This document provides calculations and diagrams for determining reactions and moments in statically determinate beams with distributed loads. It includes 6 example problems showing the process of: 1) establishing static equilibrium equations to solve for support reactions, 2) calculating shear force and bending moment diagrams, and 3) determining maximum bending moments. Diagrams are drawn and key values calculated at each step.
The document provides formulae for calculating slope and deflection of beams under different loading conditions:
1) It gives equations for calculating slope at free ends and deflection at any section for cantilever beams with various loads including concentrated load, uniform load, and uniformly varying load.
2) Equations are also provided for simply supported beams with the same loading types to determine slope at supports and maximum deflection in the center or at load location.
3) The document contains 10 sections that list the beam type, applied load, formulae for slope at free ends or supports, and expressions for deflection along the beam and point of maximum deflection.
1. The document discusses concepts in thermodynamics including classical vs statistical thermodynamics, conservation of energy, units of mass and force, properties of systems and processes.
2. It provides examples of applying concepts like Newton's laws to calculate weight on different planets, mass and weight of air in a room, and acceleration of objects.
3. Key points covered are properties of open and closed systems, intensive vs extensive properties, conditions of equilibrium, and types of processes like isothermal and isobaric.
66.-Merle C. Potter, David C. Wiggert, Bassem H. Ramadan - Mechanics of Fluid...HectorMayolNovoa
This document contains solutions to problems in a textbook on mechanics of fluids. It has 14 chapters that cover topics like fluid statics, fluids in motion, integral and differential forms of fundamental laws, dimensional analysis, internal and external flows, compressible flow, open channel flow, piping systems, and turbomachinery. Each chapter contains problems and their step-by-step solutions to aid instructors and students.
Mechanics of materials 9th edition goodno solutions manualKim92736
This document contains solutions to problems from Chapter 2 of the 9th Edition Mechanics of Materials textbook by Goodno. It lists over 300 problems from sections 2.2 through 2.12 that have been solved, along with the full textbook reference and a link to download the full solutions manual PDF. The problems cover topics in stress, strain, Hooke's law, axial loading, torsion, shear stresses, and more.
Mechanics of Materials 9th Edition Hibbeler Solutions Manualpofojufyv
A tension test was performed on a steel specimen. The data is plotted on a stress-strain diagram. The modulus of elasticity is approximated as 30.0(103) ksi. The yield stress is approximated as 11.8 kip and the ultimate stress is approximated as 19.6 kip. These values are determined from the stress-strain diagram where the yield stress corresponds to a strain of 0.002 and the ultimate stress corresponds to the highest stress on the diagram.
1) The document provides solutions to problems involving pressure distributions in fluids. It calculates normal and shear stresses on a plane cutting through a two-dimensional stress field.
2) It also calculates pressures, stresses, and fluid levels at various points in systems involving combinations of fluids such as water, oil, air and mercury.
3) The problems require use of concepts such as hydrostatic pressure, stress and normal/shear stress relationships, and properties of various fluids to determine unknown values at different points in the systems.
Este documento trata sobre la elasticidad de los materiales sólidos. Explica conceptos como esfuerzo, deformación, ley de Hooke, módulos elásticos, coeficiente de Poisson y curvas de esfuerzo-deformación. Incluye ejemplos numéricos que ilustran cómo calcular esfuerzos, deformaciones y recuperación elástica usando las fórmulas y propiedades dadas.
The document provides formulae for calculating slope and deflection of beams under different loading conditions:
1) It gives equations for calculating slope at free ends and deflection at any section for cantilever beams with various loads including concentrated load, uniform load, and uniformly varying load.
2) Equations are also provided for simply supported beams with the same loading types to determine slope at supports and maximum deflection in the center or at load location.
3) The document contains 10 sections that list the beam type, applied load, formulae for slope at free ends or supports, and expressions for deflection along the beam and point of maximum deflection.
1. The document discusses concepts in thermodynamics including classical vs statistical thermodynamics, conservation of energy, units of mass and force, properties of systems and processes.
2. It provides examples of applying concepts like Newton's laws to calculate weight on different planets, mass and weight of air in a room, and acceleration of objects.
3. Key points covered are properties of open and closed systems, intensive vs extensive properties, conditions of equilibrium, and types of processes like isothermal and isobaric.
66.-Merle C. Potter, David C. Wiggert, Bassem H. Ramadan - Mechanics of Fluid...HectorMayolNovoa
This document contains solutions to problems in a textbook on mechanics of fluids. It has 14 chapters that cover topics like fluid statics, fluids in motion, integral and differential forms of fundamental laws, dimensional analysis, internal and external flows, compressible flow, open channel flow, piping systems, and turbomachinery. Each chapter contains problems and their step-by-step solutions to aid instructors and students.
Mechanics of materials 9th edition goodno solutions manualKim92736
This document contains solutions to problems from Chapter 2 of the 9th Edition Mechanics of Materials textbook by Goodno. It lists over 300 problems from sections 2.2 through 2.12 that have been solved, along with the full textbook reference and a link to download the full solutions manual PDF. The problems cover topics in stress, strain, Hooke's law, axial loading, torsion, shear stresses, and more.
Mechanics of Materials 9th Edition Hibbeler Solutions Manualpofojufyv
A tension test was performed on a steel specimen. The data is plotted on a stress-strain diagram. The modulus of elasticity is approximated as 30.0(103) ksi. The yield stress is approximated as 11.8 kip and the ultimate stress is approximated as 19.6 kip. These values are determined from the stress-strain diagram where the yield stress corresponds to a strain of 0.002 and the ultimate stress corresponds to the highest stress on the diagram.
1) The document provides solutions to problems involving pressure distributions in fluids. It calculates normal and shear stresses on a plane cutting through a two-dimensional stress field.
2) It also calculates pressures, stresses, and fluid levels at various points in systems involving combinations of fluids such as water, oil, air and mercury.
3) The problems require use of concepts such as hydrostatic pressure, stress and normal/shear stress relationships, and properties of various fluids to determine unknown values at different points in the systems.
Este documento trata sobre la elasticidad de los materiales sólidos. Explica conceptos como esfuerzo, deformación, ley de Hooke, módulos elásticos, coeficiente de Poisson y curvas de esfuerzo-deformación. Incluye ejemplos numéricos que ilustran cómo calcular esfuerzos, deformaciones y recuperación elástica usando las fórmulas y propiedades dadas.
