The document discusses bending stresses in beams. It begins by outlining simplifying assumptions made in deriving the flexure formula to relate bending stresses to bending moments. These assumptions include plane sections remaining plane and perpendicular to the deformed beam axis. The neutral axis is defined as the axis where longitudinal fibers experience no deformation.
The derivation of the flexure formula is shown. Flexural stresses are proportional to the distance from the neutral axis and bending moment. Procedures for determining stresses at given points, as well as maximum stresses, are provided. Sample problems demonstrate applying the flexure formula and finding maximum stresses for different beam cross sections.
1. This document discusses trial sizing, design, and analysis of short columns under concentric axial loads.
2. The criteria for determining if a column is considered short is based on the slenderness ratio being less than a specified value depending on the column cross section shape.
3. A design example is provided for a 4m long square tied column and circular spiral column both carrying an axial load of 2000 kN. The design includes calculating reinforcement, checking reinforcement ratio, and detailing requirements.
- Surveying involves making field measurements on or near the Earth's surface to determine relative positions of points or establish points. It includes preliminary surveys to collect data, layout surveys to define proposed construction locations, and construction surveys to provide line and grade during construction.
- Control surveys establish horizontal and vertical reference points and lines that preliminary and construction surveys are referenced to. Horizontal control may be tied to grid monuments, property lines, or baselines while vertical control uses benchmark elevations from leveling surveys.
- Route surveys initially layout highways as a series of tangents joined by circular curves. Compound curves consist of two or more joining circular arcs between main tangents turning in the same direction. Reverse curves connect lines through
This document discusses strength of materials concepts related to shear force diagrams (SFD) and bending moment diagrams (BMD) for beams. It defines key terms like shear force, bending moment, and point of contraflexure. It also explains how to draw SFDs and BMDs for different beam types under various loading conditions and the relationships between loading, shear force, and bending moment. Application of the diagrams to reinforcement design is also mentioned.
The document provides lecture material on the equilibrium of coplanar and non-coplanar force systems. It includes objectives, definitions, conditions, procedures, and examples for determining unknown forces and reactions in static equilibrium systems. Graphical methods like the triangle and polygon laws and Lami's theorem are presented. Several practice problems and their solutions are provided to illustrate the application of the equilibrium equations to concurrent, parallel, and non-parallel/non-concurrent force systems. Homework assignments consisting of additional practice problems are also given.
Solution Manual for Structural Analysis 6th SI by Aslam Kassimaliphysicsbook
https://www.unihelp.xyz/solution-manual-structural-analysis-kassimali/
Solution Manual for Structural Analysis - 6th Edition SI Edition
Author(s): Aslam Kassimali
Solution Manual for 6th SI Edition (above Image) is provided officially. It include all chapters of textbook (chapters 2 to 17) plus appendixes B, C, D.
This document provides an introduction to strength of materials, including concepts of stress, strain, Hooke's law, stress-strain relationships, elastic constants, and factors of safety. It defines key terms like stress, strain, elastic limit, modulus of elasticity, and ductile and brittle material behavior. Examples of stress and strain calculations are provided for basic structural elements like rods, bars, and composite structures. The document also covers compound bars, principle of superposition, and effects of temperature changes.
This document discusses bending moments and shear forces in beams. It defines different types of beams such as simply supported beams, cantilever beams, and beams with overhangs. It also defines types of loads like concentrated loads, distributed loads, and couples. It explains how to calculate the shear force and bending moment at any cross-section of a beam and discusses relationships between loads, shear forces and bending moments. It provides examples of drawing shear force and bending moment diagrams. Finally, it discusses bending stresses in beams and bending of beams made of two materials.
The document discusses non-concurrent forces and how to find their resultant. It defines non-concurrent forces as those whose lines of action do not meet at a single point. It provides examples of such forces, like those on a ladder leaning against a wall. The document discusses using graphical and algebraic methods to resolve non-parallel, non-concurrent forces into components. It also addresses calculating the total moment of such force systems to find the resultant force and its location.
1. This document discusses trial sizing, design, and analysis of short columns under concentric axial loads.
2. The criteria for determining if a column is considered short is based on the slenderness ratio being less than a specified value depending on the column cross section shape.
3. A design example is provided for a 4m long square tied column and circular spiral column both carrying an axial load of 2000 kN. The design includes calculating reinforcement, checking reinforcement ratio, and detailing requirements.
- Surveying involves making field measurements on or near the Earth's surface to determine relative positions of points or establish points. It includes preliminary surveys to collect data, layout surveys to define proposed construction locations, and construction surveys to provide line and grade during construction.
- Control surveys establish horizontal and vertical reference points and lines that preliminary and construction surveys are referenced to. Horizontal control may be tied to grid monuments, property lines, or baselines while vertical control uses benchmark elevations from leveling surveys.
- Route surveys initially layout highways as a series of tangents joined by circular curves. Compound curves consist of two or more joining circular arcs between main tangents turning in the same direction. Reverse curves connect lines through
This document discusses strength of materials concepts related to shear force diagrams (SFD) and bending moment diagrams (BMD) for beams. It defines key terms like shear force, bending moment, and point of contraflexure. It also explains how to draw SFDs and BMDs for different beam types under various loading conditions and the relationships between loading, shear force, and bending moment. Application of the diagrams to reinforcement design is also mentioned.
The document provides lecture material on the equilibrium of coplanar and non-coplanar force systems. It includes objectives, definitions, conditions, procedures, and examples for determining unknown forces and reactions in static equilibrium systems. Graphical methods like the triangle and polygon laws and Lami's theorem are presented. Several practice problems and their solutions are provided to illustrate the application of the equilibrium equations to concurrent, parallel, and non-parallel/non-concurrent force systems. Homework assignments consisting of additional practice problems are also given.
Solution Manual for Structural Analysis 6th SI by Aslam Kassimaliphysicsbook
https://www.unihelp.xyz/solution-manual-structural-analysis-kassimali/
Solution Manual for Structural Analysis - 6th Edition SI Edition
Author(s): Aslam Kassimali
Solution Manual for 6th SI Edition (above Image) is provided officially. It include all chapters of textbook (chapters 2 to 17) plus appendixes B, C, D.
This document provides an introduction to strength of materials, including concepts of stress, strain, Hooke's law, stress-strain relationships, elastic constants, and factors of safety. It defines key terms like stress, strain, elastic limit, modulus of elasticity, and ductile and brittle material behavior. Examples of stress and strain calculations are provided for basic structural elements like rods, bars, and composite structures. The document also covers compound bars, principle of superposition, and effects of temperature changes.
This document discusses bending moments and shear forces in beams. It defines different types of beams such as simply supported beams, cantilever beams, and beams with overhangs. It also defines types of loads like concentrated loads, distributed loads, and couples. It explains how to calculate the shear force and bending moment at any cross-section of a beam and discusses relationships between loads, shear forces and bending moments. It provides examples of drawing shear force and bending moment diagrams. Finally, it discusses bending stresses in beams and bending of beams made of two materials.
The document discusses non-concurrent forces and how to find their resultant. It defines non-concurrent forces as those whose lines of action do not meet at a single point. It provides examples of such forces, like those on a ladder leaning against a wall. The document discusses using graphical and algebraic methods to resolve non-parallel, non-concurrent forces into components. It also addresses calculating the total moment of such force systems to find the resultant force and its location.
The document describes Maxwell's diagram, a graphical method for analyzing trusses. It involves drawing force polygons to scale for each joint to determine member forces. The process includes solving for support reactions, drawing a force polygon around the entire truss clockwise, then drawing individual force polygons for each joint with two unknown forces to measure member forces from the diagram. As an example, it provides a sample problem to determine stresses in a Fink truss using this graphical method.
