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 provides information about moments including definitions, formulas, and examples. Moment is defined as the product of a force and the perpendicular distance to the turning point.
2) Examples of moments in everyday life are shown including a beam with forces applied on either side.
3) The concept of resolving forces into perpendicular components using trigonometry and the parallelogram of forces is explained.
4) An example problem calculates the work done to hoist a 700kg skip through the first 40m of a total 120m distance by considering the weight of the skip and steel rope.
1) The document discusses shear force diagrams and bending moment diagrams for loaded beams. It provides equations that govern loaded beams and defines clockwise and anticlockwise moments.
2) An example beam with different loads is used to calculate support reactions, draw the shear force diagram, and bending moment diagram. The shear force diagram shows vertical forces plotted against distance along the beam.
3) The bending moment diagram is created by calculating the area under the shear force diagram section by section, and plots bending moment against distance along the beam. This shows how bending moment changes along the beam.
Shear Force Diagrams
Bending Moment Diagrams
Shear Force Diagrams Calculations
Bending Moment Diagrams Calculations
Moments Equation
Engineering Science
Udl
Uniformly Distributed Load
Point Load
Loaded Beam ( Udl and Point Load Combinations)
Reaction Support
Tables of BMD and SFD
Calculation of BMD (Area under the SFD Curve)
1. The document discusses bending moment diagrams (BMD) which show the variation of bending moment along the length of a beam caused by applied loads.
2. It provides procedures for drawing shear force diagrams (SFD) and BMD for beams under different loading conditions such as concentrated loads, uniform distributed loads, and combinations of loads.
3. Key relationships discussed include: the slope of the shear diagram equaling the distributed load; the slope of the moment diagram equaling the shear; and the change in moment between two points equaling the area under the shear diagram between those points.
it contains the basic information about the shear force diagram which is the part of the Mechanics of solid. there many numerical solved and whivh will give you detaild idea in S.f.d.
Shear force and bending moment diagrams are constructed for a beam subjected to various loads. The shear force is maximum at supports and zero at points of inflection, while the bending moment is maximum at points of inflection and zero at supports. For the given beam under a uniform distributed load and two concentrated loads, the shear force and bending moment diagrams are drawn showing the variations along the length of the beam. Key points of zero shear force and maximum bending moment are identified.
This document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions. It defines key terms like shear force, bending moment, sagging and hogging bending moments. It also describes the relationships between applied loads, shear forces and bending moments. Examples are provided to demonstrate how to draw SFDs and BMDs and calculate reactions, shear forces and bending moments at different sections of beams. Points of contraflexure, where the bending moment changes sign, are also identified.
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.
1) The document provides information about moments including definitions, formulas, and examples. Moment is defined as the product of a force and the perpendicular distance to the turning point.
2) Examples of moments in everyday life are shown including a beam with forces applied on either side.
3) The concept of resolving forces into perpendicular components using trigonometry and the parallelogram of forces is explained.
4) An example problem calculates the work done to hoist a 700kg skip through the first 40m of a total 120m distance by considering the weight of the skip and steel rope.
1) The document discusses shear force diagrams and bending moment diagrams for loaded beams. It provides equations that govern loaded beams and defines clockwise and anticlockwise moments.
2) An example beam with different loads is used to calculate support reactions, draw the shear force diagram, and bending moment diagram. The shear force diagram shows vertical forces plotted against distance along the beam.
3) The bending moment diagram is created by calculating the area under the shear force diagram section by section, and plots bending moment against distance along the beam. This shows how bending moment changes along the beam.
Shear Force Diagrams
Bending Moment Diagrams
Shear Force Diagrams Calculations
Bending Moment Diagrams Calculations
Moments Equation
Engineering Science
Udl
Uniformly Distributed Load
Point Load
Loaded Beam ( Udl and Point Load Combinations)
Reaction Support
Tables of BMD and SFD
Calculation of BMD (Area under the SFD Curve)
1. The document discusses bending moment diagrams (BMD) which show the variation of bending moment along the length of a beam caused by applied loads.
2. It provides procedures for drawing shear force diagrams (SFD) and BMD for beams under different loading conditions such as concentrated loads, uniform distributed loads, and combinations of loads.
3. Key relationships discussed include: the slope of the shear diagram equaling the distributed load; the slope of the moment diagram equaling the shear; and the change in moment between two points equaling the area under the shear diagram between those points.
it contains the basic information about the shear force diagram which is the part of the Mechanics of solid. there many numerical solved and whivh will give you detaild idea in S.f.d.
