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The document discusses equilibrium of concurrent forces in space and provides an example problem. A 1000 lb force acts at point A in the x-direction. Points B, C, and D define members AB, AC, and AD of a framework. The task is to determine the forces in each member, given the positions of points A, B, C, and D. Using equilibrium equations, the forces are calculated to be: 667 lb in AD, 933 lb in AB, and 333 lb in AC.
This document provides examples and problems related to static equilibrium of structures. Example 1 shows applying the equations of equilibrium to a weight suspended by a rope over a pulley. Example 2 calculates the forces in ropes supporting a weighted crate. Problem 3.7 asks the minimum force P needed for equilibrium of a crate supported by three ropes meeting at a point.
If both the ends of a beam are supported by end supports then the beam is known as Simply Supported Beam. One end of the beam is supported by roller support and the other end is supported by a hinged or pinned support. Copy the link given below and paste it in new browser window to get more information on Simply Supported Beam Examples:- http://www.transtutors.com/homework-help/mechanical-engineering/bending-moment-and-shear-force/simply-supported-beam-examples.aspx
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
This document discusses methods for determining areas, volumes, centroids, and moments of inertia of basic geometric shapes. It begins by introducing the method of integration for calculating areas and volumes. Standard formulas are provided for areas of rectangles, triangles, circles, sectors, and parabolic spandrels. Formulas are also provided for volumes of parallelepipeds, cones, spheres, and solids of revolution. The concepts of center of gravity, centroid, and center of mass are defined. Equations are given for calculating the centroids of uniform bodies, plates, wires, and line segments. Methods for finding centroids of straight lines, arcs, semicircles, and quarter circles are illustrated.
The document provides solutions to multiple problems involving calculating support reactions for beams. The problems involve drawing free body diagrams of beams, then applying equations for the sum of forces and sum of moments at the supports to solve for the unknown support reactions. Key steps include considering the absence of any horizontal forces, setting equations for the sum of vertical forces and moments equal to zero, and solving the resulting systems of equations to find the support reactions.
Engmech 06 (equilibrium of non_concurrent force system)physics101
This document discusses the equilibrium of non-concurrent coplanar forces. It provides examples of solving for tensions, reactions, and angles in systems involving rods, cables, and other objects in equilibrium under various loading conditions. Solutions are shown using free body diagrams and summing moments and forces. Key steps include reducing the system to a resultant force and couple, setting the linear and rotational components equal to zero, and solving the resulting equations for the unknown values.
In Engineering Mechanics the static problems are classified as two types: Concurrent and Non-Concurrent force systems. The presentation discloses a methodology to solve the problems of Concurrent and Non-Concurrent force systems.
The document discusses equilibrium of concurrent forces in space and provides an example problem. A 1000 lb force acts at point A in the x-direction. Points B, C, and D define members AB, AC, and AD of a framework. The task is to determine the forces in each member, given the positions of points A, B, C, and D. Using equilibrium equations, the forces are calculated to be: 667 lb in AD, 933 lb in AB, and 333 lb in AC.
This document provides examples and problems related to static equilibrium of structures. Example 1 shows applying the equations of equilibrium to a weight suspended by a rope over a pulley. Example 2 calculates the forces in ropes supporting a weighted crate. Problem 3.7 asks the minimum force P needed for equilibrium of a crate supported by three ropes meeting at a point.
If both the ends of a beam are supported by end supports then the beam is known as Simply Supported Beam. One end of the beam is supported by roller support and the other end is supported by a hinged or pinned support. Copy the link given below and paste it in new browser window to get more information on Simply Supported Beam Examples:- http://www.transtutors.com/homework-help/mechanical-engineering/bending-moment-and-shear-force/simply-supported-beam-examples.aspx
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
This document discusses methods for determining areas, volumes, centroids, and moments of inertia of basic geometric shapes. It begins by introducing the method of integration for calculating areas and volumes. Standard formulas are provided for areas of rectangles, triangles, circles, sectors, and parabolic spandrels. Formulas are also provided for volumes of parallelepipeds, cones, spheres, and solids of revolution. The concepts of center of gravity, centroid, and center of mass are defined. Equations are given for calculating the centroids of uniform bodies, plates, wires, and line segments. Methods for finding centroids of straight lines, arcs, semicircles, and quarter circles are illustrated.
The document provides solutions to multiple problems involving calculating support reactions for beams. The problems involve drawing free body diagrams of beams, then applying equations for the sum of forces and sum of moments at the supports to solve for the unknown support reactions. Key steps include considering the absence of any horizontal forces, setting equations for the sum of vertical forces and moments equal to zero, and solving the resulting systems of equations to find the support reactions.
Engmech 06 (equilibrium of non_concurrent force system)physics101
This document discusses the equilibrium of non-concurrent coplanar forces. It provides examples of solving for tensions, reactions, and angles in systems involving rods, cables, and other objects in equilibrium under various loading conditions. Solutions are shown using free body diagrams and summing moments and forces. Key steps include reducing the system to a resultant force and couple, setting the linear and rotational components equal to zero, and solving the resulting equations for the unknown values.
In Engineering Mechanics the static problems are classified as two types: Concurrent and Non-Concurrent force systems. The presentation discloses a methodology to solve the problems of Concurrent and Non-Concurrent force systems.
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
This document provides definitions and concepts related to engineering mechanics. It defines key terms like equilibrium, center of gravity, centroid, moment of inertia, and polar moment. It also explains concepts such as moment, couple, torque, Varignon's principle, transmissibility, instantaneous center, and Pappus-Guldinus theorems. Examples are provided to illustrate moment, couple, and methods to locate the instantaneous center for objects undergoing both translation and rotation.
The blocks and ladder problems can be summarized as follows:
1) The documents provide diagrams of blocks on inclined planes or ladders against walls, connected by cords or as single structures.
2) Frictional forces are calculated using coefficients of friction for each surface.
3) Force and moment sums are used to relate normal and frictional forces to weights, angles, and applied forces to determine minimum/maximum values for motion to occur.
This document discusses moments of inertia, which are a measure of an object's resistance to rotational acceleration about an axis. It defines the moment of inertia of an area and introduces key concepts like the parallel axis theorem, radius of gyration, and calculating moments of inertia through integration or for composite shapes. Examples are provided to demonstrate calculating moments of inertia for various 2D shapes, including rectangles, triangles, and composite areas, about different axes. The document also covers determining moments of inertia at the centroidal axes versus other axes using the parallel axis theorem.
