STATICALLY INDETERMINATE STRUCTURES


 Augustine Barrett
 1 years ago
 Views:
Transcription
1 STATICALLY INDETERMINATE STRUCTURES
2 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 and vertical directions), and for a roller support reaction component is only one (in vertical direction). For any truss, if the number of members forming the truss is m and the number of joints is j, then for the frame to be rigid m = 2 j 3. The equations of static equilibrium are sufficient to determine the internal forces in members of a rigid truss. If members are less than 2j 3, the truss becomes a collapsible truss and if m > (2j 3), the truss becomes a redundant truss.
3 Equations from statics, F x 0, F 0, M 0 y O i.e. summation of forces in x and ydirections, respectively, is zero and summation of moments of forces about any point is zero, are not sufficient to determine the internal forces in members of a redundant truss. A unit load method is used to analyze forces in members of a redundant truss. Structures used in bridges are generally redundant with multiple degree of redundancy. The study of such structures can be done by numerical methods; we will analyze the trusses with single degree of redundancy. There are many redundant structures in which strain compatibility condition can be used for analyzing the forces in members of trusses or frames.
4 ANALYSIS OF REDUNDANT FRAMES WITH STRAIN COMPATIBILITY CONDITION Let us consider a system of three wires of steel, brass, and aluminium supporting a rigid bar OABCD, which is hinged at end O. A load P is applied at end D of the bar as shown in Figure 1. Let s say the internal forces in bars of steel, brass, and aluminium are F A, F B, and F C, respectively. Then from the conditions of static equilibrium: P = F A + F B + F C  R O Moreover, taking moments about O, where R O is reaction at hinged end O, 4aP = a F A + 2a F B + 3a F C or 4P = F A + 2F B + 3F C Figure 1 (a) A rigid bar supported by three wires (b) Deformations in wires
5 There are four unknowns, i.e. R O, F A, F B, and F C, but only two equations, so conditions of static equilibrium cannot analyze the forces in this indeterminate structure. We have to consider the deformation produced in each bar after the application of load, or in other words strain compatibility condition has to be used. The rigid bar is going to take the new position OA B C D. From this diagram, AA = δ A deformation in bar A BB = δ B deformation in bar B CC = δ C deformation in bar C but from the strain compatibility, as is obvious from Figure 1 (b). δ B = 2 δ A δ C = 3 δ A
6 where From strain compatibility condition Let us take E steel = 2E brass = 3E al So, F B = F A and F C = F A
7 F B = F A and F C = F A Putting these values in 4P = F A + 2F B + 3F C From P = F A + F B + F C  R O, reaction R O 3F A P 2 3 P P 3 P R 0 reaction is downward.
8 Example 7.1 A frame ABCD is shown in the figure. There are three bars AD, BD, and CD hinged at D as shown. Bars AD and CD are of copper while bar BD is of steel. Length of middle bar BD is L. Area of crosssection of each bar is the same. A load P is applied at end D. Determine the forces in the three bars. Given E steel = 2E copper. From the condition of static equilibrium: P = F BD + F AD cos45 + F CD cos45 (i) Due to symmetry F AD = F CD, because both bars are identical and made of copper. So, P = F BD + 2F AD cos45 F BD = F BD F AD (ii) F AD F CD
9 We have to use conditions of strain compatibility. After the application of load P, point D moves down to D and all the bars are extended. Extension in AD is D D and extension in BD is DD. But DD sin45 = D D (iii) and where A is the area of the crosssection. Using Eq. (iii),
10 From Eqs. (iv) and (v), Force F BD F AD F CD Now from Eq. (ii)
11 Let us verify the equilibrium of forces at joint D P = F BD F AD = ( ) P = ( ) P = P F BD F AD F CD
12 Exercise 7.1 A rigid bar is suspended by three bars of steel and aluminium as shown in the figure. The bars are of equal length and equal area of crosssection. A load W is suspended on rigid bar so that the rigid bar remains horizontal after the application of load. Determine forces in aluminium and steel bars. E steel = 3E aluminium
13 DEGREE OF REDUNDANCY The total degree of redundancy or degree of indeterminacy of a frame is equal to the number by which the unknown reaction components exceed the condition of equation of equilibrium. The excess members are called redundants. Total degree of redundancy, T T = m (2 j R) where m = j = R = total number of members in a frame, total number of joints in frame, and total number of reaction components. Following criteria are applied for reaction components at supports of the frame: 1. For a roller support reaction component is For a hinged support reaction components are For a fixed support as in cantilever reaction components are 3.