Shear force and bending moment diagrams are constructed for a beam subjected to various loads. The shear force is maximum at supports and zero at points of inflection, while the bending moment is maximum at points of inflection and zero at supports. For the given beam under a uniform distributed load and two concentrated loads, the shear force and bending moment diagrams are drawn showing the variations along the length of the beam. Key points of zero shear force and maximum bending moment are identified.
This document discusses shear force and bending moment diagrams (SFD & BMD) for beams under different loading conditions. It defines key terms like shear force, bending moment, sagging and hogging bending moments. It also describes the relationships between applied loads, shear forces and bending moments. Examples are provided to demonstrate how to draw SFDs and BMDs and calculate reactions, shear forces and bending moments at different sections of beams. Points of contraflexure, where the bending moment changes sign, are also identified.
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 power point presentation includes concept of beam, types of beam, types of support, concept of shear force and bending moment diagram, concept of determinate and indeterminate beams, rules to draw SFD and BMD and numerical based on above said topic. It also includes concepts of drawing loading diagram and bending moment diagram from shear force diagram and numerical based on this concept.
This document provides an outline for a lecture on fundamental mechanics concepts including forces, force systems, moments, and moment of forces. Key points covered include:
- Definitions of concurrent and non-concurrent force systems, and parallel force systems. Concepts of resultant forces and free body diagrams.
- Definition of moment as the turning effect of a force, calculated as the product of the force and its perpendicular distance from the point of interest. Sign conventions for clockwise and anti-clockwise moments.
- Principle of transmissibility which states a force's effect is the same along its line of action.
- Several example problems demonstrating calculation of moments for single and combined force systems using the moment equation and
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.
Stucture Design-I(Bending moment and Shear force)Simran Vats
* Given: Maximum bending moment the beam can resist = 22 kNm
* Span of beam = 5 + 5 + 4 = 14 m
* Point loads = 1 + 2 + 1 = 4 kN
* To find: Maximum uniform load (w) the beam can carry
* Bending moment at center due to point loads
= (1 × 2.5) + (2 × 2.5) + (1 × 1.5) = 7.5 kNm
* Bending moment at center due to uniform load
= wl^2/8 = w × (14)^2/8 = w × 98 kNm
* Total bending moment should not exceed 22 kNm
7.5 + w
This document summarizes solutions to three theoretical questions:
1) Describes how to use measurements of gravitational redshift to determine the mass and radius of a star.
2) Explains Snell's law and how it can be used to determine the path of light rays through a medium with a linear change in refractive index.
3) Analyzes the motion of a floating cylindrical buoy, determining equations for its vertical and rotational oscillations and relating the periods.
Mechanics of materials lecture 02 (nadim sir)mirmohiuddin1
This document discusses shear force and bending moment in structural members. It defines shear force, normal force, and bending moment as the internal forces that develop in a beam due to applied loads. It presents methods for determining the shear force diagram and bending moment diagram of a beam based on the slope relationships between load, shear force, and bending moment. Several examples are worked through to demonstrate how to calculate and draw the shear force diagram and bending moment diagram for beams under different loading conditions.
This document provides guidance on calculating shear force and bending moment diagrams (SFD and BMD) for beams under different loading conditions. It begins by explaining the process for a sample problem, which involves a beam with uniform and point loads. The key steps are to determine support reactions, divide the beam into sections, then calculate the SFD and BMD for each section. Linear variation indicates a straight line SFD, while parabolic variation means a curved BMD. Interpretations are provided for different loading types and the shapes of the resulting diagrams. References for further reading are listed at the end.
This document discusses bending moment and shear force for beams. It contains 3 main sections:
1) An introduction to bending moment, shear force, and the relationship between loading, shear force and bending moment.
2) How to draw shear force diagrams and bending moment diagrams by calculating shear forces and bending moments at critical points along a beam.
3) How to calculate reactions for simply supported beams and cantilever beams by applying equations of equilibrium. Several examples of calculating reactions are provided.
The document provides step-by-step instructions for drawing the shear force and bending moment diagrams of a cantilever beam subjected to various loads. It first finds the reactions at the support, then draws the shear force diagram showing the shear force values change from -45 kN to -75 kN over the length of the beam. It then draws the bending moment diagram, indicating the bending moment values change from -33.75 kN.m to -168.75 kN.m at the end of the beam due to the applied loads and moments. Key points on both diagrams are indicated.
In this presentation you will get knowledge about shear force and bending moment diagram and this topic very useful for civil as well as mechanical engineering department students.
A graphical method to determine the shear force and bending moment distribution along a simply supported beam is given. A suitable example is used to illustrate the major steps in the process.
