The document analyzes the failure of a bike pedal spindle. It determines the location of fatigue crack origin and calculates various forces, moments, and stresses acting on the spindle. The largest stress was found to be torque stress. This agrees with observations that the spindle failed in a perpendicular direction to the torque stress. Based on the images and fracture mechanics analysis, the critical crack length and size of the plastic zone were estimated. The maximum pedal force required to cause final fracture was also calculated based on the material's fracture toughness.
This document discusses transformation of stresses and strains when an element is rotated. It defines normal and shear stresses, and shows how to calculate them based on forces and geometry. It then demonstrates how to use Mohr's circle to determine maximum and minimum stresses, and stresses and shear stresses at any angle of rotation. As an example, it also shows calculations for stresses in a thin-walled pressure vessel where shear stress is zero.
This document summarizes a student's research on modeling light bending near a hypothetical monopole field. The student developed a mathematical model using Fermat's principle and tested different coupling parameters to analyze the light bending effect. The results showed that for both attractive and repulsive cases, increasing the coupling parameter strengthened the light bending. Additionally, the deflection angle peaked at a maximum impact parameter value. The student concluded that the relationship between deflection angle and impact parameter was consistent for different coupling parameters. Diagrams and graphs were also used to illustrate the light trajectories and relationships found in the analysis.
Reliability Analysis of the Sectional Beams Due To Distribution of Shearing S...researchinventy
This paper shows the results of the Reliability Analysis of the sectional beams due to distribution of Shear Stress. It is assumed that the load was uniformly distributed over the beam. It is discussed that the distribution of shear stress over the beam. It is discussed that the average shears stress and maximum shear stress across the section of the beam for Weibull distribution. The reliability analysis of distribution of shearing stresses over sectional beams is performed. Also it is derived that the hazard functions for these types of beams. Reliability comparison has also been done for the sectional beams. It is observed that the reliability of the beam decreased when the width (b) of the beam decreases, and the load (F) is high. The reliability of the beam is increased when the height (h) of the triangular section increases , diameter(d) of the circular beam is increased and parameter 푘 decreasses
This book is intended to cover the basic Strength of Materials of the first
two years of an engineering degree or diploma course ; it does not attempt
to deal with the more specialized topics which usually comprise the final
year of such courses.
The work has been confined to the mathematical aspect of the subject
and no descriptive matter relating to design or materials testing has been
included.
ESA Module 5 Part-B ME832. by Dr. Mohammed ImranMohammed Imran
1. The document describes the moire fringes technique for experimental stress analysis. Moire fringes occur when two similar patterns are overlaid, allowing strains to be measured.
2. There are two approaches to analyzing moire fringe patterns - the geometrical approach regards fringes as intersections of grids, while the displacement approach uses fringes to determine displacements.
3. The distance between bright or dark fringes equals the master grid pitch divided by the strain. Fringes indicate loci of equal displacement and allow calculating strains from measured displacements.
- Shear stress distribution in beams takes a parabolic shape, with the maximum stress at the neutral axis and zero at the ends. In rectangular beams the stress is highest at y=0. In I-beams, most stress is carried by the web in a "top-hat" distribution.
- Circular beams have a shear stress distribution that also follows a parabolic shape, calculated using the area moment of the shaded portion.
- Principal stresses can be determined in beams using the bending and shear stresses. The bending stress is not a principal stress and the principal stresses are found using an equation involving the bending and shear stresses.
Principale of super position and maxwell reciprocal therom KEVINMISTRY8
1. The principle of superposition states that the displacements or deflections caused by individual loads or forces acting on a structure can be added to determine the total displacement caused by all loads acting together.
2. Maxwell's reciprocal theorem states that the deflection at one point on a structure due to a unit load at another point is equal to the deflection at the second point due to a unit load at the first point.
3. Betti's law states that the virtual work done by one system of forces during displacements caused by a second system of forces is equal to the virtual work done by the second system of forces during displacements caused by the first system of forces.
This document discusses transformation of stresses and strains when an element is rotated. It defines normal and shear stresses, and shows how to calculate them based on forces and geometry. It then demonstrates how to use Mohr's circle to determine maximum and minimum stresses, and stresses and shear stresses at any angle of rotation. As an example, it also shows calculations for stresses in a thin-walled pressure vessel where shear stress is zero.
This document summarizes a student's research on modeling light bending near a hypothetical monopole field. The student developed a mathematical model using Fermat's principle and tested different coupling parameters to analyze the light bending effect. The results showed that for both attractive and repulsive cases, increasing the coupling parameter strengthened the light bending. Additionally, the deflection angle peaked at a maximum impact parameter value. The student concluded that the relationship between deflection angle and impact parameter was consistent for different coupling parameters. Diagrams and graphs were also used to illustrate the light trajectories and relationships found in the analysis.
Reliability Analysis of the Sectional Beams Due To Distribution of Shearing S...researchinventy
This paper shows the results of the Reliability Analysis of the sectional beams due to distribution of Shear Stress. It is assumed that the load was uniformly distributed over the beam. It is discussed that the distribution of shear stress over the beam. It is discussed that the average shears stress and maximum shear stress across the section of the beam for Weibull distribution. The reliability analysis of distribution of shearing stresses over sectional beams is performed. Also it is derived that the hazard functions for these types of beams. Reliability comparison has also been done for the sectional beams. It is observed that the reliability of the beam decreased when the width (b) of the beam decreases, and the load (F) is high. The reliability of the beam is increased when the height (h) of the triangular section increases , diameter(d) of the circular beam is increased and parameter 푘 decreasses
This book is intended to cover the basic Strength of Materials of the first
two years of an engineering degree or diploma course ; it does not attempt
to deal with the more specialized topics which usually comprise the final
year of such courses.
