SRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK
|
|
- Vernon Ross
- 5 years ago
- Views:
Transcription
1 SRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK
2 SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS Today s Objectives: Students will be able to: a) Define a simple truss. b) Determine the forces in members of a simple truss. c) Identify zero-force members. In-Class Activities: Applications Simple Trusses Zero-force Members Group Problem Solving
3 APPLICATIONS Trusses are commonly used to support roofs. For a given truss geometry and load, how can you determine the forces in the truss members and thus be able to select their sizes? A more challenging question is, that for a given load, how can we design the trusses geometry to minimize cost?
4 APPLICATIONS Trusses are also used in a variety of structures like cranes and the frames of aircraft or the space station. How can you design a light weight structure satisfying load, safety, cost specifications, is simple to manufacture, and allows easy inspection over its lifetime?
5 SIMPLE TRUSSES A truss is a structure composed of slender members joined together at their end points. If a truss, along with the imposed load, lies in a single plane (as shown at the top right), then it is called a planar truss. A simple truss is a planar truss which begins with a triangular element and can be expanded by adding two members and a joint. For these trusses, the number of members (M) and the number of joints (J) are related by the equation M = 2 J 3.
6 ANALYSIS AND DESIGN ASSUMPTIONS When designing the members and joints of a truss, first it is necessary to determine the forces in each truss member. This is called the force analysis of a truss. When doing this, two assumptions are made: 1. All loads are applied at the joints. The weight of the truss members is often neglected as the weight is usually small as compared to the forces supported by the members. 2. The members are joined together by smooth pins. This assumption is satisfied in most practical cases where the joints are formed by bolting the ends together. With these two assumptions, the members act as two-force members. They are loaded in either tension or compression. Often compressive members are made thicker to prevent buckling.
7 STEPS FOR ANALYSIS 1. If the truss s support reactions are not given, draw a FBD of the entire truss and determine the support reactions (typically using scalar equations of equilibrium). 2. Draw the free-body diagram of a joint with one or two unknowns. Assume that all unknown member forces act in tension (pulling on the pin) unless you can determine by inspection that the forces are compression loads. 3. Apply the scalar equations of equilibrium, F X = 0 and F Y = 0, to determine the unknown(s). If the answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression). 4. Repeat steps 2 and 3 at each joint in succession until all the required forces are determined.
8 ZERO-FORCE MEMBERS If a joint has only two non-collinear members and there is no external load or support reaction at that joint, then those two members are zero-force members. In this example members DE, DC, AF, and AB are zero force members. Zero-force members can be removed (as shown in the figure) when analyzing the truss.
9 ZERO-FORCE MEMBERS If three members form a truss joint for which two of the members are collinear and there is no external load or reaction at that joint, then member DA and CA are zero force members. Please note that zero-force members are used to increase stability and rigidity of the truss, and to provide support for various different loading conditions.
10 EXAMPLE 1 Given: Loads as shown on the truss Find: The forces in each member of the truss. Plan: 1. Check if there are any zero-force members. (Note that member BD is zero-force member. F BD = 0) 2. First analyze pin D and then pin A.
11 EXAMPLE 1: SOLUTION 45 º F AD D 450 kn 45 º F CD FBD of pin D + F X = F CD cos 45 F AD cos 45 = 0 + F Y = F CD sin 45 F AD sin 45 = 0 F CD = 318 kn (Tension) or (T) and F AD = 318 kn (Compression) or (C)
12 EXAMPLE 1: SOLUTION Analyzing pin A: F AD A 45 º F AB A Y FBD of pin A + F X = F AB + ( 318) cos 45 = 0; F AB = 225 kn (T) Could you have analyzed Joint C instead of A?
13 GROUP PROBLEM SOLVING Given: Loads as shown on the truss Find: Plan: Determine the force in all the truss members (do not forget to mention whether they are in T or C). a) Check if there are any zero-force members. Is member CE zero-force member? b) Draw FBDs of pins D, C, and E, and then apply E-of-E at those pins to solve for the unknowns.
