TM Strength of Materials R.K. Kaushik Distributed by: @seismicisolation @seismicisolation @seismicisolation @seismicisolation Strength of Materials Dr. R.K. Kaushik Professor Department of Mechanical Engineering Ganga Institute of Technology & Management Kablana, Dist. Jhajjar (Haryana), India @seismicisolation @seismicisolation ©Copyright 2019 I.K. International Pvt. Ltd., New Delhi-110002. This book may not be duplicated in any way without the express written consent of the publisher, except in the form of brief excerpts or quotations for the purposes of review. The information contained herein is for the personal use of the reader and may not be incorporated in any commercial programs, other books, databases, or any kind of software without written consent of the publisher. Making copies of this book or any portion for any purpose other than your own is a violation of copyright laws. 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Trademarks: All brand names and product names used in this book are trademarks, registered trademarks, or trade names of their respective holders. Dreamtech Press is not associated with any product or vendor mentioned in this book. ISBN: 978-93-89307-31-3 EISBN: 978-93-89447-89-7 Edition: 2019 @seismicisolation @seismicisolation Dedicated to my wife dear Saroj Gaur Kaushik @seismicisolation @seismicisolation @seismicisolation @seismicisolation Preface Though in my 48 years experience, 28 years is in MNC ABB (earlier Taylor), but for 20 years I taught various subjects to Higher Diploma (equivalent to B.Tech, Mech), B.Tech subjects. Strength of Materials, Material Science and Machine Design (Mech) have been my favourite subjects. This book covers the syllabuses of most universities of B.Tech as well as Diploma in Mechanical Engineering. Enough solved examples and problems for exercise are given in the book. I wish to express my special thanks to my wife Saroj Gaur Kaushik who assisted me a lot during preparation of the book. I request students and teachers to point out mistakes, if any to give me opportunity to correct the same in next edition. R.K. Kaushik @seismicisolation @seismicisolation @seismicisolation @seismicisolation Contents Preface vii 1. Introduction 1 2. Stress and Strains Stress Deformation of a Body Due to Self Weight Extension of Tapered Rectangular Strip Bar of Uniform Strength Exercise 4 4 11 17 20 26 3. Temperature Stress and Strain Composite Tube or Bar Thermal Stresses in a Bar of Tapering Section Exercise 29 30 32 38 4. Elastic Constants Relation between Modulus of Elasticity and Modulus of Rigidity Relation between Modulus of Elasticity and Bulk Modulus Exercise 40 43 44 45 5. Principal Stresses and Strains Stresses on a Oblique Section Material Subjected to Two Perpendicular Stresses Material Subjected to Shear Stresses Material Subjected to Direct and Shear Stresses Graphical Method (Mohr’s Circle of Stresses) Graphical (Mohr’s) Method Exercise 47 47 50 54 55 59 69 77 6. Shearing Force and Bending Moment Shearing Force Bending Moment 79 79 80 @seismicisolation @seismicisolation x • Contents Shearing Force and Bending Moment Diagrams Cantilevers Exercise 82 89 94 7. Centre of Gravity Centroid for Plane Figures Centre of Gravity for Solid Bodies Centroid of Different Sections Exercise 97 97 99 101 106 8. Moment of Inertia Rectangular Section Radius of Gyration Theorem of Parallel Axis Theorem of the Perpendicular Axis Moment of Inertia of a Hollow Rectangular Section Moment of Inertia of a Circular Section Moment of Inertia of a Triangle Moment of Inertia of Semi-circular Lamina about its Centroidal Axis Product of Inertia of Rectangle Exercise 111 111 112 113 113 114 114 115 116 122 122 9. Bending of Beams Relationship Between Curvature and Strain Moment of Resistance Modulus of Section Beams of Uniform Strength Composite Beams or Flitched Beams Combined Bending and Direct Stresses Conditions for No Tension in the Section Exercise 125 125 126 128 136 138 142 146 151 10. Shear Stresses in Beams (A) Shear Stress Distribution for Beam of Rectangular Section (B) Shear Stress Distribution of a Solid Circular Section (C) Shear Stress Distribution in an I Section Exercise 157 159 160 162 170 11. Torsion Assumptions Torsional Moment of Resistance Twist of the Shaft Composite Shaft Twisting Beyond the Limit of Proportionality 175 175 176 182 185 185 @seismicisolation @seismicisolation Contents • xi Torsion of a Tapering Shaft Thin Circular Tube Subjected to Torsion Exercise 187 190 193 12. Thin Cylindrical and Spherical Shells (i) Hoop Stress or Circumferential Stress (ii) Longitudinal Stress Change in Volume Wire-Bound Thin Cylindrical Shells Exercise 196 196 197 198 205 212 13. Thick Cylinders and Spheres Solid Circular Shaft Subjected to External Pressure Thick Spherical Shells Exercise 214 219 229 231 14. Deflection of Beams Relation between Slope, Deflection and Radius of Curvature (A) Double Integration Method (B) Macaulay’s Method (C) Moment Area Method (D) Conjugate Beam Method (E) Superposition Method (F) Strain Energy Method Props Exercise 234 234 235 251 259 266 277 278 285 286 15. Strain Energy, Impact Loading and Deflection Due to Bending (A) Strain Energy Stored in a Body When the Load is Gradually Applied (B) Suddenly Applied Load (C) Strain Energy Stored in a Body, When the Load is Applied with Impact Strain Energy in Pure Shearing Strain Energy in Torsion Strain Energy Due to Bending Maxwell’s Reciprocal Theorem Betti’s Theorem of Reciprocal Deflections Exercise 291 291 292 294 305 306 307 311 312 318 16. Theories of Elastic Failure 1. Maximum Principal Stress Theory (Rankine’s Theory) 2. Maximum Shear Stress Theory (Guest or Teresa’s Theory) 3. Maximum Principal Strain Theory (St. Venant’s Theory) 320 320 321 322 @seismicisolation @seismicisolation xii • Contents 4. Maximum Strain Energy Theory (Haigh’s Theory) 5. Maximum Shear Strain Energy Theory (Von Mises and Henkey’s Theory) Exercise 323 324 339 17. Combined Stresses (Direct, Bending, Torsion) Combined Bending and Twisting Exercise 342 343 355 18. Fixed Beams Fixed Beam with a Point Load at the Centre Fixed Beam with Uniformly Distributed Load Over the Span By Area Moment Method Deflection for a Fixed Beam with Concentrated Load Anywhere on the Span Exercise 357 357 359 361 19. Continuous Beams Clapeyron’s Equation of Three Moments Exercise 377 377 387 20. Springs Wahl’s Correction Factor Close-coiled Helical Spring with Axial Couple Springs in Series and Parallel Open-coiled Helical Spring Composite Action of Axial Load and Couple Flat Spiral Spring Leaf, Laminated or Carriage Springs or Semi-elliptic Spring Quarter-Elliptic Leaf Spring Exercise 388 390 391 396 402 404 405 408 410 421 21. Columns and Struts (a) Both Ends Hinged (b) Column with One End Fixed and the Other Free (c) Both Ends Fixed (d) One End Fixed, Other End Hinged Limitation for the Use of Euler’s Theory Rankine’s Formula Columns Subjected to Eccentric Loading (Secant Formula) Perry’s Formula for Eccentrically Loaded Column Exercise 424 426 427 428 429 437 438 446 449 452 @seismicisolation @seismicisolation 362 375 Contents • xiii 22. Bending of Curved Bars Beams with Large Radius of Curvature (or Small Curvature) Beam with Small Initial Radius of Curvature (or Large Curvature) Rectangular Section Circular Section Triangular Section Trapezoidal Section Stress in a Chain Link Exercise 456 456 459 464 464 465 466 484 488 23. Unsymmetrical Bending Determination of Principal Axes and Principal Moments of Inertia Beam with Unsymmetrical Bending Moment Momental Ellipse Deflection of Beams due to Unsymmetrical Bending Method for Finding Bending Stream is Unsymmetrical Bending Shear Centre for Channel Section Exercise 491 492 493 495 496 497 508 516 24. Rotating Discs and Cylinders Rotating Ring Rotating Disc of Constant Thickness Solid Disc Hollow Disc Rotating Long Cylinder Solid Cylinder Disc of Uniform Strength Exercise 520 520 524 526 527 542 545 554 557 25. Frameworks Bow's Notation for Graphical Solution The Method of Sections Exercise 559 560 580 587 26. Dams Rectangular Dams Trapezoidal Dams with Water Face Vertical Rule of Middle Third Trapezoidal Dams with Water Face Inclined Exercise 591 591 596 599 603 608 @seismicisolation @seismicisolation xiv • Contents 27. Riveted Joints Rivet Types of Rivet Heads Material of Rivets Types of Riveted Joints Some Definitions Related to Riveted Joints Failure of a Riveted Joint Efficiency of a Riveted Joint Diamond Riveting Exercise 610 610 610 612 612 614 615 617 621 625 28. Welding Joints Classification of Welding Advantages and Disadvantages of Welded Joints over Riveted Joints Unsymmetrical Welded Section Fillet Welds Under Bending Moment Eccentrically Loaded Welded Joint Exercise 626 626 627 632 636 643 647 29. Mechanical Testing of Materials 1. Tensile Test 2. Compression Test 3. Shear Tests 4. Modulus of Rupture 5. Impact Test 6. Cupping Test 7. Modulus of Rigidity of Rubber 8. Torsion of a Round Bar 9. Verification of Macaulay’s Method for Beam Deflection 10. Verification of Maxwell’s Reciprocal Theorem 11. Hardness Testing Machines 12. Experiment: Close-coiled Helical Springs 13. Experiment: Value of the Modulus of Rigidity C of a Close-coiled Helical Spring 14. Experiment: Young’s Modulus of Elasticity of a Material Using Simply Supported Beam 15. Fatigue Testing Exercise @seismicisolation @seismicisolation 649 649 653 657 659 661 664 665 667 669 670 671 681 683 685 686 688 Contents • 30. Miscellaneous Solved Problems: Stresses and Strains Complex Stress and Strain Shear Force and Bending Moment Diagram Bending Stresses Shear Stresses Torsion Springs Application of Castigliano’s Theorem Moment of Inertia Strain Energy Theories of Failure Rotating Discs and Cylinder Bending of Curved Bars Fixed Beams Columns and Struts Combined Loading Index xv 689 697 698 699 702 703 706 710 714 716 719 726 727 728 730 732 737 @seismicisolation @seismicisolation @seismicisolation @seismicisolation C HAPTER 1 INTRODUCTION In our life we come across many steel structures such as bridges, machine tools, etc., it is impossible to construct these without thorough knowledge of strength of materials. As we know if the load reaches a maximum stress beyond permissible limits, the structure is going to fail. Hence, it is very important to study the strength of materials before their application in practical field. During one study we will come across fundamental quantities such as length, stress and time. In this book S.I. units are used everywhere for the convenience of students. Earlier units such as F.P.S., C.G.S and M.K.S. have been discarded. Following dimensions are used for various quantities: Force: N (Newton = kg m/s2 ) Pressure: Pa (Pascal = N/m2 ) Length: metre or mm Stress: Pa (1 N/m2 ) Remember 1 GN/m2 = 1 kN/mm2 , 1 MPa = 1 N/mm2 , G = 109 called giga. Mass density = kg/m3 Before we plunge into details of strength of materials it is better to understand the following abbreviations: T (tera) 1012 G (giga) 109 M (mega) 106 k (kilo) 103 Some important formulae: sin2 A + cos2 A = 1 sec2 A = 1 + tan2 A cosec2 A = 1 + cot2 A @seismicisolation @seismicisolation 2 • Strength of Materials The law of sines: In any triangle b c a = = sin A sin B sinC If a, b, and c are the sides and A, B and C their opposite angles. sin (A + B) = sin A cos B + cos A sin B sin (A − B) = sin A cos B − cos A sin B cos (A + B) = cos A + cos B − sin A sin B tan (A + B) = tan A + tan B 1 − tan A tan B tan (A − B) = tan A − tan B 1 + tan A tan B 1 − cos 2A 2 1 + cos 2A cos2 A = , sin 2A = 2 sin A cos A 2 sin2 A = 2 cos A sin B = sin (A + B) − sin (A − B), cos 2A = cos2 A − sin2 A If ax2 + bx + c = 0 Then, √ −b ± b2 − 4ac x= 2a dxn = nxn−1 dx dc = a where c is a constant. dx d (u.v) dv du = u. + v. dx dx dx d (sin x) d (cos x) = cos x, = − sin x dx dx d (tan x) d (cot x) = sec2 x, = −cosec2 x dx dx d d (sec x) = sec x tan x; (cosec x) = −cosec x · cot x dx dx @seismicisolation @seismicisolation Introduction xn dx = (ax + b)n dx = xn+1 + c, n+1 c is a constant (ax + b)n+1 (n + 1) × a logxe = 2.3(log x) (a − bx)−n+1 −n (a − bx) dx = (−n + 1)(−b) @seismicisolation @seismicisolation • 3 C HAPTER 2 STRESS AND STRAINS Stress Stress can be classified broadly in three types as described below: l dl 1. Tensile stress: It is illustrated in Fig. 2.1 where a tensile load W is applied to a uniform rod fixed at one end. Tensile stress, σ = W Figure 2.1 W W = Cross-sectional area of rod A unit is N/mm2 or MN/m2 , σ (Greek letter sigma). 2. Compressive stress: As shown in Fig. 2.2 when load W tends to compress a rod of cross-section area A, then compressive W stress = . A W Figure 2.2 3. Shear stress: If two plates are joined together with rivet as shown in Fig. 2.3. The stress in rivet is known as shear stress, it is denoted by τ (Greek letter tau), shear F stress in rivet, τ = . A F @seismicisolation @seismicisolation F Figure 2.3 Stress and Strains • 5 Behaviour of an elastic limit when subjected to a varying load. Strain: It is defined as the ratio of change in length to original length. It is denoted by Greek letter ε (epsilon). Thus strain, Change in length dl = ε= Original length l It is illustrated in Fig. 2.1. Hooke’s law: It states that stress is proportional to strain within elastic limit. ∴ Stress ∝ Strain or Stress = a constant = E Strain Modulus of elasticity: E is called modulus of elasticity or Young’s modulus. Following Table gives value of E for some important materials. E(N/mm2 ) approx. E(GN/m2 or kN/mm2 ) approximately Steel 200 × 109 200 Cast iron 115 × 109 115 Wrought iron 175 × 109 175 Brass 85 × 109 85 Aluminium alloys 70 × 109 70 Copper 120 × 109 120 Timber 10 × 109 10 Spheroidal C.I. 175 × 109 175 Rubber 0.8 × 106 − Material Stress-Strain Curve: Figure 2.4 shows a graph for plain carbon steel during a tensile test. D F C σ B C' A (Stress) ε (Strain) Figure 2.4 @seismicisolation @seismicisolation 6 • Strength of Materials It may be noted that point A is the limit of proportionality and B is the elastic limit. Between point A and B it is a curve thus not linear relationship. Therefore, actually E is constant within the limit of proportionality, though in Hooke’s law, we had mentioned, within ‘elastic limit’, because A and B are very close to each other. Let us name the important points on the graph: A: Limit of proportionality B: Elastic limit: It may be noted that on removal of load up to elastic limits, specimen comes back to its original dimension. C: Higher yield point: This is the point where yielding of the material begins. C : Lower yield point: The stress associated with the lower yield point is known as yield strength. D: Maximum stress: Here the stress is maximum because due to plastic behaviour of the material, area of cross section is very low. E: Point of fracture: At this point ‘waisting occurs’ as shown in Fig. 2.5. Cone Waisting Cap Figure 2.5 If the material is loaded beyond the elastic limit and then load is removed, a permanent extension remains, called permanent set. Proof Stress: For engineering purposes it is desirable to know the stress to which a highly ductile material such as aluminium can be loaded safely before a permanent extension takes place. This stress is known as the proof stress or offset stress and is defined as the stress at which a specified permanent extension has taken place in the tensile test. Proof stress is found from the stress-strain curve as given in Fig. 2.6. The extension specified is usually 