Statics - TAM 211
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Vector Mechanics: Statics
PDHOnline Course G492 (4 PDH) Vector Mechanics: Statics Mark A. Strain, P.E. 2014 PDH Online | PDH Center 5272 Meadow Estates Drive Fairfax, VA 22030-6658 Phone & Fax: 703-988-0088 www.PDHonline.org www.PDHcenter.com An Approved Continuing Education Provider www.PDHcenter.com PDHonline Course G492 www.PDHonline.org Table of Contents Introduction ..................................................................................................................................... 1 Vectors ............................................................................................................................................ 1 Vector Decomposition ................................................................................................................ 2 Components of a Vector ............................................................................................................. 2 Force ............................................................................................................................................... 4 Equilibrium ..................................................................................................................................... 5 Equilibrium of a Particle ............................................................................................................. 6 Rigid Bodies.............................................................................................................................. 10 Pulleys ...................................................................................................................................... -
Knowledge Assessment in Statics: Concepts Versus Skills
Session 1168 Knowledge Assessment in Statics: Concepts versus skills Scott Danielson Arizona State University Abstract Following the lead of the physics community, engineering faculty have recognized the value of good assessment instruments for evaluating the learning of their students. These assessment instruments can be used to both measure student learning and to evaluate changes in teaching, i.e., did student-learning increase due different ways of teaching. As a result, there are significant efforts underway to develop engineering subject assessment tools. For instance, the Foundation Coalition is supporting assessment tool development efforts in a number of engineering subjects. These efforts have focused on developing “concept” inventories. These concept inventories focus on determining student understanding of the subject’s fundamental concepts. Separately, a NSF-supported effort to develop an assessment tool for statics was begun in the last year by the authors. As a first step, the project team analyzed prior work aimed at delineating important knowledge areas in statics. They quickly recognized that these important knowledge areas contained both conceptual and “skill” components. Both knowledge areas are described and examples of each are provided. Also, a cognitive psychology-based taxonomy of declarative and procedural knowledge is discussed in relation to determining the difference between a concept and a skill. Subsequently, the team decided to focus on development of a concept-based statics assessment tool. The ongoing Delphi process to refine the inventory of important statics concepts and validate the concepts with a broader group of subject matter experts is described. However, the value and need for a skills-based assessment tool is also recognized. -
Fluid Mechanics
MECHANICAL PRINCIPLES HNC/D MOMENTS OF AREA The concepts of first and second moments of area fundamental to several areas of engineering including solid mechanics and fluid mechanics. Students who are not familiar with this concept are advised to complete this tutorial before studying either of these areas. In this section you will do the following. Define the centre of area. Define and calculate 1st. moments of areas. Define and calculate 2nd moments of areas. Derive standard formulae. D.J.Dunn 1 1. CENTROIDS AND FIRST MOMENTS OF AREA A moment about a given axis is something multiplied by the distance from that axis measured at 90o to the axis. The moment of force is hence force times distance from an axis. The moment of mass is mass times distance from an axis. The moment of area is area times the distance from an axis. Fig.1 In the case of mass and area, the problem is deciding the distance since the mass and area are not concentrated at one point. The point at which we may assume the mass concentrated is called the centre of gravity. The point at which we assume the area concentrated is called the centroid. Think of area as a flat thin sheet and the centroid is then at the same place as the centre of gravity. You may think of this point as one where you could balance the thin sheet on a sharp point and it would not tip off in any direction. This section is mainly concerned with moments of area so we will start by considering a flat area at some distance from an axis as shown in Fig.1.2 Fig..2 The centroid is denoted G and its distance from the axis s-s is y. -