14 Let us consider the following example: A frame ABCDEF shown in Figure 2 is hinged at end A and roller supported at end E. There are 11 members in frame. Number of members, m = 11 Number of joints, j = 6 Figure 2 Redundant frame Reaction components, R = 2 (for hinged support at A) + 1 (for roller support at E) = 3 Degree of redundancy, T = m (2j R) = 11 (2 6 3) = 11 9 = 2
15 Similarly, let us consider another frame ABCD, as shown in Figure 3, having six members and four joints. Frame is roller supported at A and hinged at end C. Let us determine its degree of redundancy. m = 6, number of members j = 4, number of joints Figure 3 Triangular frame R = number of reaction components = 2 (for hinged end C) + 1 (for roller support A) = 3 Degree of redundancy T = m (2j R) = 6 (2 4 3) = 6 5 = 1
16 Exercise 7.2 Determine the degree of redundancy for the frames shown in the figure. T = m (2j R) [For all the cases, degree of redundancy is 1].
17 ANALYSIS OF STATICALLY INDETERMINATE TRUSSES Method of virtual work or unit load method is one of the several methods available for solution of forces in members of an indeterminate truss. The following relationship is used to calculate the displacement in a member subjected to tensile or compressive load. Displacement SUL AE where S = internal force in truss members due to applied external loads (external loads on truss plus a unit load) U = axial load in truss members due to unit load applied externally at a point, E = Young s modulus of elasticity of member, L = length of member, and A = area of crosssection of member.
18 We will consider trusses with only single degree of redundancy. The following steps are taken in order to determine the forces in members of an indeterminate truss. From the indeterminate truss, remove the redundant member so as to obtain a statically determinate structure as shown in Figures 4 (a) and (b). Obtain the forces in members of the statically determinate truss as shown in Figure 4(b) by equilibrium equations of statics. Say these forces are S 0 in each member. Figure 4 (a) Statically indeterminate (b) Member BD removed statically determinate
19 Consider the same truss with one redundant member BD cut. Let the force X a = 1 is applied on this member (from both ends, no support). Find the forces in members without the external force P (Figure 5). Net force in each member S = S 0 + X a u a Now let X a be different from unity. Figure 5 Unit load along BD Let the relative displacement in redundant member be δ a. Say the area of crosssection of each member is A and Young s modulus is E.
20 or Put (rigid structure) and get the value of X a, or or
21 Forces in members of truss (putting the value of Xa) Member S Magnitude of S AB 0.707X a + 0.5P BC P 0.707X a 0.5P CD P 0.707X a 0.5P DA 0.707X a + 0.5P BD X a 0.707P AC P Figure 6 shows the forces in the members of the truss. Figure 6 Forces in members of truss Equilibrium of Truss at joints: Let us check the equilibrium of forces at joints of the truss. For that, we have to first calculate the reactions. Taking moments about A, P a = R D a Reaction, R D = P Reaction at A, R AH = P (horizontal component) R AV = P (vertical component)
22 Exercise 7.2 A redundant frame ABCD is shown in the figure. Material of all the bars is the same and the area of crosssection is also the same. Determine forces in all the members of the frame.
23 Example 7.3 For the truss shown in the figure, determine the support reactions and forces in all members. Area of crosssection of members AB, BC, and BD is 25 cm 2 and for members AD and DC it is 22.5 cm 2. Material is the same for all the members. Number of members, m = 5 Number of joints, j = 4 Both the supports are hinged. So, the number of reaction components, R = 2 2 = 4 Degree of redundancy T = m (2j R) = 5 (2 4 4) = 1
24 Vertical reactions at A and C R AV = R CV = 100 kn (due to symmetry). Let us say horizontal reaction at A and C is X a. Remove X a from both ends and calculate forces in members. Due to symmetry, we will solve for half the truss. Angles, s = 45 b = tan = sin s = coss = sin b = cos b = Forces R CV F AB cos45 = F AD cos26.56 (x direction) F AB = F AD F AB = F AD
25 Now F AD sin = F AB sin 45 F AD = F AB putting the value of F AB = (1.265 F AD ) = F AD Force, (To have equilibrium of forces at joint A, F AD is tensile and F AB will be compressive as shown). F AB = F BC = 283 kn Joint B F AB sin 45 = F BC sin 45
26 But = F BD Force, = F BD F BD = +200 kn (tensile) A Horizontal reactions Now applying the unit load X a = 1 at ends A and C as shown in the figure. F AD sin = F AB sin 45 F AD = F AB F AD = F AB (y direction) F AB = F AD cos = 0.894F AD (x direction) = F AB (putting the value of F AD ) = 1.414F AD F AB = (tension) F AD = = (comp.)
