Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
1. The document presents information from a slideshow on tacheometric surveying. It discusses various methods of tacheometric surveying including fixed hair, movable hair, tangential, and subtense bar methods.
2. Formulas are provided for calculating horizontal distance, vertical distance, and elevation of points using these different tacheometric surveying methods under various sighting conditions such as inclined or depressed lines of sight.
3. The document also discusses tacheometric constants, anallatic lenses, and procedures for conducting field work in a tacheometric survey including selecting instrument stations, taking observations of vertical angles and staff readings, and shifting to subsequent stations.
The document discusses hydrostatic forces on surfaces submerged in fluids. It defines fluid statics as dealing with fluids at rest, where the only stress is normal stress due to pressure variations from fluid weight. It describes how to calculate the resultant force and center of pressure on plane and curved surfaces for non-uniform pressures. For a plane surface, the resultant force is the pressure at the centroid multiplied by the area. The center of pressure is found by equating moments. Examples are provided to demonstrate these calculations for rectangular plates and cylindrical gates submerged in water.
The document discusses the differences between centroid and center of gravity. The centroid is defined as a point about which the entire line, area or volume is assumed to be concentrated, and is related to the distribution of length, area and volume. The center of gravity is defined as the point about which the entire weight of an object is assumed to be concentrated, also known as the center of mass, and is related to the distribution of mass. Examples are provided to illustrate the concepts of centroid and center of gravity.
Stress distribution in soils can be caused by self-weight of soil layers and surface loads. Stresses increase with depth due to self-weight and decrease radially from applied surface loads. Boussinesq developed equations to determine stresses below concentrated, line, strip and rectangular loads by representing them as point loads and using influence factors. Newmark proposed charts to simplify determining stresses below uniformly loaded areas of different shapes. Approximate methods like the 2:1 method also exist but are less accurate.
This document discusses the different types of fluid flows:
- Steady and unsteady flows, uniform and non-uniform flows, laminar and turbulent flows, compressible and incompressible flows, rotational and irrotational flows, and one, two, and three-dimensional flows. Each type of flow is defined and examples are provided. The key characteristics such as changes in velocity, density, and flow patterns with respect to time and space are outlined for each type of flow. Reynolds number criteria for laminar versus turbulent flow is also mentioned.
This document provides an overview of fluid kinematics, which is the study of fluid motion without considering forces. It discusses key concepts like streamlines, pathlines, and streaklines. It describes Lagrangian and Eulerian methods for describing fluid motion. It also covers various types of fluid flow such as steady/unsteady, laminar/turbulent, compressible/incompressible, and one/two/three-dimensional flow. Important topics like continuity equation, velocity, acceleration, and stream/velocity potential functions are also summarized. The document is intended to outline the syllabus and learning objectives for a course unit on fluid kinematics.
1. The document presents information from a slideshow on tacheometric surveying. It discusses various methods of tacheometric surveying including fixed hair, movable hair, tangential, and subtense bar methods.
2. Formulas are provided for calculating horizontal distance, vertical distance, and elevation of points using these different tacheometric surveying methods under various sighting conditions such as inclined or depressed lines of sight.
3. The document also discusses tacheometric constants, anallatic lenses, and procedures for conducting field work in a tacheometric survey including selecting instrument stations, taking observations of vertical angles and staff readings, and shifting to subsequent stations.
The document discusses hydrostatic forces on surfaces submerged in fluids. It defines fluid statics as dealing with fluids at rest, where the only stress is normal stress due to pressure variations from fluid weight. It describes how to calculate the resultant force and center of pressure on plane and curved surfaces for non-uniform pressures. For a plane surface, the resultant force is the pressure at the centroid multiplied by the area. The center of pressure is found by equating moments. Examples are provided to demonstrate these calculations for rectangular plates and cylindrical gates submerged in water.
The document discusses the differences between centroid and center of gravity. The centroid is defined as a point about which the entire line, area or volume is assumed to be concentrated, and is related to the distribution of length, area and volume. The center of gravity is defined as the point about which the entire weight of an object is assumed to be concentrated, also known as the center of mass, and is related to the distribution of mass. Examples are provided to illustrate the concepts of centroid and center of gravity.
Stress distribution in soils can be caused by self-weight of soil layers and surface loads. Stresses increase with depth due to self-weight and decrease radially from applied surface loads. Boussinesq developed equations to determine stresses below concentrated, line, strip and rectangular loads by representing them as point loads and using influence factors. Newmark proposed charts to simplify determining stresses below uniformly loaded areas of different shapes. Approximate methods like the 2:1 method also exist but are less accurate.
This document discusses the different types of fluid flows:
- Steady and unsteady flows, uniform and non-uniform flows, laminar and turbulent flows, compressible and incompressible flows, rotational and irrotational flows, and one, two, and three-dimensional flows. Each type of flow is defined and examples are provided. The key characteristics such as changes in velocity, density, and flow patterns with respect to time and space are outlined for each type of flow. Reynolds number criteria for laminar versus turbulent flow is also mentioned.
This document provides an overview of fluid kinematics, which is the study of fluid motion without considering forces. It discusses key concepts like streamlines, pathlines, and streaklines. It describes Lagrangian and Eulerian methods for describing fluid motion. It also covers various types of fluid flow such as steady/unsteady, laminar/turbulent, compressible/incompressible, and one/two/three-dimensional flow. Important topics like continuity equation, velocity, acceleration, and stream/velocity potential functions are also summarized. The document is intended to outline the syllabus and learning objectives for a course unit on fluid kinematics.
Fluid MechanicsVortex flow and impulse momentumMohsin Siddique
1. The momentum equation relates the total force on a fluid system to the rate of change of momentum as fluid flows through a control volume.
2. Forces can be resolved into components in different directions for multi-dimensional flows. The total force is equal to the sum of pressure, body, and reaction forces.
3. Examples of applying the momentum equation include calculating forces on a pipe bend, nozzle, jet impact, and curved vane due to changing fluid momentum. Setting up coordinate systems aligned with the flow is important for resolving forces into components.
This document provides an overview of turbulent fluid flow, including:
1) Turbulent flow occurs when the Reynolds number is greater than 2000 and involves irregular, random movement of fluid particles in all directions.
2) The magnitude and intensity of turbulence can be calculated based on the root mean square of turbulent fluctuations and the average flow velocity.
3) The Moody diagram relates the friction factor to the Reynolds number and relative roughness of a pipe to characterize head losses in turbulent pipe flow.