14 GROUP PROBLEM SOLVING FBD of pin D Y 5 F DE 3 4 D 600N X F CD Analyzing pin D: + F X = F DE (3/5) 600 = 0 F DE = 1000 N = 1.00 kn (C) + F Y = 1000 (4/5) F CD = 0 F CD = 800 N = 0.8 kn (T)
15 GROUP PROBLEM SOLVING FBD of pin C Y F CD = 800 N F CE C 900 N X Analyzing pin C: F BC + F X = F CE 900 = 0 F CE = 900 N = 0.90 kn (C) + F Y = 800 F BC = 0 F BC = 800 N = 0.80 kn (T)
16 GROUP PROBLEM SOLVING FBD of pin E Y 3 F DE = 1000 N E F CE = 900 N F AE F BE X Analyzing pin E: + F X = F AE (3/5) + F BE (3/5) 1000 (3/5) 900 = 0 + F Y = F AE (4/5) F BE (4/5) 1000 (4/5) = 0 Solving these two equations, we get F AE = 1750 N = 1.75 kn (C) F BE = 750 N = 0.75 kn (T)
17 THE METHOD OF SECTIONS Today s Objectives: Students will be able to determine: 1. Forces in truss members using the method of sections. In-Class Activities: Applications Method of Sections Group Problem Solving
18 APPLICATIONS Long trusses are often used to construct large cranes and large electrical transmission towers. The method of joints requires that many joints be analyzed before we can determine the forces in the middle of a large truss. So another method to determine those forces is helpful.
19 THE METHODS OF SECTIONS In the method of sections, a truss is divided into two parts by taking an imaginary cut (shown here as a-a) through the truss. Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut members will also be either tensile or compressive with the same magnitude as the forces at the joint. This result is based on the equilibrium principle and Newton s third law.
20 STEPS FOR ANALYSIS 1. Decide how you need to cut the truss. This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general). 2. Decide which side of the cut truss will be easier to work with (minimize the number of external reactions). 3. If required, determine any necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E.
21 STEPS FOR ANALYSIS 4. Draw the FBD of the selected part of the cut truss. You need to indicate the unknown forces at the cut members. Initially, you may assume all the members are in tension, as done when using the method of joints. Upon solving, if the answer is positive, the member is in tension as per the assumption. If the answer is negative, the member must be in compression. (Please note that you can also assume forces to be either tension or compression by inspection as was done in the figures above.)
22 STEPS FOR ANALYSIS 5. Apply the scalar equations of equilibrium (E-of-E) to the selected cut section of the truss to solve for the unknown member forces. Please note, in most cases it is possible to write one equation to solve for one unknown directly. So look for it and take advantage of such a shortcut!
23 EXAMPLE Given: Loads as shown on the truss. Find: Plan: The force in members KJ, KD, and CD. a) Take a cut through members KJ, KD and CD. b) Work with the left part of the cut section. Why? c) Determine the support reactions at A. What are they? d) Apply the E-of-E to find the forces in KJ, KD and CD.
24 EXAMPLE: SOLUTION 56.7 kn Analyzing the entire truss for the reactions at A, we get F X = A X = 0. A moment equation about G to find A Y results in: M G = A Y (18) 20 (15) 30 (12) 40 (9) = 0; A Y = 56.7 kn Now take moments about point D. Why do this? + M D = 56.7 (9) + 20 (6) + 30 (3) F KJ (4) = 0 F KJ = 75.1 kn or 75.1 kn ( C )
25 EXAMPLE: SOLUTION 56.7 kn Now use the x and y-directions equations of equilibrium. + F Y = (4/5) F KD = 0; F KD = 8.38 kn (T) + F X = ( 75.1) + (3/5) (8.38) + F CD = 0; F CD = 70.1 kn (T)
26 GROUP PROBLEM SOLVING Given: Loads as shown on the truss. Find: The force in members GB and GF. Plan: a) Take the cut through members GF, GB, and AB. b) Analyze the left section. Determine the support reactions at A. Why? c) Draw the FBD of the left section. d) Apply the equations of equilibrium (if possible, try to do it so that every equation yields an answer to one unknown.
27 GROUP PROBLEM SOLVING: SOLUTION 1) Determine the support reactions at A by drawing the FBD of the entire truss. + F X = A X = 0 + M D = A Y (28) (18) (10) = 0; A Y = kn Why is A x equal zero by inspection?
28 GROUP PROBLEM SOLVING: SOLUTION 2) Analyze the left section. + M B = (10) + F GF (10) = 0; F GF = 671 kn (C) + F Y = F GB = 0; F GB = 671 kn (T)
29 FRAMES AND MACHINES Today s Objectives: Students will be able to: a) Draw the free-body diagram of a frame or machine and its members. b) Determine the forces acting at the joints and supports of a frame or machine. In-Class Activities: Applications Analysis of a Frame/Machine Group Problem Solving
30 APPLICATIONS Frames are commonly used to support various external loads. How is a frame different than a truss? To be able to design a frame, you need to determine the forces at the joints and supports.
31 APPLICATIONS Machines, like those above, are used in a variety of applications. How are they different from trusses and frames? How can you determine the loads at the joints and supports? These forces and moments are required when designing the machine s members.
32 DEFINITION: FRAMES AND MACHINES Frame Machine Frames and machines are two common types of structures that have at least one multi-force member. (Recall that trusses have nothing but two-force members). Frames are generally stationary and support external loads. Machines contain moving parts and are designed to alter the effect of forces.