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) Bending moment is a measure of the bending effect on a beam due to applied forces and is measured in units of Newton-meters or foot-pounds force.
2) The bending moment equation is the algebraic sum of the moments about a section of the beam from all forces acting on one side.
3) Positive bending moments cause tension in the bottom fibers and compression in the top fibers (sagging) while negative moments cause the opposite (hogging).
1. Influence lines represent the variation of reaction, shear, or moment at a specific point on a structural member as a concentrated load moves along the member. They are useful for analyzing the effects of moving loads.
2. To construct an influence line, a unit load is placed at different points along the member and the reaction, shear, or moment is calculated at the point of interest using statics. The values are plotted to show the influence of the load.
3. Influence lines allow engineers to determine the maximum value of a response (reaction, shear, moment) caused by a moving load and locate where on the structure that maximum occurs.
This document discusses buoyancy, floatation, and the equilibrium of submerged and floating bodies. It defines buoyancy as the upward force that opposes gravity when an object is immersed in a fluid. Archimedes' principle states that the buoyant force is equal to the weight of the fluid displaced by the object. The point where the buoyant force is applied is called the center of buoyancy. For a floating body to be in stable equilibrium, the metacenter must be above the center of gravity. The distance between these two points is called the metacentric height.
This PPT covers curvilinear motion of an object in a very systematic and lucid manner. I hope this PPT will be helpful for instructor's as well as students.
The document discusses different failure modes of columns under eccentric loading. It defines equations for determining the load carrying capacity (Pn) and moment capacity (Mn) of columns. Three failure cases are analyzed: 1) pure axial load/crushing failure where only compression steel contributes, 2) balanced failure where steel yields in tension and compression, and 3) pure flexural failure where the column acts like a beam with no axial load. Equations are derived for Pn and Mn for each case.
The document classifies and describes different types of trusses. Simple trusses are made of basic triangular elements connected by additional members. Compound trusses connect two or more simple trusses together using common joints or connecting members. Complex trusses cannot be classified as simple or compound; their equilibrium equations cannot be uncoupled. The document provides an example of analyzing a complex truss using the method of substitute members.
Shear force and bending moment diagram for simply supported beam _1Psushma chinta
This document discusses shear force and bending moment diagrams for beams. It provides examples of calculating and drawing these diagrams for simple beams with various load cases, including concentrated loads, uniformly distributed loads, and combinations of loads. The maximum bending moment is identified as occurring at the point where the shear force is zero.
This document provides an overview of engineering mechanics statics. It covers topics including:
- Defining mechanics as the science dealing with bodies at rest or in motion under forces.
- Dividing mechanics into statics, dynamics, and other subfields. Statics deals with bodies at rest.
- Introducing fundamental concepts of forces, units of measurement, and representing forces as vectors that add according to the parallelogram law.
- Providing examples of adding forces graphically using the parallelogram law and triangle rule to determine the resultant force.
- Discussing problems involving determining the magnitude and direction of resultant forces from multiple forces acting on structures, stakes, and brackets
1. The document discusses shear force and bending moment diagrams. It defines shear force as a force that causes sliding, and bending moment as a force that causes rotation.
2. It provides an example of calculating the shear force and bending moment at a section for a simply supported beam with three point loads. The maximum shear force is 13.2 kN and the maximum bending moment is 39.2 kN-m.
3. The key steps to draw shear force and bending moment diagrams are outlined as calculating reactions, shear forces at sections, bending moments at sections, and then plotting the diagrams.
This document discusses structural analysis of cables and arches. It provides examples of determining tensions in cables subjected to concentrated and uniform loads. It also discusses the analysis procedure for cables under uniform loads. Examples are given for calculating tensions at different points of cables supporting bridges. Methods for analyzing fixed and hinged arches are demonstrated through examples finding internal forces at various arch sections.
Fluid Mechanics Chapter 2 Part II. Fluids in rigid-body motionAddisu Dagne Zegeye
1. The document discusses rigid-body motion of fluids, where fluid particles move together with no internal motion or deformation. It presents equations of motion relating pressure, acceleration, and gravity for fluids undergoing rigid-body translation or rotation.
2. Special cases are considered, including fluids at rest, where pressure only varies with height, and fluids in free fall or accelerated upward, where pressure gradients are altered by acceleration.
3. For fluids accelerating linearly, equations are derived showing pressure varies with both vertical position and horizontal displacement from the acceleration axis, forming parallel inclined surfaces of constant pressure.
Chapter 4-internal loadings developed in structural membersISET NABEUL
This document provides examples and explanations for determining internal forces like shear force and bending moment in structural members like beams and frames. It begins by introducing sign conventions and the procedure for analysis, which involves determining support reactions, drawing free body diagrams, and using equilibrium equations. Numerous step-by-step examples are then provided to demonstrate how to calculate and graph shear force and bending moment diagrams for beams and frames with different loading and support conditions.
This document discusses statics concepts including forces, laws of forces, resolution of forces, moments and couples of forces, and conditions for equilibrium. It provides examples and problems involving determining resultant forces, reactions, and friction forces for systems in equilibrium, including ladders, beams with hanging masses, and objects on inclined planes. Key concepts covered are the parallelogram law, triangle law, Lami's theorem, and conditions that forces and moments must balance for a body to be in equilibrium.
This document discusses basic principles of statics, which is the branch of mechanics dealing with stationary bodies under forces. It defines static equilibrium as when the sum of all forces and moments equals zero. It also discusses concepts such as static determinacy, types of forces including vectors and their addition, resolution of forces, concurrent forces, and free-body diagrams. Examples are provided to demonstrate how to determine reactions and tensions using these principles.
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
This document provides definitions and concepts related to engineering mechanics. It defines key terms like equilibrium, center of gravity, centroid, moment of inertia, and polar moment. It also explains concepts such as moment, couple, torque, Varignon's principle, transmissibility, instantaneous center, and Pappus-Guldinus theorems. Examples are provided to illustrate moment, couple, and methods to locate the instantaneous center for objects undergoing both translation and rotation.
