Unit III Theory of columns. Dr.P.Venkateswara Rao, Associate Professor, Dept. of Civil Engg., SVCE, Sriperumbudir
|
|
- Dwayne McKinney
- 6 years ago
- Views:
Transcription
1 Unit III Theory of columns 1
2 Unit III Theory of Columns References: Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi Rattan.S.S., "Strength of Materials", Tata McGraw Hill Education Pvt. Ltd., New Delhi, Rajput R.K., "Strength of Materials (Mechanics of Solids)", S.Chand & company Ltd., New Delhi, Ramamrutham S., Theory of structures Dhanpat Rai & Sons, New Delhi
3 Columns Columns are compression members. There are various examples of members subjected to compressive loads. Various names of compression members as per application: Post is a general term applied to a compression member. Strut is a compression member whose lateral dimensions are small compared to it s length. A strut may be horizontal, inclined or vertical and this term is used in trusses. (Tie is a tension member in a truss) But a vertical strut, used in buildings or frames is called column. Columns, pillars and stanchions are vertical members used in building frames. 3
4 Classification of Columns 1.Short column: Short column fails by crushing (compressive yielding) of the material. 2. Long column: Long column fails by buckling or bending. ( geometric or configuration failure) 3. Intermediate column: Intermediate column fails by combined buckling and crushing. (failure due to both material crushing and geometrical instability) 4
5 Buckling When a slender member is subjected to an axial compressive load, it may fail by a condition called buckling. Buckling is a geometric instability in which - - -:original shape : Buckled shape the lateral displacement of the axial member can suddenly become very large. 5
6 Buckled R.C.C. Columns 6
7 Buckled steel columns 7
8 Buckling examples of structural members 1. Building columns that transfer loads to the ground 2. Truss members in compression 3. Machine elements 4. Submarine hulls subjected to water pressure 8
9 Equilibrium States Three types of equilibrium: 1. Stable equilibrium 2. Unstable equilibrium 3. Neutral equilibrium 9
10 Buckling Mechanism 1. Stable equilibrium: If the load P is sufficiently small, when the force F is removed, the column will go back to its original straight condition. Gravity is the restoring force Elasticity of the column is the restoring force. 10
11 Buckling Mechanism (contd ) 2. Neutral equilibrium: When the column carries critical load P cr (Increased value of the load P) and a lateral force F is applied and removed, the column will remain in the slightly deflected position. Deflection amount depends on magnitude of force. Elastic restoring force is sufficient to prevent excessive deflection. 11
12 Buckling Mechanism (contd ) 3. Unstable equilibrium: When the column carries a load which is more than critical load P cr (Increased value of the load P) and a lateral force F is applied and removed, the column will bend considerably and it grows into excessively large deflection. Even small disturbance causes unstable. Elastic restoring force is insufficient to prevent excessive deflection. 12
13 Buckling Mechanism (contd ) Conclusion: Depending on the magnitude of force P, either column remains in straight position or in slight bent position or collapse due to crack extension. 13
14 Euler s long column theory The direct stress f 0 due to direct load is very small compared to bending stress f b due to buckling in long column. Euler derived an equation, for the buckling load of long column based on bending stress (neglecting the effect of direct stress). Buckling load cannot be used in short column. 14
15 Assumptions in the Euler s theory 1. The column is initially straight. 2. The cross section is uniform throughout. 3. The ends of the column are frictionless. 4. The material is homogeneous and isotropic. 5. The self weight of the column is neglected. 6. The line of thrust coincides exactly with the axis of the column. 7. The shortening of column due to axial compression is negligible. 8. The column failure occurs due to buckling only. 15
16 Cases of long columns based on end conditions 1. Both end pinned 2. Both ends fixed 3. One end fixed and the other end pinned 4. One end fixed and the other end free 16
17 Sign conventions for bending moments Convexity towards centre line, M x is +ve Concavity towards centre line, M x is -ve 17
18 End conditions of column Three important end conditions based on support types. (i) Pinned end: End is fixed in position only. Deflection, y = 0. (ii) Fixed end: End is fixed in position and direction. Deflection y = 0 and slope, dy dx = 0. (iii) Free end: Neither fixed in position nor in direction. 18
19 Case 1. Both ends hinged Consider a column AB of length l with its both ends free to rotate around frictionless pins and carrying a critical load P. As a result of critical loading, let the column deflect into a buckled shape AX 1 B as shown in the figure. Prior to this critical load, the column is straight. Smallest force at which buckled shape is possible is known as critical force. P 19
20 Case 1. Both ends hinged (contd ) 20
21 Case 1. Both ends hinged (contd ) 21
22 Case 1. Both ends hinged (contd ) 22
23 Case 1. Both ends hinged (contd ) 23
24 2. Both ends fixed 24
25 2. Both ends fixed (contd ) 25
