Statics of Rigid Bodies, Lecture notes of Statistics

Download Statics of Rigid Bodies and more Lecture notes Statistics in PDF only on Docsity! University of Eastern Philippines COLLEGE OF ENGINEERING Civil Engineering Department University Town, Northern Samar Prepared by: LORNA VICTORIA B. RAMIS, CE Professor 1. OVERVIEW OF THE MODULE To provide you, the student with a clear and thorough presentation of the theory and application of statics of rigid bodies, it is important to know the fundamental first. In this module we will discuss the fundamental concepts of the forces and their characteristics, the effect of a force or a force system on the body acted upon which can be both external or internal, force systems which can be a set of two or more coplanar or spatial forces, components of a force, moment of a force, and vector analysis. After learning the fundamental concepts, resultant of force systems follows. We will be discussing the resultants of the different force systems. 1.1 Purpose of the Module The purpose or objectives of this module is for you, my dear students, to learn to do the following:  How to add forces and resolve them into components using the parallelogram law.  How to express force and position in the Cartesian vector form and how to determine the vector’s magnitude and direction.  To introduce the dot product in order to determine the angle between two vectors or the projection of one vector onto another.  To determine resultants of the force systems in the coplanar systems and the spatial force systems. 1.2 Module Title and Description MODULE NO.1 - FUNDAMENTALS CONCEPTS and RESULTANTS OF FORCE SYSTEMS 1.3 Module Guide This module is an introduction to this course and so, you should take note of the following:  A review of your physics, algebra and trigonometry will help greatly you in understanding the topic at hand.  Read the learning outcomes of this module for you determine the focus of our topic.  The illustrative problems discuss the technique of solution as well as the application of principles.  Solve the exercises that are provided for you after every subtopic.  Studying the example problems helps, but the most effective way of learning the principles of the topic is to solve problems.  To be successful at this, it is important to always present the work in a logical and orderly manner, as suggested by the following sequence of steps: o Read the problem carefully and try to correlate the actual physical situation with the theory studied. o Tabulate the problem data and draw to a large scale any necessary diagrams. o Apply the relevant principles, generally in mathematical form. When writing any equations, be sure they are dimensionally homogeneous. o Solve the necessary equations, and report the answer with no more than three significant figures. o Study the answer with technical judgment and common sense to determine whether or not it seems reasonable. A representation of a three-dimensional Cartesian coordinate system with the x-axis pointing towards the observer. 3.2 Learning Objectives 3.2.1 Fundamental Concepts Before we begin our study of engineering mechanics, it is important to understand certain fundamental concepts and principles. Basic Quantities. The following four quantities are used throughout mechanics. Length. Length is used to locate the position of a point in space and thereby describe the size of a physical system. Once a standard unit of length is defined, one can then use it to define distances and geometric properties of a body as multiples of this unit. Time. Time is conceived as a succession of events. Although the principles of statics are time independent, this quantity plays an important role in the study of dynamics. Mass. Mass is a measure of a quantity of matter that is used to compare the action of one body with that of another. This property manifests itself as a gravitational attraction between two bodies and provides a measure of the resistance of matter to a change in velocity. Force. In general, force is considered as a “push” or “pull” exerted by one body on another. This interaction can occur when there is direct contact between the bodies, such as a person pushing on a wall, or it can occur through a distance when the bodies are physically separated. Examples of the latter type include gravitational, electrical, and magnetic forces. In any case, a force is completely characterized by its magnitude, direction, and point of application. Idealizations. Models or idealizations are used in mechanics in order to simplify application of the theory. Here we will consider three important idealizations. Particle. A particle has a mass, but a size that can be neglected. For example, the size of the earth is insignificant compared to the size of its orbit, and therefore the earth can be modeled as a particle when studying its orbital motion. When a body is idealized as a particle, the principles of mechanics reduce to a rather simplified form since the geometry of the body will not be involved in the analysis of the problem. Rigid Body. A rigid body can be considered as a combination of a large number of particles in which all the particles remain at a fixed distance from one another, both before and after applying a load. This model is important because the body’s shape does not change when a load is applied, and so we do not have to consider the type of material from which the body is made. In most cases the actual deformations occurring in structures, machines, mechanisms, and the like are relatively small, and the rigid-body assumption is suitable for analysis. Concentrated Force. A concentrated force represents the effect of a loading which is assumed to act at a point on a body. We can represent a load by a concentrated force, provided the area over which the load is applied is very small compared to the overall size of the body. An example would be the contact force between a wheel and the ground. Newton’s Three Laws of Motion. Engineering mechanics is formulated on the basis of Newton’s three laws of motion, the validity of which is based on experimental observation. These laws apply to the motion of a particle as measured from a nonaccelerating reference frame. They may be briefly stated as follows. First Law. A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force, Fig. 1–1a. 