Elements of Civil Engineering and Engineering Mechanics 14CIV23ELEMENTS OF CIVIL ENGINEERING AND MECHANICS (4:0:0) Sub code :14CIV13 Internal Marks: 25 Hrs/week : 4+0+0 Exam Marks : 100 Total Hours: 50 Module 1 Introduction to Civil Engineering & Engineering mechanics Introduction to Civil Engineering, Scope of different fields of Civil Engineering -Surveying, Building Materials, Construction Technology, Geotechnical Engineering, Structural Engineering, Hydraulics, Water Resources and Irrigation Engineering, Transportation Engineering, Environmental Engineering. Infrastructure: Types of infrastructure, Role of Civil Engineer in the Infrastructural Development, Effect of the infrastructural facilities on socio-economic development of a country. Roads: Type of roads, Components, Classification of Roads and their functions. Bridges: Types of Bridges and Culverts, RCC, Steel and Composite Bridges Dams: Different types of Dams based on Material, structural behaviour and functionality with simple sketches. Introduction to Engineering mechanics: Basic idealizations -Particle, Continuum and Rigid body; Force and its characteristics, types of forces, Classification of force systems; Principle of physical independence of forces, Principle of superposition of forces, Principle of transmissibility of forces; Newton's laws of motion, Introduction to SI units. Couple, Moment of a couple, Characteristics of couple, Moment of a force, Equivalent force -Couple system; Numerical problems on moment of forces and couples, on equivalent force -couple system 10 hrs Module 2 Analysis of Force Systems-Concurrent & Non Concurrent System Concurrent Force System: Composition of forces -Definition of Resultant; Composition of coplanar -concurrent force system, Parallelogram Law of forces, Principle of resolved parts; Numerical problems on composition of coplanar concurrent force systems. Non Concurrent Force System: Composition of coplanar -non-concurrent force system, Varignon's principle of moments; Numerical problems on composition of coplanar non-concurrent Force system 10 hrs Module 3 Equilibrium of Forces and Friction Equilibrium of Concurrent and Non-concurrent Forces Equilibrium of forces -Definition of Equilibrant; Conditions of static equilibrium for different force systems, Lami's theorem; Numerical problems on equilibrium of coplanar – concurrent and non-concurrent force systems. Support Reaction Types of Loads and Supports, statically determinate beams, Numerical problems on support reactions for statically determinate beams with Point load (Normal and inclined) and uniformly distributed loads. Friction Definitions: Types of friction, Laws of static friction, Limiting friction, Angle of friction, angle of repose; Dept of Civil Engg, SJBIT Page 1 Elements of Civil Engineering and Engineering Mechanics 14CIV23 Impending motion on horizontal and inclined planes; Numerical Problems on single and two blocks on inclined planes. Wedge friction; Numerical problems Ladder friction; Numerical problems. 10 hrs Module 4 Centroid and Moment of Inertia of Engineering Sections Centroids Introduction to the concept, centroid of line and area, centroid of basic geometrical figures, computing centroid for composite lines and Engineering composite sections – T, L, I and Z sections and their built up sections. Moment of inertia Introduction to the concept, Radius of gyration, Parallel axis theorem, Perpendicular axis theorem, Moment of Inertia of basic planar figures, computing moment of Inertia for Engineering composite sections – T, L, I and Z sections and their built up sections 10 hrs Module 5 Kinematics Definitions – Displacement – Average velocity – Instantaneous velocity – Speed – Acceleration -Average acceleration – Variable acceleration – Acceleration due to gravity – Newton‘s Laws of Motion, Rectilinear Motion – Numerical problems. Curvilinear Motion – Projectile Motion – Relative motion – Numerical problems. Motion under gravity – Numerical problems 10 hrs TEXT BOOKS 1. Engineering Mechanics by S.Timoshenko, D.H.Young, and J.V.Rao, TATA McGraw- Hill Book Company, New Delhi 2. Elements of Civil Engineering and Engineering Mechanics by M.N. Shesha Prakash and G.B. Mogaveer, PHI Learning (2014) 3. Engineering Mechanics-Statics and Dynamics by A Nelson, Tata McGraw Hill Education Private Ltd, New Delhi, 2009. REFERENCES 1. Elements of Civil Engineering (IV Edition) by S.S. Bhavikatti, New Age International Publisher, New Delhi, 3rd edition 2009. rd 2. Beer FP and Johnson ER, “Mechanics for Engineers-Dynamics and Statics”-3 SI Metric edition, Tata McGraw Hill. -2008 3. Shames IH, “Engineering Mechanics – Statics & Dynamics”-PHI – 2009 Dept of Civil Engg, SJBIT Page 2 Elements of Civil Engineering and Engineering Mechanics 14CIV23 Table of Contents MODULE TOPIC PAGE NO. I Introduction to civil engineering and engineering 2 mechanics II Analysis of concurrent and non concurrent force 28 system III Equilibrium of forces and friction 57 IV Centroid and Moment of inertia 87 V Kinematics 125 Dept of Civil Engg, SJBIT Page 1 Surveying helps in preparing maps and plans. Constructing bridges across streams. directions and vertical heights directly or indirectly. 9. civil engineers specialize themselves in one field of civil engineering. The role of civil engineers is seen in every walk of life or infrastructure development activity such as follows:- 1.1 Civil Engineering: It is the oldest branch of professional engineering. 1. Providing other means of transportation such as railways. 2. using data from remote sensing satellites is helping to prepare maps & plans & thus cut down the cost of surveying. deciding the location for a dam or airport or harbour) The cost of the project can also be estimated before implementing the project. consist of components like foundations. An Engineer is a person who plays a key role in such activities. However. 3.1 Introduction Engineering: It is a profession of converting scientific knowledge into useful practical applications. 2. dam. 1. Supplying safe and potable water for public & industrial uses. 7. 8. Surveying: It is a science and art of determining the relative position of points on the earth‘s surface by measuring distances. Constructing hydro-electric & thermal-power plants for generating electricity. where the civil engineers are concerned with projects for the public or civilians. Laying ordinary village roads to express highways. which help in project implementation. 5. The soil should be thoroughly checked for its suitability for construction purposes. 4. Tunneling across mountains & also under water to connect places easily & reduce distance. civil engineering is a very broad discipline that incorporates many activates in various fields. multipurpose dams & canals for supplying water to agricultural fields. where the materials & forces in nature are effectively used for the benefit of mankind. As seen above. 6. Geo-Technical Engineering (Soil Mechanics): Any building. The different fields of civil engineering and the scope of each can be briefly discussed as follows. harbour & airports. rivers and also across seas. bridge. retaining wall etc. Now-a-days. (setting out the alignment for a road or railway track or canal. SJBIT Page 2 . The study dealing Dept of Civil Engg.1. Constructing irrigation tanks. Providing shelter to people in the form of low cost houses to high rise apartments.Elements of Civil Engineering and Engineering Mechanics 14CIV23 MODULE 1 Introduction to Civil Engineering and Engineering Mechanics 1. The foundation is laid from a certain depth below the ground surface till a hard layer is reached. Protecting our environment by adopting sewage treatment & solid waste disposal techniques. The different activities should be planned properly. slabs etc. machinery & money. all sizes of vehicles. sedimentation tanks. In such cases. In case of large construction projects management Dept of Civil Engg. are also necessary. the runway & other facilities such as taxiways. the manpower. air & waterways. If the treatment plants are for away from the town or city. filter beds. The water available (surface water & ground water) may not be fit for direct consumption. For an airport. Building Materials & Construction Technology: Any engineering structure requires a wide range of materials known as building materials. 7. In a town or city. Roads to remote places should be developed. should be designed. a part of the water supplied returns as sewage. Here the role of civil engineers is to construct facilities related to each one. 4. is reducing the time and also to maintain accuracy. It becomes important for any construction engineer to be well versed with the properties & applications of the different materials. Ports & harbours should be designed to accommodate. It becomes important to determine the magnitude & direction the nature of the forces and acting all the time. For water purification. Depending upon the materials available or that can be used for construction. The solid waster that is generated in a town or locality should be systematically collected and disposed off suitably. Apart from dams & canals other associated structures like canals regulators. These components are always subjected to forces. columns. Hydro electric power generation facilities are also included under this aspect. suitable pipelines for conveying water & distributing it should also be designed. manpower. the components or the parts of the building should be safely & economically designed. weirs. aqua ducts. Structural Engineering: A building or a bridge or a dam consists of various elements like foundations. Since rainfall in an area is insufficient or unpredictable in an area. Water Supply and Sanitary Engineering (Environmental Engineering): People in every village. The choice of the materials is wide & open. water flowing in a river can be stored by constructing dams and diverting the water into the canals & conveyed to the agricultural fields. should be properly designed. barrages etc. Irrigation & Water resources engineering (Hydraulics Engineering): Irrigation is the process of supplying water by artificial means to agricultural fields for raising crops. beams. Transportation Engineering: The transport system includes roadways. control towers etc. A structured engineer is involved in such designing activity. town & city need potable water. railways. Sometimes crucial sections of railways & roads should be improved. terminal buildings. The knowledge of the geology of an area is also very much necessary. 3. so that the construction is completed in time and in an economical manner. etc. 5. materials & machinery should be optimally utilized. Any construction project involves many activities and also required many materials. SJBIT Page 3 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 with the properties & behaviour of soil under loads & changes in environmental conditions is called geo-technical engineering. This sewage should be systematically collected and then disposed into the natural environment after providing suitable treatment. segregation of materials should be done so that any material can be recycled & we can conserve our natural resources. 6. Before disposal. The use of computers in designing the members. the water should be purified and then supplied to the public. They give maximum service to one and all. Health care facilities 7. help in completing the project orderly in time. Exporting agricultural goods can fetch foreign currency. They have the following advantages when compared with other modes of transportation. It provided direct employment to many number of skilled. Housing facilities 8. Slums are created in cities. 7. plastics. 3. It leads to the growth of associated industries like cement. It improved health care & Cultural activities. It becomes a huge financial burden on the government and tax prayers. It is a basic necessity for any country or state. research & education. Dept of Civil Engg.2 ROADS Transportation of goods & people can be done by roadways. railways. Education facilities 6. 1. 5. Roads play a crucial role in any country‘s development. Exploitation of natural resources can lead to environmental disasters. Each mode of transportation has advantages and disadvantages of its own in comparison to the others. 3. glass. Power generation & transmission facilities 5. Recreation facilities The well being of a nation is dependent on the quality & the quantity of the above services that are provided to the public. 2. Some ill effects of infrastructure development can also be listed as follows: 1.1. Irrigation facilities 4. Transport facilities 2. Drinking water and sanitation facilities 3. 8. 4. Migration of people from villages to towns & cities in search of job takes place. 6. 1. semiskilled & unskilled laborers. steel. It provided housing & means of communication to people. 2.Elements of Civil Engineering and Engineering Mechanics 14CIV23 techniques of preparing bar charts & network diagrams. waterways & airways. Development of infrastructure has number of good effects which can be listed as follows. paints. 1. 1. It forms a part of business.2 Effects of Infrastructure development on the Socio-economic development of a country: The term infrastructure is widely used to denote the facilities available for the socio-economic development of a region. SJBIT Page 4 . timber. electrical goods etc. 4. It helps in increasing food production & protection from famine. The infrastructure facilities to be provided for the public include: 1. Hence. 