CHAPTER 6 Space Trusses
|
|
- Dayna Curtis
- 7 years ago
- Views:
Transcription
1 CHAPTER 6 Space Trusses
2 INTRODUCTION A space truss consists of members joined together at their ends to form a stable threedimensional structures A stable simple space truss can be built from the basic tetrahedral, formed by connecting six members with four joints
3 INTRODUCTION
4 OBJECTIVES To determine the stability and determinacy of space trusses To determine member forces of space trusses using tension coefficient analysis
5 1. Simple Space Truss TYPE OF SPACE TRUSSES This truss is constructed from a tetrahedron. The truss can be enlarged by adding three members.
6 TYPE OF SPACE TRUSSES 2. Compound Space Truss This truss is constructed by combining two or more simple truss.
7 TYPE OF SPACE TRUSSES 3. Complex Space Truss Complex truss is a truss that cannot be classified as simple truss or compound truss.
8 DETERMINACY & STABILITY Due to three dimensions, there will be three equations of equilibrium for each joint. ( F x = 0; F y = 0; F z = 0) The external stability of the space truss requires that the support reactions keep the truss in force and moment equilibrium. Generally, the least number of required reactions for stable and externally determinate is SIX If r < 6 Unstable If r > 6 Externally Indeterminate
9 DETERMINACY & STABILITY For internal determinacy, if m = number of members; j = number of joints; r = number of supports; therefore: If m = 3j + r Stable and Internally Determinate If m < 3j + r Unstable If m > 3j + r Internally Indeterminate Internal stability can sometimes be checked by careful inspection of the member arrangement.
10 DETERMINACY & STABILITY Internally m + r = 3j m + r > 3j m + r < 3j Externally r < 6 r = 6 r > 6 Determinate Truss Indeterminate Truss Unstable Truss Unstable Truss Determinate if Truss is Stable Indeterminate Truss
11 TYPES OF SUPPORT
12 EXAMPLE 1 Ball & Socket m = 3, j = 4, r = 9 m + r = 12 3j = 12 m + r = 3j Determinate Truss
13 EXAMPLE 2 m = 15, j = 10, r = 15 Ball & Socket m + r = 30 3j = 30 m + r = 3j Determinate Truss
14 EXAMPLE 3 m = 13, j = 8, r = 12 m + r = 25 3j = 24 m + r = 25 3j = 24 Ball & Socket Indeterminate Truss in the First Degree
15 ASSUMPTIONS FOR DESIGN The members are joined together by smooth pins (no friction cannot resist moment) All loadings and reactions are applied centrally at the joints The centroid for each members are straight and concurrent at a joint Therefore, each truss member acts as an axial force member: If the force tends to elongate Tensile (T) If the force tends to shorten Compressive (C)
16 THEOREM 1: ZERO FORCE MEMBERS If all members and external force except one member at a joint, (say, joint B) lie in the same plane, then, the force in member A is zero. Member A B z P y x The force in member A is zero
17 THEOREM 2: ZERO FORCE MEMBERS If all members at a joint has zero force except for two members, (say member A and B), and both members (A and B) do not lie in a straight line, then the force in member A and B are zero. Member B Member A F By F Bx F Ay FAx 0 Both members A and B has zero force because both members do not lie in a straight line. 0 0
18 ZERO FORCE MEMBERS If members A and B lie in a straight line, then, the forces in these members MIGHT NOT be zero. In fact, referring to the example below: F Ax = -F Bx F Ay = -F By Member A Member B F Ay FAx F Bx FBy
19 THEOREM 3: ZERO FORCE MEMBERS If three members at a joint do not lie in the same plane and there is no external force at that joint, then the force in the three members is zero. Member B Member A Three members connected at a joint has zero force. A plane can consists of two members, say member A and B. Thus, no force can balance the component of member C that is normal to the plane. Member C
20 Identify the members of the space truss that has zero force. Theorem 1: Joint L F 1 = 0 and F 2 = 0 Theorem 3: Joint K F 3 = F 4 = F 5 = 0 and F 1 = 0 Theorem 3: Joint M F 6 = F 7 = F 8 = 0 and F 2 = 0 Theorem 3: Joint J F 9 = F 10 = F 11 = 0 and F 5 = F 6 = 0 EXAMPLE 4
21 EXAMPLE 5 Theorem 1: Joint F Members FE, FC, FD lie in a plane, except member FG. Thus, member FG has zero force. Theorem 3: Joint F Members FE, FC, FD have zero force. Theorem 3: Joint E Members ED, EA, EH have zero force.