33 STEPS FOR ANALYZING A FRAME OR MACHINE 1. Draw a FBD of the frame or machine and its members, as necessary. Hints: a) Identify any two-force members, F AB b) Note that forces on contacting surfaces (usually between a pin and a member) are equal and opposite, and c) For a joint with more than two members or an external force, it is advisable to draw a FBD of the pin.
34 STEPS FOR ANALYZING A FRAME OR MACHINE 2. Develop a strategy to apply the equations of equilibrium to solve for the unknowns. Look for ways to form single equations and single unknowns. F AB Problems are going to be challenging since there are usually several unknowns. A lot of practice is needed to develop good strategies and ease of solving these problems.
35 EXAMPLE Given: The frame supports an external load and moment as shown. Find: The horizontal and vertical components of the pin reactions at C and the magnitude of reaction at B. Plan: a) Draw FBDs of the frame member BC. Why pick this part of the frame? b) Apply the equations of equilibrium and solve for the unknowns at C and B.
36 EXAMPLE: SOLUTION 800 N m 400 N C X 1 m F AB 45 FBD of member BC (Note AB is a 2-force member!) Please note that member AB is a two-force member. Equations of Equilibrium: Start with B 1 m 2 m M C since it yields one unknown.. + M C = F AB sin45 (1) F AB cos45 (3) N m (2) = 0 F AB = 1131 N C Y
37 EXAMPLE: SOLUTION. 800 N m 400 N C X 1 m B 1 m 2 m C Y F AB 45 FBD of member BC + F X = C X sin 45 = 0 C X = 800 N + F Y = C Y cos = 0 C Y = 400 N
38 GROUP PROBLEM SOLVING Given: A frame supports a 50-kN load as shown. Find: Plan: The reactions exerted by the pins on the frame members at B and C. a) Draw a FBD of member BC and another one for AC. b) Apply the equations of equilibrium to each FBD to solve for the four unknowns. Think about a strategy to easily solve for the unknowns.
39 GROUP PROBLEM SOLVING: SOLUTION FBDs of members BC and AC 3.5 m 50 kn C Y 6 m C X A X 8 m Applying E-of-E to member AC: A Y + M A = C Y (8) + C X (6) + 50 (3.5) = 0 (1) + F X = C X A X = 0 + F Y = 50 A Y C Y = 0
40 GROUP PROBLEM SOLVING: SOLUTION FBDs of members BC and AC 50 kn 3.5 m C Y 6 m C X A X 8 m Applying E-of-E to member BC: A Y + M B = 50 (2) 50 (3.5) + C Y (8) = 0 ; C Y = = 34.4 kn From Eq (1), C X can be determined; C X = = 16.7 kn + F X = B X = 0 ; B X = 66.7 kn + F Y = B Y = 0 ; B Y = 15.6 kn
41 The End
ENGR-1100 Introduction to Engineering Analysis. Lecture 19
ENGR-1100 Introduction to Engineering Analysis Lecture 19 SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS Today s Objectives: Students will be able to: In-Class Activities: a) Define a simple
More informationChapter 6: Structural Analysis
Chapter 6: Structural Analysis APPLICATIONS Trusses are commonly used to support a roof. For a given truss geometry and load, how can we determine the forces in the truss members and select their sizes?
More informationAnnouncements. Trusses Method of Joints
Announcements Mountain Dew is an herbal supplement Today s Objectives Define a simple truss Trusses Method of Joints Determine the forces in members of a simple truss Identify zero-force members Class
More informationSIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS
SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZERO-FORCE MEMBERS Today s Objectives: Students will be able to: a) Define a simple truss. b) Determine the forces in members of a simple truss. c) Identify zero-force
More informationStatics: Lecture Notes for Sections
Chapter 6: Structural Analysis Today s Objectives: Students will be able to: a) Define a simple truss. b) Determine the forces in members of a simple truss. c) Identify zero-force members. READING QUIZ
More informationEngineering Mechanics: Statics STRUCTURAL ANALYSIS. by Dr. Ibrahim A. Assakkaf SPRING 2007 ENES 110 Statics
CHAPTER Engineering Mechanics: Statics STRUCTURAL ANALYSIS College of Engineering Department of Mechanical Engineering Tenth Edition 6a by Dr. Ibrahim A. Assakkaf SPRING 2007 ENES 110 Statics Department
More informationLecture 20. ENGR-1100 Introduction to Engineering Analysis THE METHOD OF SECTIONS
ENGR-1100 Introduction to Engineering Analysis Lecture 20 THE METHOD OF SECTIONS Today s Objectives: Students will be able to determine: 1. Forces in truss members using the method of sections. In-Class
More informationENGR-1100 Introduction to Engineering Analysis. Lecture 20
ENGR-1100 Introduction to Engineering Analysis Lecture 20 Today s Objectives: THE METHOD OF SECTIONS Students will be able to determine: 1. Forces in truss members using the method of sections. In-Class
More informationENGR-1100 Introduction to Engineering Analysis. Lecture 23
ENGR-1100 Introduction to Engineering Analysis Lecture 23 Today s Objectives: Students will be able to: a) Draw the free body diagram of a frame and its members. FRAMES b) Determine the forces acting at
More informationLecture 23. ENGR-1100 Introduction to Engineering Analysis FRAMES S 1
ENGR-1100 Introduction to Engineering Analysis Lecture 23 Today s Objectives: Students will be able to: a) Draw the free body diagram of a frame and its members. FRAMES b) Determine the forces acting at
More informationME Statics. Structures. Chapter 4
ME 108 - Statics Structures Chapter 4 Outline Applications Simple truss Method of joints Method of section Germany Tacoma Narrows Bridge http://video.google.com/videoplay?docid=-323172185412005564&q=bruce+lee&pl=true
More information6.6 FRAMES AND MACHINES APPLICATIONS. Frames are commonly used to support various external loads.
