Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 ENGG 407:ENGG 407: Statics of Rigid Bodies Final Project Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 PROPONENTS:NGG 407: Catilo, Carl Renszey M. Ditan, Khier Allen L. Pecho, Jenn Kathleen E. Soriano, Kaye Anne S. Valderama, Alliah Eunice Z. 407: Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Engineering Mechanics Engineering Mechanics is the physical science that deals with the actions or effects of forces to the rigid bodies. The application of Mechanics is ubiquitous in our daily errands, such that when we go to our university, we can see a bridge withstanding the weight of every vehicle passing through, a car moving on the road, or simply when we are jogging on a field. The subject of mechanics is logically divided into two parts: • • Statics – concerns the equilibrium of bodies under the action of forces Dynamics – concerns the motion of bodies Basic Concepts and Principles The following are concepts and definition of fundamental terminologies needed to the study of engineering mechanics. Space. It is the geometric region occupied by bodies positions are described by linear and angular measurements relative to coordinate system. For two dimensional problems, only two coordinates are required. Time. The measure of succession of events and is a basic quantity in dynamics, however, time is not directly involved in the analysis of statics-related problems. Mass. It can be thought of as the quantity of matter in a body. Force. The action of one body on another. It can be described as the push or pull. Rigid body. A body is considered rigid when the change in distance between any two of its points is negligible for the purpose at hand. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Newton’s Laws Sir Isaac Newton (1643 – 1727) was the first to state the laws governing the motion of a particle to validate their validity. These laws are: Newton’s First Law: Inertia. A particle remains at rest or continues to move with uniform velocity (in a straight constant speed) if there is no unbalanced force acting on it. Newton’s Second Law: Force. The acceleration of a particle is proportional to the vector sum of forces acting on it and is in the direction of this vector sum. Newton’s Third Law: Action and Reaction. The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear (lying on the same line). Units There are four fundamental quantities called dimensions used in mechanics: length, mass, force, and time. However, these units of measurement cannot all be chosen independently as it must be consistent with Newton’s second law (F = ma). The two most used system of units are SI Units and U.S Customary units. Quantity Mass Length Time Force Dimensional Symbol M L T F SI Units Unit Kilogram Meter Second Newton Symbol kg m s N U.S Customary Units Unit Slug Foot Second Pound Symbol – ft sec lb SI Units (Systeme International d’Unites) – refers to the international system of unit. It is the modern version of metric system. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 English Unit (U.S. Customary Unit) – or as British System of Units, also called as the FPS (foot-pound-second) System. PREFIXES USED IN THE SI SYSTEM Prefix Symbol exa peta tera giga mega kilo hector deka deci centi milli micro nano pico femto alto E P T G M k h da d c m π n p f a Meaning 1 000 000 000 000 000 000 1 000 000 000 000 000 1 000 000 000 000 1 000 000 000 1 000 000 1 000 100 10 1 0.1 0.01 0.001 0.000001 0.000000001 0.000000000001 0.000000000000001 0.000000000000000001 Exponential Notation π₯1018 π₯1015 π₯1012 π₯109 π₯106 π₯103 π₯102 π₯101 π₯100 π₯10−1 π₯10−2 π₯10−3 π₯10−6 π₯10−9 π₯10−12 π₯10−15 π₯10−18 Important Conversion/Formula 1. Sine Rule π΄ π΅ πΆ = = sinβ‘(180 − πΌ) sinβ‘(180 − π½) sinβ‘(180 − πΎ) 2. Important Conversion 1N = 1β‘ππβ‘π₯β‘1β‘π/π 2 = 1000β‘ππβ‘π₯β‘100β‘ππ/π 2 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 g = 9.81β‘π/π 2 1 H.P. = 9735.5β‘πΎπ 1 Pa = 1β‘π/π2 1 kPa = 103 β‘π/π2 1 MPa = 106 β‘π/π2 1 GPa = 109 β‘π/π2 1 bar = 105 β‘π/π2 3. Important Trigonometrical Formulas 1. sinβ‘(Aβ‘ + β‘B) β‘ = β‘sinβ‘A. cosβ‘Bβ‘ + β‘cosβ‘A. sinβ‘Bβ‘ 2. sin(A − β‘B) = sin A . cos B − cos A . sin B 3. cosβ‘(Aβ‘ + β‘B) = β‘cosβ‘A. cosβ‘B − β‘cosβ‘A. sinβ‘B 4. cos(A − β‘B) β‘ = β‘cosβ‘A. cosβ‘Bβ‘ + β‘sinβ‘A. sinβ‘B tan π΄β‘+tan π΅ 5. tanβ‘(Aβ‘ + β‘B) β‘ = (1−tan π΄.tan π΅)β‘ tan π΄−tan π΅ 6. tanβ‘(A − β‘B) β‘ = (1+tan π΄.tan π΅) 7. π ππ2β‘π΄ = 2 sin π΄β‘. cos π΄ 8. π ππ2 π΄ + cos2 π΄ = 1 9. 1 + π‘ππ2 π΄ = sec 2 π΄ 10. 1 + πππ‘ 2 π΄ = csc 2 π΄ 11. 1 + πππ π΄ = β‘ 12. ! − cos π΄ = 2 cos2 π΄ 2 2 sin2 π΄ 2 13. 2 cos π΄. sin π΅ = sin(π΄ + π΅) − sin(π΄ − π΅) Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 14. sin(−π΄) = β‘ − sin π΄ 15. cos(−π΄) = cos π΄ 16. tan(−π΄) = −tan π΄ 17. sin(90° − π΄) = cos π΄ 18. cos(90° − π΄) = sin π΄ 19. tan(90° − π΄) = cot π΄ 20. sin(90° + π΄) = cos π΄ 21. cos(90° + π΄) = −sin π΄ 22. tan(90° + π΄) = −cot π΄ 23. sin(180° − π΄) = sin π΄. 24. cos(180° − π΄) = −cos π΄ 25. tan(180° − π΄) = −tan π΄ 26. sin(180° + π΄) = − sin π΄ 27. cos(180° + π΄) = −cos π΄ 28. tan(180° + π΄) = tan π΄ 4. Important Assumptions In mechanics, we take 1. 2. 3. 4. 5. 6. Upwards force as positive Downwards force as negative Towards Right hand force as positive Towards Left hand force as negative Clockwise moment as positive Counterclockwise moment as negative Vector and Scalar Quantities Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Scalar Quantity. It is completely specified by a single value with an appropriate unit and has no direction, hence magnitude only. Example of scalar quantities include time, volume, density, speed, energy, and mass. Vector Quantity. It is completely specified by a number with an appropriate unit plus a direction. This type of quantity includes displacement, velocity, force, and momentum. Moreover, vectors representing physical quantities can be classified as free, sliding, or fixed. A free vector is one whose action is not confined or associated with a unique line of space. This vector describes equally well the direction and magnitude of the displacement of every point in the body. A sliding vector possess a unique line of action, however not a unique point of application. A fixed vector is one which the unique point of application is specified. The action of a force on a deformable or nonrigid body must be specified by a fixed vector at the point of application of the force. Resultant. The sum of a number of vectors of a particular type is called resultant. It is a single vector that would have the same effect as all the original vectors taken together. Reference Angle Standard Angle Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Bearing Vector of Addition A. Graphical Method Parallelogram Method. he resultant of two vectors acting at any angle may be represented by a diagonal of a parallelogram. If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then their resultant is represented in magnitude and direction by the diagonal passing through the point. For instance, Consider two forces Vector P and Vector Q acting at a point O inclined at an angle θ as shown in Fig. The forces Vector P and Vector Q are represented in magnitude and direction by the sides OA and OB of a parallelogram OACB as shown in Fig. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 The resultant Vector R of the forces Vector P and Vector Q is the diagonal OC of the parallelogram. The magnitude of the resultant is π = √π2 + π 2 + 2β‘ππ cos π The direction of the resultant is πΌ = tan−1 [ π sin π ] π + ππππ π Source: Parallelogram and Triangle law of forces. (n.d.). BrainKart; Retrieved November 20, 2022, from https://www.brainkart.com/article/Parallelogram-andTriangle-law-of-forces_3109/ Polygon Method. In this method, the resultant vector can be represented as the closing side of a polygon. This method is valid for adding any number of vectors unlike the parallelogram method. For example, if we have ‘n’ number of vectors starting from a1→ to an→, then the resultant of these vectors (say R→ ) can be represented as follows: Source: What are the parallelograms and the polygon method? (2021, April 7). Vedantu: Online Courses - Best LIVE Classes For CBSE, ICSE, JEE & NEET. https://www.vedantu.com/questionanswer/are-the-parallelograms-and-the-polygon-methodclass-11-physics-cbse-606df25c477df81db103b240 Thus, we can see that the resultant vector is calculated by constructing a polygon, hence the name, polygon method of vector addition. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 B. Analytical Method Triangle Method. Draw the vectors one after another, placing the initial point of each successive vector at the terminal point of the previous vector. Then draw the resultant from the initial point of the first vector to the terminal point of one last vector. This method is also called the head-to-tail method. In other words, if two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a triangle taken in order, then the closing side of the triangle taken in the reversed order represents the resultant of the forces in magnitude and direction. Source: Parallelogram and Triangle law of forces. (n.d.). BrainKart; Retrieved November 20, 2022, from https://www.brainkart.com/article/Parallelogram-andTriangle-law-of-forces_3109/ Force Sometimes, when we push the wall or lean our back against the wall, there are no changes in the position of the wall, but it is indubitable that we apply a force. Since the applied force is insufficient to move the wall, there is no motion produced. This proves that force does not always necessarily produce a motion in a body. By this, we can define force as the agency that can change the state of rest of motion of a body. A force is the action of one body to another. In engineering mechanics, applied forces are broadly divided into two types: tensile and compressive force. Tensile Force. A force which pulls the body. Here, member AB is a tension member carrying tensile force P. Compressive Force. A force which pushes the body. A P B A P P B P Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Tensile Force Compressive Force Force System • Force system is formed when multiple forces act on a body at the same time. • A term used to describe a group of forces. • Some real-life applications for the system of forces are: "forces on the tug of war", "forces on blades of ceiling fans", "forces at a joint of truss", etc. Different types of Force system: Source: “Engineering Mechanics: LESSON 2. FORCE SYSTEM.” E-Krishi Shiksha, 13 Aug. 2013, http://ecoursesonline.iasri.res.in/mod/page/view.php?id=125327. Coplanar Force System. Forces acting on only one plane are known as coplanar force system. Or, we can say that if the effect of forces is only in one plane of object, then the force system is known as Coplanar force. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Concurrent Force System. A concurrent force system is a system of two or more forces whose lines of action ALL intersect at a common point. Concurrent forces may or may not be coplanar. Non-Coplanar Force System. When the line of action of all the forces does not lie in one plane, it is called a non-coplanar force system. Non-Coplanar Force System Collinear Force System. When the lines of action of all the forces of a system act along the same line, this force system is called collinear force system. Collinear Force System Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Parallel Force. A parallel coplanar force system consists of two or more forces whose lines of action are all parallel to one another. Parallel Forces Non-Concurrent Force System. When the forces of a system do not meet at a common point of concurrency, this type of force system is called non-concurrent force system. Parallel forces are the example of this type of force system. Non-concurrent forces may be coplanar or non-coplanar. Coplanar and Concurrent Force System. A force system in which all the forces lie in a single plane and meet at one point, For example, forces acting at a joint of a roof truss. Coplanar and Non-Concurrent force system. These forces do not meet at a common point; however, they lie in a single plane. Non-Coplanar and concurrent force system. In this system, the forces lie in different planes but pass through a single point. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Non-Coplanar and non-concurrent force system. The forces which do not lie in a single plane and do not pass through a single point are known as non-coplanar and non-concurrent forces. Example is the loads transferred through columns to the rectangular mat foundation. Free-Body Diagram A free body diagram is a sketch of a body, a portion of a body, or two or more bodies completely isolated or free from all other bodies, showing the forces exerted by all other bodies on the one being considered. It is used to show the relative magnitude and direction of all forces acting upon an object in a given situation. Free-body Diagram (FBD) can simply be defined as the isolated representation of forces. Characteristics of FBD • • • It is a diagram or sketch of a body. The body is shown completely separated from all other bodies. The action on the body of each body removed in the isolating process is shown as a force or forces on the diagram Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 For example, A block rests on the table, as shown. A light rope is attached to it and runs over a pulley. The other end of the rope is attached to a second block. The two blocks are said to be coupled. Block m2 exerts a force due to its weight, which causes the system (two blocks and a string) to accelerate. Draw a free-body diagram for each block. FBD: Resolution of a force into components The process of splitting up a given force into different components without changing its effect on the body is called the resolution of a force. Resultant Also known as net force, is the force that would produce the same effect on an object as the multiple forces applied. Forces acting in opposite directions will cancel out and forces acting in the same direction will add together. Resultant of Concurrent Forces Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 The line of action of each force in a coplanar concurrent force system is on the same plane. All of these forces meet at a common point, thus concurrent. In x-y plane, the resultant can be found by the following formulas: Resultant of Parallel Forces Parallel forces can be in the same or in opposite directions. The sign of the direction can be chosen arbitrarily, meaning, taking one direction as positive makes the opposite direction negative. The difference between a concurrent and a parallel force system is that in the former the position of the resultant is known by inspection whereas in the latter it is not. Also, employing moment is involved in parallel forces to determine the position of resultant given by the formula; πΉπ = π΄π πΉ = β‘ ∑ π = ππ + ππ + ππ … π΄π = β‘ ππ (ππ ) + ππ (ππ ) + ππ (ππ ) … Moment The moment of a force depends on the magnitude of the force and the distance from the axis of rotation. The moment of a force about a point is (the magnitude of the force) × (the perpendicular distance of the line of action of the force from the point) or M=Fd. The distance d is frequently called the moment arm of the force. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 The axis of moments, which is perpendicular to the plane of forces, appears as a point commonly called the center of moments. In the figure, point Pivot is the center of moments. Couple A couple is made up of two parallel forces that are equal in magnitude, contrary in sense, and therefore have no prevalent line of action. This plays an important role on force systems’ analysis. Couples are also known as pure moments because they exert a net moment without exerting a net force. The two equal and opposite forces exerted on this lug wrench are a couple. They exert a moment on the lug nut on this wheel without Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 exerting any net force on the wheel. Adapted from image by Steffen Heinz Caronna CC-BY-SA 3.0. A couple's moment differs from a single force's moment in that it is independent of the location you are taking the moment about. In the following example, we have a couple acting on a beam. The magnitude of each force is F, and the distance between them is d. The moment exerted by this couple is independent of the of the distance x. We now have a point A that is x distance from the first of the two forces. We can calculate the net moment about point A by taking the moment of each force about point A and adding these moments together. M=−(F∗x)+(F∗(x+d)) We can see that if we rearrange and simplify the above formula, the variable x actually disappears from the equation, leaving the net moment equal to the magnitude of the forces (F) times the distance between the two forces (d). M=−(F∗x)+(F∗x)+(F∗d) M=(F∗d) This means that no matter what value of x we have, the magnitude of the couple's moment will be the same. The magnitude of the moment caused by the couple is independent of the location of the moment. This is also applicable in two or three dimensions. The magnitude of a couple's moment Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 is always equal to the magnitude of the forces multiplied by the perpendicular distance between the two forces. Sample Problem 1 Determine the moments of the 1150 N force acting in the figure shown below about points A, B, C, and D. Generally, we use the formula, M = fd, where d is the perpendicular distance from the point on the moment axis to the line of action of the force. Hence, ππ΄ = 1150β‘πβ‘(3π + 5π) = 9200β‘π β πβ‘πΆπ ππ΅ = 1150β‘πβ‘(5π) = 5750β‘π β πβ‘πΆπ ππΆ = 0 Since the line of action of force F passes through point C, hence the moment is zero. ππ· = 1150β‘πβ‘(1π) = 1150β‘π β πβ‘πΆπΆπ The CW and CCW are clockwise and counterclockwise (anticlockwise) respectively, which indicates the sense of rotation of the moment defined by the direction of the force that orbits about each point. Sample Problem 2 Three wires exert the tensions indicated on the ring in fig (). Assuming a concurrent system, determine the force in a single wire will replace three wires. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SOLUTION Single force, which replaces all other forces, is always the resultant of the system, so first resolved all the forces in horizontal and vertical direction ∑H = Sum of Horizontal Component = 70 cos 0 °β‘ + β‘30 cos 68 °β‘ + β‘50 cos 270 ° = 81.24β‘N ∑π = Sum of Vertical Component = 70 sin 0 °β‘ + β‘30 sin 68 °β‘ + β‘50 sin 270 ° = −22.18β‘N Let R be the resultant of coplanar forces π = β‘ √∑ π» 2 + ∑ π 2 π = β‘ √(γ81.24)γ2 + (γ−22.18β‘)γ2 πΉ = ππ. ππβ‘π΅ π θβ‘ = β‘tan– 1 (π π ) π» 22.18 θβ‘ = β‘tan– 1 (81.24) πβ‘ = β‘ππ. ππ° Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Exercise Problem 1 The screw eye in Fig 1 is subjected to two forces, πΉ1 and πΉ2 Determine the magnitude and direction of the resultant force. Exercise Problem 2 \The magnitude of vertical force F is 4000 N. Resolve F into components parallel to the bars AB and AC Exercise Problem 3 The end of the boom O in fig () is subjected to three concurrent and coplanar forces. Determine the magnitude and orientation of the resultant force. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Exercise Problem 4 A force P = 1200 N is shown in figure. a. Find the y-component of P with respect to x and y axis. b. Find the y'-component of P with respect to x' and y' axis. c. Find the y-component of P with respect to x' and y axis. d. Find the y'-component of P with respect to x and y' axis. Exercise Problem 5 Four coplanar forces are acting at a point. Three forces have magnitude of 20, 50 and 20N at angles of 45°, 200° and 270° respectively. Fourth force is unknown. Resultant force has magnitude of 50N and acts along x-axis. Determine the unknown force and its direction from x-axis. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 EQUILIBRIUM To understand equilibrium in a rigid body, it is first essential to know what a rigid body is and an equilibrium point. A rigid body is a substance that has a definite shape that does not change with external force. A rigid body is a system of particles that are each distanced equally, and the distance cannot be changed. A body or particle has an equilibrium state if the motion or internal energy of the particle doesn’t change over the time. For example, body equilibrium is when the total external force or torque is zero. Conditions for Rigid Body Equilibrium For a rigid body in static equilibrium, that is a non-deformable body where forces are not concurrent, the sum of both the forces and the moments acting on the body must be equal to zero. ∑ πΉ = 0β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ ∑ ππ = β‘0 The addition of moments (as opposed to particles where we only looked at the forces) adds another set of possible equilibrium equations, allowing us to solve for more unknowns as compared to particle problems. Moments, like forces, are vectors. This means that our vector equation needs to be broken down into scalar components before we can solve the equilibrium equations. In a two-dimensional problem, the body can only have clockwise or counterclockwise rotation (corresponding to rotations about the z axis). This means that a rigid body in a two-dimensional problem has three possible equilibrium equations; that is, the sum of force components in the x and y directions, and the moments about the z axis. The sum of each of these will be equal to zero. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Equations of Equilibrium in Two Dimensions There are three alternate forms of equilibrium equations for two-dimensional problems. (i) Two component force equations (x and y) are one moment equation (z). ∑ πΉπ₯ = 0β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ ∑ πΉπ¦ β‘ = 0β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ ∑ ππ΄ = β‘0 (ii) One component force equation (x or y) and two moment equations (both about different points in the z direction). ∑ πΉπ₯ = 0β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ ∑ ππ΄ β‘ = 0β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ ∑ ππ΅ = β‘0 (iii) Three moment equations (points A, B and C cannot be collinear). ∑ ππ΄ = 0β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ ∑ ππ΅ β‘ = 0β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ ∑ ππΆ = β‘0 Finding the Equilibrium Equations The initial step in determining the equilibrium equations, as with particles, is to build a free-body diagram of the body under consideration. This diagram should depict all of the force vectors that act on the body. In the free body diagram, provide values for any of the known magnitudes, directions, and points of application for the force vectors and provide variable names for any unknowns (either magnitudes, directions, or distances). The x and y axes must then be selected. These axes must be perpendicular to one another, but they are not required to be horizontal or vertical. You will simplify further analysis if you choose coordinate axes that correspond to some of your force vectors. After you have decided on axes, you must divide all the force vectors into components along the x and y axes. Your first equation will be the sum of the magnitudes of the components in the x direction equal to zero, and your second equation will be the sum of the magnitudes of the components in the y direction equal to zero. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 You will then need to generate the moment equations. To do this, you must first decide on a focal point for the moments. Any point should work however it is usually preferable to select one that reduces the number of unknowns in the equation. Keep in mind that every force vector that passes through a particular place has no moment about that point. To write down the moment equations, just add the moments exerted by each force about the specified point and axis (x or y) and divide by zero. Once you have your equilibrium equations, you can solve these formulas for unknowns. The number of unknowns that you will be able to solve for will again be the number or equations that you have. Here is a visual example of using the equilibrium equations: If we only consider the y (vertical) direction, the 200 lbs pushing down on the beam must be balanced by the reaction forces pushing up. The two reaction forces are equivalent because the forces on top are balanced evenly between the reaction forces. If they are at different locations, we use the sum of the moment’s equation and the distances of the people to determine the size of the reaction forces. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Support for Rigid Bodies Subjected to Two-Dimensional Force Systems Types of Connection Reaction Number of Unknowns One unknown. The reaction is a tension force which acts away from the member in the direction of the cable. One unknown. The reaction is a force which acts along the axis of the link. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contract. One unknown. The reaction is a force which acts perpendicular to the slot. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contract. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 One unknown. The reaction is a force which acts perpendicular to the surface at the point of contract ` One unknown. The reaction is a force which acts perpendicular to the rod. Two unknowns. The reactions are two components of force, or the magnitude and direction ∅ at the resultant force. Two unknowns. The reactions are the couple moment and the force which acts perpendicular to the rod. Three unknowns. The reactions are the couple moment and the two force components, or the couple moment and the magnitude and direction ∅ of the resultant force. Support Reactions. Before presenting a formal procedure as to how to draw a free-body diagram, we will first consider the various types of reactions that occur at supports and points of support between bodies subjected to coplanar force systems. Generally, if a support prevents the translation of a Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 body in a given direction, then a force is developed on the body in that direction. Likewise, if rotation is prevented, a couple moment is exerted on the body. For example, let us consider three ways in which a horizontal member, such as a beam, is supported at its end. One method consists of roller or cylinder, Fig (). Since this support only prevents the beam from translating in the vertical direction, the roller can only exert a force on the beam in this direction, Fig (). The beam can be supported in a more restrictive manner by using a pin as shown in Fig (). This pin passes through a hole in the beam and two leaves which are fixed to the ground. Here the pin can prevent translation of the beam in any direction ∅, Fig(), and so the pin must exert a force F on the beam in this direction. For purposes of this analysis, it is generally easier to represent this resultant force F by its two components ππ and ππ , Fig(). If πΉπ₯ and πΉπ¦ are known, then F and ∅ can be calculated. The most restrictive way to support the beam would be to use a fixed support as shown in Fig(). This support will prevent both translation and rotation of the beam, and so to do this, a force and couple must be developed on the beam at its point of connection, Fig (). As in the case of the pin, the force is usually represented by its components ππ and ππ . Static Determinacy/Partial and Improper Constraints Static Indeterminacy. It occurs when a system has more constraints than is necessary to hold the system in equilibrium (i.e., the system is over constrained and thus has redundant reactions). Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Static Determinacy. It occurs when a system has a sufficient number of constraints to prevent motion without any redundancy. Partial Constraint. It occurs when there is an insufficient number of reaction forces to prevent motion of the system (i.e., the system is partially constrained). Improper Constraint. It occurs when a system has a sufficient number of reaction forces but one or more are improperly applied so as not to prevent motion of the system (i.e., the syste m is improperly constrained). Sample Problem 1 A truss. Loaded as shown in the figure is supported by a hinge at A and a roller support D. Find the reactions at the supports. The roller D constrains the reaction to be vertical. The reaction at the hinge A is resolved into its components Ax and Ay. These three unknowns of the FBD of the truss are determined now by a Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 straightforward application of three basic equations of equilibrium. To simplify the computations, we choose the equations of equilibrium which will each contain one unknown. ∑ πΉπ₯ = 0 π΄π₯ − 200 = 0 π¨π = πππβ‘ππ To find the vertical component of D, moment about point A would eliminate the horizontal and vertical component of A. Hence, ∑ ππ΄ = 0 27β‘π·π¦ + 300(15) − 450β‘(9) − 450(18) π«π = πππ. ππβ‘ππ To determine Ay, moment summation about point D will give a result of, ∑ ππ· = 0 27β‘π΄π¦ − 450(18) − 450β‘(9) − 300(15) π¨π = πππ. ππβ‘ππ Sample Problem 2 In the fig. () assuming clockwise moment as positive, compute the moment of force F = 7.7 KN and of force P = 5.8 KN about points A, B, C and D. Each block is of 1m2. SOLUTION: tanβ‘θ1 β‘ = β‘3/4β‘ ⇒ β‘ θ1 β‘ = β‘36.86° tanβ‘θ2 β‘ = β‘3/2β‘ ⇒ β‘ θ2 β‘ = β‘56.3° Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 First, we find the moment of force F about points A, B, C and D F = 7.7 KN (1) About point A: Mπ΄ β‘ =β‘– Fβ‘cosβ‘36.86°β‘Xβ‘3β‘– β‘Fβ‘sinβ‘36.86°β‘Xβ‘1 Mπ΄ β‘ =β‘– 7.7β‘cosβ‘36.86°β‘Xβ‘3β‘– β‘7.7β‘sinβ‘36.86°β‘Xβ‘1 ππ¨ β‘ =β‘– ππ. ππβ‘ππ − π¦ (2) About point B: Mπ΅ β‘ = β‘Fβ‘cosβ‘36.86°β‘Xβ‘3β‘ + β‘Fβ‘sinβ‘36.86°β‘Xβ‘4 Mπ΅ β‘ = β‘7.7β‘cosβ‘36.86°β‘Xβ‘3β‘ + β‘7.7β‘sinβ‘36.86°β‘Xβ‘4 ππ© β‘ = β‘ππ. ππβ‘β‘ππ − π¦ (3) About point C: MπΆ β‘ = β‘Fβ‘cosβ‘36.86°β‘Xβ‘0β‘– β‘Fβ‘sinβ‘36.86°β‘Xβ‘5 MπΆ β‘ = β‘7.7β‘cosβ‘36.86°β‘Xβ‘0β‘– β‘7.7β‘sinβ‘36.86°β‘Xβ‘5 ππͺ β‘ = β‘ −ππ. ππβ‘β‘ππ − π¦ (4) About point D: Mπ· β‘ = Fβ‘cosβ‘36.86°β‘Xβ‘3β‘– β‘Fβ‘sinβ‘36.86°β‘Xβ‘1β‘ Mπ· β‘ = 7.7β‘cosβ‘36.86°β‘Xβ‘3β‘– 7.7β‘sinβ‘36.86°β‘Xβ‘1 ππ« β‘ = β‘ππ. ππβ‘β‘ππ − π¦ Now we find the moment of force P about points A, B, C and D P = 5.8 KN (1) About point A: Mπ΄ β‘ =β‘– Pβ‘cos56.3°β‘Xβ‘3β‘ + β‘Pβ‘sinβ‘56.3°β‘Xβ‘2 Mπ΄ β‘ =β‘– 5.8β‘cos56.3°β‘Xβ‘3β‘ + β‘5.8β‘sinβ‘56.3°β‘Xβ‘2 ππ¨ β‘ = β‘π. πππβ‘β‘ππ − π¦ (2) About point B: Mπ΅ β‘ = β‘Pβ‘cosβ‘56.3°β‘Xβ‘3β‘β‘– β‘Pβ‘sinβ‘56.3°β‘Xβ‘3 Mπ΅ β‘ = β‘5.8β‘cosβ‘56.3°β‘Xβ‘3β‘β‘– β‘5.8β‘sinβ‘56.3°β‘Xβ‘3 ππ© β‘ = β‘ −π. ππππβ‘β‘ππ − π¦ (3) About point C: MπΆ β‘ =β‘– Pβ‘cos56.3°β‘Xβ‘0β‘– Pβ‘sinβ‘56.3°β‘Xβ‘4 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 MπΆ β‘ =β‘– 5.8β‘cos56.3°β‘Xβ‘0β‘– 5.8β‘sinβ‘56.3°β‘Xβ‘4 ππͺ β‘ = β‘ −ππ. ππππβ‘ππ − π¦ (4) About point D: Mπ· β‘ =β‘– Pβ‘cosβ‘56.3°β‘Xβ‘3β‘– Pβ‘sinβ‘56.3°β‘Xβ‘2 Mπ· β‘ =β‘– 5.8β‘cosβ‘56.3°β‘Xβ‘3β‘– 5.8β‘sinβ‘56.3°β‘Xβ‘2 ππ« β‘ = β‘ −ππ. πππβ‘β‘ππ − π¦ Exercise Problem 1 Determine the reactions at supports of simply supported beam of 6m span carrying increasing load of 1500N/m to 4500N/m from one end to other end. Exercise Problem 2 The frame shown supports part of the of roof a small building. Given that the tension of the cable is 170 kN, determine the reaction at the fixed end E. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Exercise Problem 3 The weight W of a traveling crane is 40 tons. To prevent the crane from tipping to the right when carrying a load P of 40 tons, a counterweight S is used. Determine the value and position of S so that the crane will remain in equilibrium both when the maximum load P is applied and when the load P is removed. : Exercise Problem 4 A fine light string ABCDE whose extremity A is fixed, has weights W1 and W2 attached to it at B and C. It passes round a small smooth peg at D carrying a weight of 60N at the free end E as shown in fig (). If in the position of equilibrium, BC is horizontal and AB and CD makes 150° and 120° with BC, find (a) Tension in the portion AB, BC, and CD of the string and (b) Magnitude of W1 and W2. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Exercise Problem 5 If the roller at A and the pin at B can support a load up to 4 kN and 8 kN, respectively, determine the maximum intensity of the distributed load W, measured in kN/m, so that failure of the supports does not occur. FRICTION Friction is defined as a force of resistance acting on a body that prevents or delays the body's slipping relative to another body or surface with which it is in contact. This force is always tangential to the surface at points of contact with other bodies and is directed to oppose the body's possible or existing motion relative to these points. There are two types of friction that happen between surfaces: Fluid Friction. It exists when the surfaces that come into contact with each other are separated by a fluid film (liquid or gas). The nature of fluid friction is studied in fluid mechanics because it is dependent on knowing the fluid's velocity and its ability to resist shear force. Skin Friction. Skin friction, also known as friction drag, is a component of the force that prevents a solid body from moving through a fluid. Internal Friction. Internal friction is related to shear deformation of solid materials under cyclical loading. Internal friction may accompany deformation that occurs during loading. Dry Friction. It describes the friction force between two unlubricated solid surfaces. Dry friction always opposes the sliding surfaces relative to one another, and it can either oppose or cause motion in bodies. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Coulomb’s Theory of Dry Friction Source: Andrew Pytel and Jaan Kuisalaas. “Engineering Mechanics – STATICS”. 4th Edition. Page 342. Accessed November 16, 2022. https://www.academia.edu/38359974/Pytel_Mechanical_Engineering_Statics_4th_pdf . 1 . Coulomb's theory is best explained by imagining two bodies in contact with each other, as illustrated in Figure (a). Although this figure depicts a single point of contact, the following discussion also applies to a finite area of contact. At the point of contact, the plane of contact shown is tangent to both bodies. The free-body diagrams of the bodies are shown in Figure 3.1(b), where N is the normal contact force and F is the friction force. The force N is perpendicular to the plane of contact, whereas the force F is perpendicular to the plane of contact. Elements of Dry Friction N = Total reaction perpendicular to the contact surface f = Friction force μ = Coefficient of friction R = Resultant of f and N Ο = angle of friction 1 https://www.academia.edu/38359974/Pytel_Mechanical_Engineering_Statics_4th_pdf Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Source: Mathalino (ND). “Friction”. Retrieved from https://mathalino.com/reviewer/engineering-mechanics/friction. Accessed November 19, 2022. Law of Dry Friction Source: Katie Bowlan (2017). “Friction”. Retrieved https://slideplayer.com/slide/3958461/ Accessed November 18, 2022. • • • • 2 from A block of weight (W) is located on a horizontal surface. The weight of the block and the surface reaction are the forces acting on it (N). A small horizontal force (P) is applied to the block. A horizontal component (F) of the surface reaction is required for the block to remain stationary, in equilibrium. (F) is a static-friction force. As (P) increases, so does the static-friction force (F), until it reaches a maximum value (Fm). Further increase in (P) causes the block to begin to move as (F) drops to a smaller kineticfriction force (Fk). Coefficient of Friction Coulomb's theory is made up of several postulates, all of which are explained in the following. A. Static Case – Coulomb proposed the following law: If no relative motion exists between two surfaces in contact, the normal force N and friction force F satisfy the following relationship. where Fmax denotes the maximum static friction force that can exist between the contacting surfaces and µs denotes the static friction coefficient. The coefficient of static 2 https://www.academia.edu/38359974/Pytel_Mechanical_Engineering_Statics_4th_pdf Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 friction is an experimental constant that depends on the contacting surfaces' composition and roughness. The equation simply states that the friction force F under static conditions (no relative motion) has an upper limit that is proportional to the normal force. B. Impending sliding – Consider the static case in which the friction force equals its limiting value; that is, The surfaces are on the verge of sliding in this case, which is known as impending sliding. The surfaces are at rest relative to each other when sliding impends. Any change that requires an increase in friction force, on the other hand, will cause sliding. Fmax's direction can be determined by the fact that Fmax always opposes impending sliding. C. Dynamic case – If the two contact surfaces are sliding relative to each other, the friction force F is postulated to be where N denotes the contact normal force, µk represents the coefficient of kinetic friction, and Fk denotes the kinetic, or dynamic, friction force. The coefficient of kinetic friction is usually smaller than its static counterpart. As in the static case, Fk always opposes sliding. Characteristics of Dry Friction As a result of experiments that pertain to the foregoing discussion, the following rules which apply to bodies subjected to dry friction may be stated. • The frictional force acts in a direction opposite to the relative motion or tendency for motion of one surface against another. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 • • • • The maximum static frictional force F that can be developed is independent of the area of contact, as long as the normal pressure is not too low or too high to severely deform or crush the bodies' contacting surfaces. For any two surfaces of contact, the maximum static frictional force is generally greater than the kinetic frictional force. However, if one of the bodies moves at a very low velocity over the surface of the other, Fk approaches Fs, i.e., µs ≈ µk. When slipping at the contact surface is imminent, the maximum static frictional force is proportional to the normal force, such that Fs = μsN. When slipping occurs at the surface of contact, the kinetic frictional force is proportional to the normal force, so Fk = µkN Four Situations can occur in Friction System: Source: Katie Bowlan (2017). “Friction”. Retrieved from https://slideplayer.com/slide/3958461/ Accessed November 18, 2022. • • • • There is no friction force; the forces applied to the body do not tend to move it along the surface of contact (Fig. a). The applied forces tend to move the body along the surface of contact but are not strong enough to set it in motion. We can calculate the static-friction force F by solving the body's equilibrium equations. Because there is no evidence that F has reached its maximum value, the equation Fm = μsN cannot be used to calculate the friction force (Fig. b). The applied forces are such that the body is on the verge of sliding. That motion is said to be imminent. The friction force F has reached its maximum value Fm, balancing the applied forces with the normal force N. Both equilibrium equations and the equation Fm = μsN can be used. It should be noted that the friction force has an opposite sense to the sense of impending motion (Fig. c). The body is sliding as a result of the applied forces, and the equilibrium equations no longer apply. F, on the other hand, is now equal to Fk, and we can use the equation Fk = μkN. The sense of Fk is diametrically opposed to the sense of motion (Fig. d). Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 3 Angle of Friction Source: Katie Bowlan (2017). “Friction”. Retrieved from https://slideplayer.com/slide/3958461/ Accessed November 18, 2022. • • • Consider a weight W block resting on a horizontal plane surface. When no horizontal force is applied to the block, the resultant R equals the normal force N. If the applied force P has a horizontal component Px that tends to move the block, force R has a horizontal component F and, thus, forms an angle ΙΈ with the normal to the surface (Fig. b). When Px is increased until motion is imminent, the angle between R and the vertical grows to its maximum value. This is known as the angle of static friction, and it is denoted by ΙΈs. Angle of Static Friction • tan ΙΈs = Fm/N = µsN/N • tan ΙΈs = µs Angle of Kinetic Friction • tan ΙΈk = Fk/N = µkN/N • tan ΙΈk = µk 4 Application of Friction 3 https://www.studocu.com/ph/document/cebu-institute-of-technology-university/dynamics-of-rigid-bodies/frictionlecture-notes-in-dynamics-of-rigids-bodies-online-class-given-by-ma-carmen/30965159 4 https://dewwool.com/what-are-the-applications-of-friction-in-our-daily-lives/ Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 • • • • • Tire Traction – The friction between your car's tires and the road surface that allows it to accelerate, decelerate, and negotiate turns and corners. Humans being able to walk – Friction helps us to walk The meteor's friction with the atmosphere heats it up, causing it to burn and shrink in size due to air resistance. Friction is used to start fires by rubbing wood against wood or stone against stone. The use of friction is used by pulleys to draw water from wells. Sample Problem 1 A 300-lb block is in contact with a plane inclined at 30 degrees to the horizontal. A force P, parallel to the acting up the plane is applied to the body. If the coefficient of statics friction is 0.20, find: a. The value of P to just cause motion to impend up the plane, and; b. Find P to just prevent motion down the plane. (a) (b) If the block has impending motion up the plane, the friction force, acting at its maximum available value, will act down the plane to resist motion. The FBD of the block is shown in (a). Thus, Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 ∑ πΉπ¦ = 0 πΉ = ππ π − 300 cos 30 = 0 πΉ = 0.2(259.81β‘ππ) π΅ = πππ. ππβ‘ππ π = ππ. πππβ‘ππ To determine P, we apply the equilibrium equation π − πΉ − 300π ππ30 = 0 π = 51.962 + 300 sin 30 π· = πππ. πππβ‘ππ When the block has impending motion down the plane, the friction force, acting at its maximum value, will act up the plane to resist its motion. The FBD is shown in (b). Hence, 300 sin 30 − π − πΉ = 0 300 sin 30 − π − 51.962 = 0 π· = ππ. πππβ‘ππ Sample Problem 2 The uniform rod having a weight W and length t is supported at its ends against the surface at A and B in Fig (). If the rod is on the verge of slipping when π = 30°, determine the coefficient of static friction ππ at A and B. Neglect the thickness of the rod for the calculation. SOLUTION Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 As shown in the FBD, Fig (), there are five unknowns: πΉπ΄ , ππ΄ , πΉπ΅ , ππ . Equations of Friction and Equilibrium: πΉ = β‘ ππ N;β‘ πΉπ΄ = β‘ ππ ππ΄ β‘ πΉπ΅ = β‘ ππ ππ΅ β‘ Using these results and applying the equations of equilibrium yields ∑ πΉπ₯ = 0;β‘β‘ππ ππ΄ + β‘ ππ ππ΅ πππ 30° − β‘ ππ΅ π ππβ‘30° = 0β‘ ∑ πΉπ¦ = 0;β‘β‘ππ΄ − π + β‘ ππ΅ πππ 30° + β‘ ππ ππ΅ π ππ30° = 0β‘ π ∑ ππ΄ = 0;β‘β‘ππ΅ π − π ( ) πππ 30° = 0β‘ 2 (1) (2) (3) ππ΅ = 0.4330β‘π From Eqs. 1 and 2 ππ ππ΄ = 0.2165π − (0.3750β‘π)ππ ππ΄ = 0.6250β‘π − (0.2165β‘π)ππ By division, 0.6250ππ − 0.2165ππ 2 = 0.2165 − 0.375ππ ππ 2 − 4.619ππ + 1 = 0 Solving for the smallest root, ππ = π. πππ Exercise Problem 1 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 The 700 lb block shown in figure has impending motion up the plane caused by the horizontal force of 1300 lb. Determine the coefficient of static friction between the contact surfaces Exercise Problem 2 A block weighing 5KN is attached to a chord, which passes over a frictionless pulley, and supports a weight of 2KN. The coefficient of friction between the block and the floor is 0.35. Determine the value of force P if (a) The motion is impending to the right (b) The motion is impending to the left. Exercise Problem 3 Determine the range of mass m for which the 125 kg block is in equilibrium. All wheels and pulleys have negligible friction. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Exercise Problem 4 Determine the value of P just sufficient to start the 10° wedge under the 40-kN block. The angle of friction is 20° for all contact surfaces. Exercise Problem 5 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Block A weighing 2000N rests over block B which weights 3000N as shown in fig (). Block A is tied to wall with a horizontal string. If the coefficient of friction between A and B is 1/4 and between B and floor is 1/3, what should be the value of P to move the block B. If (a) P is horizontal (b) P is at an angle of 300 with the horizontal. WEDGES A wedge is a simple device that serves the same function as a lever; it creates tension. It is used to convert applied forces into much larger forces. Directed at an angle to the applied force, wedges can also be used to transfer or adjust heavy loads. Wedges can be used in two different ways: Low friction wedges. Simple machines that allow users to generate large normal forces in order to move objects with relatively small input forces. High-friction wedges. Also known as self-locking wedges that direct or hold objects in place. Transferring forces with wedges is frequently discussed in terms of mechanical advantage (M.A.) as: M.A. = direct force / wedge force The numerator in this formula is the amount of direct force required to complete the task normally, and the denominator is the force applied to the wedge to complete the same task. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Consider the wedge shown below, which is being pushed into a crack by the applied force P. The angle created by the tip of the wedge is 2β; where β is known as the wedge angle. Ignoring the weight, the wedge's free-body diagram at impending sliding is shown in figure B. As before, we let φ be the angle between the contact force R and the normal n to the contact surface. Because sliding impends, φ = φs, where φs = tan−1 µs is the angle of friction. From the force diagram figure C, we see that R = P/[2 sin(φs +β)], which is substantially greater than P if the wedge angle β is small and the sides of the wedge are lubricated (giving a small value for φs). A wedge should ideally be slippery enough to be easily driven into the crack. However, it must have enough friction to remain in place when the driving force is removed. In the absence of P, the wedge transforms into a two-force body. As a result, the contact forces R must be collinear, as shown by the free-body diagram in figure below. Given that equilibrium can exist if and only if φ ≤ φs, we arrive at the conclusion that the wedge will remain in place as long as β ≤ φs. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Sample Problem 1 Determine the force P required to start the wedge shown in the figure. The angle of friction for all surfaces in contact is 15 degrees. (a) (b) From the FBD of the block to the right ∑ πΉπ¦ = 0 π 1 β‘πππ β‘15 = β‘ π 2β‘ π ππβ‘15 + 141.4β‘ππβ‘ π 1 β‘πππ β‘15 = β‘ π 2β‘ π ππ16 + 141.4 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 πΉπ = β‘ πΉπβ‘ πππππ+πππ.π πππβ‘ππ – equation 1 π 2 πππ 15 = π 1 π ππ15 + 50 π 2β‘ π ππ15+141.4 π 2β‘ πππ β‘15 = [( πππ β‘15 ) (π ππβ‘15)] + 50 – Substitute the value of R1 or the equation 1 π 2β‘ πππ 15 = (π 2β‘ π ππ15 + 141.4)π‘ππ15 + 50 π 2β‘ πππ 15 = π 2β‘ π ππ15β‘π‘ππ15 + 141.4π‘ππ15 + 50 π 2β‘ πππ 15 − π 2β‘ π ππ15β‘π‘ππ15 = 141.4π‘ππ15 + 50 πΉπβ‘ = ππ. ππβ‘ππ΅ ∴ πΉπβ‘ = πππ. ππβ‘ππ΅ From the FBD of the wedge to the left ∑ πΉπ₯ = 0 ∑ πΉπ¦ = 0 π 3 β‘πππ β‘30 = β‘ π 2β‘ πππ β‘15β‘ π = π 2 sin 15 + β‘ π 3 sin 30 π 3 β‘πππ β‘30 = β‘98.03β‘πππ β‘15 π = 98.03 sin 15 + β‘109.34 sin 30 π 3 = β‘109.34β‘ππ π· = ππ. ππβ‘ππ΅ Sample Problem 2 To adjust the vertical position of a column supporting 300-kN load, two 5° wedges are used as shown in Fig. (). Determine the force P necessary to start the wedges is the angle of friction at all contact surfaces is 25°. Neglect friction at the rollers. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SOLUTION From the FBD of the upper wedge ∑ πΉπ = 0 π 2 πππ 30° = 300 π 2 = 346.41β‘ππ From the FBD of the lower wedge ∑ πΉπ = 0 π 3 πππ 25° = π 2 πππ 30° π 3 πππ 25° = 346.41πππ 30° π 3 = 331.01β‘ππ ∑ πΉπ» = 0 π = β‘ π 2 π ππ30° + β‘ π 3 π ππ25° π = β‘346.41π ππ30° + β‘331.01π ππ25° π· = β‘πππ. ππβ‘ππ΅ Exercise Problem 1 Two 5° wedges are used to adjust the vertical position of a column supporting a 450-kN load, as shown in the figure. If the angle of friction at all contact surfaces is 25°, calculate the force P required to start the wedges. Regard the friction at the rollers. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 From the FBD of the upper wedge ΣπΉπ = 0 R 2 πππ 30° = 450 R 2 = 519.6152β‘ππ From the FBD of the lower wedge ΣπΉπ = 0 R 3 πππ 25° = R 2 πππ 30° R 3 πππ 25° = 519.6152πππ 30° R 3 πππ 25° = 450β‘ππ ΣπΉπ» = 0 π = R 2 π ππ30° + R 3 π ππ25° π = 519.6152β‘ + 450β‘ π = 969.6152β‘ππβ‘ Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Exercise Problem 2 In fig (), determine the minimum weight of block B that will keep it at rest while a force P starts blocks A up the incline surface of B. The weight of A is 200 lb and the angle of friction for all surfaces in contact is 15°. Exercise Problem 3 Determine the force P required to start the wedge shown in Fig. P-541. The angle of friction for all surfaces in contact is 15°. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Exercise Problem 4. Determine the horizontal force P in the figure to start the 500-lbs wedge (a) moving to the right and (b) moving to the left, with the angle of friction at 20º at all contact surfaces. Exercise Problem 5 Find the largest weight of the wedge to cause motion with μ = 0.5 at A and C, and μ = 0.6 at point B. Centroid Center of Gravity of a Two-Dimensional Body Consider a flat horizontal plate in the figure. We can divide plate into n small elements. The coordinates of the first element are denoted by π1 and π1 , those of the second element by π2 and π2 etc. The forces exerted by the earth on the elements of plate will be denoted, respectively, by βπ1, βπ2, …, βππ . These forces or weights are directed towards the center of the earth; however, for all practical purposes they can be assumed to be parallel. Their resultant is therefore a single Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 force in the same direction. The magnitude W of this force is obtained by adding the magnitudes of the elemental weights ∑ πΉπ§ = π = βπ1 , +βπ2 + β― , βππ .β‘ To obtain the coordinates x and y of the point G where the resultant W should be applied, we write that the moments of W about the y and x axes are equal to the sum of the corresponding moments of the elemental weights, ∑ ∑ ππ¦:β‘β‘β‘π₯π = π1 βπ1 , +β‘π2 βπ2 + β― , ππ β‘βππ β‘ ππ₯:β‘β‘β‘β‘π¦π = π¦1 βπ1 , +β‘π¦2 βπ2 + β― , π¦π β‘βππ β‘ If we now increase the number of elements into which the plate is divided and simultaneously decrease the size of each element, we obtain in the limit the following expressions: π =β‘∫ ππβ‘β‘β‘β‘β‘β‘β‘β‘π₯π = ∫ π₯ππβ‘β‘β‘β‘β‘β‘β‘β‘π¦π = ∫ π¦β‘ππ These equations define the weight W and the coordinates x and y of the center of gravity G of a flat plate. The same equations can be derived for a wire lying in the xy plane. We note that the center of gravity G of a wire is usually not located on the wire. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Centroids of Areas and Lines In the case of flat homogeneous plate of uniform thickness, the magnitude βπ of the weight of an element of the plate can be expressed as βπ = β‘πΎπ‘βπ΄ Where πΎβ‘= specific weight (weight per unit volume) of the material t = thickness of the plate βπ΄ = area of the element Similarly, we can express the magnitude W of the weight of the entire plate as π = β‘πΎπ‘π΄ Where A is the total area of the plate If U.S . customary units re used, the specific weight πΎ should be expressed in ππ/ππ‘ 3 , the thickness t in feet, and the areas βπ΄ and A in square feet. We observe that βπ and W will then be expressed in pounds. If SI units are used, πΎβ‘should be expressed in π/π3 , t in meters, and the areas βπ΄β‘πππβ‘π΄ in square meters: the weights βπ and W will then be expressed in newtons. Substituting for βπ and W in the moment equation and dividing throughout but πΎπ‘, we obtain, ∑ ππ¦:β‘β‘β‘π₯π΄ = π1 βπ΄1 , +β‘π2 βπ΄2 + β― , ππ β‘βπ΄π β‘ ∑ ππ₯:β‘β‘β‘β‘π¦π = π¦1 βπ΄1 , +β‘π¦2 βπ΄2 + β― , π¦π β‘βπ΄π β‘β‘ If we increase the number of elements into which the area A is divided and simultaneously decrease the size of each element, we obtain in the limit, π₯π΄ = β‘ ∫ π₯β‘ππ΄β‘β‘β‘β‘π¦π΄ = β‘ ∫ π¦β‘ππ΄β‘β‘ These equations define the coordinates x and y of the center of gravity of a homogeneous plate. The point whose coordinates are x and y is also known as the centroid C of the area A of the plate. If the plate is not homogeneous, these equations cannot be used to determine the center of gravity of the plate; they still define, however, the centroid of the area. In the case of homogeneous wire of uniform cross section, the magnitude βπ of the weight of an element of wire can be expressed as Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 βπ = β‘πΎπβ‘βπΏ Where πΎ = specific weight of the material π = cross-sectional area of the wire βπΏ = length of the element Centroid of an Area Centroid of a line The centroid of an object represents the average location of all particles of the object. It can be defined for objects of any dimension, such as lines, areas, volumes or even higher dimension objects. It is a purely geometrical property, in contrast to the center of mass (also called center of gravity), which takes into account the mass distribution in the object. For objects with uniform mass distribution, the centroid is also the center of mass. Simply, centroid is the geometrical center of a plane figure: a curve, area, or volume. It is the average position of all the points of an object. To find the location of the centroid of some geometrical shape, here is the tabulation of formula: Shape Area π₯ π¦ 2ππΌ 2 π ππ π ππβ‘πΌβ‘ πΌ 0 Circular Arc Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Circular Sector 2π 2 πΌ 2π π ππ π ππβ‘πΌβ‘ 3πΌ 0 ππ π 2π π ππ 2 2 r 4π 3π ππ 2 2π π 2π π Semi-circular Arc Semi-circular Area Quarter Circular Arc Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Quadrant of a Circle ππ 2 4 4π 3π 4π 3π πβ π 2 β 2 πππ 4 4π 3π 4π 3π πβ 2 2π 3 β 3 Rectangular Area Quadrant of an Ellipse Triangular Area (straight h) Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Triangular Area (slant h) πβ 2 π+π 3 β 3 πβ 3 3π 4 3β 10 2πβ 3 5π 8 2β 5 Parabolic Spandrel Quadrant of a Parabola Source: https://faculty.utrgv.edu/constantine.tarawneh/Fluid%20Mechanics/Centroid%20Locations.pdf https://faculty.mercer.edu/jenkins_he/documents/Centroids.pdf SAMPLE PROBLEM 1 Locate the centroid of the shaded region in the figure. At first glance, we can notice three shapes or areas present in the figure: quarter circle, triangle, and the whole rectangular shape. Using the formulas of each shape given, compute for its area and the centroid of each shape with respect to x and y origin. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 For rectangle: π π΄1 = πβ π1 = 2 π΄1 = 55 + 55(80) π1 = π΄1 = 4455β‘ππ2 π1 = 55β‘ππ 110 2 β π1 = 2 π1 = 80 2 π1 = 40β‘π For triangle: 1 π΄2 = 2 πβ 1 2 π2 = 55 + 3 π 2 1 π2 = 3 β 1 π΄2 = 2 (55)(80) π2 = 55 + 3 (55) π2 = 3 (80) π΄2 = 2200β‘ππ2 π2 = 91.67β‘ππ π2 = 26.67β‘ππ For quarter circle: π΄3 = π΄3 = ππ 2 4 π(40)2 4 π΄3 = 400β‘πβ‘0πβ‘1256.64β‘ππ2 4π π3 = 3π π3 = 4(40) 3π π3 = 16.98β‘ππ 4π π3 = 70 − 3π π3 = 70 − 4(40) 3π π3 = 53.02β‘ππ After getting the area of each shape, calculate for the total area covered by the figure. ∑ π΄ = β‘ π΄1 + π΄2 + π΄3 ∑ π΄ = β‘4455β‘ππ2 − 2200β‘ππ2 − 1256.02β‘ππ2 ∑ π¨ = β‘πππ. ππβ‘πππ Since the area of the rectangular shape being calculated earlier was the whole area of the figure, we have to subtract the two areas of quarter circle and triangle to determine its total area. To get the centroid of the figure @ x and y, get the summation of the area times the corresponding x and y values over total area. For instance, Centroid @ x: Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 π₯ =β‘ π₯ =β‘ π΄1 π1 + π΄2 π2 + π΄3 π3 + π΄4 π4 π΄ (4455)(55) − 2200(91.67) − 1256.64(16.98) 998.36 π₯ = β‘22.05β‘ππβ‘ Centroid @ y: π¦ =β‘ π¦ =β‘ π΄1 π1 + π΄2 π2 + π΄3 π3 π΄ (4455)(40) − 2200(53.02) − 1256.64(26.67) 998.36 π¦ = β‘52.99β‘ππβ‘ SAMPLE PROBLEM 2 Using the given figure, find the coordinates of the centroid in a slender homogeneous wire of uniform cross section that is bent into the shape. Solution: L1= 34 in. x1= 21 in. y1= 17 in. L2= π (21) = 65.97 in. x2=0 y2= 2(21) π = 13.37 in. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 L3 = 51 in. x3 = 25.5 + 25.5cos45° = 43.53 in. y3 = 25.5sin45° = 18.03 in. L = L1+L2+L3 L = 34 + 65.97 + 51 L = 150.97 in. L π₯β‘= Σlx 150.97 π₯β‘= 34(−25.5) − 150.97 (0) + 51 (43.53) π₯β‘= 1353.03 in. answer Lπ¦= Σly 150.97 π¦ = 34(17) + 150.97 (13.37) + 51 (18.03) π¦β‘= 23.29 in. answer EXERCISE PROBLEMS 1. The figure shown is made from a piece of thin, homogeneous wire. Determine the location of its center of gravity. 2. Locate the centroid of the figure below Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 3. Determine y, which locates the centroidal axis x′ for the cross-sectional area of the Tbeam, and then find the moments of inertia Ix′ and Iy′. 4. A slender homogeneous wire of uniform cross section is bent into the shape shown in the figure. Determine the coordinates of the centroid. 5. An I section has the following dimensions Top flange = 8 cm × 2 cm; Bottom flange = 12 cm × 2 cm; Web = 12 cm × 2 cm; Over all depth of the section = 16 cm. Determine the MI of the I section about two centroidal axis. Theorem of Pappus Pappus theorem, a theorem by Pappus of Alexandria, enable us to work out the volume of a solid of revolution if we know the position of the centroid of a plane area, or vice versa; or to work out the area of a surface of revolution if we know the position of the centroid of a plane curve or vice versa. It is not necessary that the plane or the curve be rotated through a full 360o. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Pappus's theorem is actually two theorems that allow us to find surface areas and volumes without using integration. 1. If a region in a plane is rotated along an axis within that plane but one that does not pass through the area, the volume swept out equals the area times the distance moved by the centroid. Formula: V = Ad Or V = 2βΌyA Where yA = yd 2. If a region in a plane is rotated along an axis within that plane but one that does not pass through the area, the area swept out equals the length times the distance moved by the centroid. Formula: A = Ld Or A= 2βΌyL Where yL = ydL SAMPLE PROBLEM 1 Find the centroid of a uniform semicircle of radius πΉ. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SOLUTION: Let π be the distance between the centroid πΊ and the axis of rotation. When the semicircle makes the full turn, the path π traversed by the centroid is equal to π = 2ππ The solid of rotation is a ball of volume 4ππ 3 π =β‘ 3 By the 2nd theorem of Pappus, we have the relationship π = π΄π, Where π΄ = β‘ Hence, ππ 3 2 is the area of the semicircle. π π = β‘ 2ππ΄ = β‘ 4ππ 3 3 ππ 3 2πβ‘π₯β‘ 2 4π β‘ = β‘ 3π = β‘β‘π. ππβ‘πΉ EXERCISE PROBLEMS 1. A triangle with the vertices π΄(1,2), π΅(2,6), πΆ(6,2) is rotated about the axis. Find the volume of the solid of revolution thus obtained. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 2. The outside diameter of a pulley is 0.8 m, and the cross section of its rim is as shown. Knowing that the pulley is made of steel and that the density of steel isπ = 7.85π₯103 β‘ππ/π3 , determine the mass and the weight of the rim. 3. Calculate the volume V of the solid generated by revolving the 60-mm right triangular area through 180° about the z-axis 4 Find the centroid of the region bounded by the parabola y=4−x2 and the x−axis. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 MOMENT OF INERTIA Introduction In this chapter, we discuss the second moments of plane areas, also known as moments of inertia, ∫ π₯ 2 β‘ππ΄ and ∫ π¦ 2 β‘ππ΄. We also introduce the product of inertia ∫ π₯π¦β‘ππ΄. Moments and products of inertia arise in the analysis of linear load distributions acting on plane areas. Such distributions occur in members subjected to bending (beams), and in circular shafts carrying twisting couples. In addition, moments and products of inertia are encountered in the determination of resultants acting on submerged surfaces. This chapter also covers the relationship between moments and products of inertia on the coordinate system's orientation. The transformation equations for moments and products of inertia, which are used to calculate the maximum and minimum moments of inertia at a place, are produced as a result of this relationship. The description of Mohr's circle, a visual representation of the transformation equations, comes at the end of this chapter. Moments of Inertia of Areas and Polar Moments of Inertia a. Moment of Inertia of Area The first moments of the area of a plane region about the x- and y-axes were defined as (1) Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 where π΄ is the area of the region and x and y are the coordinates of the differential area element ππ΄, as shown in Fig. (). Fig () The moments of inertia of the area about the x- and y-axes, respectively, are defined by π°π = β‘β‘β‘β‘β‘ ππ β‘π π¨ π°π = β‘β‘β‘β‘β‘ ππ β‘π π¨ (2) Because the distances x and y are squared, πΌπ₯ and πΌπ¦ are sometimes called the second moments of the area. The dimension for moment of inertia of area is [πΏ4 ]. Therefore, Therefore, the units are ππ.4, ππ4 , and so forth. Although the first moment of an area can be positive, negative, or zero, its moment of inertia is always positive, because both x and y in the equation (2) are squared. Caution. Recall that the first moment of an area can be obtained from ππ₯ = π΄yΜ ,β‘where yΜ is the centroidal coordinate of the area. A mistake frequently made is to assume that πΌπ₯ = π΄yΜ 2 . Although the first moment of an area equals the area times the centroidal distance, the second moment of an area is not equal to the area times the centroidal distance squared. b. Polar Moment of Inertia Referring again to Fig. (), the polar moment of inertia of the area about point O (strictly speaking about an axis through O, perpendicular to the plane of the area) is defined by π±πΆ = β‘β‘β‘β‘β‘ ππ β‘π π¨ (3) where r is the distance from O to the differential area element ππ΄. Note that the polar moment of an area is always positive, and its dimension is [πΏ4 ]. From Fig. (), we note that π 2 =β‘ π¦ 2 + β‘β‘ π₯ 2 , which gives the following relationship between polar moment of inertia and moment of inertia: Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 or π±πΆ = β‘ π°π + β‘ π°π (4) This relationship states that the polar moment of inertia of an area about a point O equals the sum of the moments of inertia of the area about two perpendicular axes that intersect at O. c. Parallel-axis Theorems There is a simple relationship between the moments of inertia about two parallel axes, provided that one of the axes passes through the centroid of the area. Referring to Fig. () (a), let C be the centroid of the area contained in the plane region A and let the π₯ ′ -axis be the centroidal axis that is parallel to the x-axis. We denote the moment of inertia about the π₯ ′ -axis by πΌΜ π₯ , which is to be read as the “moment of inertia about the centroidal x-axis” (about the axis that is parallel to the x-axis and passes through the centroid of the area). Observe that the y-coordinate of the differential area ππ΄ can be written as π¦ = β‘ π¦Μ + β‘ π¦ ′ , where π¦Μ (the centroidal coordinate of the area) is the distance between the two axes. Equations (2) yield the following expression for the moment of inertia of the area about the x-axis (note that y is constant). A (a) Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Fig. () Noting that ππ΄ = π΄ (the area of the region), π¦ ′ ππ΄ = 0 (the first moment of the area about a centroidal axis vanishes), and (π¦ ′ )2 ππ΄ = πΌΜ π₯ (the second moment of the area about the xaxis), Eq. (a) simplifies to Μ π π°π = β‘ π°Μ π + π¨π (5a) This relationship is known as the parallel-axis theorem for moment of inertia of an area. The distance π¦Μ is sometimes called the transfer distance (the distance through which the moment of inertia is to be “transferred”). It is important to remember that the theorem is valid only if πΌΜ π₯ is the moment of inertia about the centroidal x-axis. If this is not the case, the term β‘β‘β‘β‘β‘β‘π¦ ′ ππ΄ in Eq. (a) would not vanish, giving rise to an additional term in Eq. (5a). Because the direction of the x-axis in Fig. () (a) can be chosen arbitrarily, the parallel-axis theorem applies to axes of any orientation. For example, applying the theorem to the y-axis yields Μ π β‘ π°π = β‘ π°Μ π β‘ + π¨π (5b) where πΌΜ π₯ is the moment of inertia of the area about the centroidal y-axis —that is, the π¦ ′ -axis in Fig. () (a), and π₯Μ is the x-coordinate of the centroid. In general, the parallel-axis theorem can be written as π°πͺ = β‘ Μ Μ Μ π°πͺ + π¨π π (6) As illustrated in Fig. ()(b),β‘πΌπΆ is the moment of inertia about an arbitrarily oriented πΆ-axis, Μ πΌπΆ represents the moment of inertia about the parallel axis that passes through the centroid C, and d is the distance between the axes (transfer distance). By inspection of Eq. (6), we see that, given the direction of the axis, the moment of inertia of an area is smallest about the axis that passes through the centroid of the area. In other words, πΌΜ πΆ is smaller than the moment of inertia about any other axis that is parallel to the πΆ-axis. The parallel-axis theorem also applies to the polar moment of inertia. Denoting the