22 ANALYSIS METHOD: TENSION COEFFICIENT To determine member forces Based on 3D particle equilibrium L x = L cos : L y = L cos ; L z = L cos F x = F cos : F y = F cos ; F z = F cos Therefore: F x L x = F L F y L y = F L F z L z = F L
23 ANALYSIS METHOD: TENSION Tension Coefficient: t = F/L COEFFICIENT F x = t L x ; F y = t L y ; F z = t L z If t positive (tension), if t negative (compression) The actual force and length are given from: L = L x 2 + L y 2 + L z 2 F = F x 2 + F y 2 + F z 2 For 3D equilibrium: F x = 0 ; F y = 0 ; F z = 0 tl x = 0 ; tl y = 0 ; tl z = 0
24 ANALYSIS METHOD: TENSION COEFFICIENT y L, F L y, F y L z, F z x A L x, F x z The component of length, L, and force, F in the x, y, z direction
25 B y ANALYSIS METHOD: TENSION COEFFICIENT O F y F X F x A D x F x = Fcosθ x F x = F L x L = F L L x F x = t L x z E F z C Where t = F/L t = tension coefficient
26 B y ANALYSIS METHOD: TENSION COEFFICIENT O F y Y F F x A D x F y = Fcosθ y F y = F L y L = F L L y F z F y = t L y E C z
27 B y ANALYSIS METHOD: TENSION COEFFICIENT F y F z O Z F F x A D x F z = Fcosθ z F z = F L z L = F L L z F z = t L z E C z
28 EXAMPLE 6 The space truss shown in the figure has roller and socket support at joint A, B, C and D z y x i) Determine the member force for all members at joint F and G 20 kn 40 kn E 40 kn 10 kn 40 kn G ii) Determine the reaction at support C A B F C 2 m 2 m 4 m D 1 m
29 1. Start at Joint G Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) GC GE GF Force (kn) F x = 0: -4t GE + 10 = 0 F z = 0: -2t GC 40 = 0 EXAMPLE 6 Solution F y = 0: -2t GE 2t GF = 0 t GE = 2.5 kn/m t GC = -20 kn/m -2(2.5) 2t GF = 0 t GF = -2.5 kn/m
30 2. Move to Joint F Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) FC FD FE FG Force (kn) F x = 0: -4t FE = 0 F y = 0: 2t FC t FD + 2t FG = 0 F z = 0: -2t FC 2t FD 40 = 0...(ii) (i) + (ii): -3t FD -45 = 0 From (i): -2t FC (-15) 5 = 0 EXAMPLE 6 Solution t FE = 0 kn/m 2t FC t FD 5 = 0... (i) t FD = -15 kn/m t FC = -5 kn/m
31 3. Calculate the Reaction at Joint C Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) CF CG Force (kn) R Cx R Cy R Cz F x = 0: R Cx = 0 kn F y = 0: -2t CF + R Cy = 0-2(-5) + R Cy = 0 R Cy = -10 kn ( ) F z = 0: 2t CF + 2t CG + R Cz = 0 EXAMPLE 6 Solution 2(-5) + 2(50) + R Cz = 0 R Cz = 50 kn ( )
32 Slotted roller constraint in a cylinder z B C EXAMPLE 7 Short link Determine the force in each member of the space truss shown. 4 m D x A E z = 4 kn Ball & socket 4 m y 2 m E 2 m
33 EXAMPLE 7 Solution 1. Start with joints where there are only 3 unknowns force/reaction. Joint B 5 unknowns Joint C 5 unknowns Joint A 7 unknowns Joint E 4 unknowns Joint D Theorem 3: Three members at a joint and no external force. Thus, all members have zero forces. t DC = t DA = t DE = 0
34 Joint E EXAMPLE 7 Solution Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) EC EB ED EA Force (kn) F z = 0: 4t EB + 4t EC 4 = 0...(i) F x = 0: -2t EC + 2t EB 2t ED + 2t EA = 0... (ii) where t ED = 0 F y = 0: -4t EC 4t EB 4t ED 4t EA = 0...(iii) where t ED = 0 Solve Eq. (i), (ii) & (iii): t EC = 0 kn/m, t EA = -1 kn/m and t EB = 1 kn/m
35 Joint C EXAMPLE 7 Solution Joint C has three (3) unknowns, as t CE = t CD = 0 z R Cy C Therefore, by Theorem 3: t CB = t CA = R Cy = 0 B D x A y E z = 4 kn E
36 Joint C EXAMPLE 7 Solution Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) CE CA CD CB Force (kn) 0 -R Cy 0 F z = 0: 4t CA = 0 F x = 0: 4t CA + 4t CB = 0 F y = 0: -R Cy = 0 t CA = 0 kn/m t CB = 0 kn/m
37 Joint B EXAMPLE 7 Solution Joint B has three (3) unknowns. z C R By B R Bx D x A y E z = 4 kn E
