Materials Engineering Department Subject: Engineering Mechanics Class: First Lecturer: Dr. Emad AL-Hassani Date : 2009-2010 Lecture # 1 1.1. WHAT IS MECHANICS? Mechanics can be defined as that science which describes and predicts the conditions of rest or motion of bodies under the action of forces. It is divided into three parts: mechanics of rigid bodies, mechanics of deformable bodies, and mechanics of fluids. The mechanics of rigid bodies is subdivided into statics and dynamics, the former dealing with bodies at rest, the latter with bodies in motion. In this part of the study of mechanics, bodies are assumed to be perfectly rigid. Actual structures and machines, however, are never absolutely rigid and deform under the loads to which they are subjected. But these deformations are usually small and do not appreciably affect the conditions of equilibrium or motion of the structure under consideration. They are important, though, as far as the resistance of the structure to failure is concerned and are studied in mechanics of materials, which is a part of the mechanics of deformable bodies. The third division of mechanics, the mechanics of fluids, is subdivided into the study of incompressible fluids and of compressible fluids. An important subdivision of the study of incompressible fluids is hydraulics, which deals with problems involving water. Mechanics is a physical science, since it deals with the study of physical phenomena. However, some associate mechanics with mathematics, while many consider it as an engineering subject. Both these views are justified in part. Mechanics is the foundation of most engineering sciences and is an indispensable prerequisite to their study. 1 However, it does not have the empiricism found in some engineering sciences, i.e., it does not rely on experience or observation alone; by its rigor and the emphasis it places on deductive reasoning it resembles mathematics. But, again, it is not an abstract or even a pure science; mechanics is an applied science. The purpose of mechanics is to explain and predict physical phenomena and thus to lay the foundations for engineering applications. 1.2. FUNDAMENTAL CONCEPTS AND PRINCIPLES The basic concepts used in mechanics are space, time, mass, and force. These concepts cannot be truly defined; they should be accepted on the basis of our intuition and experience and used as a mental frame of reference for our study of mechanics. The concept of space is associated with the notion of the position of a point P. The position of P can be defined by three lengths measured from a certain reference point, or origin, in three given directions. These lengths are known as the coordinates of P. To define an event, it is not sufficient to indicate its position in space. The time of the event should also be given. The concept of mass is used to characterize and compare bodies on the basis of certain fundamental mechanical experiments. Two bodies of the same mass, for example, will be attracted by the earth in the same manner; they will also offer the same resistance to a change in translational motion. A force represents the action of one body on another. It can be exerted by actual contact or at a distance, as in the case of gravitational forces and magnetic forces. A force is characterized by its point of 2 application, its magnitude, and its direction; a force is represented by a vector . In Newtonian mechanics, space, time, and mass are absolute concepts, independent of each other. (This is not true in relativistic mechanics, where the time of an event depends upon its position, and where the mass of a body varies with its velocity.) On the other hand, the concept of force is not independent of the other three. Indeed, one of the fundamental principles of Newtonian mechanics listed below indicates that the resultant force acting on a body is related to the mass of the body and to the manner in which its velocity varies with time. You will study the conditions of rest or motion of particles and rigid bodies in terms of the four basic concepts we have introduced. By particle we mean a very small amount of matter which may be assumed to occupy a single point in space. A rigid body is a combination of a large number of particles occupying fixed positions with respect to each other. The study of the mechanics of particles is obviously a prerequisite to that of rigid bodies. Besides, the results obtained for a particle can be used directly in a large number of problems dealing with the conditions of rest or motion of actual bodies. The study of elementary mechanics rests on six fundamental principles based on experimental evidence. The Parallelogram Law for the Addition of Forces This states that two forces acting on a particle may be replaced by a single force, called their resultant, obtained by drawing the diagonal of the parallelogram which has sides equal to the given forces. 3 The Principle of Transmissibility This states that the conditions of equilibrium or of motion of a rigid body will remain unchanged if a force acting at a given point of the rigid body is replaced by a force of the same magnitude and same direction, but acting at a different point, provided that the two forces have the same line of action. Newton’s Three Fundamental Laws. Formulated by Sir Isaac Newton in the latter part of the seventeenth century, these laws can be stated as follows: FIRST LAW If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in motion) . SECOND LAW If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant force., this law can be stated as: F=m a ……. (1.1) Where F, m, and a represent, respectively, the resultant force acting on the particle, the mass of the particle, and the acceleration of the particle, expressed in a consistent system of units. 