polar moment of inertia of the area about the origin O by π½π , and about the centroid C by π½Μ πΆ , we have from Eqs. (4) and (5). π½π = β‘ πΌπ₯ + β‘ πΌπ¦ = (πΌΜ π₯ β‘ + β‘π΄π¦Μ 2 ) + (πΌΜ π¦ β‘ + β‘π΄π₯Μ 2 ) Using πΌΜ π₯ + β‘ πΌΜ π¦ = β‘β‘ π½Μ πΆ , this equation becomes π±πΆ = β‘ π±Μ πͺ + β‘π¨πΜ π β‘β‘ (7) Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 where πΜ = β‘ √π₯Μ 2 + β‘ π¦Μ 2 is the distance between points O and C, as shown in Fig. 9.2(a). d. Radius of Gyration In some structural engineering applications, it is common practice to introduce the radius of gyration of area. The radius of gyration of an area about the x-axis, the y-axis, and the origin O are defined as π°π π°π π±πΆ ππ = β‘ √ β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ππ = β‘ √ β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ππΆ = β‘ √ β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘ π¨ π¨ π¨ (8) The dimension of the radius of gyration is [L]. However, the radius of gyration is not a distance that has a clear-cut physical meaning, nor can it be determined by direct measurement; its value can be determined only by computation using Eqs. (8). The radius of gyration are related by the equation ππ πΆ = β‘ ππ π + ππ π (9) which can be obtained by substituting Eqs. (8) into Eq. (4). e. Integration Techniques When computing the moment of inertia of an area about a given axis by integration, we must choose a coordinate system and decide whether to use single or double integration. The differential area elements ππ΄ associated with various coordinate systems were discussed in the previous chapter. If double integration is used, the moments of inertia can be calculated from Eqs. (2) in a straightforward manner. However, in single integration we must view Eqs. (2) in the form where ππΌπ₯ and ππΌπ¦ are the moments of inertia of the area element ππ΄ about the x- and y-axes. In general, ππΌπ₯ = β‘ π¦ 2 ππ΄ only if all parts of the area element are the same distance y from the x-axis. To satisfy this condition, the area element must be either a double differential element (ππ΄β‘ = β‘ππ₯β‘ππ¦), or a strip of width dy that is parallel to the x-axis. A similar argument applies to ππΌπ¦. f. Method of Composite Areas Consider a plane region that has been divided into the subregions π1 , π2 , π3 , …. The moment of inertia of the area of π about an axis can be computed by summing the moments of inertia of the subregions about the same axis. This technique, known as the method of composite Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 areas, follows directly from the property of definite integrals: the integral of a sum equals the sum of the integrals. For example, πΌπ₯ , the moment of inertia about the x-axis, becomes which can be written as π°π = β‘ (π°π )β‘π + β‘ (π°π )β‘π + β‘ (π°π )β‘π + β―β‘ (10a) where (β‘πΌπ₯ )π is the moment of the inertia of the area of the subregion π1 with respect to the x-axis. Obviously, the method of composite areas also applies to the computation of polar moments of areas: π±πΆ = β‘ (π±πΆ )β‘π + β‘ (π±πΆ )β‘π + β‘ (π±πΆ )β‘π + β―β‘ (10b) where (β‘π½π )π is the polar moment of inertia of the subregion π1 with respect to point O. The moments of inertia of the component areas about their centroidal axes can be found in tables, such as Tables () and (). The parallel-axis theorem must then be used to convert these moments of inertia to the common axis before they can be summed. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Table (). Inertial Properties of Plane Areas: Part 1 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Table (). Inertial Properties of Plane Areas: Part 2 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SAMPLE PROBLEM 1 Determine the moment of inertia for the shaded area about the x and y axis. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 For x axis. Here π₯2 = 1 π π 2 1 π¦ 2 β‘πππβ‘π₯1 = 2 π¦ . Thus, the area of the differential element parallel to π π2 π 1 π the x axis shown shaded in the figure is ππ΄ = (π₯2 − π₯1 )ππ¦ = ( 1 π¦ 2 − π2 π¦ 2 ) ππ¦. π2 Performing integration. 0 π 1 π π πΌπ₯ = ∫π΄ π¦ 2 ππ΄ = ∫0 π¦ 2 ( 1 π¦ 2 − π2 π¦ 2 ) ππ¦ β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= π2 π 2 π 5 ∫0 π¦ ( 1 π¦ 2 π2 2π 7 π β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= ( 1 7π 2 β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= For y axis. Here π¦2 = π − π2 π¦ 4 ) ππ¦ π¦ 2 − 5π2 π¦ 5 ) ππππ ππ 1 π π 2 1 π₯ 2 β‘πππβ‘π¦1 = 2 π₯ . Thus, the area of the differential element parallel to π π2 π 1 π the y axis shown shaded in the figure is ππ΄ = (π¦2 − π¦1 )ππ₯ = ( 1 π₯ 2 − β‘ π2 π₯ 2 ) ππ₯. π2 Performing integration. 0 π πΌπ₯ = ∫ π₯ 2 ππ΄ = ∫ π₯ 2 ( π΄ 0 π π 1π₯ π2 π 5 1 2 −β‘ π 2 π₯ ) ππ¦ π2 π β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= ∫0 π₯ 2 ( 1 π₯ 2 − β‘ π2 π₯ 4 ) ππ¦ π2 β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= ( 2π 1 7π2 π 7 π π₯ 2 − 5π2 π₯ 5 ) π β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= π ππ π − β‘ π ππ π β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= ππππ ππ Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SAMPLE PROBLEM 2 Determine the moment of inertia for the shaded area about the x and y axis. 1 For x axis. Here π₯2 = π¦β‘πππβ‘π₯1 = 2 π¦ 2 . Thus, the area of the differential element parallel to the x 1 axis shown shaded in the figure is ππ΄ = (π₯2 − π₯1 )ππ¦ = (π¦ − 2 π¦ 2 ) ππ¦. Performing integration. 0 2π π¦ 2 (π¦ − 2 π¦ 2 ) ππ¦ 2π π¦ 2 (π¦ 3 − 2 π¦ 4 ) ππ¦ πΌπ₯ = ∫π΄ π¦ 2 ππ΄ = ∫0 β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= ∫0 1 π¦4 1 π¦5 β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= ( 4 − 10) β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= π. πβ‘ππ 1 For y axis. Here π¦2 = √2π₯2 β‘πππβ‘π¦1 = π₯. Thus, the area of the differential element parallel to the 1 y axis shown shaded in the figure is ππ΄ = (π¦2 − π¦1 )ππ₯ = (√2π₯2 − π₯) ππ₯. Performing integration. 0 2π πΌπ₯ = ∫ π₯ 2 ππ΄ = ∫ π΄ 1 π₯ 2 (√2π₯2 − π₯) ππ¦ 0 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 2π β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= ∫0 5 π₯ 2 (√2π₯2 − π₯ 3 ) ππ¦ 2√2 β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= ( 7 7 π₯2 − π₯4 4 ) 4 β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= β‘ 7 π4 β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘β‘= β‘π. πππβ‘ππ EXERCISE PROBLEMS 1. Find the moment of inertia about the geometric center of the given structure made up of one thin rod connecting two similar solid spheres as shown in Figure. 2. Find the moment of inertia of a disc of mass 5 kg and a radius of 60 cm about the following axes: i. Axis passing through the center perpendicular to the plane of the disc ii. Axis touching the edge and perpendicular to the plane Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 3. Two timber beams are mounted at right angles and in contact with each other at their midpoints. The upper beam A is 2 in wide by 4 in deep and simply supported on an 8-ft span; the lower beam B is 3 in wide by 8 in deep and simply supported on a 10-ft span. At their cross-over point, they jointly support a load P = 2000 lb. Determine the contact force between the beams. 4. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12β‘kg/m2 . Determine the location of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 5, A uniform lamina of mass m is bounded by the exponential curvey π¦ = expβ‘(−π₯)β‘and the coordinate axes. Find the moment of inertia of the lamina about the y−axis. Concepts of Moment of Inertia Built Up Sections Moments of inertia of areas are widely used in "strength" calculations to calculate stresses and deflections in beams. A built-up section is made up of several sections such as rectangular sections, channel sections, I-sections, and so on. Built-up beams are used when the span, load, and bending moment are so large that the rolled steel beam section is insufficient to provide the required section modulus. Here are some examples of built-up sections The figure above depicts the built-up section, which is used for heavy loads and short spans. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 The diagram above illustrates a built-up section that is used for light loads, spans, and other functions provided by Gantry girders. A built-up section is typically formed by symmetrically placing and then welding or riveting these sections together. It will be interesting to learn that a built-up section functions as a single unit. It is frequently advantageous to combine a number of smaller members to create a stronger beam or column. Adding the moments of inertia of the component parts yields the moment of inertia of such a built-up section. This is possible if and only if the moments of inertia of each component area are measured about a common axis and the resulting section acts as a single unit. When two moments of inertia reference the same axis, they are additive. That is to say: The equations above can be used in determining built-up sections. The following steps are used to determine the moment of inertia of a built-up section. 1. Determine the moment of inertia of the built-up sections about their respective centers of gravity. 2. Now, using the Theorem of Parallel Axis, transfer these moments of inertia about the required axis (say, X-X or Y-Y). Note that in most of the standard sections, their moments of inertia about their respective centers of gravity are generally given. However, if it is not provided, we must calculate it before transferring it to the appropriate axis. As sections become more complex, it is frequently easier to perform the calculations by creating tables to determine the centroid and moment of inertia. Radius of Gyration The radius of gyration is the distance from the centroid of a section at which all of the area could be concentrated without affecting the moment of inertia. It is used to examine how certain shapes behave under various conditions. The radius of gyration, k, is equal to the square root of the moment of inertia to area ratio. A shape's radius of gyration with respect to each axis is given by: Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Let's look at some examples of finding the moment of inertia of built-up sections. SAMPLE PROBLEM 1 Consider a built-up section comprised of an S12 x 50 standard section capped with an MC10 x 28.3 miscellaneous channel. Determine the centroidal moment of inertia. Knowing the location of the centroid is required to calculate any centroidal moment of inertia. Finding the centroid of a built-up section is the same as finding the centroid of a composite geometric section. • First, place an x-y coordinate axis at a location of your choice. In doing so, it is important to note that it is appropriate to recognize symmetry and use it to your advantage. The section in this example is symmetrical about the y-axis but not the x-axis. This places the centroid's xcoordinate at the center of the section; this is an excellent location for the y-axis of our imposed coordinate system. • Although the x-axis can be placed anywhere, it is usually most convenient to place it at the bottom of the structure. The axis xc represents the centroidal x-axis, whose location we want to find. • Now, go to the online text's appendix and calculate the geometry and moments of inertia for each section. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 • The centroid location for an 'I-beam' section is not explicitly stated in the tables. Because of symmetry, it is assumed that you know it is equal to half the depth of the section. • In this case, the centroid is yc1 away from our arbitrary x-axis. The measurement is 6.0". • The distance dy2 is a little trickier. The centroid of a channel section is measured from the back of the web and is given in the steel tables as distance Xc. • This distance must be calculated using the same x-axis as the S-section. As a result, yc2 equals the depth of the S-section plus the width of the channel web less the distance Xc. • This means yc2 = 12” + 0.477” - 0.933” yc2 = 11.544” • Steel tables can also be used to calculate areas. And once we have the areas and locations of the centroids, we can calculate the location of the centroid. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 • After locating the centroid, the moment of inertia is calculated using the parallel axis theorem. The distances between the composite section's centroidal axis and the individual sections are easily calculated as follows: The remainder of the calculations are expressed in the table below Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SAMPLE PROBLEM 2 A column will buckle around the axis with the lowest moment of inertia. As a result, the moment of inertia about the x- and y-axes of a column section should be roughly equal. Consider a column made up of two MC12x35 channels. Calculate the distance 'd' between the sections so that the centroidal moment of inertia about the x and y axes is equal. • As both sections have the same x-axis, we can deduce that the composite moment of inertia about the x-axis is the sum of the moments of inertia of each individual section about its own centroidal x-axis. This gives us the desired Iy. However, we must use the parallel axis theorem to calculate the distance 'd' that will result in Iy. • Since the section is symmetrical about the y-axis, we know that d1 = d2 = ½ d+ tw + xc = ½ d + 1.517 Each section's moment of inertia as it references the common centroidal y-axis must be half of the desired Iy. • Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 EXERCISE PROBLEMS 1. A column section is built up using two channels ISJC 200 and two plates 250 mm x 10 mm as shown. Calculate moment of inertia at xx and yy axes of built-up section. 