27 Joint B F BD = 2 F AB = = 2 (compressive) Total force in a member, S = S 0 + X a u a = Force in a member due to external load after removing redundant reaction + force due to load X a Also,
28 If we tabulate the values of S, u a, L, A, etc., E is the same for all members:
29 Now Reactions and forces in members or The forces are balanced at all the joints. All members are in compression.
30 Example 7.4 A triangular truss ABCD is shown in the figure. It is subjected to a horizontal load P at joint B. Determine the support reactions and forces in the members of truss. The crosssectional area of outer members AB, BC, and CA is twice the crosssectional area of inner members AD, BD, and CD. The material of all members is the same. Taking moments of forces about point A P 0.866L = R CV L R CV, reaction = 0.866P R AV = 0.866P R AH = P as shown.
31 For degree of redundancy: Number of members, m = 6 Number of joints, j = 4 Reactions components, R = 2 (for hinged end) + 1 (roller supported end) = 3 Degree of redundancy, T = m (2j R) = 6 (2 4 3) = 1
32 Let us remove one member AC from the frame, and find forces S 0 in members. Member AC is removed (figure). Joint C F BC cos 60 = F DC cos 30 F BC 0.5 = 0.866F DC F BC = F DC Moreover, F DC sin P = F BC sin F DC P = F BC 120 Putting the value of F BC 0.5 F DC P = F DC = 1.5 F DC Force, F DC = 0.866P F BC = = 1.5P
33 Joint D ADC = CDB = BDA = 120 so F DC = F DB = F DA = 0.866P (tensile) Joint A F AB cos60 + F AD cos30 = P 0.5F AB P = P 0.5F AB = (P 0.75P) F AB = 0.5P 120 Checking for vertical loads. F AB sin60 + F AD sin30 = P P = 0.866P (verified)
34 If we tabulate these values: Now consider only X a = 1 along member AC as shown in the figure. F AB sin60 = F AD sin30 F AB = F AD 0.5 F AD = 1.732F AB F AD cos30 = F AB cos F AD = 0.5F AB + 1 Putting the value of F AD F AB = 0.5F AB + 1 F AB = 1 = F BC F AD = (comp) = F DC = F DB
35 If we tabulate these u a values also Length AB = BC = CA = L, area of crosssection 2a
36 Material of all members is the same. Let us tabulate the different parameters.
37 Now or Resultant Forces in members
38 Example 7.5 Find the bending moment at any point of a semicircular arch shown in the figure. Both the supports are hinged. Flexural rigidity EI is constant throughout. For the arch, number of members, m = 1 number of joints, j = 2 number of reaction components, R = 2 2= 4 (both ends are hinged). Degree of redundancy, T = m (2j R) = 1 (2 2 4) = 1
39 There will be reactions X and at each end as shown. At a particular section aa, bending moment where k = R sin θ. Applying the principle of virtual work or unit load method, X M O M ds EI X 2 M i ds EI Taking the advantage of symmetrical loading,
40 or
41 Therefore, Bending moment at any section
42 Example 7.6 A rigid bar EF of negligible weight is suspended by four wires A, B, C, and D of the same length, same area of crosssection and same material. A load P is suspended at C as shown in the figure. Determine the forces in the four wires. From the conditions of static equilibrium, P = P A + P B + P C + P D (i) Moments about point E P A 0 + P B a + P C 2a + P D 3A = 2aP or 2P = P B + 2P C + 3P D (ii)
43 From the two equations, the values of four forces P A, P B, P C, and P D cannot be determined. We have to take the help of a compatibility condition, i.e. considering the deformation of the bar as shown in the figure. Deformation in bar A =δ A Deformation in bar B = δ A + δ Deformation in bar C = δ A + 2δ Deformation in bar D = δ A + 3δ Rigid bar EF will take the position A B C D after the application of load. Using Hooke s law, in the elastic region, deformation is directly proportional to load applied. δa δ A A Say δ A P A (load on wire A) δ P (load due to additional deformation δ)
44 So, P B = P A + P P C = P A + 2P P D = P A + 3P Total reactions = P A + P B + P C + P D = P A + P A + P + P A + 2P + P A + 3P = 4P A + 6P = P (iii) δ A Putting the values in (ii) 2P = P A + P + 2 (P A + 2P ) + 3 (P A + 3P ) 2P = P A + 2P A + 3P A + P + 4P + 9P 2P = 6P A + 14P or P = 3P A + 7P (iv)
45 From Eqs. (iii) and (iv), 4P A + 6P = 3P A + 7P P A = P (v) So, P B = 2P P C = 3P P D = 4P Total reaction P + 2P + 3P + 4P = P P = 0.10P (vi) δ A or forces in wire A, P A = P = 0.1P in wire B, P B = 0.2P in wire C, P C = 0.3P in wire D, P D = 0.4P
46 Exercise 7.3 A rigid bar EF of negligible weight is suspended by four wires A, B, C, and D of the same material, same length and same area of crosssection. A load P is suspended as shown in the figure. Determine the forces in wires A, B, C, and D.