The blocks and ladder problems can be summarized as follows:
1) The documents provide diagrams of blocks on inclined planes or ladders against walls, connected by cords or as single structures.
2) Frictional forces are calculated using coefficients of friction for each surface.
3) Force and moment sums are used to relate normal and frictional forces to weights, angles, and applied forces to determine minimum/maximum values for motion to occur.
This document discusses moments of inertia, which are a measure of an object's resistance to rotational acceleration about an axis. It defines the moment of inertia of an area and introduces key concepts like the parallel axis theorem, radius of gyration, and calculating moments of inertia through integration or for composite shapes. Examples are provided to demonstrate calculating moments of inertia for various 2D shapes, including rectangles, triangles, and composite areas, about different axes. The document also covers determining moments of inertia at the centroidal axes versus other axes using the parallel axis theorem.
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) Bending moment is a measure of the bending effect on a beam due to applied forces and is measured in units of Newton-meters or foot-pounds force.
2) The bending moment equation is the algebraic sum of the moments about a section of the beam from all forces acting on one side.
3) Positive bending moments cause tension in the bottom fibers and compression in the top fibers (sagging) while negative moments cause the opposite (hogging).
1. Influence lines represent the variation of reaction, shear, or moment at a specific point on a structural member as a concentrated load moves along the member. They are useful for analyzing the effects of moving loads.
2. To construct an influence line, a unit load is placed at different points along the member and the reaction, shear, or moment is calculated at the point of interest using statics. The values are plotted to show the influence of the load.
3. Influence lines allow engineers to determine the maximum value of a response (reaction, shear, moment) caused by a moving load and locate where on the structure that maximum occurs.
This document discusses buoyancy, floatation, and the equilibrium of submerged and floating bodies. It defines buoyancy as the upward force that opposes gravity when an object is immersed in a fluid. Archimedes' principle states that the buoyant force is equal to the weight of the fluid displaced by the object. The point where the buoyant force is applied is called the center of buoyancy. For a floating body to be in stable equilibrium, the metacenter must be above the center of gravity. The distance between these two points is called the metacentric height.
This PPT covers curvilinear motion of an object in a very systematic and lucid manner. I hope this PPT will be helpful for instructor's as well as students.
The document discusses different failure modes of columns under eccentric loading. It defines equations for determining the load carrying capacity (Pn) and moment capacity (Mn) of columns. Three failure cases are analyzed: 1) pure axial load/crushing failure where only compression steel contributes, 2) balanced failure where steel yields in tension and compression, and 3) pure flexural failure where the column acts like a beam with no axial load. Equations are derived for Pn and Mn for each case.
The document classifies and describes different types of trusses. Simple trusses are made of basic triangular elements connected by additional members. Compound trusses connect two or more simple trusses together using common joints or connecting members. Complex trusses cannot be classified as simple or compound; their equilibrium equations cannot be uncoupled. The document provides an example of analyzing a complex truss using the method of substitute members.
Shear force and bending moment diagram for simply supported beam _1Psushma chinta
This document discusses shear force and bending moment diagrams for beams. It provides examples of calculating and drawing these diagrams for simple beams with various load cases, including concentrated loads, uniformly distributed loads, and combinations of loads. The maximum bending moment is identified as occurring at the point where the shear force is zero.
This document provides an overview of engineering mechanics statics. It covers topics including:
- Defining mechanics as the science dealing with bodies at rest or in motion under forces.
- Dividing mechanics into statics, dynamics, and other subfields. Statics deals with bodies at rest.
- Introducing fundamental concepts of forces, units of measurement, and representing forces as vectors that add according to the parallelogram law.
- Providing examples of adding forces graphically using the parallelogram law and triangle rule to determine the resultant force.
- Discussing problems involving determining the magnitude and direction of resultant forces from multiple forces acting on structures, stakes, and brackets
1. The document discusses shear force and bending moment diagrams. It defines shear force as a force that causes sliding, and bending moment as a force that causes rotation.
2. It provides an example of calculating the shear force and bending moment at a section for a simply supported beam with three point loads. The maximum shear force is 13.2 kN and the maximum bending moment is 39.2 kN-m.
3. The key steps to draw shear force and bending moment diagrams are outlined as calculating reactions, shear forces at sections, bending moments at sections, and then plotting the diagrams.
This document discusses structural analysis of cables and arches. It provides examples of determining tensions in cables subjected to concentrated and uniform loads. It also discusses the analysis procedure for cables under uniform loads. Examples are given for calculating tensions at different points of cables supporting bridges. Methods for analyzing fixed and hinged arches are demonstrated through examples finding internal forces at various arch sections.
Fluid Mechanics Chapter 2 Part II. Fluids in rigid-body motionAddisu Dagne Zegeye
1. The document discusses rigid-body motion of fluids, where fluid particles move together with no internal motion or deformation. It presents equations of motion relating pressure, acceleration, and gravity for fluids undergoing rigid-body translation or rotation.
2. Special cases are considered, including fluids at rest, where pressure only varies with height, and fluids in free fall or accelerated upward, where pressure gradients are altered by acceleration.
3. For fluids accelerating linearly, equations are derived showing pressure varies with both vertical position and horizontal displacement from the acceleration axis, forming parallel inclined surfaces of constant pressure.
Chapter 4-internal loadings developed in structural membersISET NABEUL
This document provides examples and explanations for determining internal forces like shear force and bending moment in structural members like beams and frames. It begins by introducing sign conventions and the procedure for analysis, which involves determining support reactions, drawing free body diagrams, and using equilibrium equations. Numerous step-by-step examples are then provided to demonstrate how to calculate and graph shear force and bending moment diagrams for beams and frames with different loading and support conditions.
This document discusses statics concepts including forces, laws of forces, resolution of forces, moments and couples of forces, and conditions for equilibrium. It provides examples and problems involving determining resultant forces, reactions, and friction forces for systems in equilibrium, including ladders, beams with hanging masses, and objects on inclined planes. Key concepts covered are the parallelogram law, triangle law, Lami's theorem, and conditions that forces and moments must balance for a body to be in equilibrium.
This document discusses basic principles of statics, which is the branch of mechanics dealing with stationary bodies under forces. It defines static equilibrium as when the sum of all forces and moments equals zero. It also discusses concepts such as static determinacy, types of forces including vectors and their addition, resolution of forces, concurrent forces, and free-body diagrams. Examples are provided to demonstrate how to determine reactions and tensions using these principles.