2. Allen Batumbakal is an engineering student from Batangas State University. Along his way to the academe, he saw a gray colored perfectly shaped inverted T tile. He thought that it can be a good example in computing the moment of inertia of a built-up section. With that, he is to calculate the moment of inertia for an inverted T-section about its horizontal centroidal axis. Given that the size of the flange is 150 mm x 45 mm and a vertical web of 130 mm x 45 mm. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 3. Determine the inertia of a built-up section about X-X axis passing through the centre of gravity of the section. 4. Two 1-in. steel plates are welded to a rolled S section as shown. Determine the moments of inertia and the radii of gyration of the section with respect to the centroidal x and y axes. METHOD OF JOINTS The method of joints is a method for calculating unknown forces operating on truss components. The approach focuses on the members' joints or connection points, and it is generally the quickest and easiest way to solve for all the unknown forces in a truss system. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Treating the entire truss structure as a rigid body, draw a free body diagram, write out the equilibrium equations, and solve for the external reacting forces acting on the truss structure. This analysis should not differ from the analysis of a single rigid body. Assume that each of the connecting locations between the members has a pin or some other tiny piece of material. Then, for each connecting point, make a free body diagram. Remember to include the following: • • • Any external reaction or load forces that may be acting at that joint. A normal force for each two force member connected to that joint. Remember that for a two force member, the force will be acting along the line between the two connection points on the member. We will also need to guess if it will be a tensile or a compressive force. An incorrect guess now though will simply lead to a negative solution later on. A common strategy then is to assume all forces are tensile, then later in the solution any positive forces will be tensile forces and any negative forces will be compressive forces. Label each force in the diagram. Include any known magnitudes and directions and provide variable names for each unknown. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Calculate the equilibrium equations for each joint. Because the joints should be treated as particles, there will be force equations but no moment equations. You should have a high number of equations if you use two (for 2D issues) or three (for 3D problems) equations for each joint. For each joint in planar trusses, the sum of the forces in the x direction and the sum of the forces in the y direction will be zero. For each joint in space trusses, the sum of forces in the x direction, the sum of forces in the y direction, and the sum of forces in the z direction will be zero. Finally, given the unknowns, calculate the equilibrium equations. You can solve for one variable at a time algebraically, or you can utilize matrix equations to solve for everything at once. If you previously thought that all forces were tensile, negative results imply compressive forces in the members. Sample Problem: Using the method of joints, determine the force in each member of the truss shown in Fig. (a). Indicate whether the members are in tension or compression Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SOLUTION The FBD of the entire truss is shown in Fig. (b). The three unknowns (NA, Cx , and Cy ) can be computed from the three equilibrium equations. We now proceed to the computation of the internal forces by analyzing the FBDs of various pins. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 - - - The FBD of pin A, shown in Fig. (c), contains two unknowns: PAB and PAD. We can compute these two forces immediately, because two independent equilibrium equations are available from this FBD. The FBD of pin D, in Fig. (d), contains the forces PAD, PBD, and PCD. Because PAD has already been found, we have once again two equations that can be solved for the two unknowns. Figure (e) shows the FBD of pin C. With PCD previously found, the only remaining unknown is PBC, which can be easily computed. ), From the FBD of pin A, Fig. (c), From the FBD of pin D, Fig. (d), Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 From the FBD of pin C, Fig. (e), The truss pinned to the floor at D, and supported by a roller at point A is loaded as shown in Fig. T-06. Determine the force in member CG. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SOLUTION: At joint A ΣMD=0 6RA=4(100)+2(120) RA=106.67 kN ΣFV=0 313√FAG=RA 313√FAG=106.67 FAG=128.20 kN compression At joint F FFG=0 ΣFH=0 FAB=213√FAG FAB=213√(128.20) FAB=71.11 kN tension At joint B ΣFH=0 FBC=FAB FBC=71.11 kN tension ΣFV=0 FBG=0 At joint G ΣFV=0 313√FCG+100=313√FAG 313√FCG+100=313√(128.20) FCG=8.01 kN tension answer Exercise Problem: 1. Determine the forces acting in all the members of the truss shown in the figure. 2. Using the method of joints, analyze the truss shown in Figure. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 3. Find the force acting in each of the members in the truss bridge shown. Remember to specify if each member is in tension or compressive. 4. The structure in Fig. T-02 is a truss which is pinned to the floor at point A, and supported by a roller at point D. Determine the force to all members of the truss. 5. In the cantilever truss, compute the force in members AB, BE, and DE. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Method of Sections We have seen that the method of joints consists of analyzing trusses by applying the principles of equilibrium to the concurrent force systems acting at each joint. The principles of equilibrium of non-concurrent force systems may also be applied to truss analysis – the procedure known as the method of sections. Its use permits us to determine directly the force in almost any member instead of proceeding to that member by a joint-to-join analysis. In the method of sections, a portion of the truss is isolated as a free body by passing an imaginary cutting plane through the entire truss, separating it into two parts. If possible the separation should be done without cutting more than three members of the truss. We shall then have two isolated parts of the truss, each constituting a nonconcurrent system of forces in equilibrium under the action of the known loads that act on each part of the unknown forces that the cut members of one part exert on the other. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 For example, consider the truss shown in the Figure (001), if the cutting plane passes through the members DF, EF, and EG, the truss may be separated into two parts each of which is acted upon by unknown forces equivalent to the loads being transmitted by these members. Figure (001) Strictly, the method of joints is a variation of method of sections, where the section is taken around the join, thus isolating the joint as a free body. It is traditional, however, to use the expression method of joints for the equilibrium of concurrent forces at a joint, and the expression method of sections when considering the nonconcurrent force system set up by a section through the truss. These two parts are shown in Fig. (001). Each part in this figure constitutes a system of nonconcurrent forces in equilibrium. Since the unknowns in either system are the same, it is generally best to determine the unknown from the equations of equilibrium applied to a simpler system. Here, part (a) is obviously the simple system since it involves fewer internal to the free bodies shown. Observe that the action of all uncut members members on their end joints occurs in equal opposite pairs at their end joins and hence the cancels out of any calculations involving the entire free body. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Figure (002) Each segment of the truss is held in equilibrium by forces equal to those in the cut members. To simplify the calculations, use a condition of equilibrium that determines each unknown force independently of the other unknowns. A moment summation is usually the best, the center of moments being chosen at the intersection of the unknown forces to be eliminated from the moment summation. Thus, in Fig 002, • • if the force in DF is desired, select E as the center of moments to determine the force in EG, select the moment center at the intersection of DF and EF. This will be at point F’ having the location of the original point F with respect to the left section. However, the moment cannot be applied to determine EF because DF and EG being parallel, intersect at infinity. Here, we determine EF from a vertical summation of forces which eliminated the horizontal forces EG and DF. The procedure just described constitutes the method of sections. This method usually lets the force in any member ne determined without finding the forces in the other members. Merely pass the cutting plane trough the member whose force is desired. It is essential, however, that the plane does not cut more than three members whose internal forces are unknown, because the method of sections is based on the equilibrium of coplanar force systems for which only three independent equations of equilibrium are available to solve for three unknowns. An exception is the case where the cutting plane may pass through the desired number as well as several ithers, provided all these other members intersect at the common joint which is used as the center of moments. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Observe that a cutting section need to straight; its purpose is to isolate the truss segment selected as a free body without involving more than three unknowns. In this case, the truss segments ADE is isolated and exposes a force in AB, CD, and EF. Although this curved cuts through CF and BF twice, the equal opposite action of the exposed forces in these members cancels out any force of moment summation. Then, after determining the forces AB, CD, and EF from the isolated segment ADE, the method of joints may be used to find the forces in the remaining members. SAMPLE PROBLEM SOLUTION: From section to the left of a-a: ΣFV=0 529√FBE=80+60 FBE=150.78 kN tension answer Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 ΣME=0 5FBC=6(80)+2(60) FBC=120 kN compression ΣMB=0 5FDE=4(80) FDE=64 kN tension answer answer Use the method of sections to determine the force in members BD, CD, and CE of the Warren truss Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 ΣMA=0 20RE=5(1000)+10(4000)+15(3000) RE=4500 kN At section through M-M CDE is equilateral triangle, thus, CD = 10 ft. y=10sin60β=53–√ ft ΣMC=0 yFBD+5(3000)=10(4500)y 53–√FBD+5(3000)=10(4500) FBD=3464.10 lb compression answer ΣFV=0 FCDsin60β+3000=4500 FCD=1732.05 lb tension answer ΣMD=0 yFCE=5(4500) 53–√FCE=5(4500) FCE=2598.08 lb tension answer Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Example Problems: 1. Determine the force in members DC, HC, HI, ED, EH, and CH of the truss and state if the members are in tension or compression. 2. Use the method of sections to determine the force in members BD, CD, and CE of the Warren truss Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 3. Determine the force in members BE, BJ, DJ, and IJ. (ADJ and HDB are not continues, each consist of two elements.) 4. Use the method of sections to find the forces in members AC, BC, CD, and CE. Be sure to indicate if the forces are tensile or compressive. 5. The structure shown in Figure T-07 is pinned to the floor at A and H. Determine the magnitude of all the support forces acting on the structure and find the force in member BF. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Graphical method Plotting all the applied forces in the order of their application is part of this strategy. Forming a force polygon according to scale in both magnitude and direction (head-to-tail). Each force is identified by letter to ensure clarity. The direction (inclination) of each structural member is then plotted from the point of intersection of the forces adjacent to it, one at a time. Plot of the members that serve as legs of the angle between them will meet in at a common point in the resulting diagram. This point of meeting (intersection) will be the starting point of the plot for the next member that serve as a leg of the angle between it and any of the member plotted before. The process is repeated until all the members have been plotted. The magnitudes of the bar forces are then determined by measuring their plots according to the same scale used in plotting the applied forces. Exercise Problems: Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 1. A concurrent force system is shown in the figure. Using graphical method, find the resultant of a force system. 