In the preparation for the Geodetic Engineering Licensure Examination, the BSGE students must memorized the fastest possible solution for the LEVELING ADJUSTMENT such as level circuit, level net, etc. using casio fx-991 es plus calculator technique in order to save time during the said examination. note: lec 2 and above wala akong nilagay na solution para hindi makupya techniques ko. just add me on fb para ituro ko sa inyo solution. Kasi itong solution ko wala sa google, youtube, calc tech books at hindi rin itinuro sa review center.
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 provides examples and problems related to static equilibrium of structures. Example 1 shows applying the equations of equilibrium to a weight suspended by a rope over a pulley. Example 2 calculates the forces in ropes supporting a weighted crate. Problem 3.7 asks the minimum force P needed for equilibrium of a crate supported by three ropes meeting at a point.
This document provides an overview of the content covered in the Basic Civil Engineering course. It discusses the following topics:
1. Mechanics of Rigid Bodies and Mechanics of Deformable Bodies, which make up Parts I and II of the course.
2. Concepts in mechanics of solids including resultant and equilibrium of coplanar forces, centroids, moments of inertia, kinetics principles, stresses and strains.
3. Five textbooks recommended as references for the course.
4. Definitions of terms like particle, force, scalar, vector, and rigid body.
5. Methods for resolving forces into components, obtaining the resultant of coplanar forces, and solving mechanics problems
The document discusses three methods for analyzing trusses: the method of joints, method of sections, and graphical (Maxwell's diagram) method. The method of joints involves isolating each joint as a free body diagram and using equilibrium equations to solve for unknown member forces. The method of sections uses equilibrium equations applied to portions of the truss cut off by an imaginary section through several members. Maxwell's diagram method constructs force polygons representing the forces at each joint to graphically determine member forces.
The document discusses methods for determining the deflection and slope of beams, specifically the integration method and Macaulay's bracket method. It provides an example problem solved using the integration method to find the deflection at a point and slope at another point for a simply supported beam with two concentrated loads. The document also introduces Macaulay's bracket method using singularity functions as an alternative to the integration method for problems with multiple loads to avoid lengthy calculations. It provides an initial step for an example problem solved using this method.
This powerpoint presentation deals mainly about bearing stress, its concept and its applications.
Members:
BARIENTOS, Lei Anne
MARTIREZ, Wilbur
MORIONES, Jan Ebenezer
NERI, Laiza Paulene
Sir Romeo Alastre - MEC32/A1
The document provides information about bending stresses and shear stresses in beams. It includes definitions of key terms like bending moment, shear force, radius of gyration, moment of inertia. It describes the assumptions in simple bending theory and concepts of neutral layer and neutral axis. Flexural formulas for pure bending and stress distribution diagrams are presented. Formulas for moment of inertia of various cross sections and moment of resistance are provided. Two example problems are included, one calculating moment of inertia for a rectangular lamina and another finding maximum stress induced in a beam with a non-uniform cross section.
This document provides an overview of the key concepts related to equilibrium of rigid bodies, including:
- Developing free-body diagrams to represent the forces acting on a rigid body and identify unknown support reactions.
- Using the equations of equilibrium (sum of forces equals zero in x and y directions, sum of moments equals zero) to solve for unknown forces and support reactions.
- Analyzing two-force and three-force members, as well as statically determinate and indeterminate structures.
The document discusses the differences between centroid and center of gravity. The centroid is defined as a point about which the entire line, area or volume is assumed to be concentrated, and is related to the distribution of length, area and volume. The center of gravity is defined as the point about which the entire weight of an object is assumed to be concentrated, also known as the center of mass, and is related to the distribution of mass. Examples are provided to illustrate the concepts of centroid and center of gravity.
This document discusses torsion in circular shafts. It defines torque as the turning force applied to a shaft multiplied by the diameter. The angle of twist is the angle of rotation at the surface of the shaft under an applied torque. Shear stress is induced in the shaft under pure torsion. The maximum torque a shaft can transmit depends on its diameter and the allowable shear stress. Assumptions in torsion theory and the polar moment of inertia are also defined. Several examples calculating shaft dimensions, torque, power, and angle of twist are provided. Shaft couplings and keys are also discussed.
Structural Analysis (Solutions) Chapter 9 by WajahatWajahat Ullah
The document provides information about determining displacements of joints in truss structures using the method of virtual work and Castigliano's theorem. It includes the geometry, applied forces, and cross-sectional areas of sample truss problems. The user is asked to determine the vertical displacement of various joints by calculating the internal virtual work of the truss members. Solutions are provided using both the virtual work method and Castigliano's theorem.
Trusses are structures composed of straight members connected at joints. They are used to support roofs and bridges. Trusses can only experience axial loads and moments are excluded. To determine the forces in each member, assumptions are made including that loads only act at end points. The internal forces are calculated using methods like the joints method where equilibrium is applied at each node. For example, in one truss problem the reactions were first calculated and then equilibrium was applied at each node to determine the tensions and compressions in each member.
This document discusses various types of trusses and methods for analyzing truss structures. It begins by describing common types of trusses used in roofs and bridges. It then covers topics such as classifying trusses as simple, compound, or complex, and determining their stability and determinacy. The document introduces analytical methods like the method of joints and method of sections for calculating member forces in statically determinate trusses. It provides examples of applying these methods to solve for unknown member forces.
The document discusses different types of soil settlement including immediate, primary, and secondary consolidation settlements. It provides formulas to calculate settlement, defines concepts like void ratio, compression index, coefficient of consolidation, and overconsolidation ratio. It also includes sample calculations for estimating primary consolidation settlement of a clay layer under a surcharge load based on laboratory consolidation test results and given soil properties.
Mechanics of materials deals with the relationship between external loads on a body and the internal loads within the body. It involves analyzing deformations and stability when subjected to forces. Equilibrium requires balancing all forces and moments on a body. Internal resultant loads include normal forces, shear forces, torques, and bending moments. Average normal stress is calculated as force over cross-sectional area. Average shear stress is calculated as shear force over cross-sectional area. A factor of safety is used to determine allowable loads based on failure loads to account for unknown factors.
1) The documents provide examples of solving for forces in truss members by using free body diagrams and equilibrium equations.
2) The solutions involve drawing FBDs of the trusses or sections of trusses, then writing the ΣF and ΣM equations and solving the systems of equations.
3) The determined forces are then identified as either tension or compression forces in the members.
This document gives the class notes of Unit 5 shear force and bending moment in beams. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
The blocks and ladder problems can be summarized as follows:
1) The documents provide diagrams of blocks on inclined planes or ladders against walls, connected by cords or as single structures.
2) Frictional forces are calculated using coefficients of friction for each surface.
3) Force and moment sums are used to relate normal and frictional forces to weights, angles, and applied forces to determine minimum/maximum values for motion to occur.
B Ending Moments And Shearing Forces In Beams2Amr Hamed
This document discusses bending moments and shear forces in beams. It defines different types of beams such as simply supported beams, cantilever beams, and beams with overhangs. It also defines types of loads like concentrated loads, distributed loads, and couples. It explains how to calculate the shear force and bending moment at any cross-section of a beam and discusses relationships between loads, shear forces and bending moments. It provides examples of drawing shear force and bending moment diagrams. Finally, it discusses bending stresses in beams and bending of beams made of two materials.
This document discusses bending moments and shear forces in beams. It defines different types of beams such as simply supported beams, cantilever beams, and beams with overhangs. It also defines types of loads like concentrated loads, distributed loads, and couples. It explains how to calculate the shear force and bending moment at any cross-section of a beam and discusses relationships between loads, shear forces and bending moments. It provides examples of drawing shear force and bending moment diagrams. Finally, it discusses bending stresses in beams and bending of beams made of two materials.
The document describes Maxwell's diagram, a graphical method for analyzing trusses. It involves drawing force polygons to scale for each joint to determine member forces. The process includes solving for support reactions, drawing a force polygon around the entire truss clockwise, then drawing individual force polygons for each joint with two unknown forces to measure member forces from the diagram. As an example, it provides a sample problem to determine stresses in a Fink truss using this graphical method.
