Engineering Mechanics. Electrical Engineering. First stage

Transcription

1 Assist AliA. A Engineering Mechanics Electrical Engineering First stage Syllabus Static 1- General Princples 2- System of Forces 3- Composition and Resolution of Foreces Moments 5- Equilibrium 6- Trusses 7- Friction 8- Centroid and Center of Gravity 9- Moment of Inertia REFERENCFS ENGINEERING MECHANICS ENGINEERING MECHANICS HIGDON HIBBELER

2 GENERAL PRINCIPLES Chapter ((1)) Mechanics is a branch of physical sciences that is concerned with the state of rest or motion of bodies that are subjected to the action of forces. In general, this subject can be divided into three branches, Rigid-body mechanics, Deformable-body mechanics Fluid mechanics. Rigid-body mechanics is divided into two areas: 1. Static: deals with the equilibrium of bodies, that is, those that are at rest or move with a constant velocity. 2. Dynamics: is concerned with the accelerated motion of bodies. Fundamental Concepts ss, Basic Quantities: The basic quantities which used throughout mechanics are, Length, Time, Mass, and Force Force, fame is the action of one body to another. A force tends to move a body in the direction of its action. Note: A force is completely characterized by its magnitude, direction and point of application. Rigid body. A rigid body can be considered as a combination of a large number of particles in which all the particles remain at fixed distance from offearither. Weight. The weight of body is the gravitational force acting on it. Scalars and Vectors Scalar Quantities:- A scalar is a quantity that has magnitude only. Examples of scalar quantities include length, mass, and time. + Vector Quantities.- A vector is a quantity that possesses magnitude and direction. Examples of vector force, moments, displacements, velocity. Units of Measurement SI Units: The International System (SI) U.S. Customary: In the U. S. Customary system of units (or British System of units) (FPS) Name International system of US. Customary Conversion units(si) (FPS) Length Time Second See onki S,.-5.TOZAZ,tte? 15, - " 14- Mass Force Newton (N) Pound (lb) 1lb-4.448N

3 FOR E SYSTEM Forces:- A force may be defined as the action n one body on another body which changes or tend to change the motion bf the body acted on. Because of the inertia possessed by all material bodies, they react or oppose any force which acts on the (Newton's third low). Notel: F es may be considered as localized vectors and they can not be defined nless all the following characteristics men ned: Magnitude, Direction (sense and slope), Location of any p its line of action. Note2: The third characteristic shows that if two forces have e same di ection, they will produce the same external effect on a rigid boe This fact leads to the principle of transmissibility. So principle of transmissibility states that the external effect of a force in a rigid body is independent the point of application of the force along its line of action. System of Forces When several forces act in a given situation, they are called system of forces or force system. Force systems can be classified according to the arrangemen the lines of action of the forces of the system as follows: Collinear: All forces of the system have a common line of action. 3 Concurrent, Coplanar The action lines of all the forces of the system are in the same plane and intersect at a common point.

4 Parallel, Coplanar The action lines of all the forces of the system are parallel and lie in the same plane. If-7, F, -t- F, t C Nonconcurrent, Nonparallel, Coplanar: The action lines of all the forces of the system are in the same plane, but they are not all parallel and they do not intersect at a common point. Concurrent, Noncoplanar: The action lines of all the forces of the system are intersect at a common point, but they are not all in one plane. Parallel, Noncoplanar: The action lines of all the forces of the system are parallel and but they are not all in the same plane. Nonconcurent, Nonparallel, Noncoplanar: The action lines of all the forces of the system do not all intersect at a common point, they are not parallel, and they do not lie in the same plane. Resultant: The Resultant of a force system is the simplest force which can replace the original system without changing its external effect on a rigid body. The resultant of a force system can be: a single force, a pair of parallel forces having the same magnitudes but opposite sense (called a couple), or a force and a couple. If the resultant is a force and a couple, the force will not be parallel to the plane containing the couple. Composition and Resolution of Forces The process of replacing a force system by its resultant is called composition. The resultant of pair of concurrent forces can be determined by means of parallelogram law. R = /( F1) 2 + (F2) 2 2(F1 ). (F2)cos 0

5 The angle the resultant makes with either force can be determined by law of sines, for example: F 1 F2 R = sin a sin 0 sin 13 Cosine Law and Sine law: The cosine law and sine law are applicable to compute angles and sides of a triangle :- Law of cosines: C = V A2 + B 2 2A. B cos c zn A Law of sines : = sin a sin LI sin c I;?:\ Resolution: The process of replacing a force by its components is called resolution. A component of a force is any one of two or more forces having the given forces as a resultant. So the term "component' is used to mean either one of two concurrent forces or any one of three noncoplanar concurrent forces having the given iorce as a resunant -