Module 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 informationChapter 4b Axially Loaded Members
CIVL 222 STRENGTH OF MATERIALS Chapter 4b Axially Loaded Members AXIAL LOADED MEMBERS Today s Objectives: Students will be able to: a) Determine the elastic deformation of axially loaded member b) Apply
More informationIndeterminate Analysis Force Method 1
Indeterminate Analysis Force Method 1 The force (flexibility) method expresses the relationships between displacements and forces that exist in a structure. Primary objective of the force method is to
More informationLecture 8: Flexibility Method. Example
ecture 8: lexibility Method Example The plane frame shown at the left has fixed supports at A and C. The frame is acted upon by the vertical load P as shown. In the analysis account for both flexural and
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 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 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 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 11 The Force Method of Analysis: Frames Instructional Objectives After reading this chapter the student will be
More informationModule 4 : Deflection of Structures Lecture 4 : Strain Energy Method
Module 4 : Deflection of Structures Lecture 4 : Strain Energy Method Objectives In this course you will learn the following Deflection by strain energy method. Evaluation of strain energy in member under
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 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 informationIf the number of unknown reaction components are equal to the number of equations, the structure is known as statically determinate.
1 of 6 EQUILIBRIUM OF A RIGID BODY AND ANALYSIS OF ETRUCTURAS II 9.1 reactions in supports and joints of a twodimensional structure and statically indeterminate reactions: Statically indeterminate structures
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV UNIT I STRESS, STRAIN DEFORMATION OF SOLIDS PART A (2 MARKS)
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 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 informationQUESTION BANK ENGINEERS ACADEMY. Hinge E F A D. Theory of Structures Determinacy Indeterminacy 1
Theory of Structures eterminacy Indeterminacy 1 QUSTION NK 1. The static indeterminacy of the structure shown below (a) (b) 6 (c) 9 (d) 12 2. etermine the degree of freedom of the following frame (a) 1
More informationUNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation.
UNIT 1 STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define stress. When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The magnitude
More informationThe University of Melbourne Engineering Mechanics
The University of Melbourne 436291 Engineering Mechanics Tutorial Four Poisson s Ratio and Axial Loading Part A (Introductory) 1. (Problem 922 from Hibbeler  Statics and Mechanics of Materials) A short
More informationLecture 4: PRELIMINARY CONCEPTS OF STRUCTURAL ANALYSIS. Introduction
Introduction In this class we will focus on the structural analysis of framed structures. We will learn about the flexibility method first, and then learn how to use the primary analytical tools associated
More informationFLEXIBILITY METHOD FOR INDETERMINATE FRAMES
UNIT  I FLEXIBILITY METHOD FOR INDETERMINATE FRAMES 1. What is meant by indeterminate structures? Structures that do not satisfy the conditions of equilibrium are called indeterminate structure. These
More informationModule 6. Approximate Methods for Indeterminate Structural Analysis. Version 2 CE IIT, Kharagpur
Module 6 Approximate Methods for Indeterminate Structural Analysis Lesson 35 Indeterminate Trusses and Industrial rames Instructional Objectives: After reading this chapter the student will be able to
More informationUNIT I ENERGY PRINCIPLES
UNIT I ENERGY PRINCIPLES Strain energy and strain energy density strain energy in traction, shear in flexure and torsion Castigliano s theorem Principle of virtual work application of energy theorems
More informationQUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A
DEPARTMENT: CIVIL SUBJECT CODE: CE2201 QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State
More informationMethod of Consistent Deformation
Method of onsistent eformation Structural nalysis y R.. Hibbeler Theory of StructuresII M Shahid Mehmood epartment of ivil Engineering Swedish ollege of Engineering and Technology, Wah antt FRMES Method
More information14. *14.8 CASTIGLIANO S THEOREM
*14.8 CASTIGLIANO S THEOREM Consider a body of arbitrary shape subjected to a series of n forces P 1, P 2, P n. Since external work done by forces is equal to internal strain energy stored in body, by
More informationME Final Exam. PROBLEM NO. 4 Part A (2 points max.) M (x) y. z (neutral axis) beam crosssec+on. 20 kip ft. 0.2 ft. 10 ft. 0.1 ft.