The work has been confined to the mathematical aspect of the subject
and no descriptive matter relating to design or materials testing has been
included.
ESA Module 5 Part-B ME832. by Dr. Mohammed ImranMohammed Imran
1. The document describes the moire fringes technique for experimental stress analysis. Moire fringes occur when two similar patterns are overlaid, allowing strains to be measured.
2. There are two approaches to analyzing moire fringe patterns - the geometrical approach regards fringes as intersections of grids, while the displacement approach uses fringes to determine displacements.
3. The distance between bright or dark fringes equals the master grid pitch divided by the strain. Fringes indicate loci of equal displacement and allow calculating strains from measured displacements.
- Shear stress distribution in beams takes a parabolic shape, with the maximum stress at the neutral axis and zero at the ends. In rectangular beams the stress is highest at y=0. In I-beams, most stress is carried by the web in a "top-hat" distribution.
- Circular beams have a shear stress distribution that also follows a parabolic shape, calculated using the area moment of the shaded portion.
- Principal stresses can be determined in beams using the bending and shear stresses. The bending stress is not a principal stress and the principal stresses are found using an equation involving the bending and shear stresses.
Principale of super position and maxwell reciprocal therom KEVINMISTRY8
1. The principle of superposition states that the displacements or deflections caused by individual loads or forces acting on a structure can be added to determine the total displacement caused by all loads acting together.
2. Maxwell's reciprocal theorem states that the deflection at one point on a structure due to a unit load at another point is equal to the deflection at the second point due to a unit load at the first point.
3. Betti's law states that the virtual work done by one system of forces during displacements caused by a second system of forces is equal to the virtual work done by the second system of forces during displacements caused by the first system of forces.
This document summarizes the key concepts from a seminar on structural vibration analysis using finite element methods. It introduces common sources and types of vibration, including free vibration from impacts and forced vibration from repetitive external forces. It also describes using finite element analysis to model structural vibration, including modeling structures as mass-spring-damper systems and discretizing continuous structures into finite elements to analyze their vibration modes and frequencies.
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 bending stresses in beams. It describes how bending stresses are developed in beams to resist bending moments and shearing forces. The theory of pure bending is introduced, where only bending stresses are considered without the effect of shear. Equations for calculating bending stresses are derived based on the beam's moment of inertia, bending moment, and distance from the neutral axis. Several beam cross-section examples are provided to demonstrate how to calculate the maximum bending stress and section modulus.
This document provides summaries and equations for various topics in mechanics including:
1) Common equations of motion that relate displacement, velocity, acceleration, and time without including one of the variables.
2) Momentum and impulse equations and the relationship between momentum, velocity, and mass.
3) How to resolve vectors into components using trigonometry.
4) Graphical representations of displacement, velocity, and acceleration over time and their relationships.
5) Newton's laws of motion and how force equals mass times acceleration.
- The normal stress distribution in a curved beam is hyperbolic and determined using specialized formulas, as the neutral axis does not pass through the centroid due to curvature.
- To analyze a curved beam, one must first determine the cross-sectional properties and location of the neutral axis, then calculate the normal stress distribution using the appropriate formula.
- For a rectangular steel bar bent into a circular arc, the maximum moment that can be applied before exceeding the allowable stress is 0.174 kN-m, with maximum stress at the bottom of the bar. If the bar was straight, the maximum moment would be 0.187 kN-m.
This report summarizes a study on the contact mechanics of polyhedral shapes using finite element analysis (FEA). FEA models were created in Abaqus to simulate the contact between polyhedral particles and surfaces. The models varied the angle of corners and edges of triangular and pyramidal particles. Results showed that larger angles allowed for more deformation and reaction forces increased nonlinearly with displacement. The trends in the FEA models matched those of experimental models, though the FEA curves were not as smooth. Varying the geometry and mesh refinement could produce smoother force-displacement curves. The material properties and body geometry were found to influence reaction forces in polyhedral contact problems.
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 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.
Solution of Chapter- 05 - stresses in beam - Strength of Materials by SingerAshiqur Rahman Ziad
This document discusses stresses in beams, including flexural and shearing stresses. It provides formulas for calculating flexural stress based on the beam's moment of inertia, bending moment, and distance from the neutral axis. Several example problems are worked through applying these formulas. The document also discusses using economic beam sections that optimize the use of material by placing more area on the outer fibers where stresses are highest.
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 derives the flexure formula, which relates bending moment to flexural stress in a beam. It makes assumptions about the beam's material properties and deformations under load. The derivation proceeds by considering the strains and stresses in fibers of the beam under bending. Equating the internal and external forces and moments leads to the flexure formula, which expresses maximum flexural stress as a function of the bending moment and section properties. The derivation shows that the neutral axis passes through the centroid and stress is highest at the furthest point from the neutral axis.
This document provides an overview of basic engineering theory concepts covered in 8 chapters:
1) Mechanics of Materials: Stress and Strain
2) Hooke's Law
3) Failure Criteria
4) Beams
5) Buckling
6) Dynamics
7) Fluid Mechanics
8) Heat Transfer
Each chapter introduces fundamental concepts and equations within the topic area.