6.6 FRAMES AND MACHINES APPLICATIONS Frames are commonly used to support various external loads. How is a frame different than a truss? How can you determine the forces at the joints and supports of a
More informationTo show how to determine the forces in the members of a truss using the method of joints and the method of sections.
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
More informationPlane Trusses Trusses
TRUSSES Plane Trusses Trusses- It is a system of uniform bars or members (of various circular section, angle section, channel section etc.) joined together at their ends by riveting or welding and constructed
More informationREADING 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
READING QUIZ 1. One of the assumptions used when analyzing a simple truss is that the members are joined together by. A) Welding B) Bolting C) Riveting D) Smooth pins E) Super glue 2. When using the method
More informationCHAPTER 2: EQUILIBRIUM OF RIGID BODIES
For a rigid body to be in equilibrium, the net force as well as the net moment about any arbitrary point O must be zero Summation of all external forces. Equilibrium: Sum of moments of all external forces.
More informationMEE224: Engineering Mechanics Lecture 4
Lecture 4: Structural Analysis Part 1: Trusses So far we have only analysed forces and moments on a single rigid body, i.e. bars. Remember that a structure is a formed by and this lecture will investigate
More informationCHAPTER 5 ANALYSIS OF STRUCTURES. Expected Outcome:
CHAPTER ANALYSIS O STRUCTURES Expected Outcome: Able to analyze the equilibrium of structures made of several connected parts, using the concept of the equilibrium of a particle or of a rigid body, in
More informationNewton s Third Law Newton s Third Law: For each action there is an action and opposite reaction F
FRAMES AND MACHINES Learning Objectives 1). To evaluate the unknown reactions at the supports and the interaction forces at the connection points of a rigid frame in equilibrium by solving the equations
More informationENGINEERING MECHANICS STATIC
Trusses Simple trusses The basic element of a truss is the triangle, three bars joined by pins at their ends, fig. a below, constitutes a rigid frame. The term rigid is used to mean noncollapsible and
More informationENT 151 STATICS. Contents. Introduction. Definition of a Truss
CHAPTER 6 Analysis ENT 151 STATICS Lecture Notes: Mohd Shukry Abdul Majid KUKUM of Structures Contents Introduction Definition of a Truss Simple Trusses Analysis of Trusses by the Method of Joints Joints
More informationFRAMES AND MACHINES Learning Objectives 1). To evaluate the unknown reactions at the supports and the interaction forces at the connection points of a
FRAMES AND MACHINES Learning Objectives 1). To evaluate the unknown reactions at the supports and the interaction forces at the connection points of a rigid frame in equilibrium by solving the equations
More informationSTATICS VECTOR MECHANICS FOR ENGINEERS: Eleventh Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr. David F. Mazurek
Eleventh E 6 Analysis CHAPTER VECTOR MECHANICS OR ENGINEERS: STATICS erdinand P. Beer E. Russell Johnston, Jr. David. Mazurek of Structures Contents Application Introduction Definition of a Truss Simple
More informationEQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS
EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS Today s Objectives: Students will be able to: a) Apply equations of equilibrium to solve for unknowns, and b) Recognize two-force members. In-Class
More informationEQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS
EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMEBERS Today s Objectives: Students will be able to: a) Apply equations of equilibrium to solve for unknowns, and, b) Recognize two-force members. In-Class
More informationME 201 Engineering Mechanics: Statics
ME 201 Engineering Mechanics: Statics Unit 7.1 Simple Trusses Method of Joints Zero Force Members Simple Truss structure composed of slender members joined together at their end points Planar Truss Simple
More informationChapter 6: Structural Analysis
Chapter 6: Structural Analysis 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
More informationEQUATIONS OF EQUILIBRIUM & TWO-AND THREE-FORCE MEMEBERS
EQUATIONS OF EQUILIBRIUM & TWO-AND THREE-FORCE MEMEBERS Today s Objectives: Students will be able to: a) Apply equations of equilibrium to solve for unknowns, and, b) Recognize two-force members. READING
More informationThe 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
Unit 12 Centroids Page 12-1 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 (12-5) For the area shown
More informationSupplement: Statically Indeterminate Trusses and Frames
: Statically Indeterminate Trusses and Frames Approximate Analysis - In this supplement, we consider an approximate method of solving statically indeterminate trusses and frames subjected to lateral loads