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, Fig. 1–1b.* If F is applied to a particle of mass m, this law may be expressed mathematically as Third Law. The mutual forces of action and reaction between two particles are equal, opposite, and collinear, Fig. 1–1c. *Stated another way, the unbalanced force acting on the particle is proportional to the time rate of change of the particle’s linear momentum. Units of Measurements The four basic quantities—length, time, mass, and force—are not all independent from one another; in fact, they are related by Newton’s second law of motion, F = ma. Because of this, the units used to measure these quantities cannot all be selected arbitrarily. The equality F = ma is maintained only if three of the four units, called base units, are defined and the fourth unit is then derived from the equation. SI Units. The International System of units, abbreviated SI after the French “Système International d’Unités,” is a modern version of the metric system which has received worldwide recognition. As shown in Table 1–1, the SI system defines length in meters (m), time in seconds (s), and mass in kilograms (kg). The unit of force, called a newton (N), is derived from F = ma. Thus, 1 newton is equal to a force required to give 1 kilogram of mass an acceleration of 1 m/s2 (N = kg•m/s2). If the weight of a body located at the “standard location” is to be determined in newtons, then Eq. 1–3 must be applied. Here measurements give g = 9.806 65 m/s2; however, for calculations, the value g = 9.81 m/s2 will be used. Thus Therefore, a body of mass 1 kg has a weight of 9.81 N, a 2-kg body weighs 19.62 N, and so on, Fig. 1–2a. U.S. Customary. In the U.S. Customary system of units (FPS) length is measured in feet (ft), time in seconds (s), and force in pounds (lb), Table 1–1. The unit of mass, called a slug, is derived from F = ma. Hence, 1 slug is equal to the amount of matter accelerated at 1 ft/s2 when acted upon by a force of 1 lb (slug = lb•s2/ft). Fig. 1-1c Fig. 1-2 Coplanar and Non-concurrent: Spatial and Concurrent: Spatial and parallel: Spatial and non-concurrent: 3.3.1 Fundamental Axioms of Mechanics 1. The Parallelogram Law: The resultant of two forces is the diagonal of the parallelogram formed on the vectors of these forces. 2. Two forces are in equilibrium only when equal in magnitude, opposite in direction, and collinear in action. 3. A set of forces in equilibrium may be added to any system of forces without changing the effect of the original system. 4. Action and reaction forces are equal but oppositely directed. 3.3.2 Introduction to Free-Body Diagram A sketch of the isolated body which shows only the forces acting upon the body is called free-body diagram. The forces acting on the free body are the action forces also called the applied forces. Example: Equilibrium of 2D rigid body example FBD of 2D equilibrium of rigid body 3.4 Scalar and Vector Quantities Scalars. Imagine two groups of marbles, one consisting of 10 marbles and the other of 5. If a common group is formed by mixing them, the resultant number will be 15 marbles, a result obtained by arithmetical addition. Quantities which possess magnitude only and can be added arithmetically are defined as scalar quantities. Vectors. A vector of a quantity can be represented geometrically by drawing a line acting in the direction of the quantity, the length of the line representing to some scale the magnitude of the quantity. 3.4.1 Vector Addition of Forces Experimental evidence has shown that a force is a vector quantity since it has a specified magnitude, direction, and sense and it adds according to the parallelogram law. Two common problems in statics involve either finding the resultant force, knowing its components, or resolving a known force into two components. We will now describe how each of these problems is solved using the parallelogram law. Finding a Resultant Force. The two component forces F1 and F2 acting on the pin in Fig. 2–7a can be added together to form the resultant force FR = F1 + F2, as shown in Fig. 2–7b. From this construction, or using the triangle rule, Fig. 2– 7c, we can apply the law of cosines or the law of sines to the triangle in order to obtain the magnitude of the resultant force and its direction. Finding the Components of a Force. Sometimes it is necessary to resolve a force into two components in order to study its pulling or pushing effect in two specific directions. For example, in Fig. 2–8a, F is to be resolved into two components along the two members, defined by the u and v axes. In order to determine the magnitude of each component, a parallelogram is constructed first, by drawing lines starting from the tip of F, one-line parallel to u, and the other line parallel to v. These lines then intersect with the v and u axes, forming a parallelogram. The force components Fu and Fv are then established by