6 Chapter Two Force,S) stem Resultant of concurrent forces: a) Resultant of concurrent, coplanar forces: The resultant of concurrent, coplanar forces is force only, while as we will be seeing later, the resultant of noncocurrent forces is force or moment or both. Example 1: The screw in Figure (a) is subjected to two forces, F1 and F2. Determine the magnitude and direction of the resultant force. (a) Solution: By parallelogram, the resultant FR is: FR V (100)(150)cos115 = ( ) A I 6 Zr 1-2(65") - 115" (1"1 0 /100 N 90' - 25' = 65 = 212.6N E_- 213 N Applying the law of sines to determine 6, sin 115 sin sin 49= = (sin 115 ) Thus, the direction 0 of FR, measured from the horizontal, is 0 = =54.8 (B) illeehanielstaticiti Ch m;

7 Chapter Tiro Force Sistem Example 2: Resolve the horizontal 600-lb force in Figure (a) into components acting along u and v axes and determine the magnitude of these components.,-ur 3o. / i iztr--- it I. i _ / r.- 126, 600 lb J/' (c) (al Solution: Applying the law of sines, tbl F. 600 sin 120 sin 30 F,, = 600 sin 30 sin 30 c===> Fu= 1039 lb F,7= 600 lb Note: The result for Fu shows that sometimes a component can have a greater magnitude than the resultant. Example 3: Determine the magnitude of the components force F in Figure (a) and the magnitude of the resultant force FR if FR is directed along the positive y- axis. FX 4 "S 45?'. 3fr 200 lb 200 lb _Ayr (a) (b) (c) Solution: The magnitude of FR and F can be determined by applying the law of sines, 200 sin 60 sin 45 F= 245 lb Y

8 Chapter Two Foire St ~torn FR 200 sin 75 sin 45 FR= 273 lb Example 4: It is required that the resultant force acting on the eyebolt in Figure (a) be directed along the positive x-axis and that F2 has a minimum magnitude. Determine this magnitude, the angle 0, and the corresponding resultant force. F,=SOON = SOO N r ~ F1 = SOON 1 \ I F. F = 90' 172 Solution: The magnitude of F2 is a minimum or the shortest length when its line of action is perpendicular to the line of action of FR, that is, when, 0=90 Since the vector addition now forms a right triangle, the two unknown magnitudes can be obtained by trigonometry. FR = 800 cos 60 =400 N F2 = 800 sin 60 =693 N (VS) llfechanic4statielf 4 Class

9 Rectangular Components When a force is resolved into two components along the X and Y axes, the components are then called rectangular Components. Let the force (F) with the direction 0, we can resolve this force into two components:- Horizontal component Fx-Fcos Vertical component Fy=Fsin 0 Where F _.1Fx2 Fy2 - x Fx=F cos ' tan 0 = Fy Fx Instead of using the angle 0, however, the direction of Fcan also be defined using a small "slope" triangle. Since this triangle and the larger triangle are similar, the proportional length of the sides gives Fx = F F y F II b - 0, a - --o- x cos0 Resultant of Several Coplanar Forces In determination of the resultant of several forces (more than two forces), using the rectangular component is more convenient than using the parallelogram rule more than once. Ft s. It, F. F, Fr, I h I

10 Consider three forces as shown in figure below. So the resultant of these coplanar forces may be determined by the following steps: 1. Resolve each force into xand ycomponents. 2. Add the respective using scalar algebra since they are collinear FRx- F = FiY --F-)x 4- -F3x E = 4 Foy F.3y 3. The resultant force is then computed by using Pythagorean theorem, R = JFRx 2 + FRy 2 And the angle 0, which specities the direction of resultant, is determined from trigonometry: tan 0 FRy r, Rx F:, FR, FR t F. N. (4) I '1 9/1. - q

11 Chapter Two Foire,S)-stein Example 5: Determine the x and y components of F1 and F1 acting on the boom shown in Figure (a). Solution: F1 is resolved into x and y components, Fu = -200 sin30 = -100 N = 100 N.-- Fly = 200 cos30 =173 N = 173 N The force F2 is resolved into its x and y components, F2 = ) = Similarly Fey = 260( 5 13 = 100 NI Hence; F2x = 240 N=240 F2y = -100 =100 NI F 12, = 260 (-13) N X 111 sy = 260 ( - 13-pr ** = 260 N (c) (so) Meehan ielstatie! Class

12 Chapter Two Fowl? SIsteni Example 6: The link in Figure (a) is subjected to two forces F1 and F2. Determine the magnitude and direction of the resultant force. Solution: FRx=y,Fx; Fit, 600 cos sin45 =236.8 FR),=/F),; FRy= 600 sin cos45 = The resultant force has a magnitude of, and its direction, FR =, = 629N F, (582.8) = tan-1h=tan, = r ( 1) MeehanielStatielld Class