In the preparation for the Geodetic Engineering Licensure Examination, the BSGE students must memorized the fastest possible solution for the LEVELING ADJUSTMENT such as level circuit, level net, etc. using casio fx-991 es plus calculator technique in order to save time during the said examination. note: lec 2 and above wala akong nilagay na solution para hindi makupya techniques ko. just add me on fb para ituro ko sa inyo solution. Kasi itong solution ko wala sa google, youtube, calc tech books at hindi rin itinuro sa review center.
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 provides examples and problems related to static equilibrium of structures. Example 1 shows applying the equations of equilibrium to a weight suspended by a rope over a pulley. Example 2 calculates the forces in ropes supporting a weighted crate. Problem 3.7 asks the minimum force P needed for equilibrium of a crate supported by three ropes meeting at a point.
This document provides an overview of the content covered in the Basic Civil Engineering course. It discusses the following topics:
1. Mechanics of Rigid Bodies and Mechanics of Deformable Bodies, which make up Parts I and II of the course.
2. Concepts in mechanics of solids including resultant and equilibrium of coplanar forces, centroids, moments of inertia, kinetics principles, stresses and strains.
3. Five textbooks recommended as references for the course.
4. Definitions of terms like particle, force, scalar, vector, and rigid body.
5. Methods for resolving forces into components, obtaining the resultant of coplanar forces, and solving mechanics problems
The document discusses three methods for analyzing trusses: the method of joints, method of sections, and graphical (Maxwell's diagram) method. The method of joints involves isolating each joint as a free body diagram and using equilibrium equations to solve for unknown member forces. The method of sections uses equilibrium equations applied to portions of the truss cut off by an imaginary section through several members. Maxwell's diagram method constructs force polygons representing the forces at each joint to graphically determine member forces.
The document discusses methods for determining the deflection and slope of beams, specifically the integration method and Macaulay's bracket method. It provides an example problem solved using the integration method to find the deflection at a point and slope at another point for a simply supported beam with two concentrated loads. The document also introduces Macaulay's bracket method using singularity functions as an alternative to the integration method for problems with multiple loads to avoid lengthy calculations. It provides an initial step for an example problem solved using this method.
This powerpoint presentation deals mainly about bearing stress, its concept and its applications.
Members:
BARIENTOS, Lei Anne
MARTIREZ, Wilbur
MORIONES, Jan Ebenezer
NERI, Laiza Paulene
Sir Romeo Alastre - MEC32/A1
The document provides information about bending stresses and shear stresses in beams. It includes definitions of key terms like bending moment, shear force, radius of gyration, moment of inertia. It describes the assumptions in simple bending theory and concepts of neutral layer and neutral axis. Flexural formulas for pure bending and stress distribution diagrams are presented. Formulas for moment of inertia of various cross sections and moment of resistance are provided. Two example problems are included, one calculating moment of inertia for a rectangular lamina and another finding maximum stress induced in a beam with a non-uniform cross section.
This document provides an overview of the key concepts related to equilibrium of rigid bodies, including:
- Developing free-body diagrams to represent the forces acting on a rigid body and identify unknown support reactions.
- Using the equations of equilibrium (sum of forces equals zero in x and y directions, sum of moments equals zero) to solve for unknown forces and support reactions.
- Analyzing two-force and three-force members, as well as statically determinate and indeterminate structures.
The document discusses the differences between centroid and center of gravity. The centroid is defined as a point about which the entire line, area or volume is assumed to be concentrated, and is related to the distribution of length, area and volume. The center of gravity is defined as the point about which the entire weight of an object is assumed to be concentrated, also known as the center of mass, and is related to the distribution of mass. Examples are provided to illustrate the concepts of centroid and center of gravity.
This document discusses torsion in circular shafts. It defines torque as the turning force applied to a shaft multiplied by the diameter. The angle of twist is the angle of rotation at the surface of the shaft under an applied torque. Shear stress is induced in the shaft under pure torsion. The maximum torque a shaft can transmit depends on its diameter and the allowable shear stress. Assumptions in torsion theory and the polar moment of inertia are also defined. Several examples calculating shaft dimensions, torque, power, and angle of twist are provided. Shaft couplings and keys are also discussed.
Structural Analysis (Solutions) Chapter 9 by WajahatWajahat Ullah
The document provides information about determining displacements of joints in truss structures using the method of virtual work and Castigliano's theorem. It includes the geometry, applied forces, and cross-sectional areas of sample truss problems. The user is asked to determine the vertical displacement of various joints by calculating the internal virtual work of the truss members. Solutions are provided using both the virtual work method and Castigliano's theorem.
Trusses are structures composed of straight members connected at joints. They are used to support roofs and bridges. Trusses can only experience axial loads and moments are excluded. To determine the forces in each member, assumptions are made including that loads only act at end points. The internal forces are calculated using methods like the joints method where equilibrium is applied at each node. For example, in one truss problem the reactions were first calculated and then equilibrium was applied at each node to determine the tensions and compressions in each member.
This document discusses various types of trusses and methods for analyzing truss structures. It begins by describing common types of trusses used in roofs and bridges. It then covers topics such as classifying trusses as simple, compound, or complex, and determining their stability and determinacy. The document introduces analytical methods like the method of joints and method of sections for calculating member forces in statically determinate trusses. It provides examples of applying these methods to solve for unknown member forces.
The document discusses different types of soil settlement including immediate, primary, and secondary consolidation settlements. It provides formulas to calculate settlement, defines concepts like void ratio, compression index, coefficient of consolidation, and overconsolidation ratio. It also includes sample calculations for estimating primary consolidation settlement of a clay layer under a surcharge load based on laboratory consolidation test results and given soil properties.
Mechanics of materials deals with the relationship between external loads on a body and the internal loads within the body. It involves analyzing deformations and stability when subjected to forces. Equilibrium requires balancing all forces and moments on a body. Internal resultant loads include normal forces, shear forces, torques, and bending moments. Average normal stress is calculated as force over cross-sectional area. Average shear stress is calculated as shear force over cross-sectional area. A factor of safety is used to determine allowable loads based on failure loads to account for unknown factors.
1) The documents provide examples of solving for forces in truss members by using free body diagrams and equilibrium equations.
2) The solutions involve drawing FBDs of the trusses or sections of trusses, then writing the ΣF and ΣM equations and solving the systems of equations.
3) The determined forces are then identified as either tension or compression forces in the members.
This document gives the class notes of Unit 5 shear force and bending moment in beams. Subject: Mechanics of materials.
Syllabus contest is as per VTU, Belagavi, India.
Notes Compiled By: Hareesha N Gowda, Assistant Professor, DSCE, Bengaluru-78.
The blocks and ladder problems can be summarized as follows:
1) The documents provide diagrams of blocks on inclined planes or ladders against walls, connected by cords or as single structures.
2) Frictional forces are calculated using coefficients of friction for each surface.
3) Force and moment sums are used to relate normal and frictional forces to weights, angles, and applied forces to determine minimum/maximum values for motion to occur.
B Ending Moments And Shearing Forces In Beams2Amr Hamed
This document discusses bending moments and shear forces in beams. It defines different types of beams such as simply supported beams, cantilever beams, and beams with overhangs. It also defines types of loads like concentrated loads, distributed loads, and couples. It explains how to calculate the shear force and bending moment at any cross-section of a beam and discusses relationships between loads, shear forces and bending moments. It provides examples of drawing shear force and bending moment diagrams. Finally, it discusses bending stresses in beams and bending of beams made of two materials.
This document discusses bending moments and shear forces in beams. It defines different types of beams such as simply supported beams, cantilever beams, and beams with overhangs. It also defines types of loads like concentrated loads, distributed loads, and couples. It explains how to calculate the shear force and bending moment at any cross-section of a beam and discusses relationships between loads, shear forces and bending moments. It provides examples of drawing shear force and bending moment diagrams. Finally, it discusses bending stresses in beams and bending of beams made of two materials.