38 Joint B EXAMPLE 7 Solution Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) BC BA BE Force (kn) R Bx -R By 0 F x = 0: -4t BC 2t BE + R Bx = 0 F y = 0: -4t BE R By = 0 F z = 0: -4t BA 4t BE = 0 R Bx = 2 kn R By = -4 kn t BA = -1 kn/m
39 Joint A EXAMPLE 7 Solution z C B D R Ay x R Ax A R Az y E z = 4 kn E
40 Joint A EXAMPLE 7 Solution Member L x (m) L y (m) L z (m) L (m) t (kn/m) F (kn) AB AC AD AE Force (kn) -R Ax R Ay R Az F x = 0: -4t AC 4t AD 2t AE R Ax = 0 F y = 0: 4t AE + R Ay = 0 F z = 0: 4t AB + 4t AC + R Az = 0 R Ax = -2 kn R Ay = -4 kn R Az = 4 kn
41 Method 2 EXAMPLE 7 Solution Reactions at A, B and C can also be obtained from Method 2 R By z B R Cy C R Bx D R Ay x R Ax A R Az y E z = 4 kn E
42 Method 2 M y = 0-4(2) + R Bx (4) = 0 R Bx = 2 kn M z = 0 R Cy = 0 kn M x = 0 R By (4) 4(4) = 0 R By = 4 kn EXAMPLE 7 Solution F x = 0 2 R Ax = 0 R Ax = 2 kn F y = 0 R Ay 4 = 0 R Ay = 4 kn F z = 0 R Az 4 = 0 R Az = 4 kn
43 CLASS EXERCISE The space truss shown in the Figure consists of six members and is supported by a short link at A, two short links at B, and a ball and socket at D. Analyse the force in each of the members for the given loading. y 2 m 2 m A 3 m B O D z 7 m C x 1800 N
Statically Indeterminate Structure. : More unknowns than equations: Statically Indeterminate
Statically Indeterminate Structure : More unknowns than equations: Statically Indeterminate 1 Plane Truss :: Determinacy No. of unknown reactions = 3 No. of equilibrium equations = 3 : Statically Determinate
More informationVisa Smart Debit/Credit Certificate Authority Public Keys
CHIP AND NEW TECHNOLOGIES Visa Smart Debit/Credit Certificate Authority Public Keys Overview The EMV standard calls for the use of Public Key technology for offline authentication, for aspects of online
More informationCE 201 (STATICS) DR. SHAMSHAD AHMAD CIVIL ENGINEERING ENGINEERING MECHANICS-STATICS
COURSE: CE 201 (STATICS) LECTURE NO.: 28 to 30 FACULTY: DR. SHAMSHAD AHMAD DEPARTMENT: CIVIL ENGINEERING UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA TEXT BOOK: ENGINEERING
More informationFuture Trends in Airline Pricing, Yield. March 13, 2013
Future Trends in Airline Pricing, Yield Management, &AncillaryFees March 13, 2013 THE OPPORTUNITY IS NOW FOR CORPORATE TRAVEL MANAGEMENT BUT FIRST: YOU HAVE TO KNOCK DOWN BARRIERS! but it won t hurt much!
More informationENGINEERING MECHANICS STATIC
EX 16 Using the method of joints, determine the force in each member of the truss shown. State whether each member in tension or in compression. Sol Free-body diagram of the pin at B X = 0 500- BC sin
More informationModule 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method. Version 2 CE IIT, Kharagpur
Module Analysis of Statically Indeterminate Structures by the Matrix Force Method esson 11 The Force Method of Analysis: Frames Instructional Objectives After reading this chapter the student will be able
More informationStatically determinate structures
Statically determinate structures A statically determinate structure is the one in which reactions and internal forces can be determined solely from free-body diagrams and equations of equilibrium. These
More informationChapter 1: Statics. A) Newtonian Mechanics B) Relativistic Mechanics
Chapter 1: Statics 1. The subject of mechanics deals with what happens to a body when is / are applied to it. A) magnetic field B) heat C ) forces D) neutrons E) lasers 2. still remains the basis of most
More informationSOLUTION 6 6. Determine the force in each member of the truss, and state if the members are in tension or compression.
