Course Overview. Statics (Freshman Fall) Dynamics: x(t)= f(f(t)) displacement as a function of time and applied force

Transcription

1 Course Overview Statics (Freshman Fall) Engineering Mechanics Dynamics (Freshman Spring) Strength of Materials (Sophomore Fall) Mechanism Kinematics and Dynamics (Sophomore Spring ) Aircraft structures (Sophomore Spring and Junior Fall) Vibration(Senior) Statics: F 0 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 1

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 2

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 3

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 Particles - 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 4

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 force F experiences 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. F ma 5

6 Chapter 2 Force Vector Chapter Outline 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 6

7 2.1 Scalar and Vector 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 A quantity that has magnitude and direction, e.g. Position, force and moment, presented as A and its magnitude (positive quantity) as A Vector Subtraction R = A B = A + ( - B ) Finding a Resultant Force Parallelogram law is carried out to find the resultant force F R = ( F 1 + F 2 ) 7

8 2.4 Addition of a System of Coplanar Forces Scalar Notation Components of forces expressed as algebraic scalars F x F cos and F y F sin Cartesian Vector Notation Cartesian unit vectors i and j are used to designate the x and y directions with unit vectors i and j F Fi F j x y Coplanar Force Resultants F F i F j 1 1x 1y F F i F j 2 2x 2 y F F i F j 3 3x 3 y Coplanar Force Resultants Scalar notation F F F F F F i j R Rx Ry F F Rx Ry F F 1x 1y F F 2x 2 y F F 3x 3y 8

9 Example 2.6 The link is subjected to two forces F 1 and F 2. Determine the magnitude and orientation of the resultant force. Cartesian Vector Notation F 1 = { 600cos30 i + 600sin30 j } N F 2 = { -400sin45 i + 400cos45 j } N Thus, F R = F 1 + F 2 = (600cos30ºN - 400sin45ºN)i + (600sin30ºN + 400cos45ºN)j = {236.8i j}N The magnitude and direction of F R are determined in the same manner as before. 9

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 application of the parallelogram law A = A + A z A = A x + A y Combing the equations, A can be expressed as A = A x + A y + A z A A ua 10

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 A x cos A A y cos A A z cos A ua A / A ( Ax / A ) i + ( Ay / A) j + ( Az / A) k ua cosi + cos j + cos k cos + cos + cos = 1 A A u A A cosi + A cos j + A cos k Ax i + Ay j + Az k 11

12 Example 2.8 Express the force F as Cartesian vector. Solution : Since two angles are specified, the third angle is found by cos cos cos 1 cos 2 cos cos o cos 45 2 ( 0. 5 ) ( ) 2 o 1 2 ± 0. 5 o cos 1 o ( ) or cos By inspection, β= 60º since F x is in the +x direction Given F = 200N F F cos i+ Fcos jf cos k =(200cos60 N) +(200cos60 N) +(200cos45 N) ={100.0i j k}n Checking: i j k F F 2 x F 2 y F 2 z N 12

13 2.7 Position Vectors: Displacement and Force Position Displacement Vector r = xi + yj + zk F can be formulated as a Cartesian vector F = F u = F ( r / r ) 13

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: r m 2m 6m 7m 14

15 2.9 Dot Product Laws of Operation 1. Commutative law A B = B A or A T B=B T A 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) 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 Applications - The angle formed between two vectors cos ( AB / ( A B )) The components of a vector parallel and perpendicular to a line Aa A cos Au 15

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 u B Thus B B 2i 6 j 3k i j 0.429k F AB r r j i j k Fu B (0)(0.286) (300)(0.857) (0)(0.429) 257.1N F cos 16

17 Solution Since result is a positive scalar, F AB has the same sense of direction as u B. F F u AB AB AB 257.1N 0.286i j 0.429k {73.5i 220 j 110 k} N Perpendicular component F F F 300 j (73.5i 220 j 110 k) { 73.5i 80 j 110 k} N AB Magnitude can be determined from F or from Pythagorean Theorem F F FAB 300N 257.1N 155N 17

18 Chapter 3 Equilibrium of a Particle Chapter Objectives 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 18

19 3.1 Equilibrium and the Free-Body Diagram Best representation of all the unknown forces ( F) which acts on a body A sketch showing the particle free from the surroundings with all the forces acting on it 19