1) Flow through pipes connected in series was analyzed. Head loss is calculated as the sum of losses in each pipe plus local losses at connections. Total head loss (H) equals the height difference between reservoirs.
2) Flow through pipes connected in parallel was also examined. The total flow (Q) equals the sum of individual pipe flows (Q1 + Q2). If pipe characteristics are the same, head loss will be equal in each pipe.
3) An example problem demonstrated calculating flow rate (Q) through two pipes in series where diameter changes over length. Q was found to be 0.158 m3/s. A second example calculated flow rates for two parallel pipes and the diameter required to replace
This document discusses methods for computing the areas of irregularly shaped polygons in surveying. It describes four key methods: the mid-ordinate method, average ordinate method, trapezoidal rule, and Simpson's rule. The mid-ordinate method involves measuring ordinates at mid-points and calculating the average. The average ordinate method uses the average of all ordinates. The trapezoidal rule treats the shapes as trapezoids. Simpson's rule assumes the boundaries form parabolic arcs. An example problem demonstrates using the mid-ordinate method to calculate the area of an irregular shape.
Forces acting on submerged surfaces include hydrostatic forces. Hydrostatic forces form a pressure prism on plane surfaces with a base equal to the surface area and a length equal to the varying pressure. The hydrostatic force passes through the centroid of this pressure prism. For curved surfaces like circles, the hydrostatic force always passes through the center. Hydrostatic forces can be determined on multilayered fluids by considering each fluid-surface interface separately. Examples are given for forces on submerged rectangular and circular plates.
Similitude and Dimensional Analysis -Hydraulics engineering Civil Zone
This document discusses similitude and dimensional analysis for model testing in hydraulic engineering. It introduces key concepts like similitude, prototype, model, geometric similarity, kinematic similarity, dynamic similarity, dimensionless numbers, and model laws. Reynolds model law is described in detail, which states that the Reynolds number must be equal between the model and prototype for problems dominated by viscous forces, such as pipe flow. An example problem demonstrates how to calculate the velocity and flow rate in a hydraulic model based on given prototype parameters and Reynolds model law.
This document discusses the shrinkage limit test for soils. It defines shrinkage limit as the moisture content at which a saturated soil stops decreasing in volume as it dries, even though saturation remains near 100%. The test involves drying a soil sample and measuring its volume and weight changes to determine the moisture content at which further drying does not cause additional volume reduction. This limit provides important information for designing structures in expansive soils and assessing soil suitability for construction materials.
If two or more than two forces are acting on a single point then the forces are known as system of concurrent forces and if they are acting on a single plane then the forces are called as coplanar forces. Copy the link given below and paste it in new browser window to get more information on Lami's Theorem:- http://www.transtutors.com/homework-help/mechanical-engineering/force-systems-and-analysis/lamis-theorem-solved-examples.aspx
This document describes the working principle and experimental setup for calibrating a venturimeter. A venturimeter consists of an inlet section followed by a converging section, cylindrical throat, and gradually diverging cone. It works by creating a pressure difference between the inlet and throat sections due to an increase in flow velocity at the throat. This pressure difference is measured to determine the flow rate. The experiment involves taking pressure and flow rate measurements at the inlet and throat sections using a manometer and collecting water over time. The data is then used to calculate discharge coefficients and Reynolds numbers to calibrate the venturimeter.
Surveying is the technique of determining the relative position of different features on, above or beneath the surface of the earth by means of direct or indirect measurements and finally representing them on a sheet of paper known as plan or map. Please refer this pdf to learn about Circular Curves.
The document provides information about stress distribution in soil due to self-weight and surface loads. It discusses Boussinesq's formula for calculating vertical stress in soil due to a concentrated surface load. The formula shows that vertical stress is directly proportional to the load, inversely proportional to depth squared, and depends on the ratio of radius to depth. A table of coefficient values used in the formula for different ratios of radius to depth is also provided.
This document discusses flow through pipes, including:
- Laminar and turbulent flow characteristics defined by Reynolds number
- Head losses calculated using Darcy-Weisbach and minor loss equations
- Friction factors determined from Moody diagrams for laminar and turbulent flows
- Total head loss in a pipe system equals major losses in pipe sections plus minor losses from fittings
D'Alembert's Principle states that the resultant of all external forces and inertia forces acting on a body is zero for the body to be in dynamic equilibrium. Inertia forces are represented as minus mass times acceleration. The principle allows equations of static equilibrium to be applied to bodies undergoing translational motion by considering an imaginary inertia force equal and opposite to actual inertia. Several example problems are provided applying the principle to analyze motion of connected bodies over pulleys, motion on inclined planes, and motion within elevators.
The document defines incompressible flow and fluid as flow or a fluid where density remains nearly constant. It also discusses the no-slip condition, where fluid sticks to a solid surface due to viscosity. Forced flow is caused by external means like pumps, while natural flow is caused by natural means like buoyancy. A boundary layer develops near a surface due to the no-slip condition, where velocity gradients are significant. Steady flow does not change over time or at boundaries. Systems, surroundings, and boundaries separate a region of study from the outside environment. Closed systems have fixed mass, while open systems study a region of space.
This document contains 6 practice problems related to fluid mechanics:
1) Calculating the specific gravity of a fluid given tank pressures
2) Finding the angle of tilt of a tube open to the atmosphere
3) Calculating pressures at different points in a system of connected containers and cavities
4) Determining the differential height of a mercury column given an air pressure
5) Finding air pressure and equilibrium mercury levels in a tank with multiple fluids
6) Calculating the amplification factor of a manometer setup using oil instead of water.
1) The document discusses the impact of a jet of water on stationary and moving plates. It defines impact of jet as the force exerted by the jet on a plate.
2) Key factors that determine the force include the jet velocity, plate velocity, plate angle, and whether the plate is flat, curved, or includes a series of vanes.
3) Formulas are provided to calculate the force and work done on plates in different configurations based on impulse-momentum principles.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
The document discusses boundary layer separation, which occurs when the boundary layer can no longer stick to the surface of a solid body as the fluid flows over it. This is called boundary layer separation and leads to disadvantages like additional resistance to flow and loss of energy. Methods to control separation include accelerating the fluid in the boundary layer, suctioning fluid from the boundary layer, and moving the solid boundary to match the fluid velocity. Boundary layer separation tends to occur at the inner radius of bends due to pressure gradients.
This document provides information about moment of inertia including:
- Definitions of terms like center of gravity, radius of gyration, section modulus, and moment of inertia.
- Formulas for calculating moment of inertia of basic geometric sections and symmetrical/unsymmetrical sections about various axes.