8. Roads can be used by various types of vehicles but other modes of transportation can cater to a particular locomotive only. the width of the pavement for a single lane road becomes 3. The severity of accidents is related to speed. They are directly accessible to the users at all destination points.8 mts. flyovers.5 mts. 5. Other modes of transportation are dependent on roads to serve. 2. Roads can provide door to door service. It also includes the path way and other related structures like bridges. For short distance traveling. The different cross section elements of a road can be represented as follows. At the same time. In case of multi lane traffic.Elements of Civil Engineering and Engineering Mechanics 14CIV23 2. They have maximum flexibility with respect to route. The width of the road is designed according to the traffic volume on the road.1: Components or Cross section elements of a road: All roads should essentially consist of the following components:- 1. the total width of the pavement can be fixed. Depending upon the number of lanes. Roads get easily damaged in heavy rainfall areas and require frequent maintenance. Dept of Civil Engg. SJBIT Page 5 .68 mts for safety should also be provided. underpasses which make road traveling easier. the people from their terminals. 3. Pavement or Carriage way 2. It does not provide much comfort for long distance travel. the width of each lane should be at least 3. road travel is easier. the disadvantages of roads can be listed as follows:- 1. Construction of roads is easier when compared to other modes. the maximum width of a vehicle should not be more than 2. According to Indian Road Congress (IRC) specification. 6.44 mts. 3. Shoulders Pavement or Carriage way: It refers to the path over which the vehicles and other traffic can move lawfully. 7. Frequency of accidents among road users is more.2. direction and speed. 1. A side clearance of 0. 4. 4. that can be provided. The surface of the shoulder should be rough to avoid the drivers from using the shoulders frequently. The top surface of the varying course is provided a lateral slope (camber). to separate the pavements from the footpath. This layer provides a proper support for the upper layers. thus avoiding head on collision. It takes the loads directly. Kerbs: Kerb Footpath Pavement Within the city limits. SJBIT Page 6 . This layer should be moderately rough to provide good grip for the vehicles. a lean mixture of concrete with large amounts of sand and stones is used in preparing this layer. smaller stones mixed with cohesive soil or cement are provided and thoroughly compacted. Sub Soil: It refers to the natural soil or prepared soil on which the loads coming on the road are ultimately transferred. Hence the Sub Soil should be prepared by compacting it properly by rollers. 4. Much attention should be given in preparing the sub grade. 1. At the top level. 5.Elements of Civil Engineering and Engineering Mechanics 14CIV23 1. 3. Base Course: This layer is constructed in one or two layers consisting of stones mixed with gravel. This should remain dry and stable throughout the year. This layer consists of disintegrated rocks mixed with gravel. These can be provided as a yellow colour strip or steel barricade or a permanent median all along the centre lines of the road. Shoulders: The width of a road is always extended beyond the road on both sides by a width of at least 2 to 5 mts.2: Other components of roads: The following components are also essential for roads depending upon. the type of vehicles and traffic volume on the road. 2. Surface course/wearing course: It is the topmost layer of the carriage way. the places where the roads are provided. to drain off the rain water from the road surface quickly and effectively. Now a day. Bigger stones are used at the bottom. 1. The height of the kerb is normally 15 to 20 cms. Traffic separators: These are provided to separate the traffic moving in opposite directions. This layer is either made of flexible materials (bitumen or coal tar mixed with stones) or a rigid material (concrete).2. 2. The thickness of this layer depends upon. Sub grade: This layer gives support to the road structure. The shoulders should satisfy the following requirements:- They should have a sufficient bearing capacity even in wet condition. Dept of Civil Engg. The shoulders should have distinctive colour from the pavement to guide the vehicles users on the pavement only. a raised stone (kerb) is provided at the edge of the pavement. This space acts as a space for moving away any broken down vehicles or parking vehicles in an emergency. width are provided all along the length of the road. running through the length & breadth of a country. fencing is provided all along the road to prevent the cattle and people from entering the traffic zone. b. Footpaths should be provided essentially everywhere. All weather roads: These are roads which are usable in all seasons including rainy season in a year. where the traffic volume is very high. The footpath may be in level with the road surface or slightly raised higher than the road surface. Fencing: Whenever a highway or an expressway passes through urban areas. which interconnect district headquarters and also interlink national highways running through a state or neighbouring states.3 Classification or types of roads: Depending upon various criteria. separate cycle tracks of 2 mts. 2.surfaced roads: These are roads in which the topmost layer is not covered by a bituminous material or concrete but covered with a layer of stones mixed with gravel & thoroughly compacted. Guard stones and guard rails: Whenever. c. industrial areas and tourist destinations. These roads inter connect state capitals. Based on the nature of pavement surface provided: a. Based on the importance of connectivity. Footpath: Apart from vehicles. SJBIT Page 7 . 1.2. State highways: These are roads at a state level. 6.Elements of Civil Engineering and Engineering Mechanics 14CIV23 3. Dept of Civil Engg. 5. On these roads cross traffic & traffic in opposite direction is not allowed. Parking Lanes: These are usually provided or reserved on the road edges within a city limit for allowing the vehicles to be parked conveniently. 3. Un. Fencing is also provided all along the road. the road formation level is higher than the natural ground level. Based on seasonal usage: a. Cycle tracks: In some countries in urban areas. union territories. 4. National highways: These are the main network of roads. b. The parking lanes are distinctively separated by white colour strips so that moving vehicles do not enter parking lanes. guard stones or guard rails should be provided to avoid accidental fall of vehicles from the earth slope. 7. b. Throughout the length of the road medians are provided & vehicles can move at high speeds. function & traffic volume: a. Fair weather roads: These are roads which are usable during the dry seasons in a year. major ports. at the edges of the shoulders. roads can be classified as follows: 1. the pedestrians should also be provided some space for moving at the edge of the roads. Expressways: These are roads which are developed to inter connect two important cities only. Surfaced roads: These are roads in which the topmost layer is covered with a bituminous material or a rigid material like concrete. Highway running in cutting: Dept of Civil Engg. These roads may also inter connect state highways and national highways. They also serve as a link between agricultural areas and market places.Elements of Civil Engineering and Engineering Mechanics 14CIV23 d. The permissible speed and traveling comforts on such roads is lesser. but fairly leveled and covered with stones and gravel. Village roads: These are roads connecting villages & remote habitat groups with major district roads & other district roads. Other district roads: These are roads which interlink taluk headquarters and other main roads. f. Typical Cross Sections of a Highway: 1. Major district roads: These are important roads within a district. Highway running over an embankment: 2. which help in moving goods from agricultural production areas to market places. The surface of such roads may not be covered with a bituminous layer. e. SJBIT Page 8 . without closing the obstacle below. Sub Structure Dept of Civil Engg. a. A bridge may also be built for the safe passage of a canal (aqua duct). The obstacle to be crossed may be a river or stream.2 BRIDGES A bridge is a structure which provides a safe passage for a road or railway track over obstacles. a canal. Components of a bridge: A bridge basically consists of following two components. SJBIT Page 9 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 3. Super Structure b. road or a railway track. Highway running in urban areas: 1. Deck slab. a.Elements of Civil Engineering and Engineering Mechanics 14CIV23 a. guard rails. etc. Deck bridges: These are bridges in which road formation level or pavement is above the bearing level in a bridge. pavement etc. Super Structure: It refers to the part of the bridge above the bearing level. foundation. The components included in the super structure are RCC beam. piers. abutments. The components included in the substructure are bearings. wing walls. Dept of Civil Engg. bridges are classified as follows. Sub Structure: It refers to the part of the bridge below the bearing level. Classification of bridges: Depending upon the position of the road surface or road formation level with respect to the bearing level in a bridge. SJBIT Page 10 . b. SJBIT Page 11 . Through bridges: These are bridges in which the road formation level is lower than the bearing level in the bridge. b. Steel girder bridges are example for Deck bridges. Such bridges may not provide sufficient head room for all vehicles. Cable stayed bridges & truss bridges are example for through bridges. Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 In such bridges sufficient head room for all vehicles is available. c. Semi -through bridges: These are bridges in which the road formation level is at some intermediate level of the super structure. RCC beam bridges. if the road formation level is increased subsequently. If the centre line of the bridge is at right angles to the direction of flowing water in the river. the bridge is known as square bridge. the centre line of the bridge should be aligned at right angles to the direction of flowing water in the river. If the centre line of the bridge is not at right angles to the direction of flowing water in the river. Square Bridge Skew bridge Dept of Civil Engg. SJBIT Page 12 . Steel girder bridges are examples of semi -through bridges. the flowing water does not exert excessive forces on the piers and abutments.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Such bridges do not provide sufficient head room for all vehicles when the road formation level increases subsequently. In such cases. the bridge is known as skew bridge. Square bridges & skew bridges: Whenever a bridge is to be constructed over a stream or river. For controlling floods in a river 6. Dams serve the following purposes: 1. The stored water is released to the down stream side through canals and can be utilized for various uses. For inland navigation Classification of Dams: Depending upon various criteria. dams can be classified as follows. b.Elements of Civil Engineering and Engineering Mechanics 14CIV23 1. c. Storage dams: These are dams built across non perennial rivers to store water in a reservoir during excess flow. Supplying water for domestic & industrial uses 3. By Dept of Civil Engg. without making an attempt to store water. Coffer dams: These are temporary dams which are constructed during the construction of actual main dam to keep the dam site free from water. SJBIT Page 13 . Based on the purpose served: a. Diversion dams: These are irrigation structures which are constructed across a river to slightly raise the water level. 