ME 323  Final Exam Name December 15, 2015 Instructor (circle) PROEM NO. 4 Part A (2 points max.) Krousgrill 11:30AM12:20PM Ghosh 2:303:20PM Gonzalez 12:301:20PM Zhao 4:305:20PM M (x) y 20 kip ft 0.2
More informationModule 1 : Introduction : Review of Basic Concepts in Mechnics Lecture 4 : Static Indeterminacy of Structures
Module 1 : Introduction : Review of Basic Concepts in Mechnics Lecture 4 : Static Indeterminacy of Structures Objectives In this course you will learn the following Review of the concepts of determinate
More informationChapter 11. Displacement Method of Analysis Slope Deflection Method
Chapter 11 Displacement ethod of Analysis Slope Deflection ethod Displacement ethod of Analysis Two main methods of analyzing indeterminate structure Force method The method of consistent deformations
More informationChapter 2: Deflections of Structures
Chapter 2: Deflections of Structures Fig. 4.1. (Fig. 2.1.) ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 1 (2.1) (4.1) (2.2) Fig.4.2 Fig.2.2 ASTU, Dept. of C Eng., Prepared by: Melkamu E. Page 2
More informationPinJointed Frame Structures (Frameworks)
PinJointed rame Structures (rameworks) 1 Pin Jointed rame Structures (rameworks) A pinjointed frame is a structure constructed from a number of straight members connected together at their ends by frictionless
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 informationLevel 7 Postgraduate Diploma in Engineering Computational mechanics using finite element method
9210203 Level 7 Postgraduate Diploma in Engineering Computational mechanics using finite element method You should have the following for this examination one answer book No additional data is attached
More informationStructural Analysis III Compatibility of Displacements & Principle of Superposition
Structural Analysis III Compatibility of Displacements & Principle of Superposition 2007/8 Dr. Colin Caprani, Chartered Engineer 1 1. Introduction 1.1 Background In the case of 2dimensional structures
More informationChapter 2 Basis for Indeterminate Structures
Chapter  Basis for the Analysis of Indeterminate Structures.1 Introduction... 3.1.1 Background... 3.1. Basis of Structural Analysis... 4. Small Displacements... 6..1 Introduction... 6.. Derivation...
More informationStresses in Curved Beam
Stresses in Curved Beam Consider a curved beam subjected to bending moment M b as shown in the figure. The distribution of stress in curved flexural member is determined by using the following assumptions:
More informationInterstate 35W Bridge Collapse in Minnesota (2007) AP Photo/Pioneer Press, Brandi Jade Thomas
7 Interstate 35W Bridge Collapse in Minnesota (2007) AP Photo/Pioneer Press, Brandi Jade Thomas Deflections of Trusses, Beams, and Frames: Work Energy Methods 7.1 Work 7.2 Principle of Virtual Work 7.3
More informationQUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS
QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1 STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,
More informationLecture 6: The Flexibility Method  Beams. Flexibility Method
lexibility Method In 1864 James Clerk Maxwell published the first consistent treatment of the flexibility method for indeterminate structures. His method was based on considering deflections, but the presentation
More informationPurpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on Exam 3.
ES230 STRENGTH OF MTERILS Exam 3 Study Guide Exam 3: Wednesday, March 8 th inclass Updated 3/3/17 Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on
More informationContinuing Education Course #207 What Every Engineer Should Know About Structures Part B Statics Applications
1 of 6 Continuing Education Course #207 What Every Engineer Should Know About Structures Part B Statics Applications 1. As a practical matter, determining design loads on structural members involves several
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 information2012 MECHANICS OF SOLIDS
R10 SET  1 II B.Tech II Semester, Regular Examinations, April 2012 MECHANICS OF SOLIDS (Com. to ME, AME, MM) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~
More information29. Define Stiffness matrix method. 30. What is the compatibility condition used in the flexibility method?