- Examples of finding the center of gravity and moment of inertia of different cross-sections like rectangles, circles, T-sections, and L-sections.
Here are the key steps to solve this problem:
1) Draw a free body diagram of each block, showing all external forces.
2) Write the equation of motion for each block in the x and y directions: ΣFx = max, ΣFy = may
3) The tension in each cable will be the same. Substitute this into the equations of motion.
4) Solve the equations simultaneously to find the acceleration and tension.
The acceleration and tension can be determined by setting up and solving the simultaneous equations of motion for each block based on Newton's 2nd law. Friction and the coefficient of kinetic friction must be accounted for between block C and the horizontal surface.
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Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visi tus: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
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Fluid MechanicsVortex flow and impulse momentumMohsin Siddique
1. The momentum equation relates the total force on a fluid system to the rate of change of momentum as fluid flows through a control volume.
2. Forces can be resolved into components in different directions for multi-dimensional flows. The total force is equal to the sum of pressure, body, and reaction forces.
3. Examples of applying the momentum equation include calculating forces on a pipe bend, nozzle, jet impact, and curved vane due to changing fluid momentum. Setting up coordinate systems aligned with the flow is important for resolving forces into components.
This document provides an overview of turbulent fluid flow, including:
1) Turbulent flow occurs when the Reynolds number is greater than 2000 and involves irregular, random movement of fluid particles in all directions.
2) The magnitude and intensity of turbulence can be calculated based on the root mean square of turbulent fluctuations and the average flow velocity.
3) The Moody diagram relates the friction factor to the Reynolds number and relative roughness of a pipe to characterize head losses in turbulent pipe flow.
1) Flow through pipes connected in series was analyzed. Head loss is calculated as the sum of losses in each pipe plus local losses at connections. Total head loss (H) equals the height difference between reservoirs.
2) Flow through pipes connected in parallel was also examined. The total flow (Q) equals the sum of individual pipe flows (Q1 + Q2). If pipe characteristics are the same, head loss will be equal in each pipe.
3) An example problem demonstrated calculating flow rate (Q) through two pipes in series where diameter changes over length. Q was found to be 0.158 m3/s. A second example calculated flow rates for two parallel pipes and the diameter required to replace
This document discusses methods for computing the areas of irregularly shaped polygons in surveying. It describes four key methods: the mid-ordinate method, average ordinate method, trapezoidal rule, and Simpson's rule. The mid-ordinate method involves measuring ordinates at mid-points and calculating the average. The average ordinate method uses the average of all ordinates. The trapezoidal rule treats the shapes as trapezoids. Simpson's rule assumes the boundaries form parabolic arcs. An example problem demonstrates using the mid-ordinate method to calculate the area of an irregular shape.
Forces acting on submerged surfaces include hydrostatic forces. Hydrostatic forces form a pressure prism on plane surfaces with a base equal to the surface area and a length equal to the varying pressure. The hydrostatic force passes through the centroid of this pressure prism. For curved surfaces like circles, the hydrostatic force always passes through the center. Hydrostatic forces can be determined on multilayered fluids by considering each fluid-surface interface separately. Examples are given for forces on submerged rectangular and circular plates.
Similitude and Dimensional Analysis -Hydraulics engineering Civil Zone
This document discusses similitude and dimensional analysis for model testing in hydraulic engineering. It introduces key concepts like similitude, prototype, model, geometric similarity, kinematic similarity, dynamic similarity, dimensionless numbers, and model laws. Reynolds model law is described in detail, which states that the Reynolds number must be equal between the model and prototype for problems dominated by viscous forces, such as pipe flow. An example problem demonstrates how to calculate the velocity and flow rate in a hydraulic model based on given prototype parameters and Reynolds model law.
This document discusses the shrinkage limit test for soils. It defines shrinkage limit as the moisture content at which a saturated soil stops decreasing in volume as it dries, even though saturation remains near 100%. The test involves drying a soil sample and measuring its volume and weight changes to determine the moisture content at which further drying does not cause additional volume reduction. This limit provides important information for designing structures in expansive soils and assessing soil suitability for construction materials.
If two or more than two forces are acting on a single point then the forces are known as system of concurrent forces and if they are acting on a single plane then the forces are called as coplanar forces. Copy the link given below and paste it in new browser window to get more information on Lami's Theorem:- http://www.transtutors.com/homework-help/mechanical-engineering/force-systems-and-analysis/lamis-theorem-solved-examples.aspx
This document describes the working principle and experimental setup for calibrating a venturimeter. A venturimeter consists of an inlet section followed by a converging section, cylindrical throat, and gradually diverging cone. It works by creating a pressure difference between the inlet and throat sections due to an increase in flow velocity at the throat. This pressure difference is measured to determine the flow rate. The experiment involves taking pressure and flow rate measurements at the inlet and throat sections using a manometer and collecting water over time. The data is then used to calculate discharge coefficients and Reynolds numbers to calibrate the venturimeter.
Surveying is the technique of determining the relative position of different features on, above or beneath the surface of the earth by means of direct or indirect measurements and finally representing them on a sheet of paper known as plan or map. Please refer this pdf to learn about Circular Curves.
The document provides information about stress distribution in soil due to self-weight and surface loads. It discusses Boussinesq's formula for calculating vertical stress in soil due to a concentrated surface load. The formula shows that vertical stress is directly proportional to the load, inversely proportional to depth squared, and depends on the ratio of radius to depth. A table of coefficient values used in the formula for different ratios of radius to depth is also provided.
This document discusses flow through pipes, including:
- Laminar and turbulent flow characteristics defined by Reynolds number
- Head losses calculated using Darcy-Weisbach and minor loss equations
- Friction factors determined from Moody diagrams for laminar and turbulent flows
- Total head loss in a pipe system equals major losses in pipe sections plus minor losses from fittings
D'Alembert's Principle states that the resultant of all external forces and inertia forces acting on a body is zero for the body to be in dynamic equilibrium. Inertia forces are represented as minus mass times acceleration. The principle allows equations of static equilibrium to be applied to bodies undergoing translational motion by considering an imaginary inertia force equal and opposite to actual inertia. Several example problems are provided applying the principle to analyze motion of connected bodies over pulleys, motion on inclined planes, and motion within elevators.