A sample calculation for the determination of the maximum stress values is also given.
This document discusses shear and moment diagrams for beams. It provides examples of calculating reactions, shear, and bending moment at different points along simple beams subjected to various load cases including point loads and uniform loads. Key steps include determining reactions, drawing the shear diagram, and using the shear diagram to draw the moment diagram. The maximum shear and bending moment as well as points where shear and moment equal zero are identified. Overhanging ends can cause negative bending moments. Understanding shear and moment diagrams is important for beam design.
The document discusses the concepts of shear and bending moment in beams. It covers reaction forces at supports, calculating internal shear and bending moment by cutting sections of the beam, and how to construct shear and moment diagrams by sketching the beam, supports, and loads and calculating shear and moment values at points along the beam from left to right. The diagrams allow identifying locations of maximum shear and moment.
Three forces of 2P, 3P and 4P act along the three sides of an equilateral triangle with a side length of 100 mm. The resultant force is calculated to be 1.732P with a position of (-1.5P, -0.866P).
The document summarizes the method of components approach for determining the resultant force R for a two-force coplanar system. It provides an example problem and shows the step-by-step working to find: 1) the x and y components of each force, 2) the sum of the x and y components, 3) the magnitude of R, and 4) the direction and sense of R.
SFD & BMD Shear Force & Bending Moment DiagramSanjay Kumawat
The document discusses shear force and bending moment in beams. It defines key terms like beam, transverse load, shear force, bending moment, and types of loads, supports and beams. It explains how to calculate and draw shear force and bending moment diagrams for different types of loads on beams including point loads, uniformly distributed loads, uniformly varying loads, and loads producing couples or overhangs. Sign conventions and the effect of reactions, loads and geometry on the shear force and bending moment diagrams are also covered.
This document discusses shear force diagrams, bending moment diagrams, and working stress in beams. It defines beams as structural members that are long compared to their lateral dimensions and induce bending when subjected to transverse loads. The document explains that shear forces balance external loads on beam cross sections and act parallel to the cross section. Bending moments are the resistance of a beam to bending and result from the summation of shear forces. The document also describes how to draw shear force and bending moment diagrams and defines working stress as the yield point divided by the factor of safety used in design.
Problems on simply supported beams (udl , uvl and couple)sushma chinta
1) A simply supported beam is subjected to a uniformly distributed load (UDL) over part of its span and a couple moment at one end.
2) Shear force and bending moment diagrams are drawn by dividing the beam into sections and analyzing each section.
3) The maximum bending moment occurs where the shear force is zero and is calculated to be 3761.59 N-m located at 2.637 m from the left support.
This document discusses the principle of moments, which states that the moment of a force about a point is equal to the sum of the moments of its components about that point. It provides examples of calculating the moment of a force about different points using both trigonometric methods and resolving the force into Cartesian components. In one example, the moment of a 200N force acting at an angle is calculated to be 14.1 Nm about point A. In another, the moment of a 400N force is found to be -98.6 Nm about point O.
The document discusses shear force and bending moment in beams. It defines key terms like shear force, bending moment, and types of loads, supports and beams. It provides examples of different loading conditions and how to calculate and draw the shear force and bending moment diagrams for beams subjected to point loads, uniformly distributed loads, uniformly varying loads, couples and overhanging beams. The diagrams show the variations in shear force and bending moment, including locations of maximum and points of contraflexure where bending moment changes sign.
1) The document discusses vector addition of forces using the parallelogram law and trigonometry.
2) Forces can be added by constructing a parallelogram with the force vectors as sides and the resultant vector as the diagonal.
3) Trigonometric relationships like the law of sines and cosines allow determining the magnitudes and directions of resultant and component forces.
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.
This power point presentation includes concept of beam, types of beam, types of support, concept of shear force and bending moment diagram, concept of determinate and indeterminate beams, rules to draw SFD and BMD and numerical based on above said topic. It also includes concepts of drawing loading diagram and bending moment diagram from shear force diagram and numerical based on this concept.
This document provides an outline for a lecture on fundamental mechanics concepts including forces, force systems, moments, and moment of forces. Key points covered include:
- Definitions of concurrent and non-concurrent force systems, and parallel force systems. Concepts of resultant forces and free body diagrams.
- Definition of moment as the turning effect of a force, calculated as the product of the force and its perpendicular distance from the point of interest. Sign conventions for clockwise and anti-clockwise moments.
- Principle of transmissibility which states a force's effect is the same along its line of action.
- Several example problems demonstrating calculation of moments for single and combined force systems using the moment equation and
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.