0.1, 0.2 or 0.5 per cent of gauge length. Stress 0.1% Proof stress 0 0.001 Strain Figure 2.6 @seismicisolation @seismicisolation Stress and Strains • 7 The proof stress here is found on the basis of 0.1 per cent strain. Procedure: Draw a line parallel to the initial slope of the curve. The stress at the point where this line cuts the curve is the 0.1% proo f stress. The 0.2 per cent proof stress is also found in the same manner. Note: Though we define Hooke’s law to be taken without elastic limit, but strictly speaking it is applicable up to the point of proportionality B in Fig. 2.4. Brittle Materials: Fig. 2.6 shows that the stress-strain graph for brittle materials such as cast iron. The metal is almost elastic and up to fracture but does not obey Hooke’s law. A material such as this which has little plasticity or ductility and does not neck down before fracture is termed ‘brittle’. The modulus of elasticity for cast iron is not constant but depends on the portion of the curve from which it is calculated. Fracture Stress Strain Figure 2.7 Cast iron in tension The following table is very useful for mechanical properties: Material ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ Copper annealed Copper hard Aluminium soft Aluminium hard Black mild steel Bright mild steel Structural steel Cast iron Spheroidal Graphite Cast iron (annealed) Stainless steel Percentage elongation Yield stress MN/m2 0.1% proof stress MN/m2 Ultimate tensile stress MN/m2 60 4 35 5 25–26 14–17 20 − − − − − 60 320 30 140 220 400 90 150 220–250 − 430–500 − 230 − – 280–340 600 − 60 @seismicisolation @seismicisolation 8 • Strength of Materials stress strain Let F, L, A, dl be the force, length, area of cross section, and extension or contraction Now E = respectively, then E = F.L A.dl E XAMPLE 2.1: A bar of mild steel has an overall length of 2.1 m. The diameter up to 700 mm length is 56 mm, the diameter of the remaining 1.4 m is 35 mm. Calculate the extension of the bar due to a tensile load of 55 kN. E = 200 GN/m2 . S OLUTION : ∴ Remember 1 GN/m2 = 1 kN/mm2 E = 200 kN/mm2 Fl F.l We know E = ∴ dl = Adl A.E ∴ Therefore, for portion of 700 mm, the extension dl1 = 55000 × 700 × 7 × 4 5 = mm = 0.0178 mm 200000 × 22 × 56 × 56 64 Now dl2 for 1400 mm length of 35 mm dia, 55000 × 1400 × 7 × 4 = 0.4 mm 200000 × 22 × 35 × 35 Total extension = dl1 + dl2 = 0.0178 + 0.4 = 0.4178 mm dl2 = Compound bars: When two or more materials (members) are rigidly fixed together so that they share the same load and extend or compress by same amount, the two members form compound bar. Let us say that in Fig. 2.7 we have to find stress in each material and amount of compression. P Material A of ES l Material B of EB Let the outer tube of material A has outside dia as d1 and inside dia as d2 and inner tube of material B has outside dia as d3 and inside dia as d4 . Both ends are joined rigidly to make compound bar of length l. Figure 2.8 @seismicisolation @seismicisolation Stress and Strains • 9 Now the basic method is very simple as we know that load P is shared in some proportion by materials is A and B. That means σa Aa + σb Ab = P (i) Also, since the tubes are rigidly connected so they will have same strain as length l is common. stress σ = ε strain σa σb ε= = Ea Eb E= ∴ since strain ε is equal. (ii) Since the modulus of material A and B is known, so equations (i) and (ii) can be solved to find σa and σb . Now for compression, σa σ x or b strain = , pick any strain l Ea Eb σa Then x = × l = Compression Ea E XAMPLE 2.2: A column is made up of a steel tube, 70 mm inside diameter filled with concrete. If the maximum stress in the concrete is not to exceed 21 N/mm2 and the column is to carry a compressive load of 200 kN, calculate the minimum outside diameter of the tube. For concrete E = 20 kN/mm2 and for steel, E = 200 kN/mm2 . S OLUTION : Let suffixes c and s denote the concrete and steel, respectively Ac = π × 702 = 3850 mm2 4 As = π (d 2 − 702 ) mm2 4 where d in mm = outside diameter of the tube. Since the steel and concrete are of equal length of the compression of both are the same, strains are equal, then working throughout in kN and mm, we know, strain in concrete = strain in steel σs σc σ = because =E Es Ec ε Es 200 × 21 = 210 N/mm2 σs = × σc = Ec 20 Total load, P = σc Ac + σs As P = σc Ac + σs As ∴ 200000 = 21 × 3850 + 210 × @seismicisolation @seismicisolation π (d 2 − 702 ) 4 10 • Strength of Materials Hence, or d 2 = 5630 mm2 d = 75 mm E XAMPLE 2.3: A steel bar of 20 mm diameter and 400 mm long is placed concentrically inside a gunmetal tube (Fig. 2.9). The tube has inside diameter 22 mm and thickness 4 mm. The length of the tube exceeds the length of the steel bar by 0.12 mm. Rigid plates are placed on the compound assembly. Find: a) the load which will just make tube and bar of same length and b) the stresses in the steel and gunmetal when a load of 50 kN is applied. E for steel = 213 GN/m2 , E for gunmetal = 100 GN/m2 . S OLUTION : P 0.12 mm Area of gunmetal tube, Ag = π (0.032 − 0.0222 ) 4 = 0.000327 m2 Area of steel bar As = π (0.02)2 = 0.0003142 m2 4 Figure 2.9 a) For tube to compress 0.12 mm: 0.12 = 0.0003, Let σ1 be the stress in the tube 400 σ1 σ1 = 0.0003 = 0.0003, Eg 100 strain = ∴ σ1 = 0.0003 × 100 = 0.03 GN/m2 = 30000 kN/m2 Hence, load = 30000 × 0.000327 = 9.81 kN b) Load available to compress bar and tube as a compound bar is given by, let σ2 be the additional stress produced in the gunmetal tube due to this load and σs be the corresponding stress in the steel bar, then Load on compound bar = 50 − 9.81 = 40.19 kN P = σ2 Ag + σs As 40.19 = σ2 × 0.000327 + σ3 × 0.0003142 @seismicisolation @seismicisolation (i) Stress and Strains • 11 Also σ2 σs = , Eg Es σ2 = ∴ 100 σs 2100 (ii) From Eqns. (i) and (ii) σ2 = 40, 600 kN/m2 = 40.6 MN/m2 σs = 85300 kN/m2 = 85.3 MN/m2 Final stress in gunmetal = σ1 + σ2 = 40600 + 30000 = 70, 600 kN/m2 = 70.6 MN/m2 Deformation of a Body Due to Self Weight B Let us consider a bar AB which is hanging freely under its own weight (see Fig. 2.10) Let w = specific weight of the bar material Now consider a small section dx at a distance x from A. Weight of the bar for a length = w× volume = wAx (A is cross section of the bar) dx l x A Figure 2.10 Now elongation of the elementry length dx due to weight of the bar for length x, (wAx) = pl (wAx) dx wx.dx = = A.E A.E E l Total elongation = wxdx w = E E 0 ∴ elongation, dl = x.dx 0 = l 2 l w x E 2 0 wl 2 2E Because total weight of bar, W = wA.l. wAl.l Now elongation dl can be written as 2AE Wl Hence, dl = 2AE @seismicisolation @seismicisolation 12 • Strength of Materials This result also proves that the extension due to own weight is half if same weight is applied at the end (of course neglecting extension due to self weight). E XAMPLE 2.4: A steel bar ABC 18 m long is having cross-sectional area 4 mm2 weighs 22.5 N (Refer Fig. 2.11). If modulus of elasticity of wire is 210 GN/m2 , find the deflections at C and B. Deflection at C due to self weight of wire AC = dlc A 9m dlc = B 22.5 × 18000 Wl = = 0.241 mm 2AE 2 × 4 × 210000 Deflection at B: Now deflection at B is due to two reasons: i) due to self weight of AB and ii) due to weight of BC. 9m C dlB = W /2 × l/2 W /2 × l/2 + 2AE A.E dlB = W /2 × l/2 AE Figure 2.11 ∴ = 1 +1 2 22.5 × 9000 (1.5) = 0.181 mm 2 × 4 × 210000 Sometimes a machine member is a acted upon by a number of forces, some acting at outer edges while some are acting inside the body. In such cases in order to find out the total extension or contraction, the principle of superposition is applied. This has been very well made clear by the following examples: E XAMPLE 2.5: A steel bar ABC of 400 mm length and 20 mm diameter is subjected to a point load as shown in Fig. 2.12. Determine the total change in the length of bar. Take E = 200 GPa. A B 60 kN C 20 kN 200 mm 200 mm Figure 2.12 @seismicisolation @seismicisolation 40 kN Stress and Strains • 13 S OLUTION : For simplification split it into two parts as under: A C 40 kN 40 kN 400 mm A B C 20 kN 20 kN δ= 200 mm A= Pl AE π (20)2 = 314 mm2 4 δAC = 40 × 103 × 400 = 0.255 mm 314 × 200000 δAB = 20 × 103 × 200 = 0.064 mm 314 × 200000 δ = 0.255 + 0.064 = 0.319 mm Ans. Total E XAMPLE 2.6: A copper rod ABCD of 800 mm2 cross-sectional area and 7.5 m long is subjected to forces as shown in Fig. 2.13. Find the total elongation of the bar. Take E = 100 GPa 3.5 m 2.5 m 1.5 m B A C 30 kN 40 kN D 50 kN 20 kN 3.5 m 1.5 m 2.5 m S OLUTION : Splitting into three figures as shown below: A D 40 kN 40 kN 7.5 m C B 20 kN 20 kN 1.5 m B D 10 kN 10 kN 4m Figure 2.13 @seismicisolation @seismicisolation 14 • Strength of Materials S OLUTION : δ= δAD = pl ; AE 40 × 103 × 7500 800 × 100 × 10000 = 3.75 mm δBC = 20 × 103 × 1500 800 × 100 × 1000 = 0.375 mm δBD = 10 × 103 × 4000 800 × 100 × 1000 = 0.50 mm ∴ Total extention = δAD + δBC + δBD = 3.75 + 0.375 + 0.50 = 4.6 mm Ans. Extension of Tapering Rod (Circular): l p d1 d2 x dx Figure 2.14 Let the dia. of elementary strip x from dia d1 be d . ∴ let d = d1 − d1 − d2 l d1 − d2 =k l ∴ d = d1 − kx @seismicisolation @seismicisolation x p Stress and Strains Hence, cross-sectional area at distance x from larger end A = ∴ π d 2 π = (d1 − kx)2 4 4 σ = Extension of elementary length 4P Total extention of the bar = δ = πE ∴ 15 P 4P = A π (d1 − kx)2 σ 4P Strain = ε = = E π E(d1 − kx)2 Stress at this section, ∴ • l 0 dx = ε dx = 4P dx π E(d1 − kx)2 l 4P (dl − kx)−1 dx = (d1 − kx)2 π E −1 × −k o = l 4P 1 π Ek d1 − kx 0 = 4P π Ek = 4P 1 1 − π E(d1 − d2 ) d1 − d1 + d2 d1 = 4Pl π E(d1 − d2 ) = d1 − d2 4Pl · π E(d1 − d2 ) d1 d2 δ= 1 1 − d1 − kl d1 but k = d1 − d2 l 1 1 − d2 d1 4Pl π Ed1 d2 If both the diameters are equal to d. Then δ= 4Pl π Ed 2 E XAMPLE 2.7: A round steel rod of different cross-sections is loaded as shown in Fig. 2.15. Find the maximum stress induced in the rod and its deformations. Take E = 210 GPa. @seismicisolation @seismicisolation 16 • Strength of Materials A S OLUTION : Let AB be part I, BC part II and CD part III A 75 mm φ 0.9 m π (75)2 = 4415.6 mm2 4 π A2 = (45)2 = 1589.6 mm2 4 π A3 = (452 − 302 ) = 883.1 mm2 4 A1 = B B 120 kN 45 mm φ 2.2 m 60 kN C C D D To simplify the force 120 kN acting at B − B. 0.9 m 30 mm φ 20 kN Figure 2.15 80 kN A 0.9 m Tensile stress in Part I A 80 kN B 40 kN B B 80 kN C σ1 = B 2.2 m 20 kN C C C 0.9 m 40 kN D D ∴ Compressive stress in Part II PBC 40000 = 25.16 MPa σ2 = = A2 1589.6 Tensile stress in Part III PCD 20000 σ3 = = = 22.65 MPa A3 883.1 20 kN Max stress induced is in BC = 25.16 MPa. PAB 80000 = 18.1 MPa = A1 4415.6 Ans 80 × 103 × 0.9 × 1000 δ1 = P1 l1 = A1 E δ2 = 40 × 103 × 2.2 × 1000 P2 l2 = = 0.264 mm (Contr.) A2 E 1589.6 × 210000 δ3 = P3 l3 20 × 103 × 0.9 × 1000 = = 0.097 mm (Ext) A3 E 883.1 × 210000 4415.6 × 210000 = 0.0776 mm (Ext) Deformation of the rod = δ1 − δ2 + δ3 = 0.0776 − 0.264 + 0.097 = −0.0894 mm @seismicisolation @seismicisolation Contraction Ans Stress and Strains Extension of Tapered Rectangular Strip x P a P b x x dx t Figure 2.16 Consider any section x − x distant x from the bigger end Width of the section = t ∴ ∴ Area of the section = Extension of an elemental length dx = ∴ P t(a − kx) Pdx t(a − kx)E Total extension of the rod = δ = P tE l 0 dx a − kx P 1 · − loge [(a − kx)]l0 tE k P = − [loge (a − kl) − loge a] tE a P = tkE a − k =− But a−b l P.l a δ= log e Et(a − b) b k= @seismicisolation @seismicisolation • 17 18 • Strength of Materials E XAMPLE 2.8: A straight bar of steel rectangular in section is 3 m long and of thickness of 12 mm. The width of rod varies uniformly from 110 mm or one end to 35 mm at the other end. If the rod is subjected to an axial load (tensile) of 25 kN, find the extension of the rod. Take E = 200000 N/mm2 . Extension of the rod, δ = a Pl log e Et(a − b) b P = 25000 N, l = 3000 mm, t = 12 mm a = 110 mm, ab = 35 mm & E = 200000 N/mm2 ∴ δ= = 25000 × 3000 110 loge 5 35 2 × 10 × 12(110 − 35) 25000 × 3000 × 1.1452 2 × 105 × 12 × 75 = 0.477 mm Ans E XAMPLE 2.9: A rigid bar AB is attached to two vertical rods as shown in Fig. 2.17 is horizontal before the load is applied. Determine the vertical movement of P if it is of magnitude 60 kN. Steel Aluminium For aluminium L=3m A = 500 mm2 E = 75 GPa C A 3.5 m 2.5 m 60 kN Figure 2.17 @seismicisolation @seismicisolation For steel B L=4m A = 300 mm2 E = 210 GPa Stress and Strains S OLUTION : For Al, ∑ MB = 0, 6PAl = 2.5 × 60 2.5 × 60 = 25 kN = 25000 N 6 25000 × 3000 PL = = 2 mm δAl = A.E 500 × 75000 ∴ PA = For steel ∑ MA = 0 gives Pst = 3.5 × 60 3.5 × 60 = 35 kN 6 35000 × 4 × 1000 = = 2.33 mm 300 × 200000 Pst = σST A C B 2 mm C1 B1 2.33 mm Y A1 C2 B2 Figure 2.18 Now from similar triangles A1 C1 C2 and A1 , B1 , B2 B1 B2 Y = ; A1C1 A1 B1 ∴ Y 2.33 − 2 = 3.5 6 Y = 0.1925 mm Now vertical movement of P = CC2 = CC1 +Y = 2 + 0.1925 = 2.1995 mm @seismicisolation @seismicisolation Ans • 19 20 • Strength of Materials Bar of Uniform Strength As we have seen earlier that the stress due to self weight is not constant. It increases with the increase of distance from the lower end. We wish to find the shape of the bar of which the self weight is considered and is having uniform stress on all sections when subjected to an axial P. Figure 2.19 shows such a bar of uniform stress in which the area increases from the lower end to the upper end. Area A1 σ(A+dA) Area A1 A dA L dx x A σA+wAdx A2 Area A2 P (a) (b) Figure 2.19 Let L be the length of bar, having area A1 , and area A2 be cross-sectional areas of the bar at top and bottom, respectively. Let w be the specific weight of the bar material (1.2. weight per unit volume of the bar). The forces acting on the elementary stripe are: i) Weight of the strip acting downward and is equal to w× volume of strip. ii) Force on section AB due to uniform stress is equal to σ × A. This is acting downward. A is area of elementary stripe. iii) Force on section CD due to uniform (σ ) is equal to σ (A + dA). This is acting upwards. Total force acting upwards = Total force acting downwards σ (A + dA) = σ × A + wA dx σA + σ dA = σ A + wA.dx or dA w = dx A σ @seismicisolation @seismicisolation Stress and Strains • 21 Integrating, w x x σ 0 A wx ln = A2 σ wx A = A2 .e σ [ln A]AA2 = Therefore, cross-sectional area at upper side A1 = A2 e wL σ In order to calculate the extension of the bar, consider extension of the elementary stripe. Let du be the extension of small length dx. du σ = dx E σ or du = dx E Then strain = Integrating σ u= E L dx (because σ is constant in the bar) 0 ∴ u= σL E E XAMPLE 2.10: A vertical bar fixed at the upper end and of uniform strength carries an axial tensile load of 800 kN. The bar is 22 m long and having weight per unit volume as 0.000075 N/mm3 . If the area of the bar at the lower end is 450 mm2 , find the area of the bar at the upper end. S OLUTION : P = 800 kN = 800000 N L = 22 m = 22000 mm w = 0.000075 N/m3 A2 = 450 mm2 Let the area at upper end be A1 uniform stress on the bar; σ= P 800000 = 1777 N/mm2 = A2 450 @seismicisolation @seismicisolation • 22 Strength of Materials Using equation, wL σ 0.000075 × 22000 −4 1777.8 = 450e9.28×10 A1 = 450 e A1 = A2 e A1 = 450.4 mm2 Ans E XAMPLE 2.11: A steel rod of 25 mm dia passes centrally through a copper tube of 30 mm inside diameter and 40 mm outside diameter. Copper tube is 850 mm long and is closed by rigid washers of negligible thickness, which are fastened by nut threaded on the rod as shown in Fig. 2.20. The nuts are tightened till the load on the assembly is 20 kN. Calculate: i) the initial stresses on the copper