The document defines incompressible flow and fluid as flow or a fluid where density remains nearly constant. It also discusses the no-slip condition, where fluid sticks to a solid surface due to viscosity. Forced flow is caused by external means like pumps, while natural flow is caused by natural means like buoyancy. A boundary layer develops near a surface due to the no-slip condition, where velocity gradients are significant. Steady flow does not change over time or at boundaries. Systems, surroundings, and boundaries separate a region of study from the outside environment. Closed systems have fixed mass, while open systems study a region of space.
This document contains 6 practice problems related to fluid mechanics:
1) Calculating the specific gravity of a fluid given tank pressures
2) Finding the angle of tilt of a tube open to the atmosphere
3) Calculating pressures at different points in a system of connected containers and cavities
4) Determining the differential height of a mercury column given an air pressure
5) Finding air pressure and equilibrium mercury levels in a tank with multiple fluids
6) Calculating the amplification factor of a manometer setup using oil instead of water.
1) The document discusses the impact of a jet of water on stationary and moving plates. It defines impact of jet as the force exerted by the jet on a plate.
2) Key factors that determine the force include the jet velocity, plate velocity, plate angle, and whether the plate is flat, curved, or includes a series of vanes.
3) Formulas are provided to calculate the force and work done on plates in different configurations based on impulse-momentum principles.
Class notes of Geotechnical Engineering course I used to teach at UET Lahore. Feel free to download the slide show.
Anyone looking to modify these files and use them for their own teaching purposes can contact me directly to get hold of editable version.
The document discusses boundary layer separation, which occurs when the boundary layer can no longer stick to the surface of a solid body as the fluid flows over it. This is called boundary layer separation and leads to disadvantages like additional resistance to flow and loss of energy. Methods to control separation include accelerating the fluid in the boundary layer, suctioning fluid from the boundary layer, and moving the solid boundary to match the fluid velocity. Boundary layer separation tends to occur at the inner radius of bends due to pressure gradients.
This document provides information about moment of inertia including:
- Definitions of terms like center of gravity, radius of gyration, section modulus, and moment of inertia.
- Formulas for calculating moment of inertia of basic geometric sections and symmetrical/unsymmetrical sections about various axes.
- Examples of finding the center of gravity and moment of inertia of different cross-sections like rectangles, circles, T-sections, and L-sections.
Here are the key steps to solve this problem:
1) Draw a free body diagram of each block, showing all external forces.
2) Write the equation of motion for each block in the x and y directions: ΣFx = max, ΣFy = may
3) The tension in each cable will be the same. Substitute this into the equations of motion.
4) Solve the equations simultaneously to find the acceleration and tension.
The acceleration and tension can be determined by setting up and solving the simultaneous equations of motion for each block based on Newton's 2nd law. Friction and the coefficient of kinetic friction must be accounted for between block C and the horizontal surface.
coplanar forces res comp of forces - for mergeEkeeda
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visi tus: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
Coplanar forces res & comp of forces - for mergeEkeeda
Ekeeda Provides Online Video Lectures for Civil Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
Ekeeda Provides Online Civil Engineering Degree Subjects Courses, Video Lectures for All Engineering Universities. Video Tutorials Covers Subjects of Mechanical Engineering Degree.
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
1. The document discusses problems related to resultant of coplanar forces including the law of parallelogram of forces, component law of forces, Varignon's theorem, and finding the resultant and its properties for various force systems.
2. It provides 17 figures showing different force systems and problems require calculating the resultant, its magnitude and direction, and in some cases locating its position.
3. The document also includes theory questions related to definitions of force, laws of mechanics, characteristics of couples, and calculating resultants of concurrent and non-concurrent forces.
This document provides an overview of engineering statics concepts related to force systems. It defines key terms like force, vector, moment, and couple. It also describes methods for analyzing both 2D and 3D force systems, including resolving forces into rectangular components, calculating moments and couples, and determining resultant forces and wrench resultants. The examples show how to use these methods to solve static equilibrium problems involving various force combinations and configurations.
This document provides an overview of the content covered in the Basic Civil Engineering course. It discusses the following topics:
1. Mechanics of Rigid Bodies and Mechanics of Deformable Bodies, which make up Parts I and II of the course.
2. Concepts in mechanics of solids including resultant and equilibrium of coplanar forces, centroids, moments of inertia, kinetics principles, stresses and strains.
3. Five textbooks recommended as references for the course.
4. Definitions of terms like particle, force, scalar, vector, and rigid body.
5. Methods for resolving forces into components, obtaining the resultant of coplanar forces, and solving mechanics problems
This document provides an overview of coplanar non-concurrent force systems and methods for analyzing them. It defines key terms like resultant, equilibrium, and equilibrant. Examples are provided to demonstrate determining resultants and support reactions for coplanar force systems, beams under different loading conditions, and plane trusses. Methods like Lami's theorem, free body diagrams, and the principles of equilibrium are used to solve for unknown forces. Truss analysis is also briefly discussed, noting trusses are articulated structures carrying loads at joints, with members in axial tension or compression.
This document outlines the key concepts and objectives related to equilibrium conditions in statics, including:
- Defining forces and moments, and understanding them as vector quantities
- Resolving forces into components and determining the resultant force of concurrent and non-concurrent systems
- Calculating moments of individual forces and the resultant moment of multiple forces
- Understanding couples as systems with zero net force but nonzero moment
The document provides examples and practice problems for analyzing equilibrium through force components, resultants, moments, and couples.
This document outlines the course content for Mechanics of Solids. It is divided into two parts: Mechanics of Rigid Bodies and Mechanics of Deformable Bodies. The first part covers topics like forces, moments, couples, and equilibrium of force systems for rigid bodies. The second part covers stresses, strains, indeterminate problems, and shear and bending moment diagrams for deformable bodies. The document also lists several recommended reference books and provides an overview of the first lecture which introduces concepts of rigid bodies, forces, and force composition and resolution.
Three key concepts are discussed in the document:
1) Mechanics deals with the static and dynamic behavior of bodies under the influence of forces or torques. This includes rigid bodies, deformable bodies, and fluids.
2) A free body diagram shows all external forces acting on a particle or rigid body and is essential for writing equations of equilibrium.
3) The equilibrium of a particle in 2D involves applying equations that set the sum of forces in the x and y directions equal to zero to solve for unknown forces or angles.
This document discusses mechanics and statics concepts such as forces, moments, and couples. It begins by defining mechanics as the branch of physics dealing with motion and forces. It then discusses rigid bodies, deformable bodies, and fluids. The document reviews the international system of units and conversions between SI and US customary units. It introduces concepts of force systems, the parallelogram law, and the principle of transmissibility. Subsequent sections cover vector addition of forces, moments of forces, moments of couples, and developing equivalent force-couple systems. Examples are provided to demonstrate solving static mechanics problems by resolving forces into components and applying principles of moments.