4 THIRD LAW The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense . Newton’s Law of Gravitation This states that two particles of mass M and m are mutually attracted with equal and opposite forces F and −F (Fig. 1.1) of magnitude F given by the formula F=G (Mm/r2)…….. (1.2) Where r = distance between the two particles G = universal constant called the constant of gravitation Newton’s law of gravitation introduces the idea of an action exerted at a distance and extends the range of application of Newton’s third law: the action F and the reaction −F in Fig. 1.1 are equal and opposite, and they have the same line of action. Fig: (1.1) A particular case of great importance is that of the attraction of the earth on a particle located on its surface. The force F exerted by the earth 5 on the particle is then defined as the weight W of the particle. Taking M equal to the mass of the earth, m equal to the mass of the particle, and r equal to the radius R of the earth, and introducing the constant g = GM ……… (1.3) The magnitude W of the weight of a particle of mass m may be expressed as W =mg ………. (1.4) The value of R in formula (1.3) depends upon the elevation of the point considered; it also depends upon its latitude, since the earth is not truly spherical. The value of g therefore varies with the position of the point considered. As long as the point actually remains on the surface of the earth, it is sufficiently accurate in most engineering computations to assume that g equals 9.81 m/s2 or 32.2 ft/s2. 1.3. SYSTEMS OF UNITS With the four fundamental concepts introduced in the preceding section are associated the so-called kinetic units, i.e., the units of length, time, mass, and force. These units cannot be chosen independently if Eq. (1.1) is to be satisfied. Three of the units may be defined arbitrarily; they are then referred to as basic units. The fourth unit, however, must be chosen in accordance with Eq. (1.1) and is referred to as a derived unit. Kinetic units selected in this way are said to form a consistent system of units. 6 International System of Units (SI Units) In this system, which will be in universal use after the United States has completed its conversion to SI units, the base units are the units of length, mass, and time, and they are called, respectively, the meter (m), the kilogram (kg), and the second (s). All three are arbitrarily defined. The second, which was originally chosen to represent 1/86 400 of the mean solar day, is now defined as the duration of 9 192 631 770 cycles of the radiation corresponding to the transition between two levels of the fundamental state of the cesium-133 atom. The meter, originally defined as one ten-millionth of the distance from the equator to either pole, is now defined as 1 650 763.73 wavelengths of the orange-red light corresponding to a certain transition in an atom of krypton-86. The kilogram, which is approximately equal to the mass of 0.001 m3 of water, is defined as the mass of a platinum-indium standard kept at the international Bureau of Weights and Measures at Sevres, near Paris, France. The unit of force is a derived unit. It is called the Newton (N) and is defined as the force which gives an acceleration of 1 m/s2 to a mass of 1 kg (Fig. 1.2). From Eq. (1.1) we write 1 N = (l kg) (l m/s2) = l kg.m/s2 …….. (1.5) 7 The SI units are said to form an absolute system of units. This means that the three base units chosen are independent of the location where measurements are made. The meter, the kilogram, and the second may be used anywhere on the earth; they may even be used on another planet. They will always have the same significance. The weight of a body, or the force of gravity exerted on that body, should, like any other force, be expressed in newtons. From Eq. (1.4) it follows that the weight of a body of mass 1 kg (Fig. 1.3) is W =mg = (1 kg) (9.81 m/s2) =9.81N Multiples and submultiples of the fundamental SI units may be obtained through the use of the prefixes defined in Table 1.1. The multiples and submultiples of the units of length, mass, and force most frequently used in engineering are, respectively, the kilometer (km) and 8 the millimeter (mm); the megagram (Mg) and the gram (g); and the kilonewton (kN). According to Table 1.1, we have 1 km =1000 m 1 mm = 0.001 m l Mg= 1000kg 1 g = 0.001 kg 1kN = 1000 N The conversion of these units into meters, kilograms, and newtons, respectively, can be effected by simply moving the decimal point three places to the right or to the left. For example, to convert 3.82 km into meters, one moves the decimal point three places to the right: 3.82 km = 3820 m Similarly, 47.2 mm is converted into meters by moving the decimal point three places to the left: 47.2 mm = 0.0472 m 9 Table (1.1): SI Prefixes Multiplication Factor Pretixt Symbol 1000 000 000 000 =1012 tera T 1 000 000 000 = 109 giga G 1 000 000 = 106 mega M 1000 =103 kilo k 100 = 102 hecto h 10 = 101 deka da 0.1= 10−1 deci d 0.01 = 10−2 centi c 0.001 = 10−3 milli m 0.000 001 = 10−6 micro µ 0.000 000 001 = 10−9 nano n 0.000 000 000 001=10−12 pico p femto f atto a 0.000 000 000 000 001= 10−15 0.000 000 000 000 000 001 = 10−18 Using scientific notation, one may also write 3.82 km = 3.82 x 103 m 47.2 mm = 47.2 x 10−3 The multiples of the unit of time are the minute (min) and the hour (h). Since 1 min = 60 s and 1 h 60 min = 3600 s, these multiples cannot be converted as readily as the others. 