Classical Mechanics
Classical Mechanics Hyoungsoon Choi Spring, 2014 Contents 1 Introduction4 1.1 Kinematics and Kinetics . .5 1.2 Kinematics: Watching Wallace and Gromit ............6 1.3 Inertia and Inertial Frame . .8 2 Newton's Laws of Motion 10 2.1 The First Law: The Law of Inertia . 10 2.2 The Second Law: The Equation of Motion . 11 2.3 The Third Law: The Law of Action and Reaction . 12 3 Laws of Conservation 14 3.1 Conservation of Momentum . 14 3.2 Conservation of Angular Momentum . 15 3.3 Conservation of Energy . 17 3.3.1 Kinetic energy . 17 3.3.2 Potential energy . 18 3.3.3 Mechanical energy conservation . 19 4 Solving Equation of Motions 20 4.1 Force-Free Motion . 21 4.2 Constant Force Motion . 22 4.2.1 Constant force motion in one dimension . 22 4.2.2 Constant force motion in two dimensions . 23 4.3 Varying Force Motion . 25 4.3.1 Drag force . 25 4.3.2 Harmonic oscillator . 29 5 Lagrangian Mechanics 30 5.1 Configuration Space . 30 5.2 Lagrangian Equations of Motion . 32 5.3 Generalized Coordinates . 34 5.4 Lagrangian Mechanics . 36 5.5 D'Alembert's Principle . 37 5.6 Conjugate Variables . 39 1 CONTENTS 2 6 Hamiltonian Mechanics 40 6.1 Legendre Transformation: From Lagrangian to Hamiltonian . 40 6.2 Hamilton's Equations . 41 6.3 Configuration Space and Phase Space . 43 6.4 Hamiltonian and Energy . 45 7 Central Force Motion 47 7.1 Conservation Laws in Central Force Field . 47 7.2 The Path Equation . -
Bending Stress
Bending Stress Sign convention The positive shear force and bending moments are as shown in the figure. Figure 40: Sign convention followed. Centroid of an area Scanned by CamScanner If the area can be divided into n parts then the distance Y¯ of the centroid from a point can be calculated using n ¯ Âi=1 Aiy¯i Y = n Âi=1 Ai where Ai = area of the ith part, y¯i = distance of the centroid of the ith part from that point. Second moment of area, or moment of inertia of area, or area moment of inertia, or second area moment For a rectangular section, moments of inertia of the cross-sectional area about axes x and y are 1 I = bh3 x 12 Figure 41: A rectangular section. 1 I = hb3 y 12 Scanned by CamScanner Parallel axis theorem This theorem is useful for calculating the moment of inertia about an axis parallel to either x or y. For example, we can use this theorem to calculate . Ix0 = + 2 Ix0 Ix Ad Bending stress Bending stress at any point in the cross-section is My s = − I where y is the perpendicular distance to the point from the centroidal axis and it is assumed +ve above the axis and -ve below the axis. This will result in +ve sign for bending tensile (T) stress and -ve sign for bending compressive (C) stress. Largest normal stress Largest normal stress M c M s = | |max · = | |max m I S where S = section modulus for the beam. For a rectangular section, the moment of inertia of the cross- 1 3 1 2 sectional area I = 12 bh , c = h/2, and S = I/c = 6 bh . -
Exercises in Classical Mechanics 1 Moments of Inertia 2 Half-Cylinder
Exercises in Classical Mechanics CUNY GC, Prof. D. Garanin No.1 Solution ||||||||||||||||||||||||||||||||||||||||| 1 Moments of inertia (5 points) Calculate tensors of inertia with respect to the principal axes of the following bodies: a) Hollow sphere of mass M and radius R: b) Cone of the height h and radius of the base R; both with respect to the apex and to the center of mass. c) Body of a box shape with sides a; b; and c: Solution: a) By symmetry for the sphere I®¯ = I±®¯: (1) One can ¯nd I easily noticing that for the hollow sphere is 1 1 X ³ ´ I = (I + I + I ) = m y2 + z2 + z2 + x2 + x2 + y2 3 xx yy zz 3 i i i i i i i i 2 X 2 X 2 = m r2 = R2 m = MR2: (2) 3 i i 3 i 3 i i b) and c): Standard solutions 2 Half-cylinder ϕ (10 points) Consider a half-cylinder of mass M and radius R on a horizontal plane. a) Find the position of its center of mass (CM) and the moment of inertia with respect to CM. b) Write down the Lagrange function in terms of the angle ' (see Fig.) c) Find the frequency of cylinder's oscillations in the linear regime, ' ¿ 1. Solution: (a) First we ¯nd the distance a between the CM and the geometrical center of the cylinder. With σ being the density for the cross-sectional surface, so that ¼R2 M = σ ; (3) 2 1 one obtains Z Z 2 2σ R p σ R q a = dx x R2 ¡ x2 = dy R2 ¡ y M 0 M 0 ¯ 2 ³ ´3=2¯R σ 2 2 ¯ σ 2 3 4 = ¡ R ¡ y ¯ = R = R: (4) M 3 0 M 3 3¼ The moment of inertia of the half-cylinder with respect to the geometrical center is the same as that of the cylinder, 1 I0 = MR2: (5) 2 The moment of inertia with respect to the CM I can be found from the relation I0 = I + Ma2 (6) that yields " µ ¶ # " # 1 4 2 1 32 I = I0 ¡ Ma2 = MR2 ¡ = MR2 1 ¡ ' 0:3199MR2: (7) 2 3¼ 2 (3¼)2 (b) The Lagrange function L is given by L('; '_ ) = T ('; '_ ) ¡ U('): (8) The potential energy U of the half-cylinder is due to the elevation of its CM resulting from the deviation of ' from zero. -
Post-Newtonian Approximation