26 2. Both ends fixed (contd ) 26
27 2. Both ends fixed (contd ) 27
28 2. Both ends fixed (contd ) 28
29 3. One end fixed and other hinged M A = Fixed end moment at A and H= Horizontal reaction at A and B as shown in Figure. Moment at X due to critical load P, M = Py + H(l x) 29
30 3. One end fixed and other hinged (contd ) 30
31 3. One end fixed and other hinged (contd ) 31
32 3. One end fixed and other hinged (contd ) 32
33 3. One end fixed and other hinged (contd ) 33
34 3. One end fixed and other hinged (contd ) 34
35 4. One end fixed, other free 35
36 4. One end fixed, other free (contd ) 36
37 4. One end fixed, other free (contd ) 37
38 4. One end fixed, other free (contd ) 38
39 4. One end fixed, other free (contd ) 39
40 Equivalent Length of a column 40
41 Equivalent length (Effective Length) 41
42 Effective Length Physically, the effective length is the distance between points on the buckled column where the moment goes to zero, i.e., where the column is effectively pinned. Considering the deflected shape, the moment is zero where the curvature is zero (from beam theory). Zero curvature corresponds to an inflection point in the deflected shape (where the curvature changes sign). 42
43 Effective length for columns with common end conditions l e = l l l 0.5 l l 0.7l l 2l 43
44 Effective length for columns with common end conditions 44
45 Effective length for columns with common end conditions 45
46 Critical stress of a column 46
47 Critical stress of a column (contd ) 47
48 Critical stress of a column (contd ) 48
49 Limitations of Euler s formula 1. It is applicable to an ideal strut only and in practice, there is always crookedness in the column and the load applied may not be exactly co-axial. 2. It takes no account of direct stress. It means that it may give a buckling load for struts, far in excess of load which they can be withstand under direct compression. 49
50 Problems Problem. A hollow circular column of internal diameter 20 mm and external diameter 40 mm has a total length of 5m. One end of the column is fixed and the other end is hinged. Find out the crippling stress of the column if E = N/mm 2. Also findout the shortest length of this column for which Euler s formula is valid taking the yield stress equal to 250 N/mm 2. 50
51 Solution. Problems d=20 mm; D=40 mm; l = 5 mm; E= N/mm 2. Euler s crippling load for one end fixed and the Hinged, P = 2π2 EI l 2 Euler s crippling stress, p c = P c = 2π2 EI A Al 2 Area of the column, A = π(d2 d 2 ) Moment of inertia, I = π(d4 d 4 ) 4 = π( ) 4 = mm 2 64 = π( ) 64 = mm 4 51
52 Solution. Problems Euler s crippling stress, p c = P c = 2π2 EI A = N/mm 2 Yield stress= 250 N/mm 2. = 2π Al l k = 2π2 E 250 = 2π I = k = I A = = l = k = = mm Shortest length of this column= 1.4 m. 52
53 Problems Problem1. A T-section 150 mm x 120 mm x 20 mm is used as a strut of 4 m long with hinged at its both ends. Calculate the crippling load if modulus of elasticity for the material be 2.0 x 10 5 N/mm 2. 53
54 Problem figure: Problems 150 mm 20 mm 120 mm 20 mm 54
55 Problems Solution : 55
56 Solution (contd ) Problems 150 mm y =34 mm 20 mm 120 mm 20 mm 56
57 Solution (contd ) Problems 150 mm Y y =34 mm X X 20 mm 20 mm 120 mm Y 57
58 Solution (contd ) Problems 58
59 Problems Problem 2: Compare the ratio of the strength of a solid steel column to that of a hollow of the same cross-sectional area. The internal diameter of the hollow column is ¾ of the external diameter. Both the column have the same length and are pinned at both ends. 59
60 Problems Solution: 60
61 Solution (contd ): Problems 61
62 Solution (contd ): Problems 62
63 Intermediate columns: Empirical Formulae Euler s formula is valid only for long columns, i.e. for columns having L ratio greater than a certain value for a particular K material. Euler s formula though valid for L > 89 for mild steel column, k doesn t take into account the direct compressive stress. For intermediate columns, empirical formulae i.e., Rankine s formula, Gordan s formula and Johnson s formulae are appropriate. 63
64 Rankine s Formula Rankine proposed an empirical formula for very short to very long columns. 64
65 Rankine s Formula (contd ) 65
66 Rankine s Formula (contd ) 66
67 Rankine s constant for various materials 67
68 Problems Problem 3: A hollow cast iron column 4.5 m long with both ends fixed, is to carry an axial load of 250 kn under working conditions. The internal diameter is 0.8 times the outer diameter of the column. Using Rankine Gordon s formula, determine the diameters of the column adopting a factor of safety of 4. Assume f c, the compressive strength to be 550 N/mm 2 and Rankine s constant a=1/
69 d=0.8 D L e = L 2 = Problems = 2.25 m = 2250 mm Area of the column, A = π(d2 0.8D 2 ) = 0.28 D 2 mm 2 4 = mm 2 Moment of inertia, I = π(d4 0.8 D 4 ) 64 k = I A Given: f c =550 N/mm 2, a=1/1600 = D 4 mm D4 = = 0.32 D mm 0.28 D2 69
70 Problems P safe load = P = F.S 4 Rankine Load, P R = f c A 1+a l e k = N = D D = 154D D D 2 = 154D D = 154D 4 70
71 Problems D D = 0 D 2 = ± D=61.24 mm d = = mm Diameters of the hollow cylindrical column ; d=48.99 mm D= mm. 71