More information7 STATICALLY DETERMINATE PLANE TRUSSES
7 STATICALLY DETERMINATE PLANE TRUSSES OBJECTIVES: This chapter starts with the definition of a truss and briefly explains various types of plane truss. The determinancy and stability of a truss also will
More informationThe analysis of trusses Mehrdad Negahban (1999)
The analysis of trusses Mehrdad Negahban (1999) A truss: A truss is a structure made of two force members all pin connected to each other. The method of joints: This method uses the free-body-diagram of
More informationEquilibrium Equilibrium and Trusses Trusses
Equilibrium and Trusses ENGR 221 February 17, 2003 Lecture Goals 6-4 Equilibrium in Three Dimensions 7-1 Introduction to Trusses 7-2Plane Trusses 7-3 Space Trusses 7-4 Frames and Machines Equilibrium Problem
More informationPin-Jointed Frame Structures (Frameworks)
Pin-Jointed rame Structures (rameworks) 1 Pin Jointed rame Structures (rameworks) A pin-jointed frame is a structure constructed from a number of straight members connected together at their ends by frictionless
More informationENGR-1100 Introduction to Engineering Analysis. Lecture 13
ENGR-1100 Introduction to Engineering Analysis Lecture 13 EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS Today s Objectives: Students will be able to: a) Identify support reactions, and, b) Draw a free-body
More informationEQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMBERS
EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMBERS Today s Objectives: Students will be able to: a) Apply equations of equilibrium to solve for unknowns, and, b) Recognize two-force members. APPLICATIONS
More informationOutline: Frames Machines Trusses
Outline: Frames Machines Trusses Properties and Types Zero Force Members Method of Joints Method of Sections Space Trusses 1 structures are made up of several connected parts we consider forces holding
More informationIn this chapter trusses, frames and machines will be examines as engineering structures.
In the previous chapter we have employed the equations of equilibrium in order to determine the support / joint reactions acting on a single rigid body or a system of connected members treated as a single
More informationEQUILIBRIUM OF A RIGID BODY
EQUILIBRIUM OF A RIGID BODY Today s Objectives: Students will be able to a) Identify support reactions, and, b) Draw a free diagram. APPLICATIONS A 200 kg platform is suspended off an oil rig. How do we
More informationCHAPTER 5 Statically Determinate Plane Trusses
CHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS TYPES OF ROOF TRUSS ROOF TRUSS SETUP ROOF TRUSS SETUP OBJECTIVES To determine the STABILITY and DETERMINACY of plane trusses To analyse
More informationCHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS
CHAPTER 5 Statically Determinate Plane Trusses TYPES OF ROOF TRUSS 1 TYPES OF ROOF TRUSS ROOF TRUSS SETUP 2 ROOF TRUSS SETUP OBJECTIVES To determine the STABILITY and DETERMINACY of plane trusses To analyse
More informationLecture 17 February 23, 2018
Statics - TAM 20 & TAM 2 Lecture 7 ebruary 23, 208 Announcements Monday s lecture: watch for Piazza announcement over weekend for possible change Concept Inventory: Ungraded assessment of course knowledge
More informationName. ME 270 Fall 2005 Final Exam PROBLEM NO. 1. Given: A distributed load is applied to the top link which is, in turn, supported by link AC.
Name ME 270 Fall 2005 Final Exam PROBLEM NO. 1 Given: A distributed load is applied to the top link which is, in turn, supported by link AC. Find: a) Draw a free body diagram of link BCDE and one of link
More informationSTATICS. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Contents 9/3/2015
6 Analsis CHAPTER VECTOR MECHANICS OR ENGINEERS: STATICS erdinand P. Beer E. Russell Johnston, Jr. of Structures Lecture Notes: J. Walt Oler Texas Tech Universit Contents Introduction Definition of a Truss
More informationT R U S S. Priodeep Chowdhury;Lecturer;Dept. of CEE;Uttara University//TRUSS Page 1
T R U S S A truss is a structure that consists of All straight members connected together with pin joints connected only at the ends of the members and All external forces (loads & reactions) must be applied
More informationEngineering Mechanics Department of Mechanical Engineering Dr. G. Saravana Kumar Indian Institute of Technology, Guwahati
Engineering Mechanics Department of Mechanical Engineering Dr. G. Saravana Kumar Indian Institute of Technology, Guwahati Module 3 Lecture 6 Internal Forces Today, we will see analysis of structures part
More informationEng Sample Test 4
1. An adjustable tow bar connecting the tractor unit H with the landing gear J of a large aircraft is shown in the figure. Adjusting the height of the hook F at the end of the tow bar is accomplished by
More informationChapter 3 Trusses. Member CO Free-Body Diagram. The force in CO can be obtained by using section bb. Equations of Equilibrium.