1. A free body diagram shows the forces and moments acting on an isolated body. A force couple system can replace a single force with an equal force and couple at another point.
2. A couple is a pair of two equal and unlike parallel forces that tends to rotate a body. Varignon's theorem states the sum of the moments of all forces about a point equals the moment of their resultant about that point.
3. Supports include roller, hinged, and fixed. Equations of equilibrium in 2D are the sum of horizontal forces equals zero, the sum of vertical forces equals zero, and the sum of moments equals zero.
This document defines beams and support reactions. It discusses statically determinate beams and explains that support reactions can be determined using equilibrium conditions alone for these beams. The document outlines different types of beam supports including simple, pinned, roller, and fixed supports. It also defines types of beams such as simply supported, cantilever, overhang, and continuous beams. Finally, it discusses determining support reactions for statically determinate beams using equilibrium conditions and introduces the concept of virtual work.
This chapter discusses beams and support reactions. It defines statically determinate beams and describes the following topics: types of beam supports including simple, pin/hinged, roller, and fixed supports; types of beams such as simply supported, cantilever, overhang, and continuous beams; types of loading including concentrated/point loads and distributed loads such as uniform, uniformly varying, and non-uniform loads; and the procedure to find support reactions of statically determinate beams using equilibrium conditions. It also discusses compound beams and the concept of virtual work.
This document discusses concurrent force systems and concepts related to analyzing systems of forces acting at a point, including: defining concurrent forces as those with lines of action meeting at a point; the internal/external effects of forces; equilibrium; types of forces (tensile, compressive, distributed, concentrated); the principle of transmissibility; determining resultants of two or more forces using vector methods like the parallelogram law; and resolving forces into components. Examples are provided for determining tensions in cables/wires supporting objects and forces in systems involving inclined planes. Homework problems are assigned.
This document discusses concurrent force systems and concepts related to analyzing systems of forces acting at a point, including: defining concurrent forces as those whose lines of action meet at a point; describing the internal and external effects of forces; explaining equilibrium, reduction, and resolution of forces; defining characteristics of forces like magnitude and direction; discussing types of forces including concentrated and distributed loads; introducing concepts like resultants, parallelogram law, and resolving forces into components; and providing example problems demonstrating techniques for analyzing concurrent force systems.
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 outlines the course CM 154 Statics of Rigid Bodies taught by Dr. Kofi Agyekum. It will cover topics related to engineering mechanics including types of force systems, resolution of forces, and determining the resultant of concurrent coplanar forces using various methods like the triangle law and parallelogram law. Students will learn to analyze rigid bodies that are either at rest or in static equilibrium by applying the principles of static equilibrium. Recommended textbooks are also listed.
The document provides information about mechanics of solids-I, including:
1) It describes different types of supports like simple supports, roller supports, pin-joint supports, and fixed supports. It also describes different types of loads like concentrated loads, uniformly distributed loads, and uniformly varying loads.
2) It discusses shear force as the unbalanced vertical force on one side of a beam section, and bending moment as the sum of moments about a section.
3) It explains the relationship between loading (w), shear force (F), and bending moment (M) for an element of a beam. The rate of change of shear force is equal to the loading intensity, and the rate of change of bending
A force is a push or pull on an object due to its interaction with another object. There are two categories of forces: contact forces like friction and tension, and non-contact forces like gravity and magnetism. A force is measured in Newtons, with one Newton being the force needed to accelerate a 1 kg mass by 1 m/s2. Mass refers to the amount of matter in an object and remains constant, while weight varies based on location and is the force of gravity on an object. Free body diagrams depict all external forces acting on an object to solve for the system's forces and reactions.
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
engineering mechanics - statics and dynamicsVelmuruganV15
This document provides an overview of the Engineering Mechanics course 19GES28. It covers topics that will be discussed in each unit, including basics and statics of particles, equilibrium of rigid bodies, properties of surfaces and solids, and friction and dynamics of rigid bodies. Key concepts that will be examined include Newton's laws of motion, equilibrium conditions, moments and couples, area and volume calculations, and frictional forces. The course aims to apply principles of mechanics to solve common engineering problems.
Forces acting on the beam with shear force & bending momentTaral Soliya
The document discusses different types of beams and how to analyze the shear forces and bending moments in beams. It defines beams as structural members subjected to lateral loads and describes various types of beams based on their support conditions, including simply supported beams, cantilever beams, and continuous beams. It also covers types of loads beams may experience, such as concentrated loads, distributed loads, and couples. The document then explains how to determine the shear forces and bending moments in beams by using cut sections and equilibrium equations. It provides examples of analyzing shear forces and bending moments in beams with different load conditions.
This document discusses the fundamentals of mechanics of materials including external loads, internal forces, equilibrium, and supports. It defines different types of forces like surface forces from direct contact, and body forces from gravitational effects. Supports prevent movement and develop reactions forces. Equilibrium requires the sum of forces and moments on a body to be zero. Understanding loads, forces, and supports is necessary to analyze how structures deform under stress.
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.
Tension for Concurrent and Coplanar Force System | Mechanical EngineeringTransweb Global Inc
This document provides an overview of concurrent and coplanar force systems. It defines concurrent forces as multiple forces acting at a single point and coplanar forces as multiple forces acting in a single plane. Lami's theorem is introduced which states that for three coplanar forces in equilibrium, each force is proportional to the sine of the angle between the other two forces. Examples are given to demonstrate determining tensions in strings for systems with concurrent and coplanar forces, including strings supporting a lamp, a sphere tied to a wall, and a string passing over pulleys with attached weights.
This document discusses the topic of equilibrium of rigid bodies. It covers:
- Analytical and graphical conditions for equilibrium of co-planar forces.
- Different types of beam supports like simple, pinned, roller, and fixed supports.
- Free body diagrams and their application in analyzing equilibrium and determining reactions.
- Lami's theorem which states that for three forces in equilibrium, each force is proportional to the sine of the angle between the other two forces.
- Examples of problems involving cylinders, pulleys, beams, and friction on inclined planes.