1. Statics deals with bodies at rest or in uniform motion, focusing on force and equilibrium. It is a branch of mechanics.
2. A particle is considered to have mass but negligible size, allowing simplified analysis of forces acting at a single point. Forces on a particle can be determined and resolved using vector operations like addition, subtraction and resolution into rectangular components.
3. Equilibrium of a particle occurs when the net force is zero, which can be represented graphically by a closed polygon or force diagram. Free body diagrams isolate a body, showing all external forces and their points of application, to analyze force equilibrium.
moments couples and force couple systems by ahmad khanSelf-employed
To determine the resultant force acting at the top of the tower (point D), I would:
1. Resolve each cable force into its x and y components.
2. Use the parallelogram law of forces to combine the x-components of each cable force into a single x-component force. Do the same for the y-components.
3. The x and y component forces obtained from step 2 are the x and y components of the resultant force acting at D.
4. Use the Pythagorean theorem to determine the magnitude of the resultant force from its x and y components.
5. Use trigonometry to determine the direction of the resultant force relative to the x-axis
Mechanics is the physical science concerned with the behavior of bodies acted upon by forces. It has two main branches: statics, which deals with bodies at rest or in equilibrium, and dynamics, which deals with bodies in motion. Dynamics is divided into kinematics, which considers motion without forces, and kinetics, which considers motion with forces. Mechanics studies rigid and deformable bodies, as well as fluids. Forces can be analyzed using concepts such as components, resultants, and force systems.
The document discusses static force analysis in machines. It defines static forces as those transmitted by machine members without accounting for acceleration. Static force analysis determines the forces acting on machine members at a particular configuration, assuming the members are rigid bodies in static equilibrium under external forces. This analysis is a starting point for strength design and avoids over or under estimation of forces. Equilibrium conditions for members with two or three forces are described, including the need for concurrent lines of action and zero resultant force. Examples analyze forces in four-force members and determine unknown force magnitudes through force diagrams.
Definition of Force
Definition of a Vector
Vector Diagram
Graphical Representation of a Vector
Parallelogram of Forces
Resolving and inclined force
Resultant of two forces acting at a point Force
Resultant of a system of forces acting at a point
1) The document discusses force-couple systems and how a single force can be replaced by an equivalent force-couple system. It explains that a force produces both translational and rotational effects, while a couple only produces rotational effects.
2) It provides an example of replacing a force F acting at point A with two equal and opposite forces F and -F acting along parallel lines at point B, which form a couple with moment Fd. This force-couple system is equivalent to the original single force.
3) The concept of a force-couple equivalent enables the transfer of a force to another location outside its line of action, which has many applications in mechanics.
2-vector operation and force analysis.pptRanaUmair74
This document discusses vector operations and force analysis. It begins with an overview of key concepts related to vectors, including defining scalars and vectors, and methods for finding the resultant force of multiple vectors using graphical and analytical approaches. It then covers topics such as resolving forces into rectangular components, adding vectors, and determining the magnitude and direction of resultant forces. Examples are provided to demonstrate how to apply these techniques to solve force analysis problems involving both 2D and 3D systems of forces.
This document provides an overview of statics concepts including:
- Forces on particles in 2D and 3D space including addition and resolution of forces
- Equilibrium of particles and rigid bodies using free body diagrams
- Moments of forces about points and axes
- Force couples and equivalent force systems
- Example problems are provided to demonstrate applying concepts to determine tensions, components of forces, moments, and equivalent single forces.
Mechanical Engineering is the Branch of Engineering.The mechanical engineering field requires an understanding of core areas including mechanics, dynamics, thermodynamics, materials science and structural analysis,Fluid Mechanics, Metrology and Instrumentation, Dynamics of Machinery- II, Manufacturing Processes II, Industrial Drafting and Machine Design, Engineering Graphics, Power Plant Engineering. Ekeeda offers Online Mechanical Engineering Courses for all the Subjects as per the Syllabus. Visit us: https://ekeeda.com/streamdetails/stream/mechanical-engineering
Ekeeda Provides Online Video Lectures for Mechanical Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/mechanical-engineering
Ekeeda Provides Online Video Lectures for Mechanical Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/mechanical-engineering
Civil Engineering is the Branch of Engineering.The Civil engineering field requires an understanding of core areas including Mechanics of Solids, Structural Mechanics - I, Building Construction Materials, Surveying - I, Geology and Geotechnical Engineering, Structural Mechanics, Building Construction, Water Resources and Irrigation, Environmental Engineering, Transportation Engineering, Construction and Project Management. Ekeeda offers Online Mechanical Engineering Courses for all the Subjects as per the Syllabus Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
Civil Engineering is the Branch of Engineering.The Civil engineering field requires an understanding of core areas including Mechanics of Solids, Structural Mechanics - I, Building Construction Materials, Surveying - I, Geology and Geotechnical Engineering, Structural Mechanics, Building Construction, Water Resources and Irrigation, Environmental Engineering, Transportation Engineering, Construction and Project Management. Ekeeda offers Online Mechanical Engineering Courses for all the Subjects as per the Syllabus Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
Ekeeda Provides Online Video Lectures for Civil Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
Ekeeda Provides Online Video Lectures for Civil Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
Kinetics of particles impulse momentum methodEkeeda
Ekeeda Provides Online Video Lectures for Civil Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
Learn Online Courses of Subject Introduction to Civil Engineering and Engineering Mechanics. Clear the Concepts of Introduction to Civil Engineering and Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/introduction-to-civil-engineering-and-engineering-mechanics
Learn Online Courses of Subject Introduction to Civil Engineering and Engineering Mechanics. Clear the Concepts of Introduction to Civil Engineering and Engineering Mechanics Through Video Lectures and PDF Notes. Visit us: https://ekeeda.com/streamdetails/subject/introduction-to-civil-engineering-and-engineering-mechanics
Ekeeda Provides Online Video Lectures for Civil Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
This document discusses concepts related to static equilibrium of rigid bodies, including:
- Conditions for static equilibrium are that the net force and net moment are both zero
- Free body diagrams show all forces acting on a body in isolation
- Types of supports (fixed, hinge, roller) and the reactions they provide are described
- Concepts like two-force and three-force members, Lami's theorem, and finding equilibrant forces to balance unbalanced systems are explained
- Several example problems are provided to illustrate applying concepts to determine reactions and tensions in static systems
Ekeeda Provides Online Video Lectures for Civil Engineering Degree Subject Courses for All Engineering Universities. Visit us: https://ekeeda.com/streamdetails/stream/civil-engineering
Infomatica, as it stands today, is a manifestation of our values, toil, and dedication towards imparting knowledge to the pupils of the society. Visit us: http://www.infomaticaacademy.com/
1) The document discusses Gibbs phase rule, which relates the number of degrees of freedom in a system to the number of components and phases present.