This experiment investigates the relationship between the height of release and the velocity of a solid sphere rolling down an inclined plane. A solid sphere was released from varying heights on a frictionless wooden plane and its velocity was measured using a motion sensor. The data showed a linear relationship between height and the square of velocity, supporting the theoretical equation that potential energy is converted to translational and rotational kinetic energy. However, some systematic error was present due to friction and imperfections of the sphere. Increasing the number of trials and data points could improve the accuracy of the results.
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.
The document discusses various topics related to trusses:
1. It defines the key features of a truss, including that it consists of slender straight members connected at joints, with pinned connections that allow only axial forces in the members.
2. External loads must be applied at the joints, as members cannot support lateral loads. Most structures use multiple trusses joined together to form a space framework.
3. Simple trusses are constructed by adding members and joints, following the rule that m = 2n - 3, where m is the number of members and n is the number of joints. Simple trusses are internally rigid.
4. Two common methods for analyzing trusses
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.
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
This document summarizes an experiment to examine how bending moment varies with an increasing point load on a beam. Strain gauges were used to measure strain on an aluminum beam as loads from 0-500N were applied. Bending moments were calculated from the strain readings. Graphs were made comparing experimental and theoretical bending moments and strains, showing relatively close correlation between the two despite some errors. The experiment demonstrated the relationship between bending moment and strain in beams under an increasing point load.
When a solid body deforms due to external loads or temperature changes, it experiences strain. Normal strain refers to the elongation or contraction of a line segment per unit length. Shear strain is the change in angle between two originally perpendicular line segments. For most engineering applications, only small deformations and strains are considered in the analysis.
Abstract Today’s experiment objectives are to determine the st.docxannetnash8266
Abstract
Today’s experiment objectives are to determine the stress, deflection, and the strain of a simply supported beam under load. Moreover, experimentally verify the beam stress and flexure formulas. In this week’s experiment we had to use the MTS machine in order to apply a load to a simply supported beam and measure the deflection and strain that comes out from it. As a result from the graphs we plotted, we saw that whenever the load increases, the deflection and strain also increases. We used the strain to find the theoretical stress in our calculations, and we also used the moment, moment of inertia, and the neutral axis to find the experimental stress. We calculated the moment of inertia, which came out to be 0.05122 . Also, we found the neutral axis to be 0515 in , and the maximum deflection also came out to be 0.000013 in. The maximum load applied on the beam came out to be 40049.5 psi, which we calculated from the maximum stress.
Table of Contents
Abstract……………………………………………………………..………..2
Table of Contents……………………………………….……………………3
Introduction and Theory…………………………………………………….4-6
Procedure………………………………………………….……………….7-9
Summary of Important Results…………………...………………………..10-12
Sample Calculations and Error Analysis……………….………………….13
Discussion and Conclusion………………………………………………..14-15
References……………………………………..…………………………….16
Appendix……………………………………………………….……………17
Introduction and Theory
Engineers use beams to support loads over a span length. These beams are structural
members that are only loaded non-axially causing them to be subjected to bending. “A piece is said to be in bending if the forces act on a piece of material in such a way that they tend to induce compressive stresses over one part of a cross section of the piece and tensile stresses over
the remaining part” (Ref. 1). This definition of bending is illustrated below in Figure 1.
It can be seen from Figure 1 that the compressive force, C, and the tensile force, T, acting on the member are equal in magnitude because of equilibrium. Therefore, the compressive force and the tensile force form a force couple whose moment is equal to either the tensile force multiplied by the moment arm or the compressive force multiplied by the moment arm. The moment arm is denoted, e, in Figure 1.
This is why structural members usually carry the center of the load into the tensile, compressive, or transverse loads. A beam usually carries the load transversely. During today’s experiment the load will be forced onto the beam in a symmetric order. We also must know that any cross section of the beam there will be a shear force V and a moment M. When we see in the middle of the beam we realize that the shear force diagram is zero and the moment reaches its maximum constant value.
When a beam is cur in to slices we see that if we want the moment the internal forces must be equal to the moment on the outside. So, M must be equal to the internal forces applied.
Physics 161Static Equilibrium and Rotational Balance Intro.docxrandymartin91030
Physics 161
Static Equilibrium and Rotational Balance
Introduction
In Part I of this lab, you will observe static equilibrium for a meter stick suspended horizontally. In Part II, you will observe the rotational balance of a cylinder on an incline. You will vary the mass hanging from the side of the cylinder for different angles.
Reference
Young and Freedman, University Physics, 12th Edition: Chapter 11, section 3
Theory
Part I: When forces act on an extended body, rotations about axes on the body can result as well as translational motion from unbalanced forces. Static equilibrium occurs when the net force and the net torque are both equal to zero. We will examine a special case where forces are only acting in the vertical direction and can therefore be summed simply without breaking them into components:
(1)
Torques may be calculated about the axis of your choosing:
(2)
where torque is specified by the equation:
(3)
where d is the lever arm (or moment arm) for the force. The lever arm is the perpendicular distance from the line of force to the axis about which you are calculating the torque.
Normally, up is "+" and down is "-" for forces. For torques, it is convenient to define clockwise as "-" and counterclockwise as "+". Whatever you decide to do, be consistent with your signs and make sure you understand what a "+" or "-" value for your force or torque means directionally.