10 By using the appropriate multiple or submultiple of a given unit, one can avoid writing very large or very small numbers. For example, one usually writes 427.2 km rather than 427 200 m, and 2.16 mm rather than 0.002 16 m. Units of Area and Volume The unit of area is the square meter (m2), which represents the area of a square of side 1 m; the unit of volume is the cubic meter (m3), equal to the volume of a cube of side 1 m. In order to avoid exceedingly small or large numerical values in the computation of areas and volumes, one uses systems of subunits obtained by respectively squaring and cubing not only the millimeter but also two intermediate submultiples of the meter, namely, the decimeter (dm) and the centimeter (cm). Since, by definition, 1 dm = 0.1 m = 10−1m 1 cm = 0.01 m = 10−2 m 1 mm 0.001 m = 10−2 m The submultiples of the unit of area are 1dm2 = (1dm) 2= (10−1) 2=10−2 m2 1 cm2 = (1 cm) 2 = (10−2m) 2 =10−4 m2 1 mm2 = (1 mm) 2 = (10−3m) 2 = 10−6m 2 and the submultiples of the unit of volume are 1dm3 = (1 dm) 3 = (10−1m) 3 = 10−3m 3 11 1 cm3 = (1 cm) 3 = (10−2m) 3= 10−6m 3 1 mm3 = (1 mm) 3 = (10−3m) 3 = 10−9m 3 It should be noted that when the volume of a liquid is being measured, the cubic decimeter (dm3) is usually referred to as a liter (L). Other derived SI units used to measure the moment of a force, the work of a force, etc., are shown in Table 1.2. While these units will be introduced in later chapters as they are needed, we should note an important rule at this time: When a derived unit is obtained by dividing a base unit by another base unit, a prefix may be used in the numerator of the derived unit but not in its denominator. For example, the constant k of a spring which stretches 20 mm under a load of 100 N will be expressed as k = 100 N/20 mm = l00N /0.020 m =5000 N/m Or k = 5 kN /m but never as k = 5 N/mm. Table (1.2): Principal SI Units Used in Mechanics Quantity Unit Symbol Formula Acceleration Meter per second squared ……. m/s2 Angle Radian rad * Angular acceleration Radian per second squared . rad/s2 Angular velocity Rad/s Radian per second 12 . Area Square meter . m2 Density Kilogram per cubic meter . kg/m3 Energy Joule J N. m Force Newton N Kg. m/s2 Frequency Hertz Hz s−1 Impulse Newton-second .. Kg . m/s Length Meter m Mass Kilogram kg Moment of a force Newton-meter ... N. m Power Watt W J/s Pressure Pascal Pa N/m2 Stress Pascal Pa N/ m2 Time Second s Velocity Meter per second . m/s Solids Cubic meter . m3 Liquids Liter L 10−3m 3 Work Joule J N. m Volume *Supplementary unit (1 revolution = 2π rad = 360ο). 13 U.S. Customary Units Most practicing American engineers still commonly use a system in which the base units are the units of length, force, and time. These units are, respectively, the foot (ft), the pound (lb), and the second (s). The second is the same as the corresponding SI unit. The foot is defined as 0.3048 m. The pound is defined as the weight of a platinum standard, called the standard pound, which is kept at the National Institute of Standards and Technology outside Washington, the mass of which is 0.453 592 43 kg. Since the weight of a body depends upon the earth’s gravitational attraction, which varies with location, it is specified that the standard pound should be placed at sea level and at latitude of 45° to properly define a force of 1 lb. Clearly the U.S. customary units do not form an absolute system of units. Because of their dependence upon the gravitational attraction of the earth, they form a gravitational system of units. While the standard pound also serves as the unit of mass in commercial transactions in the United Sates, it cannot be so used in engineering computations, since such a unit would not be consistent with the base units defined in the preceding paragraph. Indeed, when acted upon by a force of 1 lb, that is, when subjected to the force of gravity, the standard pound receives the acceleration of gravity, g = 32.2 ft/s2 (Fig. 1.4), not the unit acceleration required by Eq. (1.1). The unit of mass consistent with the foot, the pound, and the second is the mass which receives an acceleration of 1 ft/s2 when a force of 1 lb is applied to it (Fig. 1.5). This unit, sometimes called a slug, can be derived from the equation F = ma after substituting 1 lb and 1 ft/s2 for F and a, respectively. We write 14 F=ma and obtain 1 lb= (1slug) (1ft/s2) l slug = l lb / l (ft/s2) = 1 lb.s2/ft …….. (1.6) Comparing Figs. 1.4 and 1.5, we conclude that the slug is a mass 32.2 times larger than the mass of the standard pound. The fact that in the U.S. customary system of units bodies are characterized by their weight in pounds rather than by their mass in slugs will be a convenience in the study of statics, where one constantly deals with weights and other forces and only seldom with masses. However, in the study of dynamics, where forces, masses, and accelerations are involved, the mass m of a body will be expressed in slugs when its weight W is given in pounds. Recalling Eq. (1.4), we write m= w/ g ………….. (1.7) Where g is the acceleration of gravity (g = 32.2 ft/s2). 