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
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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
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
The document discusses phase rule and its application to a one component water system. It defines phase rule and explains how it can be applied to determine the degrees of freedom in a water system containing solid, liquid, and gas phases of water. The key points are:
1) Phase rule relates the number of phases (P), components (C), and degrees of freedom (F) as F = C - P + 2.
2) In a one component water system, degrees of freedom can be 0, 1, or 2 depending on whether there are 3, 2, or 1 phases present respectively.
3) The phase diagram of a water system shows boundary curves where two phases coexist and areas where a
Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.
it describes the bony anatomy including the femoral head , acetabulum, labrum . also discusses the capsule , ligaments . muscle that act on the hip joint and the range of motion are outlined. factors affecting hip joint stability and weight transmission through the joint are summarized.
Reimagining Your Library Space: How to Increase the Vibes in Your Library No ...Diana Rendina
Librarians are leading the way in creating future-ready citizens – now we need to update our spaces to match. In this session, attendees will get inspiration for transforming their library spaces. You’ll learn how to survey students and patrons, create a focus group, and use design thinking to brainstorm ideas for your space. We’ll discuss budget friendly ways to change your space as well as how to find funding. No matter where you’re at, you’ll find ideas for reimagining your space in this session.
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Main Java[All of the Base Concepts}.docxadhitya5119
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
How to Manage Your Lost Opportunities in Odoo 17 CRMCeline George
Odoo 17 CRM allows us to track why we lose sales opportunities with "Lost Reasons." This helps analyze our sales process and identify areas for improvement. Here's how to configure lost reasons in Odoo 17 CRM
Chapter wise All Notes of First year Basic Civil Engineering.pptxDenish Jangid
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
Walmart Business+ and Spark Good for Nonprofits.pdfTechSoup
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The webinar may also give some examples on how nonprofits can best leverage Walmart Business+.
The event will cover the following::
Walmart Business + (https://business.walmart.com/plus) is a new shopping experience for nonprofits, schools, and local business customers that connects an exclusive online shopping experience to stores. Benefits include free delivery and shipping, a 'Spend Analytics” feature, special discounts, deals and tax-exempt shopping.
Special TechSoup offer for a free 180 days membership, and up to $150 in discounts on eligible orders.
Spark Good (walmart.com/sparkgood) is a charitable platform that enables nonprofits to receive donations directly from customers and associates.
Answers about how you can do more with Walmart!"
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P
INTRODUCTION
We have so far studied the three systems of forces and to find the resultant
of the system. If the resultant of the force system happens to be zero, the
system is said to be in a state of equilibrium. Various practical examples
can illustrate the state of equilibrium of a system of forces like:
1) A lamp hanging from the ceiling is an example of Concurrent Force
System in equilibrium.
2) Structures like buildings; dams etc. are examples of General Force
System in equilibrium.
3) Students sitting on a bench in an example of parallel force system in
equilibrium.
In this chapter we shall study the conditions of equilibrium and applying
these conditions with the help of free body diagrams, we shall learn to
analyse a system in equilibrium.
CONDITIONS OF EQUILIBRIUM (COE)
A body is said to be in equilibrium if it is in a state of rest or uniform
motion. This is precisely what Newton has stated in his first law of motion.
For a body to be in equilibrium the resultant of the system should be zero.
This implies that the sum of all forces should be zero i.e. ∑ F = 0 and the
sum of all moments should also be zero i.e. ∑M = O
The above two equations are the conditions of equilibrium in vector form.
For a coplanar system of forces, the scalar equations of equilibrium are:
∑FX = O ---------- sum of all forces in x direction is zero
∑FY = 0 ---------- sum of all forces in y direction is zero
∑M = O ---------- sum of moments of all forces is zero
Coplanar Forces-Equilibrium
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FREE BODY DIAGRAM (FBD)
A diagram formed by isolating the body from its surroundings and then
showing all the forces acting on it is known as a Free Body Diagram". Such a
diagram is required to be drawn for the body under analysis. For example
consider a ladder AB of weight W resting against the smooth vertical wall and
rough horizontal floor. If the ladder is under analysis then the FBD of the
ladder shall show the weight W acting through its C.G., the normal reactions
N and RB offered by the floor and the wall respectively and the friction force F
at the rough floor.
Consider a lamp of weight W, suspended from two strings AB and AC tied to
the ceiling. The FBD of the lamp will show the weight W of the lamp and the
tensions Tae and Tec in the strings AB and AC respectively.
Importance of FBD
1. It is the first step in analysis of a body is equilibrium before applying the
conditions of equilibrium.
2. It gives a clear picture of the body under analysis, and the effects of all the
active and reactive forces and couples acting on it, can be accounted.
By including the necessary dimensions in FBD, moments of the forces can
be easily taken.
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TYPES OF SUPPORTS AND REACTIONS
Whenever a body is supported, the support offers resistance, known as
reaction. For example you are sitting on a chair while reading this book. Your
weight is being supported by the chair which offers a force of resistance
(reaction) upwards. Likewise let us see the different types of supports and the
reactions they offer.
Hinge Support
A hinge allows free rotation of the body but does not allow the body to have
any linear motion. It therefore offers a force reaction which can be split into
horizontal and vertical components. Figure shows a body having a hinge
support at A.
The horizontal component HA and the vertical component VA of the support
reaction are shown.
When two bodies are connected such that the connection allows rotation
between them and behaves as a hinge then such a connection is referred to as
an internal hinge or pin connection. For example the two members of a
scissor are connected by a pin which allows rotation but allows no linear
movement.
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Roller Support
A roller support is free to ro11 on a surface on which it rests. It offers a
force reaction in a direction normal to the surface on which the roller is
supported. A roller support may be shown in any of the three symbols as
shown in fig (c).
Fixed Support
A fixed support neither allows any linear motion nor allows any rotation. It
therefore offers a force reaction which can be split into a horizontal and a
vertical component and also a moment reaction. Figure shows a beam
having a fixed support at A. In addition to the horizontal component HA
and the vertical component VA of the force reaction, there is a moment
reaction MA. Such beams with one end fixed and other end free are called
as cantilever beams.
Smooth Surface Support
A smooth surface offers a similar reaction as a roller support, i.e. a force
reaction normal to the smooth surface. Fig. 3.6 shows a sphere supported
between two smooth surfaces. Each surface offers one force reaction,
normal to the surface at contact points.
