To show how to determine the forces in the members of a truss using the method of joints and the method of sections.

Similar documents
Chapter 6: Structural Analysis

Announcements. Trusses Method of Joints

READING QUIZ. 2. When using the method of joints, typically equations of equilibrium are applied at every joint. A) Two B) Three C) Four D) Six

Lecture 20. ENGR-1100 Introduction to Engineering Analysis THE METHOD OF SECTIONS

ENGR-1100 Introduction to Engineering Analysis. Lecture 20

Chapter 6: Structural Analysis

ENGR-1100 Introduction to Engineering Analysis. Lecture 19

Engineering Mechanics: Statics STRUCTURAL ANALYSIS. by Dr. Ibrahim A. Assakkaf SPRING 2007 ENES 110 Statics

ME Statics. Structures. Chapter 4

6.6 FRAMES AND MACHINES APPLICATIONS. Frames are commonly used to support various external loads.

SRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK

ENGR-1100 Introduction to Engineering Analysis. Lecture 23

Statics: Lecture Notes for Sections

Equilibrium Equilibrium and Trusses Trusses

Lecture 23. ENGR-1100 Introduction to Engineering Analysis FRAMES S 1

Engineering Mechanics: Statics in SI Units, 12e

SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS

Newton s Third Law Newton s Third Law: For each action there is an action and opposite reaction F

Chapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd

FRAMES AND MACHINES Learning Objectives 1). To evaluate the unknown reactions at the supports and the interaction forces at the connection points of a

EQUATIONS OF EQUILIBRIUM & TWO-AND THREE-FORCE MEMEBERS

7 STATICALLY DETERMINATE PLANE TRUSSES

EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS

ENGR-1100 Introduction to Engineering Analysis. Lecture 13

Chapter 5 Equilibrium of a Rigid Body Objectives

Plane Trusses Trusses

ENGINEERING MECHANICS STATIC

Outline: Frames Machines Trusses

In this chapter trusses, frames and machines will be examines as engineering structures.

CHAPTER 5 ANALYSIS OF STRUCTURES. Expected Outcome:

ENT 151 STATICS. Contents. Introduction. Definition of a Truss

Ishik University / Sulaimani Architecture Department. Structure. ARCH 214 Chapter -5- Equilibrium of a Rigid Body

Equilibrium of a Particle

STATICS VECTOR MECHANICS FOR ENGINEERS: Eleventh Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr. David F. Mazurek

Similar to trusses, frames are generally fixed, load carrying structures.

Chapter 3 Trusses. Member CO Free-Body Diagram. The force in CO can be obtained by using section bb. Equations of Equilibrium.

CHAPTER 5 Statically Determinate Plane Trusses

CHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS

CHAPTER 2: EQUILIBRIUM OF RIGID BODIES

ME 201 Engineering Mechanics: Statics

MEE224: Engineering Mechanics Lecture 4

Method of Sections for Truss Analysis

Equilibrium of a Rigid Body. Engineering Mechanics: Statics

3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM

EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMBERS

The centroid of an area is defined as the point at which (12-2) The distance from the centroid of a given area to a specified axis may be found by

ENG202 Statics Lecture 16, Section 7.1

Calculating Truss Forces. Method of Joints

P.E. Civil Exam Review:

The analysis of trusses Mehrdad Negahban (1999)

three Point Equilibrium 1 and planar trusses ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2014 lecture

Chapter 5: Equilibrium of a Rigid Body

EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS

Pin-Jointed Frame Structures (Frameworks)

EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMBERS

When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero.

Announcements. Equilibrium of a Rigid Body

STATICS. Bodies. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Design of a support

EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS

Chapter - 1. Equilibrium of a Rigid Body

STATICS. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Contents 9/3/2015

Chapter 7 INTERNAL FORCES

T R U S S. Priodeep Chowdhury;Lecturer;Dept. of CEE;Uttara University//TRUSS Page 1

5 Equilibrium of a Rigid Body Chapter Objectives

PAT 101 FUNDAMENTAL OF ENGINEERING MECHANICS EQUILIBREQUILIBRIUM OF A RIGID BODY IUM OF A RIGID BODY

Lecture 0. Statics. Module 1. Overview of Mechanics Analysis. IDeALab. Prof. Y.Y.KIM. Solid Mechanics

PROBLEM 6.1 SOLUTION. Free body: Entire truss: (3.2 m) (48 kn)(7.2 m) = 0 = = = BC. 60 kn. Free body: Joint B: = kn T. = 144.

When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero.

Calculating Truss Forces Unit 2 Lesson 2.1 Statics

three Equilibrium 1 and planar trusses ELEMENTS OF ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SPRING 2015 lecture ARCH 614

INTERNAL FORCES Today s Objective: Students will be able to: 1. Use the method of sections for determining internal forces in 2-D load cases.