Stucture Design-I(Bending moment and Shear force)Simran Vats
* Given: Maximum bending moment the beam can resist = 22 kNm
* Span of beam = 5 + 5 + 4 = 14 m
* Point loads = 1 + 2 + 1 = 4 kN
* To find: Maximum uniform load (w) the beam can carry
* Bending moment at center due to point loads
= (1 × 2.5) + (2 × 2.5) + (1 × 1.5) = 7.5 kNm
* Bending moment at center due to uniform load
= wl^2/8 = w × (14)^2/8 = w × 98 kNm
* Total bending moment should not exceed 22 kNm
7.5 + w
This document summarizes solutions to three theoretical questions:
1) Describes how to use measurements of gravitational redshift to determine the mass and radius of a star.
2) Explains Snell's law and how it can be used to determine the path of light rays through a medium with a linear change in refractive index.
3) Analyzes the motion of a floating cylindrical buoy, determining equations for its vertical and rotational oscillations and relating the periods.
Mechanics of materials lecture 02 (nadim sir)mirmohiuddin1
This document discusses shear force and bending moment in structural members. It defines shear force, normal force, and bending moment as the internal forces that develop in a beam due to applied loads. It presents methods for determining the shear force diagram and bending moment diagram of a beam based on the slope relationships between load, shear force, and bending moment. Several examples are worked through to demonstrate how to calculate and draw the shear force diagram and bending moment diagram for beams under different loading conditions.
This document provides guidance on calculating shear force and bending moment diagrams (SFD and BMD) for beams under different loading conditions. It begins by explaining the process for a sample problem, which involves a beam with uniform and point loads. The key steps are to determine support reactions, divide the beam into sections, then calculate the SFD and BMD for each section. Linear variation indicates a straight line SFD, while parabolic variation means a curved BMD. Interpretations are provided for different loading types and the shapes of the resulting diagrams. References for further reading are listed at the end.
This document discusses bending moment and shear force for beams. It contains 3 main sections:
1) An introduction to bending moment, shear force, and the relationship between loading, shear force and bending moment.
2) How to draw shear force diagrams and bending moment diagrams by calculating shear forces and bending moments at critical points along a beam.
3) How to calculate reactions for simply supported beams and cantilever beams by applying equations of equilibrium. Several examples of calculating reactions are provided.
The document provides step-by-step instructions for drawing the shear force and bending moment diagrams of a cantilever beam subjected to various loads. It first finds the reactions at the support, then draws the shear force diagram showing the shear force values change from -45 kN to -75 kN over the length of the beam. It then draws the bending moment diagram, indicating the bending moment values change from -33.75 kN.m to -168.75 kN.m at the end of the beam due to the applied loads and moments. Key points on both diagrams are indicated.
In this presentation you will get knowledge about shear force and bending moment diagram and this topic very useful for civil as well as mechanical engineering department students.
A graphical method to determine the shear force and bending moment distribution along a simply supported beam is given. A suitable example is used to illustrate the major steps in the process.
A sample calculation for the determination of the maximum stress values is also given.
This document discusses shear and moment diagrams for beams. It provides examples of calculating reactions, shear, and bending moment at different points along simple beams subjected to various load cases including point loads and uniform loads. Key steps include determining reactions, drawing the shear diagram, and using the shear diagram to draw the moment diagram. The maximum shear and bending moment as well as points where shear and moment equal zero are identified. Overhanging ends can cause negative bending moments. Understanding shear and moment diagrams is important for beam design.
The document discusses the concepts of shear and bending moment in beams. It covers reaction forces at supports, calculating internal shear and bending moment by cutting sections of the beam, and how to construct shear and moment diagrams by sketching the beam, supports, and loads and calculating shear and moment values at points along the beam from left to right. The diagrams allow identifying locations of maximum shear and moment.
Three forces of 2P, 3P and 4P act along the three sides of an equilateral triangle with a side length of 100 mm. The resultant force is calculated to be 1.732P with a position of (-1.5P, -0.866P).
The document summarizes the method of components approach for determining the resultant force R for a two-force coplanar system. It provides an example problem and shows the step-by-step working to find: 1) the x and y components of each force, 2) the sum of the x and y components, 3) the magnitude of R, and 4) the direction and sense of R.
SFD & BMD Shear Force & Bending Moment DiagramSanjay Kumawat
The document discusses shear force and bending moment in beams. It defines key terms like beam, transverse load, shear force, bending moment, and types of loads, supports and beams. It explains how to calculate and draw shear force and bending moment diagrams for different types of loads on beams including point loads, uniformly distributed loads, uniformly varying loads, and loads producing couples or overhangs. Sign conventions and the effect of reactions, loads and geometry on the shear force and bending moment diagrams are also covered.