6 6. etermine the force in each member of the truss, and state if the members are in tension or compression. 600 N 4 m Method of Joints: We will begin by analyzing the equilibrium of joint, and then proceed
More informationBASIC CONCEPTS AND CONVENTIONAL METHODS OF STUCTURAL ANALYSIS (LECTURE NOTES)
BASIC CONCEPTS AND CONVENTIONA METHODS OF STUCTURA ANAYSIS (ECTURE NOTES) DR. MOHAN KAANI (Retired Professor of Structural Engineering) DEPARTMENT OF CIVI ENGINEERING INDIAN INSTITUTE OF TECHNOOGY (BOMBAY)
More informationBedford, Fowler: Statics. Chapter 4: System of Forces and Moments, Examples via TK Solver
System of Forces and Moments Introduction The moment vector of a force vector,, with respect to a point has a magnitude equal to the product of the force magnitude, F, and the perpendicular distance from
More informationCopyright 2011 Casa Software Ltd. www.casaxps.com. Centre of Mass
Centre of Mass A central theme in mathematical modelling is that of reducing complex problems to simpler, and hopefully, equivalent problems for which mathematical analysis is possible. The concept of
More information4.2 Free Body Diagrams
CE297-FA09-Ch4 Page 1 Friday, September 18, 2009 12:11 AM Chapter 4: Equilibrium of Rigid Bodies A (rigid) body is said to in equilibrium if the vector sum of ALL forces and all their moments taken about
More informationShear Forces and Bending Moments
Chapter 4 Shear Forces and Bending Moments 4.1 Introduction Consider a beam subjected to transverse loads as shown in figure, the deflections occur in the plane same as the loading plane, is called the
More informationPLANE TRUSSES. Definitions
Definitions PLANE TRUSSES A truss is one of the major types of engineering structures which provides a practical and economical solution for many engineering constructions, especially in the design of
More informationAnalysis of Statically Determinate Trusses
Analysis of Statically Determinate Trusses THEORY OF STRUCTURES Asst. Prof. Dr. Cenk Üstündağ Common Types of Trusses A truss is one of the major types of engineering structures which provides a practical
More informationStructural Axial, Shear and Bending Moments
Structural Axial, Shear and Bending Moments Positive Internal Forces Acting Recall from mechanics of materials that the internal forces P (generic axial), V (shear) and M (moment) represent resultants
More informationSERVER CERTIFICATES OF THE VETUMA SERVICE
Page 1 Version: 3.4, 19.12.2014 SERVER CERTIFICATES OF THE VETUMA SERVICE 1 (18) Page 2 Version: 3.4, 19.12.2014 Table of Contents 1. Introduction... 3 2. Test Environment... 3 2.1 Vetuma test environment...
More informationDHL EXPRESS CANADA E-BILL STANDARD SPECIFICATIONS
DHL EXPRESS CANADA E-BILL STANDARD SPECIFICATIONS 1 E-Bill Standard Layout A B C D E F G Field/ DHL Account Number Billing Customer Name Billing Customer Address Billing Customer City Billing Customer
More informationCROSS REFERENCE. Cross Reference Index 110-122. Cast ID Number 110-111 Connector ID Number 111 Engine ID Number 112-122. 2015 Ford Motor Company 109
CROSS REFERENCE Cross Reference Index 110-122 Cast ID Number 110-111 Connector ID Number 111 112-122 2015 Ford Motor Company 109 CROSS REFERENCE Cast ID Number Cast ID Ford Service # MC Part # Part Type
More informationThe aerodynamic center
The aerodynamic center In this chapter, we re going to focus on the aerodynamic center, and its effect on the moment coefficient C m. 1 Force and moment coefficients 1.1 Aerodynamic forces Let s investigate
More informationP4 Stress and Strain Dr. A.B. Zavatsky MT07 Lecture 3 Statically Indeterminate Structures
4 Stress and Strain Dr... Zavatsky MT07 ecture 3 Statically Indeterminate Structures Statically determinate structures. Statically indeterminate structures (equations of equilibrium, compatibility, and
More informationIntroduction to Statics
Introduction to Statics.PDF Edition Version 0.95 Unit 19 Trusses: Method of Sections Helen Margaret Lester Plants Late Professor Emerita Wallace Starr Venable Emeritus Associate Professor West Virginia
More informationHow To Solve Factoring Problems
05-W4801-AM1.qxd 8/19/08 8:45 PM Page 241 Factoring, Solving Equations, and Problem Solving 5 5.1 Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring
More informationSTRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION
Chapter 11 STRESS AND DEFORMATION ANALYSIS OF LINEAR ELASTIC BARS IN TENSION Figure 11.1: In Chapter10, the equilibrium, kinematic and constitutive equations for a general three-dimensional solid deformable
More informationCIRCLE COORDINATE GEOMETRY
CIRCLE COORDINATE GEOMETRY (EXAM QUESTIONS) Question 1 (**) A circle has equation x + y = 2x + 8 Determine the radius and the coordinates of the centre of the circle. r = 3, ( 1,0 ) Question 2 (**) A circle
More informationAnalyze and Evaluate a Truss
#3 Learning Activity #3: Analyze and Evaluate a Truss Overview of the Activity In this learning activity, we will analyze and evaluate one of the main trusses from the Grant Road Bridge. We will create
More informationMethod of Joints. Method of Joints. Method of Joints. Method of Joints. Method of Joints. Method of Joints. CIVL 3121 Trusses - Method of Joints 1/5
IVL 3121 Trusses - 1/5 If a truss is in equilibrium, then each of its joints must be in equilibrium. The method of joints consists of satisfying the equilibrium equations for forces acting on each joint.