This document provides information about the 12th edition of the textbook "Engineering Mechanics: Dynamics" by R.C. Hibbeler, published by Prentice Hall. It also provides a link to a blog that contains free downloads of solution manuals for this and other university textbooks. The blog states it has clear, explained solutions to all the problems in the textbooks.
This chapter discusses the thermal properties of gases and the gas laws. It introduces concepts like the relationship between pressure, volume, temperature and amount of gas based on the ideal gas law. Several sample problems demonstrate how to use the gas laws to calculate pressure, volume, temperature or amount of gas given initial and final states. The chapter also covers the mole concept and relationships between mass, moles and molecules for different gases.
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.
Este documento presenta 10 problemas de hidrostática. Los problemas involucran conceptos como presión, flotación, densidad y fuerzas. Se pide calcular aceleraciones, masas, densidades, áreas y profundidades utilizando principios como la suma de fuerzas, equilibrio de cuerpos sumergidos y relaciones entre presiones y alturas de líquidos.
The document provides dimensional and mechanical properties for various standard steel shapes from the American Institute of Steel Construction (AISC) 13th edition. It includes tables listing properties like cross-sectional area, moments of inertia, radii of gyration, and weights per foot for wide flange beams, channels, angles, and other standard sections. Accompanying notes provide definitions and explanations of the terms and properties.
1) The document contains 14 problems solving for forces on surfaces due to fluid pressures.
2) The problems calculate forces using equations relating pressure, area, depth, and density.
3) Sample calculations determine the total force on a door from vacuum pressure difference or on a tank wall from liquid depth and density.
The document describes several problems involving the calculation of normal stress and strain in structural elements like posts, rods, wires, and beams. The problems involve circular and rectangular cross-sections under compression, tension or a combination of forces. Diagrams are provided and the geometry, forces and material properties are used to calculate stress, strain or unknown forces through equilibrium equations. Solutions show the relevant equations and step-by-step workings to arrive at the final numerical answers.
The document provides information about angular motion problems involving rotating disks, gears, and other mechanisms. It gives the initial conditions, equations of motion, and steps to solve for quantities like angular velocity and acceleration at various points in the systems. Quantities like radial distance, initial angular velocity, and angular acceleration are used in the equations of motion to determine velocities and accelerations at specific times.
Este documento presenta información sobre el lenguaje algebraico. Explica que el lenguaje algebraico traduce situaciones verbales a expresiones con símbolos y números, como ecuaciones. Da como ejemplo la traducción de "Lo que gasté en dulces fue el precio de cada dulce por el número de dulces que compré" a la expresión algebraica "G = P·N". El documento también incluye un enlace para descargar un archivo con una explicación animada sobre el tema del lenguaje algebraico.
Este documento trata sobre cuerpos rígidos y sus propiedades de movimiento rotacional. Explica conceptos como velocidad angular, aceleración angular, momento de inercia y su cálculo para diferentes objetos. También cubre la relación entre fuerza, torque y aceleración angular de un cuerpo rígido.
El documento promociona el sitio web www.elsolucionario.net, el cual ofrece solucionarios gratuitos de libros universitarios. Los solucionarios contienen todas las respuestas y explicaciones de los ejercicios de los libros de forma clara. Se invita a los lectores a visitar el sitio para descargar los solucionarios gratuitamente.
This is completed downloadable version of Solution Manual for Fundamentals of Engineering Thermodynamics 8th Edition by Michael J. Moran, Howard N. Shapiro and Daisie D. Boettner
Solution Manual for Fundamentals of Engineering Thermodynamics 8th Edition by Moran
Solution Manual for Fundamentals of Engineering Thermodynamics 8th Edition by Moran
Link to download Full chapter + answers sample:
https://goo.gl/s1EjyD
Click link bellow to view sample chapter of Fundamentals of Engineering Thermodynamics 8th Edition by Moran Solution Manual
https://getbooksolutions.com/wp-content/uploads/2017/06/Solution-manual-Fundamentals-of-Engineering-Thermodynamics-8th-Edition-by-Moran.pdf
Fundamentals of Engineering Thermodynamics by Moran, Shapiro, Boettner and Bailey solution manul continues its tradition of setting the standard for teaching students how to be effective problem solvers. Now in its eighth edition, this market-leading text emphasizes the authors collective teaching expertise as well as the signature methodologies that have taught entire generations of engineers worldwide. Integrated throughout the text are real-world applications that emphasize the relevance of thermodynamics principles to some of the most critical problems and issues of today, including a wealth of coverage of topics related to energy and the environment, biomedical/bioengineering, and emerging technologies.
Este documento presenta un esquema de una viga sometida a una carga uniformemente distribuida. Se pide determinar (a) la reacción en el punto A y el momento en A, (b) el diagrama del momento flector, y (c) el diagrama de la fuerza cortante. El documento muestra los cálculos para determinar las reacciones, fuerzas cortantes y momentos en varios puntos de la viga.
The document provides solutions to 19 problems involving the calculation of shear force diagrams (SFD) and bending moment diagrams (BMD) for various beams with different loadings. Each problem shows the given beam configuration and loading, the calculation steps to determine the reactions and maximum bending moment, and the resulting SFD and BMD diagrams. The solutions utilize the concepts of equilibrium of forces and moments to solve for support reactions and then draw the shear and bending moment variations along the beams.
This document contains solutions to problems involving the calculation of shear stresses in beams. It determines shear stresses at specific points of beams by using the shear formula and calculating the shear force resisted by various beam components. The maximum shear stress in several beams is also calculated. Cross-sectional properties like moment of inertia are used. Shear stresses are indicated on volume elements and shear force diagrams are sketched.
Para mantener un cubo de latón de 6 pulgadas de lado y 67 libras en equilibrio bajo el agua, se necesita una boya de espuma de 1.02 pies cúbicos. La ecuación de equilibrio muestra que este volumen de boya proporcionará flotabilidad neutra para el cubo.