Part II: Any round object when placed on an incline has tendency of rotating towards the bottom of an incline. If the downward force that causes the object to accelerate down the slope is canceled by another force, the object will remain stationary on the incline. Figure 1 shows the configuration of the setup. In order to have the rubber cylinder in static equilibrium we should satisfy the following conditions:
(4)
Figure 1: Experimental setup for Part II
The condition that the net force along the x-axis (which is conveniently taken along the incline) must be zero yields the relationship. (Prove this!)
Without static friction the cylinder would slide down the incline; the presence of friction causes a torque in clockwise (negative) direction. In order to have static equilibrium we need to balance that torque with a torque in counterclockwise direction. This is achieved by hanging the appropriate mass m.
Applying the last condition to the center of the cylinder will result in:
where r, the radius of the small cylinder (PVC fitting), is the moment arm for the mass m and R, the radius of the rubber cylinder, is the moment arm for the frictional force which accounts for M and m. Combining this equation with the equation for Ffr from above will result in:
(5)
(6)
By adjusting the mass m, we can observe how the equilibrium can be achieved.
Procedure
Part I: Static Equilibrium
Figure 2: Diagram of Torque Experiment Setup
1. Weigh the meter stick you use, including the metal hangers.
2. Attach .
This document summarizes the key concepts from a seminar on structural vibration analysis using finite element methods. It introduces common sources and types of vibration, including free vibration from impacts and forced vibration from repetitive external forces. It also describes using finite element analysis to model structural vibration, including modeling structures as mass-spring-damper systems and discretizing continuous structures into finite elements to analyze their vibration modes and frequencies.
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 bending stresses in beams. It describes how bending stresses are developed in beams to resist bending moments and shearing forces. The theory of pure bending is introduced, where only bending stresses are considered without the effect of shear. Equations for calculating bending stresses are derived based on the beam's moment of inertia, bending moment, and distance from the neutral axis. Several beam cross-section examples are provided to demonstrate how to calculate the maximum bending stress and section modulus.
This document provides summaries and equations for various topics in mechanics including:
1) Common equations of motion that relate displacement, velocity, acceleration, and time without including one of the variables.
2) Momentum and impulse equations and the relationship between momentum, velocity, and mass.
3) How to resolve vectors into components using trigonometry.
4) Graphical representations of displacement, velocity, and acceleration over time and their relationships.
5) Newton's laws of motion and how force equals mass times acceleration.
- The normal stress distribution in a curved beam is hyperbolic and determined using specialized formulas, as the neutral axis does not pass through the centroid due to curvature.
- To analyze a curved beam, one must first determine the cross-sectional properties and location of the neutral axis, then calculate the normal stress distribution using the appropriate formula.
- For a rectangular steel bar bent into a circular arc, the maximum moment that can be applied before exceeding the allowable stress is 0.174 kN-m, with maximum stress at the bottom of the bar. If the bar was straight, the maximum moment would be 0.187 kN-m.
This report summarizes a study on the contact mechanics of polyhedral shapes using finite element analysis (FEA). FEA models were created in Abaqus to simulate the contact between polyhedral particles and surfaces. The models varied the angle of corners and edges of triangular and pyramidal particles. Results showed that larger angles allowed for more deformation and reaction forces increased nonlinearly with displacement. The trends in the FEA models matched those of experimental models, though the FEA curves were not as smooth. Varying the geometry and mesh refinement could produce smoother force-displacement curves. The material properties and body geometry were found to influence reaction forces in polyhedral contact problems.
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 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.
Solution of Chapter- 05 - stresses in beam - Strength of Materials by SingerAshiqur Rahman Ziad
This document discusses stresses in beams, including flexural and shearing stresses. It provides formulas for calculating flexural stress based on the beam's moment of inertia, bending moment, and distance from the neutral axis. Several example problems are worked through applying these formulas. The document also discusses using economic beam sections that optimize the use of material by placing more area on the outer fibers where stresses are highest.
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 derives the flexure formula, which relates bending moment to flexural stress in a beam. It makes assumptions about the beam's material properties and deformations under load. The derivation proceeds by considering the strains and stresses in fibers of the beam under bending. Equating the internal and external forces and moments leads to the flexure formula, which expresses maximum flexural stress as a function of the bending moment and section properties. The derivation shows that the neutral axis passes through the centroid and stress is highest at the furthest point from the neutral axis.
This document provides an overview of basic engineering theory concepts covered in 8 chapters:
1) Mechanics of Materials: Stress and Strain
2) Hooke's Law
3) Failure Criteria
4) Beams
5) Buckling
6) Dynamics
7) Fluid Mechanics
8) Heat Transfer
Each chapter introduces fundamental concepts and equations within the topic area.
This experiment investigates the relationship between the height of release and the velocity of a solid sphere rolling down an inclined plane. A solid sphere was released from varying heights on a frictionless wooden plane and its velocity was measured using a motion sensor. The data showed a linear relationship between height and the square of velocity, supporting the theoretical equation that potential energy is converted to translational and rotational kinetic energy. However, some systematic error was present due to friction and imperfections of the sphere. Increasing the number of trials and data points could improve the accuracy of the results.
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.
The document discusses various topics related to trusses:
1. It defines the key features of a truss, including that it consists of slender straight members connected at joints, with pinned connections that allow only axial forces in the members.
2. External loads must be applied at the joints, as members cannot support lateral loads. Most structures use multiple trusses joined together to form a space framework.
3. Simple trusses are constructed by adding members and joints, following the rule that m = 2n - 3, where m is the number of members and n is the number of joints. Simple trusses are internally rigid.
