TRANSILVANIA UNIVERSITY OF BRASOV MECHANICAL ENGINEERING FACULTY DEPARTMENT OF MECHANICAL ENGINEERING ONLY FOR STUDENTS

Similar documents
BENDING OF BEAM. Compressed layer. Elongated. layer. Un-strained. layer. NA= Neutral Axis. Compression. Unchanged. Elongation. Two Dimensional View

7.2.1 Basic relations for Torsion of Circular Members

2 Governing Equations

Static equilibrium requires a balance of forces and a balance of moments.

COLD STRANGLING HOLLOW PARTS FORCES CALCULATION OF CONICAL AND CONICAL WITH CYLINDRICAL COLLAR

PHYS 1444 Lecture #5

LINEAR PLATE BENDING

STRENGTH OF MATERIALS 140AU0402 UNIT 3: BEAMS - LOADS AND STRESSES

STUDY ON 2-D SHOCK WAVE PRESSURE MODEL IN MICRO SCALE LASER SHOCK PEENING

ME 210 Applied Mathematics for Mechanical Engineers

CHAPTER 25 ELECTRIC POTENTIAL

Numerical Integration

CBN 98-1 Developable constant perimeter surfaces: Application to the end design of a tape-wound quadrupole saddle coil

Chapter 2: Basic Physics and Math Supplements

Chapter Introduction to Finite Element Methods

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

Review: Electrostatics and Magnetostatics

ESCI 342 Atmospheric Dynamics I Lesson 3 Fundamental Forces II

Chapter 7-8 Rotational Motion

The Deformation Analysis of the Curved Box Girder Bridges under Different Radius

Qualifying Examination Electricity and Magnetism Solutions January 12, 2006

Chapter 5 Force and Motion

Uniform Circular Motion

working pages for Paul Richards class notes; do not copy or circulate without permission from PGR 2004/11/3 10:50

Chapter 5 Force and Motion

Roof Support 1. Stand-Up Time (RMR):

Electrostatics (Electric Charges and Field) #2 2010

2. Plane Elasticity Problems


Unit 7: Sources of magnetic field

See the solution to Prob Ans. Since. (2E t + 2E c )ch - a. (s max ) t. (s max ) c = 2E c. 2E c. (s max ) c = 3M bh 2E t + 2E c. 2E t. h c.

1) Consider an object of a parabolic shape with rotational symmetry z

TUTORIAL 9. Static magnetic field

Physics 2B Chapter 22 Notes - Magnetic Field Spring 2018

Computational Methods of Solid Mechanics. Project report

Force between two parallel current wires and Newton s. third law

Supplementary Figure 1. Circular parallel lamellae grain size as a function of annealing time at 250 C. Error bars represent the 2σ uncertainty in

Right-handed screw dislocation in an isotropic solid

Problem 1. Part b. Part a. Wayne Witzke ProblemSet #1 PHY 361. Calculate x, the expected value of x, defined by

The physics of induction stoves

Flux. Area Vector. Flux of Electric Field. Gauss s Law

PHYSICS NOTES GRAVITATION

Solving Problems of Advance of Mercury s Perihelion and Deflection of. Photon Around the Sun with New Newton s Formula of Gravity

Cross section dependence on ski pole sti ness

PH 221-3A Fall EQUILIBRIUM and. Lectures Chapter 12 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)

Chapters 5-8. Dynamics: Applying Newton s Laws

Physics 181. Assignment 4

Liquid gas interface under hydrostatic pressure

5. Pressure Vessels and

EFFECTS OF FRINGING FIELDS ON SINGLE PARTICLE DYNAMICS. M. Bassetti and C. Biscari INFN-LNF, CP 13, Frascati (RM), Italy

Objectives: After finishing this unit you should be able to:

ELASTIC ANALYSIS OF CIRCULAR SANDWICH PLATES WITH FGM FACE-SHEETS

arxiv: v1 [physics.pop-ph] 3 Jun 2013

Three dimensional flow analysis in Axial Flow Compressors

Stress, Cauchy s equation and the Navier-Stokes equations

Introduction: Vectors and Integrals

Physics 506 Winter 2006 Homework Assignment #9 Solutions

Physics 111 Lecture 5 Circular Motion

Objects usually are charged up through the transfer of electrons from one object to the other.