tube and steel rod and ii) also calculate increase in the stresses, when one nut is tightened by onequarter of a turn relative to the other. Take pitch of the thread as 1.5 mm. E for copper = 100 GPa, E for steel = 100 GPa S OLUTION : Steel rod Washer on each side Copper tube Figure 2.20 Let σs = Stress in steel rod σc = Stress in copper rod i) π π (Ds )2 = (25)2 = 156.25π mm2 4 4 π 2 π Ac = (D − d 2 ) = (402 − 302 ) = 175π mm2 4 4 As = Tensile rod on steel = Compressive load on copper tube σs = Ac 175π × σc = × σc As 156.25π @seismicisolation @seismicisolation Stress and Strains ∴ • 23 σs = 1.12 σc P = 20000 N 20000 = σs As + σc Ac 20000 = 156.25π × 1.12 σc + σc × 175π 20000 = 350πσc σc = 20000 = 18.2 MPa 350π Ans. σs = 1.12σc , σs = 1.12 × 18.2 = 20.38 MPa Ans. σs1 = Increase in stress in steel rod ii) σc1 = Increase in the stress in the copper rod Increase in the length of steel rod, δ ls = σs1 l 1.12σc1 × 850 = 4.76 × 10−3 σc1 = Es 200, 000 Decrease in the length of copper rod δ lc = σc1 l σc1 × 850 = = 8.5 × 10−3 σc1 Ec 100000 Since the nut is tightened by 1/4 of the turn then its axial advancement 1 × pitch 4 1 = × 1.5 = 0.375 mm 4 = Since the axial advancement of the nut is equal to the decrease in the length of the tube plus increase in the length of the rod, therefore 0.374 = 4.76 × 10−3 σc1 + 8.5 × 10−3 σc1 = 13.26 × 10−3 σc1 0.374 × 103 = 28.2 MPa (Compressive) Ans. 13.26 σs1 = 1.12 × σc1 = 1.12 × 28.2 = 31.584 MPa (Tensile) σc1 = Ans. E XAMPLE 2.12: Three long parallel wires equal in length are supporting a rigid bar connected at their bottoms as shown in Fig. 2.21. If the cross-sectional area of each wire is 100 mm2 , calculate the stresses in each wire. Take EB = 100 GPa, ES = 200 GPa. @seismicisolation @seismicisolation 24 • Strength of Materials Steel Brass Brass 10 kN Figure 2.21 S OLUTION : σs × 100 + 100σb + 100σb (i) 100σs + 200σb = 10000 σs + 2σb = 100 N/mm2 (ii) σs σb = 3 200 × 10 100 × 103 ∴ σs = 2σb (iii) substituting for σs in (ii) 2σb + 2σb = 100; σb = 100 = 25 MPa Ans. 4 σs = 2 × 25 = 50 MPa Ans. E XAMPLE 2.13: Two steel rods and one copper rod each of 20 mm diameter together support a load of 50 kN as shown in Fig. 2.22. Find the stress in each rod. Take Es = 200 GPa, Eb = 100 GPa 50 kN Copper 2m Brass Brass Figure 2.22 @seismicisolation @seismicisolation 1.5 m Stress and Strains • 25 π (20)2 = 314 mm2 4 Total area of steel A s + 314 × 2 = 628 mm2 Ac = As = σs As + σc Ac = 50000 628σs + 314σc = 50000 2σs + σc = 159.24 (i) σs ls σc lc = ; Es Ec σs × 2000 σc × 1500 = 200000 100000 σs = 1.5σc (ii) Substituting for σs from Eqn. (ii) in Eqn. (i) 2 × 1.5σc + σc = 159.24 ∴ σc = 39.81 MPa Ans. σs = 1.5 × 39.81 = 59.7 MPa Ans. E XAMPLE 2.14: A uniform bar ABCD has built-in ends A&D. It is subjected to two point loads P1 and P2 equal to 80 kN and 40 kN at B and C as shown in Fig. 2.23. Find values of reactions at A and D. A B C P1 500 mm D P2 1000 mm 500 mm S OLUTION : A B RA RA 500 mm RB RB 1000 mm RC RC 500 mm Figure 2.23 @seismicisolation @seismicisolation 26 • Strength of Materials RA + RB = P1 = 80 kN (i) RC − RB = P2 = 40 kN (ii) Adding Eqns. (i) and (ii) RA + RC = 120 kN RA × 500 RB × 1000 RC × 500 − − A×E A×E A×E δ = RA − 2RB − RC = 0 0=δ = substituting, (120 − RC ) − 2(RC − 40) − RC = 0 200 = 50 kN Ans. 4 RA + RC = 120; solving ∴ RC = RA = 120 − 50 = 70 kN Ans. Note: RA and RC are reactions at A and D respectively. Exercise 2.1 A bar of mild steel has an overall length of 2100 mm. The diameter of 700 mm length is 56 mm. The diameter of the remaining 1400 mm is 35 mm. Calculate the extension of the bar due to a tensile load of 55 kN. E = 200 GN/m2 . [Ans Extension = 0.478 mm] 2.2 The block of weight W hangs from the point A, the bars AB and AC are pinned to the support at B and C. The cross-sectional area for AB is 800 mm2 and for AC is 400 mm2 . Neglecting the weight of the bars, determine the maximum safe value of W , if the stress in AB is limited to 110 MPa and in AC to 120 MPa (Figure 2.24). B C 40° A 60° W [Ans Figure 2.24 81.7 kN] 2.3 A rectangular bar of 2 m length and 12.5 mm thickness uniformly tapers from 100 mm at one end to 20 mm at the other. If the bar is subjected to a tensile force of 25 kN, find its deformation. Take E as 200 GPa. [Ans 0.4 mm] @seismicisolation @seismicisolation Stress and Strains • 27 2.4 A brass bar having cross-sectional area of 900 mm2 is subjected to axial forces as shown in Fig. 2.25 in which AB = 0.6 m, BC = 0.8 m and CD = 1.0 m. A 40 kN B C D 50 kN 20 kN 10 kN Figure 2.25 Find the total elongation of the bar. Take E = 100 GPa. [Ans. 0.11mm] 2.5 A 1200 mm long composite rod consists of a steel tube of 50 mm external diameter and 40 mm internal diameter. A copper rod of 30 mm dia is placed coaxially into the steel tube. The assembly is held between to rigid plates and is subjected to an axial compressive force of 200 kN. Find the stress induced in each material and the contraction produced. Take Es = 200 GPa, Ec = 100 GPa. [Ans σs = 188.63 MPa, σc = 94.31 MPa Contraction = 1.132 mm] 2.6 A 10 m long and 10 mm thick flat steel bar tapers from 60 mm at one end to 20 mm at the other. Determine the change in length of the bar when a tensile force P = 12 kN is acting along its axis. E = 200 GN/m2 [Ans 1.648 mm] 2.7 Two elastic rods, A and B, of equal length hang vertically 0.6 m apart and support a rigid bar horizontally. The bar remains horizontal when a vertical load of 60 kN is applied to the bar 0.2 m from A. If the stress in A is 100 MN/m2 , find the stress in B and the cross-sectional areas of the two rods. EA = 200 GN/m2 , EB = 130 GN/m2 . [Ans 65 MN/m2 , 400 mm2 , 307.5 mm2 ] 2.8 If the modulus of elasticity of steel is twice that of the modulus of elasticity of brass, calculate the outside diameter of a brass tube that should sheath a steel bar of diameter 50 mm so that the steel and brass equally share an axial compressive load. [Ans 86.6 mm] 2.9 A tubular steel tie-rod has an outside diameter of 50 mm, an inside diameter of 40 mm, and is subjected to a tensile load of 88000 N. Taking E = 200 GN/m2 , calculate the extension over a length of 2.7 m. If corrosion now reduces the effectual outside diameter of the tube to 49 mm and increases the effectual inside diameter to 41 mm, calculate the increase in extension, over the length of 2.7 m, due to corrosion. [Ans 1.68 mm, 0.42 mm] @seismicisolation @seismicisolation 28 • Strength of Materials 2.10 Figure 2.26 shows a rigid bar ABC hinged at A and suspended at two points B and C by two bars BD and CE, made of aluminium and steel, respectively. The bar carries a load of 20 kN midway between B and C. The cross-sectional area of aluminium bar BD is 3 mm2 and that of steel bar CE is 2 mm2 . Determine the loads taken by the two bars BD and CE. D E B C 1000 mm 500 mm 1000 mm 1000 mm 20 kN [Ans Figure 2.26 @seismicisolation @seismicisolation Pa = 3.481 kN, Ps = 13.26 kN] C HAPTER 3 TEMPERATURE STRESS AND STRAIN When the temperature of a body is raised (specially, metallic bodies), the dimensions of the body tend to increase, if cooled the dimensions of the bodies tend to decrease. If we prevent this increase or decrease in dimensions due to rise or fall in temperature, the thermal stresses are bound to occur. Let a steel rod having linear coefficient of expansion (α ), shown in Fig. 3.1 be clamped or fixed at both ends. Now if its temperature is raised by t ◦ c L Figure 3.1 Then prevented expansion = α l t Now since this expansion has been prevented so thermal stress of compressive nature will be induced. α lt strain = = αt l stress E= , so stress will be E α t. strain Remember that thermal strain is = Prevented expansion or contraction Original length E XAMPLE 3.1: A steel spacer of length 400 mm at 20◦ C has its temperature increased to 80◦ C. If the spacer has a diameter of 56 mm, calculate the compressive load in the spacer in newtons, if it is assumed that: a) the expansion is completely prevented b) the spacer has a length of 400.1 mm at 80◦ C, Take E = 200 GN/m2 , α = 11 × 10−6 /C◦ , a) For completely restricted expansion: E = 200 GN/m2 = 200000 N/mm2 , t = 80 − 20 = 60◦ C @seismicisolation @seismicisolation 30 • Strength of Materials Temperature stress = E α t = 200000 × 11 × 10−6 × 60 = 132 N/mm2 Load = stress × area π = 132 × (56)2 = 32495.2 N 4 b) For partially restricted expansion: Ans. Natural expansion = l α t = 400 × 11 × 10−6 × 60 = 0.264 mm Permitted expansion = 400.1 − 400 = 0.1 mm Compression effect = 0.264 − 0.1 = 0.164 mm x 0.164 = 4.1 × 10−4 Compressive strain = = l 400 = 0.00041 mm stress E= strain ∴ stress = E × strain = 200000 × 0.00041 = 82 N/mm2 Load = stress × area π = 82 × (56)2 = 201864 N Ans. 4 It is obvious from the above answers that the stress is more when expansion is completly prevented. Composite Tube or Bar Imagine we have a copper tube fitted with a steel rod inside. And this assembly is brazed at both the ends. Now if its temperature is raised by t ◦ C we will analyse what happens: xc αslt A xs A αclt l Copper Figure 3.2 @seismicisolation @seismicisolation Steel Temperature Stress and Strain • 31 Let αs and αc be the coefficient of linear expansion for steel and copper, respectively. When we heat, steel will expand (if allowed) less than copper as shown in Fig. 3.2. Since copper expands more than steel and both materials are bound together so copper will pull steel upward and steel will push the copper downward. Let line A − A be where both settle eventually xc + xs = αc lt − αs lt xc xs + = αc t − αs t = (αc − αs ) t l l εc + εs = (αc − αt ) (i) Also push on copper = pull on steel σc Ac = σs As σc = Thermal stress in copper σs = Thermal stress in steel As = Cross-sectional area of steel Ac = Cross-sectional area of copper (ii) Solving equations (i) and (ii) σs and σc can be found out. E XAMPLE 3.2: A copper flat measuring 60 mm × 40 mm is brazed to another steel flat 60 mm × 50 mm as shown in Fig. 3.3. If this composite flat is heated through 140◦ C, determine: i) the stress produced in each of the bar ii) shear force which tends to rupture the brazing iii) shear force. Take αc = 18.5 × 10−6 /◦ C αs = 12 × 106 /◦ C Ec = 110 GN/m2 Es = 220 GN/m2 length of each flat = 450 mm 40 mm Copper 50 mm Steel 450 mm Figure 3.3 S OLUTION : i) Stress produced in each of the bar. @seismicisolation @seismicisolation • 32 Strength of Materials We shall solve this problem by the method derived from the last article. εs + εc = (αc − αs ) t αs αc + = (13.5 − 12) + 10−6 × 140 220000 110000 αs αc + = 910 × 10−6 220000 110000 Pull on steel = Push on copper (i) αs × As = αc Ac αs × 60 × 50 = αc 60 × 40 or αs = 0.8αc (ii) Substituting for σs in Eqn. (i) αc 0.8αc + = 910 × 10−6 220000 110000 Multiplying the whole equation by 220000, we get ∴ 2.8σc = 200.2 0.8σc + 2σc = 200.2; or σc = 71.5 MPa Ans. σc = 0.8σc = 0.8 × 71.5 = 57.2 MPa Ans. ii) Shear force: Shear force = σs As = σc Ac = 71.5 × 60 × 40 = 171600 N = 171.6 kN Ans. iii) 171600 Shear force = Shear area 450 × 60 = 6.36 MN/m2 Shear stress = Thermal Stresses in a Bar of Tapering Section Consider a tapered round bar whose temperature is raised by Δt. A B d2 d1 A L Figure 3.4 @seismicisolation @seismicisolation Temperature Stress and Strain • 33 As the temperature increases the bar will expand. If the tapered bar is fixed at both ends, the stress will develop in tapered bar. Since we know δ L = L α Δ t (i) and also δL = 4 PL π Ed1 d2 (ii) where P is the load required to bring the deformed bar to the original. From Eqns. (i) and (ii) π E d1 d2 α Δt 4 PL ∴ P= π Ed1 d2 4 P π Ed1 d2 α Δt = Maximum stress, σ t = π π d22 4 × d22 4 4 E. α . Δt.d1 d1 (σt )max = = E α Δt d2 d2 L α Δt = ∴ ∴ When bar is of uniform cross section d1 = d2 σt = E α Δt E XAMPLE 3.3: A steel rod of 320 mm2 cross-sectional area and a coaxial copper tube of 800 mm2 cross-sectional area are rigidly bounded together at their ends. An axial compressive load of 40 kN is applied to the composite bar, and the temperature is then raised by 100◦ C. Determine the stresses in the copper and steel after heating as stated above. The modulii of elasticity for steel and copper are 200 GN/m2 and 100 GN/m2 and the coefficients of linear expansion are 12 × 10−6 /◦ C and 16 × 10−6 /◦ C, respectively. S OLUTION : as = 320 mm2 40 kN ac = 800 mm2 t = 100◦ C Steel Es = 200 GN/m2 Copper Ec = 100 GN/m2 αs = 12 × 10−6 /◦ C Figure 3.5 αc = 16 × 10−6 /◦ C @seismicisolation @seismicisolation 34 • Strength of Materials S OLUTION : Due to 40 kN load, let the stresses developed be σs and σc . σs as + σc ac = 40000 320 σs + 800 σc = 40000 σs σc = Es Ec ∴ (i) σs σ = c 200 100 σs = 2σc (ii) Substituting Eqn. (i) from Eqn. (ii) 2 × 320σc + 800σc = 40000 ∴ σc = 27.78 MPa(Compressive) Hence, σs = 27.78 × 2 = 55.56 N/mm2 (Compressive) Now due to temperature: εs + εc = (αc − αs )t Let σs & σc be the stresses due to thermal effect. σs σc + = (16 − 12)100 × 10−6 Es Ec σs σc + = 400 × 10−6 200000 100000 σs + σc = 400 × 10−6 2 Also 320σs = 800 σc ∴ σs + 2σc = 80 (iii) ∴ σs = 2.5σc (iv) Substituting Eqn. (iv) in Eqn. (iii), 2.5σc + 2σc = 80 ∴ σc = 17.78 N/mm2 σs = 2.5σc = 2.5 × 17.78 = 44.45 N/m2 σc = 44.45 = 17.78 N/mm2 2.5 (Compressive) (Tensile) (Compressive) Net stresses: σs = −55.56 + 44.45 = 11.11 N/mm2 (Compressive) Ans. σc = −17.78 − 27.78 = 45.56 N/mm2 (Compressive) Ans. @seismicisolation @seismicisolation Temperature Stress and Strain • 35 E XAMPLE 3.4: A composite bar shown in Fig 3.6 is rigidly attached to the end supports. The temperature of the composite system is raised by 65◦ C. Find out the stresses in three portions of the bar: a) if the supports are rigid and b) the supports yield by 0.5 mm. Es = 200 GPa; Ea = 90 GPa; αs = 12 × 10−6 /◦ C; 40 mm dia Ec = 100 GPa αa = 20 × 10−6 /◦ C; 90 mm dia Copper Aluminium 300 mm 450 mm αc = 16 × 10−6 /◦ C 40 mm dia Steel 300 mm Figure 3.6 S OLUTION : Free elongation of copper section = αctlc = 16 × 10−6 × 65 × 300 = 0.312 mm Free elongation of aluminium = 20 × 10−6 × 65 × 450 = 0.585 mm Free elongation of steel = 12 × 10−6 × 65 × 300 = 0.234 mm Total free extension of composite bar = 0.312 + 0.585 + 0.234 = 1.131 mm (i) Because the extension is prevented by the rigid supports, therefore compressive stresses will set up in the bar. Let F be the compressive force in the bar in N. ∴ Stress are F 4F = = 7.962 × 10−4 F N/mm2 Ac π (40)2 P 4F σa = = = 1.573 × 10−4 F N/mm2 Aa π (90)2 P 4F σs = = = 7.962 × 10−4 F N/mm2 As π (40)2 σc = @seismicisolation @seismicisolation 36 • Strength of Materials Strains are: σc 7.962 × 10−4 F = = 7.962 × 10−9 F Ec 100000 σa 1.573 × 10−4 F = 1.75 × 10−9 F εa = = Ea 90000 σs 7.962 × 10−4 F εs = = = 3.981 × 10−9 F Es 200000 εc = Extensions δ lc = εc × lc = 7.962 × 10−9 F × 300 = 2.389 × 10−6 F δ la = 1.75 × 10−9 F × 450 = 0.7875 × 10−6 F δ ls = 3.981 × 10−9 F × 300 = 1.194 × 10−6 F Total extension = (2.389 × 10−6 + 0.7875 × 10−6 + 1.194 × 10−6 )F = 4.3705 × 10−6 F Also, δ lc + δ la + δ ls = 1.131 mm from Eqn. (i) ∴ 4.3705 × 10−6 F = 1.131 ∴ F = 258780 N = 258 kN 4 × 258000 = 205.4 N/mm2 = 205.4 MPa π × 1600 4 × 258000 = 40.6 N/mm2 = 40.6 MPa σa = π × 8100 4 × 258000 σs = = 205.4 MPa π × 1600 σc = b) When the supports yield by 0.5 mm, then δ la + δ lc + δ ls = 1.131 − 0.5 = 0.631 mm 43.05 × 10−6 F = 0.631 F = 144.4 kN σs = σc = 7.962 × 10−4 × 144400 = 115 N/mm2 = 115 MPa σa = 1.573 × 10−4 × 144400 = 22.7 N/mm2 = 22.7 MPa E XAMPLE 3.5: A compound bar is made up by connecting a steel member and a copper member rigidly fixed at their ends as shown in Fig. 