This document discusses shear force diagrams, bending moment diagrams, and working stress in beams. It defines beams as structural members that are long compared to their lateral dimensions and induce bending when subjected to transverse loads. The document explains that shear forces balance external loads on beam cross sections and act parallel to the cross section. Bending moments are the resistance of a beam to bending and result from the summation of shear forces. The document also describes how to draw shear force and bending moment diagrams and defines working stress as the yield point divided by the factor of safety used in design.
Problems on simply supported beams (udl , uvl and couple)sushma chinta
1) A simply supported beam is subjected to a uniformly distributed load (UDL) over part of its span and a couple moment at one end.
2) Shear force and bending moment diagrams are drawn by dividing the beam into sections and analyzing each section.
3) The maximum bending moment occurs where the shear force is zero and is calculated to be 3761.59 N-m located at 2.637 m from the left support.
This document discusses the principle of moments, which states that the moment of a force about a point is equal to the sum of the moments of its components about that point. It provides examples of calculating the moment of a force about different points using both trigonometric methods and resolving the force into Cartesian components. In one example, the moment of a 200N force acting at an angle is calculated to be 14.1 Nm about point A. In another, the moment of a 400N force is found to be -98.6 Nm about point O.
The document discusses shear force and bending moment in beams. It defines key terms like shear force, bending moment, and types of loads, supports and beams. It provides examples of different loading conditions and how to calculate and draw the shear force and bending moment diagrams for beams subjected to point loads, uniformly distributed loads, uniformly varying loads, couples and overhanging beams. The diagrams show the variations in shear force and bending moment, including locations of maximum and points of contraflexure where bending moment changes sign.
1) The document discusses vector addition of forces using the parallelogram law and trigonometry.
2) Forces can be added by constructing a parallelogram with the force vectors as sides and the resultant vector as the diagonal.
3) Trigonometric relationships like the law of sines and cosines allow determining the magnitudes and directions of resultant and component forces.
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.
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 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. https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
Solutions manual for statics and mechanics of materials 5th edition by hibbel...zaezo
Solutions manual for statics and mechanics of materials 5th edition by hibbeler ibsn 9780134301006
download at: https://goo.gl/rTKRQ1
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1) Cartesian vectors allow representation of forces and displacements in 3D space using their x, y, z components.
2) Vectors can be added by adding their corresponding x, y, z components. This allows determining resultant forces and displacements.
3) The direction of a vector can be specified by its direction cosines or by defining a unit vector in its direction.
4) Position vectors define the location of a point in space relative to a reference origin. Force vectors can be defined along a line by using the position vector between two points and the force magnitude.
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.
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.
The cable produces a total pull of 220 kN at the top of the anchorage. Let's break this problem down step-by-step:
1) Resolve the 220 kN force into horizontal (RAH) and vertical (RAV) components:
RAH = 220 cos 30° = 191 kN
RAV = 220 sin 30° = -196 kN (note the negative sign indicates downward direction)
2) The horizontal and vertical equilibrium equations are satisfied:
ΣH = RAH - 0 = 0
ΣV = RAV - (-196) = 0
Therefore, the support reactions are:
RAH = 191 kN
RAV = -196 kN
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.
This document discusses representing and calculating vectors in 3 dimensions using Cartesian coordinates. It begins by reviewing 2D vector concepts like resolving vectors into components and adding vectors. It then introduces 3D vector notation and terminology for defining a vector's magnitude, direction angles, and direction cosines in a Cartesian coordinate system. Examples are provided for writing vectors in Cartesian form, adding vectors, and determining the magnitude and direction of a resultant vector. The document also discusses representing position vectors and force vectors directed along a line in 3D space using Cartesian coordinates.
This document discusses the composition of forces and moments. It defines key terms like resultant force and moment of a force. It describes the parallelogram, triangle and polygon laws for combining concurrent coplanar forces into a single resultant force. It also explains Varignon's principle of moments, which states that the algebraic sum of the moments of individual forces equals the moment of the resultant force about the same point. Several example problems are provided to illustrate how to use these principles to find the magnitude and direction of resultant forces and moments in systems of coplanar 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.