15 Other U.S. customary units frequently encountered in engineering problems are the mile (mi), equal to 5280 ft; the inch (in.), equal to (1/12) ft; and the kilo pound (kip), equal to a force of 1000 lb. The ton is often used to represent a mass of 2000 lb but, like the pound, must be converted into slugs in engineering computations. The conversion into feet, pounds, and seconds of quantities expressed in other U.S. customary units is generally more involved and requires greater attention than the corresponding operation in SI units. If, for example, the magnitude of a velocity is given as v = 30 mi/h, we convert it to ft/s as follows. First we write v = 30 mi/h Since we want to get rid of the unit miles and introduce instead the unit feet, we should multiply the right-hand member of the equation by an expression containing miles in the denominator and feet in the numerator. But, since we do not want to change the value of the right-hand member, the expression used should have a value equal to unity. The quotient (5280 ft)/ (1 mi) is such an expression. Operating in a similar way to transform the unit hour into seconds, we write v = (30 mi/h) (5280ft/1mi) (1 h/3600s) Carrying out the numerical computations and canceling out units which appear in both the numerator and the denominator, we obtain v = 44 ft /s 16 1.4. CONVERSION FROM ONE SYSTEM OF UNITS TO ANOTHER There are many instances when an engineer wishes to convert into SI units a numerical result obtained in U.S. customary units or vice versa. Because the unit of time is the same in both systems, only two kinetic base units need be converted. Thus, since all other kinetic units can be derived from these base units, only two conversion factors need be remembered. Units of Length By definition the U.S. customary unit of length is 1 ft =0.3048 m ……. (1.8) It follows that 1 mi = 5280 ft = 5280(0.3048m) = 1609 m or 1 mi = 1.609 km ………… (1.9) also 1 in. = 1/12 ft = 1/12 (0.3048 m) = 0.0254 m Or 1 in. = 25.4 mm …………. (1.10) Units of Force Recalling that the U.S. customary unit of force (pound) is defined as the weight of the standard pound (of mass 0.4536 kg) at sea level and at a latitude of 45° (where g = 9.807 m /s2) and using Eq. (1.4), we write W = mg 1 lb = (0.4536 kg)(9.807 m/s2) = 4.448 kg• m/s2 17 or, recalling Eq. (1.5), l Ib = 4.448N …………… (1.11) Units of Mass The U.S. customary unit of mass (slug) is a derived unit. Thus, using Eqs. (1.6), (1.8), and (1.11), we write l slug =1lb.s2/ft =1 Ib/ 1 ft/s2 = 4.448N/ 0.3048 m/s2 and, recalling Eq. (1.5), 1 slug = 1 Ib. s2/ft = 14.59 kg ………….. (1.12) Although it cannot be used as a consistent unit of mass, we recall that the mass of the standard pound is, by definition, 1 pound mass = 0.4536 kg ……………… (1.13) This constant may be used to determine the mass in SI units (kilograms) of a body which has been characterized by its weight in U.S. customary units (pounds). To convert a derived U.S. customary unit into SI units, one simply multiplies or divides by the appropriate conversion factors. For example, to convert the moment of a force which was found to be M = 47 lb. in. into SI units, we use formulas (1.10) and (1.11) and write M = 47 lb. in. = 47(4.448 N)(25.4 mm) = 5310 N. mm =5.31 N. m 18 The conversion factors given in this section may also be used to convert a numerical result obtained in SI units into U.S. customary units. For example, if the moment of a force was found to be M = 40 N. m, we write, following the procedure used in the last paragraph of Sec. 1.3, M= 40N.m = (40N.m)( 1 Ib/4.448 N) ( 1 ft /0.3048 m ) Carrying out the numerical computations and canceling out units which appear in both the numerator and the denominator, we obtain M = 29.5 lb.ft The U.S. customary units most frequently used in mechanics are listed in Table 1.3 with their SI equivalents. 19 20 Materials Engineering Department Class: First Date : 2009-2010 Subject: Engineering Mechanics Lecturer: Dr. Emad AL-Hassani Lecture # 2 Introduction • The objective for the current chapter is to investigate the effects of forces on particles: replacing multiple forces acting on a particle with a single equivalent or resultant force, - Relations between forces acting on a particle that is in a state of equilibrium • The focus on particles does not imply a restriction to miniscule bodies. Rather, the study is restricted to analyses in which the size and shape of the bodies is not significant so that all forces may be assumed to be applied at a single point. Resultant of Two Forces • .Force: action of one body on another; characterized by its point of application, magnitude, line of action, and sense • Experimental evidence shows that the combined effect of two forces may be represented by a single resultant force. • The resultant is equivalent to the diagonal of a parallelogram which contains the two forces in adjacent legs. • Force is a vector quantity 1 Vectors • Vector: parameter possessing magnitude and direction which add according to the parallelogram law. Examples: displacements, velocities, accelerations. • Scalar: parameter possessing magnitude but not direction. Examples: mass, volume, temperature • Vector classifications: - Fixed