3.7. @seismicisolation @seismicisolation Temperature Stress and Strain 1 mm dia copper bar 37 Es = 200 GN/m2 , 15 dia αs = 12 × 10−6 /◦ C 20 dia Steel 200 mm • Ec = 100 GN/m2 , 200 mm αc = 16 × 10−6 /◦ C Figure 3.7 If the composite bar’s temperature is raised by 70◦ C, determine the stresses in copper end two areas of steel. S OLUTION : Let σs be the stress in 15 mm dia of steel & σs is the stress in 20 mm dia of steel σc is the stress in copper bar of dia 15 mm. Total tension in the steel bar is the same as the compression in the copper bar. π π π σc × (15)2 = σs × (15)2 = σs × (20)2 4 4 4 225σc = 225σs = 400σs ∴ σc = σs = 1.78σs If l in the actual length of the compound bar due to rise in temperature, then σc ×l Ec σs σ × 200 + s × 200 For steel rod l − l(1 + αst) = Es Es l(1 + αct) − l = For copper rod Adding the two eqs., σc σs σ × l + × 200 + s × 200 Ec Es Es 1.786s 1.786s × 400 + × 200 l(αct − αst) = 100000 200000 σs + × 200 200000 400 × 10−6 (16 × 70 − 12 × 70) = 7.12 × 10−3 σs + 1.78 × 10−3 σs l(1 + αct) − l(1 + αst) = + σs × 1 × 10−3 0.112 = 9.9σs × 10−3 ∴ ∴ σs = 11.31 N/mm2 = 11.31 MPa σc = σs = 1.78 × σs σc = σs = 1.78 × 11.31 = 20.1318 MPa (Compressive in copper & tensile in steel) @seismicisolation @seismicisolation (i) (ii) 38 • Strength of Materials Exercise 3.1 A steel bar of 100 mm diameter is rigidly clamped at both ends so that all axial extension is prevented. A hole of 40 mm diameter is drilled out at one third of the length. If the bar is raised in temperature by 30◦ C above that of the clamps, calculate the maximum axial stress in the bar. E = 210 GN/m2 , α = 0.000012/◦ C [Ans 84.7 MN/m2 ] 3.2 Two steel bars are connected together so as to form a rod of total length 625 mm. One of mild steel is 225 mm long and 25 mm diameter. If the bar is heated to 50◦ C above 50◦ C room temperature calculate the stress in each part of the rod. For stainless steel, E = 175 GN/m2 , α = 18 × 10−6 /◦ C. For mild steel E = 200 GN/m2 , α = 12 × 10−6 /◦ C [Ans Mild steel, 44.7 MN/m2 , stainless steel 195 MN/m2 ] 3.3 A steel bar of 50 mm diameter is placed between two stops, with an end clearance of 0.05 mm. The temperature of the bar is raised by 60◦ C and the stops are found to have been forced apart a distance of 0.05 mm. Calculate the maximum stress in the bar if its total length is 250 mm and there is a hole of 25 mm diameter drilled along the length for a distance of 100 mm. E = 200 kN/m2 , α = 12 × 10−6 /◦ C. [Ans 75.3 MN/m2 ] 3.4 A compound tube is formed by a stainless steel outer tube of 50 mm outside diameter and 47 mm inside diameter, together with a concentric mild steel inner tube of wall thickness 6 mm. The radial clearance between inner and outer tubes is 2 mm. The two tubes are welded together at their ends, the compound tube being free to expand when heated. Calculate the stress in each tube due to a rise of 50◦ C. For stainless, E = 175 kN/m2 , α = 18 × 10−6 /◦ C. For mild steel, E = 200 kN/m2 , α = 12 × 10−6 /◦ C. [Ans Mild steel, 13.4 MN/m2 tensile; stainless steel, 41 MN/m2 comp] 3.5 A stainless steel bar of 25 mm diameter is placed inside and concentric with a mild steel tube of 30 mm inside diameter and 50 mm outside diameter. The tube and bar are welded together at the ends but are otherwise free to expand. Calculate the stress in each part of the compound bar so formed due to a temperature rise of 25◦ C. If the length of the compound bar is 250 mm what is the extension? For stainless steel E = 170 kN/m2 , α = 18 × 10−6 /◦ C. For mild steel E = 196 kN/m2 , α = 12 × 10−6 /◦ C. [Ans Steel bar, −19.05 MN/m2 , tube 7.42 MN/m2 ; 0.0845 mm] 3.6 A compound bar is made up of steel plate 50 mm wide, 10 mm thick clad on both sides by copper plates each 50 mm wide, 5 mm thick. The plates are bonded together along their length and at room temperature the original length is 1.2 m. Find the load taken by each plate and the increase in length when the temperature rises to 100◦ C. For steel, E = 200 kN/m2 , α = 12 × 10−6 /◦ C. For copper E = 110 kN/m2 , α = 18 × 10−8 /◦ C. [Ans Steel +21.28 kN, copper −10.64 kN; 1.41 mm] @seismicisolation @seismicisolation Temperature Stress and Strain • 39 3.7 In the arrangement shown in Fig 3.8, the steel bar is 30 mm in diameter and the bronze bar 50 mm in diameter. The bars are of equal length and just fit between the end fixings at room temperature. The distance between the fixings cannot change. Calculate the stresses produced in the steel and bronze by a temperature rise of 100◦ C. For steel, E = 200 GN/m2 , α = 12 × 10−6 /◦ C. For bronze E = 110 kN/m2 , α = 18 × 10−6 /◦ C. Bronze Steel Figure 3.8 [Ans −363 MN2 /m2 , bronze, −130.5 MN/m2 , steel] @seismicisolation @seismicisolation C HAPTER 4 ELASTIC CONSTANTS Whenever a tensile load is applied axially to a bar, there will be a longitudinal or linear strain. But, naturally there will be compressive strain on lateral side (sometimes called secondary strain). The ratio of lateral strain to longitudinal strain is known as Poisson’s ratio. or, Poisson’s ratio = Lateral strain = Constant Linear strain This is denoted by Greek letter μ (mu). Material Poisson’s ratio (μ ) Steel Cast iron Copper Brass Aluminium Concrete Rubber Stainless steel Wrought iron Bronze 0.25 to 0.33 0.23 to 0.27 0.31 to 0.34 0.32 to 0.42 0.32 to 0.38 0.08 to 0.18 0.45 to 0.50 0.305 0.278 0.350 E XAMPLE 4.1: A metal bar 60 mm × 60 mm in section is subjected to an axial compressive load of 600 kN. If the contraction of a 200 mm gauge length was found to be 0.65 mm and the increase in thickness as 0.05 mm, find the values of Young’s modulus and Poisson’s ratio for the bar material. P.L 600000 × 200 = = 0.65 A.E 60 × 60 × E 600000 × 200 = 51282 N/mm2 ∴ E= 0.65 × 60 × 60 = 51.3 GN/m2 approx. Ans. 0.65 Linear strain = = 0.00325 200 0.05 = 0.000833 Lateral strain = 60 δl = @seismicisolation @seismicisolation Elastic Constants • 41 lateral 0.000833 = = 0.256 linear 0.00325 ∴ μ = 0.256 Ans. Change in volume Volumetric strain = Original volume δv εv = v μ = Poisson’s ratio = σy σz σx σy σx σz Figure 4.1 Referring to Fig. 4.1. new length of side x = x(1 + εx ) new length of side y = y(1 + εy ) and new length of side z = z(1 + εz ) ∴ new volume = xyz (1 + εx )(1 + εy )(1 + εz ) = xyz (1 + εx + εy + εz ) (Neglecting products of strains) xyz (1 + εx + εy + εz ) − xyz ∴ εv = xyz i.e., volumetric strain = sum of perpendicular strains ∴ εv = εx + εy + εz Alternately, v = xyz ∴ log v = log x + log y + log z Differentiating w.r.t. x ∴ 1 dv 1 1 dy 1 dz · = + · + · v dx x y dx z dx Multiplying Eqn. (i) by dx ∴ dv dx dy dz = + + v x y z @seismicisolation @seismicisolation (i) 42 • Strength of Materials εv = εx + εy + εz or E XAMPLE 4.2: A rectangular bar 600 m long and 120 mm × 60 mm in cross section is subjected to forces as shown in Fig. 4.2. What is the change in the volume of bar? Take E = 200 GPa, μ = 0.3. 320 kN 180 N 150 kN 150 kN 60 mm 180 kN 320 kN 120 mm 600 mm Figure 4.2 Original volume = l × b × t = 600 × 120 × 60 = 4320000 mm3 Stress in x-x direction σx = Px 150000 = 20.83 N/mm2 (Tension) = Ax 120 × 60 σy = 180000 = 5 N/mm2 (Tension) 600 × 60 σz = 320000 = 4.44 N/mm2 (Compressive) 600 × 120 Now resultant strain in each direction σx μσy μσz 20.83 0.25 × 5 0.25 × 4.44 20.69 − + = − + = E E E 200000 200000 200000 200000 σy μσx μσz 5 0.25 × 20.83 0.25 × 4.44 0.9025 εy = + − + = − + = E E E 200000 200000 200000 200000 4.44 0.25 × 20.83 0.25 × 5 10.9 σz μσx μσy − =− − − = εz = + − E E E 200000 200000 200000 200000 20.69 0.9025 10.9 δv = + − Volumetric strain = v 200000 200000 200000 10.69 = 5.345 × 10−5 = 200000 εx = + ∴ δv = v × 5.345 × 10−5 = 4320000 × 0.5345 × 10−5 = 2309.04 mm3 . Ans. @seismicisolation @seismicisolation Elastic Constants • 43 Bulk Modulus: When a body is subjected to three mutually perpendicular stresses of equal intensity, the ratio of direct stress to the corresponding volumetric strain is known as bulk modulus. It is denoted by K. Direct stress σ K= = δv Volumetric strain v Direct stress occurs, for example, as hydrostatic pressure. Principle of shear stress If we apply shear stress in clockwise direction, then automatically shear stress in anticlockwise will set as complementary. τ′ τ τ τ′ Figure 4.3 Relation between Modulus of Elasticity and Modulus of Rigidity Modulus of rigidity = Shear stress Shear strain τ' Consider a cube τ AD2 + AD2 = BD2 ∴ 2AD2 = BD2 √ BD = AD 2 τ D D1 C1 C φ τ' Before distortion P D2 φ A After distortion B Figure 4.4 Strain of BD = BD1 − BD D1 D2 DD1 cos 45 φ √ = = = BD BD 2 AD 2 Linear strain of the diagonal BD, φ τ = 2 2C τ = shear stress C = modulus of rigidity = (i) We know that the effect of this stress will cause tensile stress on the diagonal BD and compressive stress on the diagonal AC. Therefore, tensile strain on the diagonal BD due to tensile stress on the diagonal BD is given by τ BD = (ii) E @seismicisolation @seismicisolation 44 • Strength of Materials And the tensile strain on the diagonal BD due to compressive stress on the diagonal AC = μ × Z E (iii) The combined effect of the above two stresses on the diagonal BD = τ τ τ + μ = (1 + μ ) E E E (iv) Equating Eqns. (i) & (iv), τ τ = (1 + μ ) or 2C E E 2C = 1 1+μ or 1 1 = (1 + μ ) 2C E C= E 2(1 + μ ) C = Shear modulus or modulus of rigidity Relation between Modulus of Elasticity and Bulk Modulus Let us take a cube of each side l and subject it to direct stress (hydrostatic pressure) on the faces of the cube. Initial volume of cube v = l 3 By differentiation, ∴ ∴ dv = 3l 2 dL dv Volumetric strain, εv = v 3l 2 dl l2 dl = 3εl εv = 3 l εv = ∴ Now linear strain dl of any side of cube is, l dl σ σ σ = −μ −μ l E E E σ εL = (1 − 2μ ) E 3σ εv = (1 − 2μ ) E σ K= εv εl = ∴ ∴ @seismicisolation @seismicisolation (i) (ii) Elastic Constants ∴ εv = • 45 σ K Substituting in Eqn. (i) ∴ σ 3σ = (1 − 2μ ) K E ∴ E = 3K(1 − 2μ ) Relation between E, C & K Since C = E ; E = 2C(1 + μ ) 2(1 + μ ) ∴ E −1 = μ 2C (i) Also E = 3K(1 − 2μ ) E − 1 = −2μ 3K or μ= 1 E − 2 6K (ii) Equating Eqns. (i) & (ii) 1 E E −1 = − 2C 2 6K 1 1 3 1 + = +1 = E 2C 6K 2 2 1 1 E + =3 C 3K E= 9CK C + 3K Exercise 4.1 A bar of 50 mm diameter and 300 mm length is made of a material having E = 200 GPa and μ = 0.3, calculate its modulus of rigidity. Also determine the change in the volume of the bar when subjected to a hydrostatic stress of 100 MPa. [Ans C = 76.92 GPa, 353 mm3 ] 4.2 A square piece of steel 150 mm long by 25 mm square is subjected to a compressive load of 100 kN. Find the change in length of the piece if all lateral strain is presented by the application of uniform lateral external pressure of suitable intensity. E = 200 GPa, μ = 0.25. [Ans 0.1 mm] 4.3 The gauge length marked on a steel rod of diameter of 10 mm is 60 mm. When this rod is subjected to tension test, the gauge length increases to 80 mm. The rod yields at 40 kN and maximum load applied is 80 kN after which the rod breaks at 45 kN. Determine: a) percentage elongation, b) yield strength, c) ultimate strength and d) breaking strength. [Ans 33.3%, 509.3 MPa, 1018.6 MPa, 572.9 MPa] @seismicisolation @seismicisolation 46 • Strength of Materials 4.4 A bar of cross-section 10 × 10 mm is subjected to an axial pull of 8000 N. The lateral dimension of the bar is found to be changed to 9.9985 mm × 9.9985 mm. If the modulus of rigidity of the material is 0.8 × 105 N/mm2 , determine the Poisson’s ratio and modulus of elasticity. [Ans 0.43, 229 GPa] 4.5 A 50 mm diameter steel bar is subjected to a tensile load of 100 kN. The extension over its 300 mm length was found to be 0.08 mm and change of its diameter was 0.0035 mm. Determine the modulus of rigidity of bar. [Ans 75.6 GPa] 4.6 A rectangular block is subjected to stresses 25 MPa (tensile), 20 MPa (comp) and 30 MPa (tensile) in the directions x, y and z, respectively. Determine the strains in three directions and value of the bulk and the rigidity modulii. E = 200 GPa, μ = 0.28. [Ans 0.083, 0.0001, 0.00015 K = 151.5 GPa,C = 78.125 GPa] @seismicisolation @seismicisolation 5 C HAPTER PRINCIPAL STRESSES AND STRAINS While designing a component of a machine, the essential feature in mechanical engineering and design is based on the knowledge of principal stresses at the critical section of a component. While designing a crankshaft in an engine which converts the translatory motion of the piston into rotary m, there are complex stresses induced, the study of the same is very important from design purpose of view. Principal Planes: In any strained material, there are three planes mutually perpendicular to each other which carry direct stresses only and no shear stress. Out of these three one plane carries the maximum stress, the other minimum stress and the third plane carries intermediate stress (which is not much required in design calculations). Principal Stress: The magnitude of direct stress, across a principal plane, is known as principal stress. The determination of principal planes and principal stress is an important consideration in the design of a machine component or structures. For determination of stresses of an oblique section of a strained body two methods are normally used i.e., (i) analytical methods (ii) graphical method employing Mohr’s circle. While deriving equations we shall take tensile stresses and strains as positive and compressive stresses and strain an negative. The shear stress which tends to rotate the element in clockwise direction is taken as positive and the one tensing to rotate in anticlockwise direction as negative. There are two methods used (i) Analytical Method (ii) Graphical Method Stresses on a Oblique Section Figure 5.1(a) shows a piece of material subjected to a tensile force P. If the cross-sectional area is P a, the tensile stress on the cross-section, σ = . a C C θ P A θ P A B (a) θ T B (b) Figure 5.1 @seismicisolation @seismicisolation N P 48 • Strength of Materials The part ACB is in equilibrium under the forces acting upon it, so that the resultant force on CB is equal and opposite to the applied force P. This can be resolved into normal and tangential components, N and T , producing direct and shear stresses σθ and τθ respectively. The area of the oblique section CB is a sec θ , so that N P cos θ = = σ cos2 θ a sec θ a sec θ T P sin θ σ τθ = = = σ sin θ cos θ = sin 2θ a sec θ a sec θ 2 σθ = (5.1) (5.2) σθ C α σr zθ B Figure 5.2 The resultant stress σθ is given by σθ = σθ2 + τθ2 since each of the stresses acts on the same τθ area, and α = tan−1 . Where in the simple case considered above α = θ . σθ has a maximum σθ value σ when θ = 0 and τθ has a maximum value σ /2 when θ = 45◦ . Thus, any material whose ultimate shear stress is less than half the ultimate stress in tension or compression, will fail due to shear when subjected to a tensile or compressive load. It is usual to work in terms of the applied and induced stresses rather than forces and to assume the material to be of unit thickness. E XAMPLE 5.1: A steel bar is subjected to a tensile stress of 45 MPa. What will be the values of normal and shear stresses across a section, which makes an angle of 60◦ with the tensile stress. S OLUTION : C 30º P P 60º A B Figure 5.3 @seismicisolation @seismicisolation Principal Stresses and Strains • 49 σθ = σ cos2 θ = 45 cos2 30 = 40 × 0.7499 = 33.75 MPa Ans. σ 45 sin 2 × 30 = 22.5 × 0.866 τθ = sin 2θ = 2 2 = 19.485 MPa Ans. E XAMPLE 5.2: Two wooden rods of cross section 150 mm × 110 mm are joined together along a line AB as shown in Fig. 5.4. Find the maximum force which can be applied if