CLASS: III YEAR / VI SEMESTER CIVIL SUBJECTCODE AND NAME: CE 2351  STRUCTURAL ANALYSISII UNIT1 FLEXIBILITY MATRIX METHOD. PART A 1. What is meant by indeterminate structures? 2. What are the conditions
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 informationBeams. Beams are structural members that offer resistance to bending due to applied load
Beams Beams are structural members that offer resistance to bending due to applied load 1 Beams Long prismatic members Nonprismatic sections also possible Each crosssection dimension Length of member
More informationM.S Comprehensive Examination Analysis
UNIVERSITY OF CALIFORNIA, BERKELEY Spring Semester 2014 Dept. of Civil and Environmental Engineering Structural Engineering, Mechanics and Materials Name:......................................... M.S Comprehensive
More informationLecture 11: The Stiffness Method. Introduction
Introduction Although the mathematical formulation of the flexibility and stiffness methods are similar, the physical concepts involved are different. We found that in the flexibility method, the unknowns
More informationReg. No. : Question Paper Code : B.Arch. DEGREE EXAMINATION, APRIL/MAY Second Semester AR 6201 MECHANICS OF STRUCTURES I
WK 4 Reg. No. : Question Paper Code : 71387 B.Arch. DEGREE EXAMINATION, APRIL/MAY 2017. Second Semester AR 6201 MECHANICS OF STRUCTURES I (Regulations 2013) Time : Three hours Maximum : 100 marks Answer
More informationPh.D. Preliminary Examination Analysis
UNIVERSITY OF CALIFORNIA, BERKELEY Spring Semester 2014 Dept. of Civil and Environmental Engineering Structural Engineering, Mechanics and Materials Name:......................................... Ph.D.
More informationProblem " Â F y = 0. ) R A + 2R B + R C = 200 kn ) 2R A + 2R B = 200 kn [using symmetry R A = R C ] ) R A + R B = 100 kn
Problem 0. Three cables are attached as shown. Determine the reactions in the supports. Assume R B as redundant. Also, L AD L CD cos 60 m m. uation of uilibrium: + " Â F y 0 ) R A cos 60 + R B + R C cos
More informationCHENDU COLLEGE OF ENGINEERING &TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING SUB CODE & SUB NAME : CE2351STRUCTURAL ANALYSISII UNIT1 FLEXIBILITY
CHENDU COLLEGE OF ENGINEERING &TECHNOLOGY DEPARTMENT OF CIVIL ENGINEERING SUB CODE & SUB NAME : CE2351STRUCTURAL ANALYSISII UNIT1 FLEXIBILITY METHOD FOR INDETERMINATE FRAMES PARTA(2MARKS) 1. What is
More informationTheory of Structures
SAMPLE STUDY MATERIAL Postal Correspondence Course GATE, IES & PSUs Civil Engineering Theory of Structures C O N T E N T 1. ARCES... 314. ROLLING LOADS AND INFLUENCE LINES. 159 3. DETERMINACY AND INDETERMINACY..
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 informationFree Body Diagram: Solution: The maximum load which can be safely supported by EACH of the support members is: ANS: A =0.217 in 2
Problem 10.9 The angle β of the system in Problem 10.8 is 60. The bars are made of a material that will safely support a tensile normal stress of 8 ksi. Based on this criterion, if you want to design the
More informationMECHANICS OF MATERIALS
Third CHTR Stress MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University and Strain xial oading Contents Stress & Strain: xial oading Normal
More informationME 323 Examination #2 April 11, 2018
ME 2 Eamination #2 April, 2 PROBLEM NO. 25 points ma. A thinwalled pressure vessel is fabricated b welding together two, openended stainlesssteel vessels along a 6 weld line. The welded vessel has an
More informationDownloaded from Downloaded from / 1
PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION2002 LEVEL : B. E. (Civil) SUBJECT: BEG256CI, Strength of Material Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates are required to give their
More informationUNIT IV FLEXIBILTY AND STIFFNESS METHOD
SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code : SAII (13A01505) Year & Sem: IIIB.Tech & ISem Course & Branch: B.Tech
More informationChapter 2 Examples of Optimization of Discrete Parameter Systems
Chapter Examples of Optimization of Discrete Parameter Systems The following chapter gives some examples of the general optimization problem (SO) introduced in the previous chapter. They all concern the
More informationSTRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS
1 UNIT I STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF STRESS 1. Define: Stress When an external force acts on a body, it undergoes deformation. At the same time the body resists deformation. The
More information7.4 The Elementary Beam Theory
7.4 The Elementary Beam Theory In this section, problems involving long and slender beams are addressed. s with pressure vessels, the geometry of the beam, and the specific type of loading which will be
More informationModule 3. Analysis of Statically Indeterminate Structures by the Displacement Method
odule 3 Analysis of Statically Indeterminate Structures by the Displacement ethod Lesson 14 The SlopeDeflection ethod: An Introduction Introduction As pointed out earlier, there are two distinct methods
More informationMechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection
Mechanics of Materials II Chapter III A review of the fundamental formulation of stress, strain, and deflection Outline Introduction Assumtions and limitations Axial loading Torsion of circular shafts
More informationThe problem of transmitting a torque or rotary motion from one plane to another is frequently encountered in machine design.