Mathematics for the Trades A Guided Approach Canadian 2nd Edition Carman Test...JudithLandrys
Full download : https://alibabadownload.com/product/mathematics-for-the-trades-a-guided-approach-canadian-2nd-edition-carman-test-bank/ Mathematics for the Trades A Guided Approach Canadian 2nd Edition Carman Test Bank
This document summarizes the design of a cantilever stub pier with a 65cm wide and 40cm high bridge deck that transmits a 400kg/m load. Key details include:
- The foundation level is 6.5m below grade.
- Design considers soil properties, loads, and structural checks.
- Reinforcement is designed for the stub pier, including checking capacity, development length, and distribution.
- Design of the heel includes moment, shear, and reinforcement sizing.
- Joint design considers vertical loads only.
1. The document provides the design calculations for a retaining wall with a cantilever section. Input parameters such as material strengths, soil properties, geometry, and loading are specified.
2. Preliminary calculations are shown to determine factors such as the soil pressure coefficient and active and passive pressures. Stability is checked against sliding and overturning.
3. The design of the wall reinforcement is then shown, with calculations for the vertical, horizontal, and shear capacities of different sections of the wall. Reinforcement amounts and spacing are sized to meet design requirements.
This document provides information about the 12th edition of the textbook "Engineering Mechanics: Dynamics" by R.C. Hibbeler, published by Prentice Hall. It also provides a link to a blog that contains free downloads of solution manuals for this and other university textbooks. The blog states it has clear, explained solutions to all the problems in the textbooks.
This chapter discusses the thermal properties of gases and the gas laws. It introduces concepts like the relationship between pressure, volume, temperature and amount of gas based on the ideal gas law. Several sample problems demonstrate how to use the gas laws to calculate pressure, volume, temperature or amount of gas given initial and final states. The chapter also covers the mole concept and relationships between mass, moles and molecules for different gases.
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.
Este documento presenta 10 problemas de hidrostática. Los problemas involucran conceptos como presión, flotación, densidad y fuerzas. Se pide calcular aceleraciones, masas, densidades, áreas y profundidades utilizando principios como la suma de fuerzas, equilibrio de cuerpos sumergidos y relaciones entre presiones y alturas de líquidos.
The document provides dimensional and mechanical properties for various standard steel shapes from the American Institute of Steel Construction (AISC) 13th edition. It includes tables listing properties like cross-sectional area, moments of inertia, radii of gyration, and weights per foot for wide flange beams, channels, angles, and other standard sections. Accompanying notes provide definitions and explanations of the terms and properties.
1) The document contains 14 problems solving for forces on surfaces due to fluid pressures.
2) The problems calculate forces using equations relating pressure, area, depth, and density.
3) Sample calculations determine the total force on a door from vacuum pressure difference or on a tank wall from liquid depth and density.
The document describes several problems involving the calculation of normal stress and strain in structural elements like posts, rods, wires, and beams. The problems involve circular and rectangular cross-sections under compression, tension or a combination of forces. Diagrams are provided and the geometry, forces and material properties are used to calculate stress, strain or unknown forces through equilibrium equations. Solutions show the relevant equations and step-by-step workings to arrive at the final numerical answers.
The document provides information about angular motion problems involving rotating disks, gears, and other mechanisms. It gives the initial conditions, equations of motion, and steps to solve for quantities like angular velocity and acceleration at various points in the systems. Quantities like radial distance, initial angular velocity, and angular acceleration are used in the equations of motion to determine velocities and accelerations at specific times.
Este documento presenta información sobre el lenguaje algebraico. Explica que el lenguaje algebraico traduce situaciones verbales a expresiones con símbolos y números, como ecuaciones. Da como ejemplo la traducción de "Lo que gasté en dulces fue el precio de cada dulce por el número de dulces que compré" a la expresión algebraica "G = P·N". El documento también incluye un enlace para descargar un archivo con una explicación animada sobre el tema del lenguaje algebraico.
Este documento trata sobre cuerpos rígidos y sus propiedades de movimiento rotacional. Explica conceptos como velocidad angular, aceleración angular, momento de inercia y su cálculo para diferentes objetos. También cubre la relación entre fuerza, torque y aceleración angular de un cuerpo rígido.
El documento promociona el sitio web www.elsolucionario.net, el cual ofrece solucionarios gratuitos de libros universitarios. Los solucionarios contienen todas las respuestas y explicaciones de los ejercicios de los libros de forma clara. Se invita a los lectores a visitar el sitio para descargar los solucionarios gratuitamente.
This is completed downloadable version of Solution Manual for Fundamentals of Engineering Thermodynamics 8th Edition by Michael J. Moran, Howard N. Shapiro and Daisie D. Boettner
Solution Manual for Fundamentals of Engineering Thermodynamics 8th Edition by Moran
Solution Manual for Fundamentals of Engineering Thermodynamics 8th Edition by Moran
Link to download Full chapter + answers sample:
https://goo.gl/s1EjyD
Click link bellow to view sample chapter of Fundamentals of Engineering Thermodynamics 8th Edition by Moran Solution Manual
https://getbooksolutions.com/wp-content/uploads/2017/06/Solution-manual-Fundamentals-of-Engineering-Thermodynamics-8th-Edition-by-Moran.pdf
Fundamentals of Engineering Thermodynamics by Moran, Shapiro, Boettner and Bailey solution manul continues its tradition of setting the standard for teaching students how to be effective problem solvers. Now in its eighth edition, this market-leading text emphasizes the authors collective teaching expertise as well as the signature methodologies that have taught entire generations of engineers worldwide. Integrated throughout the text are real-world applications that emphasize the relevance of thermodynamics principles to some of the most critical problems and issues of today, including a wealth of coverage of topics related to energy and the environment, biomedical/bioengineering, and emerging technologies.
Este documento presenta un esquema de una viga sometida a una carga uniformemente distribuida. Se pide determinar (a) la reacción en el punto A y el momento en A, (b) el diagrama del momento flector, y (c) el diagrama de la fuerza cortante. El documento muestra los cálculos para determinar las reacciones, fuerzas cortantes y momentos en varios puntos de la viga.
The document provides solutions to 19 problems involving the calculation of shear force diagrams (SFD) and bending moment diagrams (BMD) for various beams with different loadings. Each problem shows the given beam configuration and loading, the calculation steps to determine the reactions and maximum bending moment, and the resulting SFD and BMD diagrams. The solutions utilize the concepts of equilibrium of forces and moments to solve for support reactions and then draw the shear and bending moment variations along the beams.