13 Chapter Two Form fir stein Example 7: The end of the boom 0 in Figure (a) is subjected to three concurrent and coplanar forces. Determine the magnitude and direction of the resultant force. Solution: Each force is resolved into its x and y components, Figure (b). Summing the x components, we have Fi= 20P N FR,-=EFx ; Fkr= sin (45) = = N.-- The negative sign indicates that F acts to the left, i.e., in the negative x direction, as noted by small arrow. Summing the y components yields (b) +1 FRY=E-Fy; FRy= 250 cos (-35) =296.8 N The resultant force has a magnitude of, and its direction, FR =11(-383.2) = 485N F =tan-`( =tan-`i 383.2= 37.8 Note: Application of this method is more convenient, compared to using two applications of parallelogram law, first to add F1 and F2 then adding F3 to this resultant. 42) /Wiwi/auk IStatie Miss

14 Chapter Two Force System b) Resultant of Concurrent, Noncoplanar Forces: The resultant of concurrent, noncoplanar forces is force only, while as we will be seeing later, the resultant of noncocurrent forces is force or moment or both. Rectangular component in space (in three dimensions): Frequently it is convenient to resolve a force in space in three mutually perpendicular component parallel to three coordinate axes. The resultant force F could first be resolved into two components along ad and ae by means of parallelogram law, and the component ad could then be further resolved into components along ac and af. From the figure it is apparent that where Fx= Fcosa = Fcosa...(2-6a) Fy= FcosOy= Fcosfl P; = Fcosa = Fcosy F V F2 + Fy + F:...(2-6b)...(2-6c) a / r Fx x/c F, The angles a (or a), By, (or,a, and 9Z (or 7), are the angles between the resultant force F and the positive coordinate axes. The cosine of these angles are called direction cosine. The components can be also determined by setting the edges of a box represent component forces to the same scale as the diagonal representing the resultant force. The resultant force divided by the length of the diagonal gives a scale which, when multiplied by the length of an edge, gives the magnitude of the component along that edge. z le mechanicstaticip Class

15 Chapter Two Loire System Example 8: Express the F shown in Figure as a Cartesian components Solution: Since only two coordinate direction angles are specified, the third angle a must be determined from the equation; cos' a + cos2 cos2y= 1 cos2a + cos260 *cos245 = 1 cos a = 111 cos2 60" cos2 45 = ±0.5 Hence two possibilities exist, namely a = cos-i(0.5) = 60 or a = cos '(-0.5) = 1200 By inspection it is necessary that a = 60, since FY must be in +x direction. Fx = F cosa =200 cos60 = 100N Fy = F, cos/3 =200 cos60 = 100N FZ = Fl cosy=200 cos45 = 141.4N Show that indeed the magnitude of F = 200 N. Example 9: Determine the magnitude and coordinate direction angles of the resultant force acting on the ring in Figure (a). Solution: Resolve each force in the Cartesian F, =150 lb F1 =100 lb components. To do this calculate the proportional length 4 of each force,.14 r= Ax2 + Ay2 + Az2 (A) rl = V = 5 r2 V (_5) =7.5 And the components of resultant force are: 0.5 FR, = EFL, =_ 000)+ (150) = 50 lb Fey = EFy= 5 (100) + (150) = 40 lb FR, = EFiz =.1000)+_ 5 (150) =180 lb ihrhanie;statio; t' Class

16 Chapter %irn Pbree.51stent The magnitude of FR is, FR = (-40) =191.0 lb The coordinate direction angles a, /3, rare, cosa = cos )8 =- Fkr = = Ft, 191 = -40 = FR 191 a = 74.8" fi =102" cos- f = = = y =19.6" FR 191 Example 10: Express the force shown in Figure (a) as a Cartesian components it+) Solution: The angles of 60 and 45 defining the direction of F are not coordinate direction angles. Two successive applications of parallelogram law are needed to resolve F into its x, y, z components. IfX) lb First F= F' + F then F' = Fx + Fy, Figure (b). By trigonometry, the magnitudes of the components are, (a) = 100 sin60 = 86.6 lb F' = 100 cos60 = 50 lb F= 10011, F, = P cos45 = 100 cos45 = 35.4 lb Fy = -F sin sin45 = lb To show that the magnitude of this force indeed 100 ib, apply the following equation, F= + + = (-35.4) =100/b fi= 100lb The coordinate direction angles a, rare, F 35.4 cosa = = FR 100 a = 69.3" (c)

17 Chapter Two Foie System F 35.4 cost = = /3=111 F R 100 Example 11: The man shown in Figure (a) pulls on the cord with a force of 70 lb. Represent this force action on the support A as Cartesian components and determine its direction. Solution: Find the Cartesian projections of the r: Ax = x2 x =12 0 =12 ft Ay = y2 y,, = 0-8 = 8ft Az = z2 z = 30 6 = 24 ft r J(12)2 (_8)2 + (-24)2 = 28 ft and the Cartesian components equal: Fx = F = 12 *70 = 30 lb r 28 Fy = Ay F = 8 * 70 = 20 lb r 28 = Az F = 24 *70 = 60 lb r 28 The coordinate direction angles are measured between r (or F) and the positive axes of a localized coordinate system (a) F 70 lb with origin placed at A, Figure (b). a = cos _1( - 12 ) = fl =cos- 1( 8)= (b) y = cos-1( -24 = Cs) Meehan ielstatielr Class