Post-Newtonian gravity and gravitational-wave astronomy Polarization waveforms in the SSB reference frame Relativistic binary systems Effective one-body formalism Post-Newtonian Approximation Piotr Jaranowski Faculty of Physcis, University of Bia lystok,Poland 01.07.2013 P. Jaranowski School of Gravitational Waves, 01{05.07.2013, Warsaw Post-Newtonian gravity and gravitational-wave astronomy Polarization waveforms in the SSB reference frame Relativistic binary systems Effective one-body formalism 1 Post-Newtonian gravity and gravitational-wave astronomy 2 Polarization waveforms in the SSB reference frame 3 Relativistic binary systems Leading-order waveforms (Newtonian binary dynamics) Leading-order waveforms without radiation-reaction effects Leading-order waveforms with radiation-reaction effects Post-Newtonian corrections Post-Newtonian spin-dependent effects 4 Effective one-body formalism EOB-improved 3PN-accurate Hamiltonian Usage of Pad´eapproximants EOB flexibility parameters P. Jaranowski School of Gravitational Waves, 01{05.07.2013, Warsaw Post-Newtonian gravity and gravitational-wave astronomy Polarization waveforms in the SSB reference frame Relativistic binary systems Effective one-body formalism 1 Post-Newtonian gravity and gravitational-wave astronomy 2 Polarization waveforms in the SSB reference frame 3 Relativistic binary systems Leading-order waveforms (Newtonian binary dynamics) Leading-order waveforms without radiation-reaction effects Leading-order waveforms with radiation-reaction effects Post-Newtonian corrections Post-Newtonian spin-dependent effects 4 Effective one-body formalism EOB-improved 3PN-accurate Hamiltonian Usage of Pad´eapproximants EOB flexibility parameters P. Jaranowski School of Gravitational Waves, 01{05.07.2013, Warsaw Relativistic binary systems exist in nature, they comprise compact objects: neutron stars or black holes. These systems emit gravitational waves, which experimenters try to detect within the LIGO/VIRGO/GEO600 projects. -
Kinematics, Kinetics, Dynamics, Inertia Kinematics Is a Branch of Classical
Course: PGPathshala-Biophysics Paper 1:Foundations of Bio-Physics Module 22: Kinematics, kinetics, dynamics, inertia Kinematics is a branch of classical mechanics in physics in which one learns about the properties of motion such as position, velocity, acceleration, etc. of bodies or particles without considering the causes or the driving forces behindthem. Since in kinematics, we are interested in study of pure motion only, we generally ignore the dimensions, shapes or sizes of objects under study and hence treat them like point objects. On the other hand, kinetics deals with physics of the states of the system, causes of motion or the driving forces and reactions of the system. In a particular state, namely, the equilibrium state of the system under study, the systems can either be at rest or moving with a time-dependent velocity. The kinetics of the prior, i.e., of a system at rest is studied in statics. Similarly, the kinetics of the later, i.e., of a body moving with a velocity is called dynamics. Introduction Classical mechanics describes the area of physics most familiar to us that of the motion of macroscopic objects, from football to planets and car race to falling from space. NASCAR engineers are great fanatics of this branch of physics because they have to determine performance of cars before races. Geologists use kinematics, a subarea of classical mechanics, to study ‘tectonic-plate’ motion. Even medical researchers require this physics to map the blood flow through a patient when diagnosing partially closed artery. And so on. These are just a few of the examples which are more than enough to tell us about the importance of the science of motion. -
Relationship of Structure and Stiffness in Laminated Bamboo Composites
Construction and Building Materials 165 (2018) 241–246 Contents lists available at ScienceDirect Construction and Building Materials journal homepage: www.elsevier.com/locate/conbuildmat Technical note Relationship of structure and stiffness in laminated bamboo composites Matthew Penellum a, Bhavna Sharma b, Darshil U. Shah c, Robert M. Foster c,1, ⇑ Michael H. Ramage c, a Department of Engineering, University of Cambridge, United Kingdom b Department of Architecture and Civil Engineering, University of Bath, United Kingdom c Department of Architecture, University of Cambridge, United Kingdom article info abstract Article history: Laminated bamboo in structural applications has the potential to change the way buildings are con- Received 22 August 2017 structed. The fibrous microstructure of bamboo can be modelled as a fibre-reinforced composite. This Received in revised form 22 November 2017 study compares the results of a fibre volume fraction analysis with previous experimental beam bending Accepted 23 December 2017 results. The link between fibre volume fraction and bending stiffness shows that differences previously attributed to preservation treatment in fact arise due to strip thickness. Composite theory provides a basis for the development of future guidance for laminated bamboo, as validated here. Fibre volume frac- Keywords: tion analysis is an effective method for non-destructive evaluation of bamboo beam stiffness. Microstructure Ó 2018 The Authors. Published by Elsevier Ltd. This is an open access article under the CC BY license (http:// Mechanical properties Analytical modelling creativecommons.org/licenses/by/4.0/). Bamboo 1. Introduction produce a building material [5]. Structural applications are cur- rently limited by a lack of understanding of the properties. -
