Engineering Mechanics: Statics Course Overview


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1 Engineering Mechanics: Statics Course Overview Statics (Freshman Fall) Dynamics (Freshman Spring) Engineering Mechanics 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 Chapter 1 General Principles Basic quantities and idealizations of mechanics Newton s Laws of Motion and gravitation Principles for applying the SI system of units Chapter Outline Mechanics Fundamental Concepts The International System of Units Numerical Calculations General Procedure for Analysis 2
3 1.1 Mechanics Mechanics can be divided into:  Rigidbody Mechanics  Deformablebody Mechanics  Fluid Mechanics Rigidbody Mechanics deals with  Statics Equilibrium of bodies; at rest or 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 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 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 Addition of a System of Coplanar Forces 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 Vector Subtraction R = A B = A + (  B ) A 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 in 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 F F F F F F i j R Rx Ry Scalar notation 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 = (600cos30º  400sin45º)i + (600sin30º + 400cos45º)j = {236.8i j}N 9
10 2.5 Cartesian Vectors RightHanded Coordinate System A righthanded rectangular or Cartesian coordinate system. 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 A = A + A z A = A x + A y A = A x + A y + A z A A ua 10
11 2.5 Direction Cosines of a Cartesian Vector A x cos A A y cos A A z cos A ua A / A ( A / A ) i + ( A / A) j + ( A / A) k x y z 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. Since two angles are specified, the third angle is found by cos cos 2 2 cos cos cos cos 60 o 2 cos o 1 o ( 0. 5 ) 60 cos 1 o or cos By inspection, Given F = 200N Checking: = 60º since F x is in the +x direction F F cos i F cos j F cos k (200cos60 )i (200cos60 )j (200cos 45 )k i j141. 4k N x y z F F F F N 12
13 2.7 Position Vector: 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. r r F m 2m 6m 7m = 3/7i 2/7j 6/7k F 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) 4. Cartesian Vector Formulation  Dot product of 2 vectors A and B AB = AxBx AyBy AzBz 5. Applications  The angle formed between two vectors 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 = cos ( AB / ( A B )) The components of a vector parallel and perpendicular to a line Aa A cos Au A T u 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 freebody diagram for a particle Solve particle equilibrium problems using the equations of equilibrium Chapter Outline Condition for the Equilibrium of a Particle The FreeBody Diagram Coplanar Systems ThreeDimensional Force Systems 18
19 3.2 The FreeBody 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 3. Identify each of the forces 19
20 Example 3.1 The sphere has a mass of 6kg and is supported. Draw a freebody diagram of the sphere, the cord CE and the knot at C. FBD at Sphere Cord CE FBD at Knot 20
21 21
22 3.4 ThreeDimensional 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, 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 22
23 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 F B = F C = 23.6kN F D = 15.0kN 23
24 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 nonconcurrent force systems Reduce a simple distributed loading to a resultant force having a specified location Chapter Outline Moment of a Force Scalar Formation Moment of Force Vector Formulation Moment of a Force about a Specified Axis Moment of a Couple Simplification of a Force and Couple System Reduction of a Simple Distributed Loading 24
25 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 25
26 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 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
27 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 27
28 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 ( r F r F ) i ( r F r F ) j ( r F r F ) k y z z y x z z x x y y x 0 rz ry Fx rz 0 r x F y ry rx 0 Fz 28
29 r r AB AB rab or = r AB
30 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 30
31 4.4 Principles of Moments 4.5 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, u u u ax ay az M u ( r F) r r r a ax x y z F F F x y z 31
32 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
33 r CD F r OC r r CD CD M roc F M OA T 100 T 1 roa M roa
34 4.6 Moment of a Couple 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 34
35 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 35
36 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,then the 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 36
37 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 37
38 38
39 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 39
40 Example 4.21 Determine the magnitude and location of the equivalent resultant force acting on the shaft. For the differential area element, 2 da wdx 60x dx For resultant force 2 2 FR da 60x dx A x N For location of line of action, x 2 0 xda x(60 x ) dx A 0 0 x 1.5m da A
41 Chapter 5 Equilibrium of a Rigid Body Objectives Equations of equilibrium for a rigid body Concept of the freebody diagram for a rigid body Outline Conditions for Rigid Body Equilibrium FreeBody Diagrams Two and ThreeForce Members Equations of Equilibrium Constraints and Statical Determinacy 41
42 5.1 Conditions for RigidBody Equilibrium The equilibrium of a body is expressed as F R M R O F 0 M O 0 Consider summing moments about some other point, such as point A, we require MA r FR MR O 0 42
43 5.2 Free Body Diagrams 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. 43
44 5.2 Free Body Diagrams 44
45 5.2 Free Body Diagrams 45
46 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 46
47 Example 5.1 Draw the freebody diagram of the uniform beam. The beam has a mass of 100kg. FreeBody Diagram Support at A is a fixed wall Two forces acting on the beam at A denoted as A x, A y, with moment M A For uniform beam, Weight, W = 100(9.81) = 981N acting through beam s center of gravity 47
48 5.4 Two and ThreeForce Members When forces are applied at only two points on a member, the member is called a twoforce member Only force magnitude must be determined ThreeForce Members When subjected to three forces, the forces are concurrent or parallel 48
49 5.5 3D FreeBody Diagrams Types of Connection Reaction Number of Unknowns (1) F One unknown. The reaction is a force which acts away from the member in the known direction of the cable. cable (2) smooth surface support F One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. (3) One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. (4) roller F z F Three unknown. The reaction are three rectangular force components. ball and socked F y F x (5) single journal bearing M x F x M z F z Four unknown. The reaction are two force and two couple moment components which acts perpendicular to the shaft. Note: The couple moments are generally not applied of the body is supported elsewhere. See the example. 49
50 5.7 Constraints for a Rigid Body Redundant Constraints More support than needed for equilibrium Statically indeterminate: more unknown loadings than equations of equilibrium 50
51 5.7 Constraints for a Rigid Body Improper Constraints Instability caused by the improper constraining by the supports When all reactive forces are concurrent at this point, the body is improperly constrained 51
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