1. Aquaculture 5. Storing water for irrigation 2. Supplying water for hydroelectric power generation 4.3 DAMS A Dam is an obstruction or barrier or a hydraulic structure which is constructed across a river or stream to store water on the upstream side as an artificial lake or reservoir. the streams or tributaries which join the main river. Based on the hydraulic design: a. b. Overflow dams: Any dam is designed or constructed to store water up to a certain maximum height only. Barrages are examples of diversion dams. In overflow dams. the excess water entering the reservoir is not allowed to overtop the entire length of the dam. water is directly diverted into the canals. d. Dept of Civil Engg. Non-over flow dams: In majority of the dams. 2. the excess water should be discharged to the down stream side safely. e. SJBIT Page 14 . The excess water is released to the down stream side through a separate spill way & such dams are called non over flow dams. The spill may be included in the main portion of dam or through a separate spill way section. When the water level exceeds the maximum level. entry of silt & debris can be controlled & the useful life of the reservoir can be increased.Elements of Civil Engineering and Engineering Mechanics 14CIV23 increasing the water level. Detention dams: These are dams which are constructed to store water temporarily only during floods. the excess water is allowed to overtop the body of the dam. The water is then released to the down stream side when the floods recede. Debris dams: These are small dams which are built across. By constructing these dams across streams or tributaries on the upstream side of the main dam. The cross sections of such dams are very large & also have enormous self weight. Some amount of water also seeps through the bottom of the dam. b. Gravity dams: These are dams which are built of rigid materials like concrete & stone masonry. SJBIT Page 15 . Dept of Civil Engg. wind pressure. The body of the dam should be able to resist all such forces and different dams resist these forces in different ways. Arch dams: These are dams which are also constructed of rigid material like concrete or stone masonry. a. wind pressure etc. The destabilizing forces like hydrostatic pressure. The down stream face of the dam is exposed to wind pressure. These dams are curved in plan. The cross section of such dams is slender when compared to gravity dams.Elements of Civil Engineering and Engineering Mechanics 14CIV23 3. and wave pressure are resisted by the self weight of the dam only. to the banks of the river by arch action. the stored water exerts forces on the upstream face of the dam. The waves that are generated at the top of the reservoir also exert forces on the dam. This seeping water exerts (applies) uplift pressure on the dam. These dams transfer the water pressure. Based on the resisting action to external forces: When water is stored in a dam or reservoir. uplift pressure. SJBIT Page 16 . which is supported over a series of buttresses (piers or columns). in height. stone masonry. Gravity dams. The loads are transferred from the deck slab to the buttresses and then to the bed of the river. Dept of Civil Engg. Such dams cannot exceed 30 mtrs. 4. Rigid dams: These are dams which are constructed of rigid materials like concrete. Non rigid dams: These are dams which are constructed art of non rigid materials like earth fill & rock fill. Based on materials used: a. Arch dams and Buttress dams are examples of rigid dams.Elements of Civil Engineering and Engineering Mechanics 14CIV23 c. steel sheets. b. Buttress dams: These are dams in which the water pressure from the stored water acts on a thin deck slab. or size or the distance between any two points on the body changes on the application of external forces. Engineering Mechanics Mechanics of Solids Mechanics of Fluids Mechanics of Rigid bodies Mechanics of Deformable bodies Statics Dynamics Concepts of: Physical quantity.81 N or 10 N. The absolute (SI) unit of force is the Newton and is denoted as ‗N‘. or size or the distance between any two points on the body does not change on the application of external forces. and Vector quantity Particle: A particle is a body of infinitely small volume and the entire mass of the body is assumed to be concentrated at a point. Force: According to Newton‘s I law. which does not alter its shape. Units of force: The gravitational (MKS) unit of force is the kilogram force and is denoted as ‗kgf‘.81m/s2) Therefore 1 kgf = 9. Rigid body: It is one. . Note: 1 kgf = ‗g‘ N (But g = 9. or else it is considered as a deformable body. Engineering Mechanics deals with the application of principles of mechanics and different laws in a systematic manner.4 Engineering Mechanics It is a branch of applied sciences that describes and predicts the state of rest or of uniform motion of bodies under the action of forces.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Engineering Mechanics 1. force is defined as an action or agent. which changes or tends to change the state of rest or of uniform motion of a body in a straight line. the body considered is rigid as long as the distance between the points A and B remains the same before and after application of forces. SJBIT Page 17 . Deformable body: It is one. Scalar quantity. . Dept of Civil Engg. F A B F In the above example. which alters its shape. Ex. 2) Point of application: It indicates the point on the body on which the force acts. Line of action Body A B 10 N The characteristics of the force acting on the body are 1) Magnitude is 10 N. 3) Line of action: The arrowhead placed on the line representing the direction represents it. representing a force and also distinguishing one force from one another. which can be listed as follows. 3) Line of action is A to B or AB. which help in understanding a force completely. 1) Magnitude: It can be denoted as 10 kgf or 100 N. A force is a vector quantity.1: Consider a body being pushed by a force of 10 N as shown in figure below. It has four important characteristics. 4) Direction: It is represented by a co-ordinate or cardinal system.Elements of Civil Engineering and Engineering Mechanics 14CIV23 1. Dept of Civil Engg. SJBIT Page 18 .5 Characteristics of a force These are ones. Ex. 2) Point of application is A. 4) Direction is horizontally to right. on which a person weighing 750 N stands on the ladder at a point C on the ladder.2: Consider a ladder AB resting on a floor and leaning against a wall. . Classification of force systems: Depending upon their relative positions.. 1) Collinear forces: It is a force system. the different force systems can be classified as follows.Elements of Civil Engineering and Engineering Mechanics 14CIV23 The characteristics of the force acting on the ladder are 1) Magnitude is 750 N. F 2 Ex.: The forces or loads and the support reactions in case of beams. in which all the forces have the same line of action. They are as follows. 3) Line of action is C to D or CD. Y X Dept of Civil Engg. 1) A body consists of continuous distribution of matter. 4) A force acts through a very small point. in which all the forces are lying in the same . Y X Ex. in which all the forces are lying in the same plane and lines of action meet a single point. 3) A particle has mass but not size. points of applications and lines of actions. . a number of ideal conditions are assumed. 3) Coplanar Concurrent forces: It is a force system. Line of action F1 . Idealization or assumptions in Mechanics: In applying the principles of mechanics to practical problems. SJBIT Page 19 . 2) Coplanar parallel forces: It is a force system.: Forces in a rope in a tug of war. 2) The body considered is perfectly rigid. . plane and have parallel lines of action. 4) Direction is vertically downward. 2) Point of application is C. : The forces acting on a tripod when a camera is mounted on a tripod. Dept of Civil Engg. 5) Non. in which all the forces are lying in the different planes and still have common point of action. SJBIT Page 20 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Ex. 4) Coplanar non-concurrent forces: It is a force system. in which all the forces are lying in the different planes and still have parallel lines of action. when a ladder rests on a floor and leans against a wall.coplanar parallel forces: It is a force system.: Forces on a ladder and reactions from floor and wall. in which all the forces are lying in the same plane but lines of action do not meet a single point. Y X Z Ex.: The forces in the rope and pulley arrangement. Y X Z Ex: The forces acting and the reactions at the points of contact of bench with floor in a classroom.coplanar concurrent forces It is a force system. 6) Non. Y X Ex. unless it is compelled to do so by force acting on it.‖ This law helps in defining a unit force as one which produces a unit acceleration in a body of unit mass.‖ The significance of this law can be understood from the following figure. The body exerts a force W on the plane and in turn the plane exerts an equal and opposite reaction on the body. Consider a body weighing W resting on a plane.coplanar non-concurrent forces: It is a force system. in which all the forces are lying in the different planes and also do not meet a single point. ―Every body continues in its state of rest or of uniform motion in a straight line.‖ This law helps in defining a force. Dept of Civil Engg. SJBIT Page 21 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 7) Non. Y X Z Ex. a 3) Newton’s III law: It states. 1) Newton‘s I law 2) Newton‘s II law 3) Newton‘s III law 4) Principle or Law of transmissibility of forces 5) Parallelogram law of forces.6 Fundamental Laws in Mechanics Following are considered as the fundamental laws in Mechanics. 1) Newton’s I law: It states. ―The rate of change of momentum is directly proportional to the applied force and takes place in the direction of the impressed force. 1. 2) Newton’s II law: It states. thus deriving the relationship F = m . ―For every action there is an equal and opposite reaction.: Forces acting on a building frame. In the first case the body tends to compress and in the second case it tends to elongate.” Body A B A B F F Line of action From the above two figures we see that the effect of the force F on the body remains the same when the force is transmitted through any other point on the line of action of the force. F1 F2 R O F3 In the above figure R can be called as the resultant of the given forces F1. Resultant Force: Whenever a number of forces are acting on a body. SJBIT Page 22 . ―The state of rest or of Uniform motion of a rigid body is unaltered if the point of application of the force is Transmitted to any other point along the line of action of the force. it is possible to find a single force. F2 and F3. Explanation of limitation: Body F1 F2 F2 F1 A B A B Fig 1 Fig 2 Line of action In the example if the body considered is deformable. Thus principle of transmissibility is not applicable to deformable bodies or it is applicable to rigid bodies only. we see that the effect of the two forces on the body are not the same when they are shifted by principle of transmissibility. which can produce the same effect as that produced by the given forces acting together. Such a single force is called as resultant force or resultant. This law has a limitation that it is applicable to rigid bodies only.Elements of Civil Engineering and Engineering Mechanics 14CIV23 4) Principle or Law of transmissibility of forces: It states. Dept of Civil Engg. Parallelogram law of forces: This law is applicable to determine the resultant of two coplanar concurrent forces only.1 Let be the inclination of the resultant with the direction of the F1. F1. Let OA and OB represent forces F1 and F2 respectively both in magnitude and direction. then the resultant of the two forces is represented both in magnitude and direction by the diagonal of the parallelogram passing through the same point. The resultant R of F1 and F2 can be obtained by completing a parallelogram with OA and OB as the adjacent sides of the parallelogram. The diagonal OC of the parallelogram represents the resultant R both magnitude and direction. SJBIT Page 23 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 The process of determining the resultant force of a given force system is known as Composition of forces. Dept of Civil Engg. In analytical methods two different principles namely: Parallelogram law of forces and Method of Resolution of forces are adopted.2 F1 + F2. F2.cos Equation 1 gives the magnitude of the resultant and Equation 2 gives the direction of the resultant.e R = F12 + F2 2 + 2.cos --------.” B C R F2 F2sin F2 D O F1 A F2cos Let F1 and F2 be two forces acting at a point O and be the angle between them. then = tan-l F2 sin ------. This law states ―If two forces acting at a point are represented both in magnitude and direction by the two adjacent sides of a parallelogram. The resultant force of a given force system can be determining by Graphical and Analytical methods. From the figure OC = OD2 + CD2 = (OA + AD)2 + CD2 = ( F1 + F2 cos )2 + (F2 sin )2 i. Elements of Civil Engineering and Engineering Mechanics 14CIV23 Different cases of parallelogram law: For different values of . R = [ F1 . we can have different cases such as follows: Case 1: When = 900: R F2 R F2 O F1 F1 R= F12 + F22 = tan-l F2 F1 = 1800: Case 2: When . O F1 F2 R = [ F1 + F2 ] = 00 Dept of Civil Engg. SJBIT Page 24 .F2 ] F2 O F1 = 00 Case 3: When = 00 : . SJBIT Page 25 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 26 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 27 . let fx1. Fx2. such a force system is known as a F2 R F1 θ2 α θ1 θ3 θ4 F4 F3 coplanar concurrent force system Let F1. F4 represent a coplanar concurrent force system.Elements of Civil Engineering and Engineering Mechanics 14CIV23 MODULE 2 Analysis of Force Systems-Concurrent & Non Concurrent System 2. It is required to determine the resultant of this force system. Fx4 be the forces in the X- direction. ∑ Fy = fy1+ fy2+ fy3+ fy4 Dept of Civil Engg. F3 . F2. fx2. SJBIT Page 28 . fx3. These two sums are then combines using parallelogram law to get the resultant of the force systems. fx4 be the components of Fx1. In the ∑fig. It can be done by first resolving or splitting each force into its component forces in each direction are then algebraically added to get the sum of component forces. Let ∑ Fx be the algebraic sum of component forces in an x-direction ∑ Fx = fx1+ fx2+ fx3+ fx4 Similarly.1 Introduction If two or more forces are acting in a single plane and passing through a single point . Fx3. ∑ Fy R α ∑ Fx The magnitude os the resultant is given as R= √(∑ Fx)2 +(∑ Fy)2 The direction of resultant can be obtained if the angle α made by the resultant with x direction is determined here. ∑ Fx) and also in the y. Determine the direction of the resultant using the formula R= √(∑ Fx)2 +(∑ Fy)2 3. Determine the direction of the resultant using the formula α = tan-1( ) 2.Elements of Civil Engineering and Engineering Mechanics 14CIV23 By parallelogram law.2 Sign Conventions: Problems Dept of Civil Engg. Calculate the algebraic sum of all the forces acting in the x- direction (ie. SJBIT Page 29 . 