B Ending Moments And Shearing Forces In Beams2Amr Hamed
This document discusses bending moments and shear forces in beams. It defines different types of beams such as simply supported beams, cantilever beams, and beams with overhangs. It also defines types of loads like concentrated loads, distributed loads, and couples. It explains how to calculate the shear force and bending moment at any cross-section of a beam and discusses relationships between loads, shear forces and bending moments. It provides examples of drawing shear force and bending moment diagrams. Finally, it discusses bending stresses in beams and bending of beams made of two materials.
1. The document discusses bending deformation that occurs in a straight beam made of a homogeneous material when subjected to bending. It describes how the beam's cross-section and longitudinal lines distort under an applied bending moment.
2. It notes that the bottom of the beam stretches and the top compresses, with a neutral surface in between that does not change length. Key assumptions are presented about how the stress deforms the material.
3. Equations are developed relating normal strain and stress to the beam's geometry and applied bending moment, with stress varying linearly from compression at the top to tension at the bottom. The location of the neutral axis is also described.
This document provides an introduction to beams and beam mechanics. It discusses different types of beams and supports, how to calculate beam reactions and internal forces like shear force and bending moment, shear force and bending moment diagrams, theories of bending and deflection, and methods for analyzing statically determinate beams including the direct method, moment area method, and Macaulay's method. The key objectives are determining the internal forces in beams, establishing procedures to calculate shear force and bending moment, and analyzing beam deflection.
This document provides an overview of stresses in beams due to bending. It defines key terms like beam, bending moment, neutral axis, and radius of curvature. It then derives the bending formula that relates stress, bending moment, moment of inertia, and distance from the neutral axis. Several examples are worked through to demonstrate calculating stress in standard beam cross sections given bending moment values. The document concludes with self-assessment exercises for the reader.
Module 4 flexural stresses- theory of bendingAkash Bharti
This document provides an overview of flexural stresses and the theory of simple bending. It discusses key concepts such as:
- Assumptions in the derivation of the bending equation relating bending moment (M) to curvature (1/R) and stress (f)
- Determining the neutral axis where bending stress is zero
- Calculating bending stresses in beams undergoing simple bending and pure bending
- Deriving Bernoulli's bending equation relating stress (f) to distance from the neutral axis (y) and bending moment (M)
- Using the bending equation to locate the neutral axis and design beam cross-sections based on permissible stresses
Worked examples are provided to illustrate calculating load capacity based on beam geometry and material properties
This document discusses transverse shear stresses in beams. It begins by explaining how shear stresses develop within beams subjected to transverse loads and defines the internal shear force V. It then discusses how shear stresses cause shear strains that distort the beam's cross-section. The document proceeds to derive the shear formula that relates the shear stress to the internal shear force V and the beam's geometry. It provides examples of applying the shear formula to compute shear stresses in different beam cross-sections.
This document discusses the derivation of stress equations for curved beams. It outlines the assumptions made, defines relevant terms like the neutral axis and centroidal axis, and derives an equation showing that the stress distribution in a curved beam is hyperbolic. It provides the maximum stress equations and notes they apply for pure bending. Formulas are also provided for calculating stresses in common beam cross sections like rectangular, trapezoidal, T-shaped, and round beams.
This document discusses unsymmetric bending, where the resultant internal moment does not act about one of the principal axes of a cross section. It explains that the moment should first be resolved into components directed along the principal axes. The flexure formula can then be used to determine the normal stress caused by each moment component. Finally, the resultant normal stress at each point can be determined using the principle of superposition. It also provides an equation to determine the orientation of the neutral axis for an unsymmetrically loaded cross section.
This document provides an overview of mechanics of solids unit 2, which covers stresses in beams, deflection of beams, and torsion. It discusses key topics like pure bending, normal and shear stresses in beams, composite beams, deflection equations, and combined bending and torsion. The main assumptions and theories of simple beam bending are explained, including the relationship between bending moment and stress, neutral axis, and modulus of rupture. Beams of uniform strength and variable width/depth beams are also covered.
This document provides an overview of mechanics of solids unit 2, which covers stresses in beams, deflection of beams, and torsion. It discusses key topics like pure bending, normal and shear stresses in beams, composite beams, deflection equations, and combined bending and torsion. The main assumptions and theories of simple bending are explained, including the relationship between bending moment and stress, neutral axis, and modulus of rupture. Beams of uniform strength and varying width/depth are also covered.
Shear Force And Bending Moment In Beams22Amr Hamed
The document discusses concepts related to shear force and bending moment in beams including:
1) Definitions of bending, beams, planar bending, and different types of beams.
2) Methods for simplifying beams, loads, and supports for calculation.
3) Concepts of internal forces including shear force, bending moment, and sign conventions.
4) Relations between shear force, bending moment, and distributed loads.
5) Methods for plotting shear force and bending moment diagrams including using superposition.
6) Application to planar rigid frames.
The document discusses concepts related to shear force and bending moment in beams, including:
- Definitions of bending, beams, planar bending, and types of beams including simple, cantilever, and overhanging beams.
- Calculation sketches simplify beams, loads, and supports for analysis.
- Internal forces in bending include shear force and bending moment. Relations and diagrams relate these to external loads.
- Equations define shear force and bending moment at each beam section. Diagrams illustrate variations along the beam.
The document discusses determining internal forces in structural members using statics. It provides objectives of showing how to use the method of sections to find internal loadings and formulate equations to describe shear and moment throughout a member. Key steps are outlined, including making a section cut, drawing a free body diagram, and applying equilibrium equations to solve for the normal force, shear force and bending moment. Sign conventions are also defined. Shear and moment diagrams are then explained as plots of these internal forces along the length of a beam, with examples provided to demonstrate the full procedure.
This document provides information about bending moment in a presentation on pre-stress concrete design. It defines bending moment as a measure of bending forces acting on a beam, measured in terms of force and distance. Shear and moment diagrams can show the bending moment and shear force functions along a beam. Bending moment at a section is the sum of moments of all forces on one side and is represented in a bending moment diagram. Positive bending moment results in tension on the bottom fibers while negative bending moment results in compression. Bending moment units are Newton-meters or foot-pounds. Assumptions of simple bending theory and the differences between shear force and bending moment are also outlined.
This document provides information about bending moment in a presentation on pre-stress concrete design. It defines bending moment as a measure of bending forces acting on a beam, measured in terms of force and distance. Shear and moment diagrams can show the bending moment and shear force functions along a beam. Bending moment at a section is the sum of moments of all forces on one side and can be represented in a bending moment diagram. Positive bending moment results in tension on the bottom fibers while negative bending moment results in compression. Bending moment is measured in units of Newton-meters or foot-pounds. Simple bending theory makes assumptions about beam properties and behavior.
1) The document discusses the finite element method for analyzing beams. It covers elementary beam theory, defining the beam element and degrees of freedom, deriving the beam stiffness matrix, and accounting for distributed loads.
2) Distributed loads on a beam can be represented by equivalent nodal forces and moments chosen to produce the same strain energy as the actual distributed load.
3) The work equivalence method is used to determine equivalent nodal loads, ensuring the work done by the distributed and equivalent nodal loads is equal for any displacement field.
This document discusses shear force and bending moment in beams. It defines different types of beams, loads, and supports. Equations for calculating shear force and bending moment are presented for various beam configurations under different loading conditions, including cantilever beams with point loads and uniform loads, and simply supported beams with point and uniform loads. Diagrams illustrating the variation of shear force and bending moment along beams are shown as examples.