More informationStructural Analysis - II Prof. P. Banerjee Department of Civil Engineering Indian Institute of Technology, Bombay. Lecture - 02
Structural Analysis - II Prof. P. Banerjee Department of Civil Engineering Indian Institute of Technology, Bombay Lecture - 02 Good morning. Today is the second lecture in the series of lectures on structural
More informationTOPIC 4: DERIVATIVES
TOPIC 4: DERIVATIVES 1. The derivative of a function. Differentiation rules 1.1. The slope of a curve. The slope of a curve at a point P is a measure of the steepness of the curve. If Q is a point on the
More informationWarm-up Theorems about triangles. Geometry. Theorems about triangles. Misha Lavrov. ARML Practice 12/15/2013
ARML Practice 12/15/2013 Problem Solution Warm-up problem Lunes of Hippocrates In the diagram below, the blue triangle is a right triangle with side lengths 3, 4, and 5. What is the total area of the green
More informationChapter 4.1 Parallel Lines and Planes
Chapter 4.1 Parallel Lines and Planes Expand on our definition of parallel lines Introduce the idea of parallel planes. What do we recall about parallel lines? In geometry, we have to be concerned about
More informationMechanics of Materials. Chapter 4 Shear and Moment In Beams
Mechanics of Materials Chapter 4 Shear and Moment In Beams 4.1 Introduction The term beam refers to a slender bar that carries transverse loading; that is, the applied force are perpendicular to the bar.
More informationDATING YOUR GUILD 1952-1960
DATING YOUR GUILD 1952-1960 YEAR APPROXIMATE LAST SERIAL NUMBER PRODUCED 1953 1000-1500 1954 1500-2200 1955 2200-3000 1956 3000-4000 1957 4000-5700 1958 5700-8300 1959 12035 1960-1969 This chart displays
More informationDesign MEMO 54a Reinforcement design for RVK 41
Page of 5 CONTENTS PART BASIC ASSUMTIONS... 2 GENERAL... 2 STANDARDS... 2 QUALITIES... 3 DIMENSIONS... 3 LOADS... 3 PART 2 REINFORCEMENT... 4 EQUILIBRIUM... 4 Page 2 of 5 PART BASIC ASSUMTIONS GENERAL
More informationProblem 1: Computation of Reactions. Problem 2: Computation of Reactions. Problem 3: Computation of Reactions
Problem 1: Computation of Reactions Problem 2: Computation of Reactions Problem 3: Computation of Reactions Problem 4: Computation of forces and moments Problem 5: Bending Moment and Shear force Problem
More informationv v ax v a x a v a v = = = Since F = ma, it follows that a = F/m. The mass of the arrow is unchanged, and ( )
Week 3 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution
More informationStatics of Structural Supports
Statics of Structural Supports TYPES OF FORCES External Forces actions of other bodies on the structure under consideration. Internal Forces forces and couples exerted on a member or portion of the structure
More informationPattern Co. Monkey Trouble Wall Quilt. Size: 48" x 58"
.............................................................................................................................................. Pattern Co..........................................................................................
More information0242-1. HSR TRAINING COURSE REQUIREMENTS HSR Training Course Guidance Booklet 2
0242-1 HSR TRAINING COURSE REQUIREMENTS HSR Training Course Guidance Booklet 2 SafeWork SA 2 Contents Introduction... 4 Learning resources... 4 PART 1 UNDERPINNING PRINCIPLES FOR THE DEVELOPMENT OF A SAFEWORK
More informationSERVER CERTIFICATES OF THE VETUMA SERVICE
Page 1 Version: 3.5, 4.11.2015 SERVER CERTIFICATES OF THE VETUMA SERVICE 1 (18) Page 2 Version: 3.5, 4.11.2015 Table of Contents 1. Introduction... 3 2. Test Environment... 3 2.1 Vetuma test environment...
More informationDesign MEMO 60 Reinforcement design for TSS 102
Date: 04.0.0 sss Page of 5 CONTENTS PART BASIC ASSUMTIONS... GENERAL... STANDARDS... QUALITIES... 3 DIMENSIONS... 3 LOADS... 3 PART REINFORCEMENT... 4 EQUILIBRIUM... 4 Date: 04.0.0 sss Page of 5 PART BASIC
More informationIntermediate Value Theorem, Rolle s Theorem and Mean Value Theorem
Intermediate Value Theorem, Rolle s Theorem and Mean Value Theorem February 21, 214 In many problems, you are asked to show that something exists, but are not required to give a specific example or formula
More informationDeflections. Question: What are Structural Deflections?