4. Two common methods for analyzing trusses
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.
Learn Online Courses of Subject Engineering Mechanics of First Year Engineering. Clear the Concepts of Engineering Mechanics Through Video Lectures and PDF Notes. https://ekeeda.com/streamdetails/subject/Engineering-Mechanics
This document summarizes an experiment to examine how bending moment varies with an increasing point load on a beam. Strain gauges were used to measure strain on an aluminum beam as loads from 0-500N were applied. Bending moments were calculated from the strain readings. Graphs were made comparing experimental and theoretical bending moments and strains, showing relatively close correlation between the two despite some errors. The experiment demonstrated the relationship between bending moment and strain in beams under an increasing point load.
When a solid body deforms due to external loads or temperature changes, it experiences strain. Normal strain refers to the elongation or contraction of a line segment per unit length. Shear strain is the change in angle between two originally perpendicular line segments. For most engineering applications, only small deformations and strains are considered in the analysis.
Abstract Today’s experiment objectives are to determine the st.docxannetnash8266
Abstract
Today’s experiment objectives are to determine the stress, deflection, and the strain of a simply supported beam under load. Moreover, experimentally verify the beam stress and flexure formulas. In this week’s experiment we had to use the MTS machine in order to apply a load to a simply supported beam and measure the deflection and strain that comes out from it. As a result from the graphs we plotted, we saw that whenever the load increases, the deflection and strain also increases. We used the strain to find the theoretical stress in our calculations, and we also used the moment, moment of inertia, and the neutral axis to find the experimental stress. We calculated the moment of inertia, which came out to be 0.05122 . Also, we found the neutral axis to be 0515 in , and the maximum deflection also came out to be 0.000013 in. The maximum load applied on the beam came out to be 40049.5 psi, which we calculated from the maximum stress.
Table of Contents
Abstract……………………………………………………………..………..2
Table of Contents……………………………………….……………………3
Introduction and Theory…………………………………………………….4-6
Procedure………………………………………………….……………….7-9
Summary of Important Results…………………...………………………..10-12
Sample Calculations and Error Analysis……………….………………….13
Discussion and Conclusion………………………………………………..14-15
References……………………………………..…………………………….16
Appendix……………………………………………………….……………17
Introduction and Theory
Engineers use beams to support loads over a span length. These beams are structural
members that are only loaded non-axially causing them to be subjected to bending. “A piece is said to be in bending if the forces act on a piece of material in such a way that they tend to induce compressive stresses over one part of a cross section of the piece and tensile stresses over
the remaining part” (Ref. 1). This definition of bending is illustrated below in Figure 1.
It can be seen from Figure 1 that the compressive force, C, and the tensile force, T, acting on the member are equal in magnitude because of equilibrium. Therefore, the compressive force and the tensile force form a force couple whose moment is equal to either the tensile force multiplied by the moment arm or the compressive force multiplied by the moment arm. The moment arm is denoted, e, in Figure 1.
This is why structural members usually carry the center of the load into the tensile, compressive, or transverse loads. A beam usually carries the load transversely. During today’s experiment the load will be forced onto the beam in a symmetric order. We also must know that any cross section of the beam there will be a shear force V and a moment M. When we see in the middle of the beam we realize that the shear force diagram is zero and the moment reaches its maximum constant value.
When a beam is cur in to slices we see that if we want the moment the internal forces must be equal to the moment on the outside. So, M must be equal to the internal forces applied.
Physics 161Static Equilibrium and Rotational Balance Intro.docxrandymartin91030
Physics 161
Static Equilibrium and Rotational Balance
Introduction
In Part I of this lab, you will observe static equilibrium for a meter stick suspended horizontally. In Part II, you will observe the rotational balance of a cylinder on an incline. You will vary the mass hanging from the side of the cylinder for different angles.
Reference
Young and Freedman, University Physics, 12th Edition: Chapter 11, section 3
Theory
Part I: When forces act on an extended body, rotations about axes on the body can result as well as translational motion from unbalanced forces. Static equilibrium occurs when the net force and the net torque are both equal to zero. We will examine a special case where forces are only acting in the vertical direction and can therefore be summed simply without breaking them into components:
(1)
Torques may be calculated about the axis of your choosing:
(2)
where torque is specified by the equation:
(3)
where d is the lever arm (or moment arm) for the force. The lever arm is the perpendicular distance from the line of force to the axis about which you are calculating the torque.
Normally, up is "+" and down is "-" for forces. For torques, it is convenient to define clockwise as "-" and counterclockwise as "+". Whatever you decide to do, be consistent with your signs and make sure you understand what a "+" or "-" value for your force or torque means directionally.
Part II: Any round object when placed on an incline has tendency of rotating towards the bottom of an incline. If the downward force that causes the object to accelerate down the slope is canceled by another force, the object will remain stationary on the incline. Figure 1 shows the configuration of the setup. In order to have the rubber cylinder in static equilibrium we should satisfy the following conditions:
(4)
Figure 1: Experimental setup for Part II
The condition that the net force along the x-axis (which is conveniently taken along the incline) must be zero yields the relationship. (Prove this!)
Without static friction the cylinder would slide down the incline; the presence of friction causes a torque in clockwise (negative) direction. In order to have static equilibrium we need to balance that torque with a torque in counterclockwise direction. This is achieved by hanging the appropriate mass m.