The document discusses shear force and bending moment diagrams. It defines shear force and bending moment, explaining that shear force acts perpendicular to the beam's axis while bending moment acts to bend the beam. It outlines the procedure to determine shear force and bending moment diagrams: (1) calculate support reactions, (2) divide the beam into segments based on loading, (3) draw free body diagrams and calculate expressions for each segment. As an example, it analyzes a simply supported beam with two loads to derive the shear force and bending moment expressions and diagrams.
This document contains an assignment for a Strength of Materials course. It includes 4 questions related to analyzing stresses and torques in circular shafts and bars. Question 1 involves determining the peak shearing stress and tensile stress in a steel shaft loaded by a twisting moment at its midpoint. Question 2 involves determining the reactive torques at the fixed ends of a circular shaft loaded by couples. Question 3 involves determining the maximum shearing stress and relative angle of twist in a solid circular shaft delivering power to machines at its ends. Question 4 involves determining the inside and outside diameters of a hollow steel shaft transmitting a given torque such that the angle of twist and shearing stress do not exceed allowable limits.
This document contains instructions for an assignment on Strength of Materials (SOM). It includes 5 questions related to Course Outcome 1 (CO-1) on topics like axial elongation of tapered bars, error in using mean diameter to calculate Young's modulus, stresses under biaxial loading, principal stresses and planes, and material properties like Young's modulus, Poisson's ratio, modulus of rigidity and bulk modulus based on experimental data for extension and lateral contraction of a loaded bar. The assignment is due on April 10, 2020 and carries a maximum of 50 marks.
This document contains instructions for an assignment on strength of materials (SOM) given by the instructor Suryakant Kumar. It has 4 questions related to course outcome 2 assessing beams with different loading and support conditions to draw shear force and bending moment diagrams, and determine reactions and shear stress distribution. Students are asked to attempt all questions on plain A4 paper with diagrams drawn using pencils by the due date of April 14, 2020, for a maximum of 50 marks.
This document contains instructions for an assignment on strength of materials (SOM). It lists 5 questions about calculating deflections (EIδ) in beams under different loading conditions. The questions cover calculating deflections at specific points for an overhanging beam, simply supported beam, and cantilever beam with distributed loads. Students are instructed to show their work and use diagrams to solve the problems.
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The document appears to be a scanned receipt from a grocery store listing various food and household items purchased totaling $123.45. It includes the store name, date, time of purchase, payment method (credit card), and a thank you message at the bottom.
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Stresses in beams
1. Strength of Materials
Unit-2
Stresses in Beams
SURYAKANT KUMAR
Assistant Professor
Mechanical Engineering Department
Gaya College of Engineering, Gaya
2. 1.1 Bending Stress
a. Simplifying assumptions
The stresses caused by the bending moment are known as
bending stress, or flexure stresses. The relationship between
these stresses and the bending moment is called the flexure
formula.
In deriving the
flexure formula,
make the following
assumptions:
The beam has an
axial plane of
symmetry, which we
take to be the xy-
plane (see Fig. 5.1).
Figure 5.1 Symmetrical beam with loads
lying in the plane of symmetry.
3. The applied loads (such as F1,F2 and F3 in Fig.5.1) lie in the
plane of the symmetry and are perpendicular to the axis of the
beam (the x-axis).The axis of the beam bends but does not
stretch ( the axis lies some where in the plane of symmetry; its
location will be determined later).
Plane sections of
the beam remain
plane (do not
warp) and
perpendicular
to the deformed
axis of the beam.
Change in the
cross-sectional
dimensions of the
beam are
negligible.
Figure 5.1 Symmetrical beam
4. Because the shear stresses caused by the vertical shear force will
distort (warp) an originally plane section, we are limiting our
discussion here to the deformations caused by the bending
moment alone.
the deformations due to the vertical shear force are negligible in
the slender beams compared to the deformations caused by
bending .
5. The above assumptions lead us to the following conclusion:
Each cross section of the beam rotates as a rigid entity about a
line called the neutral axis of the cross section.
The neutral axis passes through the axis of the beam and is
perpendicular to the plane of symmetry, as shown in Fig. 5.1.
The xz-plane that contains the neutral axes of all the cross
sections is known as the neutral surface of the beam.
6. Figure 5.2 Deformation of an infinitesimal beam segment.
b. Compatibility
The neutral surface becomes curved upon deformation, as
indicated in Fig.5.2.
The longitudinal fibers lying on the neutral surface are
undeformed, whereas the fibers above the surface are compressed
and the fibers below are stretched.
7. d ab
The fiber form are arc a' b' of radius (-y) , subtended by the
angle dθ, its deformedlength is a ' b ' yd
The original length of this fiber is ab dx d. The normal
strain of the fiber
a'b' ab
yd d
y
ρ
8. Assuming that the stress is less than the proportional limit of the
material we can obtain the normal stress in the fiber ab from
Hook's law:
(5.1) E
E
y
Equation (5.1) shown that the normal stress of a longitudinal
fiber is proportional to the distance y of the fiber from the neutral
surface.
The negative sign indicates that
positive bending moment causes
compressive stress when y is
positive (fiber above the neutral
surface) and tensile stress when y
is negative (fiber below the neutral
surface).
9. c. Equilibrium
Figure 5.3 shows the normal
force acting on the
infinitesimal area dA of the
cross section is dP = dA.
Substituting = - (E/ )y,
(a)
Where y is the distance of dA
from the neutral axis (NA).
The resultant of the normal
stress distribution over the
cross section must be equal to
the bending moment M acting
about the neutral axis (z-axis).
Figure 5.3 Calculating the
resultant of the Normal
stress acting on the cross
section. Resultant is a
couple Equal to the internal
bending moment of M.
dP
E
ydA
10. In other work, A
ydp M
where the integral is taken over
the entire cross-sectional area A
the resultant axial force and the
resultant bending moment about
the y-axis must be zero; that is,
anddP 0
A A
zdP 0
These three equilibrium equations
are developed in detail below.
Resultant Axial Force Must Vanish The condition for zero
axial force is
E
A
dp
A
ydA 0
11. Because E / ≠ 0, this equation can be satisfied only if
(b)
The integral in Eq.(b) is the first moment of the cross-sectional
area about the neutral axis. It can be zero only if the neutral axis
passes through centroid C of the cross-sectional area.
Resultant Moment About y-Axis Must Vanish
This condition is
A
ydA 0
(c)
The integral in Eq.(b) is the
product of inertia of the cross-
sectional area.
AA
zdP
E
zydA 0
12. Resultant Moment About the Neutral Axis Must Equal M
Equating the resultant moment about the z-axis to M
Recognizing that is the moment of inertia of thecross-
sectional area about the neutral axis ( the z-axis), we obtain the
moment curvature relationship
(5.2a)
A convenient form of this equation is
(5.2b)
A A
y dA M ydp 2E
A
y dA I2
M
EI
EI
1
M
13. (5.4a)
where c is the distance from the
neutral axis to the outermost point
of the cross section.
d. Flexure formula; section modulus
Substituting the expression for 1/ from Eq.(5.2) into Eq. (5.1),
we get the flexure formula :
(5.3)
Note that a positive bending moment M causes negative
(compressive) stress above the neutral axis and positive ( tensile)
stress below the neutral axis
The maximum value of bending
stress without regard to its sign is
given by
I
My
I
max
Mmax c
14. Equation (5.4a) is frequently written in the form
(5.4b)
S
max
Mmax
where S = I / c is called the section modulus of the beam. The
dimension of S is [L3], so that its units are in.3, mm3, and so on.
The formulas for the section moduli of common cross sections are
given in Fig. 5.4.
Figure 5.4
Section moduli
of simple cross
sectional
shapes.
15. The section
moduli of
standard
structural
shapes are
listed in
various
handbooks; an
abbreviated list
is given in
Appendix B.
Figure 5.4 Section
moduli of
simple cross
sectional
shapes.
16. e. Procedures for determining bending stresses
Stress at a Given Point
• Use the method of sections to determine the bending moment M
at the cross section containing the given point.
• Determine the location of the neutral axis.