or bound vectors have well defined points of application that cannot be changed without affecting an analysis. - Free vectors may be freely moved in space without changing their effect on an analysis. - Sliding vectors may be applied anywhere along their line of action without affecting an analysis. • Equal vectors have the same magnitude and direction. • Negative vector of a given vector has the same magnitude and the opposite direction 2 Addition of Vectors • Trapezoid rule for vector addition • Triangle rule for vector addition 2 2 2 = + − 2 PQ cos B R P Q • Law of cosines, r r r R = P+Q C B C B • Law of sines, sin A sin B sin C = = Q R A r r r r P+Q =Q+ P • Vector addition is commutative • Vector subtraction • Addition of three or more vectors through repeated application of the triangle rule 3 • The polygon rule for the addition of three or more vectors. • Vector addition is associative, r r r r r r r r r P + Q + S = (P + Q ) + S = P + (Q + S ) • Multiplication of a vector by a scalar Resultant of Several Concurrent Forces • Concurrent forces: set of forces which all pass through the same point. A set of concurrent forces applied to a particle may be replaced by a single resultant force which is the vector sum of the applied forces. • Vector force components: two or more force vectors which, together, have the same effect as a single force vector. 4 Sample Problem The two forces act on a bolt at A. Determine their resultant. SOLUTION: • Graphical solution - construct a parallelogram with sides in the same direction as P and Q and lengths in proportion. Graphically evaluate the resultant which is equivalent in direction and proportional in magnitude to the diagonal. • Trigonometric solution - use the triangle rule for vector addition in conjunction with the law of cosines and law of sines to find the resultant. • Graphical solution - A parallelogram with sides equal to P and Q is drawn to scale. The magnitude and direction of the resultant or of the diagonal to the parallelogram are measured, R = 98 N = 35° 5 • Graphical solution - A triangle is drawn with P and Q head-to-tail and to scale. The magnitude and direction of the resultant or of the third side of the triangle are measured, R = 98 N α = 35° • Trigonometric solution - Apply the triangle rule. From the Law of Cosines, R 2 = P 2 + Q 2 − 2 PQ cos B = (40 N ) + (60 N ) − 2(40 N )(60 N ) cos155° 2 2 R = 97.73N From the Law of Sines, sin A sin B = Q R sin A = sin B Q R = sin 155° 60 N 97.73N A = 15.04° α = 20° + A α = 35.04° 6 Rectangular Components of a Force: Unit Vectors • May resolve a force vector into perpendicular components r r so that the resulting parallelogram is a rectangle. Fx and Fy are referred to as rectangular vector components and r r r F = Fx + Fy • Define perpendicular unit vectors i and j which are parallel to the x and y axes. Vector components may be expressed as products of the unit vectors with the scalar magnitudes of the vector components. r Fx and Fy are referred to as the scalar components of F 7 Addition of Forces by Summing Components • Wish to find the resultant of 3 or more concurrent forces, r r r r R = P+Q+ S Resolve each force into rectangular components r r r r r r r r Rx i + R y j = Px i + Py j + Qx i + Q y j + S x i + S y j r r = (Px + Qx + S x )i + (Py + Qy + S y ) j The scalar components of the resultant are equal to the sum of the corresponding scalar components of the given forces and to find the resultant magnitude and direction, Rx = Px + Qx + S x = ∑ Fx R = Rx2 + R y2 R y = Py + Q y + S y = ∑ Fy θ = tan −1 Ry Rx 8 Sample Problem Four forces act on bolt A as shown. Determine the resultant of the force on the bolt. SOLUTION: • Resolve each force into rectangular components. • Determine the components of the resultant by adding the corresponding force components. • Calculate the magnitude and direction of the resultant. SOLUTION: • Resolve each force into rectangular components forc mag r F1 150 r F2 80 r F3 110 r F4 100 x − comp + 129.9 y − comp + 75.0 − 27.4 + 75.2 0 − 110.0 + 96.6 − 25.9 Determine the components of the resultant by adding the corresponding force components. Calculate the magnitude and direction. tan α = Ry Rx = 14.3 N α = 4.1° 199.1 N 9 α = 4.1° R= 14.3 N = 199.6 N sin Equilibrium of a Particle • When the resultant of all forces acting on a particle is zero, the particle is in equilibrium. • Newton’s First Law: If the resultant force on a particle is zero, the particle will remain at rest or will continue at constant speed in a straight line. • Particle acted upon by two forces: - equal magnitude - same line of action - opposite sense • Particle acted upon by three or more forces: - graphical solution yields a closed polygon - algebraic solution r r R = ∑F = 0 ∑ Fx = 0 ∑ Fy = 0 10 Free-Body Diagrams Space Diagram: A sketch showing the physical conditions of the problem. Free-Body Diagram: A sketch showing only the forces on the selected particle Rectangular Components in Space The vector F is contained in the plane OBAC. • Resolve into F horizontal and vertical components. Fy = F cos θ y Fh = F sin θ y 11 Resolve Fh into rectangular components Fx = Fh cos φ = F sin θ y cos φ Fy = Fh sin φ = F sin θ y sin φ With the angles between F and the axes, r F Fx = F cosθ x Fy = F cosθ y Fz = F cosθ z r r r r F = Fx i + Fy j + Fz k r r r = F cosθ x i + cosθ y j + cosθ z k r = Fλ r r r r