This document contains solutions to problems involving the calculation of shear stresses in beams. It determines shear stresses at specific points of beams by using the shear formula and calculating the shear force resisted by various beam components. The maximum shear stress in several beams is also calculated. Cross-sectional properties like moment of inertia are used. Shear stresses are indicated on volume elements and shear force diagrams are sketched.
Para mantener un cubo de latón de 6 pulgadas de lado y 67 libras en equilibrio bajo el agua, se necesita una boya de espuma de 1.02 pies cúbicos. La ecuación de equilibrio muestra que este volumen de boya proporcionará flotabilidad neutra para el cubo.
Mathematics for the Trades A Guided Approach Canadian 2nd Edition Carman Test...JudithLandrys
Full download : https://alibabadownload.com/product/mathematics-for-the-trades-a-guided-approach-canadian-2nd-edition-carman-test-bank/ Mathematics for the Trades A Guided Approach Canadian 2nd Edition Carman Test Bank
This document summarizes the design of a cantilever stub pier with a 65cm wide and 40cm high bridge deck that transmits a 400kg/m load. Key details include:
- The foundation level is 6.5m below grade.
- Design considers soil properties, loads, and structural checks.
- Reinforcement is designed for the stub pier, including checking capacity, development length, and distribution.
- Design of the heel includes moment, shear, and reinforcement sizing.
- Joint design considers vertical loads only.
1. The document provides the design calculations for a retaining wall with a cantilever section. Input parameters such as material strengths, soil properties, geometry, and loading are specified.
2. Preliminary calculations are shown to determine factors such as the soil pressure coefficient and active and passive pressures. Stability is checked against sliding and overturning.
3. The design of the wall reinforcement is then shown, with calculations for the vertical, horizontal, and shear capacities of different sections of the wall. Reinforcement amounts and spacing are sized to meet design requirements.
This document discusses the structuring and measurement of loads for a building. It describes classifying loads into dead, live, and accidental loads from wind or seismic activity. Dead loads include the weight of the building and fixed equipment. Live loads include wind forces, seismic movements, vibrations, furniture and stored materials. Dead loads maintain a constant magnitude. The objectives are to correctly structure and pre-dimension structural elements, measure loads, and design foundations for columns and walls. It provides data on the terrain, floor heights, and material weights. Dimensioning of slabs, beams, and foundations is presented along with load calculations for stairs.
This document summarizes the planning and design calculations for a pre-stressed concrete beam with the following parameters:
1. The required bending moment (Mt) is 350 ton-meters. The concrete compressive strength (f'c) is 47 MPa.
2. The initial dimensions of the beam are calculated as 200 cm height (h) and 4339.6 cm^2 cross-sectional area (Ab).
3. The final design meets the required bending moment of 350 ton-meters with a uniform prestress force (q) of 2285.71 kg/m distributed over the beam length. Stresses in the concrete are calculated to remain below the allowable limits.
This document details the design and calculation of a concrete bridge with plate girders. It includes the dimensions of the bridge components, loading assumptions, and structural analysis. The bridge is designed to carry 3 traffic lanes and a C-40 truck load, with a 19m span. Structural checks are performed for the plate, girders, and reinforcement sizing. Reinforcement is designed for critical moments in the supports, interior spans, and overhang.
Calculo de momentos en una viga continua cuando se encuentranCNEL
1) The document describes the calculation of moments in a continuous beam structure with fixed ends. It provides the stiffness values, load values, and calculations of girder factors for each node.
2) Moment calculations are shown for each segment based on the beam's weight and applied loads.
3) The results of six load cycles are presented, showing the calculated moments at each node until equilibrium is reached and the moment values repeat.
This document summarizes the design of a lightweight slab. It includes:
- Calculations of dead and live loads on the slab
- Determination of design load as 1.4 times dead load plus 1.7 times live load
- Analysis of moments on the slab using SAP2000 software
- Calculation of minimum and maximum reinforcement areas needed
- Selection of rebar diameters to satisfy moment and shear requirements
- Consideration of temperature reinforcement with a minimum of 1 cm2/m and spacing of 0.25 m or less.
The document provides calculations for load distributions on beams supporting multiple slabs in a building. It calculates the permanent and variable actions on each slab based on the self-weight of materials. These loads are then distributed to the beams below. Bending moments and shear forces are calculated for each beam span under maximum loading conditions using a finite element method. Reactions and uniform loading values are also determined for each span.
This document contains physics exercises on motion by Alonso Acosta presented by Widmar Aguilar in February 2023. It includes over 30 problems covering concepts such as average velocity, accelerated motion, decelerated motion, free fall, and projectile motion. The problems are presented in Spanish and involve calculating distances, times, velocities and accelerations using kinematic equations for various scenarios like cars moving along a road, objects falling, and projectiles launched at angles. Step-by-step workings are shown for each problem.
This document contains exercises from Maiztegui Sabato's physics of motion in composite-circular motion. It includes several problems solving kinematic equations for circular and composite motions like airplanes, boats, cars, wheels, and bullets. The document was written by Msc. Widmar Aguilar in April 2023 and contains solutions to various circular and composite motion problems involving velocity, time, distance, and angular displacement calculations.
This document discusses momentum and the conservation of momentum. It provides examples of calculating momentum using the formula momentum (p) = mass (m) x velocity (v). It also applies the law of conservation of momentum to example problems of objects colliding or a cannon firing, calculating the momentum before and after to find the unknown velocity.
Structural Analysis of a Bungalow Reportdouglasloon
Taylor's University Lakeside Campus
School of Architecture, Building & Design
Bachelor of Science (Hons) in Architecture
Building Structures (ARC 2523 / BLD 60103)
Project 2: Structural Analysis of a Bungalow
Structural analysis of a bungalow reportChengWei Chia
The document presents the structural analysis of a bungalow conducted by three students. It includes architectural plans, quantities of dead and live loads, structural plans, load distribution diagrams, tributary area diagrams, and individual analyses of structural components by each student. Student 1 analyzes beams and columns on the ground floor. Student 2 analyzes a beam spanning from the ground floor to the first floor. Student 3 analyzes point loads applied to beams. Calculations are shown for load quantities, load diagrams, and ultimate loads on structural elements.