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Rope/ String/Cable Support
It offers a pull force in a direction away from
the body. This force is commonly referred to as
the tension force. Fig. shows a lamp suspended
by two strings, each of them offers a tension
force.
Types of Loads
The following types of loads can act on bodies.
1. Point load
This load is concentrated at a point. Fig. shows
point loads Fr , Fz, Fs acting on the beam.
2. Uniformly distributed load(u.d.l)
In this loading the load of uniform intensity is spread over a length. u.d.l.
can be converted into an equivalent point load by multiplying the load
intensity with the length. This equivalent point load would act at center of
the spread. Fig. Shows a u.d.l of intensity w N/m spread over a length AB
of L miters. The equivalent point load of w x L would therefore act at L/2
from A.
3. Uniformly varying load.(u.v.l)
In this loading, the load of uniformly varying intensity is spread over a
length. u.v.l. can be converted into an equivalent load, which is equal to
the area under the load diagram. The equivalent point load would act at
the centroid of the load diagram. Refer Fig. 3.11 which shows a u.v.l. of
intensit5r varying from zero to w N/m over a spread length of L metres.
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4. Trapezoidal load
In this loading pattern, the load intensity varies uniformly from a lower
intensity of w1 N/m to a higher intensity of w2 N/m over a span of L
metres. This loading is therefore a combination of a u.d.l. of intensity
w1 and a u.v.l of intensity varying from zero to (w2 – w1) N/m.
The u.d.I. portion is replaced by a point load of w1 x L, acting at L/2
from A. The u.v.l. portion is replaced by a point load of acting at L/3
from B. Thus a trapezoidal loading is replaced by two point loads as
shown in fig.
5. Couple load
We have studied about 'Couple' earlier in article 2.L2. A couple load
acting on a body tends to cause rotation of the body. A couple load's
location on the body is of no significance because couples are free
vectors.
Figure shows a three couples loads M1, M2 and M3 acting on the beam.
Couple loads are represented by curved arrows.
6. Varying load
In this loading, the loading intensity varies as some relation. Figure
shows a varying distributed load of parabolic nature. The equivalent
point load is the area under the curve acting at the C.G. of the area.
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EQUILIBRIUM OF A TWO FORCE BODY
The concept of two forced body states “If only two forces act on a member
and the member is in equilibrium then the two forces would be of equal
magnitude, opposite in direction and collinear”. Members in equilibrium
and subjected to two forces are referred to as two force members and their
identification is useful in the solution of equilibrium problems.
Fig. a (a) shows a frame consisting of three members AF, BC and DE.
Member BC is isolated and shown in Fig. (b). Let RB and Rc be the pin
reactions at B and C respectively. Since only two forces act on member
BC, it is a two force member. Therefore RB : RC in magnitude, opposite in
direction and are collinear (i.e. both are directed along line BC).
EQUILIBRIUM OF A THREE FORCE BODY
The concept of equilibrium of three force body states "If three coplanar forces
act on a member and the member is in equilibrium, then the forces would be
either Concurrent or Parallel”. Fig. (a) shows a uniform rod AB of weight W,
one end of which is resting against a smooth vertical wall, while the other end
is supported by a string. The member AEI is acted upon by three forces viz., a
horizontal reaction Ra at A, the self-weight W acts through the C.G. of the
rod, "while the tension T in the string acts at B. These three forces keep the
rod in equilibrium and as per the concept of three force body, should therefore
be concurrent. Refer Fig. (b).
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Fig. (a) shows a beam AB which is hinge supported at A and roller
supported at B. Let a vertical load P be applied on the beam at C. we know
that the reaction RB would be vertical. Since the beam is in equilibrium,
hinge reaction Re would also be a vertical force. This is therefore a case of
three parallel forces in equilibrium. Refer Fig. (b).
LAMI'S THEOREM
Lami's theorem deals with a particular case of equilibrium involving three
forces only. It states "If three concurrent forces act on a body keeping it in
equilibrium, then each force is proportional to sine of the angle between
the other two forces".
Fig. (a) shows a lamp held by two cables. The two tensile forces T1 and
T2 in the string and the weight W of the lamp form a system of three
forces in equilibrium. The forces would form a concurrent system. If is
the angle between T2 and W, is the angle between T1 and W and is
the angle between T1 and T2 then according to Lami‟s theorem,
1 2
sin sin sin
T T W
Or in general for a system of three forces P, Q and R as shown in Fig. (c)we
write Lami's equation as
sin sin sin
P Q R
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Note that while using Lami's theorem, the three concurrent forces should
either act towards the point of concurrence or act away from it. If this is
not the case then using the principle of transmissibility they can be made
in the required form. Fig. (a) shows such a case for a sphere resting
against smooth surfaces. The reactions Ra and Re act I to smooth
surfaces. To apply Lami's equation the forces have been arranged acting
concurrence as shown in Fig. (b). away from the point of concurrence as
shown in fig (b)
Proof: Let P, Q and R be the three concurrent forces in equilibrium as
shown in fig (a)
Since the forces are vectors they are added vectorially by head and tail
connections. We get a closed triangular polygon as shown in Fig. (b).
Applying sine rule we get,
sin(180 ) sin(180 ) sin(180 )
P Q R
sin sin sin
P Q R
…………proved
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EQUILIBRANT FORCE
An unbalanced force system can be brought to equilibrium by adding an
equilibrant force in the system. The equilibrant force has magnitude,
direction and point of application as of the resultant of the system but has
a sense opposite to that of Resultant.
To find the Equilibrant of a force system we first find magnitude, direction
and. Location of the resultant of the force system shall therefore of the
same magnitude, direction and location as of the resultant having an
opposite sense to that of the Resultant.
EXERCISE 1
1. Draw neat FBDs for the following supported bodies in equilibrium. Take all
plane surfaces as smooth.
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2. A beam AB is loaded as shown. Find support reactions.
3. For the beam shown in the figure find the reactions.
4. Find analytically the support reaction at B d load P for the beam shown in
figure if reaction at support A is Zero.
5. Figure shows beam hinged at A and roller supported at B. The L shaped
portion DEF is an extended part beam AB. For the loading shown, find
support reactions.
6. Find support reactions. Portion CD is an extended part of the beam AB
7. Determine the actions developed in cantilever beam as shown in figure.
8. Find the reactions the supports of the am loaded as shown in figure.
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9. A simply supported am AB of span 3m, overhanging on both side is loaded
as down in figure. Find the support reactions.