Mechanics of Materials

Engineering Mechanics: Statics in SI Units, 12e

Vector Mechanics: Statics

Unit M1.4 (All About) Trusses

Engineering Mechanics Statics

EQUILIBRIUM OF A RIGID BODY

1. Determine the Zero-Force Members in the plane truss.

Engineering Mechanics Department of Mechanical Engineering Dr. G. Saravana Kumar Indian Institute of Technology, Guwahati


NAME: Section: RIN: Tuesday, May 19, :00 11:00. Problem Points Score Total 100

Supplement: Statically Indeterminate Trusses and Frames

MT225 Homework 6 - Solution. Problem 1: [34] 5 ft 5 ft 5 ft A B C D. 12 kips. 10ft E F G

three point equilibrium and planar trusses Equilibrium Equilibrium on a Point Equilibrium on a Point

Theory of structure I 2006/2013. Chapter one DETERMINACY & INDETERMINACY OF STRUCTURES

LOVELY PROFESSIONAL UNIVERSITY BASIC ENGINEERING MECHANICS MCQ TUTORIAL SHEET OF MEC Concurrent forces are those forces whose lines of action

Physics 8 Friday, November 8, 2013

EQUILIBRIUM OF RIGID BODIES IN TWO DIMENSIONS

ENGI 1313 Mechanics I

FE Sta'cs Review. Torch Ellio0 (801) MCE room 2016 (through 2000B door)

STATICS. FE Review. Statics, Fourteenth Edition R.C. Hibbeler. Copyright 2016 by Pearson Education, Inc. All rights reserved.

Engineering Mechanics

Support Idealizations

Truss Analysis Method of Joints. Steven Vukazich San Jose State University

1. Determine the Zero-Force Members in the plane truss.

CIV100: Mechanics. Lecture Notes. Module 1: Force & Moment in 2D. You Know What to Do!

Lecture 17 February 23, 2018

If the number of unknown reaction components are equal to the number of equations, the structure is known as statically determinate.

Combined Stress. Axial Stress. Axial vs. Eccentric Load Combined Stress Interaction Formulas

Lecture 14 February 16, 2018

Transcription:

5 Chapter Objectives To show how to determine the forces in the members of a truss using the method of joints and the method of sections. To analyze the forces acting on the members of frames and machines composed of pin-connected members.

In-Class Activities 1. Reading Quiz 2. Simple Trusses 3. The Method of Joints 4. Zero-Force Members 5. The Method of Sections 6. Frames and Machines 7. Concept Quiz

SIMPLE TRUSSES A truss composed of slender members joined together at their end points Planar Trusses Planar trusses are used to support roofs and bridges Roof load is transmitted to the truss at joints by means of a series of purlins

SIMPLE TRUSSES (cont) Planar Trusses The analysis of the forces developed in the truss members is 2D Similar to roof truss, the bridge truss loading is also coplanar

SIMPLE TRUSSES (cont) Assumptions for Design 1. All loadings are applied at the joint - Weight of the members neglected 2. The members are joined together by smooth pins - Assume connections provided the center lines of the joining members are concurrent

SIMPLE TRUSSES (cont) Simple Truss Form of a truss must be rigid to prevent collapse The simplest form that is rigid or stable is a triangle

METHOD OF JOINTS Please refer to the website for the animation: Free-body Diagram for an A-Frame For truss, we need to know the force in each member Forces in the members are internal forces For external force members, equations of equilibrium can be applied Force system acting at each joint is coplanar and concurrent F x = 0 and F y = 0 must be satisfied for equilibrium

METHOD OF JOINTS (cont) Procedure for Analysis Draw the FBD with at least 1 known and 2 unknown forces Find the external reactions at the truss support Determine the correct sense of the member Orient the x and y axes Apply F x = 0 and F y = 0 Use known force to analyze the unknown forces

EXAMPLE 1 Determine the force in each member of the truss and indicate whether the members are in tension or compression.

EXAMPLE 1 (cont) Solution 2 unknown member forces at joint B 1 unknown reaction force at joint C 2 unknown member forces and 2 unknown reaction forces at point A For Joint B, F 500N F F F BC y x cos45 0; BC sin 45 0; N F BA N 0 F 0 F BA BC 707.1N ( C) 500N( T )

EXAMPLE 1 (cont) Solution For Joint C, F F F C y CA y x 0; 707.1cos45 0; 707.1sin 45 N N 0 0 C y F CA 500N( T ) 500N For Joint A, F 500N F 500N y x 0; Ax 0 0; A 0 y A A x y 500N 500N

EXAMPLE 1 (cont) Solution FBD of each pin shows the effect of all the connected members and external forces applied to the pin FBD of each member shows only the effect of the end pins on the member

ZERO-FORCE MEMBERS Method of joints is simplified using zero-force members Zero-force members are supports with no loading In general, when 3 members form a truss joint, the 3 rd member is a zero-force member provided no external force or support reaction is applied to the joint

EXAMPLE 2 Using the method of joints, determine all the zero-force members of the Fink roof truss. Assume all joints are pin connected.