Determining solar characteristics using planetary data

r cos, and y r sin with the origin of coordinate system located at

PROBLEM SET #3A. A = Ω 2r 2 2 Ω 1r 2 1 r2 2 r2 1

DESIGN OF BEAMS FOR MOMENTS

Lifting Surfaces. Lifting Surfaces

Lecture 5. Torsion. Module 1. Deformation Pattern in Pure Torsion In Circular Cylinder. IDeALab. Prof. Y.Y.KIM. Solid Mechanics

THE LAPLACE EQUATION. The Laplace (or potential) equation is the equation. u = 0. = 2 x 2. x y 2 in R 2

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion

Analysis of high speed machining center spindle dynamic unit structure performance Yuan guowei

Physics 2212 GH Quiz #2 Solutions Spring 2016

Euclidean Figures and Solids without Incircles or Inspheres

Chapter 13 Gravitation

Math 1105: Calculus I (Math/Sci majors) MWF 11am / 12pm, Campion 235 Written homework 3

1 Equations of linear elasticity

, and the curve BC is symmetrical. Find also the horizontal force in x-direction on one side of the body. h C

B. Spherical Wave Propagation

MECHANICAL PULPING REFINER MECHANICAL PULPS

- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.

Lifting Surfaces. Lifting Surfaces

Physics 122, Fall October 2012

Magneto-Elastic Analysis of an Annular FGM Plate Based on Classical Plate Theory Using GDQ Method

( ) ( )( ) ˆ. Homework #8. Chapter 27 Magnetic Fields II.

Chapter 22 The Electric Field II: Continuous Charge Distributions

(Sample 3) Exam 1 - Physics Patel SPRING 1998 FORM CODE - A (solution key at end of exam)

Construction Figure 10.1: Jaw clutches

THERMODYNAMICS OF SURFACES AND INTERFACES

Vectors, Vector Calculus, and Coordinate Systems

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

LINEAR AND NONLINEAR ANALYSES OF A WIND-TUNNEL BALANCE

SAMPLE LABORATORY SESSION FOR JAVA MODULE B. Calculations for Sample Cross-Section 2

ME311 Machine Design

m1 m2 M 2 = M -1 L 3 T -2

MAGNETIC FIELD AROUND TWO SEPARATED MAGNETIZING COILS

Written as per the revised syllabus prescribed by the Maharashtra State Board of Secondary and Higher Secondary Education, Pune.

EFFECTS OF GFRP REINFORCING REBARS ON SHRINKAGE AND THERMAL STRESSES IN CONCRETE

! E da = 4πkQ enc, has E under the integral sign, so it is not ordinarily an

dq 1 (5) q 1 where the previously mentioned limit has been taken.

THE NAVIER-STOKES EQUATION: The Queen of Fluid Dynamics. A proof simple, but complete.

EM-2. 1 Coulomb s law, electric field, potential field, superposition q. Electric field of a point charge (1)

Inverse Square Law and Polarization

Describing Circular motion

Transcription:

TNSILVNI UNIVSITY OF BSOV CHNICL NGINING FCULTY DPTNT OF CHNICL NGINING Couse 9 Cuved as 9.. Intoduction The eams with plane o spatial cuved longitudinal axes ae called cuved as. Thee ae consideed two classes of polems: a) initiall cuved eams whee the depth of coss-section can e consideed small in elation to the initial adius of cuvatue. ) those eams whee the depth of coss-section and initial adius of cuvatue ae appoximatel of the same ode, i.e. deep eams with high cuvatue. c) The high cuved as ending theo was developed mile Winkle. 9.. Initiall cuved slende eams In this case the ation of 5... 6, whee is the cuvatue o the a and h is the height h of the coss section. Let consideed the cuved a fom Figue 9.. Unde the complex load consisting of foces and moments geneic denoted with P i ( i,,..., n,..., q,...), in the coss sections ae P n developed tensile and shea foces N, and T and ending moments. G s is known the tensile foce N geneate N T a nomal stess unifoml distiuted on the coss section suface: P k O Figue 7. P q N, (9.) whee is the value of the coss section aea. The shea foce T develop a shea stess that is calculated with Juavski elationship: T S I. (9.) STNGTH OF TILS - PT II Pof.d.ing. Ioan Calin OSC