72 Problems Problem 4: A round steel rod of diameter 15 mm and length 2m is subjected to a gradual increasing axial compressive load. Using Euler s formula find the buckling load. Find also the maximum lateral deflection corresponding to the buckling condition. Both ends of the rod may be taken as hinged. Take E = 2.1 x 10 5 N/mm 2 and the yield stress of steel = 240 N/mm 2. 72
73 Solution: Problems (contd ) 73
74 Solution (contd ) Problems (contd ) 74
75 Solution (contd ) Problems (contd ) 75
76 Problems (contd ) Problem 5: A cast iron hollow cylindrical column 3 m in length when hinged at both ends, has a critical buckling load of P kn. When this column is fixed at both the ends, its critical load rises to P+300 kn. If the ratio of external diameter to internal diameter is 1.25 and E = 1.0x10 5 N/mm 2. Determine the external diameter of the column. 76
77 Solution: Problems (contd ) 77
78 Solution (contd ) Problems (contd ) 78
79 Problems (contd ) Problem 5: A hollow cylindrical cast iron column is 4 m long, both ends being fixed. Design the column to carry an axial load of 250 kn. Use Rankine s formula and adopt a factor of safety of 5. The internal diameter may be taken as 0.8 times the external diameter. Take f c = 550 N/mm 2 and a =
80 Problems (contd ) Solution: 80
81 Solution (contd ) Problems (contd ) 81
82 Solution (contd ) Problems (contd ) 82
83 Problems (contd ) Problem 6: A column of 9 m long has a cross section shown in figure. The column is pinned at both ends. If the column is subjected to an axial load equal in value ¼ of the Euler s critical load for the column. Determine the factor of safety on the Rankine s ultimate stress value. Take f c = 326 N/mm 2, Rankine s constant a = 1, E =200 Gpa. Properties of one RSJ area = mm 2. I xx = cm 4, I yy = cm 4, Thickness of the web=6.7 mm. 83
84 Problems (contd ) 84
85 Solution: Problems (contd ) 85
86 Solution (contd ) Problems (contd ) l e for ends pinned = l = 9000 mm 86
87 Solution (contd ) Problems (contd ) 87
88 Solution (contd ) Problems (contd ) 88
89 Problems (contd ) Problem 7: A compound stanchion is made up of two ISMC 250 placed back to back with a gap between adjacent flat surfaces. Two 360 mm x 12 mm plates are riveted to the flanges, so as to form a symmetrical box section. Detremine the amount of gap if the column is to carry the maximum load. Properties of one ISMC 250 are: Area = 3867 mm 2, Max MI (I xx1 ) = mm 4 Max MI (I yy1 ) = mm 4 Distance of the centroid from the back = 23 mm. 89
90 Problems (contd ) If the effective length of the stanchion is 8.5 m, calculate the safe maximum load, the working stress being interpolated from the following table. 90
91 Problems (contd ) 91
92 Solution: Problems (contd ) 92
93 Solution (contd ) Problems (contd ) 93
94 Solution (contd ) Problems (contd ) 94
95 Short Columns P Axial loading A compression member such as column may be subjected to an axial load which pass through the geometrical axis of the member, thus causing direct stress. The axial force will cause a direct compressive stress given by, f 0 = P A f 0 Direct stress distribution 95
96 Eccentric Loading: Short Columns Some times a compression member such as column may be subjected to an axial load which may not pass through the geometrical axis of the member, thus causing bending as well as direct stress. In all such cases position of neutral layer change or altogether vanishes. The axial force will cause a direct compressive stress given by, f 0 = P A The bending couple will cause longitudinal tensile and compressive stresses. 96
97 Eccentric Loading: Short Columns The fiber stress f b at any distance y from the N.A. is given by, f b = M Y = P e y ( tensile or compressive). I x I x The extreme fibre stress due to bending is given by, f b = M Z x = P e Z x Therefore the total stress at any section of the column is given by, f = f 0 + f b = P A ± P e y I x Hence the maximum total stresses are given by, f max = f 0 + f b = P A + M Z x and f min = f 0 f b = P A M Z x 97
98 Short Columns P Eccentric loading e f max = f 0 + f b = P A + M Z x f min = f 0 f b = P A M Z x e f 0 f b f 0 f 0 < f b f 0 = f b f 0 > f b Stress distribution f b 98
99 Eccentric loading: Short Columns If f 0 is greater than f b the stress throughout the section will be of the same sign. When f 0 = f b, f max = 2f 0 and f min = 0 If, however, f 0 is less than f b, the stress will change sign, being partly tensile and partly compressive across the section. f 0 f b f 0 f 0 < f b f 0 = f b f 0 > f b Stress distribution f b 99
100 Middle third Rule: Short Columns In masonry and concrete structures, the development of tensile stress in the section is not desirable, as they are weak in tension. This limits the eccentricity e to a certain value which will be investigated for different sections as follows. In order that the stress may not change sign from compression to tensile, we have f 0 f b P A My I P Ped ; A 2Ak 2 2k2 e ; d where K is radius of gyration of the section with regard to N.A. and d is the depth of the section. f 0 f b f 0 f 0 < f b f 0 = f b f 0 > f b Stress distribution f b 100
101 Short Columns Ex: Rectangular section: e 2k2 I = bd3 12 and A = bd k 2 = I bd3 = 12 = d2 A bd 12 Since e 2k2 d e 2 d2 12 d e d 6 d b d and hence, e max = d 6 101
102 Ex: Rectangular section: Short Columns e max = d 6 The stress will be of the same sign throughout the b section if the load line is within the middle third of the section. In the case of rectangular section, the maximum intensities of extreme stresses are given by f = P A ± Pe Z = p bd ± 6Pe bd 2 f = P bd 1 ± 6e d d 102