2) It provides examples of one-component systems like water and explains how the phase diagram changes with the number of phases present.
3) Key terms like phase, component, and degree of freedom are defined and illustrated using common chemical systems like water, sulfur, and salt solutions.
Infomatica, as it stands today, is a manifestation of our values, toil, and dedication towards imparting knowledge to the pupils of the society. Visit us/l http://www.infomaticaacademy.com/
Infomatica, as it stands today, is a manifestation of our values, toil, and dedication towards imparting knowledge to the pupils of the society. Visit us: http://www.infomaticaacademy.com/
Infomatica, as it stands today, is a manifestation of our values, toil, and dedication towards imparting knowledge to the pupils of the society. Visit us: http://www.infomaticaacademy.com/
Ekeeda Provides Online Engineering Subjects Video Lectures and Tutorials of Mumbai University (MU) Courses. Visit us: https://ekeeda.com/streamdetails/University/Mumbai-University
Ekeeda Provides Online Engineering Subjects Video Lectures and Tutorials of Mumbai University (MU) Courses. Visit us: https://ekeeda.com/streamdetails/University/Mumbai-University
A review on techniques and modelling methodologies used for checking electrom...nooriasukmaningtyas
The proper function of the integrated circuit (IC) in an inhibiting electromagnetic environment has always been a serious concern throughout the decades of revolution in the world of electronics, from disjunct devices to today’s integrated circuit technology, where billions of transistors are combined on a single chip. The automotive industry and smart vehicles in particular, are confronting design issues such as being prone to electromagnetic interference (EMI). Electronic control devices calculate incorrect outputs because of EMI and sensors give misleading values which can prove fatal in case of automotives. In this paper, the authors have non exhaustively tried to review research work concerned with the investigation of EMI in ICs and prediction of this EMI using various modelling methodologies and measurement setups.
Low power architecture of logic gates using adiabatic techniquesnooriasukmaningtyas
The growing significance of portable systems to limit power consumption in ultra-large-scale-integration chips of very high density, has recently led to rapid and inventive progresses in low-power design. The most effective technique is adiabatic logic circuit design in energy-efficient hardware. This paper presents two adiabatic approaches for the design of low power circuits, modified positive feedback adiabatic logic (modified PFAL) and the other is direct current diode based positive feedback adiabatic logic (DC-DB PFAL). Logic gates are the preliminary components in any digital circuit design. By improving the performance of basic gates, one can improvise the whole system performance. In this paper proposed circuit design of the low power architecture of OR/NOR, AND/NAND, and XOR/XNOR gates are presented using the said approaches and their results are analyzed for powerdissipation, delay, power-delay-product and rise time and compared with the other adiabatic techniques along with the conventional complementary metal oxide semiconductor (CMOS) designs reported in the literature. It has been found that the designs with DC-DB PFAL technique outperform with the percentage improvement of 65% for NOR gate and 7% for NAND gate and 34% for XNOR gate over the modified PFAL techniques at 10 MHz respectively.
Hierarchical Digital Twin of a Naval Power SystemKerry Sado
A hierarchical digital twin of a Naval DC power system has been developed and experimentally verified. Similar to other state-of-the-art digital twins, this technology creates a digital replica of the physical system executed in real-time or faster, which can modify hardware controls. However, its advantage stems from distributing computational efforts by utilizing a hierarchical structure composed of lower-level digital twin blocks and a higher-level system digital twin. Each digital twin block is associated with a physical subsystem of the hardware and communicates with a singular system digital twin, which creates a system-level response. By extracting information from each level of the hierarchy, power system controls of the hardware were reconfigured autonomously. This hierarchical digital twin development offers several advantages over other digital twins, particularly in the field of naval power systems. The hierarchical structure allows for greater computational efficiency and scalability while the ability to autonomously reconfigure hardware controls offers increased flexibility and responsiveness. The hierarchical decomposition and models utilized were well aligned with the physical twin, as indicated by the maximum deviations between the developed digital twin hierarchy and the hardware.
DEEP LEARNING FOR SMART GRID INTRUSION DETECTION: A HYBRID CNN-LSTM-BASED MODELgerogepatton
As digital technology becomes more deeply embedded in power systems, protecting the communication
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represents a multi-tiered application layer protocol extensively utilized in Supervisory Control and Data
Acquisition (SCADA)-based smart grids to facilitate real-time data gathering and control functionalities.
Robust Intrusion Detection Systems (IDS) are necessary for early threat detection and mitigation because
of the interconnection of these networks, which makes them vulnerable to a variety of cyberattacks. To
solve this issue, this paper develops a hybrid Deep Learning (DL) model specifically designed for intrusion
detection in smart grids. The proposed approach is a combination of the Convolutional Neural Network
(CNN) and the Long-Short-Term Memory algorithms (LSTM). We employed a recent intrusion detection
dataset (DNP3), which focuses on unauthorized commands and Denial of Service (DoS) cyberattacks, to
train and test our model. The results of our experiments show that our CNN-LSTM method is much better
at finding smart grid intrusions than other deep learning algorithms used for classification. In addition,
our proposed approach improves accuracy, precision, recall, and F1 score, achieving a high detection
accuracy rate of 99.50%.
6th International Conference on Machine Learning & Applications (CMLA 2024)ClaraZara1
6th International Conference on Machine Learning & Applications (CMLA 2024) will provide an excellent international forum for sharing knowledge and results in theory, methodology and applications of on Machine Learning & Applications.
Introduction- e - waste – definition - sources of e-waste– hazardous substances in e-waste - effects of e-waste on environment and human health- need for e-waste management– e-waste handling rules - waste minimization techniques for managing e-waste – recycling of e-waste - disposal treatment methods of e- waste – mechanism of extraction of precious metal from leaching solution-global Scenario of E-waste – E-waste in India- case studies.
KuberTENes Birthday Bash Guadalajara - K8sGPT first impressionsVictor Morales
K8sGPT is a tool that analyzes and diagnoses Kubernetes clusters. This presentation was used to share the requirements and dependencies to deploy K8sGPT in a local environment.