Theoretical Mechanics | Forces and Other Vector Quantities |MOVIESandsongsAndMOR
Theoretical Mechanics | Forces and Other Vector Quantities |
CHARACTERISTICS OF FORCES
Lecture Outline
1-NEWTON S THIRD LAW
2-FORCE RESULTANTS AND FORCE COMPONENTS
3-RECTANGULAR COMPONENTS OF A VECTOR
4-POLYGON LAW OF FORCES
5-NUMERICAL PROBLEMS
This document discusses the analysis of shear stresses in beams. It begins by explaining how bending causes sliding between layers in a built-up beam, making it weaker than a solid beam. It then derives the formula for calculating the horizontal shear stress in a beam based on the vertical shear force and moment of inertia. The derivation shows that horizontal and vertical shear stresses are equal. It applies the formula to a rectangular beam section, showing the shear stress is highest at the neutral axis and distributed parabolically. The assumptions and limitations of treating shear stress as uniform across a section are also addressed.
The document provides an overview of a lesson on 2-D vector addition. It includes objectives to resolve vectors into components and add vectors using Cartesian notation. Example problems are shown to resolve vectors into x and y components, add the components, and calculate the magnitude and angle of the resulting vector. Key concepts covered are scalars versus vectors, vector operations including addition and subtraction, and using the parallelogram law or Cartesian notation to perform vector math.
Solution of Chapter- 04 - shear & moment in beams - Strength of Materials by ...Ashiqur Rahman Ziad
This document discusses shear and moment in beams. It defines statically determinate and indeterminate beams. It describes types of loading that can be applied to beams including concentrated loads, uniform loads, and varying loads. It discusses how to calculate and draw shear and moment diagrams for beams with different loading conditions. It explains the relationship between load, shear, and moment and how the slope of the shear and moment diagrams relates to one another. It also addresses moving loads and how to calculate the maximum shear and moment for beams with moving single or multiple loads.
This document discusses equilibrium of particles and free body diagrams. It provides an example of drawing a free body diagram for a cylinder suspended by two cables, and using the equations of equilibrium to solve for the unknown tensions in the cables. It also discusses 3D equilibrium, giving an example problem of finding an unknown force on a particle given its position and four other forces.
6161103 2.6 addition and subtraction of cartesian vectorsetcenterrbru
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This document provides information about isometric and orthographic projections used in engineering drawings. It includes examples of isometric views of objects and their corresponding orthographic projections. Key points covered include:
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Engineering Maths N4
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1) The document discusses shear force diagrams (SFD) and bending moment diagrams (BMD) for a loaded beam.
2) Key equations for a loaded beam include the sum of vertical forces equals zero, and the sum of clockwise and anticlockwise moments equals zero.
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Level 3 NCEA - NZ: A Nation In the Making 1872 - 1900 SML.pptHenry Hollis
The History of NZ 1870-1900.
Making of a Nation.
From the NZ Wars to Liberals,
Richard Seddon, George Grey,
Social Laboratory, New Zealand,
Confiscations, Kotahitanga, Kingitanga, Parliament, Suffrage, Repudiation, Economic Change, Agriculture, Gold Mining, Timber, Flax, Sheep, Dairying,
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...PECB
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The chapter Lifelines of National Economy in Class 10 Geography focuses on the various modes of transportation and communication that play a vital role in the economic development of a country. These lifelines are crucial for the movement of goods, services, and people, thereby connecting different regions and promoting economic activities.
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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
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Iván Bornacelly, Policy Analyst at the OECD Centre for Skills, OECD, presents at the webinar 'Tackling job market gaps with a skills-first approach' on 12 June 2024
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In this webinar, participants learned how to utilize Generative AI to streamline operations and elevate member engagement. Amazon Web Service experts provided a customer specific use cases and dived into low/no-code tools that are quick and easy to deploy through Amazon Web Service (AWS.)
1. Resultant of a System of forces
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2. Resultant of a System of forces
1) Definition of Force
2) Definition of a Vector
3) Vector Diagram
4) Graphical Representation of a Vector
5) Parallelogram of Forces
6) Resolving and inclined force
7) Resultant of two forces acting at a point
Force
8) Resultant of a system of forces acting at a
point
3. 1.Definition of a Force.
A force, F Newtons, can be defined in terms of
magnitude and direction as a pull or a push such
that it will displace an object from its resting point
from point A to another point B as illustrated in
the diagram.
Yverticaldistance
Point A Point B
(x,y = 0,0) (x, y = 5 meters, 0 meters) x horizontal distance
Fig. 1
F
4. This pull or push, F Newtons, can be also be
applied to a simple lever in order to do positive
work as shown.
In this case, the Force, F, applied will cause a
turning moment of magnitude M = F Newtons X
0.65 meters at the turning point, T.