Chapter 3 Trusses Procedure for analysis 1 Free body diagram: make a decision as to how to cut or section the truss through the members where forces are to be determined. 2 Equation of equilibrium: apply
More informationChapter 04 Equilibrium of Rigid Bodies
Chapter 04 Equilibrium of Rigid Bodies Application Engineers designing this crane will need to determine the forces that act on this body under various conditions. 4-2 Introduction For a rigid body, the
More informationEquilibrium of a Particle
ME 108 - Statics Equilibrium of a Particle Chapter 3 Applications For a spool of given weight, what are the forces in cables AB and AC? Applications For a given weight of the lights, what are the forces
More informationAnnouncements. Equilibrium of a Rigid Body
Announcements Equilibrium of a Rigid Body Today s Objectives Identify support reactions Draw a free body diagram Class Activities Applications Support reactions Free body diagrams Examples Engr221 Chapter
More informationSTATICS. Bodies. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Design of a support
4 Equilibrium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University of Rigid Bodies 2010 The McGraw-Hill Companies,
More informationINTERNAL FORCES Today s Objective: Students will be able to: 1. Use the method of sections for determining internal forces in 2-D load cases.
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. In-Class Activities: Check Homework, if any Reading Quiz Applications
More informationSTATICALLY INDETERMINATE STRUCTURES
STATICALLY INDETERMINATE STRUCTURES INTRODUCTION Generally the trusses are supported on (i) a hinged support and (ii) a roller support. The reaction components of a hinged support are two (in horizontal
More informationEQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMBERS
EQUATIONS OF EQUILIBRIUM & TWO- AND THREE-FORCE MEMBERS Today s Objectives: Students will be able to: a) Apply equations of equilibrium to solve for unknowns b) Identify support reactions c) Recognize
More informationTheory of structure I 2006/2013. Chapter one DETERMINACY & INDETERMINACY OF STRUCTURES
Chapter one DETERMINACY & INDETERMINACY OF STRUCTURES Introduction A structure refers to a system of connected parts used to support a load. Important examples related to civil engineering include buildings,
More informationChapter - 1. Equilibrium of a Rigid Body
Chapter - 1 Equilibrium of a Rigid Body Dr. Rajesh Sathiyamoorthy Department of Civil Engineering, IIT Kanpur hsrajesh@iitk.ac.in; http://home.iitk.ac.in/~hsrajesh/ Condition for Rigid-Body Equilibrium
More informationUnit M1.4 (All About) Trusses
Unit M1.4 (ll bout) Trusses Readings: DL 1.9 16.001/002 -- Unified Engineering Department of eronautics and stronautics Massachusetts Institute of Technology LERNING OBJETIVES FOR UNIT M1.4 Through participation
More information3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM
3.1 CONDITIONS FOR RIGID-BODY EQUILIBRIUM Consider rigid body fixed in the x, y and z reference and is either at rest or moves with reference at constant velocity Two types of forces that act on it, the
More informationModule 2. Analysis of Statically Indeterminate Structures by the Matrix Force Method
Module 2 Analysis of Statically Indeterminate Structures by the Matrix Force Method Lesson 10 The Force Method of Analysis: Trusses Instructional Objectives After reading this chapter the student will
More informationTruss Analysis Method of Joints. Steven Vukazich San Jose State University
Truss nalysis Method of Joints Steven Vukazich San Jose State University General Procedure for the nalysis of Simple Trusses using the Method of Joints 1. raw a Free Body iagram (FB) of the entire truss
More informationVector 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
More informationthree Equilibrium 1 and planar trusses ELEMENTS OF ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SPRING 2015 lecture ARCH 614
ELEMENTS OF ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SPRING 2015 lecture three equilibrium and planar trusses Equilibrium 1 Equilibrium balanced steady resultant of forces
More informationCalculating Truss Forces. Method of Joints
Calculating Truss Forces Method of Joints Forces Compression body being squeezed Tension body being stretched Truss truss is composed of slender members joined together at their end points. They are usually
More informationthree Point Equilibrium 1 and planar trusses ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2014 lecture
ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2014 lecture three point equilibrium http:// nisee.berkeley.edu/godden and planar trusses Point Equilibrium 1 Equilibrium balanced
More information1. Determine the Zero-Force Members in the plane truss.