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P
INTRODUCTION
We have so far discussed and worked with a coplanar all system of forces,
wherein the forces in a system would lie in a single plane i.e. it was a two
dimensional force system. In this chapter we deal with" a system of forces
lying in different planes forming a three dimensional force system. Such a
system is also referred as a space Force System and a vector approach is
required to deal with these problems. The chapter begins with the study of
basic operations with forces using a vector approach. Later on we will learn to
find the resultant of a space force system and finally we shall deal with
equilibrium problems.
BASIC OPERATIONS USING VECTOR APPROACH
Since the forces have three dimensions, we employ a vector approach, which
simplifies the working. Here we will learn to represent a force vectorially, to
find vectorially moment of a force about a point and about a line, to find
magnitude and direction of a force given in vector form and other basic
operations useful in solution of a space force problem.
Force in vector form
Fig. shows a force of magnitude F in space passing through A (x1, y1, z1) and.
B (x2,y2, z2). The force in vector form is
AB
ˆF F.e
2 1 2 1 2 1
2 2 2
2 1 2 1 2 1
F F
x x i y y j z z k
x x y y z z
x y zF F i F j F k
Note: i, j and k printed in bold type denote unit
vectors along the x, y and z axis respectively.
Space Forces
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MAGNITUDE AND DIRECTION OF FORCE
Fig. shows a force x y zF F i+F j+F k making angles
x y and z, with the x, y
and z axis respectively.
Here Fx is the component of force in the x
direction. Similarly Fy and Fz are the force
components in the y and z direction.
The magnitude of the force is
2 2 2
x y zF F F F
Also,
x x
y y
z z
F F cos θ
F Fcosθ
F Fcos θ
Here x y and z are known as the force directions, the value of which lies
between 0 and 180. There is an important identity which relates them.
2 2 2
cos cos cos 1x y z
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MOMENT OF A FORCE ABOUT A POINT
This is a very important operation while dealing with forces. For coplanar
forces, moment about g point was the product of the force and the
distance. Here if the force is in space the moment calculation requires a
vector approach.
Fig. shows a force F in space passing through points A (x1, y1, z1) and B (x2,
y2, z2) on its line of action. Let C (x3, y3, z3) be the moment centre i.e. the point
about which we have to find the moment. The procedure of finding the
moment of the force about the point is as follows.
Step 1: Put the force in vector form i.e.
x y zF F i+F j+F k
Step 2: Find the position vector extending from the moment centre to any
point on the force i.e. x y zr r i+r j+r k .
Step 3: Perform the cross product of the position vector and the force vector
to get the moment vector i.e.
F
point
x y z
x y z
M
i j k
r r r
F F F
r F
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EXERCISE 1
1. A force of 50 N acts parallel to the y axis in the -ve direction. Put the force
in vector form.
2. A 130 KN force acts at B (12, O, O) and passes through C (0, 3, 4). Put the
force in vector form.
3. A force F = (3i -4j 12 + k) N acts at a point A (1,-2, 3) m. Find
a. moment of the force about origin.
b. moment of the force about point B (2, l,2)m.
4. A force F = 80 i + 50 j - 60 k passes through a point A (6, 2, 6). Compute its
moment about a point B (8, 1, 4) (M. U. Dec12)
5. A 700 N force passes through two points A (-5,-1, 4) towards B (1, 2, 6) m.
Find moment of the force about a point C (2,-2, 1)m.
6. A force of magnitude 1200 N passes from A (2, -4, 6) to B (-3, 2, 3).
Calculate the moment of this force about the origin. Also determine the
component of this force along the line AC where C has the position vector
defined by cr i 2j 3kN .
7. A rectangular plate is supported by
brackets to the wall at A and B by wire CD
as shown in figure. Knowing that tension in
wire is 2OO N determine the moment about
point A, of the force exerted by wire on
point C. All dimensions are in mm.
8. Find moment of force P = 500 N about point H.
9. Three forces act on a rectangular box as shown in figure. Determine the
moment of each force about the origin.
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RESULTANT OF CONCURRENT SPACE FORCE SYSTEM
Resultant of a concurrent space force system is a single force R , which acts
through the point of concurrence. Fig. (a) shows a concurrent system at point
P. The resultant of the system is shown 1n Fig. (b) and calculated as,
RESULTANT OF PARALLEL SPACE FORCE SYSTEM
The resultant of a parallel space force system is a single force R which acts
parallel to the force system. The location of the resultant can be found out
using Varignon's theorem.
Figure (a) shows a parallel system of three forces F1, F2 and F3. The resultant
of the system is shown in figure (b) and is calculated as,
1 2 3R F F F
The resultant acts at P. The coordinates (x, y, z) of the point P can be
calculated by using Varignon's theorem, the moments for which can be taken
about any convenient point like point O. The equation of Varignon's theorem
for space forces is
F R
O OM M
0
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RESULTANT OF GENERAL SPACE FORCE SYSTEM
A general space force system is neither a concurrent nor a parallel system.
The resultant of such a system is a single force R and a moment M at any
desired point. Since the resultant contains one force and one moment, it is
also known as a Force Couple System. Fig. (a) shows a general system of four
forces F1, F2, F3 and F4. If it is desired to have the resultant at point O, then
as per figure (b).
The single force 41 2 3R F F F F
and the single moment 31 2 4FF F F
O O O OM M M M M
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EXERCISE 2
1. A force P1 = 10 N in magnitude acts along direction AB whose co-ordinates
of points A and B are (3, 2, -l) and (8, 5, 3) respectively. Another force P2 = 5
N in magnitude acts along BC where C has co-ordinates (-2, 11, -5).
Determine
a. The resultant of Pr and Pz.
b. The moment of the resultant about a point D (1, 1, 1).
2. Knowing that the tension in AC = 20 kN, determine the required values of
TAB and TAD so that the resultant of the three forces applied at A is
vertical. Also find the resultant.
3. A plate foundation is subjected to five vertical forces as shown. Replace
these five forces by means of a single vertical force and find the point of
replacement.
4. Determine the loads to be applied at A and F, if the resultant of all six loads
is to pass through the centre of the
foundation of hexagonal shape of side 3 m.
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5. Knowing that the resultant of the three parallel forces is 500 N acting in
positive y direction and passing through the centre of the rectangular plate,
determine the forces F1 , F2 and F3.
6. A square foundation mat supports four columns as shown in figure.
Determine magnitude and point of application of resultant of four loads.
7. A rectangular parallelepiped carries four
forces as shown in the figure. Reduce the
force system to a resultant force applied at
the origin and moment around the origin.