10.Find the reactions at the supports of
the beam AB loaded as shown.
11.Find the support reactions for the
beam loaded and supported as shown in I figure.
12.Determine support reactions for the beam shown.
13.The figure shows a 10 kg lamp supported by two cables AB and AC. Find
the tension in each cable
14.A sphere of 500 N weight is lodged between two smooth surfaces as shown.
Find reactions at contact points A and B.
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15.A roller of weight W = 1000 N rest on a smooth incline plane. It is kept from
rolling down the plane by a string AC. Find the tension in the string and
reaction at the point of contact D.
16.A small boat is held in static by means of three taut ropes OA, OB and OC
as shown. The water in the river exerts a force on the boat in the direction
of flow
a) If the tension in OA and OB are 1 kN an 0.6 kN respectively determine
the force F, exerted by the flow on the boat and the tension in rope OC.
b) If rope OC breaks, will the boat remain equilibrium? What is the new
tension ropes OA and OB after OC breaks?
17.A straight vertical mast 4 m long is pinned to the ground and stayed by
mean a cable at a distance of 3 m from the bottom as shown. If a horizontal
force of 5 kN acts at the top, determine the tension in the cable and
reaction at the hinge.
18.Figure shows a crank lever ABC with a tension spring (T). The lever weighs
0.2N/mm determine the tension developed in the spring, when a load of
1OON is applied at A.
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19.Determine the tension T the cable and the
reaction at r support A for the beam loaded
as shown. Neglect weight the beam and the
size of the pulley.
20.A uniform wheel of 800 mm diameter weighing 6 kN rests against a rigid
rectangular block of 150 mm height as shown in figure. Find the least pull,
through the center of the wheel, required just to turn the wheel over the
edge A of the block. Also find the 4 reaction on the block. Take all the
surface to be smooth.
21.A uniform circular wire of radius 50 cm and weight of 0.1 N/cm of length is
suspended from a hinge at A. For the given equilibrium position i.e.
diameter AB being vertical, determine the horizontal
force P required to keep the bar in this position.
Hint: The weight of the wire acts through its centre of
gravity G. Refer table in Chapter 6 for location of G.
22.A non-uniform heavy rod AB of length 3 m libs on
horizontal ground. To lift the end B off the ground needs a vertical force of
200 N. To lift A end off the ground needs a force of 160 N. Find the weight
of the rod and the position of centre of mass.
23.A weightless bar is placed in a horizontal position on the smooth inclines as
shown. Find x at which the 200 N force should be placed from point B to
keep the bar horizontal.
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24.The force system shown has neither a resultant force nor a resultant
couple. Find magnitude of forces P, Q and R. Hint: The system is in
equilibrium.
25.Forces act on the plate ABCD as shown in figure. The distance AB is 4 m.
Given that the plate is in equilibrium find. (i) force F (ii) angle , and (iii)
the distance AD
26.Determine the equilibrant of the force system shown in figure.
27.A man raises a 12 kg joist of length 4m by pulling the rope. Find the
tension in the rope and the reaction at A.
28.A concrete dam has rectangular cross section
of height h and width b and is subjected to a
water pressure on one side. Determine the
minimum width 'b' of the dam if the dam is
not to overturn about the point B when h =
4m. Assume density of water = 1OOO kg/m3
an density of concrete : 24OO kg/m,
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EQUILIBRIUM OF CONNECTED BODIES
When two or more rigid bodies are connected to each other, they form a
system of connected bodies. Though the COE can be applied to the entire
system, the individual bodies can also be isolated from the internal
connections and conditions of equilibrium can be applied to them too.
In a system of connected bodies if there are more than three unknowns in
the system we isolate the connected system from the internal connection
and apply COE to isolated bodies as well as COE to the system, thus
getting sufficient equations for the many unknowns in the system. The
internal connection could be a hinge (commonly referred to as pin), roller,
smooth surface or a rope.
At the internal connection say of a hinge, the sense of components of
reaction are assumed on any one of the bodies initially and the opposite
sense is assumed on the other body (since an internal force, when
exposed, occurs in pair, having the same magnitude, collinear and
opposite in sence)
Fig. (a) shows a system of connected bodies. There are two bodies AC and
BC in the system. The external supports are the hinge at A and roller at B.
The internal connection is a hinge
at C. Fig. (b) show the FBD of the
whole system. The two bodies can
be isolated from the internal hinge
C. Fig. 3.20 (c) shows the FBD's of
the isolated bodies. Note that the sense of Hc and Vc on the member AC is
opposite to that on member BC.
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The following examples will help us understand the concept behind a
system of connected bodies, the way we isolate the bodies, show the
internal forces in pair, in opposite sense at the internal connection and
finally apply COE to the system as well as to the isolated bodies, to get the
unknowns.
EXERCISE 2
1. A two span beam ABCD is loaded as shown. Calculate support reaction.
2. A beam ABC, fixed at A and roller supported at C is internally connected by
a pin at B. Determine the support reactions.
3. Find the reactions of the beam shown in figure.
4. The beam is loaded as shown in figure. Determine the reactions at fixed
support roller support B and reaction at internal hinge C.
5. Find the maximum weight W that can be lifted by the crane without tipping
over the wheel B'as shown in figure. Calculate the corresponding reaction
at 'P' and 'Q'. Weight of crane is 1OO KN acting through „C‟
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6. A two bar mechanism is internally pinned at B. It is loaded .1 supported as
shown. Calculate force P required at C to maintain configuration.
7. A beam is supported and loaded as shown. Find reactions at the supports
using equations of equilibrium. C is an internal hinge.
8. Two smooth spheres of weight 100 N and of radius 250 mm each are in
equilibrium in horizontal channel of width 900 mm as shown. Find the
reactions at the surface of contact A, I and D assuming all smooth surfaces.
9. A 60ON cylinder is supported by the frame- I as shown in figure 4. The
frame is hinged D. Determine the reactions developed at .act points A, B, C
and D. Neglect the weight of frame and assume all contact surfaces are
smooth
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10. Sphere A = 1000 N rests on two spheres Band C of weight 900 N each. The
spheres B and C are connected by an inextensible string of length L = 6OO
mm. Assuming smooth contacts and radius of spheres to be 20O mm,
determine the reactions at all contact points 1 to 4 and also the force in the
string.