1. α = tan-1( ) The steps to solve the problems in the coplanar concurrent force system are. ∑ Fy) 2. therefore as follows.direction ( ie. 75N 200N 700 300 45 350 0 150 N 100 N ∑ Fy R α ∑ Fx Let R be the given resultant force system Let α be the angle made by the resultant with x. The magnitude of the resultant is given as R= √(∑ Fx)2 +(∑ Fy)2 and α = tan-1( ) Dept of Civil Engg.direction. SJBIT Page 30 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Determine the magnitude & direction of the resultant of the coplanar concurrent force system shown in figure below. 72) = 3.direction.7N ∑ Fy = 200sin 300+ 75sin700-100sin450 .21N α = tan-1( ) α = tan-1(13.72 N R= √(∑ Fx)2 +(∑ Fy)2 R= 200.75cos700-100cos450 + 150cos 350 ∑ Fx = 199. Determine the resultant of the concurrent force system shown in figure. 500kN 700k N 70 400 0 45 2 0 1 300 150k 100kN N Let R be the given resultant force system Let α be the angle made by the resultant with x. The magnitude of the resultant is given as R= √(∑ Fx)2 +(∑ Fy)2 and α = tan-1( ) Dept of Civil Engg.72/ 199. SJBIT Page 31 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ∑ Fx = 200cos 300.930 2.150sin 350 ∑ Fy = 13. 11 kN ∑ Fy = 700sin 400+ 500sin700-800sin600 .55/ 144.500cos700-800cos600 + 200cos 26.direction.11) = 43.21N α = tan-1( ) α = tan-1(137.66 3.560 ∑ Fx = 144. SJBIT Page 32 .560 ∑ Fy = 137.55kN R= √(∑ Fx)2 +(∑ Fy)2 R= 199.200sin 26. Determine the resultant of a coplanar concurrent force system shown in figure below 100N 800N 200 350 1 3 500 N Let R be the given resultant force system Let α be the angle made by the resultant with x. Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 ∑ Fx = 700cos 400. SJBIT Page 33 .600 ∑ Fy = 110. Dept of Civil Engg.90 N R= √(∑ Fx)2 +(∑ Fy)2 R= 1101.48 N ∑ Fy = 800sin 350+ 100sin700+ 500sin600 .48) = 5.Elements of Civil Engineering and Engineering Mechanics 14CIV23 The magnitude of the resultant is given as R= √(∑ Fx)2 +(∑ Fy)2 and α = tan-1( ) ∑ Fx = 800cos 350. The Magnitude and direction of the resultant of the resultant of the coplanar concurrent force system shown in figure.90/ 1095. 20k N 52kN 300 350 10k N 600 60kN N Let R be the given resultant force system Let α be the angle made by the resultant with x.780 4.100cos700+ 500cos600 + 0 ∑ Fx = 1095.direction.08 N α = tan-1( ) α = tan-1(110. SJBIT Page 34 .68 Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 The magnitude of the resultant is given as R= √(∑ Fx)2 +(∑ Fy)2 and α = tan-1( ) ∑ Fx = 20cos 600.52cos300+ 60cos600 + 10 ∑ Fx = 7.570.130.60sin600 + 0 ∑ Fy = -8.641 /7.641 kN R= √(∑ Fx)2 +(∑ Fy)2 R= 11.130.374 N α = tan-1( ) α = tan-1(289.404 kN ∑ Fy = 20 sin 600+ 52sin300.400cos53.direction.404) = -49. θ1= tan-1(1 /2) = 26.13 Let R be the given resultant force system Let α be the angle made by the resultant with x.57 θ2= 53. Determine the Magnitude and direction of the resultant of the resultant of the coplanar concurrent force system shown in figure.50cos600 + 100cos 500 ∑ Fx = -21.379 N α = tan-1( ) α = tan-1(-8.50sin600 -100 sin 500 ∑ Fy = 289.844 kN ∑ Fy = 200 sin 26.552 /21. The magnitude of the resultant is given as R= √(∑ Fx)2 +(∑ Fy)2 and α = tan-1( ) ∑ Fx = 200cos 26.570+ 400sin53.844) = -85.552 kN R= √(∑ Fx)2 +(∑ Fy)2 R= 290.40 5. 196m Dept of Civil Engg.20 7.26 N α = tan-1( ) α = tan-1(65. BC= 6m In Δle AOC Sin 600= OC/ 6 OC = 6sin600 OC = 5. Let R be the given resultant force system ∑ Fx =? ∑ Fy =? Let α be the angle made by the resultant with x.direction.71 kN ∑ Fy = 80 sin 250+ 50sin 50 0.direction. The magnitude of the resultant is given as R= √(∑ Fx)2 +(∑ Fy)2 and α = tan-1( ) ∑ Fx = 80 cos 250+ 50cos50 0+ 10cos 450 ∑ Fx = 111. A hook is acted upon by 3 forces as shown in figure. Determine the resultant force on the structure.04 /111. as shown in fig. In Δle AOC Cos 600= AC/ 6 AC = 6cos600 AC =3 m.10cos 450 ∑ Fy = 65.71) = 30. Determine the resultant force on the hook.Elements of Civil Engineering and Engineering Mechanics 14CIV23 6. Two forces are acting on a structure at a point ‗ O‖ . SJBIT Page 35 .04 kN R= √(∑ Fx)2 +(∑ Fy)2 R= 129. Let R be the given resultant force system Let α be the angle made by the resultant with x. Determine the magnitude & Direction of third unknown force. such that the resultant of all the three forces has a magnitude of 1000N.e The algebraic sum of all vertical component forces is equal to the vertical component of the resultant.89 0 ∑ Fx = 346.e The algebraic sum of all horizontal component forces is equal to the horizontal component of the resultant. θ = tan-1(OC/BC) = tan-1(5.axis as shown F3= ? θ3= ? We know that R cos α = ∑ Fx 1000cos 450 = 500cos 300+ 1000cos 600+ F3 cos θ3 F3 cos θ3= -225. Two forces of magnitude 500N & 100N are acting at a point as shown in fig below.890 ∑ Fx = 800 – 600 cos40.76 /346.(1) R sin α = ∑ Fy Dept of Civil Engg. Let F3 be the required third unknown force. ∑ Fy = R sin α i. SJBIT Page 36 .906N ------------.600 sin 40.76 kN R= √(∑ Fx)2 +(∑ Fy)2 R= 523.41) = 48.89 0 ∑ Fy = . we can write ∑ Fx = R cos α i.7 N α = tan-1( ) α = tan-1(-392.58 Note: From the above two figures.392.41 N ∑ Fy = 0 . 8.Elements of Civil Engineering and Engineering Mechanics 14CIV23 In Δle OBC. making an angle of 45 0 as shown. which makes angle θ 3 with x.19 /6 ) = 40. (1) R sin α = ∑ Fy 1000sin 450 = 500sin 300+ 100sin 600+ F3 sin θ3 F3 sin θ3= 370.08 = -467. we have –ve values from both F3 cos θ3 and F3 sin θ3 ( X & Y components of force F3 ).50N ---------------. making an angle of 450 & lying in the second quadrant.08 From (1) F3 cos θ3= -225. Two forces of magnitude 500N and 100N are acting at a point as shown in fig below. We known that R cos α = ∑ Fx -1000cos 450 = 500cos 300+ 100cos 600+ F3 cos θ3 F3 cos θ3= -1190. Thus the current direction for force F3 is represented as follows.810 θ3= tan-1(1. Let us assume F3 to act as shown in fig.(2) Dividing the Equation (2) by (1) i. F3 sin θ3/ F3 cos θ3= -408.906 Tan θ3 = 1.3113 Dept of Civil Engg. θ3=? Let F3 be a required third unknown force making an angle θ3 with the x.810) = 61.e.91N ---------------.906 / cos 61.91/ -225.axis to satisfy the given condition.14 N Here .(2) Dividing the Equation (2) by (1) i.906N F3 =-225.50/ -1190. 9. F3 sin θ3/ F3 cos θ3= 370.119 Tan θ3 = 0.Elements of Civil Engineering and Engineering Mechanics 14CIV23 1000sin 450 = 500sin 300+ 1000sin 600+ F3 sin θ3 F3 sin θ3= -408. F3=? .119N ------------. SJBIT Page 37 .e. Determine the magnitude & direction of a 3rd unknown force such that the resultant of all the three forces has a magnitude of 1000N. SJBIT Page 38 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 θ3= tan-1(0.50/sin 17. This can be understood from the following figures. Moment of Force: It is defined as the rotational effect caused by a force on a body.29 From (2) F3 sin θ3= 370.29 = 1246. such a force system is known as coplanar non concurrent force system. Dept of Civil Engg. Let‖ O‖ be a point or particle in the same plane.63N 2.3113) = 17. Hence the units for moment can be ―Nm‖ or ―KNm‖ or ― N mm‖ etc. Thus the moment of the force about the point ―O‖ is given as Mo= F x d Moment or rotational effect of a force is a physical quantity dependent on the units for force and distance.50N F3 = 370. Let ― F‖ be a force acting in a plane. but not passing through the single point. The moment produced by a force about differences points in a plane is different. Let ―d ‖ be the perpendicular distance of the line of action of the force from the point ―O‖ . Mathematically Moment is defined as the product of the magnitude of the force and perpendicular distance of the point from the line of action of the force from the point.3 COMPOSITION OF COPLANAR NONCONCURRENT FORCE SYSTEM If two or more forces are acting in a single plane. Elements of Civil Engineering and Engineering Mechanics 14CIV23 Let ― F‖ be a force in a plane and O1.4 Couple Two forces of same magnitude separated by a definite distance. O2. (acting parallely) in aopposite direction are said to form a couple. Hence the units for magnitude of a couple can be N m. A clockwise moment ( ) is treated as positive and an anticlockwise moment ( ) is treated as negative. Mo1= F x d1 Let moment of the force ― F‖ about point O2 is Mo. The points O1. As such the moment due to a couple is also denoted as M.e The magnitude of a moment due to a couple is the product of force constituting the couple & the distance separating the couple . Let S be the distance separating the couple. Mo= 0x F The given force produces a clockwise moment about point O1 and anticlockwise moment about O2 . O2. Dept of Civil Engg. Note. Let d1 & d2 be the perpendicular distance of the lines of action of the forces from the point o. N mm etc. Thus the magnitude of the moment due to the couple is given a s Mo = (Fx d1 ) + (F x d2 ) Mo = F x d i. O3 about which the moments are calculated can also be called as moment centre. 2. and O3 be different points in the same plane Let moment of the force ― F‖ about point O1 is Mo. A couple has a tendency to rotate a body or can produce a moment about the body. Let us consider a point O about which a couple acts. Mo2= F x d2 Let moment of the force ― F‖ about point O3 is Mo. SJBIT Page 39 . kN m . The procedure is as follows: 1. the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point. PROOF: For example. consider only two forces F1 and F2 represented in magnitude and direction by AB and AC as shown in figure below.e. SJBIT Page 40 . about which the moments are taken. in magnitude and Direction. Moment of force R about O=Moment of force F1 about O + Moment of force F2 about O Similarly. the resultant of two forces F1 and F2. this principle can be extended for any number of forces.. Let O be the point.Elements of Civil Engineering and Engineering Mechanics 14CIV23 2. Construct the parallelogram ABCD and complete the construction as shown in fig. By geometrical representation of moments the moment of force about O=2 Area of triangle AOB the moment of force about O=2 Area of triangle AOC the moment of force about O=2 Area of triangle AOD But. Area of triangle AOD=Area of triangle AOC + Area of triangle ACD Also. Area of triangle ACD=Area of triangle ADB=Area of triangle AOB Area of triangle AOD=Area of triangle AOC + Area of triangle AOB Multiplying throughout by 2. By using the principles of resolution composition & moment it is possible to determine Analytically the resultant for coplanar non-concurrent system of forces. Dept of Civil Engg. we obtain 2 Area of triangle AOD =2 Area of triangle AOC+2 Area of triangle AOB i. let R be the resultant force. the diagonal AD represents.5Varignon’s principle of moments: If a number of coplanar forces are acting simultaneously on a particle. By the parallelogram law of forces. Select a Suitable Cartesian System for the given problem. Compute the moments of resolved components about any point taken as the moment Centre O. SJBIT Page 41 . Hence find ∑M0 Dept of Civil Engg. Compute ∑fxi and ∑fyi 4.Elements of Civil Engineering and Engineering Mechanics 14CIV23 2. Resolve the forces in the Cartesian System 3. 1.Problems ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 42 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 2. Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 43 . Determine the resultant of the force system acting on the plate. ∑Fx = 5cos300 + 10cos600 + 14.Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- 2.14cos450 Dept of Civil Engg. SJBIT Page 44 . As shown in figure given below with respect to AB and AD. 10sin600 + 14.Elements of Civil Engineering and Engineering Mechanics 14CIV23 = 19.14sin450 = -16.2N θ= Tan-1(∑Fy/ ∑Fx) θ= Tan-1(16.83 y = 5.89 = y/6.890 Tracing moments of forces about A and applying varignon‘s principle of moments we get +16.16N R = √(∑Fx2 + ∑Fy2) = 25. 3.99/16.16X = 20x4 + 5cos300x3 -5sin300 x4 + 10 + 10cos600x3 x = 107. Dept of Civil Engg.33) = 39. Determine the magnitude .16 = 6. direction and the point of application of the resultant force.33N ∑Fy = 5sin300 .586m. The system of forces acting on a crank is shown in figure below.16/19.683m Also tan39. SJBIT Page 45 . 300 (Direction) Tracing moments of forces about O and applying varignon‘s principle of moments we get -2633x x= -500x sin600x300-1000x150+1200x150cos600 -700x300sin600 x = -371769. ∑Fy = +100 -200 -50 +400 = +250N ie.15/-2633 x= 141. 4. R = ∑Fy =250N ( ) Since ∑Fx = 0 Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 ∑Fx = 500cos600 – 700 = 450N ∑Fy = 500sin600 = -26. determine the magnitude of the resultant and also its position from A .20mm from O towards left (position). SJBIT Page 46 .For the system of parallel forces shown below.19N (Magnitude) θ = Tan-1(∑Fy/ ∑Fx) θ = Tan-1(2633/450) θ = 80.33N R = √(∑Fx2 + ∑Fy2) =(-450)2 + (-2633)2 R = 267. Given tan θ= 15/8 sin θ = 15/ 17 cos θ = 8/17 tanα= 3/4 sin α = 3/5 cos α = 4/5 ∑ Fx = 4 +5 cos α – 17 cos θ = 4 + 5 x 4/5 – 17 x 8/17 ∑ Fx = 0 Dept of Civil Engg. Find the magnitude. direction and the position of the resultant force. A beam is subjected to forces as shown in the figure given below.