Truss Example in Two-Dimensions. Ambar K. Mitra

Similar documents
Chapter 13 Exercise 13.1

Worksheet A ALGEBRA PMT

Section 5.3 Practice Exercises Vocabulary and Key Concepts

Math 101, Basic Algebra Author: Debra Griffin

Downloaded from

Math1090 Midterm 2 Review Sections , Solve the system of linear equations using Gauss-Jordan elimination.

Polynomial is a general description on any algebraic expression with 1 term or more. To add or subtract polynomials, we combine like terms.

3.1 Factors and Multiples of Whole Numbers

Algebra I EOC 10-Day STAAR Review. Hedgehog Learning

Decomposing Rational Expressions Into Partial Fractions

Factoring Quadratics: ax 2 + bx + c

MATH 181-Quadratic Equations (7 )

ICAP. Question Bank. Quantitative Methods

( ) 4 ( )! x f) h(x) = 2cos x + 1

Sandringham School Sixth Form. AS Maths. Bridging the gap

Linear function and equations Linear function, simple interest, cost, revenue, profit, break-even

795-kcmil, 3M TM Composite Conductor Compression Dead-end Connector Sustained Load Test in Accordance with ANSI C119.4

TERMINOLOGY 4.1. READING ASSIGNMENT 4.2 Sections 5.4, 6.1 through 6.5. Binomial. Factor (verb) GCF. Monomial. Polynomial.

Chapter 5 Polynomials 5.1 Multiplying Polynomials

Multiplication of Polynomials

Exercises. 140 Chapter 3: Factors and Products

Chapter 10: Mixed strategies Nash equilibria, reaction curves and the equality of payoffs theorem

Chapter 5 Self-Assessment

Introduction to Macroeconomics

Quadratic Algebra Lesson #2

Keynesian Theory (IS-LM Model): how GDP and interest rates are determined in Short Run with Sticky Prices.

ANSWER: We can find consumption and saving by solving:

Lecture 15 - General Equilibrium with Production

Eastern Mediterranean University Faculty of Business and Economics Department of Economics Fall Semester. ECON 101 Mid term Exam

BARUCH COLLEGE MATH 2003 SPRING 2006 MANUAL FOR THE UNIFORM FINAL EXAMINATION

2.4 - Exponential Functions

Annex 1: Background: The Oil and Gas Sector in Somalia

Chapter 4 Partial Fractions

MA 109 College Algebra EXAM 3 - REVIEW

Bulk Upload Standard File Format

Intermediate Macroeconomics: Economics 301 Exam 1. October 4, 2012 B. Daniel

Chapter 6: Quadratic Functions & Their Algebra

Rewriting the Income Tax Act: Exposure Draft. Foreword

The Ohio State University Department of Economics Econ 601 Prof. James Peck Extra Practice Problems Answers (for final)

ALGEBRAIC EXPRESSIONS AND IDENTITIES

Integrating rational functions (Sect. 8.4)

Econ Principles of Microeconomics - Assignment 2

1.4. Arithmetic of Algebraic Fractions. Introduction. Prerequisites. Learning Outcomes

MA 162: Finite Mathematics - Chapter 1

Name: Algebra Unit 7 Polynomials

Unit 8 Notes: Solving Quadratics by Factoring Alg 1

FE Review Economics and Cash Flow

UNIT 5 QUADRATIC FUNCTIONS Lesson 2: Creating and Solving Quadratic Equations in One Variable Instruction

Name. 5. Simplify. a) (6x)(2x 2 ) b) (5pq 2 )( 4p 2 q 2 ) c) (3ab)( 2ab 2 )(2a 3 ) d) ( 6x 2 yz)( 5y 3 z)

Lecture Notes 1 Part B: Functions and Graphs of Functions

Econ Review Set 3 - Answers

25 Increasing and Decreasing Functions

Project Management. Project Mangement. ( Notes ) For Private Circulation Only. Prof. : A.A. Attarwala.

ACCUPLACER Elementary Algebra Assessment Preparation Guide

8.1 Functions Practice Problems

2-4 Completing the Square

Graphing Calculator Appendix

chapter: >> Income and Expenditure WHAT YOU WILL LEARN IN THIS CHAPTER Krugman/Wells The Multiplier: An Informal Introduction

POD. Combine these like terms: 1) 3x 2 4x + 5x x 7x ) 7y 2 + 2y y + 5y 2. 3) 5x 4 + 2x x 7x 4 + 3x x

Topic 4: Analysis of Equilibrium.