103 Short Columns The core of a section: If the line of action of the stress is on neither of the centre lines of the sections, the bending is unsymmetrical. x A D d However, there is certain area within which the line of action of the force P must cut the cross section if the stress is not to become tensile. The area is known as core or kernal of the section. y y P B C x b 103
104 Short Columns The core of a Rectangular section : Let the point of application of the load P have the coordinates (x,y), with reference to the axis shown in above figure in which x is positive when measured to the right of origin o and y is positive when measured upwards. The stress at any point having coordinates (x,y ) is given by, f = P bd + Pxx bd 3 + Pyy db x A o y P B x b D d y C 104
105 Short Columns The core of a Rectangular section (contd..): f = P bd + Pxx bd 3 + Pyy db = 12P 1 bd 12 + xx d 2 + yy b 2 At D, x = d and 2 y = b 2 at D, we have f = 6P bd 1 6 x d y b and therefore f will be minimum. Thus x A y o B P x b D d y C 105
106 Short Columns The core of a Rectangular section (contd..): f = 6P 1 bd 6 x d y b The value of f reaches zero when x + y = 1 or 6x + 6y = 1 or d b 6 d b x d + y b = 1, which is a straight line equation, whose intercept 6 6 on the axes are respectively d 6 and b 6. Similar limits will apply in other quadrants and the stress will be y wholly compressive throughout the section. A B P b x x D d y C 106
107 Short Columns The core of a Rectangular section (contd..): If the line of action of P fall within rhombus ghjk, the diagonals of which are of length d and b respectively. 3 3 x A D d y d/3 y b/3 B C x b This rhombus is called the core of the rectangular section. 107
108 Short Columns The core of a solid circular section: For no tension, e 2k2 d I = πd4 64, A = πd2 4 k 2 = I πd 4 A = 64 πd 2 4 e 2 d2 16 d = d2 16 d/4 e d 8 The core is the circle with the same centre and diameter d/4 108
109 Short Columns The core of a hollow circular section: For no tension, e 2k2 d I = π(d4 d 4 ) 64, A = π(d2 d 2 ) 4 k 2 = I π(d 4 d 4 ) A = 64 π(d 2 d 2 ) 4 e 2 (D2 + d 2 ) 16 D = (D2 + d 2 ) 16 e (D2 + d 2 ) 8D The core is the circle with the same centre and diameter (D2 +d 2 ) 4D 109
110 A.U. Question Paper Problems State the Euler s assumption in column theory. And derive a relation for the Euler s crippling load for a columns with both ends hinged (Nov/Dec 2014). 110
111 A.U. Question Paper Problems A short length of tube having Internal diameter and external diameter are 4 cm and 5 cm respectively, which failed in compression at a load of 250 kn. When a 1.8 m length of the same tube was tested as a strut with fixed ends, the load failure was 160 kn. Assuming that σ c in Rankin s formula is given by the first test, find the value of the constant α in the same formula. What will be the crippling load of this tube if it is used as a strut 2.8 m long with one end fixed and the other hinged? (Nov/Dec 2014) 111
112 2 marks questions and Answers 1. State the middle third rule? 2. What is known as crippling load? 112
MAHALAKSHMI ENGINEERING COLLEGE
MAHALAKSHMI ENGINEERING COLLEGE TIRUCHIRAALLI - 6113. QUESTION WITH ANSWERS DEARTMENT : CIVIL SEMESTER: V SUB.CODE/ NAME: CE 5 / Strength of Materials UNIT 3 COULMNS ART - A ( marks) 1. Define columns
More informationCOURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5
COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4017 COURSE CATEGORY : A PERIODS/WEEK : 6 PERIODS/ SEMESTER : 108 CREDITS : 5 TIME SCHEDULE MODULE TOPICS PERIODS 1 Simple stresses
More informationColumns and Struts. 600 A Textbook of Machine Design
600 A Textbook of Machine Design C H A P T E R 16 Columns and Struts 1. Introduction.. Failure of a Column or Strut. 3. Types of End Conditions of Columns. 4. Euler s Column Theory. 5. Assumptions in Euler
More information: APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4021 COURSE CATEGORY : A PERIODS/ WEEK : 5 PERIODS/ SEMESTER : 75 CREDIT : 5 TIME SCHEDULE
COURSE TITLE : APPLIED MECHANICS & STRENGTH OF MATERIALS COURSE CODE : 4021 COURSE CATEGORY : A PERIODS/ WEEK : 5 PERIODS/ SEMESTER : 75 CREDIT : 5 TIME SCHEDULE MODULE TOPIC PERIODS 1 Simple stresses
More informationDEPARTMENT OF CIVIL ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF CIVIL ENGINEERING SUBJECT: CE 2252 STRENGTH OF MATERIALS UNIT: I ENERGY METHODS 1. Define: Strain Energy When an elastic body is under the action of external
More informationEngineering Science OUTCOME 1 - TUTORIAL 4 COLUMNS
Unit 2: Unit code: QCF Level: Credit value: 15 Engineering Science L/601/10 OUTCOME 1 - TUTORIAL COLUMNS 1. Be able to determine the behavioural characteristics of elements of static engineering systems
More informationENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL 4 COLUMNS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL 4 H1 FORMERLY UNIT 21718P
ENGINEERING SCIENCE H1 OUTCOME 1 - TUTORIAL COLUMNS EDEXCEL HNC/D ENGINEERING SCIENCE LEVEL H1 FORMERLY UNIT 21718P This material is duplicated in the Mechanical Principles module H2 and those studying
More informationneeded to buckle an ideal column. Analyze the buckling with bending of a column. Discuss methods used to design concentric and eccentric columns.