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Taking moments of forces about A and applying varignon‘s principle of moments -250 x = -400 x 3. If the resultant R=600 N and is acting at a distance of 4.find the magnitude of force F and position of F with respect to A Let x be the distance from A to the point of application of force F Here R = ∑ Fy 600=100+F+300 F = 200 N Taking moments of forces about A and applying varignon‘s principle of moments.5 m from A . SJBIT Page 47 .3m 5.5 + 200 x 1 – 100 x 0 X = -1075/ -250 = 4.5 + 50 x 2.F and 300 N are acting as shown in figure below.5 = 300 x 7 + F x 200 x = 600 x 4. X = 600/200 = 3m from A 6.5 -300 x 7. We get 600 x 4. The three like parallel forces 100 N. SJBIT Page 48 .89m from A (towards left) Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 ∑ Fy = 5 sin α -10 + 20 – 10 + 17 sin θ = 5 x 3/5 -10+ 20 – 10 + 17 x 15/17 ∑ Fy = 18 kN ( ) Resultant force R = (∑Fx) 2 + (∑Fy)2 = 0+182 R = 18 kN( ) Let x = distance from A to the point of application R Taking moments of forces about A and applying Varignon‘s theorem of moments -18 x = -5 x sin α x 8 +10 x 7 -20 x 5 + 10 x 2 = -3 x 8 +10 x7 – 20 x 5 + 10 x 2 X = -34/-18 = 1. Elements of Civil Engineering and Engineering Mechanics 14CIV23 -------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 49 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 50 . SJBIT Page 51 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 52 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 53 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 54 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 55 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 MODULE 3 EQUILIBRIUM OF FORCES AND FRICTION Dept of Civil Engg. SJBIT Page 56 . SJBIT Page 57 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 58 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. At contact point 1. Applying lami‘s theorem to the system of forces. determine the reactions developed at contact surfaces.Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- 1.A 200 N sphere is resting in at rough as shown in fig. the surface contact is making 60 0 to horizontal. Similarly the reaction R2 at contact point 2 makes 450 to the vertical. Hence the reaction R1 which is normal to it makes 600 with vertical. FBD as shown in figure. Assume all contact surfaces are smooth. we get Dept of Civil Engg. SJBIT Page 59 . Soln. T1 sin 30 = 0 T2 sin = T1 sin 30 --------------(i) ∑ Fy = 0 -T2 cos + T1 cos 30 – 20 = 0 T2 cos = T1 cos 30 – 20 --------------(ii) Considering the equilibrium of point C. and CD of the rope and also the inclination of BC to the vertical. ∑ Fx = 0 T3 sin 60 – T2 sin = 0 T2 sin = T3 sin 60 --------------(iii) ∑ Fy = 0 T3 cos 60+ T2 cos – 25= 0 T2 cos = . Soln.Elements of Civil Engineering and Engineering Mechanics 14CIV23 R1/ sin (180 – 45) = R2/ sin (180 –60)= 400/ sin ( 60+45) R1= 292.6N ---------------------------------------------------------------------------------------------------------------- A wire is fixed at A and D as shown in figure. Considering equilibrium o fpoint B. SJBIT Page 60 .T3 cos 60 +25 --------------(iv) From equations (i) and (ii) . BC. we get T1 sin 30 = T3 sin 60 T1= T3 From equations (ii) and (iv) . Determine the tension in the segments AB. we get ∑ Fx = 0 T2 sin .8N R2= 358. we get T1 cos 30 – 20 = . Free body diagrams of the point B and C are shown in figures respectively. Weights 20 kN and 25kN are supported at B and C respectively.T3 cos 60 +25 Dept of Civil Engg. When equilibrium is reached it is found that inclination o fAB is 300 and that of CD I s600 to the vertical. Resting on a smooth floor and leaning on a smooth wall.780 T2= 23. Let F be the horizontal force required to be applied to prevent slipping.Elements of Civil Engineering and Engineering Mechanics 14CIV23 T3= 22. For the beam to be in equilibrium.3N BEAMS A beam is a structural member or element. The force system is developed due to the loads or forces acting on the beam and also due to the support reactions developed at the supports for the beam.5cot 60= 0 RB= 298. one dimension (length) is considerably larger than the other two dimensions (breath & depth).RB=0 F. Determine the horizontal force required at floor level to prevent it from slipping when a man weighing 700 N is at 2 m above floor level.416 = 54. Self weight of 100N acts through centre point of ladder in vertical direction. In a beam.84kN ---------------------------------------------------------------------------------------------------------------- A ladder weighing 100N is to be kept in the position shown in figure. Then ∑ MA= 0 -RB x 3 + 700x2 cot 60+ 100 x 1. Dept of Civil Engg. Ra is vertical and Rb is horizontal because the surface of contact is smooth. the reactions developed at the supports the should be equal and opposite to the loads. The smaller dimensions are usually neglected and as such a beam is represented as a line for theoretical purposes or for analysis. which is in equilibrium under the action of a non-concurrent force system.RB= 298. Free body diagram of the ladder is as shown in figure.3N ∑ Fx = 0 F.5kN T1= 38.97kN From equation (i) and( ii) Tan = 1. SJBIT Page 61 . is developed. At such a support. SJBIT Page 62 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Types of Supports for beams: Supports are structures which prevent the beam or the body from moving and help to maintain equilibrium. Dept of Civil Engg. The beam is free to move only horizontally and also can rotate about the support. which are frictionless. 1) Simple support: This is a support where a beam rests freely on a support. 3) Hinged support: This is a support in which the beam is attached to a support by means of a hinge or pin. In such a support a horizontal reaction and a vertical reaction will develop. In such a support one reaction. which is perpendicular to the plane of support. Here one reaction which is perpendicular to the plane of rollers is developed. A beam can have different types of supports as follows. 2) Roller support: This is a support in which a beam rests on rollers. the beam is free to move horizontally and as well rotate about the support. The support reactions developed at each support are represented as follows. The beam is not free to move in any direction but can rotate about the support. RH M RV Types of beams Depending upon the supports over which a beam can rest (at its two ends). It is a beam where one end of the beam is hinged to a support and the other end rests on a roller support. SJBIT Page 63 . beams can be classified as follows.Elements of Civil Engineering and Engineering Mechanics 14CIV23 4) Fixed support: This is a support which prevents the beam from moving in any direction and also prevents rotation of the beam. Such a beam can carry any type of loads. Dept of Civil Engg. vertical reaction and a Fixed End Moment are developed to keep the beam in equilibrium. A beam is said to be simply supported when both ends of the beam rest on simple supports. Such a beam can carry or resist vertical loads only. 1) Simply supported beam. A B RB RA 2) Beam with one end hinged & other on rollers. In such a support a horizontal reaction. RAH A B RBH RAV RBV 4) Over hanging beam : It is a beam which projects beyond the supports.Elements of Civil Engineering and Engineering Mechanics 14CIV23 A B RAH RAV RBV 3) Hinged Beam: It is a beam which is hinged to supports at both ends. SJBIT Page 64 . B A B A RA RB RA RB A B RA RB Dept of Civil Engg. Such a beam can carry loads is any direction. A beam can have over hanging portions on one side or on both sides. A B RAH A B M M RAV RB 7) Continuous beam: It is a beam which rests over a series of supports at more than two points. RH M RV 6) Propped cantilever: It is a beam which has a fixed support at one end and a simple support at the other end. with one end fixed and other and free. Such a beam can carry loads in any directions. SJBIT Page 65 . MB A B C RC RA RB Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 5) Cantilever Beams: It is a beam. The UDL should be replaced by an equivalent point load or total load acting through the mid point of the loaded length. ΣFy = 0. As such. SJBIT Page 66 . it is assumed to act at the mid point of the loaded length and such a loading is termed as Point load or Concentrated load. then such a loading is called UDL. They need additional special conditions for analysis and as such. The magnitude of the point load or total load is equal to the product of the intensity of loading and the loaded length (distance). L w/ unit length Dept of Civil Engg. 1) Point load or Concentrated load: If a load acts over a very small length of the beam.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Note: The support reactions in case of simply supported beams. ΣM = 0). over hanging beams. such beams are known as Statically Indeterminate Beams Types of loads: The various types of loads that can act over a beam can e listed as follows. can be determined by conditions of equilibrium only (Σ Fx = 0. W=w . P A B RA RB 2) Uniformly distributed load (UDL): If a beam is loaded in such a manner that each unit length of the beam carries the same intensity of loading. Propped Cantilever and Continuous Beams the support reactions cannot be determined using conditions of equilibrium only. In beams such as Hinged Beams. beam with one end hinged and other on rollers. A UDL cannot be considered in the same manner for applying conditions of equilibrium on the beam. such beams are known as Statically Determinate Beams. and cantilever beams. L W=w . In applying conditions of equilibrium. a given UVL should be replaced by an equivalent point load or total load acting through the centroid of the loading diagram (right angle triangle). X a+x/2 b+x/2 w/ unit length x/2 w/ unit length L/2 L/2 a X b L L 3) Uniformly varying load (UVL): If a beam is loaded in such a manner. W=w. SJBIT Page 67 . The magnitude of the equivalent point load or total load is equal to the area of the loading diagram. that the intensity of loading varies linearly or uniformly over each unit distance of the beam. then such a load is termed as UVL.Elements of Civil Engineering and Engineering Mechanics 14CIV23 G L/2 L/2 L W=w .L/2 G w/ unit length 2L/3 L/3 L Dept of Civil Engg. x/2 2 x/3 x/3 w/ unit length x b a L 4) External moment: A beam can also be subjected to external moments at certain points as shown in figure. Elements of Civil Engineering and Engineering Mechanics 14CIV23 W=w . L (w2-w1)/unit length w1/ unit length w2/ unit length 2L/3 L/3 L/2 L/2 L Dept of Civil Engg. SJBIT Page 68 . L W2= {(w2-w1)/2}. W1= w1. These moments should be considered while calculating the algebraic sum of moments of forces about a point on the beam B A M RB RA Note : A beam can also be subject to a load as shown in figure below. SJBIT P Page 69 . such on opposing force developed is called friction or frictional resistance.Elements of Civil Engineering and Engineering Mechanics 14CIV23 In such a case. The frictional resistance is developed due to the interlocking of the surface irregularities at the contact surface b/w two bodies Consider a body weighing W resting on a rough plane & subjected to a force ‗P‘ to displace the body. FRICTION Whenever a body moves or tends to move over another surface or body. Limiting friction value F 45 Dept of Civil Engg. the UVL can be split into a UDL with a uniform intensity of w1/unit length another UVL with a maximum intensity of (w2-w1) /unit length. W Body of weight ‘W ’ Rough plane P F N Where P = Applied force N = Normal reaction from rough surface F = Frictional resistance W = Weight of the body The body can start moving or slide over the plane if the force ‗P‘ overcomes the frictional ‗F‘ The frictional resistance developed is proportional to the magnitude of the applied force which is responsible for causing motion upto a certain limit. a force which opposes the motion of the body is developed tangentially at the surface of contact. If the magnitude of the applied force is greater than the limiting friction value the body starts moving over the surface. F It is defined by the relationship N Where μ – Represents co-efficient of friction F – Represents frictional resistance N – Represents normal reaction.e. the magnitude of limiting friction bears a constant ratio to normal reaction between the two this ratio is called as co-efficient of friction. However F cannot increase beyond a certain limit. Dept of Civil Engg. If the magnitude of the applied force is less than the limiting friction value. W W P N F F N CO-EFFICIENT OF FRICTION: It ha been experimentally proved that between two contacting surfaces. It can range between a zero to limiting fraction value. SJBIT Page 70 . Beyond this limit (Limiting friction value) the frictional resistance becomes constant for any value of applied force.Elements of Civil Engineering and Engineering Mechanics 14CIV23 From the above graph we see that as P increases. F also increases. The friction experienced by a body when it is moving is called dynamic friction. The friction experienced by a body when it is at rest or in equilibrium is known as static friction. In our discussion on friction all the surface we consider will be dry sough surfaces. the frictional resistance developed at such surface can be called dry friction & wet friction (fluid friction) respectively. dry surface & wet surface. The dynamic friction experienced by a body as it roles over surface as shown in figure is called rolling friction. The dynamic friction experienced by a body as it slides over a plane as it is shown in figure is called sliding friction. Note: Depending upon the nature of the surface of contact i. the body remains at rest or in equilibrium. μ ) 5) The magnitude of the frictional resistance developed is exactly equal to the applied force till limiting friction value is reached or where the bodies is about to move. such an angle ‗θ‘ is called the Angle of friction As P increases. 