9/16/ (1) Review of Factoring trinomials. (2) Develop the graphic significance of factors/roots. Math 2 Honors - Santowski

Jacob: What data do we use? Do we compile paid loss triangles for a line of business?

Topic 12 Factorisation

Issues in International Finance Exchange rates review. UW Madison // Fall 2018

Worksheet 1 Laws of Integral Indices

Math 116: Business Calculus

ECONOMICS QUALIFYING EXAMINATION IN ELEMENTARY MATHEMATICS

Final Exam Review - Business Calculus - Spring x x

ST. DAVID S MARIST INANDA

The Examiner's Answers F1 - Financial Operations March 2014

Exercise 1 Output Determination, Aggregate Demand and Fiscal Policy

Finite Element Method

Time Value of Money and Economic Equivalence

CAS Ratemaking Seminar Call Paper IRR, ROE, and PVI/PVE. Ira Robbin, PhD AVP and Senior Pricing Actuary Endurance US Insurance Operations

WEB APPENDIX 8A 7.1 ( 8.9)

Polynomials. Factors and Greatest Common Factors. Slide 1 / 128. Slide 2 / 128. Slide 3 / 128. Table of Contents

Proportional Relationships Unit

Slide 1 / 128. Polynomials

DEPARTMENT OF ECONOMICS. University of New Hampshire. ECON 401 Principles of Macroeconomics FINAL EXAM. O. Kozlova. Spring 2011

Writing Exponential Equations Day 2

Multiply the binomials. Add the middle terms. 2x 2 7x 6. Rewrite the middle term as 2x 2 a sum or difference of terms. 12x 321x 22

Writing Exponential Equations Day 2

Section 3.1 Relative extrema and intervals of increase and decrease.

Math Final Examination STUDY GUIDE Fall Name Score TOTAL Final Grade

We can solve quadratic equations by transforming the. left side of the equation into a perfect square trinomial

ECON Intermediate Macroeconomic Theory

INTRODUCING RISK MODELING IN CORPORATE FINANCE

Competitiveness, Income Distribution and Economic Growth in a Small Economy

::Solutions:: Problem Set #2: Due end of class October 2, 2018

(8m 2 5m + 2) - (-10m 2 +7m 6) (8m 2 5m + 2) + (+10m 2-7m + 6)

GOOD LUCK! 2. a b c d e 12. a b c d e. 3. a b c d e 13. a b c d e. 4. a b c d e 14. a b c d e. 5. a b c d e 15. a b c d e. 6. a b c d e 16.

Name Date Student id #:

Exercise 2 Short Run Output and Interest Rate Determination in an IS-LM Model

Cosumnes River College Principles of Macroeconomics Problem Set 6 Due April 3, 2017

Factoring. (5) Page 600 #21 43 Right **********Quiz Tomorrow********** (10) Page #20 32 Right; #35 47 Right *****Quiz tomorrow****

MATH 105 CHAPTER 2 page 1

Quantitative Techniques (Finance) 203. Derivatives for Functions with Multiple Variables

LESSON 2 INTEREST FORMULAS AND THEIR APPLICATIONS. Overview of Interest Formulas and Their Applications. Symbols Used in Engineering Economy

Foreign Direct Investment I

Factoring is the process of changing a polynomial expression that is essentially a sum into an expression that is essentially a product.

Transcription:

Truss Example in Two-Dimensions Ambar K. Mitra This document contains screen-shots from the software Statics-Power. Visit www.actuspotentia.com for details. What is a truss? A truss is an assembly of two force members (rods or cables) that are joined at their ends with other members by pins. The entire assembly is then supported with either two pins or with one roller and one pin. What is truss analysis? Determine the support forces from the pins/roller. Determine the force on each two-force member. Determine the nature of the force, i.e. tension/compression, on each twoforce member. Whole Truss Free-Body-Diagram (FBD) Pin-Roller Supported Truss Draw FBD of whole truss. Write three equilibrium equations and determine Ax, Ay, Cy (or Cx). Actus Potentia, Inc. (www.actuspotentia.com) - 1 -

Pin-Pin Supported Truss In general, the four unknowns, Ax, Ay, Bx, By, cannot be determined from three equilibrium equations. However, in some special situations, you can determine two out of the four unknowns. Joint FBD A pin or joint is a point mass; therefore, we can write two force balance equations for equilibrium. Find a pin or joint that has two unknown forces. Determine the two unknown forces from the two equilibrium equations. Determine the forces in all the members by enforcing equilibrium conditions at a series of pins that have two unknown forces. Sign Convention for the Forces Consider that three members meet at a joint. The FBD of the members and the pin are Actus Potentia, Inc. (www.actuspotentia.com) - 2 -