CHAPTER OBJECTIVES Discuss the behavior of columns. Discuss the buckling of columns. Determine the axial load needed to buckle an ideal column. Analyze the buckling with bending of a column. Discuss methods
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 informationCIVIL DEPARTMENT MECHANICS OF STRUCTURES- ASSIGNMENT NO 1. Brach: CE YEAR:
MECHANICS OF STRUCTURES- ASSIGNMENT NO 1 SEMESTER: V 1) Find the least moment of Inertia about the centroidal axes X-X and Y-Y of an unequal angle section 125 mm 75 mm 10 mm as shown in figure 2) Determine
More informationCE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR
CE6306 STRENGTH OF MATERIALS TWO MARK QUESTIONS WITH ANSWERS ACADEMIC YEAR 2014-2015 UNIT - 1 STRESS, STRAIN AND DEFORMATION OF SOLIDS PART- A 1. Define tensile stress and tensile strain. The stress induced
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 informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad -00 04 CIVIL ENGINEERING QUESTION BANK Course Name : STRENGTH OF MATERIALS II Course Code : A404 Class : II B. Tech II Semester Section
More informationChapter 12 Elastic Stability of Columns
Chapter 12 Elastic Stability of Columns Axial compressive loads can cause a sudden lateral deflection (Buckling) For columns made of elastic-perfectly plastic materials, P cr Depends primarily on E and
More informationElastic Stability Of Columns
Elastic Stability Of Columns Introduction: Structural members which carry compressive loads may be divided into two broad categories depending on their relative lengths and cross-sectional dimensions.
More informationR13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A
SET - 1 II B. Tech I Semester Regular Examinations, Jan - 2015 MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (Part-A and Part-B)
More informationOnly for Reference Page 1 of 18
Only for Reference www.civilpddc2013.weebly.com Page 1 of 18 Seat No.: Enrolment No. GUJARAT TECHNOLOGICAL UNIVERSITY PDDC - SEMESTER II EXAMINATION WINTER 2013 Subject Code: X20603 Date: 26-12-2013 Subject
More informationComparison of Euler Theory with Experiment results
Comparison of Euler Theory with Experiment results Limitations of Euler's Theory : In practice the ideal conditions are never [ i.e. the strut is initially straight and the end load being applied axially
More informationPERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK
PERIYAR CENTENARY POLYTECHNIC COLLEGE PERIYAR NAGAR - VALLAM - 613 403 - THANJAVUR. DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Sub : Strength of Materials Year / Sem: II / III Sub Code : MEB 310
More informationto introduce the principles of stability and elastic buckling in relation to overall buckling, local buckling
to introduce the principles of stability and elastic buckling in relation to overall buckling, local buckling In the case of elements subjected to compressive forces, secondary bending effects caused by,
More informationDownloaded from Downloaded from / 1
PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION-2002 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 informationEMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 10 Columns
EMA 370 Mechanics & Materials Science (Mechanics of Materials) Chapter 10 Columns Columns Introduction Columns are vertical prismatic members subjected to compressive forces Goals: 1. Study the stability
More informationUNIT-I STRESS, STRAIN. 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2
UNIT-I STRESS, STRAIN 1. A Member A B C D is subjected to loading as shown in fig determine the total elongation. Take E= 2 x10 5 N/mm 2 Young s modulus E= 2 x10 5 N/mm 2 Area1=900mm 2 Area2=400mm 2 Area3=625mm
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 informationSRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA
SRI CHANDRASEKHARENDRA SARASWATHI VISWA MAHAVIDHYALAYA (Declared as Deemed-to-be University under Section 3 of the UGC Act, 1956, Vide notification No.F.9.9/92-U-3 dated 26 th May 1993 of the Govt. of
More informationMECHANICS OF MATERIALS
CHTER MECHNICS OF MTERILS 10 Ferdinand. Beer E. Russell Johnston, Jr. Columns John T. DeWolf cture Notes: J. Walt Oler Texas Tech University 006 The McGraw-Hill Companies, Inc. ll rights reserved. Columns
More informationUNIT- I Thin plate theory, Structural Instability:
UNIT- I Thin plate theory, Structural Instability: Analysis of thin rectangular plates subject to bending, twisting, distributed transverse load, combined bending and in-plane loading Thin plates having
More informationPES Institute of Technology
PES Institute of Technology Bangalore south campus, Bangalore-5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject
More informationCritical Load columns buckling critical load
Buckling of Columns Buckling of Columns Critical Load Some member may be subjected to compressive loadings, and if these members are long enough to cause the member to deflect laterally or sideway. To
More informationMechanics of Materials Primer
Mechanics of Materials rimer Notation: A = area (net = with holes, bearing = in contact, etc...) b = total width of material at a horizontal section d = diameter of a hole D = symbol for diameter E = modulus