CHAPER ORSION ORSION orsion refers to the twisting of a structural member when it is loaded by moments/torques that produce rotation about the longitudinal axis of the member he problem of transmitting
More informationProblem d d d B C E D. 0.8d. Additional lecturebook examples 29 ME 323
Problem 9.1 Two beam segments, AC and CD, are connected together at C by a frictionless pin. Segment CD is cantilevered from a rigid support at D, and segment AC has a roller support at A. a) Determine
More information2 marks Questions and Answers
1. Define the term strain energy. A: Strain Energy of the elastic body is defined as the internal work done by the external load in deforming or straining the body. 2. Define the terms: Resilience and
More information1 Static Plastic Behaviour of Beams
1 Static Plastic Behaviour of Beams 1.1 Introduction Many ductile materials which are used in engineering practice have a considerable reserve capacity beyond the initial yield condition. The uniaxial
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 zeroforce members Class
More informationPURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC.
BENDING STRESS The effect of a bending moment applied to a crosssection of a beam is to induce a state of stress across that section. These stresses are known as bending stresses and they act normally
More informationFIXED BEAMS CONTINUOUS BEAMS
FIXED BEAMS CONTINUOUS BEAMS INTRODUCTION A beam carried over more than two supports is known as a continuous beam. Railway bridges are common examples of continuous beams. But the beams in railway bridges
More informationPart IB Paper 2: Structures. Examples Paper 2/3 Elastic structural analysis
ISSUEB 011 15 NOV 2013 1 Engineering Tripos Part IB SECOND YEAR Part IB Paper 2: Structures Examples Paper 2/3 Elastic structural analysis Straightforward questions are marked by t; Tripos standard questions
More informationStructural Analysis I Chapter 4  Torsion TORSION
ORSION orsional stress results from the action of torsional or twisting moments acting about the longitudinal axis of a shaft. he effect of the application of a torsional moment, combined with appropriate
More informationSub. Code:
Important Instructions to examiners: ) The answers should be examined by key words and not as wordtoword as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationMechanics of Structure
S.Y. Diploma : Sem. III [CE/CS/CR/CV] Mechanics of Structure Time: Hrs.] Prelim Question Paper Solution [Marks : 70 Q.1(a) Attempt any SIX of the following. [1] Q.1(a) Define moment of Inertia. State MI
More informationtechietouch.blogspot.com DEPARTMENT OF CIVIL ENGINEERING ANNA UNIVERSITY QUESTION BANK CE 2302 STRUCTURAL ANALYSISI TWO MARK QUESTIONS UNIT I DEFLECTION OF DETERMINATE STRUCTURES 1. Write any two important
More informationSRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK
SRSD 2093: Engineering Mechanics 2SRRI SECTION 19 ROOM 7, LEVEL 14, MENARA RAZAK SIMPLE TRUSSES, THE METHOD OF JOINTS, & ZEROFORCE MEMBERS Today s Objectives: Students will be able to: a) Define a simple
More informationMECHANICS OF MATERIALS
CHATR Stress MCHANICS OF MATRIALS and Strain Axial Loading Stress & Strain: Axial Loading Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced
More information[7] Torsion. [7.1] Torsion. [7.2] Statically Indeterminate Torsion. [7] Torsion Page 1 of 21
[7] Torsion Page 1 of 21 [7] Torsion [7.1] Torsion [7.2] Statically Indeterminate Torsion [7] Torsion Page 2 of 21 [7.1] Torsion SHEAR STRAIN DUE TO TORSION 1) A shaft with a circular cross section is
More informationName :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CENEW)/SEM3/CE301/ SOLID MECHANICS
Name :. Roll No. :..... Invigilator s Signature :.. 2011 SOLID MECHANICS Time Allotted : 3 Hours Full Marks : 70 The figures in the margin indicate full marks. Candidates are required to give their answers
More informationMechanics of Solids. Mechanics Of Solids. Suraj kr. Ray Department of Civil Engineering
Mechanics Of Solids Suraj kr. Ray (surajjj2445@gmail.com) Department of Civil Engineering 1 Mechanics of Solids is a branch of applied mechanics that deals with the behaviour of solid bodies subjected
More informationENGINEERING MECHANICS SOLUTIONS UNITI
LONG QUESTIONS ENGINEERING MECHANICS SOLUTIONS UNITI 1. A roller shown in Figure 1 is mass 150 Kg. What force P is necessary to start the roller over the block A? =90+25 =115 = 90+25.377 = 115.377 = 360(115+115.377)
More informationName (Print) ME Mechanics of Materials Exam # 1 Date: October 5, 2016 Time: 8:00 10:00 PM
Name (Print) (Last) (First) Instructions: ME 323  Mechanics of Materials Exam # 1 Date: October 5, 2016 Time: 8:00 10:00 PM Circle your lecturer s name and your class meeting time. Gonzalez Krousgrill
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 informationStatics Chapter II Fall 2018 Exercises Corresponding to Sections 2.1, 2.2, and 2.3
Statics Chapter II Fall 2018 Exercises Corresponding to Sections 2.1, 2.2, and 2.3 2 3 Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the
More informationUNIT III DEFLECTION OF BEAMS 1. What are the methods for finding out the slope and deflection at a section? The important methods used for finding out the slope and deflection at a section in a loaded
More informationTHEME IS FIRST OCCURANCE OF YIELDING THE LIMIT?