the shear stress along AB is 2 MPa. C P P 70° 150 mm B 110 mm Figure 5.4 S OLUTION : θ = 90◦ − 70◦ = 30◦ Cross-sectional area = 150 × 110 = 16500 mm2 Let σ be the safe stress in the rods. σ sin 2θ 2 σ 2.5 = sin 2 × 30◦ = σ × 0.433 2 2.5 = 5.77 MPa Ans. Safe stress σ = 0.433 τθ = ∴ E XAMPLE 5.3: A tension member is formed by joining two wooden rectangular rods 250 mm × 120 mm in cross section as shown in Fig. 5.5. Determine the safe value of force F which can be applied if allowable normal and shear stresses in the joint are 0.7 MPa and 1.4 MPa respectively. C P 25° 65° D P 250 mm 120 mm Figure 5.5 @seismicisolation @seismicisolation 50 • Strength of Materials S OLUTION : 90◦ − 65◦ = 25◦ Let σ be the safe stress in N/mm2 . Area of cross section = 250 × 120 = 30, 000 mm2 Normal stress, σθ = σ cos2 θ σ= Shear stress τθ = ∴ 0.7 = σ cos2 25 = σ × 0.821 0.7 = 0.853 MPa 0.821 σ σ sin 2θ , 1.4 = sin 2 × 25 = 0.383σ 2 2 τθ = 1.4 = 3.655 MPa 0.383 We can see that 0.853 MPa is least of two. Hence, safe stress = 0.853 MPa ∴ Safe load = 0.853 × 30000 = 25590 MPa = 25.59 kN Ans. Material Subjected to Two Perpendicular Stresses Figure 5.6 shows an element of unit thickness which is subjected to perpendicular tensile stresses σx and σy . The wedge ABC is in equilibrium under the forces acting upon it, so that, resolving forces are normal to AC, σθ × AC = σx × AB cos θ + σy × BC sin θ ∴ σθ = σx cos2 θ + σy sin2 θ σx + σy σx − σy = + cos 2θ 2 2 (5.3) The maximum value is σx or σy , whichever is greater and the minimum value is σx or σy whichever is smaller. @seismicisolation @seismicisolation Principal Stresses and Strains • 51 σy A 6θ θ σx σx zθ B C σy Figure 5.6 Resolving forces parallel to AC, τθ × AC = σx × AB sin θ − σy × BC cos θ ∴ τθ = (σx − σy ) sin θ cos θ σx − σy = sin 2θ 2 (5.4) The maximum value is σx − σy when θ = 45◦ . If σx = σy , τθ = 0 for all values of θ . E XAMPLE 5.4: In a strained structure, at a point an element is subjected to two mutually perpendicular tensile stresses of 250 MPa and 130 MPa. Determine the intensities of normal, shear and resultant stresses on a plane inclined at 25◦ with the axis of minor tensile stress. S OLUTION : Let tensile stress along x − x axis = 250 MPa and along y − y axis = 130 MPa σx + σy σx − σy + cos 2θ 2 2 250 + 130 250 − 130 = + cos 2 × 25 2 2 250 = 190 + 60 × 0.643 = 190 + 38.57 = 228.57 MPa Ans. σx − σy 250 − 130 sin 2θ = sin 2θ × 25 Shear stress τθ = 2 2 = 60 sin 50 = 45.96 MPa Ans. Resultant stress σr = σθ2 + τθ2 = 228.572 + 45.962 √ = 52244.2 + 2112.3 = 233.14 MPa Ans. 130 Normal stress, σθ = @seismicisolation @seismicisolation 25° 250 MPa 130 Mpa Figure 5.7 52 • Strength of Materials E XAMPLE 5.5: At a certain point of a machine component, the mutually perpendicular stresses are 200 MPa and 75 MPa, both tensile. Determine the normal, shear, resultant stresses on a plane inclined at an angle of 60◦ with the axis of major tensile stress. Also find the magnitude of the maximum shear stress in the component. S OLUTION : 75 30° 200 200 MPa 60° 75 MPa Figure 5.8 σx + σy σx − σy + cos 2θ 2 2 200 + 75 200 − 75 + cos 2 × 30 = 2 2 1 = 137.5 + 62.5 × 2 Normal stress, σθ = = 168.75 MPa Ans. Shear stress, τθ = σx − σy sin 2 × 30 2 τθ = 62.5 × 0.866 = 54.13 MPa Ans. Resultant stress = σθ2 + τθ2 = 168.752 + 54.132 = 177.22 MPa Ans. τmax , Shear stress = σx − σy sin 2θ 2 τmax will be when 2θ = 90◦ ∴ τmax = ± or θ = 45◦ σx − σy 200 − 75 =± = ±62.5 MPa Ans. 2 2 @seismicisolation @seismicisolation Principal Stresses and Strains • 53 E XAMPLE 5.6: In a structural element, the stresses at a point are 150 MPa (tensile) and 80 MPa (compressive). Find the magnitude of the normal and shear stresses on a plane inclined at 55◦ with tensile stress. Also find the direction of the resultant stress and the magnitude of the maximum shear stress. S OLUTION : 80 C B 35° 150 150 MPa 55° D A 80 MPa Figure 5.9 Normal stress, σθ = = σx + σy σx − σy + cos 2θ 2 2 150 + (−80) 150 − (−80) + cos 2 × 35 2 2 = 35 + 115 × 0.342 = 74.33 MPa Shear stress, τθ = = Ans. σx − σy sin 2θ 2 150 − (−80) sin 2 × 35 = 115 × 0.939 2 = 107.98 MPa Ans. Resultant stress, σr = σθ2 + τθ2 = 74.332 + 107.982 = 131.1 MPa τmax = ± Ans. σx − σy 150 − (−80) = = 115 MPa Ans. 2 2 @seismicisolation @seismicisolation 54 • Strength of Materials Material Subjected to Shear Stresses When a material is subjected to put shear stress on one plane, an equal shear stress is induced on the perpendicular plane to present rotation of the element, such a state is shown in Fig. 5.10. τ A θ σθ B τ τθ τ τ C Figure 5.10 Resolving forces normal to AC, σθ × AC = τ × AB sin θ + τ × BC cos θ ∴ σθ = τ cos θ sin θ + τ sin θ cos θ σθ = τ sin 2θ (5.5) The maximum value is τ when θ = 45◦ Resolving forces parallel to AC, τθ × AC = −τ × AB cos θ + τ BC sin θ ∴ τθ = −τ cos2 θ + τ sin2 θ = −τ cos 2θ (5.6) The maximum value is τ when θ = 0 or 90◦ . Note: Shear stress induces numerically equal tensile and compressive stresses on planes at 45◦ to the planes of the shear stress as shown in Fig. 5.11. τ τ τ τ τ τ τ τ Figure 5.11 E XAMPLE 5.7: A steel block is fixed in a vice so that about half of the height is in the vice. Determine the normal stress and shear stress at a oblique section inclined at 30◦ to the vertical as shown in Fig. 5.12. When 25 MPa shear stress is applied. @seismicisolation @seismicisolation Principal Stresses and Strains • 55 25 MPa A 30° Vice B 25 MPa Figure 5.12 Normal stress, σθ = τ sin 2θ = 25 sin 2 × 30◦ = 21.65 MPa Ans. Shear stress, τθ = −τ cos 2θ = −25 cos 2 × 30◦ = −12.5 MPa Ans. Material Subjected to Direct and Shear Stresses This is the most general case of stresses in two dimensions, since any system of stresses in two dimensions can be reduced to this form. Refer Fig. 5.13. σy τ A τ θ σx σx τ B C τ σy Figure 5.13 From the combinations of previous formulae, σx + σy σx − σy + cos 2θ + τ sin 2θ 2 2 σx − σy τθ = sin 2θ − τ cos 2θ 2 d σθ =0 For σθ to be maximum or minimum dθ σθ = ∴ − (σx − σy ) sin 2θ + 2τ cos 2θ = 0 2τ tan 2θ = σx − σy @seismicisolation @seismicisolation (5.7) (5.8) (5.9) 56 • Strength of Materials σy τ y)2 + 4τ 2 σ2 τ (σ x – σ 2τ σx σ1 τ σx θ σ1 σ2 τ 2θ σy σx – σ y Figure 5.15 Figure 5.14 2τ Therefore, from Fig. 5.14, sin 2θ = (σx − σy )2 + 4τ 2 σx − σy and cos 2θ = (σx − σy )2 + 4τ 2 (5.10) (5.11) For maximum and minimum values of σθ , substituting the value of sin 2θ and cos 2θ in Eqn. (5.7), σx + σy σx − σy σx − σy 2τ +τ + · 2 2 2 2 (σx − σy ) + 4τ (σx − σy )2 + 4τ 2 1 = (σx + σy ) ± (σx − σy )2 + 4τ 2 2 σθ = (5.12) Equation (5.12) gives principal stresses σ1 and σ2 . It is a very important and useful equation. The planes upon which these stresses are mutually perpendicular, one of these stresses acts on one plane and the other stress acts on the perpendicular plane. The association between the stresses and planes is usually obvious by considering the equilibrium of the wedge ABC but when in doubt, a unique rotation for θ can be obtained from either of Equations (5.13) or (5.14), derived as follows: (Refer Fig. 5.10). Since there is no shear stress on a principal plane, then if AC is such a plane, the only stress acting on it is the principal stress, σ . Hence, resolving forces horizontally, σ × AC cos θ = σx × AB + τ × BC ∴ σ = σx + τ tan θ σ − σx = τ tan θ (5.13) σ × AC sin θ = σy × BC + τ × AB ∴ σ = σy + τ cot θ (5.14) Resolving vertically, @seismicisolation @seismicisolation Principal Stresses and Strains • 57 Now substituting the values of sin 2θ and cos 2θ from Eqns. (5.10) and (5.11) into expression for τθ Eqn. (5.8) gives τθ = 0 on these planes. Such planes are referred to as principal planes and the direct stresses acting on them are the principal stresses. Thus, at any point in a stressed material, there are always mutually perpendicular planes upon which the stresses are wholly tensile or compressive and these are respectively the greatest and least stresses at that point. Such an arrangement is shows in Fig. 5.13 where σ1 and σ2 are the principal stresses and θ is the angle given by Eqn. (5.9). Since this system of stress is identical with that shown in Fig. 5.6, it follows from Eqn. (5.4) that the maximum shear stress in the body is given by: σ1 − σ2 acting on planes at 45◦ to the principal. 2 1 1 2 2 Thus, τmax = (σx + σy ) + (σx − σy ) + 4τ 2 2 1 − (σx + σy ) − (σx − σy )2 + 4τ 2 2 1 ∴ τmax = (σx − σy )2 + 4τ 2 2 τmax = (5.15) In case of one direct stress only, σ= 1 σx + σx2 + 4τ 2 2 2τ σx 1 τmax = σx2 + 4τ 2 2 (5.16) tan 2θ = (5.17) E XAMPLE 5.8: In a complex stress system, an element has σx = 30 MPa compressive, σy = 50 MPa tensile and τ = 15 MPa. Determine the principal stresses at the point. S OLUTION : 1 2 2 Principal stresses σ1 σ2 = (σx + σy ) ± (σx − σy ) + 4τ 2 1 2 2 σ1 or σ2 = (−30 + 50) ± (−30 − 50) + 4 × 15 2 1 20 ± (−80)2 + 4 × 225 = 2 = 1 {20 ± 85.44} 2 @seismicisolation @seismicisolation 58 • Strength of Materials σ1 = 52.72 MPa (Tensile) σ2 = −32.72 MPa (Compressive) tan 2θ = = 2τ σx − σy 2 × 15 30 3 = = − = −0.375 −30 − 50 −80 8 tan 2θ = −0.375 ∴ or 2θ = −20.55 θ = 10.28◦ 79.72◦ or Ans. E XAMPLE 5.9: A component has a plane element which is subjected to 120 MPa tensile accompanied by a shear stress of 25 MPa. Find: i) the normal and shear stresses on a plane inclined at an angle of 25◦ with the tensile stress and ii) the maximum shear stress on the plane. S OLUTION : θ with vertical axis = 90 − 25 = 65◦ 25 120 65° 120 25° 25 Figure 5.16 σx + σy σx − σy + cos 2θ + 2 sin 2θ 2 2 120 + 0 120 − 0 = + cos 2 × 65 + 25 sin 2 × 65 2 2 = 60 + 60 × (−0.643) + 25 × 0.766 = 40.57 MPa σx − σy sin 2θ − τ cos 2θ = 60 × 0.766 − 25 × (−0.643) τθ = 2 = 62.07 MPa Ans. 1 τmax = (σx − σy )2 + 4τ 2 2 1 1 = 1202 + 4 × 252 = 1202 + 4 × 625 2 2 = 65 MPa Ans. σθ = @seismicisolation @seismicisolation Principal Stresses and Strains • 59 E XAMPLE 5.10: An element in a stress material has tensile stress of 600 MN/m2 and a compressive stress of 300 MN/m2 acting on two mutually perpendicular planes and equal shear stresses of 120 MN/m2 on these planes. Determine principal stresses and position of the principal planes. Also find maximum shear stress. S OLUTION : 1 2 2 σ1 & σ2 = (σx + σy ) ± (σx − σy ) + 4τ 2 1 2 2 600 + (−300) ± (600 − (−300)) + 4 × 120 = 2 = 1 300 ± 9002 + 4 × 14400 2 = 1 {300 ± 931.45} 2 = 615.7 MPa (Tensile) or − 315.7 MPa (Compressive) Ans. Position of principal planes θ1 , θ2 tan 2θ = 2τ 2 × 120 = σx − σy 600 − (−300) tan 2θ = 240 = 0.2666 900 ∴ 2θ = 14.93 ∴ θ1 = 7.46◦ , θ2 = 187.46◦ Ans. Max. stress (shear) τmax = σ1 − σ2 615.7 − (−315.7) = 2 2 = 465.7 MPa Ans. Graphical Method (Mohr’s Circle of Stresses) This method is easier than analytical method. Here, we will adopt a sign correction that: i) all the angles in the anticlockwise direction to the x − x axis are taken as negative and for the angles in the clockwise direction an positive; ii) the tensile stresses are drawn towards right from origin and compressive towards left and iii) the measurements above x − x axis and right of y − y axis are positive and those below x − x axis and left to y − y axis are taken as negative. Since this method involves drawing so small errors are bound to occur. @seismicisolation @seismicisolation 60 • Strength of Materials E XAMPLE 5.11: At a point in a stressed body the stresses are shown in Fig. 5.17. Determine the principal and maximum shearing stresses with the help of analytical method and Mohr’s circle method. Also find principal planes. S OLUTION : 30 MPa τ 25 MPa 60 MPa 60 MPa τ τ = 25 MPa τ 30 MPa Figure 5.17 Analytical Method: 1 2 2 Principal stress σ1 or σ2 = (σx + σy ) ± (σx − σy ) + 4τ 2 1 (60 + 30) ± (60 − 30)2 + 4(25)2 = 2 = 1 90 ± 302 + 2500 2 σ1 = 74.15 MPa σ2 = 15.85 MPa Maximum shear stress τmax = 1 2 (σx − σy )2 + 4τ 2 Substituting resultant values τmax = 29.15 MPa Principal planes: tan 2θ = 50 2τ 2 × 25 = = σx − σy 60 − 30 30 = 1.667 ∴ 2θ = 59◦ θ1 = 29.5◦ , θ2 = 119.5◦ with the plane AB of the element. @seismicisolation @seismicisolation Principal Stresses and Strains • 61 Mohr’s Circle Method +τ τmax 0 σ2 τ –τ σy By measuerment and correction σ1 = 75 MPa tensile σ2 = 17 MPa tensile τmax = 29 MPa σx 2θp = 60°, θp1 = 30° θp2 = 120° Anticlockwise σ1 Figure 5.18 E XAMPLE 5.12: A rectangular block of material is subjected to a tensile stress 100 MPa on a plane and a tensile stress of 40 MPa at right angles to the former together with a shear stress of 60 MPa on the same plane. Determine: a) analytically and b) by Mohr’s circle method i) the direction of principal planes ii) magnitude of principal stress iii) magnitude of the maximum shear stress and the corresponding plane. S OLUTION : 40 MPa 60 MPa 100 MPa 100 MPa 60 MPa 40 MPa Analytical solution: Figure 5.19 Principal stresses: 1 2 2 (σx + σy ) ± (σx − σy ) + 4τ σ1 & σ2 = 2 @seismicisolation @seismicisolation 62 • Strength of Materials 1 2 2 (100 + 40) ± (100 − 40) + 4 × 60 = 2 √ 1 1 = 140 ± 3600 + 14400 = {140 ± 134.16} 2 2 i) & ii) = 137.08 MPa (tensile), 2.22 MPa (Tensile) Ans. 1 1 2τ 2 × 60 θ = tan−1 = tan−1 2 σx − σy 2 100 − 40 1 1 = tan−1 2 = × 63.43 = 31.7◦ , 121.7◦ Ans. 2 2 1 (σx − σy )2 + 4τ 2 iii) τmax = 2 1 134.16 (100 − 40)2 + 4 × 602 = = = 67.08 MPa Ans. 2 2 60 MPa = τ C 0 Principal stress σ1 = 3 MPa B E 20 = 63° A 60 MPa = τ τmax = 66 MPa Mohr’s circle solution: D σy = 40 MPa 2θ = 63° θ = 315° σx = 100 MPa Principal stress σ2 = 135 MPa Figure 5.20 E XAMPLE 5.13: Various stresses acting on a block are shown in Fig. 5.21. Find analytically and using Mohr’s circle method, the principal and maximum shearing stresses, and the angles of the principal planes. 