1. Determine the Zero-orce Members in the plane truss. 1 . Determine the force in each member of the loaded truss. Use the Method of Joints. 3. Determine the force in member GM by the Method of Section.
More informationMechanics of Materials
Mechanics of Materials 2. Introduction Dr. Rami Zakaria References: 1. Engineering Mechanics: Statics, R.C. Hibbeler, 12 th ed, Pearson 2. Mechanics of Materials: R.C. Hibbeler, 9 th ed, Pearson 3. Mechanics
More informationWhen a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero.
When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero. 0 0 0 0 k M j M i M M k R j R i R F R z y x z y x Forces and moments acting on a rigid body could be
More informationMethod of Sections for Truss Analysis
Method of Sections for Truss Analysis Notation: (C) = shorthand for compression P = name for load or axial force vector (T) = shorthand for tension Joint Configurations (special cases to recognize for
More informationEquilibrium. Rigid Bodies VECTOR MECHANICS FOR ENGINEERS: STATICS. Eighth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.
Eighth E 4 Equilibrium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University of Rigid Bodies Contents Introduction
More informationStatic Equilibrium. University of Arizona J. H. Burge
Static Equilibrium Static Equilibrium Definition: When forces acting on an object which is at rest are balanced, then the object is in a state of static equilibrium. - No translations - No rotations In
More informationSimilar to trusses, frames are generally fixed, load carrying structures.
Similar to trusses, frames are generally fixed, load carrying structures. The main difference between a frame and a truss is that in a frame at least one member is a multi force member (çoklu kuvvet elemanı).
More information1. Determine the Zero-Force Members in the plane truss.
1. Determine the Zero-orce Members in the plane truss. 1 . Determine the forces in members G, CG, BC, and E for the loaded crane truss. Use the Method of Joints. 3. Determine the forces in members CG and
More informationEngineering Mechanics Statics
Mechanical Systems Engineering _ 2016 Engineering Mechanics Statics 7. Equilibrium of a Rigid Body Dr. Rami Zakaria Conditions for Rigid-Body Equilibrium Forces on a particle Forces on a rigid body The
More informationSTATICS. Bodies VECTOR MECHANICS FOR ENGINEERS: Ninth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.
N E 4 Equilibrium CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler Texas Tech University of Rigid Bodies 2010 The McGraw-Hill Companies,
More informationIshik University / Sulaimani Architecture Department. Structure. ARCH 214 Chapter -5- Equilibrium of a Rigid Body
Ishik University / Sulaimani Architecture Department 1 Structure ARCH 214 Chapter -5- Equilibrium of a Rigid Body CHAPTER OBJECTIVES To develop the equations of equilibrium for a rigid body. To introduce
More informationNAME: Section: RIN: Tuesday, May 19, :00 11:00. Problem Points Score Total 100
RENSSELAER POLYTECHNIC INSTITUTE TROY, NY FINAL EXAM INTRODUCTION TO ENGINEERING ANALYSIS (ENGR-1100) NAME: Section: RIN: Tuesday, May 19, 2015 8:00 11:00 Problem Points Score 1 20 2 20 3 20 4 20 5 20
More informationEQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS
Today s Objectives: Students will be able to: EQUILIBRIUM OF A RIGID BODY & FREE-BODY DIAGRAMS a) Identify support reactions, and, b) Draw a free-body diagram. In-Class Activities: Check Homework Reading
More informationSimilar to trusses, frames are generally fixed, load carrying structures.
Similar to trusses, frames are generally fixed, load carrying structures. The main difference between a frame and a truss is that in a frame at least one member is a multi force member (çoklu kuvvet elemanı).
More informationEquilibrium of a Rigid Body. Engineering Mechanics: Statics
Equilibrium of a Rigid Body Engineering Mechanics: Statics Chapter Objectives Revising equations of equilibrium of a rigid body in 2D and 3D for the general case. To introduce the concept of the free-body
More informationP.E. Civil Exam Review:
P.E. Civil Exam Review: Structural Analysis J.P. Mohsen Email: jpm@louisville.edu Structures Determinate Indeterminate STATICALLY DETERMINATE STATICALLY INDETERMINATE Stability and Determinacy of Trusses
More informationLecture 14 February 16, 2018
Statics - TAM 210 & TAM 211 Lecture 14 February 16, 2018 SoonTrending.com Announcements Structured office hours of working through practice problems will be held during Sunday office hours, starting Sunday
More informationSTATICS. FE Review. Statics, Fourteenth Edition R.C. Hibbeler. Copyright 2016 by Pearson Education, Inc. All rights reserved.