• Compute the moment of inertia I of the cross- sectional area
about the neutral axis. ( If the beam is standard structural shape,
its cross- sectional properties are listed in Appendix B. P501)
• Determine the y-coordinate of the given point. Note that y is
positive if the point lies above the neutral axis and negative if it
lies below the neutral axis.
• Compute the bending stress from σ = -My / I. If correct sign
are used for M and y, the stress will also have the correct sign
(tension positive compression negative).
17. Maximum Bending Stress: Symmetric Cross Section
If the neutral axis is an axis of symmetric of the cross section, the
maximum tensile and compression bending stresses are equal in
magnitude and occur at the section of the largest bending
moment. The following procedure is recommended for
determining the maximum bending stress in a prismatic beam:
• Draw the bending moment diagram by one of the methods
described in SFD & BMD Unit. Identify the bending moment
Mmax that has the largest magnitude (disregard the sign)
• Compute the moment of inertia I of the cross- sectional area
about the neutral axis. ( If the beam is a standard structural shape,
its cross- sectional properties are listed in Appendix B.)
• Calculate the maximum bending stress from σmax = [Mmax]c / I,
where c is the distance from the neutral axis to the top or bottom
of the cross section .
18. Maximum Tensile and Compressive Bending Stresses:
Unsymmetrical Cross Section
If the neutral axis is not an axis of symmetry of the cross
section, the maximum tensile and compressive bending
stresses may occur at different sections.
• Draw the bending moment diagram. Identify the largest
positive and negative bending moments.
• Determine the location of the neutral axis and record the
distances ctop and cbot from the neutral axis to the top and
bottom of the cross section.
• Compute the moment of inertia I of the cross section about
the neutral axis.
19. • Calculate the bending stresses at the top and bottom of the
cross section where the largest positive bending moment
occurs from σ = -My / I.
At the top of the cross section, where y = ctop,we obtain σtop =
-Mctop/ I.
At the bottom of the cross section, we have y = - cbot, so that
σbot = Mcbop/I.
• Repeat the calculations for the cross section that carries the
largest negative bending moment.
• Inspect the four stresses thus computed to determine the
largest tensile (positive) and compressive (negative) bending
stresses in the beam.
20. Note on Units
the units of terms in the flexure formula σ = -My / I.
In the U.S. Customary system, M is often measured in pound-feet
and the cross sectional properties in inches, It is recommended
that you convert M into lb·in. and compute σ in lb/in.2 (psi).
Thus, the units in the flexure formula become
Iin.4
In SI system, M is usually expressed in N · m, whereas thecross-
sectional dimensions are in mm. To obtain σ in N/m2 (Pa), he
cross sectional properties must be converted to meters, so that the
units in the flexure equation are
lb/ in.2
M lb in.yin.
N/ m 2
M Nmym
I m4
21. Sample Problem 1.1
The simply supported beam in Fig. (a) has a rectangular cross
section 120 mm wide and 200 mm high. (1) Compute the
maximum bending stress in the beam. (2) Sketch the bending
stress distribution over the cross section on which the maximum
bending stress occurs. (3) Compute the bending stress at a point
on section B that is 25 mm below the top of the beam.
22. Solution
Preliminary Calculations
The shear force and bending moment
diagrams. M max = +16 kN·m,
occurring at D. The neutral axis (NA)
is an axis of symmetry of the cross
section as shown in Fig. (a). The
moment of inertia of the cross section
about the neutral axis is
12 12
bh3
0.120.23
6
m4
I 80010
and the distance c
between the neutral axis
and the top (or bottom)
of the cross section is c
= 100 mm = 0.1 m.
23. Part 1
The maximum bending stress in the beam on the cross section that
carries the largest bending moment, which is the section at D.
Answer
Part 2
The stress distribution on the cross section at D is shown in Fig. (d)
(i) The bending stress varies linearly with distance from the neutral
axis;
(ii) Because M max is positive, the top half of the cross section is in
compression and the bottom half is in tension.
(iii)Due to symmetry of the cross section about the neutral axis, the
maximum tensile and compressive stresses are equal in
magnitude.
6
max 20.010 Pa 20.0MPa
Mmaxc
1610 0.13
I 80.0106
24. Part 3
From Fig. (c) we see that the bending moment at section B is M = +
9.28 kN·m. The y-coordinate of the point that lies 25 mm below the
top of the beam is y = 100 -25 = 75 mm = 0.075 m.
Answer
The negative sign indicates that this bending stress is compressive,
which is expected because the bending moment is positive and the
point of interest lie above the neutral axis.
0.075 10 6
3
9.28
8.7010 Pa8.70MPa
My
I 80.0106
25. Sample Problem 1.2
The simply supported beam in Fig. (a) has the T-shaped cross
section shown. Determine the values and locations of the maximum
tensile and compressive bending stresses.
26. Solution
Preliminary Calculations
Find the largest positive
and negative bending
moment. The results are
shown in Fig. (a)-(c). From
Fig.(c), the largest positive
and negative bending
moment are 3200 lb·ft and
4000 lb·ft respectively.
27. As shown in Fig.(d), the cross section to be composed of the two
rectangles with areas A1 = 0.8(8) = 6.4 in.2 and A2 = 0.8 (6) = 4.8
Compute the moment of inertia I of the cross-sectional area about
is the moment of inertia of a rectangle about its
own centroidal axis Thus,
422
1212
87.49in.
60.83
4.8 8.45.886
0.883
6.445.886I
in.2 . The centroidal coordinates of the areas are y1 4in. and y2 8.4in.,
, measured from the bottom of the cross section. The coordinate y
of the centroid C of the cross section is
y
A1 y1 A2 y2
6.44 4.88.4 5.886in.
A1 A2 6.4 4.8
2
Ai yi y,the neutral axis. Using the parallel-axis theorem, I = Ii
i i i
where I b h3
/12
28. Maximum Bending stresses
The distances from the neutral axis to the top and the bottom of
the cross section arectop 8.8 y 8.85886 2.914in. and cbot y 5.886in.,
as shown in Fig.(c). Because these distances are different, we
must investigate stresses at two locations: at x = 4 ft (where the
largest positive bending moment occurs) and at x = 10 ft (where
the largest negative bending moment occurs).
Stresses at x = 4 ft The bending moment at this section is M =
+3200 lb.ft causing compression above the neutral axis and
tension below the axis. The resulting bending stresses at the top
and bottom of the cross section are
top
Mctop
I
bot
3200122.914 1279psi
87.49
3200125.886 2580psi
87.49
I
Mcbot
29. Stresses at x = 10 ft The bending moment at this section is M = -
4000lb.ft, resulting in tension the neutral axis and compression
below the neutral axis. The corresponding bending stresses at the
extremities of the cross section are
Inspecting the above results, we conclude that the maximum tensile
and compressive stresses in the beam are
( σ T )max = 2580 psi ( bottom of the section at x = 4 ft )
( σ c )max = 3230 psi ( bottom of the section at x = 10 ft )
top
4000122.914 1599psi
Mctop
bot
I 87.49
87.49
400012 5.886 3230psi
I
Mcbot
30. Sample Problem 1.3
The cantilever beam in
Fig. (a) is composed of
two segments with
rectangular cross
sections. The width of
the each section is 2 in.,
but the depths are
different, as shown in the
figure. Determine the
maximum bending stress
in the beam.
31. Solution
Because the cross section
of the beam is not
constant, the maximum
stress occurs either at the
section just to the left of B
(MB = - 8000 lb.ft) or at
the section at D (MD = -
16000 lb.ft). the section
moduli of the two
segments are
6
242
5.333in.3bh 2
AB
AB S
66
12.0in.3
6
bh 2
BD 262
BD S
32. From Eq. (5.4b) the maximum bending stresses on the two cross
sections of the interest are
Comparing the above values, we find that the maximum bending
stress in the beam is
σmax = 18000 psi (on the cross section just to the left of B)
Answer
This is an example where the maximum bending stress occurs
on a cross section at the bending moment is not maximum.