2. The figure shows a pin-jointed frame carrying vertical loads of 1 kN each at points B and G, and a horizontal load of 4 kN at point D. Graphically, find the forces in the various members of the truss. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 3. Determine the force in each member of the truss using the graphical method. You can check your answer in graphical method by comparing the answer in the method of joints also provided below. 4. Determine the magnitude and nature of the forces in all the members of the truss by graphical method by constructing the Maxwell diagram. Parabolic Cables Introduction The Golden Gate Bridge is a well-known structure that is supported by cables. Such cables are frequently utilized in bridges and other types of structures for load transfer. This lecture series serves as an introduction to cable analysis. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Figure 1: The Golden Gate Bridge The Golden Gate Bridge is a suspension bridge, as shown in the image above. Its deck is supported by a set of vertical hangers that are connected to a pair of main wires. This configuration enables the main cables, which in turn convey the strain to the towers at the bridge's ends, to bear most of the bridge's weight. Given the near closeness of the hangers in this situation, we can conclude that the main cables are under a dispersed load. Furthermore, since the weight of each cable is insignificant compared to the load it must carry, we can neglect the cable’s own weight when we analyze the system. A line drawing illustrating one of the bridge’s main cables. Figure 2: A line drawing characterization of a main cable in a suspension bridge Let’s now consider the pedestrian bridge shown in the figure below. Note how the surface of the bridge deck follows the geometric shape of the cable. In this case, we can conceptualize the entire system as a cable hanging freely from its ends, carrying its own weight. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Figure 3: Charles Kuonen pedestrian bridge and its line drawing characterization Before we proceed any further, we should distinguish between a weightless cable that supports a linear load and a cable that solely carries its own weight. The shapes of the two cables mentioned above differ: when a cable carries a linear load spread down the horizontal axis (for example, the major cables of the Golden Gate Bridge), its shape can be specified using a parabola. When a wire hangs freely under its own weight, it takes the shape of a catenary (for example, the Charles Kuonen pedestrian bridge). Let's dig deeper into this distinction. Consider the wire depicted in Figure 4. It is suspended from both ends, providing a configuration similar to the Golden Gate Bridge. For this cable, we assume that the bridge deck applies a uniformly distributed load of π€ to the cable along the x-axis. Figure 4: A cable subjected to a uniformly distributed load If we place the origin of the coordinate system at the lowest point of the cable, we can draw the free-body diagram of the segment just to the right of the origin. This free-body diagram is shown in Figure 6. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Figure 5: The free-body diagram of a segment of a cable subjected to a uniformly distributed load In the above diagram, ππ is the tension force in the cable at its lowest point, π is the tension force at the right end of the segment, and π denotes the angle that the cable makes with the horizontal axis at its right end. Since the segment has to remain in equilibrium, the sum of the forces in the horizontal and vertical directions must be zero. Therefore, we can write two equilibrium equations, as shown below. ∑ πΉπ₯ = 0 β‘ → ππππ π = β‘ ππ β‘ (1) ∑ πΉπ¦ = 0 β‘ → ππ πππ = β‘π€π₯ (2) Dividing Equation [2] by Equation [1], we get the following: ππ πππ π€π₯ π€π₯ =β‘ β‘ → π‘πππ = β‘ π₯ ππππ π ππ ππ (3) Since tanπ can be expressed as the change in y with respect to the change in x (i.e., dy dx /), Equation [3] can be rewritten as follows: ππ¦ π€ =β‘ π₯ ππ₯ ππ (4) This first order differential equation can be easily solved for y. Let’s rewrite Equation [4] as shown below. ππ¦ = β‘ π€ π₯ππ₯ ππ (5) Applying the integral operator to both sides of Equation [5], we get the following: (6) Nation Leading Innovations, Transforming Lives, Building the Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 ∫ ππ¦ = β‘ ∫ π€ π€ 2 π₯ππ₯β‘ → π¦ = β‘ π₯ + πΆβ‘ ππ 2ππ The integration constant (C) in Equation [6] can be determined using a boundary condition. In this case, we know that at the origin, where x = 0, y is also zero. Substituting zero for x and y in the above equation, we can determine C. π¦ =β‘ π€ 2 π€ π₯ + πΆβ‘ → 0 = β‘ (0)2 + πΆ = 0 2ππ 2ππ (7) Therefore, Equation [6] can be written as follows: π¦ =β‘ π€ 2 π₯ 2ππ (8) Equation [8] is a parabolic function that describes the shape of the cable shown in Figure 4. Now let's examine the case in which the cable hangs freely under its own weight. Consider the cable shown in Figure 6. Figure 6: A cable hanging freely from its ends and subjected to its own weight Note that the weight of the cable is not distributed along the x-axis but rather along the arc length of the cable. Let’s place our origin at the lowest point of the cable just like we did before. The free-body diagram for the segment of the cable to the right of the origin is shown in Figure 7. Figure 7: The free-body diagram of a cable subjected to its own weight Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Note that the arc length of the segment in the figure above is denoted by s. We can write two equilibrium equations for the free-body diagram shown in Figure 7 as follows. ∑ πΉπ₯ = 0 β‘ → ππππ π = β‘ ππ β‘ (9) ∑ πΉπ¦ = 0 β‘ → ππ πππ = β‘π€π (10) Dividing Equation [10] by Equation [9], we obtain Equation [11]. ππ πππ π€π π€ ππ¦ π€ = β‘ β‘ → π‘πππ = β‘ π β‘ → = β‘ π β‘ ππππ π ππ ππ ππ₯ ππ (11) In order to solve the above differential equation, we need to replace s with x and y. We can do this using the Pythagorean theorem. Note that ds can be viewed as the hypotenuse of a right triangle with height ππ¦ and base ππ₯. Therefore, we can write the following: ππ 2 = β‘ ππ₯ 2 + β‘ ππ¦ 2 β‘ → ππ = β‘ √ππ₯ 2 + ππ¦ 2 (12) If we take the derivative of both sides of Equation [11] with respect to x, we get the following equation. π ππ¦ π€ π2π¦ π€ ππ ( ) = β‘ ( π ) β‘ → β‘ 2 = β‘ β‘ ππ₯ ππ₯ ππ π π₯ ππ ππ₯ (13) Substituting Equation [12] into Equation [13], we get the following second-order differential equation. π2 π¦ π€ √ππ₯ 2 + ππ¦ 2 π€ ππ¦ 2 √ =β‘ β‘ = β‘ β‘β‘ 1 + β‘ ( ) π2π₯ ππ ππ₯ ππ ππ₯ (14) When solved, the above equation yields the following solution. π¦ =β‘ ππ π€ ππ πππ β ( β‘π₯) − π€ ππ π€ (15) Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 The above equation is that of a catenary. It is used to describe the shape of a cable hanging freely under its own weight. To summarize, when analyzing cable systems subjected to distributed loads, depending on the source and nature of the loads, we may have to use a different mathematical function to describe the shape of each cable. This lecture series deals with the analysis of such cable systems. SAMPLE PROBLEM A parabolic arch is subjected to two concentrated loads. Determine the support reactions and draw the bending moment diagram for the arch. Solution: Entire arch. Ma = 0 Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 Ey (40) - 50(8) - (20 (35) = 0 Ey = 27.5 kips β¬ Fy = 0 Ay + 27.5 - 50 - 20 = 0 Ay = 42.5 kips β¬ Arch segment EC. For the horizontal reactions, sum the moments about the hinge at C. Mc = 0 27.5(20) - Ex (10) - 20(15) = 0 Ex = 25 kips β¬ Entire arch again. Fx = 0 -25 + Ax = 0 Ax = 25 kips β‘ Bending moment at the locations of concentrated loads. To find the bending moments at sections of the arch subjected to concentrated loads, first determine the ordinates at these sections using the equation of the ordinate of a parabola, which is as follows: y = 4fxL² (L - x) yx = 8 ft = 4(10)(8)(40)² (40-8) = 6.4 ft yx = 5 ft = 4(10)(5)(40)² (40-5) = 4.375 ft When considering the beam, the bending moments at B and D can be determined as follows: MB = Ay (8) - Ax (6.4) Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 = 42.4 (8) - 25 (6.4) = 180 k.ft MD = Ey (5) - Ex (4.375) = 27.5(5) - 25 (4.375) = 28.13 k.ft Exercise Problems: 1. The 36-m cable shown in the figure weighs 1.5 kN/m. Determine the sag H and the maximum tension in the cable. 2. A cable carries a uniformly distributed load W = 900 N/m. The horizontal span between supports is 2475 m. The sag of the cable at midspan is 112.5 m. Determine the maximum tension and the length of the cable. 3. A suspension cable of span 40m and central dip of 9m carries a uniformly distributed load of 15 kN/m. Determine (a) the magnitude of the load (b) the total length of the cable from A to B. 4. The uniform cable weighing 15 N/m is suspended from points A and B. The force in the cable at B is known to be 500 N. a. Find the tension at A, in N. b. Calculate the span of the cable Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 5. A parabolic arch is subjected to two concentrated loads. Determine the support reactions for the arch. Catenary Cables The term 'catenary' refers to the curve formed when wire or chain is supported at both ends. So, simply put, catenary cable is any cable that is fixed in this format, such as wires used in decorative lighting, electrical cable suspension, train lines, and fiber optics. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 In catenary cable analysis, the load (w) is assumed to be uniformly distributed along the cable itself. Thus, the total load (W) = ws. Tension in such a cable is lowest at the lowest point and highest at the highest point. A FBD of one side of the cable from its lowest point should show only three forces: the minimum tension (ππ ), the maximum tension (T) at the point of support, and the total (resultant) load (W) acting midway along the cable's length s. The catenary constant "c," the distance from the origin to the cable's lowest point, is required to illustrate a catenary cable. π Numerically, π = π€π , so ππ = π€π The general equation of catenary cable is ππ = ππ + ππ To relate horizontal distance x to cable length s, To relate x and y cable coordinates, which is the equation of a catenary. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 SAMPLE PROBLEM Two cables of the same gauge are attached to a transmission tower at B. Since the tower is slender, the horizontal component of the forces exerted by the cables at B is to be zero. Knowing that the mass per unit length of the cables is 0.4 kg/m, determine the sag and the maximum tension in each cable. The weight W of cable AB between an arbitrary point D and its rightmost point, B, is π = wXB. The sum of moments relative to point B is such that Σππ΅ = 0 → To yb - wxb (xb/2) = 0 The horizontal component T0 is then To = wxB2yB (I) For cable AB, we have xB = 45 m and T0 follows as π0 = w(45²)2h = 1012.5 wh (II) Next, for cable BC, with xB = 30 m and yB = 3 m, we have π0 = w(30²)2(3) = 150 w (III) Because the horizontal component of the forces exerted by the cables is to be zero, Equations (II) and (III) should yield the same result; that is, (To) II = (To) III β‘ 1012.5 wh = 150 w ∴ β = 1012.5150 = 6.75 m Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 The required sag is 6.75 m. Next, we appeal to the following expression for tension Tm, Tm= To2 + W² (IV) For cable AB, the distributed weight is w = 0.4(9.81) = 3.92 N/m. Then, substituting xB = 45 m and yB = h = 6.75 m into Equation (I), we obtain To = wx²B2yB = 3.92 (45²)2(6.75) = 588 N Now, the weight W is such that W = wxB = 3.92(45) = 176.4 N. Substituting these results into Equation (IV), we have Tm= 588² + 176.4² = 613.9 N The maximum tension is close to 614 N. Exercise Problems: 1. A cable is attached to two power lines A and B that can carry a uniformly distributed load of 10 kg/m. If its sag is 200 m, compute for its maximum tension and the length of the cable. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 2. A suspension cable hangs between two points of span 100m and sag of 20m carries a uniformly distributed load of 50 N/m. Determine the (a) total length of the cable from A to B, (b) minimum tension, and (c) maximum tension. 3. The cable supports a distributed load w = 12,000 lb/ft. Determine the maximum tension in the cable. 4. A catenary cable has a weight of 50 N/m along the horizontal. It has a distance of 100 m and a sag of 20 m. Compute the length of the cable, the tension and the maximum tension of the catenary hanging cable. Leading Innovations, Transforming Lives, Building the Nation Republic of the Philippines BATANGAS STATE UNIVERSITY The National Engineering University Alangilan Campus Alangilan, Batangas City, Batangas, Philippines 4200 5. A 40-m cable is strung between two buildings as shown. The maximum tension is found to be 350 N, and the lowest point of the cable is observed to be 6 m above the ground. Determine the minimum distance between the buildings and the total mass of the cable. 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