Applying the last condition to the center of the cylinder will result in:
where r, the radius of the small cylinder (PVC fitting), is the moment arm for the mass m and R, the radius of the rubber cylinder, is the moment arm for the frictional force which accounts for M and m. Combining this equation with the equation for Ffr from above will result in:
(5)
(6)
By adjusting the mass m, we can observe how the equilibrium can be achieved.
Procedure
Part I: Static Equilibrium
Figure 2: Diagram of Torque Experiment Setup
1. Weigh the meter stick you use, including the metal hangers.
2. Attach .
11 - 3
Experiment 11
Simple Harmonic Motion
Questions
How are swinging pendulums and masses on springs related? Why are these types of
problems so important in Physics? What is a spring’s force constant and how can you measure
it? What is linear regression? How do you use graphs to ascertain physical meaning from
equations? Again, how do you compare two numbers, which have errors?
Note: This week all students must write a very brief lab report during the lab period. It is
due at the end of the period. The explanation of the equations used, the introduction and the
conclusion are not necessary this week. The discussion section can be as little as three sentences
commenting on whether the two measurements of the spring constant are equivalent given the
propagated errors. This mini-lab report will be graded out of 50 points
Concept
When an object (of mass m) is suspended from the end of a spring, the spring will stretch
a distance x and the mass will come to equilibrium when the tension F in the spring balances the
weight of the body, when F = - kx = mg. This is known as Hooke's Law. k is the force constant
of the spring, and its units are Newtons / meter. This is the basis for Part 1.
In Part 2 the object hanging from the spring is allowed to oscillate after being displaced
down from its equilibrium position a distance -x. In this situation, Newton's Second Law gives
for the acceleration of the mass:
Fnet = m a or
The force of gravity can be omitted from this analysis because it only serves to move the
equilibrium position and doesn’t affect the oscillations. Acceleration is the second time-
derivative of x, so this last equation is a differential equation.
To solve: we make an educated guess:
Here A and w are constants yet to be determined. At t = 0 this solution gives x(t=0) = A,
which indicates that A is the initial distance the spring stretches before it oscillates. If friction is
negligible, the mass will continue to oscillate with amplitude A. Now, does this guess actually
solve the (differential) equation? A second time-derivative gives:
Comparing this equation to the original differential equation, the correct solution was
chosen if w2 = k / m. To understand w, consider the first derivative of the solution:
−kx = ma
a = −
k
m
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟
x
d 2x
dt 2
= −
k
m
x x(t) = A cos(ωt)
d 2x(t)
dt 2
= −Aω2 cos(ωt) = −ω2x(t)
James Gering
Florida Institute of Technology
11 - 4
Integrating gives
We assume the object completes one oscillation in a certain period of time, T. This helps
set the limits of integration. Initially, we pull the object a distance A from equilibrium and
release it. So at t = 0 and x = A. (one.
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1. MTRL 485 – Case 5
Spindle Failure
Group 4
Masseir, Alex (55761118)
May, Michael (47957105)
McGrew, Kennan (55398101)
Mei, Helen (31494115)
Muhammad, Harith (18204115)
2. Task 1
Figure 1: Fatigue Surface
The location for the fatigue crack origin was determined by the colour difference of the sample as well as
the beach marks surrounding the origin. Beach marks are the curved lines on the sample which indicates
fatigue propagation and they are caused by the change of cyclic loads in a sample. Using these markings, we
can determine the fatigue crack origin. The dark coloured surface is suspected to be the origin of the fatigue
crack because it most likely grew over periods of days or months, leading to oxidation on the surface. The
region that appears to be brightest is the fast fracture region where the sample failed rapidly due to the
sufficiently large increase of the crack. The fast fracture region is also verified by the rough section of the
sample. It is presumed that the stress intensity exceeded the critical value, creating the rapid failure. Figure
1 also shows the hardened layers at the edges of the sample which are formed from case hardening.
Run # C, wt% Mn, wt% Si, wt% Fe, wt%
1 - 1.1 0.4 98.5
2 34.3 0.6 0.1 65.0
3 37.7 0.5 61.8
4 - - 100
5 37.7 - 62.3
Table 1: EDX analysis of the fracture surface
Referring to Table 1, the composition of the samples can be determined however the data is rather
scattered especially for the carbon and iron content. Therefore, it is assumed that the analysis have been
performed incorrectly.
3. Task 2
Forces
The three main forces in the bike pedal spindle example were calculated using the three equations below
and F0 = 500N
𝐹𝜃 = 𝐹°sin( 𝜃)
𝐹𝑛 = 𝐹° ∗ si𝑛2 ( 𝜃) => 𝐹𝑛 = 𝐹𝜃 ∗ cos( 𝜃)
𝐹𝑟 = 𝐹° ∗ sin( 𝜃) ∗ cos( 𝜃) => 𝐹𝑟 = 𝐹𝜃 ∗ sin( 𝜃)
Its was expected that Fθ and Fn would be equal at θ=90° because at this point the forces are pointing in the
same direction as Fθ is perpendicular at this point like Fn always is. The assumption that the two forces
would also be greatest at point are correct because when θ = 90° both Fn and Fθ are pointing in the same
downward direction as the initial for applied by the foot of the biker F°. Fθ is also larger at all time except
90° because it is always counting Fo a downward force and Fn is always perpendicular to the spot of the
pedal compared to the rotation of the crank. Fr becomes negative at any angle above 90° because sin(θ) is
negative between 90°-180° and because below 90° Fr is pointing down and 0° - just above 90° it points up.
