixed support Mx

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