TNSILVNI UNIVSITY OF BSOV CHNICL NGINING FCULTY DPTNT OF CHNICL NGINING The nomal stess geneated the ending moment is found out with the classical Navie s elationship:. (9.3) I 9.3. Deep eams with high initial cuvatue In case of a ation of 5... 6, whee is the cuvatue o the a and h is the height of h the coss section, the as ae consideed to have high initial cuvatue (i.e. small adius of cuvatue). In this case the ending stess has to e calculate with the theo developed mile Winkle. The theo is ased on the following assumptions: The longitudinal axes ae situated in a single plane; This plane is, in the same time, a smmet plane of the a; ll the loads ae applied in the same plane that is the smmet plane; The coss section is consideed to e constant along the a; The mateial satisf the Hooke s law; It is espected the Benoulli s assumption which states that the plane coss sections ae nomal to the longitudinal axes efoe and afte defomation (the shape of the coss section is changed, unde loads, in a neglected atio and so one can conside that emains the same); It is neglected the compession developed on adial diection the ending moment etween the fies. It is consideed a pat of a plane cuved a defined angle d (Figue 9.). Thee ae made the following Figue 9. notations: - inne fie adius; - extenal fie adius; - distance fom the cuvatue cente C to the cente of gavit G ; - is the distance to the neutal axis O O ; - is the cuvatue adius of a fie. STNGTH OF TILS - PT II Pof.d.ing. Ioan Calin OSC

TNSILVNI UNIVSITY OF BSOV CHNICL NGINING FCULTY DPTNT OF CHNICL NGINING Unde the action of the ending moment the end sections (end edges) of the consideed element otate one to the othe one with an angle equal with d. To simplif the calculation one can conside that onl one end edge is otated aound the neutal axe. Thee ae made the following specifications: The fie that is situated on the neutal axe dose not changes its length. The neutal axe divides the coss section in two pats: one whee the nomal stess is positive (tensile), and the othe one whee the value of stess is negative one (compession); In the case of cuved as the neutal axe is not the same with the axe of the coss sections centes of gavit (longitudinal axe) and esults that it is necessa to find out the position of neutal axe. It is consideed a fie that has the length equal with ds that is situated at a distance fom the neutal axe. The length of the fie can e calculate, ased on Figue 9., as: ds d. (9.4) Unde the ending load, the fie length gowth with a quantit Benoulli s assumption, is: ds that, accoding with ds d. (9.5) s it can e seen in Figue (9.), etween the two quantities and exists the elationship:. (9.6) s was mentioned, it is consideed that the mateial satisfies the Hooke s law. This assumption leads to the possiilit to wite, ased on elationships (9.4) (9.6) the stain mateial as: ds ds d d d, (9.7) d and the nomal stess as: ds d. (9.8) d d Osevations: Fom elationship (9.8) esults that the vaiation of the nomal stess, on the coss sections, is epesented a hpeolic function; The highest values ae developed in the fies that ae situated at the exteme edges of the coss section; In the neutal axe o 0 the nomal stess is eo 0. 3 STNGTH OF TILS - PT II Pof.d.ing. Ioan Calin OSC

TNSILVNI UNIVSITY OF BSOV CHNICL NGINING FCULTY DPTNT OF CHNICL NGINING s it is known thee ae two elationships of equivalenc witten as: d 0 ; d, (9.9) that, ased on (9.8) ecome: dx dx dx dx d 0 ; d. (9.0) Fom the fist equation of (9.0) one can otain the geometic position of neutal axe: d d 0, (9.) that leads to: d, (9.) and the integal fom denominato has diffeent values accoding with the coss section shape. Fom the second elationship of (9.0) is otained the elationship of stain: d d. (9.3) d The value of integal fom elationship (9.3) is: d d d d Intoducing (9.4) in (9.3) one can otain: e. (9.4) d, (9.5) d e that comined with (9.8) leads to the nomal stess fomula:, e 4 STNGTH OF TILS - PT II Pof.d.ing. Ioan Calin OSC

TNSILVNI UNIVSITY OF BSOV CHNICL NGINING FCULTY DPTNT OF CHNICL NGINING o,. (9.6) e Consideing the geometical notations thee ae otained the following elationships fo the stesses developed in extemel edges: inne edge: oute edge: ; (9.7) e. (9.8) e : In the case of a tensile load that is supeposed on the ending load, the total stess is given N. (9.9) e 5 STNGTH OF TILS - PT II Pof.d.ing. Ioan Calin OSC