40 MPa τ τ 30 MPa 30 MPa τ τ = 20 MPa 40 MPa Figure 5.21 @seismicisolation @seismicisolation Principal Stresses and Strains S OLUTION : Analytical Method 1 2 2 (σx + σy ) ± (σx − σy ) + 4τ Principal stresses σ1 or σ2 = 2 1 2 2 (−30−40)± (−30 − 40) +4(20) = 2 σ1 = −4.59 MPa (Compressive) σ2 = −65.41 MPa (Compressive) 1 Maximum shear stress τmax = (σx − σy )2 + 4τ 2 2 By substituting max shear stress = 30.41 MPa tan 2θ = 60 2τ 60 = =6 = σx − σy −30 − (−40) 10 2θ p = 80.53 θ1 = 40.27, θ2 = 130.27 Ans. Principal stresses: σ1 = −4.1 MPa, σ2 = −65 MPa Ans. Max shear stress: τmax = 30.5 MPa Ans. Angles of the principal planes θ1 = 40.5◦ and θ2 = 130.5◦ (anticlockwise) Ans. +τaxis τmax σy = 40 σx = −30 –σaxis B C A τ σ1 σ2 –τaxis Figure 5.22 @seismicisolation @seismicisolation +σaxis 0 • 63 64 • Strength of Materials E XAMPLE 5.14: On a machine bracket the stresses on two mutually perpendicular planes at a point are 350 MPa tensile and 280 MPa tensile. The shear stress across these planes is 160 MPa. Determine graphically and analytically the magnitude and directions of principal stresses and maximum shear stress. S OLUTION : 1 2 2 σ1 or σ2 = (σx + σy ) ± (σx − σy ) + 4τ 2 1 = (350 + 280) ± (350 − 280)2 + 4 × 1602 2 = 1 630 ± 702 + 102400 2 = 1 {630 ± 327.6} 2 σ1 = 478.8 MPa (Tensile) Ans. σ2 = 151.2 MPa (Tensile). Ans. For direction of principal stresses θ1 , θ2 : tan 2θ = 2τ 2 × 160 320 = = = 4.57 σx − σy 350 − 280 70 ∴ 2θ = 77.66 ∴ θ1 = 38.83◦ Maximum shear stress, τmax = = θ2 = 128.83◦ σ1 − σ2 2 478.8 − 151.2 2 = 163.8 MPa @seismicisolation @seismicisolation Ans. Ans. Principal Stresses and Strains • 65 Graphical Method E σ2 = 151 MPa Ans. 0 T τ τmax M N S 2θ L 2θ = 76° & 256° Ans. P X τ F σy = 280 MPa σx = 350 MPa σ1 = 478 MPa Ans Figure 5.23 E XAMPLE 5.15: In a structure the principal stresses at a point are 80 MN/m2 (tensile) and 30 MN/m2 (tensile). Find the normal, tangential stresses and the resultant stress and its obliquity on a plane at 25◦ with the major principal plane. S OLUTION : Analytical Method: σx = 80 MN/m2 (Tensile) σy = 30 MN/m2 (Tensile) @seismicisolation @seismicisolation 66 • Strength of Materials 30 N/mm2 σx = 80 N/mm2 80 N/mm2 25° 30 N/mm2 Figure 5.24 Normal stress = σθ = σθ = Normal stress, σx + σy σx − σy + cos 2θ 2 2 80 + 30 80 − 30 + cos(2 × 25) 2 2 σθ = 55 + 25 × 0.6428 = 55 + 16.07 = 71.07 N/mm2 (Tensile). Ans. Shear stress or tangential stress, τθ = = σx − σy sin 2θ 2 80 − 30 sin(2 × 25) 2 = 25 × 0.766 = 19.15 N/mm2 Hence, resultant stress = Ans. σ 2 + τθ2 = σr 71.072 + 19.152 √ = 5050.9 + 366.7 σr = ∴ σr = 73.6 N/mm2 @seismicisolation @seismicisolation Ans. Principal Stresses and Strains • 67 Angle of obliquity, tan φ = = τθ σθ 18.15 71.17 = 0.269 ∴ φ = 15.06◦ Ans. Graphical (Mohr’s) Method: D 2θ = 50° φ O σy = A C E B X 30 N/mm2 бx = 80 N/mm2 Figure 5.25 plot OA = σy = 30 N/mm2 scale 10 N/mm2 = 1 cm and OB = σx = 80 N/mm2 Bisect AB to locate centre C. Draw CD at angle 2θ (2 × 25◦ ) = 50◦ Now by measurement Normal stress, σθ = OE = 71 N/mm2 Ans. Shear stress τθ = DE = 19.8 N/mm2 Ans. Resultant stress σr = 72.8 N/mm2 Ans. Angle of obliquity, φ = 15◦ Ans. Note: Slight difference is due to small mistakes in drawing Mohr’s circle. E XAMPLE 5.16: At a point in a machine member the principal stresses are 120 MN/m2 (tensile) and 70 MN/m2 (compressive). Determine the normal stress and the shear stress on a plane inclined at 50◦ to the axis of major principal stress. Also determine the maximum shear stress at the point. @seismicisolation @seismicisolation 68 • Strength of Materials S OLUTION : σx = +120 N/mm2 σy = −70 N/mm2 (−ve sign for compression) θ = 55◦ Analytical Method: Normal stress, σθ = σx + σy σx − σy + cos 2θ 2 2 70 MN/m2 120 MN/m2 50º 70 MN/m2 Figure 5.26 σθ = 120 − 70 120 − (−70) + cos(2 × 50◦ ) 2 2 = 25 + 95 × (−0.174) = 25 − 16.53 = 8.47 N/mm2 (Tensile) Ans. Shear stress or tangential stress, τθ = = σx − σy sin 2θ 2 120 − (−70) sin(2 × 50) 2 = 95 × 0.985 = 93.575 N/mm2 Ans. Maximum shear stress = τmax = σx − σy 2 = 120 − (−70) 2 = 95 N/mm2 Ans. @seismicisolation @seismicisolation 120 MN/m2 Principal Stresses and Strains • 69 M L P τθ = 92 N/mm2 C 100˚ B O τmax = 95.5 N/mm2 A N C σx = 120 N/mm2 σy = 70 N/mm2 Tensile Compressive Figure 5.27 Graphical (Mohr’s) Method Plot OA = 120 N/mm2 2 OB = 70 N/mm (Tensile) (Compressive) Bisect BA to get centre C and draw a circle as shown. Draw CM at an angle 2θ i.e. 100◦ . Drop perpendicular MN. On measurement, we find τθ = 92 N/mm2 (OP) τθ = 92 N/mm2 (OP) Max shear stress, Normal stress, τmax = 95.5 N/mm2 (CL) σθ = 8 N/mm2 (ON) E XAMPLE 5.17: On a steel bridge, there is a point which is subjected to perpendicular stresses of 60 MN/m2 tensile and 40 MN/m2 tensile. Compute the normal, tangential stresses and resultant stress with its angle of obliquity on a plane making an angle of 35◦ with the axis of the second stress. S OLUTION : σx = 60 MN/m2 (Tensile), θ = 90◦ − 35◦ = 55◦ σy = 40 MN/m2 (Tensile) @seismicisolation @seismicisolation 70 • Strength of Materials Analytical Method: σx + σy σx − σy + cos 2θ 2 2 60 + 40 60 − 40 = + cos(2 × 55) 2 2 = 50 + 10 × (−0.342) Normal stress = σθ = 2 = 50 − 3.42 = 46.6 N/mm 40 55° Ans. 60 60 40 Figure 5.28 Shear or tangential stress, σx − σy 60 − 40 = 2 2 τθ = 10 N/mm2 Ans. τθ = Resultant stress, σθ2 + τθ2 = 46.62 + 102 √ = 2171.56 + 100 σr = ∴ = 47.67 N/mm2 Ans. τ tan φ = θ σθ 10 = 0.2146 = 46.6 φ (obliquity) = 12.11◦ Ans. Graphical Method: Plot OA = 40 N/mm2 and OB = 60 N/mm2 Bisect AB and draw circle as shown. @seismicisolation @seismicisolation Principal Stresses and Strains • 71 σθ = 46.2 N/mm2 M O φ A 110° =2θ Q C B τθ = 9 N/mm2 40 N/mm2 σx = 60 N/mm2 Figure 5.29 On measurement: σθ = OQ = 46.2 N/mm2 τθ = MQ = 9 N/mm2 σr = 47 N/mm2 φ = 10.8◦ Ans. E XAMPLE 5.18: At a point in a member of a machine, the stresses on two mutually perpendicular planes are 40 N/mm2 tensile and 20 N/mm2 tensile. The shear stress across the plane is 10 N/mm2 . Find the magnitude and direction of the resultant stress on a plane making an angle of 35◦ with the plane of the first stress. Find also the normal and tangential stresses on the planes. S OLUTION : Analytical Method: σx = 40 N/mm2 σy = 20 N/mm2 τxy = 10 N/mm2 20 N/mm2 10 N/mm2 40 N/mm2 40 N/mm2 35° 10 N/mm2 20 N/mm2 Figure 5.30 σx + σy σx − σy + cos 2θ + τxy sin 2θ 2 2 40 + 20 40 − 20 = + cos(2 × 35◦ ) + 10 sin(2 × 35◦ ) 2 2 Normal stress, σθ = @seismicisolation @seismicisolation 72 • Strength of Materials = 30 + 10 × 0.342 + 10 × 0.9396 = 30 + 3.42 + 9.4 = 42.82 N/mm2 Ans. Shear stress on tangential stress: σx + σy sin 2θ − τxy cos 2θ 2 40 − 20 τθ = sin (2 × 35◦ ) − 10 cos (2 × 35◦ ) 2 = 10 × 0.94 − 10 × 0.342 = 9.4 − 3.42 = 5.98 N/mm2 τθ = Resultant stress: (42.82)2 + (5.98)2 √ = 1833.55 + 35.76 σr = = 43.23 N/mm2 5.98 42.88 tan φ = 0.139 Angle of obliquity = 7.9◦ Ans. tan φ = ∴ τθ σθ Ans. tan φ = ∴ Plot OA = 40 N/mm2 and OB = 20 N/mm2 Drop perpendicular LA and MB each 10 N/mm2 as shown join ML to get C. Draw Mohr’s circle taking MC as radius and C as centre. Draw CP at 70◦ to CL. P M O φ = 20° 20 N/mm2 C B 2θ = 70° N A 10 N/mm2 L 40 N/mm2 Figure 5.31 @seismicisolation @seismicisolation Principal Stresses and Strains • 73 By measurement: σθ = ON = 40 N/mm2 tensile Ans. σr = OP = 41.5 N/mm2 τ = PN = 11 N/mm2 φ = 9◦ Ans. Ans. Ans. Difference is due to errors in plotting the diagram. Principal Strains: The principal strains are the strains in the direction of the principal stresses. If the principal stresses on an element are σx and σy (Refer Fig. 5.32). σy σx σx σy Figure 5.32 Then strain in direction of σx due to σx = σx E and strain in direction of σx due to σy = −μ σy E σy σx −μ (5.18) E E σy σx Similarly, resultant stain in y direction, εy = −μ (5.19) E E If these strains are measured and it is required to find the corresponding stresses, then multiplying Eqn. (5.19) by μ and adding to Eqn. (5.18), ∴ resultant stain in x direction, εx = σx (1 − μ 2 ) E E σx = (εx + μεy ) 1 − μ2 E σy = (εy + μεx ) 1 − μ2 εx + μεy = ∴ (5.20) (5.21) Strains on an Oblique Section: Figure 5.33 shows an element ABCD which is subjected to pure strains εx and εy , the distorted shape relative to the point F being shown dotted. The line FG, inclined at θ to AB moves to the position FG, the displacements of G in the x and y directions being dx and dy respectively. @seismicisolation @seismicisolation 74 • Strength of Materials εy D C dr εx G εx E r θ A F dθ x θ dy y dx B εy Figure 5.33 Since the movement of G is very small, the distance GG may be regarded as the change in the dr length of FG, so that the strain on FG, εθ = r r2 = x2 + y2 ∴ 2r dr = 2x dx + 2y dy ∴ dr dx x = r x r 2 + dy y y r 2 εθ = εx cos2 θ + εy sin2 θ = εx + εy εx + εy + cos 2θ 2 2 Similarly, it can be shown that the strain on FE is (5.22) εx + εy εx − εy − cos 2θ 2 2 y x x dy − y dx sec2 θ d θ = x2 dy y dx y = · − · y x x x = (εy − εx ) tan θ ∴ d θ = (εy − εx ) sin θ cos θ εx − εy =− sin 2θ 2 tan θ = ∴ The negative sign indicates that when εx > εy , as assumed in Fig. 5.33, the angle θ decreases. @seismicisolation @seismicisolation Principal Stresses and Strains • 75 A similar analysis on the triangle AFE shows that FE also rotates through the same angle as FG, so that the total change in the right angle EFG is 2 d θ . This is the shear strain in direction inclined at angle θ to the faces of ABCD. i.e., φθ = (εx − εy ) sin 2θ (5.23) Equations (5.22) and (5.23) are similar to Eqns. (5.3) and (5.4) and they can therefore be represented by a similar graphical construction. Figure 5.34 shows Mohr’s strain circle, in which OA represents εx , OB represents εy and the angle AQC is 2θ . Then OD and CD represents respectively, εθ and φθ /2. εθ C 2θ B Q 0 εx+εy 2 φθ 2 D A εx−εy 2 εy εx Figure 5.34 These days for measurements electric resistance strain ganges are used very extensively. Ellipse of Stress: In order to draw an ellipse of stress, the procedure is given below with steps of construction: 1. Take O as centre and draw two circles σx and σy respectively. 2. Through O draw OM perpendicular to the FN as shown, to cut the inner circle in P and after circle in M. 3. Through M draw a line MR perpendicular to OX and through P, draw line PQ perpendicular to OY to cut the line MR in Q. Join OQ. Refer diagram: OR = σx cos θ , QR = σy sin θ OQ = OR2 + QR2 = σx2 cos2 θ + σy2 sin2 θ = Resultant stress σr ∴ tan α = QR OR tan α = σy sin θ σy = tan θ σx cos θ σx @seismicisolation @seismicisolation 76 • Strength of Materials The point Q may be drawn for different values of θ . The locus of Q will give an ellipse. y σy F σx σx θ σy M F o P x θ β N N θ α R x Q y Figure 5.35 E XAMPLE 5.19: A point in a strained component is subjected to two mutually perpendicular tensile stresses of 270 MPa and 150 MPa. Determine the resultant stress on a plane which makes 35◦ with the major principal stress by ellipse of stress. Resultant stress σr = OQ = 303.5 MPa α = 40.5◦ Y M F 150 MPa θ= ° 55 Q θ = 55° X α = 40.5° o R N Y 270 MPa Figure 5.36 @seismicisolation @seismicisolation X Principal Stresses and Strains • 77 Exercise 5.1 The state of stress of a machine component is shown in Fig. 5.37. Calculate the principal stresses and the maximum shear stress. 100 N/cm2 300 N/cm2 500 N/cm2 Figure 5.37 [Ans σ1 = 600 N/cm2 inclined at 18◦ 26 to stress of 500 N/cm2 σ2 = −400 N/cm2 perpendicular to σ1 maximum shear stress τmax = ±500 N/cm2 . It is inclined to σ1 at +45◦ and converges towards σ1 ] 5.2 A bar of uniform cross section 20 × 30 mm is subjected to an axial pull of 54 kN, applied at each end of the bar. Determine the normal and shear components of a force acting on a plane inclined at 20◦ to the line of action of the load and also find normal and shear stresses acting on the same plane. [Ans Pn = 18.47 kN, Ps = 50.77 kN σθ = 10.53 MPa, τθ = 28.83 MPa] 5.3 At a certain point in a strained material, the principal stresses are 100 MN/m2 and 40 MN/m2 , both tensile. Using and explaining Mohr’s circle of stress, find the normal, tangential and resultant stresses across a point in the plane of 48◦ to major principal plane. [Ans σθ = 66.86 MPa, τθ = 29.84 MPa σr = 73.22 MPa] 5.4 At a certain point in a structural member the value of σx = 45 MPa, σy = 75 MPa and τ = 45 MPa. Calculate the principal stresses and locate exactly the planes on which (Fig. 5.38) these exist. Represent the stress system by means of Mohr’s circle. Also find the maximum shearing stresses and the planes on which they act. τ σx σy τ τ τ σy Figure 5.38 σx [Ans σ1 = 90 MPa, σ2 = −60 MPa θ1 , θ2 = 18◦ 26 , 108◦ 20 τmax = 75 MPa, θ3 , θ4 = 63◦ , 26◦ , 153◦ 20 ] @seismicisolation @seismicisolation 78 • Strength of Materials 5.5 Direct stresses of 120 MPa (tensile) and 90 MPa (compressive) exist on two perpendicular planes at a certain point in a body. They are also accompanied by shear stress on the planes. The greater principal stress at the point due to these is 150 MPa. Determine: i) the shear stress on these planes and ii) also find the maximum shear stress at the point. [Ans τ = 84.85 N/mm2 , τmax = 135 N/mm2 ] 5.6 A point in a load carrying member is subjected to the following stress condition: σx = −120 MN/m2 , σy = 180 MPa, τxy = 80 MPa counterclockwise. Then a) Draw the initial stress element b) Draw complete Mohr’s circle c) Draw complete principal stress element d) Draw the complete shear stress element 180 MPa τ = 80 MPa CCW 120 MPa 120 MPa 180 MPa Figure 5.39 [Ans b) σr = 170 MPa, α = 28.07◦ θ = 75.96◦ ccw, θ = 59.04◦ cw σ1 = 200 MPa, σ2 = 140 MPa] 5.7 Prove that in a two-dimensional stress system, if εx and εy are the strains in two particular directions, the normal stresses in these directions are given by σx = E E (εx + μεy ), σy = (εy + μεx ) 2 1+μ 1 − μ2 By means of strain gauges, the strains in two perpendicular directions at 30◦ to the directions of the principal stresses are found to be 425 × 10−6 and 86 × 10−6 (both tensile) respectively, both tensile. Determine the normal stresses in the direction of the measured strains, and also the principal stresses. E = 200 GN/m2 , μ = 0.3. [Ans 98.8 and 86 MPa, 125.2 and 19.6 MPa] @seismicisolation @seismicisolation 6 C HAPTER SHEARING FORCE AND BENDING MOMENT Shearing Force Consider a horizontal beam AB (Fig. 6.1) which is simply supported at each end and which carries a single load W at the centre. Such a load which is assumed to act at a point, is called a concentrated or point load. Since the beam is in equilibrium, the reactions RA and RB of the supports at A and B, respectively, will each be W /2 units. A W x− L 2 L 2 X B x W 2 X L−x W 2 L RB = W 2 RA = W 2 Figure 6.1 Now let us consider any section XX of the beam, distance x from the left-hand end A. It will be noticed that two forces are acting on the part of the beam to the left of the section XX. These forces are: i) The reaction RA = W /2 acting vertically upwards ii) The concentrated load W acting vertically downwards There is, therefore a resultant force of W− W W = 2 2 tending to shear the left hand part of the beam downwards across X − X. This resultant force is called the shearing force of the beam at the section X − X. @seismicisolation @seismicisolation 80 • Strength of Materials For equilibrium, the resultant of the forces acting on the part of the beam to the right of X − X must be equal in magnitude, but part of the beam to the left of X − X. As a matter of fact, it will be noticed that the only force acting on the right-hand side of section X − X of the beam is the reaction RB = W /2 tending to shear this part of the beam upwards across. We may, therefore conclude that: ‘The shearing force at a particular section of a beam is the resultant force, of all the forces acting on either side of that section.’ Bending Moment Now let us think again of the part of beam to the left of the section X − X (Fig. 6.1). Taking moments about X − X, we get W x (clockwise) 2 L (anticlockwise) Moment due to load W = W x − 2 Moment due to reaction RA = There is therefore, a resultant moment at X − X, L W = x −W x − 2 2 = W (L − x) (clockwise) 2 Acting upon the beam in a clockwise direction, and the left-hand part of the beam will tend to rotate about XX in the direction of this resultant moment. The resultant moment is called the bending moment at the section X − X. For equilibrium, the resultant moment of all the forces acting on the part of the beam to the right of X − X must be equal in magnitude, but opposite in direction to the resultant moment of all the forces acting on the part of the beam to the right of X − X. It will be noticed that the only force acting on this part of the beam in the reaction RB = W /2 acting upwards at the end B, its distance being (L − x) from X − X. Now taking moments about X − X, we get: Moment due to reaction RB = W (L − x) (anticlockwise). 2 Hence, the right-hand part of the beam will tend to rotate about X − X in the direction of this moment. It should therefore be emphasised that there can only be one value for the bending moment (and also for the shearing force) at a particular section of a beam regardless of which side of the section is considered. From the foregoing discussion we may then conclude that: ‘The bending moment at a particular section of a beam is the resultant moment about a section of all the forces acting on either side of the section.’ @seismicisolation @seismicisolation Shearing Force and Bending Moment • 81 Sign convention for the shearing force and bending moment: i) Shear Force There are several conventions in use for mathematical sign with the magnitude of a shearing force and with the magnitude of a bending moment, but the system given below appears to be most popular. The shearing force (SF) which tends to slide the left-hand side of the section of the beam upwards relative to the right-hand side will be considered positive, and that which tends to slide the left-hand side of the section of the beam downwards relative to the right-hand side will be considered negative (see Fig. 6.2). Positive SF (left up, right down) Negative SF (left down, right up) Figure 6.2 Hence, the rule is: Left up, right down is positive Left down, right up is negative ii) Bending Moment: The bending moment (BM) which tends to cause the beam concave upwards, sometimes called sagging, will be considered positive and that which tends to cause the beam concave downwards, sometimes called hogging, will be considers negative (see Fig. 6.3). Notice that this diagram also gives the sign convention for cantilever beams. W Positive BM (sagging) Negative BM (hogging) Figure 6.3 @seismicisolation @seismicisolation 82 • Strength of Materials Shearing Force and Bending Moment Diagrams Diagrams indicating the value of the shearing force and the bending moment at any section along a loaded beam or cantilever are called Shear Force Diagram and Bending Moment Diagram. These diagrams are drawn immediately underneath the loading diagram to same horizontal scale. In each case positive values are plotted upwards and negative values downwards and, in addition, the principal values should be shown on both diagrams. Also scale for drawing (for example, 5 kN = 10 mm to suit the size of paper) and for Bending Moment Diagram (for example, 15 kNm = 10 mm to suit the size of paper) should be shown beneath each diagram. Method: In case of beams always start from the left hand end and for cantilevers from the free end. E XAMPLE 6.1: A beam ABCDE, 6 m long is simply supported at A and E. It carries concentrated loads as shown in Fig. 6.4. Draw Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) showing principal values. S OLUTION : 200 N A 400 N B 1.5 m 300 N C 1.5 m D 2 mm E 1m Figure 6.4 6m (a) Loading Diagram 400 N 400 N 200 N 200 N o o (b) S.F.D. 200 N 500 N 500 N 900 Nm 600 Nm 500 Nm (c) B.M.D. o o Figure 6.5 @seismicisolation @seismicisolation Shearing Force and Bending Moment • 83 Calculations: Beam Reactions Taking moments about A, and working in N and m, we get RE × 6 = (200 × 1.5) + (400 × 3) + (300 × 5) = 300 + 1200 + 1500 = 3000 ∴ RE = 3000 = 500 N 6 RA + RE = Sum of loads Now, RA = 200 + 400 + 300 − 500 = 400 N SFD Calculations: Starting from left point A, SF at A = 0 rising to 400 N SF at B = 400 falls to 400 − 200 = 200 N SF at C = 200 N falls to + 200 − 400 = −200 N SF at D = −200 N falls to − 200 − 300 = −500 N SF at E = −500 N rises to − 500 + 500 = 0 BMD Calculations: BM at A = 0 BM at B = +(400 × 1.5) = +600 Nm BM at C = +(400 × 3) − (200 × 1.5) = 1200 − 300 = +900 Nm BM at D = +(400 × 5) − (200 × 3.5) − (400 × 2) = +2000 + 700 − 800 = +500 Nm BM at E = +(400 × 6) − (200 × 4.5) − (400 × 3) = −(300 × 1) = +2400 − 900 − 1200 − 300 = 0 Note: Scales for drawing SFD and BMD should be chosen such that it suits the size of the paper and place allotted for diagrams. @seismicisolation @seismicisolation 84 • Strength of Materials E XAMPLE 6.2: An overhang beam (refer Fig. 6.6) is loaded as shown. Draw to seek the shearing force and bending moment diagram and mark all the important points. S OLUTION : Reactions: Taking moments about A and working in kN and m units RB × 3 + ×0.3 = (10 × 0.9) + (15 × 2.1) + (10 × 3.6) RB × 3 = 9 + 31.5 + 36 − 1.5 = 75 75 = 25 kN 3 RA + RB = Sum of loads RB = Now RA + 25 = 5 + 10 + 15 + 10 = 40 kN ∴ RA = 40 − 25 = 15 kN 10 kN 5 kN A C 0.3 m 15 kN D E 0.9 m 3 m 10 kN F B 1.2 m 0.9 m 0.6 m Figure 6.6 (a) Loading Diagram 10 kN 10 kN o 10 kN o 10 kN o o (b) SF Diagram Values in kN 15 kN 15 kN 7.5 kNm 7.5 kNm o o 1.5 kNm BM Diagram Values in kNm 6 kNm Figure 6.7 SFD Calculations: At C = 0 falls to − 5 kN At A = −5 rising to − 5 + 15 = +10 kN At D = +10 kN falls to + 10 − 10 = 0 @seismicisolation @seismicisolation Shearing Force and Bending Moment • 85 At E = 0 falls, to 0 − 15 = −15 kN At B = −15 kN rises to − 15 + 25 = +10 kN At F = +10 falls to + 10 − 10 = 0 BMD Calculations: At C = 0 At A = −5 × 0.3 = −1.5 kNm At D = (−5 × 1.2) + (15 × 0.9) = +7.5 kNm At E = (−5 × 2.4) + (15 × 2.1) − (10 × 1.2) = +7.5 kNm At B = (−5 × 3.3) + (15 × 3) − (10 × 2.1) − (15 × 0.9) = −6 kNm At F = (−5 × 3.9) + (15 × 3.6) − (10 × 2.7) − (15 × 1.5) + (25 × 0.6) = 0 The above values are then plotted to scale (for both SFD & BMD) and indicating beneath diagram beams carrying uniformly distributed loads. E XAMPLE 6.3: A 20 m long beam is simply supported between B and F. It carries uniformly distributed loads of 200 N/m along overhang AB and 100 N/m run over the length EF as shown in Fig. (6.8(a)). It also carries concentrated loads (refer Fig. 6.8(b)). Draw Shear Force and Bending Moment diagrams for the loaded beam and mark all important values. 600 N 200 N/m A 800 N C B D E 2m P 2m 400 N 100 N/m 2m 4m 1100 N 4m 1100 N 500 N 4m 2m 4m 500 N 300 SFD Values in N m Figure 6.8(a) o o 800 N F 300 700 1100 N 4800 Nm 3600 2800 BMD Values in N m 2000 o 0 400 1600 Nm Figure 6.8(b) @seismicisolation @seismicisolation 86 • Strength of Materials Beam Reactions: Taking moments about F and working in N and m units, we get: RB × 16 = (200 × 4 × 18) + (600 × 12) + (800 × 8) + (400 × 4) + (100 × 4 × 2) = 14400 + 7200 + 6400 + 1600 + 800 = 30400 30400 = 1900 N 16 RB + RF = Sum of loads ∴ Now, RB = 1900 + RF = (200 × 4) + 600 + 800 + 400 + (100 × 4) ∴ RF = 3000 − 1900 = 1100 N SFD Calculations: At A = 0 For uniformly distributed load, let us take mid-point P and Q so that we know exactly whether it will be a straight line, concave or convex. At P = −200 × 2 = −400 N At B = −200 × 4 = −800 N At C = (−200 × 4) + 1900 = 1100 N falls to 1100 − 600 = 500 N At D = 500 N falls to 500 − 800 = −300 N At E = −300 N falls to − 300 − 400 = −700 N At Q = −700 − 100 × 2 = −900 N At F = −900 N − 100 × 2 = −1100 N rises to − 1100 + 1100 = 0 BMD Calculations: At A = 0 At P = −200 × 2 × 2 = −400 Nm At B = −200 × 4 × 2 = −1600 Nm At C = −200 × 4(2 + 4) + 1900 × 4 = 19200 + 7600 = +2800 Nm At D = −200 × 4(2 + 8) + 1900 × 8 − 600 × 4 = +4800 Nm At E = −200 × 4(2 + 12) + 1900 × 12 − 600 × 8 − 800 × 4 = +3600 Nm At Q = −200 × 4(2 + 14) + 1900 × 14 − 600 × 10 − 800 × 6 − 400 × 2 − 100 × 2 × 1 = +2000 Nm At F = 0 @seismicisolation @seismicisolation Shearing Force and Bending Moment • 87 E XAMPLE 6.4: For the loaded beam in Fig. 6.9, draw BM and SF diagrams and show important points. 8 kN 0.5 m A B C 5m 3m Figure 6.9(a) This beam (6.9b) is equivalent to shown in (6.9a) 8 kN A C RA = 3.5 kN 4 kNm 5m B RB = 4.5 kN 3m Figure 6.9(b) Reactions: Taking moments about B, RA × 8 = 8 × 3 + 4 = 28 ∴ RA = 28 = 3.5 kN 8 RB = 8 − 3.5 = 4.5 kN 3.5 kN 3.5 kN o o 4.5 kN 4.5 kN 17.5 kNm 13.5 kNm o o Figure 6.9(c) @seismicisolation @seismicisolation 88 • Strength of Materials SFD Calculations: At A = 0 rises to + 3.5 kN At C = 3.5 to 3.5 − 8 = −4.5 kN At B = −4.5 rises to − 4.5 + 4.5 = 0 BMD Calculations: At A = 0 At C = +3.5 × 5 = 17.5 kNm Also at C = 17.5 − 4 = 13.5 kNm Remember: i) where shear force is zero or change sign, bending moment is maximum; ii) where B.M. is zero that point on Bending Moment Diagram is known as point of contraflexure or point of inflexion. Relation between intensity of loading, shearing force and bending moment: Consider a short length, dx of a beam Fig. 6.10, carrying a uniformly distributed load w per unit length. Over this length, let the shear force change from F to F + dF and the bending moment change from M to M + dM. Equating vertical force on the element, F + wdx = F + dF ∴w= dF dx (i) Taking moments about the right-hand end of the element, w/unit length F + dF M M + dM dx + F Figure 6.10 @seismicisolation @seismicisolation Shearing Force and Bending Moment M + F.dx + w.dx • 89 dx = M + dM 2 ignoring the second order of small quantities F.dx = dM ∴ F= dM dw (ii) Therefore, intensity of loading is the rate of change of shearing force and shearing force is the rate of change of bending moment. This latter relation shows that the maximum bending moment occurs where the shearing force is zero. Combining Eqns. (i) and (ii) d2M dF . = dx dx2 w= Cantilevers E XAMPLE 6.5: Draw SF and BM diagrams for the cantilever loaded as shown in Fig. (6.11) and show important values on both diagrams. 25 kN 15 kN C 15 kN/m D E B A F 1.5 m 1.5 m 2m Figure 6.11 1m 1m o o 15 kN SFD 15 kN 45 kN 70 kN 70 kN 15 kNm 37.5 kNm BMD 75 kNm m .5 47 kN 142.5 kNm 2 Figure 6.12 @seismicisolation @seismicisolation 90 • Strength of Materials SF Calculations: SF at B = 0 falling to − 15 kN SF at E = −15 kN SF at D = −15 kN − 15 × 2 = −45 kN SF at C = −45 kN falls to − 45 − 25 = −70 kN SF at A = −70 kN BM Calculations: BM at = 0 BM at E = −15 × 1 = −15 kNm BM at D = −15 × 3 − 15 × 2 × 1 = −75 kNm BM at C = −15 × 4.5 − 15 × 2 × 2.5 = −142.5 kNm BM at A = −15 × 6 − 15 × 2 × 4 − 25 × 1.5 = −247.5 kNm Take extra mid-point F on u.d.l. to find the nature of curve BM at F = −15 × 2 − 15 × 1 × 1 = −37.5 kNm 2 E XAMPLE 6.6: Figure 6.10 shows a partial variable loaded cantilever. Draw SF and BM diagrams. D X 4 kN/m A B 2m E F X x C 3m Figure 6.13(a) Consider a section X − X between B and C at a distance x from the free end C. ΔBCD is similar to ΔCEF For Rate of loading at X − X FC EF x EF = ; = DB BC 4 3 4 EF = x kN/m 3 @seismicisolation @seismicisolation Shearing Force and Bending Moment • 91 Now, Shear force at X − X = Total load from C to F = 2x2 1 4x × ×x = kN 2 3 3 (i) For SF at C, put x = 0 in Eqn. (i) ∴ SF at C = 0 SF at B, put x = −3 in Eqn. (i) ∴ SF at B = 2 × 32 = −6 kN 3 SF at A = −6 kN BM Calculations: BM at X − X = − 2x2 x −2x3 × = kN/m 3 3 9 Equation (ii) is that of a cubic parabola BM at C = 0 BM at B, put x = 3 in Eq. (ii) = −2(3)3 = −6 kNm 9 3 BM at A = Total load on BC × 2 + 3 1 =− × 4 × 3 × (2 + 1) 2 = −18 kNm Between A and B, BMD will be a straight line. o o SFD 6 kN o o BMD Cubic parabola 6 kNm 18 kNm Figure 6.13(b) @seismicisolation @seismicisolation (ii) 92 • Strength of Materials E XAMPLE 6.7: In a gradually varying loaded beam as shown in Fig. 6.14, the span of simply supported beam is 6 m. Load at B is 1.6 kN/m run and gradually increases to 4 kN/m run at the other end. Determine the position and amount of maximum bending moment. Also draw the shear force and bending moment diagram. 4 kN/m 1.6 kN/m A B 6m Figure 6.14 Let us split the beam in two parts: i) a rectangular of length 6 m and height 1.6 kN/m run. ii) a triangular zero at B and 4 − 1.6 = 2.4 kN/m at A. Let RA and RB be reactions due to part (i) RA = RB = 1.6 × 6 = 4.8 kN 2 Let RA and RB be reactions due to part (ii) Taking moments about B, RA × 6 = 2.4 × 6 2 × ×6 2 3 ∴ RA = 4.8 kN RA + RB = Total triangular beam (part ii) 4.8 + RB = 2.4 × 6 2 ∴ RB = 2.4 kN Net reactions: RA = RA + R = 4.8 + 4.8 = 9.6 kN RB = RB + RB = 4.8 + 2.4 = 7.2 kN SFD Calculations: S.F. at A = 0 rises to 9.6 kN S.F. at B = 7.2 rises to 7.2 − 7.2 = 0 @seismicisolation @seismicisolation Shearing Force and Bending Moment BMD Calculations: Obviously, BM at A = 0 BM at B = 0 Maximum bending moment occurs where SF changes sign. Let P be the point where maximum bending moment occurs and S.F. changes sign. Let x be the distance between P and B. As, SF is zero, therefore, 0 = −7.2 + 1.6x + 2.4x 1 ×x× 6 2 0 = −7.2 + 1.6x + 0.2x2 0.2x2 + 1.6x − 7.2 = 0 x2 + 8x − 36 = 0 √ −8 ± 64 + 4 × 36 −8 ± 14.42 x= = 2 2 = 3.21 m. Maximum BM = +7.2 × 3.21 − 1.6 × 3.21 × 3.21 2.4 × 3.21 3.21 − × 2 6 3 − 23.112 − 8.24 − 1.37 = 13.5 kNm Ans. 4 kN/m 1.6 kN/m B A 6m RB = 7.2 kN RA = 9.6 kN 9.6 o SFD P o 7.2 kN 13.5 kNm BMD o o Figure 6.15 @seismicisolation @seismicisolation • 93 94 • Strength of Materials Exercise 6.1 Draw and SF and BM diagrams for the loaded cantilever as shown is in Fig. (6.16) and state BM at D. 40 kN 2m 30 kN 1m 20 kN 10 kN/m 1m [Ans Figure 6.16 (310 kNm)] 6.2 A beam AB of length 8 m is simply supported at A and B. It carries a load which varies from 2 kN/m at A to 5 kN at B. Draw SF and BM diagrams. State what is maximum bending moment and its position. Show all principal values on SF and BM on diagrams [Ans (28.1 kNm) at 4.3m from A] 6.3 A beam AB is hinged at A and simply supported at B. A rigid bracket is attached to the middle point C of the beam is loaded as shown in Fig. 6.17. Draw SF and BM diagrams showing important values. What is the maximum BM and its position. 0.8 m Hinge 1000 N B C A 2m 2m [Ans Figure 6.17 Max. BM 1400 Nm at 2 m from A] 6.4 For the beam loaded as shown in Fig. 6.18, at 2 m from A draw SF and BM diagrams. Indicate maximum BM and its position. C A B 30 kNm 6m 4m [Ans Figure 6.18 @seismicisolation @seismicisolation 18 Nm, 6 m from A] Shearing Force and Bending Moment • 95 6.5 Draw the SF and BM diagram for beam shown in Fig. 6.19. Locate the point of contraflexure. 1 kN/m 6 kN B 30° A 2m C 3 kN E D 2m 4m 1m [Ans 0.8 m from D] Figure 6.19 6.6 For the loaded beam shown in Fig. 6.20. Draw SF and BM diagrams and indicate all principal values on both diagrams. Also find maximum bending moment and its position. 4 kN 8 kN 2 kN/m 1 kNm 6 kN C D A 5m 5m E 5m B F 5m 5m 20 m [Ans Figure 6.20 65 kNm 5 m from F] 6.7 Draw SF and BM diagrams for a cantilever loaded as shown in Fig. 6.21. Show all the principal values on both diagrams. 20 kN 20 kN/m C A 10 kN D E B 1m 1m 2m 1m Figure 6.21 @seismicisolation @seismicisolation [Ans Max BM = 190 kNm] • 96 Strength of Materials 6.8 Draw SF and BM diagrams for loaded cantilever as shown in Fig. 6.22 and indicate position and value of maximum BM. D 4 kN/m A B 2m C 3m [Ans Figure 6.22 18 kNm at A] 6.9 A simply supported beam of 5 m span carries a triangular load of total 30 kN. Draw SF and BM diagrams and indicate the maximum value of BM and its position. [Ans 25 kNm at centre] 6.10 Draw the SF and BM diagrams for the cantilever beam shown in Fig. 6.23. The vertical load of 2 kN at C is partly supported by a force of 3 kN at the prop B. State: a) the reaction at the built in end; b) the maximum BM and its position and c) where the BM is zero. 700 mm 250 mm A 2 kN C B 3 kN Figure 6.23 [Ans a) 1 kN; b) 0.5 kNm; c) At B downward] 6.11 Draw SF and BM diagrams for the beam shown in Fig. 6.24. Determine the points of contraflexure. What is the maximum BM. 100 N 100 N/m C A 1m D B 6m Figure 6.24 2m [Ans @seismicisolation @seismicisolation 7.79 m and 3.38 m from free end D] C HAPTER 7 CENTRE OF GRAVITY Centroid: This is the point at which the whole area of a plane figure is assumed to be concentrated. It is applicable for plane geometrical figures such as rectangle, square, triangle, etc. Centre of Gravity: The centre of gravity of a body is the point in the body through which the whole weight of the body acts. Centroid for Plane Figures a) A circle: G π d2 Area = 4 Centroid is at centre O d 2 Figure 7.1 A b) A triangular plane figure: 1 Area = × b × h 2 G h h 3 B C b Figure 7.2 c) Square: Area = A2 G A centroid at from each side. 2 A A Figure 7.3 Y d) Rectangle: Area = A × L L A X̄ = , Ȳ = 2 2 A A 2 L 2 L Figure 7.4 @seismicisolation @seismicisolation X • Strength of Materials e) Semicircle: π r2 2 4r from X − X axis Ȳ = 3π Area = G r Y X X Figure 7.5 f) Trapezium: a 1 Area = c(a + b) 2 a + 2b c X̄ = a+b 3 b X c Figure 7.6 Y g) Qudrant of Circle: 1 Area = π r2 4 4r Ȳ = X̄ = 3π G X X o X r Figure 7.7 h) Sector of a Circle: O r2 θ Area = 2 Y θ is in radius θ r G 2r sin α Ȳ = 3 α Figure 7.8 In denominator α is to be taken into radians. l i) Circular Segment: Y Area = A Length of area = l Angle in degrees = α Radius = r Height of segment = h G h C r 98 α 1 C = 2 h(2r − h) A = [rl −C(r − h)] 2 @seismicisolation @seismicisolation Figure 7.9 Centre of Gravity j) Ellipse: Area = π ab G b a Figure 7.10 Centroid is where major axis and minor axis intersect. Ȳ = b k) Parabola: x 4 Area = xy 3 3x X̄ = 5 y x Figure 7.11a For one half of parabola: Area = 2xy 3 X̄ = 3x 5 and Ȳ = 3y 8 G Y X x Figure 7.11b l) Semicircular arc: Ȳ = 2r π y G r Figure 7.12 Centre of Gravity for Solid Bodies a) Cylinder: G h 2 Volume = π r h h ȳ = r 2 x̄ = 0 (since cylinder is symmetrical about y axis). Figure 7.13 2 @seismicisolation @seismicisolation y h • 99 100 • Strength of Materials b) Rod: π d2 V= ·l 4 l ȳ = 0 (because of symmetry) x̄ = 2 y x x l y Figure 7.14 c) Hemisphere: v= 1 4 3 × πr 2 3 y 3r 3r from the base so ȳ = 8 8 x̄ = 0 (due to symmetry) C.G. is at G 3r 8 x r Figure 7.15 d) Sphere: 4 V = π r3 for solid sphere 3 4 = π (r13 − r23 ) for hollow sphere 3 r2 r1 C.G. is at centre. Figure 7.16 e) Qudrent of a sphere: 3 x̄ = r, 8 Y 3 ȳ = r 8 G 3 8 r 3 8 X r Figure 7.17 f) Solid cone: 1 V = π r2 h 3 h from the base Ȳ = 4 G Figure 7.18 @seismicisolation @seismicisolation Y x Centre of Gravity g) Hollow cone: h Ȳ = 3 • 101 from the base G Y Figure 7.19 Centroid of Different Sections E XAMPLE 7.1: Determine the centroid of 120 mm × 180 mm × 30 mm T section. 