Question: What are Structural Deflections? Answer: The deformations or movements of a structure and its components, such as beams and trusses, from their original positions. It is as important for the
More informationApproximate Analysis of Statically Indeterminate Structures
Approximate Analysis of Statically Indeterminate Structures Every successful structure must be capable of reaching stable equilibrium under its applied loads, regardless of structural behavior. Exact analysis
More information1.3 Algebraic Expressions
1.3 Algebraic Expressions A polynomial is an expression of the form: a n x n + a n 1 x n 1 +... + a 2 x 2 + a 1 x + a 0 The numbers a 1, a 2,..., a n are called coefficients. Each of the separate parts,
More informationINCIDENCE-BETWEENNESS GEOMETRY
INCIDENCE-BETWEENNESS GEOMETRY MATH 410, CSUSM. SPRING 2008. PROFESSOR AITKEN This document covers the geometry that can be developed with just the axioms related to incidence and betweenness. The full
More informationTHREE DIMENSIONAL GEOMETRY
Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,
More informationUSB HID to PS/2 Scan Code Translation Table
Key Name HID Usage Page HID Usage ID PS/2 Set 1 Make* PS/2 Set 1 Break* PS/2 Set 2 Make PS/2 Set 2 Break System Power 01 81 E0 5E E0 DE E0 37 E0 F0 37 System Sleep 01 82 E0 5F E0 DF E0 3F E0 F0 3F System
More informationService Instruction. 1.0 SUBJECT: ECi Accessory Cases for Lycoming 4-Cylinder engines with single magneto configurations and TITAN 361 Engines
Title: Service Instruction ECi Accessory Cases Installed on Engines S.I. No.: 03-1 Page: 1 of 7 Issued: 2/28/2003 Revision: 2 (4/13/2009) Technical Portions are FAA DER Approved. 1.0 SUBJECT: ECi Accessory
More information2. Axial Force, Shear Force, Torque and Bending Moment Diagrams
2. Axial Force, Shear Force, Torque and Bending Moment Diagrams In this section, we learn how to summarize the internal actions (shear force and bending moment) that occur throughout an axial member, shaft,
More informationQuadrilateral Geometry. Varignon s Theorem I. Proof 10/21/2011 S C. MA 341 Topics in Geometry Lecture 19
Quadrilateral Geometry MA 341 Topics in Geometry Lecture 19 Varignon s Theorem I The quadrilateral formed by joining the midpoints of consecutive sides of any quadrilateral is a parallelogram. PQRS is
More informationMath 312 Homework 1 Solutions
Math 31 Homework 1 Solutions Last modified: July 15, 01 This homework is due on Thursday, July 1th, 01 at 1:10pm Please turn it in during class, or in my mailbox in the main math office (next to 4W1) Please
More informationNew approaches in Eurocode 3 efficient global structural design
New approaches in Eurocode 3 efficient global structural design Part 1: 3D model based analysis using general beam-column FEM Ferenc Papp* and József Szalai ** * Associate Professor, Department of Structural
More informationCM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation
CM2202: Scientific Computing and Multimedia Applications General Maths: 2. Algebra - Factorisation Prof. David Marshall School of Computer Science & Informatics Factorisation Factorisation is a way of
More informationWORK DONE BY A CONSTANT FORCE
WORK DONE BY A CONSTANT FORCE The definition of work, W, when a constant force (F) is in the direction of displacement (d) is W = Fd SI unit is the Newton-meter (Nm) = Joule, J If you exert a force of
More informationThe elements used in commercial codes can be classified in two basic categories:
CHAPTER 3 Truss Element 3.1 Introduction The single most important concept in understanding FEA, is the basic understanding of various finite elements that we employ in an analysis. Elements are used for
More information3. Reaction Diffusion Equations Consider the following ODE model for population growth
3. Reaction Diffusion Equations Consider the following ODE model for population growth u t a u t u t, u 0 u 0 where u t denotes the population size at time t, and a u plays the role of the population dependent
More informationChapter 7 WORK, ENERGY, AND Power Work Done by a Constant Force Kinetic Energy and the Work-Energy Theorem Work Done by a Variable Force Power
Chapter 7 WORK, ENERGY, AND Power Work Done by a Constant Force Kinetic Energy and the Work-Energy Theorem Work Done by a Variable Force Power Examples of work. (a) The work done by the force F on this
More informationProjective Geometry - Part 2
Projective Geometry - Part 2 Alexander Remorov alexanderrem@gmail.com Review Four collinear points A, B, C, D form a harmonic bundle (A, C; B, D) when CA : DA CB DB = 1. A pencil P (A, B, C, D) is the
More informationIntroduction to Engineering Analysis - ENGR1100 Course Description and Syllabus Monday / Thursday Sections. Fall '15.
Introduction to Engineering Analysis - ENGR1100 Course Description and Syllabus Monday / Thursday Sections Fall 2015 All course materials are available on the RPI Learning Management System (LMS) website.