Note that we assumed that all the members are in tension. It is a good idea to stick to this convention. When we know that a force is compression, while writing the equilibrium equations, we insert a negative numerical value for this force. Visual Clue Tensile forces in members show up as arrows pointing outward from a joint. Zero-Force Member No force is acting on the joint P. Three members (PQ, PR, PS) meet at the joint. Two out of three members (PQ and PR) are aligned with one straight line. Force on the third member (PS), F(PS) = 0. F(RP) = F(PQ) Example Actus Potentia, Inc. (www.actuspotentia.com) - 3 -

Figure-1a A truss is supported with two pins at A and F and is loaded as shown. Determine the support forces at the pins and the force on each member. Identify the forces in the members as tension or compression. No force is acting on joint B. Three members (AB, BC, and BG) meet at the joint. Two out of three members (AB and BC) are aligned with one straight line. Force on the third member, F(BG) = 0. F(AB) = F(BC) By removing the zero-force member BG from the truss, we arrive at the truss of Figure-1b. Actus Potentia, Inc. (www.actuspotentia.com) - 4 -

Figure-1b No force is acting on joint G. Three members (AG, GF, and GC) meet at the joint. Two out of three members (AG and GF) are aligned with one straight line. Force on the third member, F(GC) = 0. F(AG) = F(GF) By removing the zero-force member GC from the truss, we arrive at the truss of Figure-1c. Actus Potentia, Inc. (www.actuspotentia.com) - 5 -

Figure-1c No force is acting on joint C. Three members (BC, CD, and CF) meet at the joint. Two out of three members (BC and CD) are aligned with one straight line. Force on the third member, F(CF) = 0. F(BC) = F(CD) By removing the zero-force member CF from the truss, we arrive at the truss of Figure-1d. Actus Potentia, Inc. (www.actuspotentia.com) - 6 -

Figure-1d Note: By identifying and removing one zero-force member in a truss, you may start a chain reaction that creates other zero-force members and removal of new zero-force members greatly simplifies the analysis of the truss. Problem Solution We choose to enforce the equilibrium conditions at joint E. Actus Potentia, Inc. (www.actuspotentia.com) - 7 -

Figure-1e Joint label: E Joint coordinate: (7.5,6) Joint force: (0,-2000) Pin/roller at joint: None Members meet at this joint: 2 (ED and EF) Members with unknown forces: 2 (ED and EF) Member data: o Member ED Label: D Coordinate of D: (6,6) F(ED): unknown o Member EF Actus Potentia, Inc. (www.actuspotentia.com) - 8 -

Label: F Coordinate of F: (6,4) F(EF): unknown The equilibrium equations for joint E are: By solving the equations, we find: Figure-1f F(EF) = -2500lb (2500lb compression) F(ED) = 1500lb (1500lb tension) Next, we choose to enforce the equilibrium conditions at joint F. Actus Potentia, Inc. (www.actuspotentia.com) - 9 -

Figure-1g Joint label: F Joint coordinate: (6,4) Joint force: (0,0) Pin/roller at joint: Pin/roller Fx at pin: unknown Fy at pin: unknown Members meet at this joint: 3 (EF, DF, and FG) Members with unknown forces: 2 (DF and FG) Actus Potentia, Inc. (www.actuspotentia.com) - 10 -

Figure-1h There are four unknowns at this joint, namely, Fx, Fy, F(DF), and F(FG), and two equilibrium equations. Therefore, this joint is not solvable. Next, we choose to enforce the equilibrium conditions at joint D. Actus Potentia, Inc. (www.actuspotentia.com) - 11 -

Figure-1i Joint label: D Joint coordinate: (6,6) Joint force: (0,-1200) Pin/roller at joint: None Members meet at this joint: 3 (DC, DF, and DE) Members with unknown forces: 2 (DC and DF) Member data: o Member DC Label: C Coordinate of D: (5,5) F(DC): unknown o Member DF Label: F Actus Potentia, Inc. (www.actuspotentia.com) - 12 -

Coordinate of F: (6,4) F(DF): unknown o Member DE Label: E Coordinate of F: (7.5,6) F(DE): known = 1500lb The equilibrium equations for joint D are: By solving the equations we find: Figure-1j F(DC) = 2121lb = 2121lb (tension) F(DF) = -2700lb = 2700lb (compression) Actus Potentia, Inc. (www.actuspotentia.com) - 13 -