λ = cosθ x i + cosθ y j + cosθ z k ( ) λ is a unit vector along the line of action of F and cosθx, cosθy, and cosθz are the direction cosines for F Direction of the force is defined by the location of two points M (x1 , y1 , z1 ) and N ( x2 , y2 , z 2 ) r d = vector joining M and N r r r = d xi + d y j + d z k d x = x2 − x1 d y = y2 − y1 d z = z 2 − z1 r r F = Fλ r r 1 r r λ = d xi + d y j + d z k d Fd y Fd x Fd z Fx = Fy = Fz = d d d ( ) 12 Materials Engineering Department Class: First Date : 2009-2010 Subject: Engineering Mechanics Lecturer: Dr. Emad AL-Hassani Lecture # 3 Rigid Bodies: Equivalent Systems of Forces Introduction • Treatment of a body as a single particle is not always possible. In general, the size of the body and the specific points of application of the forces must be considered. • Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the conditions of equilibrium or motion of the body. • Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with a simpler equivalent system. - Moment of a force about a point - Moment of a force about an axis - Moment due to a couple • Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple. External and Internal Forces • Forces acting on rigid bodies are divided into two groups: - External forces Internal forces • External forces are shown in a free-body diagram. 1 • If unopposed, each external force can impart a motion of translation or rotation, or both. Principle of Transmissibility: Equivalent Forces • Principle of Transmissibility Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F’ are equivalent forces. • Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck. • Principle of transmissibility may not always apply in determining internal forces and deformations. 2 Vector Product of Two Vectors • Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product. • Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions: 1. Line of action of V is perpendicular to plane containing P and Q. 2. Magnitude of V is V = PQ sin θ 3. Direction of V is obtained from the righthand rule • Vector products: - are not commutative, Q × P = −(P × Q ) - are distributive, P × (Q1 + Q 2 ) = P × Q1 + P × Q 2 -are not associative, (P × Q )× S ≠ P × (Q × S ) Vector Products: Rectangular Components • Vector products of Cartesian unit vectors, r r i ×i = 0 r r r i× j =k r r r i ×k = − j r r r r r r j × i = −k k × i = j r r r r r j× j =0 k × j = −i r r v r r j ×k = i k ×k = 0 3 • Vector products in terms of rectangular coordinates ( )( ) r r r r r r r V=Pxi +Py j +Pzk ×Qxi +Qy j +Qzk (P = y Q (P + z x r i r j r k = Px Qx Py Qy Pz Qz − PzQ Q y r y )i − PyQ + x (P z Q r )k x − PxQ r z )j Moment of a Force About a Point • A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application. • The moment of F about O is defined as M O = r × F • The moment vector MO is perpendicular to the plane containing O and the force F. • Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO. M O = rF sin θ = Fd • The sense of the moment may be determined by the right-hand rule. 4 • Any force F’ that has the same magnitude and direction as F, is equivalent if it also has the same line of action and therefore, produces the same moment. • Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure. • The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane • If the force tends to rotate the structure clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive. • If the force tends to rotate the structure counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative Varignon’s Theorem • The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same point O. ( ) r r r r r r r r × F1 + F2 + L = r × F1 + r × F2 + L • Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F. 5 Rectangular Components of the Moment of a Force • The moment of F about O, r r r MO = r × F, r r r r r = xi + yj + zk r r r r F = Fx i + Fy j + Fz k r r r r M O = M xi + M y j + M z k r i r j r k = x Fx y Fy z Fz r r r = ( yFz − zFy )i + ( zFx − xFz ) j + (xFy − yFx )k • The moment of F about B, r r r M B = rA / B × F r r r rA / B = rA − rB r r r = ( x A − xB ) i + ( y A − y B ) j + ( z A − z B ) k r r r r F = Fx i + Fy j + Fz k r i r M B = ( x A − xB ) Fx r j ( y A − yB ) Fy r k (z A − z B ) Fz 6 • For two-dimensional structures r r M O = (xFy − yFz )k MO = MZ = xFy − yFz r r M O = ( x A − x B )F y − ( y A − y B )Fz k [ ] MO = MZ = ( x A − x B )F y − ( y A − y B )Fz Sample Problem A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O. Determine: a) moment about O, b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, d) location for a 240-lb vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original force 7 a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper M O = Fd d = (24 in.) cos 60° = 12 in. M O = (100 lb )(12 in.) M O = 1200 lb ⋅ in b) Horizontal force at A that produces the same moment, d = (24 in.)sin 60° = 20.8 in. M O = Fd 1200 lb ⋅ in. = F (20.8 in.) 1200 lb ⋅ in. F= 20.8 in. F = 57.7 lb c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA. M O = Fd 1200 lb ⋅ in. = F (24 in.) 1200 lb ⋅ in. F= 24 in. F = 50 lb d) To determine the point of application of a 240 lb force to produce the same moment, M O = Fd 1200 lb ⋅ in. = (240 lb )d 1200 lb ⋅ in. = 5 in. 240 lb OB cos60° = 5 in. d= OB = 10 in. 8 e)Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force. Sample Problem The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C. SOLUTION: The moment MA of the force F exerted by the wire is obtained by evaluating the vector product, r r r r r rC A = rC − rA = (0.3 m )i + (0.08 m ) j r r r M A = rC A × F 9 r r r rC D F = Fλ = (200 N ) rC D r r r − (0.3 m )i + (0.24 m ) j − (0.32 m )k = (200 N ) 0.5 m r r r = −(120 N ) i + (96 N ) j − (128 N )k r v r r M A = −(7.68 N ⋅ m ) i + (28.8 N ⋅ m ) j + (28.8 N ⋅ m )k Scalar Product of Two Vectors • The scalar product or dot product between two vectors P and Q is defined as r r P • Q = PQ cos θ (scalar result ) • Scalar products: r r r r Q•P -are commutative,r P •r Q = r r r ( ) r r • + = • + • Q2 P Q Q P Q P 1 2 1 -are distributive, ( ) r r r P -are not associative • Q • S = undefined 10 • Scalar products with Cartesian unit components ( )( r r r r r r r r P • Q = Px i + Py j + Pz k • Qx i + Q y j + Qz k r r r r r r r r i •i =1 j • j =1 k •k =1 i • j = 0 r r P • Q = Px Qx + Py Q y + Pz Qz r r P • P = Px2 + Py2 + Pz2 = P 2 11 ) r r v r j •k = 0 k •i = 0 Materials Engineering Department Class: First Date : 2009-2010 Subject: Engineering Mechanics Lecturer: Dr. Emad AL-Hassani Lecture # 4 Equilibrium of Rigid Bodies Introduction • For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body. • The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces form a system equivalent to zero, r F ∑ =0 ( ) r r r M = r ∑ O ∑ ×F =0 • Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium, ∑ Fx = 0 ∑ Fy = 0 ∑ Fz = 0 ∑Mx = 0 ∑M y = 0 ∑Mz = 0 Free-Body Diagram • First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram. • Select the extent of the free-body and detach it from the ground and all other bodies 1 • Indicate point of application, magnitude, and direction of external forces, including the rigid body weight • Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body • Include the dimensions necessary to compute the moments of the forces Reactions at Supports and Connections for a Two-Dimensional Structure • Reactions equivalent to a force with known line of action. 2 • Reactions equivalent to a force of unknown direction and magnitude. • Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude Equilibrium of a Rigid Body in Two Dimensions • For all forces and moments acting on a two-dimensional structure Fz = 0 M x = M y = 0 M z = M O 3 • Equations of equilibrium become ∑F x =0 ∑F y =0 ∑M A =0 where A is any point in the plane of the structure • The 3 equations can be solved for no more than 3 unknowns • The 3 equations can not be augmented with additional equations, but they can be replaced ∑F x =0 ∑M A ∑M =0 B =0 Statically Indeterminate Reactions More unknowns than equations Fewer unknowns than equations, partially constrained 4 Equal number unknowns and equations but improperly constrained Sample Problem A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. SOLUTION: • Create a free-body diagram for the crane • Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A. • Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components • Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero. • Create the free-body diagram • Determine B by solving the equation for the sum of the moments of all forces about A. ∑M A = 0 : + B(1.5m ) − 9.81 kN(2m ) − 23.5 kN(6m ) = 0 B = +107.1 kN 5 • Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces ∑ Fx = 0 : Ax + B = 0 Ax = −107.1 kN ∑F y = 0: A y − 9 . 81 kN − 23 . 5 kN = 0 A y = + 33 .3 kN • Check the values obtained Equilibrium of a Rigid Body in Three Dimensions • Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three dimensional case. ∑ Fx = 0 ∑ Fy = 0 ∑ Fz = 0 ∑Mx = 0 ∑My = 0 ∑Mz = 0 • These equations can be solved for no more than 6 unknowns which generally represent reactions at supports or connections • The scalar equations are conveniently obtained by applying the vector forms of the conditions for equilibrium r r r r ∑ F = 0 ∑ M O = ∑ (r × F ) = 0 6 Reactions at Supports and Connections for a Three-Dimensional Structure 7 Sample Problem A sign of uniform density weighs 270 lb and is supported by a ball-andsocket joint at A and by two cables. Determine the tension in each cable and the reaction at A. SOLUTION: • Create a free-body diagram for the sign • Apply the conditions for static equilibrium to develop equations for the unknown reactions • Create a free-body diagram for the sign. Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading. r r r rD − rB TBD = TBD r r rD − rB r r r − 8i + 4 j − 8k = TBD 12 r r 2r 2 1 = TBD − 3 i + 3 j − 3 k r r r rC − rE TEC = TEC r r rC − rE r r r − 6i + 3 j + 2 k = TEC 7 r r 2r 6 3 = TEC − 7 i + 7 j + 7 k ( ) ( ) 8 • Apply the conditions for static equilibrium to develop equations for the unknown reactions. r r r r r ( ) F = A + T + T − 270 lb j =0 ∑r BD EC i : Ax − 23 TBD − 76 TEC = 0 r j : Ay + 13 TBD + 73 TEC − 270 lb = 0 r k : Az − 23 TBD + 72 TEC = 0 r r r r r r r ∑ rM A = rB × TBD + rE × TEC + (4 ft )i × (− 270 lb) j = 0 j : 5.333TBD − 1.714 TEC = 0 r k : 2.667 TBD + 2.571TEC − 1080 lb = 0 Solve the 5 equations for the 5 unknowns, TBD = 101.3 lb TEC = 315 lb r r r v A = (338 lb )i + (101.2 lb ) j − (22.5 lb )k 9 Materials Engineering Department Class: First Subject: Engineering Mechanics Lecturer: Dr. Emad AL-Hassani Centroids and Centers of Gravity Introduction: • The earth exerts a gravitational force on each of the particles forming a body. These forces can be replace