AB
B
S 5.333max
MB
800012
18000psi
BD
D
S 12.0max
MD
1600012
16000psi
33. Sample Problem 1.4
The wide- flange section W 14×30 is use as a cantilever beam, as
shown in Fig.(a). Find the maximum bending stress in the beam.
Solution
The largest bending moment is
∣Mm a x ∣= 15000 lb · ft acting
just to the left of section B.
From the tables in Appendix B,
we find that the section
modulus of a W14×30 (P520)
section is S = 42.0 in.3.
Therefore, the maximum
bending stress in the beam is
S 42.0
max
15000 12
4290psi
Mmax
34. 1.2 Economic Sections
The portions of a located near the neutral surface are
understressed compared with at the top or bottom. Therefore,
beams with certain cross- sectional shape ( including a
rectangle and circle) utilize the material inefficiently because
much of the cross section contributes little to resisting the
bending moment.
Consider, for example, in Fig. 5.5(a) The
section modulus has increased to S = bh2/6 =
2(6)2/6 = 12 in.3. If working stress is σ w =
18 ksi, the maximum safe bending moment
for the beam is M =σw · S =18 (12) = 216
kip·in.
Figure 5.5 Different ways to distribute the 12-in.2 cross-
sectional area in (a) without changing the depth.
35. In Fig. 5.5(b), we have rearranged the area of the cross section but
kept the same overall depth. It can be shown that the section that
the section modulus has increased to S = 25.3 in.3(the parallel-
axis theorem). Thus, the new maximum allowable moment is M
= 18 (25.3) = 455 kip·in., which is more than twice the allowable
moment for the rectangular section of the same area.
The section in Fig. 5.5(b) is not practical because its two parts,
called the flanges. As in Fig. 5.5(c). The vertical connecting piece
is known as the web of the beam. The web functions as the main
shear-carrying component of the beam.
36. a. Standard structural shapes
Figure 5.5 (c) is similar to a wide-flange beam, referred to as a
W-shape. Another “slimmer”version of the shape is the I-beam
(referred to as an S-shape) shown in Fig. 5.5(d). The I-beam
preceded the wide- flange beam, but because it is not as
efficient, it has largely been replaced by the wide- flange beam.
37. Properties of W-and S-shapes are given in Appendix B.
in SI units, the designation W610×140 indicates a wide-flange
beam with a nominal depth of 610mm and a nominal mass per
unit length of 140 kg/m. The tables in Appendix B indicates the
actual depth of the beam is 617 mm and the actual mass is 140.1
kg/m.
In U.S. Customary units, a W36×300 is a wide-flange beam with
a nominal depth 36 in. that weighs 300 lb/ft. The actual depth of
this section is 36.74 in.
Referring to Appendix B, in addition to listing the dimensions,
tables of structural shapes give properties of the cross-sectional
area, such as moment of inertia (I), section modulus (S), and
radius of gyration (r)4 for each principal axis of the area.
38. When a structural section is selected to be used as a beam. The
section modulus must be equal to or greater than section
modulus determined by the flexure equation; that is,
(5.5)
w
the section modulus of the selected beam must be equal to or
greater than the ratio of the bending moment to the working
stress.
If a beam is very slender (large L/r), it may fail by lateral
bucking before the working stress is reached. I-beams are
particularly vulnerable to lateral bucking because of their low
torsional rigidity and small moment of inertia about the axis
parallel to the web.
S
Mmax
39. b. Procedure for selecting standard shapes
A design engineer is often required to select the lightest
standard structural shape ( such as a W-shape) that can carry a
given loading in addition to the weight of the beam. Following
is an outline of the selection process;
.Neglecting the weight of the beam, draw the bending moment
diagram to find the largest bending moment Mmax..
.Determine the minimum allowable section modulus from Smin =
|Mmax.︱/σw , is the working stress.
.Choose the lightest shape from the list of structural shapes (such
as a Appendix B) for which S≥Smin and note its weight.
.Calculate the maximum bending stress σmax in the selected
beam caused by the prescribed loading plus the weight of the
beam. Ifσmax≤σw, the selection is finished. Otherwise, the
second-lightest shape with S≥Smin must be considered and the
maximum bending stress recalculated. The process must be
repeated unit a satisfactory shape is found.
40. Sample Problem 1.5
What is the lightest W-shape beam that will support the 45-kN
load shown in Fig. (a) without exceeding a bending stress of 120
MPa?Determine the actual bending stress in the beam.
Solution
Finding the reactions
shown in Fig.(a), and
sketch the shear force
and bending moment
diagrams in Figs. (b)
and (c).
41. The minimum bending acceptable section modulus that can carry
this moment is
3M max
Smin
w
60 103
500 106
m3
500 103
mm
120 106
Referring to the table of Properties of W-shape (Appendix B SI
Unit) and find that the following are the lightest beams in each size
group that satisfy the requirement S≥Smin: (P508)
Section S(mm3) Mass(kg/m)
W200×52 512×103 52.3
W250×45 534×103 44.9
W310×39 549×103 38.7
Our first choice is the W310×39
section with S = 549×10-6 m3.
The reason is that although
the lightest beam is the
cheapest on the basis of the
weight alone, headroom
clearances frequently require
a beam with less depth than
the lightest one.
42. The weight of the beam for the W310×39 section is
wo = (38.7 kg/m)×(9.81 m/s2) = 380 N/m = 0.380 kN/m
From (d) shows the beam supporting both the 45-kN load and
the weight of the beam. The maximum bending moment is
found to be Mmax = 61.52 kN·m, again occurring under the
concentrated load.
Therefore, the maximum
bending stress in the
selected beam is
6
max 112.110 pa 112.1MPa
61.52 103
M max
S 549106
Because this stress is less than the allowable stress of 120 MPa,
the lightest W-shape that can safely support the 45-kN load is
(with σmax =112.1MPa)W310×39 Answer
43. 1.4 Shear Stress in Beams
a. Analysis of flexure action
In Fig. 5.6, The separate layers would slide past one another, and
the total bending strength of the beam would be the sum of the
strength of the individual layers. Such a built-up beam would be
considerably weaker than a solid beam of equivalent dimensions.
From the above observation, we conclude that the horizontal
layers in a solid beam are prevented from sliding by shear stresses
that act between the layers.
Figure 5.6 Bending of a layered beam with no adhesive
between the layers.
44. In Fig. 5.7. We isolate the shaded portion of the beam by
using two cutting planes: a vertical cut along section 1
and horizontal cut located at the distance y’above the
neutral axis.
Figure 5.7 Equilibrium of the shaded portion of the beam
requires a longitudinal shear force F = P, where
P is the resultant of the normal stress acting on
area A’ of section (1).
45. Calculate P using Fig. 5.8. The axial force acting on the area
element dA of the cross section is dP = σdA.
If M is the bending moment acting at section 1 of the beam,
the bending stress is given by Eq. (5.3): σ = - My/I, where y is
the distance of the element from the neutral axis, and I is the
moment of inertia of the entire cross-sectional area of the
beam about the neutral axis.
Figure 5.8
Calculating the
resultant force
of the normal
stress over a
portion of the
cross-sectional
area.
46. Integrating over the area A’, we get
(5.6)
Where
(5.7a)
is the first moment of area A’ about the neutral axis. The negative
sign in Eq. (5.6) indicates that positive M results in forces P and F
that are directed opposite to those shown in Fig. 5.7.
dP
My
dA
I
Q A`
ydA
A`A`
ydA
MQ
II
P dp
M
47. (5.7b).Q A`y
Denoting the distance between the neutral axis and centroid
C` of the area A’by y
, we can write Eq. (5.7) as
In Eqs. (5.7b), Q represents the first
moment of the cross-sectional area
that lies above y’. Because the first
moment of the total cross-sectional
area about the neutral axis is zero,
that first moment of the area below
y’is - Q. Therefore, the magnitude
of Q can be computed by using the
area either above or below y’,
whichever is more convenient.
y
48. Figure 5.9 Variation of the first moment Q of area A’about
the neutral axis for a rectangular cross section.