Fn and Fθ never become negative because between 0° - 180° in this example they are always pointing
downward. A graph of the force vs. degrees in below. All values can be found in the appendix
Graph 1: Forces vs. Degrees
Reaction Forces
The equations below were used to calculate the reaction forces
Σ𝑁𝑖 = 0
𝑁𝑥 + 0 = 0 => 𝑁𝑥 = 0
𝑁𝑦 + 𝐹𝑟 = 0 => 𝑁𝑦 = −𝐹𝑟
𝑁𝑧 − 𝐹𝑛 = 0 => 𝑁𝑧 = 𝐹𝑛
4. It is expected that Nx will equal zero because there are no forces in the Nx direction. As we can see looking
at chart 2 below and chart one above reaction force Ny is the opposite if Fr because the Ny reaction forces
are acting in the Y direction and are equal and opposite. Figure one will gives a good visual representation
of the relation ship between Ny and Fr. Also from looking at the two charts and figure one we see is acting
in the Z direction and has the same for as values as Fn. All values can be found in the appendix
Figure 1
Moments
The Mx, My, and Mz moments were calculated using the formulas below.
Σ𝑀𝑖 = 0
𝑀𝑥 − 𝐹𝑛 ∗ 𝑏 = 0 => 𝑀𝑥 = 𝐹𝑛 ∗ 𝑏
𝑀𝑦 − 𝐹𝑛 ∗ 𝑎 = 0 => 𝑀𝑦 = 𝐹𝑛 ∗ 𝑎
𝑀𝑧 = 𝐹𝑟 ∗ 𝑎 = 0 => 𝑀𝑧 = 𝐹𝑟 ∗ 𝑎
𝑎 = 0.055𝑏 = 0.17
The moments were calculated using a = 0.055 m and b = 0.17 m. Looking at figure two it is easy to see the
moments present on the crank. We can see the moment in the X direction will in general be the largest
because it is the only one that has the large b distance in its equation. Moments in the Y and Z direction will
in general be small because their distance used in their momentum equations is a, which is significantly
smaller. Mz moment is smaller than My and becomes negative because it is dependant on Fr which is
smaller and becomes negative past 90°. All values can be found in the appendix.
5. Figure 2
Graph 2: Moment and Reaction Forces vs. Degrees
-300.000
-200.000
-100.000
0.000
100.000
200.000
300.000
400.000
500.000
600.000
0 50 100 150 200
MomentandReactionForces
Degrees
Moment and ReactionForces Vs. Degrees
Mx
My
Mz
Nx
Ny
Nz
6. Stresses
The bending, shear, and torque stresses were calculated using the equations below and previous
information.
𝜎𝐵𝑒𝑛𝑑𝑖𝑛𝑔 =
𝑀𝑜𝑚𝑒𝑛𝑡
𝐼
∗ 𝑍
𝜎𝑆ℎ𝑒𝑎𝑟 =
𝑆ℎ𝑒𝑎𝑟𝑓𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
𝜎𝑆ℎ𝑒𝑎𝑟−𝑇𝑜𝑟𝑞𝑢𝑒 =
𝑇𝑜𝑟𝑞𝑢𝑒
𝐽
∗ 𝑟
𝐴𝑟𝑒𝑎 = 𝜋 ∗ 𝑟2
𝐼 =
𝜋
4
∗ 𝑟4
𝐽 =
𝜋
2
∗ 𝑟4
𝑟 = 0.006
The bending stresses were calculated at y = ± R & z = 0 and at y = 0 and z = ± R. When y = R and z = 0 the
shear stress was calculated using the Mz moment as the moment and the first moment of inertia I, as
calculated below, z in the bending stress equation is r (.006). When y = -R and z = 0 the answer is the same
but negative cause z is negative. When y = 0 and z = R the bending stress is calculated using the My moment
as well as I and r the same as previously. When y = 0 and z = -R the answer is the same but negative. Doing
this leads to the bending stresses being greatest when y = 0 and z = ± R. The shear stresses xy and xy were
calculated using above equations for shear stress and area. In the xy shear reaction force Ny was used as
the shear force, in the xz shear stress the Nz reaction force was used. This leads to the xz shear being larger.
Finally the torque stress was found using the second moment of inertia and the equation above. Moment
Mx was used as a value for torque and this lead to the largest about of stress coming by means of torque.
All values can be found in the appendix.
Graph 3: Stresses vs. Degrees
-200.000
-150.000
-100.000
-50.000
0.000
50.000
100.000
150.000
200.000
250.000
300.000
0 20 40 60 80 100 120 140 160 180 200
Stresses
Degrees
Stresses vs. Degrees
Bending y=R, z=0
Bending y=-R, z=0
Bending y=0, z=R
Bending y=0, z=-R
Shear xy
Shear xz
Torque
7. Task 3
The material properties presented and measured show a large variability. Try to bound each property
between an upper and lower bound.
What does case hardening do to a material and its properties?
Case hardening increases the hardness of the outer layer while the core remains relatively soft. The
combination of a hard shell with a soft core allows the material to withstand higher stresses and fatigue.
The shell would have a higher wear resistance and fatigue strength while the core would have a higher
impact strength.
Where would you expect yielding to initiate?
Failure always occurs perpendicular to the direction of the largest stress. In this case this is where we would
expect yielding to initiate. In our results the largest stress occurs in torque. Since torque stress is the largest
stress observed in our analysis we would expect yielding to occur perpendicular to the torque stress. We
expect the yielding to happen at a 45-degree angle as it did in the example.