Turning Point, T
Fig. 2
F
5. Generally, this formula is written thus:
Moment, M = Fd
where F = Force applied to the beam or lever in
newtons
d= perpendicular distance between the force
applied and the turning point as illustrated above.
2. Definition of a Vector
The forces as illustrated in Figure 1 and Figure 2
both have magnitude in Newtons and have
direction. They are thus referred to as vectors.
The direction of the force is usually measured in
terms of the angle in degrees that the force makes
with the positive x axis.
6. 3. Vector Diagram
Assuming that the force, F, applied to displace the block
in figure 3 is inclined at an angle 45 degrees to the
horizontal, this force would have to be resolved into
vertical and horizontal forces.
Figure 6 shows a vector diagram in which the
diagonal,DB, of the diagram ABCD is the given force, F,
which displaces the block from point A to point B.
Vector diagrams are drawn in a tail- head, tail – head
sequence.
In Fig 6, the vector FV is drawn vertically from the tail of
the given force F. Secondly, a horizontal line is drawn to
represent FH from the head of the inclined force F.
The intersection of FV and FH will give the point A.
7. 4. Graphical Representation of a Vector
Given an force of 600N, inclined to the negative x
axis at an angle 450 , FV and FH can be drawn
graphically by applying a scale of , say, 50N =
10mm.
The following diagrams illustrates the graphical
solution of resolving the inclined force of 600N.
Fig 1 a Step 1 Fig 1b Step 2
D
B x
600 600
D
B xC
y y
8. Fig 1 c Step 3 Fig 1 d Step 4
600
600
4.1 Graphical Solution of a vector.
Procedure (Fig 1a – d)
Step 1 Mark of the point B by drawing two lines such
that they intersect to mark point B. These two lines
are vertical line (y axis) and horizontal line (x axis)as
shown above.
Step 2 Using the negative x axis as reference,
measure an angle 45 degrees from this horizontal
line.
D
BC
D
BC
9. Step 3 Based on the scale of 50N = 10mm, mark
off point D by measuring 120mm along the 45
degree line to represent the magnitude of this
force. (600N /50N) X 10mm = 120mm
Step 4 Draw a vertical line downward from the
point D.
Step 5 Extend the horizontal negative x axis so
that it intersects the vertical line from point D.
mark this point A.
Step 6. Measure FH and FV in millimetres.
Step 7 Convert the distance of FH and FV in
millitres to force in Newtons by applying the
formula:
(distance in mm)/10mm X 50N
10. 5. Parallelogram of Forces
Parallelogram of Forces may also be called
Parallelogram of Vectors.
A parallelogram is a plane shape bounded by
four sides. The two pairs of opposite sides of the
rectangle in Fig 6, DA and CB, are parallel and
have the same magnitude and direction. This is
also true for the pair of opposite sides as shown,
namely, DC and AB.
Rectangles and squares are special types of
parallelograms because they have four equal
angles which are 90 degrees.
Most parallelograms look like the plane shape in
fig. 9
11. 6. Resolving an inclined force.
(analytically)
As stated in section 3 the inclined force, F, applied to
displace the block in figure 3 through a distance of 5
metres is inclined at an angle 45 degrees to the
horizontal. This force, F, has a vertical component, FV,
and horizontal component, FH.
The Forces, FV and FH may be calculated or resolved
by applying trigonometrical functions as shown on
page 14.
It will be very important to note here that the
resolved forces are of the same magnitude and
direction as FH and FV as obtained in the graphical
solution of section 6.
12. Yverticaldistanc
Point A Point B
(x,y = 0,0) (x, y = 5 meters, 0 meters) x horizontal distance
Fig. 3
F
450
Turning Point, T x
Fig. 4
F y
300 1500
14. z
Fig 3 2 units Fig 4 2(2)1/2
units
600 600
300300
450 450
Trigonometric
Ratio
Angle 60 Angle 45 Angle 30
Sine Ө opp/hyp 30.5/2 20.5/2 1/2
Cos Ө Adj/hyp 1/2 20.5/2 30.5/2
Tan Ө Opp/adj 30.5 1 30.5/3
Opp. Opposite to the
angle
Adj. Adjacent to the angle
Hyp. The inclined side of a
right angle triangle
Table 1
31/2units
15. FV = the subscript V denotes vertical component of
the inclined force.
FH= the subscript H denotes horizontal The Fig 3 , Fig
4 and table 1 above illustrates clearly the fact that FV
and FH can be calculated by applying the sine, cosine
or tan of the angle the inclined force makes with the
positive or negative x axis.