More informationLecture Slides. Chapter 4. Deflection and Stiffness. The McGraw-Hill Companies 2012
Lecture Slides Chapter 4 Deflection and Stiffness The McGraw-Hill Companies 2012 Chapter Outline Force vs Deflection Elasticity property of a material that enables it to regain its original configuration
More informationCOLUMNS: BUCKLING (DIFFERENT ENDS)
COLUMNS: BUCKLING (DIFFERENT ENDS) Buckling of Long Straight Columns Example 4 Slide No. 1 A simple pin-connected truss is loaded and supported as shown in Fig. 1. All members of the truss are WT10 43
More informationMarch 24, Chapter 4. Deflection and Stiffness. Dr. Mohammad Suliman Abuhaiba, PE
Chapter 4 Deflection and Stiffness 1 2 Chapter Outline Spring Rates Tension, Compression, and Torsion Deflection Due to Bending Beam Deflection Methods Beam Deflections by Superposition Strain Energy Castigliano
More informationName :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CE-NEW)/SEM-3/CE-301/ 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 informationSub. Code:
Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationChapter 5 Compression Member
Chapter 5 Compression Member This chapter starts with the behaviour of columns, general discussion of buckling, and determination of the axial load needed to buckle. Followed b the assumption of Euler
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 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 cross-section 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 information9.5 Compression Members
9.5 Compression Members This section covers the following topics. Introduction Analysis Development of Interaction Diagram Effect of Prestressing Force 9.5.1 Introduction Prestressing is meaningful when
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 information18.Define the term modulus of resilience. May/June Define Principal Stress. 20. Define Hydrostatic Pressure.
CE6306 STREGNTH OF MATERIALS Question Bank Unit-I STRESS, STRAIN, DEFORMATION OF SOLIDS PART-A 1. Define Poison s Ratio May/June 2009 2. What is thermal stress? May/June 2009 3. Estimate the load carried
More informationENG2000 Chapter 7 Beams. ENG2000: R.I. Hornsey Beam: 1
ENG2000 Chapter 7 Beams ENG2000: R.I. Hornsey Beam: 1 Overview In this chapter, we consider the stresses and moments present in loaded beams shear stress and bending moment diagrams We will also look at
More informationmportant nstructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationDesign of Beams (Unit - 8)
Design of Beams (Unit - 8) Contents Introduction Beam types Lateral stability of beams Factors affecting lateral stability Behaviour of simple and built - up beams in bending (Without vertical stiffeners)
More informationChapter 6: Cross-Sectional Properties of Structural Members
Chapter 6: Cross-Sectional Properties of Structural Members Introduction Beam design requires the knowledge of the following. Material strengths (allowable stresses) Critical shear and moment values Cross
More informationME 354, MECHANICS OF MATERIALS LABORATORY COMPRESSION AND BUCKLING
ME 354, MECHANICS OF MATERIALS LABATY COMPRESSION AND BUCKLING PURPOSE 01 January 2000 / mgj The purpose of this exercise is to study the effects of end conditions, column length, and material properties
More informationUnit IV State of stress in Three Dimensions
Unit IV State of stress in Three Dimensions State of stress in Three Dimensions References Punmia B.C.,"Theory of Structures" (SMTS) Vol II, Laxmi Publishing Pvt Ltd, New Delhi 2004. Rattan.S.S., "Strength
More informationDETAILED SYLLABUS FOR DISTANCE EDUCATION. Diploma. (Three Years Semester Scheme) Diploma in Architecture (DARC)
DETAILED SYLLABUS FOR DISTANCE EDUCATION Diploma (Three Years Semester Scheme) Diploma in Architecture (DARC) COURSE TITLE DURATION : Diploma in ARCHITECTURE (DARC) : 03 Years (Semester System) FOURTH
More informationChapter 4 Deflection and Stiffness
Chapter 4 Deflection and Stiffness Asst. Prof. Dr. Supakit Rooppakhun Chapter Outline Deflection and Stiffness 4-1 Spring Rates 4-2 Tension, Compression, and Torsion 4-3 Deflection Due to Bending 4-4 Beam
More informationSTRENGTH OF MATERIALS-I. Unit-1. Simple stresses and strains
STRENGTH OF MATERIALS-I Unit-1 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 informationUnit I Stress and Strain
Unit I Stress and Strain Stress and strain at a point Tension, Compression, Shear Stress Hooke s Law Relationship among elastic constants Stress Strain Diagram for Mild Steel, TOR steel, Concrete Ultimate
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 informationSTRESS, STRAIN AND DEFORMATION OF SOLIDS
VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, MADURAI 625009 DEPARTMENT OF CIVIL ENGINEERING CE8301 STRENGTH OF MATERIALS I -------------------------------------------------------------------------------------------------------------------------------
More informationCOURSE TITLE : THEORY OF STRUCTURES -I COURSE CODE : 3013 COURSE CATEGORY : B PERIODS/WEEK : 6 PERIODS/SEMESTER: 90 CREDITS : 6
COURSE TITLE : THEORY OF STRUCTURES -I COURSE CODE : 0 COURSE CATEGORY : B PERIODS/WEEK : 6 PERIODS/SEMESTER: 90 CREDITS : 6 TIME SCHEDULE Module Topics Period Moment of forces Support reactions Centre
More information3 Hours/100 Marks Seat No.