20 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 forces 20

21 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/s 2 ) = 58.9N and the force on cord CE. Cord CE Two forces acting: sphere and knot For equilibrium, F CE = F EC FBD at Knot 3 forces acting: cord CBA, cord CE and spring CD 21

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23 3.4 Three-Dimensional Systems For particle equilibrium F = 0 Resolving into i, j, k components F x i + F y j + F z k = 0 Procedure for Analysis Free-body Diagram -Establish the x, y, z axes -Label all known and unknown force Equations of Equilibrium - Apply F x = 0, F y = 0 and F z = 0 - Negative results indicate that the sense of the force is opposite to that shown in the FBD. 23

24 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, F B = F B (r B / r B ) = F B i 0.424F B j F B k F C = F C (r C / r C ) = F C i 0.424F C j F C k F D = F D i and W = -40k 24

25 Solution For equilibrium, F = 0; F B + F C + F D + W = F B i 0.424F B j F B k 0.318F C i 0.424F C j F C k + F D i - 40k = 0 F x = 0; 0.318F B 0.318F C + F D = 0 F y = 0; 0.424F B 0.424F C = 0 F z = 0; 0.848F B F C 40 = 0 Solving, F B = F C = 23.6kN F D = 15.0kN 25

26 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 26

27 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 F x or simple moment (M o ) z Magnitude For magnitude of M O, M O = Fd (Nm) where d = perpendicular distance from O to its line of action of force Direction Direction using right hand rule Resultant Moment M Ro = Fd 27

28 4.2 Cross Product Cross product of two vectors A and B yields C, which is written as C = A x B = (AB sinθ)u C 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 28

29 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 AB i j k A A A x y z B B B x y z Moment of force F about point O can be expressed using cross product M O = r x F For magnitude of cross product, M O = rf sinθ Treat r as a sliding vector. Since d = r sinθ, M O = rf sinθ = F (rsinθ) = Fd 29

30 4.3 Moment of Force - Vector Formulation For force expressed in Cartesian form, i j k MO r F rx ry r z F x Fy Fz with the determinant expended, M0 ( ry Fz rz Fy ) i ( rx Fz rz Fx ) j ( rx Fy ry Fx ) k 0 rz ry Fx rz 0 r x F y ry rx 0 Fz 30

31 r r AB AB rab or = r AB

32 Example 4.4 Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector. r A 5 j m rb 4i 5 j 2 k m The resultant moment about O is M r F r F r F O A B i j k i j k i 40 j 60 k kn m 32

33 4.4 Principles of Moments Since F = F 1 + F 2, M O = r x F = r x (F 1 + F 2 ) = r x F 1 + r x F Moment of a Force about a Specified Axis Vector Analysis For magnitude of M A, M A = M O cosθ = M O u a where u a = unit vector In determinant form, M u ( r F) u u u ax ay az r r r a ax x y z F F F x y z 33

34 Example 4.8 Determine the moment produced by the force F which tends to rotate the rod about the AB axis rc 0, 0 F M r F rb 0.2, B 0.2 u T M AB ub M

35 r CD F r Solution: OC r r CD CD T M roc F T MOA roa M roa

36 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 36

37 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 = r A x (-250k) + r B x (250k) = (0.8j) X (-250k) + (0.66cos30ºi + 0.8j 0.6sin30ºk) x (250k) = {-130j}N.cm Take moment about point A M = r AB 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 37

38 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, R R O O F F M M M R x x R y y R O O F F F F M M M 38

39 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 FR F 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 39

40 4.8 Simplification of a Force and Couple System Consists of forces that are all parallel to the z axis 40

41 4.9 Distributed Loading Large surface area of a body may be subjected to distributed loadings, often defined as pressure measured in Pascal (Pa): 1 Pa = 1N/m 2 Magnitude of Resultant Force Magnitude of df is determined from differential area da under the loading curve. For length L, F wxdx da A Location of Resultant Force R L df produces a moment of xdf = x w(x) dx about O xf R xw( x) dx L Solving for x A x L L xw( x) dx w( x) dx 41

42 Example 4.21 Determine the magnitude and location of the equivalent resultant force acting on the shaft. Solution For the colored differential area element, 2 da wdx 60x dx For resultant force F R F; 2 2 FR da 60x dx A x N For location of line of action, x x 2 0 x(60 x ) dx A 0 0 A xda da 1.5m 42