11. Two spheres A and B of weight 1OOO N and 750 N respectively are kept as
shown in figure. Determine the reactions at all contact points 1 2, 3 and 4.
Radius of A = 4OO mm and radius of B = 300 mm.
12. Determine the reactions at points of contact 1, 2
and 3. Assume smooth surfaces. Take
ma=1kg,m"=4kg.
13. Find the reactions at A, B, C and D. Neglect friction. (notion stands for
diameter)
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14. A cylinder of diameter 1 m weighing 1000 N and another block weighing
500 N supported by a beam of length 7 m and weighing 25.r the help of a
cord as shown the surfaces of contact are frictionless, determine tension in
cord and reaction at point of contacts.
15. Two homogenous solid cylinders of identical weight of l0 N and radius of
0.4 m are resting against inclined wall and, ring ground as shown.
Assuming smooth surfaces find reactions at A, B and C of the contact
points.
16. Two identical rollers each of weight 500 N and radius r are kept on a right
angle frame ABC having negligible weight. Assuming smooth surfaces, find
the reactions induced at all contact surfaces.
17. A 30 kg pipe is supported at „A‟ by a system of five
chords determine the force in each chord for
equilibrium.
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ANALYSIS OF PIN JOINTED FRAMES
Frames are structures which support loads. Frames are a system of
connected bodies where the members are interconnected by hinges (pins).
Loads can be applied anywhere on the member of the frame. Since frames
are connected by pins, we are required to find out the forces acting on the
connecting pins on a loaded frame. These forces are found out by applying
COE to the entire frame and also to the individual members after isolating
the members. On isolating at the connecting hinge (pin) the forces occur in
pair, equal in magnitude, collinear and opposite in sense. The following
examples explain the method of analysis of frames.
EXERCISE 3
1) Calculate support reactions at A and D in the frame. All pin connections
are frictionless.
2) An X frame is loaded and Find the horizontal and components of reaction at
A and C.
3) Determine the reactions at A, B and C for the pin-jointed frame loaded as
shown.
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4) Determine the reactions at supports A, B and C for the frame loaded as
shown.
5) Find the reactions at and E if the frame shown loaded at A by a load of is
the force on pin at C. lkN supports D in figure is 1 kN. What is the force on
pin at C.
6) For the frame and loading shown determine the components of all forces
acting on member ABD.
7) Find the reactions at the supports A and E of the frame shown in figure.
Also the force in pin at B.
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8) Figure shows a frame in which the pulley at D has a mass of 2OO kg.
Neglecting the weights of the bars find out the components of hinge
reactions at A, B and c.
9) Frame shown supports horizontal forces. Determine the forces acting on
pins A,B,C and D
10)Determine support reactions and C for the pin connected frame.
11)Determine reactions at A, B and C 'the frame shown in figure.
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12)For a system of connected bodies find the support reactions.
EXERCISE 4
THEORY QUESTIONS
Q.1. What are the necessary conditions for a body to be in a state of
equilibrium.
Q.2. Define the term Free Body Diagram. What is the significance of drawing a
free body diagram.
Q.3. List and explain different types of supports and their reactions.
Q.4. What is equilibrium of two force body.
Q.5. What is equilibrium of three force body.
Q.6. State and prove Lami's theorem of three forces
Q.7. Define equilibrant.
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UNIVERSITY QUESTIONS
1. Find the reactions at „A‟. (10 Marks)
2. Three cylinders are piled up in a rectangular channel as shown in figure.
Determine 10 the reactions between cylinders A, B and C with channel
surfaces. Assume all smooth surfaces. (10 Marks)
150 , 4
400 6
200 5
A A
B B
c c
w N r cm
w N r cm
w N r cm
3. Two spheres rests on a smooth surfaces as shown in figure. Find the
reactions at points of contact 1,2,3 and 4. (10 Marks)
500 , 200 , 0.25
0.20
A B A
B
w N w N r m
r m
4. State and prove Lami‟s Theorem. (4 Marks)
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5. A member ABC is loaded by distributed load and pure moment as shown in
the fig. Find the i) magnitude and ii) position along AC of the resultant.
(6 Marks)
6. A cylinder weighing 1000 N and 1.5 m diameter is
supported by a beam AB of length 6 m and weight 400
N as shown in fig. Neglecting friction at the surface of
contacts, determine i) Wall reaction at D. ii) Tension in
the cable BC and iii) Hinged reaction at support A.
(8 Marks)
7. Find the support reactions at A and B for the beam shown in fig. (8 Marks)
8. A cylinder ,, 1000B NB W dia. 40 cm, hangs by a cable AB = 40 cm rests
against a smooth wall. Find out reaction at C and TAB.
(4 Marks)
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9. Find out resultant of given (lever) force system w.r.t. „„B‟‟. (4 Marks)
10. Two identical cylinders dia 100 mm weight 200 N are placed as shown. All
contacts are smooth. Find out reactions at A, B and C. (8 Marks)
11. Find the support reactions at Hinge A and Roller B. (8 Marks)
12. A cylinder of weight 500 N is kept on two inclined places as shown in the
fig. Determine the reactions at the contact points A and B (4 Marks)
13. A Cylinder of weight 300 N is help in equilibrium as shown in fig. given
below. Determine the tension in the string AD and reaction at C and B. The
length of AE = 750 mm. (8 Marks)
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14. Find the reactions at supports B and F for the beam loaded as shown in the
fig. below. (8 Marks)
15. Two cylilnders are kept in a channel as shown in figure. Determine the
reactions at all the contact points A,B,C and D. Assume all surfaces
smooth. (8 Marks)
16. Find the support reaction at B and the load P, for the beam shown in figure
if the reaction at support A is zero. (8 Marks)
17. Determine the tensions in cords AB & BC for equilibrium of 30kg block
(8 Marks)
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18. Replace the force system by a single force w.r.to point C. (6 Marks)
19. A bar of 3m. length & negligible weight rests in horizontal position on two
smooth inclined planes Determine the distance x at which the load Q = 150
N should be placed from point 8 to keep the bar horizontal. (8 Marks)
20. Find the support reaction for the beam. (8 Marks)