simply joining the tail of F to the intersection points on the u and v axes, Fig. 2– 8b. This parallelogram can then be reduced to a triangle, which represents the triangle rule, Fig. 2–8c. From this, the law of sines can then be applied to determine the unknown magnitudes of the components. Addition of Several Forces. If more than two forces are to be added, successive applications of the parallelogram law can be carried out in order to obtain the resultant force. For example, if three forces F1, F2, F3 act at a point O, Fig. 2–9, the resultant of any two of the forces is found, say, F1 + F2—and then this resultant is added to the third force, yielding the resultant of all three forces; i.e., FR = (F1 + F2) + F3. Using the parallelogram law to add more than two forces, as shown here, often requires extensive geometric and trigonometric calculation to determine the numerical values for the magnitude and direction of the resultant. Instead, problems of this type are easily solved by using the “rectangular component method,” which is explained in the next topic. NOTE: The result for Fu shows that sometimes a component can have a greater magnitude than the resultant. 3. Resolve the horizontal 600-lb force in figure shown into components acting along the u and v axes and determine the magnitudes of these components. SOLUTION: The parallelogram law of addition is shown in Fig. 1 below, and the triangle rule is shown in Fig. 2 below. Fig. 2 Fig. 1 The magnitudes of FR and F are the two unknowns. They can be determined by applying the law of sines. 4. It is required that the resultant force acting on the eyebolt in figure shown be directed along the positive x axis and that F2 have a minimum magnitude. Determine this magnitude, the angle θ, and the corresponding resultant force. SOLUTION: (a) (b) The triangle rule for FR = F1 + F2 is shown in Fig. (a). Since the magnitudes (lengths) of FR and F2 are not specified, then F2 can actually be any vector that has its head touching the line of action of FR, Fig. (b). However, as shown, 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 Since the vector addition now forms the shaded right triangle, the two unknown magnitudes can be obtained by trigonometry. Note: It is strongly suggested that you test yourself on the solutions to these examples, by covering them over and then trying to draw the parallelogram law, and thinking about how the sine and cosine laws are used to determine the unknowns. 3.4.3 EXERCISES: (ACTIVITY 1) 1. Determine the magnitude of the resultant force acting on the screw eye and its direction measured clockwise from the x axis. 2. Resolve the 30-lb force into components along the u and v axes, and determine the magnitude of each of these components. 3. Two forces act on the hook. Determine the magnitude of the resultant force. 4. If force F is to have a component along the u axis of Fu = 6 kN, determine the magnitude of F and the magnitude of its component Fv along the v axis. Applications. The dot product has two important applications in mechanics.  The angle formed between two vectors or intersecting lines. The angle θ between the tails of vectors A and B in Fig. 2–40 can be determined from Eq. 2–12 and written as Here A • B is found from Eq. 2–13. In particular, notice that if A • B = 0, θ = cos-1 0 = 90° so that A will be perpendicular to B.  The components of a vector parallel and perpendicular to a line. The component of vector A parallel to or collinear with the line aa in Fig. 2–40 is defined by Aa where Aa = A cos θ. This component is sometimes referred to as the projection of A onto the line, since a right angle is formed in the construction. If the direction of the line is specified by the unit vector ua, then since ua = 1, we can determine the magnitude of Aa directly from the dot product (Eq. 2–12); i.e., Hence, the scalar projection of A along a line is determined from the dot product of A and the unit vector ua which defines the direction of the line. Notice that if this result is positive, then Aa has a directional sense which is the same as ua, whereas if Aa is a negative scalar, then Aa has the opposite sense of direction to ua. The component Aa represented as a vector is therefore The component of A that is perpendicular to line aa can also be obtained, Fig. 2-41. Since There are two possible ways of obtaining One way would be to determine θ from the dot product, then Alternatively, if A is known, then by Pythagorean’s theorem we can also write 3.5.1 SAMPLE PROBLEMS: 1. Determine the magnitudes of the projection of the force F in Fig. 2–42 onto the u and v axes. Solution: Projections of Force. The graphical representation of the projections is shown in Fig. 2–42. From this figure, the magnitudes of the projections of F onto the u and v axes can be obtained by trigonometry: (Fu)proj = (100 N) cos 45° = 70.7 N (Fv)proj = (100 N) cos 15° = 96.6 N NOTE: These projections are not equal to the magnitudes of the components of force F along the u and v axes found from the parallelogram law. They will only be equal if the u and v axes are perpendicular to one another. 2. The frame shown in Fig. 2–43a is subjected to a horizontal force F = {300j} N. Determine the magnitudes of the components of this force parallel and perpendicular to member AB. 3.5.2 EXERCISES (ACTIVITY 2) 1. Find the magnitude of the projected component of the force along the pipe AO. 2. Determine the angle θ between the force and the line AB 3. Express the force F in Cartesian vector form if it acts at the midpoint B of the rod. 4. Determine the magnitude of the projection of the force F1 along cable AC.