*17304* 17304 14115 3 Hours/100 Marks Seat No. Instructions : (1) All questions are compulsory. (2) Illustrate your answers with neat sketches wherever necessary. (3) Figures to the right indicate full
More informationVALLIAMMAI ENGINEERING COLLEGE
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF CIVIL ENGINEERING QUESTION BANK IV SEMESTER CE6402 STRENGTH OF MATERIALS Regulation 2013 Academic Year 2017 18 Prepared by
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 informationLecture 15 Strain and stress in beams
Spring, 2019 ME 323 Mechanics of Materials Lecture 15 Strain and stress in beams Reading assignment: 6.1 6.2 News: Instructor: Prof. Marcial Gonzalez Last modified: 1/6/19 9:42:38 PM Beam theory (@ ME
More informationAdvanced Structural Analysis EGF Section Properties and Bending
Advanced Structural Analysis EGF316 3. Section Properties and Bending 3.1 Loads in beams When we analyse beams, we need to consider various types of loads acting on them, for example, axial forces, shear
More informationGATE SOLUTIONS E N G I N E E R I N G
GATE SOLUTIONS C I V I L E N G I N E E R I N G From (1987-018) Office : F-16, (Lower Basement), Katwaria Sarai, New Delhi-110016 Phone : 011-65064 Mobile : 81309090, 9711853908 E-mail: info@iesmasterpublications.com,
More informationSECTION 7 DESIGN OF COMPRESSION MEMBERS
SECTION 7 DESIGN OF COMPRESSION MEMBERS 1 INTRODUCTION TO COLUMN BUCKLING Introduction Elastic buckling of an ideal column Strength curve for an ideal column Strength of practical column Concepts of effective
More information[8] Bending and Shear Loading of Beams
[8] Bending and Shear Loading of Beams Page 1 of 28 [8] Bending and Shear Loading of Beams [8.1] Bending of Beams (will not be covered in class) [8.2] Bending Strain and Stress [8.3] Shear in Straight
More informationMECHANICS OF MATERIALS Sample Problem 4.2
Sample Problem 4. SOLUTON: Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. ya ( + Y Ad ) A A cast-iron machine part is acted upon by a kn-m couple.
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 informationLongitudinal buckling of slender pressurised tubes
Fluid Structure Interaction VII 133 Longitudinal buckling of slender pressurised tubes S. Syngellakis Wesse Institute of Technology, UK Abstract This paper is concerned with Euler buckling of long slender
More informationAccordingly, the nominal section strength [resistance] for initiation of yielding is calculated by using Equation C-C3.1.
C3 Flexural Members C3.1 Bending The nominal flexural strength [moment resistance], Mn, shall be the smallest of the values calculated for the limit states of yielding, lateral-torsional buckling and distortional
More informationStructural Mechanics Column Behaviour
Structural Mechanics Column Behaviour 008/9 Dr. Colin Caprani, 1 Contents 1. Introduction... 3 1.1 Background... 3 1. Stability of Equilibrium... 4. Buckling Solutions... 6.1 Introduction... 6. Pinned-Pinned
More informationChapter 3. Load and Stress Analysis
Chapter 3 Load and Stress Analysis 2 Shear Force and Bending Moments in Beams Internal shear force V & bending moment M must ensure equilibrium Fig. 3 2 Sign Conventions for Bending and Shear Fig. 3 3
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 informationSample Question Paper
Scheme I Sample Question Paper Program Name : Mechanical Engineering Program Group Program Code : AE/ME/PG/PT/FG Semester : Third Course Title : Strength of Materials Marks : 70 Time: 3 Hrs. Instructions:
More informationCHAPTER 6: ULTIMATE LIMIT STATE
CHAPTER 6: ULTIMATE LIMIT STATE 6.1 GENERAL It shall be in accordance with JSCE Standard Specification (Design), 6.1. The collapse mechanism in statically indeterminate structures shall not be considered.
More informationCivil Engineering Design (1) Design of Reinforced Concrete Columns 2006/7
Civil Engineering Design (1) Design of Reinforced Concrete Columns 2006/7 Dr. Colin Caprani, Chartered Engineer 1 Contents 1. Introduction... 3 1.1 Background... 3 1.2 Failure Modes... 5 1.3 Design Aspects...