This document discusses key concepts in physics related to motion, forces, energy, and sound. It covers Aristotle and Galileo's early contributions, followed by Newton's laws of motion and universal law of gravitation. Key topics include linear and circular motion, forces, work, energy, heat transfer, and properties of sound waves like frequency, wavelength, and the Doppler effect. Equations for acceleration, velocity, forces, energy, heat, and sound are provided alongside examples to illustrate their application.
This document provides details on the structural design of a building, including load assumptions, member dimensions, and seismic load calculations according to SNI 1726:2012. It summarizes the seismic load calculations for two orthogonal directions and evaluates the structure's capacity through pushover analysis. Plastic hinges form at various displacement steps. The document concludes with a redesign of the structure according to SRPMK seismic provisions.
1. The document provides examples of calculating consolidation parameters such as void ratio, coefficient of consolidation, and primary consolidation settlement from given soil testing data.
2. Parameters like initial void ratio, applied pressure, and thickness of soil layers are used to determine the change in stress and void ratio to then calculate settlement.
3. Several methods are presented to calculate the average effective stress and stress change at different points to then determine the consolidation settlement under different boundary conditions, stress histories, and soil properties.
Roof Truss Design (By Hamza Waheed UET Lahore )Hamza Waheed
This presentation defines, describes and presents the most effective and easy way to design a roof truss with all the necessary steps and calculations based on Allowable Stress Design. Soft-wares like MD Solids, Truss Analysis have been used. It is most convenient way to design a roof truss which is being the most important structural components of All types of steel bridges.
Similar to Resistencia de los materiales (ejercicios- equilibrio estático, diagrama de corte y momento flector) (20)
International Conference on NLP, Artificial Intelligence, Machine Learning an...gerogepatton
International Conference on NLP, Artificial Intelligence, Machine Learning and Applications (NLAIM 2024) offers a premier global platform for exchanging insights and findings in the theory, methodology, and applications of NLP, Artificial Intelligence, Machine Learning, and their applications. The conference seeks substantial contributions across all key domains of NLP, Artificial Intelligence, Machine Learning, and their practical applications, aiming to foster both theoretical advancements and real-world implementations. With a focus on facilitating collaboration between researchers and practitioners from academia and industry, the conference serves as a nexus for sharing the latest developments in the field.
Electric vehicle and photovoltaic advanced roles in enhancing the financial p...IJECEIAES
Climate change's impact on the planet forced the United Nations and governments to promote green energies and electric transportation. The deployments of photovoltaic (PV) and electric vehicle (EV) systems gained stronger momentum due to their numerous advantages over fossil fuel types. The advantages go beyond sustainability to reach financial support and stability. The work in this paper introduces the hybrid system between PV and EV to support industrial and commercial plants. This paper covers the theoretical framework of the proposed hybrid system including the required equation to complete the cost analysis when PV and EV are present. In addition, the proposed design diagram which sets the priorities and requirements of the system is presented. The proposed approach allows setup to advance their power stability, especially during power outages. The presented information supports researchers and plant owners to complete the necessary analysis while promoting the deployment of clean energy. The result of a case study that represents a dairy milk farmer supports the theoretical works and highlights its advanced benefits to existing plants. The short return on investment of the proposed approach supports the paper's novelty approach for the sustainable electrical system. In addition, the proposed system allows for an isolated power setup without the need for a transmission line which enhances the safety of the electrical network
Understanding Inductive Bias in Machine LearningSUTEJAS
This presentation explores the concept of inductive bias in machine learning. It explains how algorithms come with built-in assumptions and preferences that guide the learning process. You'll learn about the different types of inductive bias and how they can impact the performance and generalizability of machine learning models.
The presentation also covers the positive and negative aspects of inductive bias, along with strategies for mitigating potential drawbacks. We'll explore examples of how bias manifests in algorithms like neural networks and decision trees.
By understanding inductive bias, you can gain valuable insights into how machine learning models work and make informed decisions when building and deploying them.
Introduction- e - waste – definition - sources of e-waste– hazardous substances in e-waste - effects of e-waste on environment and human health- need for e-waste management– e-waste handling rules - waste minimization techniques for managing e-waste – recycling of e-waste - disposal treatment methods of e- waste – mechanism of extraction of precious metal from leaching solution-global Scenario of E-waste – E-waste in India- case studies.
Using recycled concrete aggregates (RCA) for pavements is crucial to achieving sustainability. Implementing RCA for new pavement can minimize carbon footprint, conserve natural resources, reduce harmful emissions, and lower life cycle costs. Compared to natural aggregate (NA), RCA pavement has fewer comprehensive studies and sustainability assessments.
TIME DIVISION MULTIPLEXING TECHNIQUE FOR COMMUNICATION SYSTEMHODECEDSIET
Time Division Multiplexing (TDM) is a method of transmitting multiple signals over a single communication channel by dividing the signal into many segments, each having a very short duration of time. These time slots are then allocated to different data streams, allowing multiple signals to share the same transmission medium efficiently. TDM is widely used in telecommunications and data communication systems.
### How TDM Works
1. **Time Slots Allocation**: The core principle of TDM is to assign distinct time slots to each signal. During each time slot, the respective signal is transmitted, and then the process repeats cyclically. For example, if there are four signals to be transmitted, the TDM cycle will divide time into four slots, each assigned to one signal.
2. **Synchronization**: Synchronization is crucial in TDM systems to ensure that the signals are correctly aligned with their respective time slots. Both the transmitter and receiver must be synchronized to avoid any overlap or loss of data. This synchronization is typically maintained by a clock signal that ensures time slots are accurately aligned.
3. **Frame Structure**: TDM data is organized into frames, where each frame consists of a set of time slots. Each frame is repeated at regular intervals, ensuring continuous transmission of data streams. The frame structure helps in managing the data streams and maintaining the synchronization between the transmitter and receiver.
4. **Multiplexer and Demultiplexer**: At the transmitting end, a multiplexer combines multiple input signals into a single composite signal by assigning each signal to a specific time slot. At the receiving end, a demultiplexer separates the composite signal back into individual signals based on their respective time slots.
### Types of TDM
1. **Synchronous TDM**: In synchronous TDM, time slots are pre-assigned to each signal, regardless of whether the signal has data to transmit or not. This can lead to inefficiencies if some time slots remain empty due to the absence of data.