We shall now proceed to calculate the values of FV
and FH as shown on Fig 6 based on the table on the
previous page.
Example 1. Calculate FV and FH for the inclined force F
in Fig. 3 given that F is 600N.
16. X
300
600
300
300
Z
WY
Fig. 8 Fig 9 Parallelogram
F
600
Solution: Based on the table 1
Sin 450 = FV/F Equation 1
Sin 450 = 0.70711 Equation 2
Therefore, FV/F = 0.70711
FV = 0.70711 X F
FV = 0.70711 X 600N = 424.2641N
Direction is in the negative y axis.
17. Similarly,
Sin 450 = FH/F Equation 1
Sin 450 = 0.70711 Equation 2
Therefore, FH/F = 0.70711
FH = 0.70711 X F
FH = 0.70711 X 600N = 424.2641N
Direction is in the positive x axis.
In order to verify our answer, we may apply
Pythagoras Theorem to the rectangle or system of
forces in Fig 6 as follows.
DB = √DA2 + AB2
DB = √FV
2 + FH
2
DB = √(424.2641N)2 + (424.2641N)2 =600N
18. Example 2. Calculate the FH and FV for the Fig 5
and Fig 7. The force, F, as applied on the lever is
750N.
Solution: Considering the vector diagram of fig.
7 and applying the trigonometric ratios:
cos 600 = FH /F Equation 1
cos 600 = 0.5 Equation 2
FH /F = 0.5
FH = 0.5F
FH = 0.5 X 750N = 375N
Likewise, FV = 750N X sin 60 = 649.5191N
Confirming our answer: (3752 + 649.51912)1/2 =
750N
19. 7. The Resultant of two forces acting at a point.
Given two forces FV, 250N, acting vertically
upwards and FH, 620N, acting in the positive x
direction as shown, the resultant of the two
forces is the diagonal of the corresponding
parallelogram of forces.
The resultant R = √F2
H + F2
V
y
x
FV = 250N
Fig 10 FH = 620N
20. y
x
FV = 250N
Fig 11 FH = 620N
The magnitude and direction of the resultant is given by the
Pythagoras theorem formula and the arctan formula as
shown above will give the direction of the resultant.
The resultant of two forces acting at a point is that single
force that will represent the two forces in magnitude and
direction.
There is, thus, a relationship between the resolution of an
inclined force and the resultant of a force.
tan -1 (FV/FH)
21. The Example 3. By applying the Pythagoras
Theorem and the arctan formula, calculate
the resultant of the force in Fig 11 in terms
of magnitude and direction.
Solution:
The resultant R = √F2
H + F2
V
And Ө = tan-1 (FV/FH)
Substituting the values of FV = 250N and FH
= 620N into the equations above, we have:
R = √ 2502 + 6202 =
and
Ө = tan-1 (250N/620N)= 21.9610
22. 8. The Resultant of a system of forces
acting at a point.
y
x
FA = 250N
Fig 12 Calculating the resultant of a system of forces
FE = 250N
FB = 250N
FC = 250N
FD = 250N
23. Each of the forces, FA, FB, FC, FD and FE has to be
resolved into the vertical and horizontal
components as illustrated in the next page.
All vertical forces acting downward in the
negative y direction are configured to be
negative. Likewise, vertical upward forces in
the positive y direction are positive.
This configuration is applied to the horizontal
components of the inclined forces.
The table 2 summarizes the resolution of these
five forces acting at a point.
Example 4. Calculate the resultant of the
system of forces as shown in Fig 12.
24. y
x
FA = 85N
Fig 13 FH = 620N
FE = 350N
FB = 150N
FC = 250N
FD = 350N
600
200300
350
26. y
x
Fig 13 Resultant in terms of magnitude and direction
FH = -155.3728N 240.9980
40.9980 FV = -135.0584N
R= √ (155.3728)2 + (135.0584)2
R = 205.8676N and Ө = tan-1 (FV/FH)= 40.9980
Based on the configurations as explained above, the
sums of FV and FH respectively are -135.0584N and -
155.3728N.
The resultant, R, has a direction and magnitude as
illustrated in Fig 13 below.
27. The direction of the resultant, Ө, may be stated based
on two perspectives. A) the
positive x axis
In this case, the direction is stated thus: (1800 + 40.9980)
= 240.9980
B)The negative x axis.
Ө = 40.9980
The two answers above answers are correct. However,
it is important to understand the fact that A) has been
stated in terms of a positive angle. On the other hand
B) has been stated in terms of a reference angle. All
positive angles are measured with respect to the
positive x axis in the counter clockwise direction.