OA = 5 m, OB = 2m, OC = 4 m. (M. U
Dec 12)
8. Figure shows a rectangular
parallelepiped subjected to four
forces in the direction shown.
Reduce them to a resultant force at
the origin and a moment.
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9. A tetrahedron A B C D is loaded by forces
F1 = 100 N at A along DA, F2 = 2OO N at B
along CB and F3 = 300 N at C along DC as
shown in the figure. Replace the three
force system by a single resultant force R
at B and a single resultant moment vector
M at B. Take the co-ordinates in metre
units.
10. The following forces act on the block shown in
figure. F1 = 40 kN at point C along CD, F2 = 3O
kN at point D along DB and F3 = 1OO kN at point
D along DE. Find the resultant force and
resultant moment of these forces acting at O.
(SPCE Nov 12)
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EQUILIBRIUM OF SPACE FORCES
When the resultant of a system is zero, the system is said to be in
equilibrium. For the resultant to be zero, the resultant force F and the
resultant moment M should be zero.
This gives us six scalar equations of equilibrium viz.
0xF 0xM
0yF 0yM
0zF 0zM
SUPPORTS FOR SPACE STRUCTURE
a) Ball and Socket Support
This support allows rotation in all the three
direction, unlike a hinge which allows rotation
about one axis only. A ball and socket support
does not allow any linear movement in any
direction. This results in a reaction force having
components along x, y and z axis. Fig. shows a
ball and socket support offering three components of reaction force.
b) Fixed End Support
A fixed end will not allow rotation in any
direction, nor will allow any linear movement.
This results in a reaction moment and a
reaction force having components along x, y
and z axis.
Refer Fig.
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Equilibrium of Concurrent Space Force System
For a concurrent system, only the resultant force needs to be zero. This is the
necessary and sufficient condition of equilibrium. This gives us three scalar
equations of equilibrium, viz.
0xF
0yF
0zF
We can also use the other three scalar equations of equilibrium viz.
0xM
0yM
0zM
EQUILIBRIUM OF NON-CONCURRENT SPACE FORCE SYSTEM
For equilibrium of both parallel and general space force systems, all the six
scalar equations of equilibrium are applicable viz.
0xF 0xM
0yF 0yM
0zF 0zM
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EXERCISE 3
1. A boom AB supports a load of 1000 N as shown. Neglect weight of the
boom. Determine tension in each cable and the reaction at A.
2. A vertical load of 1100 N is supported by the three
rods shown in figure. Find the force in each rod.
Points C, O and D are in XZ plane while point B
lies 5 m above this plane.
3. A vertical load of 1OOO N is supported by three
bars as shown. Find the force in each bar. Point
C, O and D are in the x-z plane while B is 1.5 m
above this plane.
4. A vertical tower DC shown is subjected to a horizontal force P = 5O kN at its
top and is anchored by two similar guy wires BC and AC. Calculate
a. Tension in the guy wires.
b. Thrust in the tower pole.
Co-ordinates of the points are as below,
O (0, 0, 0), B (0, 0, -4), D (3, 0, 0), A (0, 0, 4), C (3, 20, 0)
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5. A uniform thin plate ABCD, square in shape is
suspended by strings GA (2a m length), GD (2a m
length), GE 3amlength and GH (a m length) so that it
remains in horizontal plane. If the size of the plate is
2a x 2a and E is the midpoint of BC, find the tension in
each of the string if the plate weighs 400 N.
6. A square steel Plate 2400 mm 2400 mm has a mass of 1800 kg with
mass centre at G. Calculate the tension in each of the three cables with
which the plate is lifted while remaining horizontal. Length DG = 2400 mm.
7. A crate is supported by three cables as shown. Determine the weight of the
crate, if the tension in the cable AB is 750 N.
8. Determine the tension in cable BC and BD and reactions at the ball socket
joint A for the mast as shown in figure.
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9. A vertical mast OD is having base ‘0’ with ball and socket. Three cables DA,
DB and DC keep the mast in equilibrium. If tension in the cable DA is 100
kN, find tensions in the cables DB and DC and force in the mast. Refer
figure.
10. A load of 500 N is held in equilibrium by means of two strings CA and CB
and by a force p as shown in figure. Determine tensions in strings and
magnitude of P.
11. A uniform equilateral triangular plate weighing 2.4
kN is held in horizontal plane by means of three
cables at A, B and C. An additional load of 1.2 kN
acts at midpoint of edge AC of the plate. Determine
the tensions in the three cables.
12. A plate ACED 1O mm thick weighs 76O0 kg/ms. It is held in horizontal
plane by three wires at A, B and C. find tensions in the wires.
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13. Figure shows steel Pipe ABCD of length 1.8 m having 900 bend at B. The
Pipe weighs 30 N/m and supported by three vertical wires attached to the
Pipe at A, C and D. Find tension in wires.
EXERCISE 4
THEORY QUESTIONS
Q.1 How is a space force represented in magnitude and direction.
Q.2 Write the equations of equilibrium to be satisfied for non co-planar
concurrent force system. (SPCE Nov 12)
Q.3 State conditions of equilibrium for non-concurrent forces in space.
UNIVERSITY QUESTIONS
1. A pole is held in place by three cables. If the force of each cable acting on
the pole is as shown in fig. determine the resultant. (6 Marks)
2. A force of 10 kN acts at a point P(2,3,5) m and has its line of action passing
through Q(10, -3, 4)m. Calculate moment of this force about a point
S(1,-10,3) m. (4 Marks)
3. Explain conditions for equilibrium for forces in spare. (6 Marks)
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4. Force F = (3i – 4j + 12k) N acts at point A (1,-2,3) Find -- (4 Marks)
(i) Moment of force about origin
(ii) Moment of force about point B (2,1,2)m.
5. A rectangular parallelepiped carries Three forces shown in fig. Reduce the
force system to a resultant force applied at the origin and a moment around
origin. (6 Marks)
6. The lines of action of three forces concurrent at origin ‘O’ pass respectively
through points A(-1,2,4), B (3,0,-3) and C(2,-2,4)m. The magnitude of
forces are 40N, 10N and 30N respectively. Determine the magnitude and
direction of their resultant. (6 Marks)
7. A force of 1200N acts along PQ, P(4,5,-2) and Q (-3,1,6)m. Calculate its
moment about a point A(3,2,0)m (4 Marks)
8. A force 80 50 60F i j k passes through a point A Compute its moment
about point B (8,1,4). (6 Marks)