CIE309 : PLASTICITY THEME IS FIRST OCCURANCE OF YIELDING THE LIMIT? M M  N N + + σ = σ = + f f BENDING EXTENSION Ir J.W. Welleman page nr 0 kn Normal conditions during the life time WHAT HAPPENS DUE TO
More informationRODS: STATICALLY INDETERMINATE MEMBERS
RODS: STTICLLY INDETERMINTE MEMERS Statically Indeterminate ackground In all of the problems discussed so far, it was possible to determine the forces and stresses in the members by utilizing the equations
More informationPDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics
Page1 PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [2910601] Introduction, Fundamentals of Statics 1. Differentiate between Scalar and Vector quantity. Write S.I.
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 informationMETHOD OF LEAST WORK
METHOD OF EAST WORK 91 METHOD OF EAST WORK CHAPTER TWO The method of least work is used for the analysis of statically indeterminate beams, frames and trusses. Indirect use of the Castigliano s nd theorem
More informationMethods of Analysis. Force or Flexibility Method
INTRODUCTION: The structural analysis is a mathematical process by which the response of a structure to specified loads is determined. This response is measured by determining the internal forces or stresses
More informationEquilibrium Equilibrium and Trusses Trusses
Equilibrium and Trusses ENGR 221 February 17, 2003 Lecture Goals 64 Equilibrium in Three Dimensions 71 Introduction to Trusses 72Plane Trusses 73 Space Trusses 74 Frames and Machines Equilibrium Problem
More informationSTRENGTH OF MATERIALSI. Unit1. Simple stresses and strains
STRENGTH OF MATERIALSI Unit1 Simple stresses and strains 1. What is the Principle of surveying 2. Define Magnetic, True & Arbitrary Meridians. 3. Mention different types of chains 4. Differentiate between
More informationChapter 5 CENTRIC TENSION OR COMPRESSION ( AXIAL LOADING )
Chapter 5 CENTRIC TENSION OR COMPRESSION ( AXIAL LOADING ) 5.1 DEFINITION A construction member is subjected to centric (axial) tension or compression if in any cross section the single distinct stress
More informationFIXED BEAMS IN BENDING
FIXED BEAMS IN BENDING INTRODUCTION Fixed or builtin beams are commonly used in building construction because they possess high rigidity in comparison to simply supported beams. When a simply supported
More informationMechanical Design in Optical Engineering
OPTI Buckling Buckling and Stability: As we learned in the previous lectures, structures may fail in a variety of ways, depending on the materials, load and support conditions. We had two primary concerns:
More informationEMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain  Axial Loading
MA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain  Axial Loading MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain  Axial Loading Statics
More informationTYPES OF STRUCUTRES. HD in Civil Engineering Page 11
E2027 Structural nalysis I TYPES OF STRUUTRES H in ivil Engineering Page 11 E2027 Structural nalysis I SUPPORTS Pin or Hinge Support pin or hinge support is represented by the symbol H or H V V Prevented:
More informationOUTCOME 1  TUTORIAL 3 BENDING MOMENTS. You should judge your progress by completing the self assessment exercises. CONTENTS
Unit 2: Unit code: QCF Level: 4 Credit value: 15 Engineering Science L/601/1404 OUTCOME 1  TUTORIAL 3 BENDING MOMENTS 1. Be able to determine the behavioural characteristics of elements of static engineering
More information