More informationCHAPTER 4: BENDING OF BEAMS
(74) CHAPTER 4: BENDING OF BEAMS This chapter will be devoted to the analysis of prismatic members subjected to equal and opposite couples M and M' acting in the same longitudinal plane. Such members are
More informationISHIK UNIVERSITY DEPARTMENT OF MECHATRONICS ENGINEERING
ISHIK UNIVERSITY DEPARTMENT OF MECHATRONICS ENGINEERING QUESTION BANK FOR THE MECHANICS OF MATERIALS-I 1. A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. If the modulus
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 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 informationMECHANICS OF SOLIDS. (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University
MECHANICS OF SOLIDS (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University Dr. S.Ramachandran, M.E., Ph.D., Mr. V.J. George, M.E., Mr. S. Kumaran,
More informationStructural Steelwork Eurocodes Development of A Trans-national Approach
Structural Steelwork Eurocodes Development of A Trans-national Approach Course: Eurocode Module 7 : Worked Examples Lecture 0 : Simple braced frame Contents: 1. Simple Braced Frame 1.1 Characteristic Loads
More informationε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram
CHAPTER NINE COLUMNS 4 b. The modified axial strength in compression is reduced to account for accidental eccentricity. The magnitude of axial force evaluated in step (a) is multiplied by 0.80 in case
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 in-class Updated 3/3/17 Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on
More informationProperties of Sections
ARCH 314 Structures I Test Primer Questions Dr.-Ing. Peter von Buelow Properties of Sections 1. Select all that apply to the characteristics of the Center of Gravity: A) 1. The point about which the body
More information71- Laxmi Nagar (South), Niwaru Road, Jhotwara, Jaipur ,India. Phone: Mob. : /
www.aarekh.com 71- Laxmi Nagar (South), Niwaru Road, Jhotwara, Jaipur 302 012,India. Phone: 0141-2348647 Mob. : +91-9799435640 / 9166936207 1. Limiting values of poisson s ratio are (a) -1 and 0.5 (b)
More informationPLATE GIRDERS II. Load. Web plate Welds A Longitudinal elevation. Fig. 1 A typical Plate Girder
16 PLATE GIRDERS II 1.0 INTRODUCTION This chapter describes the current practice for the design of plate girders adopting meaningful simplifications of the equations derived in the chapter on Plate Girders
More informationENCE 455 Design of Steel Structures. III. Compression Members
ENCE 455 Design of Steel Structures III. Compression Members C. C. Fu, Ph.D., P.E. Civil and Environmental Engineering Department University of Maryland Compression Members Following subjects are covered:
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 informationCHAPTER 4. ANALYSIS AND DESIGN OF COLUMNS
4.1. INTRODUCTION CHAPTER 4. ANALYSIS AND DESIGN OF COLUMNS A column is a vertical structural member transmitting axial compression loads with or without moments. The cross sectional dimensions of a column
More informationSimulation of Geometrical Cross-Section for Practical Purposes
Simulation of Geometrical Cross-Section for Practical Purposes Bhasker R.S. 1, Prasad R. K. 2, Kumar V. 3, Prasad P. 4 123 Department of Mechanical Engineering, R.D. Engineering College, Ghaziabad, UP,
More informationMODULE C: COMPRESSION MEMBERS
MODULE C: COMPRESSION MEMBERS This module of CIE 428 covers the following subjects Column theory Column design per AISC Effective length Torsional and flexural-torsional buckling Built-up members READING:
More information7.3 Design of members subjected to combined forces
7.3 Design of members subjected to combined forces 7.3.1 General In the previous chapters of Draft IS: 800 LSM version, we have stipulated the codal provisions for determining the stress distribution in
More informationQuestion 1. Ignore bottom surface. Solution: Design variables: X = (R, H) Objective function: maximize volume, πr 2 H OR Minimize, f(x) = πr 2 H
Question 1 (Problem 2.3 of rora s Introduction to Optimum Design): Design a beer mug, shown in fig, to hold as much beer as possible. The height and radius of the mug should be not more than 20 cm. The
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 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 informationStress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy
Stress Analysis Lecture 4 ME 76 Spring 017-018 Dr./ Ahmed Mohamed Nagib Elmekawy Shear and Moment Diagrams Beam Sign Convention The positive directions are as follows: The internal shear force causes a
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 information675(1*7+ 2) 0$7(5,$/6
675(1*7+ 2) 0$7(5,$/6 (MECHANICS OF SOLIDS) (As per Leading Universities Latest syllabus including Anna University R2013 syllabus) Dr. S.Ramachandran, Professor and Research Head Faculty of Mechanical
More information12 Bending stresses and direct stresses corn bined
12 Bending stresses and direct stresses corn bined 12.1 Introduction Many instances arise in practice where a member undergoes bending combined with a thrust or pull. If a member carries a thrust, direct
More informationCompression Members. ENCE 455 Design of Steel Structures. III. Compression Members. Introduction. Compression Members (cont.)
ENCE 455 Design of Steel Structures III. Compression Members C. C. Fu, Ph.D., P.E. Civil and Environmental Engineering Department University of Maryland Compression Members Following subjects are covered:
More information3. BEAMS: STRAIN, STRESS, DEFLECTIONS
3. BEAMS: STRAIN, STRESS, DEFLECTIONS The beam, or flexural member, is frequently encountered in structures and machines, and its elementary stress analysis constitutes one of the more interesting facets
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 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 information