2. **Asynchronous TDM (or Statistical TDM)**: Asynchronous TDM addresses the inefficiencies of synchronous TDM by allocating time slots dynamically based on the presence of data. Time slots are assigned only when there is data to transmit, which optimizes the use of the communication channel.
### Applications of TDM
- **Telecommunications**: TDM is extensively used in telecommunication systems, such as in T1 and E1 lines, where multiple telephone calls are transmitted over a single line by assigning each call to a specific time slot.
- **Digital Audio and Video Broadcasting**: TDM is used in broadcasting systems to transmit multiple audio or video streams over a single channel, ensuring efficient use of bandwidth.
- **Computer Networks**: TDM is used in network protocols and systems to manage the transmission of data from multiple sources over a single network medium.
### Advantages of TDM
- **Efficient Use of Bandwidth**: TDM all
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Resistencia de los materiales (ejercicios- equilibrio estático, diagrama de corte y momento flector)
1. -SEÑALAR Y CALCULAR LAS REACCIONES PRODUCIDAS EN C/U DE LOS APOYOS.
-CALCULAR LOS DIAGRAMAS DE CORTE Y MOMENTO.
EJERCICIO #1
Equilibrio Estático son 3 condiciones
Ah = 0
– (1599 Kg/m * 2 m) – (1299 Kg/m * 3 m) – (1099Kg/m * 2,5 m) + (Av) + (Bv) = 0
– 3198 Kg – 3897 Kg – 2747,5 Kg + Av + Bv = 0
– 9842,5 Kg + Av + Bv = 0
Av + Bv = 9842,5 Kg
(3198 Kg * 1m) + (3897 Kg * 4,5m) + (2747,5 Kg * 8,25 m) – (Bv * 9,5m) = 0
3198 Kg*m + 17536,5 Kg*m + 22666,875 Kg*m – 9,5mBv = 0
43401,375 Kg*m – 9,5mBv = 0
– 9,5mBv = – 43401,375 Kg*m
Bv = – 43401,375 Kg*m = 347211 kg = 4568,57
– 9,5m 76
(De la formula) Av + Bv = 9842,5 Kg Se Despeja Av
Av = 9842,5 Kg – Bv
Av = 9842,5 Kg – 347211 kg = 400819 kg = 5273,93
76 76
Av = 400819 kg = 5273,93
76
2. CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = 0 + (400819/76 kg) = (400819/76 Kg) = 5273,93 Kg
V (0-2) = (400819/76 Kg) – (1599 Kg/m * 2m) = (157771/76 Kg) = 2075,93 Kg
V (2-3) = (157771/76 Kg) + 0 = (157771/76 Kg) = 2075,93 Kg
V (3-6) = (157771/76 Kg) – (1299 Kg/m * 3m) = - (138401/76 Kg) = - 1821,07 Kg
V (6-7) = - (138401/76 Kg) + 0 = - (138401/76 Kg) = - 1821,07 Kg
V (7-9,5) = - (138401/76 Kg) – (1099 Kg/m * 2,5m) = (347211/76 Kg) = - 4568,57 Kg
V (9,5) = - (138401/76 Kg) + (138401/76 Kg) = 0
Se Grafican y se calculan las áreas =
CALCULO DE LAS ÁREAS
Para hallar las distancias de los triángulos que se Forman al cortar el eje X se procede a
una relación de triangulo
(1299 Kg/m * 3m) (157771 / 76 Kg)
3m X1 = 1,598101779 m
(1299 Kg/m * 3m) (138401 / 76))
3m X2 = 1,401898221 m
A1 =
퐻푀+ 퐻푚
2
∗ 퐵 =
400819
76
157771
+
76
2
∗ 2 = (279295 / 38) = 7349,87
A2 = 퐵 ∗ 퐻 = 1 ∗
157771
76
= (157771 / 76) = 2075,93
A3 =
퐻∗퐵
2
=
1,598101779 ∗ (157771 / 76)
2
= 1658,777077
A4 =
퐻∗퐵
2
=
138401
1,401898221 ∗
76
2
= 1276,474446
A5 = 퐵 ∗ 퐻 = 1 ∗ 138401 / 76 = (138401 / 76) = 1821,07
A6 =
퐻푀+ 퐻푚
2
∗ 퐵 =
(
347211
76
) + (
138401
76
)
2
∗ 2,5 = (607015 / 76) = 7987,04
4. EJERCICIO #2
Equilibrio Estático son 3 condiciones
Ah = 0
– (799 Kg) – (699 Kg) – (599 Kg) + (Av) + (Bv) = 0
– (2087 Kg) + Av + Bv = 0
Av + Bv = 2097 Kg
(699 Kg * 4m) + (599 Kg * 9m) – (Bv * 9m) = 0
5. 2796 Kg*m + 5391 Kg*m – 9mBv = 0
8187 Kg*m – 9mBv = 0
– 9mBv = – 8187 Kg*m
Bv = – 8187 Kg*m = 2729 kg = 909,67
– 9m 3
(De la formula) Av + Bv = 2097 Kg Se Despeja Av
Av = 2097 Kg – Bv
Av = 2097 Kg – 2729 kg = 3562 kg = 1187,33
3 3
CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = (3562 / 3 Kg) - (799 Kg) = (1165 / 3 Kg) = 388,33 Kg
V (0-4) = (1165 / 3 Kg) + 0 = (1165 / 3 Kg) = 388,33 Kg
V (4) = (1165 / 3 Kg) – 699 Kg = - (932 / 3 Kg) = -310,67 Kg
V (4-9) = - (932 / 3 Kg) + 0 = - (932 / 3 Kg) = -310,67 Kg
V (9) = - (932 / 3 Kg) – 599 Kg + (2729/3 Kg ) = 0