Centrally Loaded Columns
Ziemian c03.tex V1 - 10/15/2009 4:17pm Page 23 CHAPTER 3 CENTRALLY LOADED COLUMNS 3.1 INTRODUCTION The cornerstone of column theory is the Euler column, a mathematically straight, prismatic, pin-ended, centrally loaded1 strut that is slender enough to buckle without the stress at any point in the cross section exceeding the proportional limit of the material. The buckling load or critical load or bifurcation load (see Chapter 2 for a discussion of the significance of these terms) is defined as π 2EI P = (3.1) E L2 where E is the modulus of elasticity of the material, I is the second moment of area of the cross section about which buckling takes place, and L is the length of the column. The Euler load, P E , is a reference value to which the strengths of actual columns are often compared. If end conditions other than perfectly frictionless pins can be defined mathemat- ically, the critical load can be expressed by π 2EI P = (3.2) E (KL)2 where KL is an effective length defining the portion of the deflected shape between points of zero curvature (inflection points). In other words, KL is the length of an equivalent pin-ended column buckling at the same load as the end-restrained 1Centrally loaded implies that the axial load is applied through the centroidal axis of the member, thus producing no bending or twisting. 23 Ziemian c03.tex V1 - 10/15/2009 4:17pm Page 24 24 CENTRALLY LOADED COLUMNS column. For example, for columns in which one end of the member is prevented from translating with respect to the other end, K can take on values ranging from 0.5 to l.0, depending on the end restraint. -
New Rotational Dynamics- Inertia-Torque Principle and the Force Moment the Character of Statics Guagsan Yu* Harbin Macro, Dynamics Institute, P
Computa & tio d n ie a l l p M Yu, J Appl Computat Math 2015, 4:3 p a Journal of A t h f e o m l DOI: 10.4172/2168-9679.1000222 a a n t r ISSN: 2168-9679i c u s o J Applied & Computational Mathematics Research Article Open Access New Rotational Dynamics- Inertia-Torque Principle and the Force Moment the Character of Statics GuagSan Yu* Harbin Macro, Dynamics Institute, P. R. China Abstract Textual point of view, generate in a series of rotational dynamics experiment. Initial research, is wish find a method overcome the momentum conservation. But further study, again detected inside the classical mechanics, the error of the principle of force moment. Then a series of, momentous the error of inside classical theory, all discover come out. The momentum conservation law is wrong; the newton third law is wrong; the energy conservation law is also can surpass. After redress these error, the new theory namely perforce bring. This will involve the classical physics and mechanics the foundation fraction, textbooks of physics foundation part should proceed the grand modification. Keywords: Rigid body; Inertia torque; Centroid moment; Centroid occurrence change? The Inertia-torque, namely, such as Figure 3 the arm; Statics; Static force; Dynamics; Conservation law show, the particle m (the m is also its mass) is a r 1 to O the slewing radius, so it the Inertia-torque is: Introduction I = m.r1 (1.1) Textual argumentation is to bases on the several simple physics experiment, these experiments pass two videos the document to The Inertia-torque is similar with moment of force, to the same of proceed to demonstrate. -
Lecture Notes for PHY 405 Classical Mechanics
Lecture Notes for PHY 405 Classical Mechanics From Thorton & Marion's Classical Mechanics Prepared by Dr. Joseph M. Hahn Saint Mary's University Department of Astronomy & Physics November 25, 2003 revised November 29 Chapter 11: Dynamics of Rigid Bodies rigid body = a collection of particles whose relative positions are fixed. We will ignore the microscopic thermal vibrations occurring among a `rigid' body's atoms. The inertia tensor Consider a rigid body that is composed of N particles. Give each particle an index α = 1 : : : N. Particle α has a velocity vf,α measured in the fixed reference frame and velocity vr,α = drα=dt in the reference frame that rotates about axis !~ . Then according to Eq. 10.17: vf,α = V + vr,α + !~ × rα where V = velocity of the moving origin relative to the fixed origin. Note that the particles in a rigid body have zero relative velocity, ie, vr,α = 0. 1 Thus vα = V + !~ × rα after dropping the f subscript 1 1 and T = m v2 = m (V + !~ × r )2 is the particle's KE α 2 α α 2 α α N so T = Tα the system's total KE is α=1 X1 1 = MV2 + m V · (!~ × r ) + m (!~ × r )2 2 α α 2 α α α α X X 1 1 = MV2 + V · !~ × m r + m (!~ × r )2 2 α α 2 α α α ! α X X where M = α mα = the system's total mass. P Now choose the system's center of mass R as the origin of the moving frame.