EXAMPLE 2 (cont) Solution For Joint G, F y 0 F 0 GC GC is a zero-force member. For Joint D, F x 0 F 0 DF

EXAMPLE 2 (cont) Solution For Joint F, F 90 y, F 0 F FC 0 FC cos 0 For Joint B, F F BH x 0 2kN 2kN F BH 0

EXAMPLE 2 (cont) Solution F HC satisfies Fy = 0 and therefore HC is not a zero-force member.

THE METHOD OF SECTIONS Used to determine the loadings within a body If a body is in equilibrium, any part of the body is in equilibrium To find forces within members, an imaginary section is used to cut each member into 2 and expose each internal force as external

THE METHOD OF SECTIONS Consider the truss and section a-a as shown Please refer to the website for the animation: Free-Body Diagram for a Truss Member forces are equal and opposite to those acting on the other part Newton s Law

THE METHOD OF SECTIONS Procedure for Analysis Free-Body Diagram Decide the section of the truss Determine the truss s external reactions Use equilibrium equations to solve member forces at the cut session Draw FBD of the sectioned truss which has the least number of forces acting on it Find the sense of an unknown member force Equations of Equilibrium Summed moments about a point Find the 3 rd unknown force from moment equation

EXAMPLE 3 Determine the force in members GE, GC, and BC of the truss. Indicate whether the members are in tension or compression.

EXAMPLE 3 (cont) Solution Choose section a-a since it cuts through the three members Draw FBD of the entire truss F M A F y x 0; 0; 0; 400N 1200N(8m) 400N(3m) D A y A x 0 A 1200N 900N x 400N 0 y A (12m) y 300N 0 D y 900N

EXAMPLE 3 (cont) Solution Draw FBD for the section portion M M G C F 0; 0; y 0; 300N(4m) 400N(3m) F 300N(8m) F 300N 3 5 F GC GE (3m) 0 F 0 F GC BC (3m) 0 F GE 500N( T) BC 800N( C) 800N( T)

FRAMES AND MACHINES Composed of pin-connected multi-force members Frames are stationary Apply equations of equilibrium to each member to determine the unknown forces

FRAMES AND MACHINES (cont) Free-Body Diagram Isolate each part by drawing its outlined shape show all forces and couple moments that act on the part identify each known and unknown force and couple moment indicate any dimensions apply equations of equilibrium assume sense of unknown force or moment draw FBD

EXAMPLE 3 For the frame, draw the free-body diagram of (a) each member, (b) the pin at B and (c) the two members connected together.

EXAMPLE 3 Solution Part (a) BA and BC are not two-force AB is subjected to the resultant forces from the pins

EXAMPLE 3 Solution Part (b) Pin at B is subjected to two forces, force of the member BC and AB on the pin For equilibrium, forces and respective components must be equal but opposite Bx and By shown equal and opposite on members AB

EXAMPLE 3 Solution Part (c) Bx and By are not shown as they form equal but opposite internal forces Unknown force at A and C must act in the same sense Couple moment M is used to find reactions at A and C

Example 4: Solution: By inspection it can be seen that AB is a two force member. The FBD is

Example 5:

Solution:

Example 6: Determine the components of the forces acting on each member of the frame shown. Solution:

Example 6: (cont)

Example 7

Solution 1 Solution 2

CONCEPT QUIZ 1) Truss ABC is changed by decreasing its height from H to 0.9H. Width W and load P are kept the same. Which one of the following statements is true for the revised truss as compared to the original truss? a) Forces in all its members have decreased. b) Forces in all its members have increased. c) Forces in all its members have remained the same. d) None of the above.

CONCEPT QUIZ (cont) 2) For this truss, determine the number of zero-force members. a) 0 F b) 1 F c) 2 d) 3 e) 4

CONCEPT QUIZ (cont) 3) Can you determine the force in member ED by making the cut at section a-a? Explain your answer. a) No, there are 4 unknowns. b) Yes, using MD = 0. c) Yes, using ME = 0. d) Yes, using MB = 0.

CONCEPT QUIZ (cont) 4) If you know F ED, how will you determine F EB? a) By taking section b-b and using ME = 0 b) By taking section b-b, and using FX = 0 and FY = 0 c) By taking section a-a and using MB = 0 d) By taking section a-a and using MD = 0

CONCEPT QUIZ (cont) 5) As shown, a cut is made through members GH, BG and BC to determine the forces in them. Which section will you choose for analysis and why? a) Right, fewer calculations. b) Left, fewer calculations. c) Either right or left, same amount of work. d) None of the above, too many unknowns.

CONCEPT QUIZ (cont) 6) When determining the force in member HG in the previous question, which one equation of equilibrium is best to use? a) MH = 0 b) MG = 0 c) MB = 0 d) MC = 0

CONCEPT QUIZ (cont) 7) When determining the reactions at joints A, B, and C, what is the minimum number of unknowns for solving this problem? a) 3 b) 4 c) 5 d) 6

CONCEPT QUIZ (cont) 8) For the same problem, imagine that you have drawn a FBD of member AB. What will be the easiest way to write an equation involving unknowns at B? a) MC = 0 b) MB = 0 c) MA = 0 d) FX = 0