by a single equivalent force equal to the weight of the body and applied at the center of gravity for the body • The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid. • Determination of the area of a surface of revolution and the volume of a body of revolution are accomplished with the Theorems of PappusGuldinus Center of Gravity of a 2D Body • Center of gravity of a plate ∑ M y x W = ∑ x ∆W = ∫ x dW ∑ M y yW = ∑ y∆W = ∫ y dW • Center of gravity of a wire 1 Centroids and First Moments of Areas and Lines • Centroid of an area x W = ∫ x dW x (γAt ) = ∫ x (γt )dA x A = ∫ x dA = Q y = first moment with respect to y yA = ∫ y dA = Qx • = first moment with respect to x • Centroid of a line x W = ∫ x dW x (γ La ) = ∫ x (γ a )dL x L = ∫ x dL yL = ∫ y dL First Moments of Areas and Lines • An area is symmetric with respect to an axis BB’ if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and is divided into two equal parts by BB’. 2 • The first moment of an area with respect to a line of symmetry is zero. • If an area possesses a line of symmetry, its centroid lies on that axis • If an area possesses two lines of symmetry, its centroid lies at their intersection • An area is symmetric with respect to a center O if for every element dA at (x,y) there exists an area dA’ of equal area at (-x,-y). • The centroid of the area coincides with the center of symmetry 3 Centroids of Common Shapes of Areas Centroids of Common Shapes of Lines 4 Composite Plates and Areas • Composite plates X ∑W = ∑ x W Y ∑W = ∑ y W • Composite area X ∑ A = ∑ xA Y ∑ A = ∑ yA Sample Problem For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid. SOLUTION: • Divide the area into a triangle, rectangle, and semicircle with a circular cutout. • Calculate the first moments of each area with respect to the axes • Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout 5 • Compute the coordinates of the area centroid by dividing the first moments by the total area. • Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout. Qx = +506.2 ×103 mm3 Q y = +757.7 × 103 mm3 • Compute the coordinates of the area centroid by dividing the first moments by the total area ∑ x A = + 757.7 ×10 mm X= ∑ A 13.828 ×10 mm 3 3 3 2 X = 54.8 mm ∑ y A = + 506.2 ×10 mm Y = ∑ A 13.828 ×10 mm 3 3 Y = 36.6 mm 6 3 2 Materials Engineering Department Subject: Engineering Mechanics Class: First Lecturer: Dr. Emad AL-Hassani Friction Introduction • In preceding chapters, it was assumed that surfaces in contact were either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent relative motion between surfaces). • Actually, no perfectly frictionless surface exists. For two surfaces in contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other. • However, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied. • The distinction between frictionless and rough is, therefore, a matter of degree. • There are two types of friction: dry or Coulomb friction and fluid friction. Fluid friction applies to lubricated mechanisms. The present discussion is limited to dry friction between nonlubricated surfaces. The Laws of Dry Friction. Coefficients of Friction • Block of weight W placed on horizontal surface. Forces acting on block are its weight and reaction of surface N. • Small horizontal force P applied to block. For block to remain stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force. 1 • As P increases, the static-friction force F increases as well until it reaches a maximum value Fm. Fm = µ s N • Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force Fk. Fk = µ k N • Maximum static-friction force: Fm = µ s N • Kinetic-friction force: Fk = µ k N µ k ≅ 0.75µ s 2 • Four situations can occur when a rigid body is in contact with a horizontal surface: No friction, (Px = 0) No motion ,(Px < Fm) • Motion impending, (Px = Fm) Motion, (Px > Fm) 3 Angles of Friction • It is sometimes convenient to replace normal force N and friction force F by their resultant R: No friction No motion • tan φ s = Fm µ s N = N N tan φ s = µ s • Motion impending Motion • tan φ k = Fk µ k N = N N tan φ k = µ k 4 • Consider block of weight W resting on board with variable inclination angle θ. • No friction No motion • Motion impending Motion 5 Problems Involving Dry Friction • All applied forces known • Coefficient of static friction is known • Determine whether body will remain at rest or slide • All applied forces known • Motion is impending • Determine value of coefficient of static friction. • Coefficient of static friction is known • Motion is impending • Determine magnitude or direction of one of the applied forces 6 Sample Problem A 450 N force acts as shown on a 1350 N block placed on an inclined plane. The coefficients of friction between the block and plane are ms = 0.25 and mk = 0.20. Determine whether the block is in equilibrium and find the value of the friction force. SOLUTION: • Determine values of friction force and normal reaction force from plane required to maintain equilibrium. • Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide. • If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force. ∑F x = 0: 450 N − 53 (1350 N ) − F = 0 F = −360 N ∑F y = 0: N − 54 (1350 N ) = 0 N = 1080 N Fm = µ s N Fm = 0.25(1080 N ) = 270 N The block will slide down the plane Factual = Fk = µ k N = 0.20(1080 N ) Factual = 216 N 7

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