Fatigue cracks tend to initiate and grow perpendicular tothe maximum principal stress. Do your analysis
agree with experimental observations?
In our analysis we found that the largest stress in the spindle is from torque. For this analysis the direction
that the spindle would fail would be in a perpendicular direction of the torque, this is in a 45-degree angle
into the depth of the spindle. Looking at the images of the actual failure we see that this is the way that the
spindle cracked and eventually broke.
Perform a fracture analysis based on the crack length at final fracture from the images in this report.
a) Crack Growth
Rate of crack growth depends on the stress intensity at the head of our fatigue crack. Hypothesis can be
made that the higher the stress intensity, the higher the crack growth rate. When Kmax (maximum stress
intensity factor) reaches the value fracture toughness Kc of the material and thickness of the spindle, crack
propagation is expected to occur when the length of crack is equivalent to ac that is critical for brittle
fracture. The following formula is used to calculate the critical crack length of the material.
8. 𝑎 𝑐 =
1
𝜋
(
𝐾𝑐
𝐹𝜎 𝑚𝑎𝑥
)2
In addition, crack growth causes a loss of cross-sectional area, and thus an increase in the stress on the
remaining uncracked (net) area. Depending on the material and the member geometry and size, fully plastic
yielding occurs before Kmax = Fracture toughness value.
b) Plastic Zone Size
Based on the crack length at final fracture from the images and fracture mechanics, it is estimated that the
critical length is 9.1765 mm. The material starts to fail at stress𝜎𝑓𝑎𝑖𝑙 =
𝐾𝐼𝐶
√ 𝜋𝑎
.
For the material that deforms plastically, the local stresses eventually exceed the yields strength. The stress
cannot continue to rise according to linear elastic behavior. We can estimate the size of the plastic zone,
the region denoted rp where the stresses are predicted based on the elastic solution to be above the yield
stress. This plastic zone depends on the ratio of the applied stress to the yield stress as well on the crack
size. With this, we can define the effective stress intensity factor where the crack size is assumed to be
larger by a factor equal to the plastic zone size.
Plastic Zone Size is obtained by this formula:
c) Fracture
9. Before the point at the material maintains its shape and undergoes the plastic deformation. However, it
reaches the fracture limit as the curve drops exponentially according to the fracture toughness equation
shown as it weakens the materials and causes the crack propagation to occur and finally fails. Having this
plastic zone shows us that the material yields when the crack length is very small before reaching the value
of critical crack length.
According to the Paris Law, region 3 represents the period where the unstable crack growth leads the
material to fracture.
How large has the pedal force has to be to cause final fracture based on the KIC = 60 MPam0.5?
The question asks for the maximum pedal force to initiate the crack propagation thus leads to the fracture.
The KIC given is 60MPam0.5. A given material can resist a crack without brittle fracture occurring as long as
this K is below a critical value called as facture toughness This fracture toughness is dependent on the
thickness of the material. Thicker material has lower KIC. KIC is defined as a measure of given ability to resist
fracture in the presence of crack.
Method 1: Using KIC = 60MPam0.5
KIC = 𝑌 ∗ 𝜎 ∗ √ 𝜋 ∗ 𝑎 𝑐𝑎𝑠𝑖𝑓𝑎 ≪ 𝑏
Y = geometric factor
𝜎 = applied stress
ac = critical crack length
Area = 0.000113097 m^2
Initially we solve the applied stress to obtain the maximum force. To do this, you need to have the critical
crack length of the material. Based on the image given in the notes, the critical crack length is about 9.1765
mm.
Solving for stress and maximum force gives us that the maximum force is about 39.966 kN.
Method2: Assuming 𝜎max = 565MPa
We know that 𝜎max = 565 MPa based on the material properties. The formula below can be used to
calculate the critical crack length:
𝑎 𝑐 =
1
𝜋
(
𝐾𝐼𝐶
𝐹𝜎 𝑚𝑎𝑥
)2
Where F is the geometrical factor and 𝜎max is the maximum stress.
𝜎 𝑚𝑎𝑥 =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
10. By assuming 𝜎max = 565 MPa, we calculate the maximum force to cause fracture by using the formula
above.
𝐹𝑜𝑟𝑐𝑒 = 𝜎 𝑚𝑎𝑥 ∗ 𝐴𝑟𝑒𝑎
= 565 ∗ 106
∗ 0.000113097
= 63899.8𝑁
How are modern spindles designed?
Modern spindles are design as a square taper because this is a simple and durable system for modern bikes.
The square spindle can be made from various materials; however aluminum and stainless steel are the
materials being used on the market.
Other spindles that are found for modern spindles designs are the Octalink Interface and the ISIS bottom
bracket interface. The Octalink interface and the ISIS bottom bracket interface are eight grooved splines
that are in lined with eight inverse splines on the crank [1]. The two designs provide more power transfer
from the legs of the rider into the mechanical system of the bike [1]. For high end bikes, the spindle is
designed as an external bottom bracket to improve the performance of the bike performance while
maintaining its durability. The power and the stiffness of the bottom bracket are improved by increasing the
diameter and the size of the spindle [1]. However, with this design, the lifespan of the product will decrease.
What parameters do we know well and what parameters don’t we know well in this problem?
The parameters that we are aware of are the dimensions of the spindle and the crank system. We can also
calculate the forces and moments applied on the system. The parameters that we are unaware of are the
number of cycles which caused the failure and the environmental effects on the spindle.