More informationwww.sakshieducation.com
LENGTH OF THE PERPENDICULAR FROM A POINT TO A STRAIGHT LINE AND DISTANCE BETWEEN TWO PAPALLEL LINES THEOREM The perpendicular distance from a point P(x 1, y 1 ) to the line ax + by + c 0 is ax1+ by1+ c
More informationAP Physics - Chapter 8 Practice Test
AP Physics - Chapter 8 Practice Test Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A single conservative force F x = (6.0x 12) N (x is in m) acts on
More informationNotes from February 11
Notes from February 11 Math 130 Course web site: www.courses.fas.harvard.edu/5811 Two lemmas Before proving the theorem which was stated at the end of class on February 8, we begin with two lemmas. The
More informationm i: is the mass of each particle
Center of Mass (CM): The center of mass is a point which locates the resultant mass of a system of particles or body. It can be within the object (like a human standing straight) or outside the object
More informationFactoring Polynomials and Solving Quadratic Equations
Factoring Polynomials and Solving Quadratic Equations Math Tutorial Lab Special Topic Factoring Factoring Binomials Remember that a binomial is just a polynomial with two terms. Some examples include 2x+3
More informationSolutions to Practice Problems
Higher Geometry Final Exam Tues Dec 11, 5-7:30 pm Practice Problems (1) Know the following definitions, statements of theorems, properties from the notes: congruent, triangle, quadrilateral, isosceles
More informationVehicle Identification Numbering System 00.03
Vehicle Identification Numbering System 00.03 IMPORTANT: See Subject 050 for the vehicle identification numbering system for vehicles built before May 1, 2000. Federal Motor Vehicle Safety Standard 115
More informationAdvanced Encryption Standard by Example. 1.0 Preface. 2.0 Terminology. Written By: Adam Berent V.1.7
Written By: Adam Berent Advanced Encryption Standard by Example V.1.7 1.0 Preface The following document provides a detailed and easy to understand explanation of the implementation of the AES (RIJNDAEL)
More informationC relative to O being abc,, respectively, then b a c.
2 EP-Program - Strisuksa School - Roi-et Math : Vectors Dr.Wattana Toutip - Department of Mathematics Khon Kaen University 200 :Wattana Toutip wattou@kku.ac.th http://home.kku.ac.th/wattou 2. Vectors A
More informationMATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set.
MATH 304 Linear Algebra Lecture 9: Subspaces of vector spaces (continued). Span. Spanning set. Vector space A vector space is a set V equipped with two operations, addition V V (x,y) x + y V and scalar
More informationA DIVISION OF THE MENO. Meno proposes a question: whether virtue can be taught. Three conversations or discussions following question
A DIVISION OF THE MENO 70A 70B-100B Meno proposes a question: whether virtue can be taught Three conversations or discussions following question 70B-80D Conversation on a question before Meno's: what is
More informationLecture 14: Section 3.3
Lecture 14: Section 3.3 Shuanglin Shao October 23, 2013 Definition. Two nonzero vectors u and v in R n are said to be orthogonal (or perpendicular) if u v = 0. We will also agree that the zero vector in
More informationIdeal Cable. Linear Spring - 1. Cables, Springs and Pulleys
Cables, Springs and Pulleys ME 202 Ideal Cable Neglect weight (massless) Neglect bending stiffness Force parallel to cable Force only tensile (cable taut) Neglect stretching (inextensible) 1 2 Sketch a
More informationAdvanced Encryption Standard by Example. 1.0 Preface. 2.0 Terminology. Written By: Adam Berent V.1.5
Written By: Adam Berent Advanced Encryption Standard by Example V.1.5 1.0 Preface The following document provides a detailed and easy to understand explanation of the implementation of the AES (RIJNDAEL)
More informationKALE: A High-Degree Algebraic-Resistant Variant of The Advanced Encryption Standard
KALE: A High-Degree Algebraic-Resistant Variant of The Advanced Encryption Standard Dr. Gavekort c/o Vakiopaine Bar Kauppakatu 6, 41 Jyväskylä FINLAND mjos@iki.fi Abstract. We have discovered that the
More informationCLASSICAL STRUCTURAL ANALYSIS
Table of Contents CASSCA STRUCTURA ANAYSS... Conjugate beam method... External work and internal work... 3 Method of virtual force (unit load method)... 5 Castigliano s second theorem... Method of consistent
More informationA UNIVERSAL METHOD OF SOLVING QUARTIC EQUATIONS
International Journal of Pure and Applied Mathematics Volume 71 No. 011, 51-59 A UNIVERSAL METHOD OF SOLVING QUARTIC EQUATIONS Sergei L. Shmakov Saratov State University 83, Astrakhanskaya Str., Saratov,
More informationLuxembourg (Luxembourg): Trusted List
Luxembourg (Luxembourg): Trusted List Institut Luxembourgeois de la Normalisation, de l'accréditation de la Sécurité et qualité des produits et services Scheme Information TSL Version 4 TSL Sequence Number
More informationAcceptance Page 2. Revision History 3. Introduction 14. Control Categories 15. Scope 15. General Requirements 15
Acceptance Page 2 Revision History 3 Introduction 14 Control Categories 15 Scope 15 General Requirements 15 Control Category: 0.0 Information Security Management Program 17 Objective Name: 0.01 Information
More informationRigid and Braced Frames
Rigid Frames Rigid and raced Frames Rigid frames are identified b the lack of pinned joints within the frame. The joints are rigid and resist rotation. The ma be supported b pins or fied supports. The
More informationThe Method of Partial Fractions Math 121 Calculus II Spring 2015
Rational functions. as The Method of Partial Fractions Math 11 Calculus II Spring 015 Recall that a rational function is a quotient of two polynomials such f(x) g(x) = 3x5 + x 3 + 16x x 60. The method
More informationSection 16: Neutral Axis and Parallel Axis Theorem 16-1
Section 16: Neutral Axis and Parallel Axis Theorem 16-1 Geometry of deformation We will consider the deformation of an ideal, isotropic prismatic beam the cross section is symmetric about y-axis All parts
More informationENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P
ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P This material is duplicated in the Mechanical Principles module H2 and those
More informationexpression is written horizontally. The Last terms ((2)( 4)) because they are the last terms of the two polynomials. This is called the FOIL method.