The maximum value of Q occurs at the neutral axis where
y’= 0. It follows that horizontal shear force F is largest on
the neutral surface. The variation of Q with y’for a
rectangular cross section is illustrated in Fig. 5.9.
49. b. Horizontal shear stress
Consider Fig. 5.10. A horizontal plane located a distance y’above
the neutral axis of the cross section. If the bending moment at
section1 of the beam is M, the resultant force acting on face 1 of the
body is given by Eq. (5.6):
Figure 5.10 Determining the longitudinal shear stress from the
free-body diagram of a beam element.
P M
Q
I
50. The bending moment acting at section 2 is M+dM, where dM is
the infinitesimal change in M over the distance dx. Therefore,
the resultant normal force acting on face 2 of the body is
(a)
I
p dP M dMQ
I
I
P dPP M dMQ
M
Q
Σ F = 0 : (P +dP)-P + τ b dx = 0
I
P M
Q
I
Equilibrium can exist only if there is an equal and opposite shear
force dF acting on the horizontal surface. If we letτbe the
average shear stress acting on the horizontal surface, its
resultant is dF = τbdx. Where b is the width of the crosssection
at y = y`, as shown in Fig. 5.10. The equilibrium requirement for
the horizontal forces is
dM
Q
51. Substituting for(P+dP) - P from Eq. (a), we get
(b)
Recalling the relationship V = dM/dx between the shear force and
the bending moment we obtain for the average horizontal shear
stress τ
(5.8)
I
dM
Q
bdx 0
dx Ib
dM Q
Ib
VQ
52. c. Vertical shear stress
shear stress is always accompanied by a complementary shear
stress of equal magnitude, the two stresses acting on mutually
perpendicular plane.
Ib
Eq. (5.8)
VQ
( a plane parallel to the neutral surface ).A
Figure 5.11 The vertical stress τ ’ acting at a point on a cross
section equals the longitudinal shear stress τ acting at
the same point.
53. course, the shear force V;.
In a beam, the complementary stress τ ’ is a vertical shear stress
that acts on the cross section of the beam, as illustrated in Fig.
5.11 (a). Because τ = τ ’ , Eq.(5.8) can be used to compute the
vertical as well as the horizontal shear stress at a point in a
beam.
The resultant of the vertical shear stress on
the cross-sectional area A of the beam is,of
A
V dA
To prove that τ = τ ` , consider Fig. 5.11(b).
The horizontal and vertical forces are τdxdz
and τ’dydz, respectively. These forces from
two couples of opposite sense. For rotational
equilibrium, the magnitudes of the couples
must be equal; that is, (τdxdz) dy =
(τ`dydz) dx, which yields τ = τ ’ .
54. d. Discussion and limitations of the shear stress formula
• The shear stress formula τ = VQ/(Ib) predicts that the largest
shear stress in a prismatic beam occurs at the cross section
that carries the largest vertical shear force V.
• The location ( the value of y’) of the maximum shear stress
within that section is determined by the ratio Q/b. Because Q
is always maximum at y’= 0, the neutral axis is usually a
candidate for the location of the maximum shear stress.
• However, If the width b at the neutral axis is larger than at
other parts of the cross section, it is necessary to compute τ
at two or more values of y’before its maximum value can be
determined.
55. τ should be considered at the average shear stress. This
restriction is necessary because the variation of the shear
stress across the width b the cross section is often unknown.
Equation (5.8) is sufficiently accurate for rectangular cross
sections and for cross sections that are composed of
rectangles, such as W and S-shapes.
Let us consider as an example the
circular cross section in Fig. 5.12.
Figure 5.12 Shear stress
distribution along a
horizontal line of a
circular cross section.Ib
VQ
Ib
VQ
When deriving the shear stress formula, Eq. (5.8),
56. It can be shown that the shear
stress at the periphery of the
section must be tangent to the
boundary, as shown in the figure.
The direction of shear stresses at
interior points is unknown, except
at the centerline, where the stress is
vertical due to symmetry. To
obtain an estimate of the maximum
shear stress, the stresses are
assumed to be directed toward a
common center B, as shown.
Figure 5.12 Shear stress
distribution along a
horizontal line of a
circular cross section.
For other cross- sectional shapes, however, the formula for τ
must be applied with caution. Let us consider as an example
the circular cross section in Fig. 5.12.
57. The vertical components of these shear stresses are assumed
to be uniform across the width of the section and are
computed from Eq. (5.8). Under this assumption, the shear
stress at the neutral axis is 1.333V/ (πr2 ). (4/3)(V/ πr2 )
A more elaborate analysis shows that the shear stress actually
varies from 1.23 V/ (πr2 ) at the edges to 1.38 V/ (πr2 ) at the
center.
Shear stress, like normal stress, exhibits stress concentrations
near shape corners, fillets and holes in the cross section. The
junction between the web and the flange of a W-shape is also
an area of stress concentration.
58. a. Procedure for analysis of shear stress:
.Use equilibrium analysis to determine the vertical shear force
V acting on the cross section containing the specified point (
the construction of a shear force diagram is usually a good
idea).
.Locate the neutral axis and compute the moment of inertia I
of the cross- sectional area about the neutral axis (If the
beam is a standard structural shape, its cross- sectional
properties are listed in Appendix B.)(Andrew Pytel)
.Compute the first moment Q of the cross- sectional area that
lies above (or below)the specified point.
.Calculate the shear stress from τ = VQ/(Ib), where b is the
width of the cross section at the specified point.
59. The maximum shear stressτmax on a given cross section
occurs where Q/b is largest.
If the width b is constant, then τm a x occurs at theneutral
axis because that is where Q has its maxmum value.
If b is not constant, it is necessary to compute the shear stress
at more than one point in order to determine its maximum
value.
In the U.S. Customary system,
Im4
bm
N /m2
VNQm3
In the SI system,
3
2
Iin.4
bin.
VlbQin. lb/in .
60. Sample Problem 1.6
The simply supported wood beam in Fig.(a) is fabricated by gluing
together three 160-mm by 80-mm plans as shown. Calculate the
maximum shear stress in (1) the glue; and (2) the wood.
61. Solution
From the shear force diagram in Fig. (b), the maximum shear force
in the beam is Vmax = 24 kN, occurring at the supports. The moment
of inertia of the cross-sectional area of the beam about the neutral
axis is
12 12
bh3
1602403
184.32106
mm4
184.32106
m4
I
1.024103
m3
Part 1
The shear stress is the glue corresponds to
the horizontal shear stress. Its maximum
value can be computed from Eq. (5.8):
τm a x = Vmax Q/(Ib), where Q is the first
moment of the area A’ shown in Fig.(c);
that is,
Q A`y` 1608080 1.024 106
62. Therefore, the shear stress in the glue, which occurs over either
support, is
Answer
0.160
3
max 6
3
184.3210
2410 1.02410 max
Ib
V Q
8.33103
Pa 8.33kPa
938kPa
2 A 2 0.1600.240
3
max 93810 Pa
3 24103
3 Vmax
Part 2
Because the cross section is
rectangular, the maximum shear
stress in the wood can be calculated
from Eq. (5.9):
63. The same result can be obtained
from Eq. (5.8), where now A’ is
the area above the neutral axis,
as indicated in Fig. (d). The first
moment of this area about the
neutral axis is
1.152103
m3
Q A`y` 16012060 1.152 106
mm3
Equation (5.8)this becomes
0.160
max 6
3 3
2410 1.15210
Ib 184.3210
Vma
xQ
938 103
Pa 938kPa
which agrees with the previous result.