3) The frictional resistance is independent of the area of contact between the two bodies. 4) The ratio of the limiting friction value (F) to the normal reaction (N) is a constant (co- efficient of friction. SJBIT Page 71 . 2) Frictional resistance acts in a direction opposite to the motion of the body. Let θmax = α Where α represents angle of limiting friction F tan θmax = tanα = N F But =μ N Dept of Civil Engg. F cannot increase beyond the limiting friction value and as such ‗θ‘ can attain a maximum value only. Let ‗R‘ be resultant of F & N. frictional resistance reaches limiting friction value. Let ‗P‘ be an applied force required to just move the body such that. They are as follows.Elements of Civil Engineering and Engineering Mechanics 14CIV23 LAWS OF DRY FRICTION: (COLUMB’S LAWS) The frictional resistance developed between bodies having dry surfaces of contact obey certain laws called laws of dry friction. F also increases and correspondingly ‗θ‘ increases. 1) The frictional resistance depends upon the roughness or smoothness of the surface. However. Let ‗θ‘ be the angle made by the resultant with the direction of N. ANGLE OF FRICTION W Limiting friction value Body P F F 45 θ N R P Consider a body weighing ‗W‘ placed on a horizontal plane. SJBIT Page 72 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Therefore μ = tanα i. W N F θ 90 .θ) – F cosθ = 0 N sinθ = F cosθ F tanθ = N Dept of Civil Engg. on which a body free from external forces can rest without sliding is called angle of repose. a stage reaches when the body tends to slide down the plane The maximum inclination of the plane with the horizontal. which makes an angle ‗θ‘ with the horizontal. When ‗θ‘ value is small. ∑Fx = 0 N cos(90. co-efficient of friction is equal to the tangent of the angle of limiting friction ANGLE OF REPOSE: Body of weight ‘W’ Rough plane θ Consider a body weighing ‗w‘ placed on a rough inclined plane. If ‗θ‘ is gradually increased.e. the body is in equilibrium or rest without sliding. Let us draw the free body diagram of the body before it slide.θ Applying conditions of equilibrium. Let θmax = Ф Where Ф = angle of repose When = angle of repose. all the surface that bee consider are rough surfaces. such that. If ‗P‘ is rotated through 360o.e. We can consider the body to be at rest or in equilibrium & we can still apply conditions of equilibrium on the body to calculate unknown force.Elements of Civil Engineering and Engineering Mechanics 14CIV23 tanθmax = tan Ф F but =μ N μ = tanα tan Ф = tanα Ф=α i. Dept of Civil Engg. angle of repose is equal to angle of limiting friction W CONE OF FRICTION Body of weight ‘W’ Rough plane P F Nθ R N Consider a body weighting ‗W‘ resting on a rough horizontal surface. when the body tends to move frictional resistance opposing the motion comes into picture tangentially at the surface of contact in all the examples. Let ‗R‘ be the resultant of ‗F‘ & ‗N‘ making an angel with the direction of N. SJBIT Page 73 . If the direction of ‗P‘ is changed the direction of ‗F‘ changes and accordingly ‗R‘ also changes its direction. R also rotates through 360o and generates an imaginary cone called cone of friction. Note: In this discussion. Let ‗P‘ be a force required to just move the body such that frictional resistance reaches limiting value. the body considered is at the verge of moving such that frictional resistance reaches limiting value. SJBIT Page 74 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 75 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 76 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 77 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 78 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 79 . SJBIT Page 80 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 81 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 82 . SJBIT Page 83 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 84 . SJBIT Page 85 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. whereas the centroid is a point in a plane area such that the moment of areas about an axis through the centroid is zero Plane area ‘A’ G g2 g1 d2 d1 a2 a1 Note: In the discussion on centroid. however the entire weight of a body is assumed to act through a single point and such a single point is called centre of gravity. Centre of gravity refer to bodies with mass and weight whereas.2 Centroid In case of plane areas (bodies with negligible thickness) such as a triangle quadrilateral.. 2.1 Centre of Gravity Everybody is attracted towards the centre of the earth due gravity. centre of gravity is a point is a point in a body through which the weight acts vertically downwards irrespective of the position. circle etc. The force of attraction is proportional to mass of the body. centroid refers to plane areas. the area of any plane figure is assumed as a force equivalent to the centroid referring to the above figure G is said to be the centroid of the plane area A as long as a1d1 – a2 d2 = 0. the total area is assumed to be concentrated at a single point and such a single point is called centroid of the plane area. 1. Every body has one and only centre of gravity. Dept of Civil Engg. The term centre of gravity and centroid has the same meaning but the following differences. 4. Everybody consists of innumerable particles. SJBIT Page 86 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 MODULE 4 Centroid and Moment of Inertia 4. a bottommost horizontal line or a horizontal line through the bottommost point can be made as the X – axis and a leftmost vertical line or a vertical line passing through the leftmost point can be made as Y. While locating the centroid of plane areas. The distance of centroid G from vertical reference axis or Y axis is denoted as X and the distance of centroid G from a horizontal reference axis or X axis is denoted as Y. X and Y axis.axis.Elements of Civil Engineering and Engineering Mechanics 14CIV23 4.3 Location of centroid of plane areas Y Plane area ‘A’ X G Y The position of centroid of a plane area should be specified or calculated with respect to some reference axis i.e. SJBIT Page 87 . Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 In some cases the given figure is symmetrical about a horizontal or vertical line such that Y lies on the line of symmetry. SJBIT Page 88 . The centroid of a circle is centre of the circle itself. Y=? The centroid of plane geometric area can be located by one of the following methods a) Graphical methods b) Geometric consideration c) Method of moments The centroid of simple elementary areas can be located by geometric consideration. where the three medians intersect.4 METHOD OF MOMENTS TO LOCATE THE CENTROID OF PLANE AREAS Dept of Civil Engg. 4. The centroid of a square is a point where the two diagonals bisect each other. the centroid of the plane area X G Y X b b/2 The above figure is symmetrical about a vertical line such that G lies on the line of symmetry. The centroid of a triangle is a point. Thus X= b/2. x3 . Let g1 .. Let G be the centroid of the plane area. x2 + a3 . as shown in figure.. a 1 .axis i. x1 + a2 . Let us divide the given area A into smaller elemental areas a1. Similarly a1.dA Y= or Y = A A Dept of Civil Engg.axis is =A X --(1) The sum of the moments of the elemental areas about Y axis is a1 . to calculate X and Y. x3 + ………. etc.. y1 a2 . x1 + a2 . g3…… be the centroids of elemental areas a1. x2. x is the distance of elemental area from Y axis. X= A (ax) x. It is required to locate the position of centroid G with respect to the reference axis like Y.axis and Xi.(2) Equating (1) and (2) A . a3 ……. X = a1 . x3 + ………………. a3 ……. a2. x2 + a3 . y3 . x2 a 3 . Y= A (a. SJBIT Page 89 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Let us consider a plane area A lying in the XY plane.g2. y ) y. y2 a3 . from Y. x1 a 2 . x3 etc be the distance of the centroids g1 g2 g3 etc. Let x1.. a2.dA X= or X = A A Where a or dA represents an elemental area in the area A ..e. SJBIT Page 90 . let g be the centroid of the rectangle. Let us consider an elemental area dA of breadth b and depth dy lying at a distance of y from the X axis. Dept of Civil Engg. Let us consider the X and Y axis as shown in the figure.Elements of Civil Engineering and Engineering Mechanics 14CIV23 TO LOCATE THE CENTROID OF A RECTANGLE FROM THE FIRST PRINCIPLE (METHOD OF MOMENTS) Y b dy X G d y Y X Let us consider a rectangle of breadth b and depth d. T d d y. dy . W.dy Y = 0 = d 0 A d 1 y2 A=b. SJBIT Page 91 .K.d d Y = 2 Similarly Dept of Civil Engg. 1 d2 d Y = d 2 y.(b.dA 1 Y y.d Y = d 2 0 dA = b .dy ) Y = 0 b.Elements of Civil Engineering and Engineering Mechanics 14CIV23 . dy) 0 Y= [as x varies b1 also varies] b.h A= 2 dA = b1 . Let us consider the X. lying at a distance y from X-axis.K. W. Elements of Civil Engineering and Engineering Mechanics 14CIV23 Centroid of a triangle Let us consider a right angled triangle with a base b and height h as shown in figure.axis and Y. SJBIT Page 92 .h 2 Dept of Civil Engg.T h y. Let G be the centroid of the triangle. dy h y.axis as shown in figure.dA 0 Y= A b.(b1 . Let us consider an elemental area dA of width b1 and thickness dy. Elements of Civil Engineering and Engineering Mechanics 14CIV23 h 2 y2 Y= .h Y= 6 h Y= similarly X =b 3 3 Dept of Civil Engg. y dy h 0 h h 2 y2 y3 Y= h 2 3. SJBIT Page 93 .h 0 2 h2 h3 Y= h 2 3.h 2 h2 h2 Y= h 2 3 1 1 Y = 2h 2 3 2. d 3 0 . SJBIT Page 94 .let ‗G‘ be centroid of the semi-circle.r 2 A= 2 2r = cos 0 3 y. Neglecting the curvature.d 3 Y= A 2r = sin . Let us consider an elemental area ‗dA‘ with centroid ‗g‘ as shown in fig.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Centroid of a semi circle X =0 Let us consider a semi-circle.d WKT 2 2 y.dθ and height ‗r‘. Let ‗O‘ be the centre of the semi-circle . sin .sin 3 r2 dA = .r.r 2r 2 Here y = .dA 3 3 Y= A Dept of Civil Engg.dA 2r Y= = [1+1] A 3 4r 2r Y= .d .dA = r sin . with a radius ‗r‘. the elemental area becomes an isosceles triangle with base r. Let y be the distance of centroid ‗g‘ from x axis. 1 dA = . Let us consider the x and y axes as shown in figure. sin .d 3 0 .r 2 dA = . Let ‗y‘ be the distance of centroid ‗g‘ from x axis. sin 2r 3 .d . Let us consider an elemental area ‗dA‘ with centroid ‗g‘ as shown in fig. . the elemental area becomes an isosceles triangle with base r.dA 2r /2 3 = cos 0 WKT 3 Y= A 4r y.d A= 2 2 Similarly Y= y. SJBIT 4r Page 95 = sin . Neglecting the curvature. sin .r 2 4 /2 Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Centroid of a quarter circle y Let us consider a quarter circle with radius r.dA 4r 2 X = 2r r 3 Y= .dθ and height ‗r‘.r.r 3 dA = Y= 2 A 4r 2 Y= r 3 .dA 1 = [0+1] . 2r Here y = . Let ‗O‘ be the centre and ‗G‘ be the centroid of the quarter circle. Let us consider the x and y axes as shown in figure.d A 3 2 . SJBIT Page 96 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. its moment about an axis can be found by multiplying the area with distance of its centroid from the reference axis. use of sections which are built up of many simple sections is very common. Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 4. SJBIT Page 97 . one need not go for the first principle (method of integration). To locate the centroid of composite sections. Such sections may be called as built-up sections or composite sections. After determining moment of each area about reference axis.5 Centroid of Composite Sections In engineering practice. The given composite section can be split into suitable simple figures and then the centroid of each simple figure can be found by inspection or using the standard formulae listed in the table above. Assuming the area of the simple figure as concentrated at its centroid. SJBIT Page 98 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section. ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. PROBLEMS: Q) Locate the centroid of the T-section shown in fig. ----- ------------------------------------------------------------------------------------------------------------ Dept of Civil Engg.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Q) Find the centroid of the unequal angle 200 ×150 × 12 mm. shown in Fig. SJBIT Page 99 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 Q) Locate the centroid of the I-section shown in Fig. SJBIT Page 100 . ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. 44 333322. Y = 35.2 X = 34. SJBIT Page 101 .66 mm Dept of Civil Engg. wrt O.88 mm.99 136973.98 42. (Jan 2011) Component Area (mm2 ) X (mm) Y (mm) aX aY Quarter circle 7853.7 ∑ aX = ∑ aY = ∑a= 3926.99 50 21.97 16.82 X = 42. Y = 63.08 mm ----------------------------------------------------------------------------------------------------------------- Find the centroid of the shaded area shown in fig.97 -21325.9 333322.82 71674.9 Semi circle -3926. obtained by cutting a semicircle of diameter 100mm from the quadrant of a circle of radius 100mm.4 249992.2 Triangle 900 40 50 36000 45000 Rectangle 2400 30 20 72000 48000 ∑ aX = ∑ aY = ∑a= 2043.64 16.2 -21325. Component Area (mm2 ) X (mm) Y (mm) aX Ay Quarter circle -1256.22 -196350 -83330.Elements of Civil Engineering and Engineering Mechanics 14CIV23 1.36 86674.42 mm.44 42.Determine the centroid of the lamina shown in fig. Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 102 . SJBIT Page 103 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 MOMENT OF INERTIA Dept of Civil Engg. SJBIT Page 104 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg, SJBIT Page 105 Elements of Civil Engineering and Engineering Mechanics 14CIV23 Perpendicular Axis Theorem Theorem of the perpendicular axis states that if IXX and IYY be the moment of inertia of a plane section about two mutually perpendicular axis X-X and Y-Y in the plane of the section, then the moment of inertia of the section I ZZ about the axis Z-Z, perpendicular to the plane and passing through the intersection of X-X and Y-Y is given by: IZZ = IXX + IYY The moment of inertia IZZ is also known as polar moment of inertia. Determination of the moment of inertia of an area by integration Dept of Civil Engg, SJBIT Page 106 Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg, SJBIT Page 107 SJBIT Page 108 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 109 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Properties of plane areas Dept of Civil Engg. SJBIT Page 110 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 111 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 112 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 113 . Elements of Civil Engineering and Engineering Mechanics 14CIV23 Moment of inertia of composite areas Dept of Civil Engg. SJBIT Page 114 . SJBIT Page 115 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 116 . SJBIT Page 117 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 Dept of Civil Engg. SJBIT Page 118 . SJBIT Page 119 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 120 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 121 . SJBIT Page 122 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. SJBIT Page 123 .Elements of Civil Engineering and Engineering Mechanics 14CIV23 ----------------------------------------------------------------------------------------------------------------- Dept of Civil Engg. Elements of Civil Engineering and Engineering Mechanics 14CIV23 MODULE 5 KINEMATICS INTRODUCTION TO DYNAMICS Dynamics is the branch of science which deals with the study of behaviour of body or particle in the state of motion under the action of force system. The first significant contribution to dynamics was made by Galileo in 1564. Later, Newton formulated the fundamental laws of motion. Dynamics branches into two streams called kinematics and kinetics. Kinematics is the study of relationship between displacement, velocity, acceleration and time of the given motion without considering the forces that causes the motion, or Kinematics is the branch of dynamics which deals with the study of properties of motion of the body or particle under the system of forces without considering the effect of forces. Kinetics is the study of the relationships between the forces acting on the body, the mass of the body and the motion of body, or Kinetics is the branch of dynamics which deals with the study of properties of motion of the body or particle in such way that the forces which cause the motion of body are mainly taken into consideration. TECHNICAL TERMS RELATED TO MOTION Motion: A body is said to be in motion if it is changing its position with respect to a reference point. Path: It is the imaginary line connecting the position of a body or particle that has been occupied at different instances over a period of time. This path traced by a body or particle can be a straight line/liner or curvilinear. Displacement and Distance Travelled Displacement is a vector quantity, measure of the interval between two locations or two points, measured along the shortest path connecting them. Displacement can be positive or negative. Distance is a scalar quantity, measure of the interval between two locations measured along the actual path connecting them. Distance is an absolute quantity and always positive. A particle in a rectilinear motion occupies a certain position on the straight line. To define this position P of the particle we have to choose some convenient reference point O called origin (Figure 5.1). The distance x1 of the particle from the origin is called displacement. Figure 5.1 Let, Dept of Civil Engg, SJBIT Page 124 Elements of Civil Engineering and Engineering Mechanics 14CIV23 P —> Position of the particle at any time t1 x1—> Displacement of particle measured in +ve direction of O x2—> Displacement of particle measured in -ve direction of O In this case the total distance travelled by a particle from point O to P to P 1 and back to O is not equal to displacement. Total distance travelled = x1+ xl + x2 + x2 = 2(x1 + x2). Whereas the net displacement is zero. Velocity: Rate of change of displacement with respect to time is called velocity denoted by v. Mathematically v = dx/dt Average velocity: When an object undergoes change in velocities at different instances, the average velocity is given by the sum of the velocities at different instances divided by the number of instances. That is, if an object has different velocities v1, v2, v3, ... , vn, at times t = t1, t2, t3, ..., tn, then the average velocity is given by V = (v1+v2+v3+….vn)/n Instantaneous velocity: It is the velocity of moving particle at a certain instant of time. To calculate the instantaneous velocity Δx is considered as very small. Instantaneous velocity v = Δt0 Δx/Δt , Speed: Rate of change of distance travelled by the particle with respect to time is called speed. Acceleration: Rate of change of velocity with respect to time is called acceleration Mathematically a = dv/dt Average Acceleration Consider a particle P situated at a distances of x from O at any instant of time t having a velocity v. Let Pl be the new position of particle at a distance of (x + Δx) from origin with a velocity of (v + Δv). See Figure 5.2. Figure 5.2 Dept of Civil Engg, SJBIT Page 125 Elements of Civil Engineering and Engineering Mechanics 14CIV23 Average acceleration over a time t, is given by aavg = Δv/Δt Acceleration due to gravity: Each and everybody is attracted towards the centre of the earth by a gravitational force and the acceleration with which the body is pulled towards the centre of the earth due to gravity is denoted by 'g'. The value of g is normally taken as 9.81 m/s 2. Newton's Laws of Motion Newton's first law: This law states that 'everybody continues in its state of rest or of uniform motion, so long as it is under the influence of a balanced force system'. Newton's second law: This law states that 'the rate of change momentum of a body is directly proportional to the impressed force and it takes place in the direction of force acting on it. Newton's third law: This law states that 'action and reaction are equal in magnitude but opposite in direction'. Types of Motion 1. Rectilinear motion 2. Curvilinear motion 3. Projectile motion Graphical representation: The problems in dynamics can be analysed both analytically and graphically without compromising on the accuracy. Most of the times graphical representations can lead to simpler solutions to complicated problems. Using the simple terms defined in the initial portions of the section, we can draw different types of graphs. Displacement-time graph: The representation with graph in Figure 5.3 shows that the displacement is uniform with time. Hence it is understood that the body is under rest as the displacement is constant with respect to time. The representation with graph in Figure 5.4 shows that the plot is having a constant slope and the variation of displacement is uniform with time. The slope indicates the ratio. of displacement to time which is equal to velocity of the body; Hence it is understood that the body is moving under uniform velocity. Figure 5.5 shows variation of displacement with time as a curve. The tangent to this curve at any point indicates the velocity of the body at that instant. As can be seen the slope of the tangent is changing with respect to time and ever increasing, it indicates that the velocity is changing with respect to time and also indicates that the velocity is increasing with respect to time. This increasing velocity with respect to time is termed acceleration. Dept of Civil Engg, SJBIT Page 126 3 Variation of displacement with time.6.5 Variation of displacement with time.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Figure 5. In case of Figure 5.4 Variation of displacement with time Figure 5. the curvature is decreasing. SJBIT Page 127 . Dept of Civil Engg. and the slope of the tangent is decreasing with respect to time and rate change of velocity is decreasing. Figure 5. This is termed as deceleration. .Elements of Civil Engineering and Engineering Mechanics 14CIV23 Figure 5.7 Variation of velocity with time. In case of Figure 5. SJBIT Page 128 . the velocity is varying uniformly with respect to time as seen from sloped straight line. the area under V-T graph will produce the distance traveled by the body/particle from time t1 to t2. Velocity-time graph: A plot of velocity with respect to time is termed as velocity-time graph Figure 5.8. Unit of velocity = v = LT-l Unit of time = T Velocity x Time = LT-l xT=L Distance Hence.6 Variation of displacement with time.) = vt . (i) This is applicable only when the velocity is uniform. s = v* (t2 –t1. Dept of Civil Engg. 5(v2-v1)t But acceleration = a = (v2-v1)/t Substituting.5 x at2 or ut + 0.6).8 Variation of velocity with time. SJBIT Page 129 . In case of Figure 5. the total distance traveled is given by the area under the curve and hence the area is given as S = v1*t + 0. The slope of the line is gives acceleration a = (v2-v1)/(t2-t1) (v2-v1) = a(t2-t1) v2 = v1+a(t2-t1) = u + at (1) where v = final velocity.5at 2 where u is the initial velocity or velocity at time t 1 Acceleration-time graph: It is a plot of acceleration versus time graph as shown in Figure 5. wherein the velocity variation is constant.10. The coordinates in acceleration-time graph show the area under the velocity-time curve. The same can be connected to velocity-time graph (Figure 5.9 It is seen that die acceleration is constant with respect to time t. it shows the variation of acceleration to be uniform. it is seen that the acceleration line in acceleration-time plot. u = initial velocity and t = (t 2-t1) As seen from earlier graph. Dept of Civil Engg. we get S = v1 x t + 0.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Figure 5. The curve in velocity-time graph will be simplified as a straight line in acceleration-time graph.10 Variation of velocity with time. Dept of Civil Engg. we substitute for t from Eq.u2 = 2as …(3) Rectilinear Motion When a particle or a body moves along a straight line path. Using Eqs (1) and (2). then it is called linear motion or rectilinear motion. 2.5[(v-u)/a]2 v2 . SJBIT Page 130 . to get an equation without tim.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Figure 5. we get S = u(v-u)/a + 0. Figure 5.9 Variation of acceleration with time. 1 in Eq. (1). (2). we get a =12 m/s2 Example 2: The motion of a particle is defined by the relation x = t 3.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Equation of motion along a straight line v= u + at v2 .5at Example 1: The motion of a particle is given by the equation x = t 3 – 3t2 -9t + 12.9t2 + 24t . (2) and (3). (2). we get t = -1 or t = 3 s (negative value of t can be discarded) Substitute t = 3 s in (1). Determine (i) Retardation (ii) Distance travelled by the car after applying the brakes. we get v= dx/dt = 3t2-6t-9 (2) when v = 0 The above equation is in the form of ax2+bx+c=0 and the solution is x= -b±√(b2-4ac)/2a (3) substituting the respective values in Eq. Solution X = t3-3t2-9t+l2 (1) Differentiating Eq. distance travelled and acceleration of particle when velocity becomes zero. we get dx/dt=v = 3t2-18t + 2 (2) Differentiating Eq. Solution x=t3-9t2+24t-6 (1) Differentiating Eq. Determine the position. (1). (3).u2 = 2as s = ut + 0. we get d2x/dt2 = a =6t -18 Substitute t = 5 s in Eqs.6. (1) with respect to 'x'. SJBIT Page 131 . we get x = 14 m v = 9 m/s a =12 m/s2 Example 3: A car is moving with a velocity of 15 m/s. Determine the time. we get x = -15 m Differentiating Eq. Dept of Civil Engg. The car is brought to rest by applying brakes in 5 s. velocity and acceleration when t = 5 s. the equations of motion are v = u-gt v2 = u2 . When a body is allowed to fall freely.5gt2 (ii) When a body is projected vertically upward.81 m/s 2 and is always directed towards the centre of earth.5at2 s= 15x5+ 0.2gh h = ut -0. the value of g is considered as positive and if the body is projected vertically upwards. - Solution Consider the first ball. The force of attraction of the earth that pulls all bodies towards the centre of earth with uniform acceleration is known as acceleration due to gravity. The value of acceleration due to gravity is constant in general and its value is considered to be 9.5gt2 30 = 35t – 0. We know that s = ut + 0. Evidently.116 = 0 t = 6. we know that h = u1t-0. SJBIT Page 132 .81*t2 t2 – 7.138 s Consider the second ball Dept of Civil Engg.5*9. under the action of gravity. under the action of gravity.5gt2 Example 4: A ball is thrown vertically upward into air with an initial velocity of 35 m/s. it is acted upon by acceleration due to gravity and its velocity goes on increasing until it reaches the ground. Acceleration due to gravity is generally denoted by 'g'. When the body is moving vertically downwards. After 3 s another ball is thrown vertically.135t + 6. the equations of motion are v = u + gt v2 = u2 + 2gh h = ut + 0. What initial velocity must be the second ball has to pass the first ball at 30 m from the ground.5m MOTION UNDER GRAVITY We know that everybody on the earth experiences a force of attraction towards the centre of the earth is known as gravity.