STATICS FE Review 1. Resultants of force systems VECTOR OPERATIONS (Section 2.2) Scalar Multiplication and Division VECTOR ADDITION USING EITHER THE PARALLELOGRAM LAW OR TRIANGLE Parallelogram Law: Triangle
More informationFigure 9.1 (a) Six performers in the circus; (b) free-body diagram of the performers / Alan Thornton/Stone/Getty Images
Creatas In this chapter we use equilibrium analysis to look at loads internal to three types of systems: frames, machines, and trusses. By the end of this chapter, you will be able to systematically find
More informationWhen a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero.
When a rigid body is in equilibrium, both the resultant force and the resultant couple must be zero. 0 0 0 0 k M j M i M M k R j R i R F R z y x z y x Forces and moments acting on a rigid body could be
More informationSpring 2018 Lecture 28 Exam Review
Statics - TAM 210 & TAM 211 Spring 2018 Lecture 28 Exam Review Announcements Concept Inventory: Ungraded assessment of course knowledge Extra credit: Complete #1 or #2 for 0.5 out of 100 pt of final grade
More informationEngineering Mechanics: Statics in SI Units, 12e
Engineering Mechanics: Statics in SI Units, 12e 5 Equilibrium of a Rigid Body Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body
More informationLecture 17 February 23, 2018
Statics - TAM 20 & TAM 2 Lecture 7 ebruary 23, 208 Announcements Monday s lecture: watch for Piazza announcement over weekend for possible change Concept Inventory: Ungraded assessment of course knowledge
More informationChapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd
Chapter Objectives To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using
More informationEquilibrium of Rigid Bodies
Equilibrium of Rigid Bodies 1 2 Contents Introduction Free-Bod Diagram Reactions at Supports and Connections for a wo-dimensional Structure Equilibrium of a Rigid Bod in wo Dimensions Staticall Indeterminate
More informationChapter 5 Equilibrium of a Rigid Body Objectives
Chapter 5 Equilibrium of a Rigid Bod Objectives Develop the equations of equilibrium for a rigid bod Concept of the free-bod diagram for a rigid bod Solve rigid-bod equilibrium problems using the equations
More informationThe case where there is no net effect of the forces acting on a rigid body
The case where there is no net effect of the forces acting on a rigid body Outline: Introduction and Definition of Equilibrium Equilibrium in Two-Dimensions Special cases Equilibrium in Three-Dimensions
More informationDetermine the resultant internal loadings acting on the cross section at C of the beam shown in Fig. 1 4a.
E X M P L E 1.1 Determine the resultant internal loadings acting on the cross section at of the beam shown in Fig. 1 a. 70 N/m m 6 m Fig. 1 Support Reactions. This problem can be solved in the most direct
More informationFE Sta'cs Review. Torch Ellio0 (801) MCE room 2016 (through 2000B door)
FE Sta'cs Review h0p://www.coe.utah.edu/current- undergrad/fee.php Scroll down to: Sta'cs Review - Slides Torch Ellio0 ellio0@eng.utah.edu (801) 587-9016 MCE room 2016 (through 2000B door) Posi'on and
More informationName ME 270 Summer 2006 Examination No. 1 PROBLEM NO. 3 Given: Below is a Warren Bridge Truss. The total vertical height of the bridge is 10 feet and each triangle has a base of length, L = 8ft. Find:
More informationD : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown.
D : SOLID MECHANICS Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown. Q.2 Consider the forces of magnitude F acting on the sides of the regular hexagon having
More informationES230 STRENGTH OF MATERIALS
ES230 STRENGTH OF MATERIALS Exam 1 Study Guide. Exam 1: Wednesday, February 8 th, in-class Updated 2/5/17 Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will
More informationIntroduction. 1.1 Introduction. 1.2 Trigonometrical definitions
Introduction 1.1 Introduction Stress analysis is an important part of engineering science, as failure of most engineering components is usually due to stress. The component under a stress investigation
More informationEQUILIBRIUM OF A PARTICLE, THE FREE-BODY DIAGRAM & COPLANAR FORCE SYSTEMS
EQUILIBRIUM OF PRTICLE, THE FREE-BODY DIGRM & COPLNR FORCE SYSTEMS Today s Objectives: Students will be able to : a) Draw a free body diagram (FBD), and, b) pply equations of equilibrium to solve a 2-D
More informationME 201 Engineering Mechanics: Statics. Final Exam Review
ME 201 Engineering Mechanics: Statics inal Exam Review inal Exam Testing Center (Proctored, 1 attempt) Opens: Monday, April 9 th Closes : riday, April 13 th Test ormat 15 Problems 10 Multiple Choice (75%)
More information