A polynomial of degree n (in one variable, with real coefficients) is an expression of the form: a n x n + a n 1 x n 1 + a n 2 x n 2 + + a 2 x 2 + a 1 x + a 0 where a n, a n 1, a n 2, a 2, a 1, a 0 are
More informationThe ASCII Character Set
The ASCII Character Set The American Standard Code for Information Interchange or ASCII assigns values between 0 and 255 for upper and lower case letters, numeric digits, punctuation marks and other symbols.
More informationGeometry. Relationships in Triangles. Unit 5. Name:
Geometry Unit 5 Relationships in Triangles Name: 1 Geometry Chapter 5 Relationships in Triangles ***In order to get full credit for your assignments they must me done on time and you must SHOW ALL WORK.
More informationThe following sketches show the plans of the two cases of one-way slabs. The spanning direction in each case is shown by the double headed arrow.
9.2 One-way Slabs This section covers the following topics. Introduction Analysis and Design 9.2.1 Introduction Slabs are an important structural component where prestressing is applied. With increase
More information1 Determinants and the Solvability of Linear Systems
1 Determinants and the Solvability of Linear Systems In the last section we learned how to use Gaussian elimination to solve linear systems of n equations in n unknowns The section completely side-stepped
More information4. How many integers between 2004 and 4002 are perfect squares?
5 is 0% of what number? What is the value of + 3 4 + 99 00? (alternating signs) 3 A frog is at the bottom of a well 0 feet deep It climbs up 3 feet every day, but slides back feet each night If it started
More informationSection 8.8. 1. The given line has equations. x = 3 + t(13 3) = 3 + 10t, y = 2 + t(3 + 2) = 2 + 5t, z = 7 + t( 8 7) = 7 15t.
. The given line has equations Section 8.8 x + t( ) + 0t, y + t( + ) + t, z 7 + t( 8 7) 7 t. The line meets the plane y 0 in the point (x, 0, z), where 0 + t, or t /. The corresponding values for x and
More informationSolving Simultaneous Equations and Matrices
Solving Simultaneous Equations and Matrices The following represents a systematic investigation for the steps used to solve two simultaneous linear equations in two unknowns. The motivation for considering
More informationOnline EFFECTIVE AS OF JANUARY 2013
2013 A and C Session Start Dates (A-B Quarter Sequence*) 2013 B and D Session Start Dates (B-A Quarter Sequence*) Quarter 5 2012 1205A&C Begins November 5, 2012 1205A Ends December 9, 2012 Session Break
More informationAlgebraic Properties and Proofs
Algebraic Properties and Proofs Name You have solved algebraic equations for a couple years now, but now it is time to justify the steps you have practiced and now take without thinking and acting without
More informationCOUNCIL OF INTERNATIONAL SCHOOLS (CIS) SCHOOL IMPROVEMENT THROUGH ACCREDITATION
COUNCIL OF INTERNATIONAL SCHOOLS (CIS) SCHOOL IMPROVEMENT THROUGH ACCREDITATION Guide to School Evaluation and Accreditation - 8 th Edition (Version 8.2) STANDARDS AND INDICATORS SECTION A SCHOOL GUIDING
More informationLecture L2 - Degrees of Freedom and Constraints, Rectilinear Motion
S. Widnall 6.07 Dynamics Fall 009 Version.0 Lecture L - Degrees of Freedom and Constraints, Rectilinear Motion Degrees of Freedom Degrees of freedom refers to the number of independent spatial coordinates
More informationStresses in Beam (Basic Topics)
Chapter 5 Stresses in Beam (Basic Topics) 5.1 Introduction Beam : loads acting transversely to the longitudinal axis the loads create shear forces and bending moments, stresses and strains due to V and
More informationFactoring Polynomials
Factoring Polynomials Factoring Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 x 2 are 2x + 1 and 3x 2. In this section, we will be factoring
More informationThe University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY. Wednesday, January 28, 2015 9:15 a.m. to 12:15 p.m.
GEOMETRY The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY Wednesday, January 28, 2015 9:15 a.m. to 12:15 p.m., only Student Name: School Name: The possession or use of any
More informationwww.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x
Mechanics 2 : Revision Notes 1. Kinematics and variable acceleration Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx differentiate a = dv = d2 x dt dt dt 2 Acceleration Velocity
More information