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Solution (i) Retardation We know that v = u + at 0= 15 + ax5 a = -3 m/s2 (ii) Distance travelled by the car after applying the brakes. all equations of motion are applicable except by replacing uniform acceleration V with acceleration due to gravity 'g1 and are written as (i) When a body is projected vertically downward.5x(-3)x(5)2 s = 37. then acceleration due to gravity is considered as negative. Elements of Civil Engineering and Engineering Mechanics 14CIV23 t2 = (6.138 -3) = 3. 138s h = u2t2 – 0.5gt22 h = 30 m u2 = 24.91m/sec CURVILINEAR MOTION Introduction When a moving particle describes a path other than a straight line is said be a particle in curvilinear motion. If the curved path lies in a single plane is called plane curvilinear motion. Most of the motions of particles encountered in engineering practices are of this type. Curvilinear Motion in Cartesian Coordinates In Cartesian coordinates two axes of reference will be chosen. To define the position of particle at any instant of time we have to choose a reference axis namely A and y. Let, P be the position of particle at any instant of time t 'P1' be the new position at an instant of time (t + Δt) from origin. Join O to P and O to P1 Let r be the position vector of P having magnitude and direction. r1 be the position vector P1 Δr be the rate of change in displacement amount over a time Δt Average velocity over a time Δt = Δr/Δt Figure 5.11 Velocity of particle is vector tangent to the path of panicle' Let, Δx be the distance travelled in x direction Δy be the distance travelled in y direction Dept of Civil Engg, SJBIT Page 133 Elements of Civil Engineering and Engineering Mechanics 14CIV23 Velocity in 'x' direction = vx = dx/dt Velocity in y direction = vy = dv/dt Resultant velocity = v = \/(vx2 + vu2) Normal and tangential component of acceleration: Velocity of moving particle is always vector tangential to the path of particle. But acceleration is not tangential to path. But it is convenient to resolve the acceleration along tangential and normal direction. Figure 5.11 Tangential acceleration = at = dv/dt Normal acceleration = an = (v2/ρ)ρ = r Where 'ρ' is the radius of curvature. From the above expression it is evident that tangential component of acceleration is equal the rate of change of velocity with respect to time. Normal component of acceleration is equal to the square of velocity divided by the radius of curvature. Example 6: The motion of a particle is described by the following equation x = 2(t + 1 ) 2 , y = 2(t + 1) -2. Show that path travelled by the particle is rectangular hyperbola. Also find the velocity and acceleration of particle at t = 0 Solution To find the path travelled, we know that x = 2(t+1)2 y = 2(t+1)2 Multiplying the two equation xy = 2 [xy = constant] This represents a rectangular hyperbola We know x = 2(t + 1)2 Component of velocity in x direction vx= 2 x 2(t + 1) Component of acceleration in x direction a x = d2x/db2 = 2x2 = 4 m/s2 When t = 0, vx = 4 m/s2 ax = 4 m/s2 Dept of Civil Engg, SJBIT Page 134 Elements of Civil Engineering and Engineering Mechanics 14CIV23 We know y = 2(t + 1)-2 Component of velocity in y direction vy = dy/dt = 2(-2)(t+1)-3 = -4(t+1)-3 Cornponent of acceleration in y direction ay = d2 y/dt2 = -12(t+1)-4 When t = 0 v y = 4m/s ay = 12m/s velocity = v = √(vx2 + vy2) tan Ѳ = vy/vx = -1 Ѳ = 45o Acceleration = a = √(ax2 + ay2) = 12.65 m/s2 α = 71.6o CURVILINEAR MOTION IN POLAR COORDINATES The curvilinear motion of particle can be expressed in terms of rectangular components and components along the tangent and normal to the path of particle. In certain problems the position of particle is more conveniently described by its polar coordinates. In that case it is much simpler to resolve the velocity and acceleration of particle into components that are parallel and perpendicular to the position vector V of the particle. These components are called radial and transverse components. Figure 5.12 Consider a collar P sliding outward along a straight rod OA, which itself is rotating about fixed point O. It is much convenient to define the position of collar at any instant in terms of Dept of Civil Engg, SJBIT Page 135 Ѳ).. Then r reduces to rotation along circular path.Elements of Civil Engineering and Engineering Mechanics 14CIV23 distance r from the point 'O' and angular position 'Ѳ' of rod OA with x axis. r = constant r=r=0 vѲ = rѲ vr = r = 0 Dept of Civil Engg.vѲѲ ar = vѲ . Transverse components of velocity and acceleration are taken to the positive if pointing towards the increasing value of Ѳ. It can be shown that the radial and transverse components of velocity are v r = r (Radial component directed along position vector r) Transverse component vѲ = rѲ (Transverse component directed along the normal to the position vector r) Total velocity = v = √(vr2 + vѲ2) Radial component of acceleration ar = r. + 2 rѲ Total acceleration = a = √(ar2 +aѲ2) The component of velocity and acceleration are related as ar = vr . SJBIT Page 136 .vrѲ From the above equation it can be seen that 'ar' is not equal to vr and aѲ' is not equal to vѲ It would be noted that radial component of velocity and acceleration are taken to the positive in the same sense of position vector r..13 Transverse component of acceleration aѲ = rѲ.r(Ѳ)2 Figure 5. To understand the physical significance of above results let us assume the following two situations. (i) If r is of constant length and Ѳ varies. . Thus polar coordinates of point P these are (r. dѲ/dt.3t2 dѲ/dt = 2*0. it travels in the air and traces a parabolic path and falls on the ground point (target) other than the point of projection.3t d2Ѳ/dt2 = 0.6 rad d2Ѳ/dt2 = 0. The particle itself is called projectile and the path traced by the projectile is called trajectory.6 rad/s2 r = 2.6 At t = 1s Ѳ = 0. only r varies and Ѳ constant it then resolves a rectilinear motion along a fixed direction Ѳ Ѳ = constant Ѳ=0 vѲ = 0 vr = raѲ = 0 ar = r Example 7: The plane curvilinear motion of particle is defined in polar coordinates by r = t3/3 + 2t and Ѳ = 0. acceleration of path when t = 1 s.Elements of Civil Engineering and Engineering Mechanics 14CIV23 aѲ = rѲ ar = -r(Ѳ)2 (-ve sign indicates that ar is directed opposite to the sense of position vector 'r' or towards 'O') (ii) If.33*0.3t2 Evaluating Ѳ.33m dr/dt = 3m/s d2r/dt2 = 2m/s2 Velocity Vr = dr/dt = 3m/s vѲ = rѲ = 2. d2Ѳ/dt2 Ѳ =0.6 PROJECTILES Whenever a particle is projected upwards with some inclination to the horizontal (but not vertical). SJBIT Page 137 . Dept of Civil Engg.3t2 Find the magnitude of velocity.3 rad dѲ/dt = 0. Solution: Equations of motion are r = t3/3 + 2t & Ѳ = 0. Time of flight (T): It is the total time required for the projectile to travel from the point of projection to the point of target. say at B: y = 0 at t = T Above equation becomes 0 = (u sin α)t.5gt2 When the projectile hits the ground. Horizontal range (R): It is the horizontal distance between the point of projection and target point.5gt T = (2u sin α)/g Horizontal range of the projectile: During the time of flight. 5. SJBIT Page 138 . We know that the vertical ordinate at any point on the path of projectile after a time T is given by y = (u sin α)t – 0. 2. Angle of projection (α): It is the angle with which the projectile is projected with respect to horizontal. Some relations Time of flight: Let T be the time of flight. 4.14 Terms used in projectile 1.5gt2 (u sin α) = 0.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Figure 5. 3.0. the horizontal component of velocity of projectile = u cos α {Horizontal distance of the projectile} = R= {Horizontal component of velocity of projection} {Time of flight} = u cos α x T Dept of Civil Engg. Vertical height (h): It is the vertical distance/height reached by the projectile from the point of projection. Velocity of projection (u): It is the velocity with which projectile is projected in the upward direction with some inclination to the horizontal. 39.Elements of Civil Engineering and Engineering Mechanics 14CIV23 R = (u cos α *2u sin α)/g = (u2 sin (2α))/g sin (2 α) will be maximum only when sin 2 α = 1 sin 2 α = sin 90 or α = 45° Hence maximum horizontal range is given by Rmax = (u2 sin 90)/g = u2/g Maximum height attained by the projectile: When the projectile reaches its maximum height.gt t = u sin α / g Motion of projectile: Let a particle be projected upward from a point O at an angle a with horizontal with an initial velocity of u m/s as shown in Figure 11.15 The vertical component of velocity is always affected by acceleration due to gravity. SJBIT Page 139 . we get ux = u sinα uy = u cosα Figure 5. vertical component of velocity of projection becomes zero. The particle will reach the maximum height when vertical component becomes zero. V2 – u2 = 2gs 0 – u2 sin2 α = -2ghmax hmax = u2 sin2 α/ 2g Time required to reach the maximum height is given by v = u + at 0 = u sin α . Dept of Civil Engg. Now resolving this velocity into two components. The combined effect of horizontal and vertical components of velocity will move the particle along some path in air and then fall on the ground other than the point of projection. The horizontal component of velocity will remains constant since there is no effect of acceleration due to gravity. The particle will move along a certain path OPA in the air and will fall down at A.81 m/s2 and α = 60° To find: u and hmax We know that R = (u2 sin 2α)/g …………(1) Substituting the known values in Eq. x = u cosαt y = u sin αt – 0. R = 5 km = 5000 m. Find (i) Velocity of projection (ii) Maximum height attained by the particle Solution Data given. (2). g = 9. (1) t = x/(u cos α) substitute in Eq.5g[x/ucosα]2 y = x tanα – [gx2/(2u2cos2α)] Figure 5. let a particle reach any point 'P' with x and y as coordinates as shown in Figure 5.5gt2 From Eq. horizontal component of velocity of projection = u cosα Vertical component of velocity of projection = u sin α Therefore. we get y = u sinα[x/(u cosα)] – 0.16 We know that. The horizontal range of particle is 5 km.Elements of Civil Engineering and Engineering Mechanics 14CIV23 Equation for the path of projectile (Trajectory equation): Let a particle is projected at a certain angle from point O.16 Example 8: A particle is projected at an angle of 60° with horizontal. SJBIT Page 140 . we get Dept of Civil Engg. (1). Let u = velocity of projection α = angle of projection After t seconds. 17 (ii) The horizontal distance travelled by the body R = (horizontal component of velocity at B)t R = ut (iii) The vertical component of velocity at point A is obtained from the equation v = u + gt or v = gt Resultant Velocity at A = R = √(u2 + v2) Example 9: An aircraft is moving horizontally at a speed of 108 km/h at an altitude of 1000 m Dept of Civil Engg. SJBIT Page 141 .98 m/s Again. As the body moves in the air towards the ground.9m Motion of a body thrown horizontally from a certain height into air The figure shows a body thrown horizontally from certain height 'H' into air. maximum height attained by the particle hmax = (u2 sin α)/2g = 2164. Resultant velocity = R = √(u2 + v2) & Ѳ = tan-1(v/u) (i) Vertical downward distance travelled by the body is given by H = (vertical component of velocity at B)t + 0. At 'B' there is only horizontal component of velocity. But the vertical component of velocity in the downward direction will be subjected to gravitational force and hence will not be a constant. the body has both horizontal and vertical components of velocity.Elements of Civil Engineering and Engineering Mechanics 14CIV23 u = 237.5gt2 Figure 5. The horizontal component of velocity from B to A remains constant and will be equal to u. Elements of Civil Engineering and Engineering Mechanics 14CIV23 towards a target on the ground releases a bomb which hits the target.278 = 428. SJBIT Page 142 . Solution Speed of aircraft = (108xl00)/(60x30) = 30m/s Horizontal velocity of bomb = u = 200 m/s Height H= 1000m Let t be the time required for the bomb to hit the target We know that H = 0.9 m/s Horizontal component of velocity = u = 30 m/s Resultant velocity = R = √(u2 + v2)= 143.278 = 1 39. We know that R = u x t = 30 x 14.5gt2 1000 . Vertical component of velocity = v = gt = 9.08 m/s Direction = Ѳ= tan-1 (v/u) = 77 Dept of Civil Engg. Calculate also the direction and velocity with which bomb hits the target.57 m (ii) Velocity with which bomb hits the target.81 x t2 or t = 14.278 s (i) Horizontal distance of aircraft from the target when it releases a bomb.5 x 9.8 1 x 14. Estimate the horizontal distance of aircraft from the target when it release a bomb.