CALCULO DE LAS ÁREAS
A1 = 퐵 ∗ 퐻 = 4 ∗ (1165 / 3 ) = (4660 / 3) = 1553,33
A2 = 퐵 ∗ 퐻 = 5 ∗ (932 / 3 ) = (4660 / 3) = 1553,33
CALCULO DE LOS MOMENTOS FLECTORES
Msig = Manterior ± Área de la carga
M(0) = 0 + 0 = 0
M(0-4) = 0 + (4660 / 3) = (4660 / 3) = 1553,33
M(2-3) = (4660 / 3) - (4660 / 3) = 0
Se grafica los momentos
6. EJERCICIO #3
Las fuerzas y los momentos vendrán dados por el peso propio de la viga (Wv)
Equilibrio Estático son 3 condiciones
Ah = 0
Av + Wv = 0
Av = Wv
Ma = - (Wv*2m)
7. EJERCICIO #4
Equilibrio Estático son 3 condiciones
Bh = 0
– ((4099 Kg/m * 3,5 m)/2) – (599 Kg)+ (Av) + (Bv) = 0
– 7173,25 Kg – 599 Kg + Av + Bv = 0
– 7772,25 Kg + Av + Bv = 0
Av + Bv = 7772,25 Kg
– (599 Kg*3m) – (((4099 Kg/m * 3,5m)/2) * ((2/3*3,5) + 3)m) + (Av * 6,5m) = 0
– (1797 Kg*m) – ((7173,25 Kg) * (7/3 + 3)m) + (Av * 6,5m) = 0
– (1797 Kg*m) – ((7173,25 Kg) * (16/3 m)) + (Av * 6,5m) = 0
– (1797 Kg*m) – (114772/3 Kg*m) + (Av * 6,5m) = 0
– (120163/3 Kg*m) + (Av * 6,5m) = 0
(Av * 6,5m) = 120163/3 Kg*m
Av = 120163/3 Kg*m = 240326 kg = 6162,21 kg
6,5m 39
(De la formula) Av + Bv = 7772,25 Kg Se Despeja Bv
Bv = 7772,25 Kg – Av
Bv = 6574,25 Kg – 240326 kg = 251167 kg = 1610,05
39 156
8. CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = 0 + (240326/39 Kg) = (240326/39 Kg) = 6162,21 Kg
V (0-3,5) = (240326/39 Kg) - ((4099 Kg/m * 3,5m)/2) = - (157723/156 Kg) = -1011,04 Kg
V (3,5) = - (157723/156 Kg) – 599 Kg = - (251167 / 156 Kg) = -1610,04 Kg
V (3,5-6,5) = - (251167 / 156 Kg) + 0 = - (251167 / 156 Kg) = -1610,04 Kg
V (9) = - (251167 / 156 Kg) + (251167 / 156 Kg) = 0
CALCULO DE LOS MOMENTOS FLECTORES
Se Seccionan Tramos Para Conseguir Los Momentos actuantes
– (7173,25 Kg* 2/3*3,5m) + (6162,21 Kg* 3,5m) + (Ma-a) = 0
– (7173,25 Kg* 7/3m) + (6162,21 Kg* 3,5m) + (Ma-a) = 0
– (16737,58 Kg*m) + (21567,74 Kg*m) + (Ma-a) = 0
– (4830,16 Kg*m) + (Ma-a) = 0
(Ma-a) = (4830,16 Kg*m)
9. Se grafica los momentos
EJERCICIO #5
Y = Sen(30) * 599 kg = 299,50kg
X = Cos(30) * 599 kg = 518,7492169
Equilibrio Estático son 3 condiciones
Ah = 518,75 Kg
10. – (299,5 Kg) + (Av) = 0
Av = 299,5 Kg
(299,50 Kg * 2,5m) + Ma = 0
(748,75 Kg*m) + Ma = 0
- Ma = (748,75 Kg*m)
CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = 0 + 299,5 Kg = 299,5 Kg
V (0-2,5) = 299,5 Kg + 0 = 299,5 Kg
V (2,5) = 299,5 Kg – 299,5 Kg = 0
V (2,5-5) = 0 + 0 = 0
CALCULO DE LAS ÁREAS
A1 = 퐵 ∗ 퐻 = 2,5 ∗ 299,5 kg = 748,45
CALCULO DE LOS MOMENTOS FLECTORES
Msig = Manterior ± Área de la carga
M(0) = 0 + -748,75 Kg = -748,75 Kg
M(0-2,5) = -748,75 Kg + 748,75 Kg = 0
M(2,5 - 5) = 0 + 0 = 0
11. Se grafica los momentos
EJERCICIO #6
Y1 = Sen(45) * 1599 kg = 1130,663743 Kg
X1 = Cos(45) * 1599 kg = 1130,663743 Kg
Y2 = Sen(25) * 6599 kg = 2788,857909 Kg
X2 = Cos(25) * 6599 kg = 5980,725087 Kg
Y3 = Sen(30) * 799 kg = 399,50 Kg
X3 = Cos(30) * 799 kg = 691,9542976 Kg
Equilibrio Estático son 3 condiciones
–1130,663743 Kg - 5980,725087 Kg - 691,9542976 Kg + Bh = 0
–7803,343128 Kg + Bh = 0
Bh = 7803,343128 Kg
12. – 1130,663743 Kg – 2788,857909 Kg – 399,50 Kg + Av + Bv = 0
– 4319,021652 Kg + Av + Bv = 0
Av + Bv = 4319,021652 Kg
+ (1130,663743 Kg * 0m) – (2788,857909 Kg * 3,2m) – (399,50 Kg * 7,2m) + (Av * 7,20m) = 0
+ 0 – 8924,345309 Kg*m – 2876,40 Kg*m + (Av * 7,20m) = 0
– 11800,74531 Kg*m + (Av * 7,20m) = 0
Av = 11800,74531 Kg*m = 1638,992404 Kg
7,20m
Se despeja b
Av + Bv = 4319,021652 Kg
Bv = 4319,021652 Kg - 1638,992404 Kg
Bv = 2680,029248 Kg
CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = 0 + 1638,992404 - 399,50 Kg = 1239,492404 Kg
V (0-4) = 1239,492404 Kg + 0 = 1239,492404 Kg
V (4) = 1239,492404 Kg – 2788,85709 Kg = - 1549,365505 Kg
V (4- 7,2) = -1549,365505 Kg + 2680,029248 Kg - 1130,663743 Kg = 0
CALCULO DE LAS ÁREAS
A1 = 퐵 ∗ 퐻 = 4,0 ∗ 1239,492404 = 4957,969616
A2 = 퐵 ∗ 퐻 = 3,2 ∗ 1549,365505 = 4957,969616
CALCULO DE LOS MOMENTOS FLECTORES
Msig = Manterior ± Área de la carga
M(0) = 0 + 0 = 0
M(0- 4) = 0 + 4957,969616 = 4957,969616
M(4-7,2) = 4957,969616 - 4957,969616 = 0
Se grafica los momentos
13. EJERCICIO #7
Equilibrio Estático son 3 condiciones
Ah = 0
– ((2799 Kg/m*3,5m)/2) – ((5599 Kg/m *4m)/2) + Av + Bv = 0
– (4898,25 Kg) – (11198 Kg) + Av + Bv = 0
– (16096,25 Kg) + Av + Bv = 0
Av + Bv = 16096,25 Kg*m
14. + (4898,25 Kg * (1/3*3,5m)) + (11198 Kg* ((2/3*4m)+3,5m)) - 7,5m *Bv = 0
+ (4898,25 Kg * (3,5/3m)) + (11198 Kg* ((8/3m)+3,5m)) - 7,5m *Bv = 0
+ (4898,25 Kg * 7/6m) + (11198 Kg*37/6m) - 7,5m *Bv = 0
+ (5714,625 Kg*m) + (69054,33333 Kg*m) - 7,5m *Bv = 0
+ (74768,95833 Kg*m) - 7,5m *Bv = 0
Bv = 74768,95833 Kg*m = 9969,194444 Kg
7,5m
Se despeja a
Av + Bv = 16096,25 Kg*m
Av = 16096,25 Kg*m - 9969,194444 Kg
Av = 6127,05556 Kg
CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = 0 + 6127,05556 = 6127,05556
V (0-3,5) = 6127,05556 – (2799*3,5)/2 = 1228,805556
V (3,5-7,5) = 1228,805556 – (5599*4)/2 = - 9969,194444
V (7,5) = - 9969,194444 + 9969,194444 = 0
CALCULO DE LOS MOMENTOS FLECTORES
Se Seccionan Tramos Para Conseguir Los Momentos actuantes
16. – (2099 Kg*5,6m) – (1099 Kg*3m) + Av + Bv = 0
– 11754,4 Kg – 3297 Kg + Av + Bv = 0
– 15051,4 Kg + Av + Bv = 0
Av + Bv = 15051,4 Kg
- (11754,4 Kg * (5,6/2 m)) - (3297 Kg * (3/2 m)) + 5,6m *Av = 0
- (11754,4 Kg * 2,8m) - (3297 Kg * 1,5 m) + 5,6m *Av = 0
- (32912,32 Kg*m) - (4945,50 Kg*m) + 5,6m *Av = 0
- (37857,82 Kg*m) + 5,6m *Av = 0
Av = 37857,82 Kg*m = 6760,325 Kg
5,6m
Se despeja b
Av + Bv = 15051,4 Kg
Bv = 15051,4 Kg - 6760,325 Kg
Bv = 8291,075 Kg.
CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = 0 + 6760,325 = 6760,325
V (0-2,6) = 6760,325 – (2099 * 2,6) = 1302,925
V (2,6- 5,6) = 1302,925 – (2099*3) – (1099*3) = -8291,075
V (5,6) = -8291,075 + 8291,075 = 0
CALCULO DE LAS ÁREAS
Para hallar las distancias de los triángulos que se Forman al cortar el eje X se procede a una
relación de triangulo
(3198 Kg/m * 2,6m) (1302,925 Kg)
2,6m X1 = 0,4074186992m
(3198 Kg/m * 2,6m) (8291,075 Kg)
2,6m X2 = 2,59258130 m
A1 = ((퐻푀 + 퐻푚)/2) ∗ 퐵) = (
6760,325+1302,925
2
) ∗ 2,6 = 10482,225
A2 = ((퐻 ∗ 퐵)/2) = (
0,4074186992 ∗ 1302,925
2
) = 265,4180043
A3 = (
퐻∗퐵
2
) = (
2,59258130 ∗ 8291,075
2
) = 10747,643
17. CALCULO DE LOS MOMENTOS FLECTORES
Msig = Manterior ± Área de la carga
M(0) = 0 + 0 = 0
M(0-2,6) = 0 + 10482,225 = 10482,225
M(2,6-3,01) = 10482,225 - 265,4180043 = 10747,643
M(3,01-5,6) = 10747,643- 10747,643 = 0
Se grafica los momentos
EJERCICIO #9
Equilibrio Estático son 3 condiciones
Ah = 0
– (2099 Kg) + Av + Bv = 0
Av + Bv = 2099 Kg
18. (47199 Kg*m) + (2099 Kg * 12m) + 12m *Bv = 0
(72387 Kg*m) + 12m *Bv = 0
Bv = 72387 Kg*m = 6032,25 Kg
12m
Se despeja A
Av + Bv = 2099 Kg
Av = 2099 Kg - 6032,25 Kg
Av = -3933,25 Kg
El sentido asumido de A lo cambio a negativo
CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = 0 - 3933,25 = -3933,25
V (0-12) = - 3933,25 + 0 = - 3933,25
V (12) = -3933,25 –2099 +6032,25 = 0
CALCULO DE LAS ÁREAS
A1 = (퐵 ∗ 퐻) = (5 ∗ 3933,25) = 19666,26
A2 = (퐵 ∗ 퐻) = (7 ∗ 3933,25) = 27535,75
CALCULO DE LOS MOMENTOS FLECTORES
Msig = Manterior ± Área de la carga
M(0) = 0 + 0 = 0
M(0-5) = 0 - 19666,26 = -19666,26
M(5) = - 19666,26 + 47199 = 27535,75
M(5-12) = 27535,75 - 27535,75 = 0
Se grafica los momentos
19. EJERCICIO # 10
Equilibrio Estático son 3 condiciones
Ah = 0
Av = 0
- (14099 Kg*m) + Ma = 0
Ma = 14099 Kg*m
CÁLCULOS DE LOS MOMENTOS CORTANTES
Vsig = Vanterior ± Área de la Puntual.
V (0) = 0
V (0-5,2) = 0
V (5,2) = 0
20. CALCULO DE LOS MOMENTOS FLECTORES
Msig = Manterior ± Área de la carga
M(0) = 0 + 14099 = 14099
M(0-5,2) = 14099 + 0 = 14099
M(5,2) = 14099 – 14099 = 0
Se grafica los momentos