120 mm 30 1 180 2 A y B 30 mm Figure 7.20 S OLUTION : In such cases divide the section into two different smaller sections, e.g., 1 and 2. Since it is symmetrical about Ȳ axis so X̄ = 0. a1 = 120 × 30 = 3600 mm2 y1 = 165, a2 = 150 × 30 = 150 = 75 mm 4500 mm2 , y2 = 2 A, total area = 3600 + 4500 = 8100 mm2 Aȳ = a1 y1 + a2 y2 a1 y1 + a2 y2 A 3600 × 165 + 4500 × 75 = 115 mm = 8100 E XAMPLE 7.2: Determine centroid of plane Fig. 7.21. S OLUTION : ∴ ȳ = C 25 mm 1 300 mm x G y 25 mm 2 A B 200 mm Figure 7.21 @seismicisolation @seismicisolation from AB Ans 102 • Strength of Materials Divide the sections into two rectangular strips 1 and 2 ., Now a1 = 275 × 25 = 6875, y1 = 162.5, x1 = 12.5, a2 = 200 × 25 = 5000 mm2 , x2 = 100 mm, y2 = 12.5 mm A = a1 + a2 = 6875 + 5000 = 11875 mm2 a1 y1 + a2 y2 A 6875 × 162.5 + 5000 × 12.5 Ȳ = 11875 1117187.5 + 62500 = 11875 Ȳ = = 99.34 mm X̄ = (from AB) 6875 × 12.5 + 5000 × 100 11875 = 49.34 mm Ans E XAMPLE 7.3: Determine the centroid of plane Fig. 7.22. S OLUTION : Let us divide the section into simple rectangles 1, 2 and 3. x̄ = 0 because of symmetry 120 mm 1 30 mm 2 a1 = 200 × 30 = 6000 mm2 y1 = 15 + 250 + 40 = 305 mm 250 mm G 30 mm y a2 = 250 × 30 = 7500 mm2 , 250 y2 = + 40 = 165 mm 2 3 A 350 mm 40 = 20 mm Figure 7.22 2 A = a1 + a2 + a3 = 6000 + 7500 + 14000 = 27500 mm2 a1 y1 + a2 y2 + a3 6000 × 305 + 7500 × 165 + 14000 × 20 ȳ = = A 27500 ȳ = 121.73 mm from AB Ans a3 = 350 × 40 = 14000 mm2 , y3 = E XAMPLE 7.4: Find the centroid of section shown in Fig. 7.23 y1 140 mm 3 60 mm 1 2 x 200 mm y Figure 7.23 @seismicisolation @seismicisolation 40 mm B 60 mm 55 mm x Centre of Gravity • 103 S OLUTION : 55 π 552 = 1187.31 mm2 y1 = = 27.5 mm 4×2 2 4r 4 × 27.5 x1 = 27.5 − = 27.5 − x1 = 15.82 mm 3π 3 × 3.14 a2 = 200 × 55 = 11000 mm2 , y2 = 27.5 mm, x2 = 100 + 27.5 = 127.5 mm 60 × 60 60 = 1800 mm2 , y3 = + 55 = 75 mm, x3 = 30 + 140 + 27.5, x3 = 197.5 mm a3 = 2 3 A = a1 + a2 + a3 = 1187.31 + 11000 + 1800 = 13987.31 a1 x1 + a2 x2 + a3 x3 1187.31 × 15.82 + 11000 × 127.5 + 1800 × 197.5 x̄ = = A 13987.31 = 126.98 mm a1 y1 + a2 y2 + a3 y3 1187.31 × 27.5 + 11000 × 27.5 + 1800 × 75 ȳ = = A 13987.31 ȳ = 33.613 mm Ans a1 = E XAMPLE 7.5: Determine the centroid of the shaded area shown in Fig. 7.24. C 75 mm 75 mm D 180 mm m 0m 10 A 150 mm B 150 mm Figure 7.24 The figure is symmetrical about y-axis so x = 0. Take whole trapezium as 1 including whole and take hole as 2 alone. 180 {(75 + 75) + (150 + 150)} = 60 (150 + 300) = 27000 mm2 2 180 300 + 2 × 150 y1 = × = 80 mm 3 300 + 150 a1 = π r2 3.14 × 1002 = = 15700 mm2 2 2 4r 4 × 100 y2 = = = 63.694 mm 3π 2 × 3.14 a2 = @seismicisolation @seismicisolation 104 • Strength of Materials In this case A = a1 − a2 = 27500 − 15700 = 11800 mm2 In this case ȳ = = a1 y1 − a2 y2 A 27500 × 80 − 15700 × 63.694 = 101.695 mm from AB 11800 Ans. E XAMPLE 7.6: A frustum of a solid right circular cone has from bottom an axial hole of 60 cm diameter as shown in Fig. 7.25. Find the centre of gravity of the body. Figure 7.25 S OLUTION : The frustum is symmetrical about vertical axis. Therefore, its centre of gravity lies on this axis. A cross-sectional view is shown in Fig. 7.26 of the frustum. Due to similar triangle: i) Cone OAB. a x 0 = 1.7 0.8 a = 1.6 a 1.8 + a 0.8 m = 1.7 0.8 C D x = 3.4 m (1.8 + a) 0.8 = 1.7a 1.8 m 1.44 + 0.8a = 1.7a 1.44 = 0.9a A 1.44 B 0.6 ∴ a= = 1.6 m left AB be the 1.7 m 0.9 reference Figure 7.26 @seismicisolation @seismicisolation Centre of Gravity • 105 Taking AB as reference, i) OAB π 2 π 1.7 2 × 3.4 = 2.5711 m3 R H= 3 3 2 3.4 = 0.085 m y1 = 4 ii) Right circular cone OCD π π V2 = r2 h = (0.4)2 × 1.6 = 0.268 m3 3 3 1.6 y2 = + 1.8 = 2.2 m 4 iii) Circular hole π 1.8 = 0.9, V3 = (0.6)2 × 1.8 y3 = 2 4 3 = 0.5087 m ; V1 = V = 2.5711 − 0.268 − 5087 = 1.7944 m3 V1 y1 +V2 y2 −V3 y3 Ȳ = V 2.5711 × 0.85 + 0.268 × 2.2 − 0.5087 × 0.9 = 1.7944 2.185 − 0.5896 − 0.45783 = 1.7944 = 0.634 m from AB Ans Total volume, E XAMPLE 7.7: A hemisphere and a cone have their bases joined together, the two being of same size of dia 300 mm. If the height of cone is 500 mm, find the centre of gravity from the bottom of hemisphere (Refer Fig. 7.27). D 500 mm 1 A B 2 C 300 mm Figure 7.27 @seismicisolation @seismicisolation 150 mm 106 • Strength of Materials Let cone be 1 and hemisphere be 2. Let C.G. be calculated from bottom of hemisphere C. 1. Cone: Volume, π 2 π R H = (150)2 × 500 = 11775000 mm3 3 3 500 y1 = + 150 = 275 mm 4 V1 = 2. Hemisphere: Volume, V2 = 1 4 3 2 × π R = × 3.14 × (150)3 = 7064929, mm3 2 3 3 y2 = 150 − 3 × 150 = 93.75 mm, 8 Total volume = 11775000 + 7064929 = 18839929 ȳ = V1 y1 +V2 y2 V = 11775000 × 275 + 7064929 × 93.75 18839929 = 3238125000 + 662337094 18893329 = 207.03 mm from C Ans Exercise 7.1 A circular hole of 60 mm diameter is cut out from a circular disc of 120 mm diameter as shown in Fig. 7.28. Find the centre of gravity of the section from A. 120 mm A B 60 mm [Ans: 50 mm from A] Figure 7.28 @seismicisolation @seismicisolation Centre of Gravity • 107 7.2 Find the centre of gravity from AB of channel section shown in Fig. 7.29. 100 20 300 mm 20 mm 20 [Ans: 27.39 mm from AB] Figure 7.29 7.3 Find the centroid of area of figure shown in Fig. 7.30. A 100 mm B 20 mm G H 180 mm C D 20 mm F 120 mm E Figure 7.30 [Ans: x̄ = 106.74 from DE ȳ = 95.26 from FE] 7.4 Determine the centroid of T section as shown in Fig. 7.31. 160 mm 10 mm 150 mm 10 mm [Ans: ȳ = 43.71 mm] Figure 7.31 @seismicisolation @seismicisolation 108 • Strength of Materials 7.5 Determine the centroid of figure shown below: 100 mm 30 15 70 20 mm 20 50 mm [Ans: 44.2 mm from top surface] Figure 7.32 7.6 A cylinder with a hemispherical cavity and a conical cap is shown in Fig. 7.33. All dimensions are in mm. Determine the centre of gravity of composite volume if the cylinder is made of steel and conical cap is made of aluminium. Assume density of steel = 7870 kg/m3 and density of aluminium = 2770 kg/m3 . A 420 mm C 500 mm B 600 mm [Ans: Figure 7.33 371.13 mm from AB] 7.7 A square hole is punched out of a circular lamina as shown in Fig. 7.34. The diagonal of the square which is punched out is equal to the radius of circle. Find the centroid of remaining laminar. Y X X a Y [Ans: x̄ = 0.095 a from y axis] Figure 7.34 @seismicisolation @seismicisolation Centre of Gravity • 109 7.8 Find the centroid of the cross-sectional area of a Z section as shown in Fig. 7.35. Q 10 cm B 2.5 cm E 2.5 cm 20 cm 5 cm A B 20 cm Figure 7.35 [Ans: x̄ = 13.557 cm from GE ȳ = 7.69 cm from AB] 7.9 Locate the centroid of a shaded area in Fig. 7.36. Y R2 = 6 cm R2 R1 = 2 cm R1 X 3 cm 3 cm [Ans: 2.417 cm, 3.03 cm] Figure 7.36 7.10 Determine the centroid of the cross-sectional area of an I section shown in Fig. 7.37. Y 20 cm 5 cm 15 cm [Ans: X 30 cm Y 5 cm X Figure 7.37 @seismicisolation @seismicisolation ȳ = 10.96 cm] 110 • Strength of Materials 7.11 A hemisphere of radius 40 mm is cut out from a right circular cylinder of diameter 80 mm and height 160 mm as shown in Fig. 7.38. Find the centre of gravity of the body from the base AB. 160 mm A B 80 mm [Ans: 77.2 mm] Figure 7.38 7.12 Determine the centre of gravity of the component shown in Fig. 7.39. 4 cm dia 2.5 cm A 6 cm dia 3.5 cm B [Ans 1.517 cm from bottom] Figure 7.39 @seismicisolation @seismicisolation C HAPTER 8 MOMENT OF INERTIA Measurement using moment of inertia with respect to an axis is an important property, which gives a measure of resistance to bending in the case of thin plates or plane areas. Y x dA y X 0 Figure 8.1 Let dA be a small elementary area of plane figure shown in Fig. 8.1. Now elementary moment of inertia of dA about X axis will be dA.y × y = dA y2 . Then total moment of inertia of whole area will be: IXX = ∑ dA y2 Similarly, IYY = ∑ dA x2 Rectangular Section D Moment of inertia (second moment of area) of elementary strip dy = b.dy.y2 Y dy C h 2 Y X h h 2 A Y b B Figure 8.2 @seismicisolation @seismicisolation X 112 • Strength of Materials +h/2 by2 dy Total moment of inertia = −h/2 =b +h/2 y3 3 −h/2 ⎡ ⎢ = b⎣ h 2 3 3 − − h2 3 3 ⎤ ⎥ ⎦ bh3 12 hb3 = 12 = Similarly, IYY Radius of Gyration The second moment of area has dimensions of length to the fourth power, it can be expressed as the product of area ‘A’ and length ‘k’ squared. Thus I = Ak2 k is known as radius of gyration. Product of Inertia: Let x and y be the distances from y-axis and x-axis of an elemental area dA. Y x dA y X Figure 8.3 The moment of this area about x-axis is dA.y. And the moment of dAy about y-axis is xydA. This term xydA is called the product of inertia of area dA with respect to x-axis and y-axis. And xydA is called the product inertia of the entire area, denoted as IXY . IXY = xydA @seismicisolation @seismicisolation Moment of Inertia • 113 Theorem of Parallel Axis It states that if the moment of inertia of a plane over an axis in the same plane through the centre of gravity of the area be represented as IG , then the moment of inertia of a given plane area about parallel axis AB in the plane of area at a distance h from C.G. of the area is given by Plane Area A IAB = IG + Ah2 G x x h A B Figure 8.4 Theorem of the Perpendicular Axis It states that if IXX and IYY are the moment of inertia of a plane section about two mutually perpendicular axis X − X and Y −Y in the planes of the section, then the moment of inertia of the section IZZ perpendicular to the plane and passing through the intersection of X − X and Y −Y is given by: Z IZZ = IXX + IYY x 0 X y dA Y Figure 8.5 Moment of inertia of the rectangular section about a line passing through the base b A bd 3 IGG = 12 Applying the parallel axis theorem, IDC = IGG + Ah2 2 d bd 3 bd 3 bd 3 + bd + IDC = = 12 2 12 4 Y G X d D bd 3 IDC = 3 Y Figure 8.6 @seismicisolation @seismicisolation B X C 114 • Strength of Materials b Moment of Inertia of a Hollow Rectangular Section b1 IXX = bd 3 b1 d13 − 12 12 d d1 X X Figure 8.7 Moment of Inertia of a Circular Section Consider an elementary circular ring of radius r and thickness dr. Area of circular ring = 2π rdr Let us first find the moment of inertia about an axis passing through O (polar axis), perpendicular to the plane of the paper. Figure 8.8 Moment of inertia of the circular ring about an axis passing through O (polar axis), = (Area of ring) × (Radius of ring)2 = (2π rdr) × r2 = 2π r3 dr R 2π r3 dr Total moment of inertia = 0 = = R 2π r4 4 π d4 32 = 0 π R4 2 = where d = 2r π d 2 4 2 (i) Now according to the perpendicular axis theorem, IZZ = IXX + IYY In circular section, IXX = IYY @seismicisolation @seismicisolation (ii) • Moment of Inertia 115 Substituting from (i) & (ii) π d4 = 2IXX 32 π d4 Also IYY = 64 ∴ π d4 64 IXX = where, d = diameter = 2r The moment of inertia of a hollow circular section Y Z π D4 π d 4 Polar moment of inertia = − 32 32 π 4 = (D − d 4 ) 32 π D4 π d 4 − IYY = IXX = 64 64 4 πd (D4 − d 4 ) = 64 X d X D Z Y Figure 8.9 Moment of Inertia of a Triangle A a) About its base: Consider an elemental strip at y from A, of thickness dy. y dy Area of strip = DE · dy ∴ ∴ B y DE = BC h h C.G. h 3 b X Now triangles ADE and ABC are similar. E D Figure 8.10 y by ∴ DE = BC · = h h by Area of stripe = · dy h Therefore, moment of inertia of the stripe about the base BC, = Area of stripe × (h − y)2 = by · dy(h − y)2 h Moment of inertial of whole triangle about the base BC, IBC = h by 0 h (h − y)2 dy = @seismicisolation @seismicisolation b h h 0 y(h − y)2 dy X C 116 • Strength of Materials b h 2 y(h + y2 − 2hy) dy h 0 b h 2 = (yh + y3 − 2hy2 ) dy h 0 h b y2 h2 y4 2hy3 + − = h 2 4 3 0 4 4 4 b h h 2h = + − h 2 4 3 = Hence, IBC = bh3 12 b) About an axis passing through C.G. and parallel to the base: Using parallel axis theorem, IBC = IG + A 2 h 3 Ah2 bh3 bh h2 = − × 9 12 2 9 bh3 bh3 = − 12 18 bh3 IGG = 36 or IG = IBC − Hence, Moment of Inertia of Semi-circular Lamina about its Centroidal Axis IG = 0.00686d 4 X R G X 4R 3π d Figure 8.11a Moment of inertia of a quarter of a circle: IXX = 0.00343d 4 X G R d 2 Figure 8.11b @seismicisolation @seismicisolation X 4R 3π Moment of Inertia • 117 E XAMPLE 8.1: Find the IXX for the figure shown: S OLUTION : 100 mm 10 mm 15 mm 130 mm X X α= 100–15 = 42.5 mm 2 α 10 mm 100 mm Figure 8.12 In this case we can find IX by subtracting IX of hollow figure in to IX of whole figure (100 × 130 mm). 100 × 1303 42.5 × 1103 Required IX = − ×2 12 12 = 18308333.3 − 9427916.7 = 8880416.6 mm4 Ans E XAMPLE 8.2: Find the second moment of area and the radius of gyration about axis XX passing through centroid shown in figure. (The section is symmetrical about Y -axis.) S OLUTION : 80 20 mm I 20 2 X C X 120 mm Y 3 A 30 mm B 160 mm Figure 8.13 @seismicisolation @seismicisolation 118 • Strength of Materials S OLUTION : To find centre of gravity, divide the section in the three simple sections. a1 = 80 × 20 = 1600 mm2 , y1 = 160 mm a2 = 120 × 20 = 2400 mm , y2 = 90 mm a3 = 160 × 30 = 4800 mm2 , y3 = 15 mm 2 A = a1 + a2 + a3 = 1600 + 2400 + 4800 = 8800 mm2 1600 × 160 + 2400 × 90 + 4800 × 15 ȳ = 8800 = 61.82 mm, Now using the parallel axis theorem, 80 × 203 20 × 1203 + 1600 (160 − 61.82)2 + IXX = 12 12 160 × 303 + 4800 (15 − 61.82)2 + 2400 (90 − 61.82) + 12 = [53333.3 + 15422899.8 + 2880000 + 67632 + 360000 + 10522139.5] 2 = 29306004.6 mm4 Now I = Ak Ans 2 where k is radius of gyration I 29306004.6 = k= A 8800 = 57.71 mm Ans ∴ E XAMPLE 8.3: Find the IX and IY of the Fig. 8.14 shown below: 5 mm 80 mm X Let us take x & y from AB & AC. Y C x G X y 1 2 A 5 mm B 45 mm Y Figure 8.14 @seismicisolation @seismicisolation Moment of Inertia • 119 Let us divide the area in to two, i.e., 1 & 2 a1 = 80 × 5 = 400 a2 = 40 × 5 = 200 x1 = 2.5, y1 = 40 x2 = 25, y2 = 2.5 Total area = 400 + 200 = 600 mm2 400 × 2.5 + 200 × 2.5 = 10 mm x̄ = 600 400 × 40 + 200 × 2.5 ȳ = = 27.5 mm 600 Using parallel axis theorem, IX = IG + Ah2 5 × 803 40 × 53 IX = + 400 (40 − 27.5)2 + + 200 (2.5 − 27.5)2 12 12 = [213333.3 + 62500 + 416.67 + 125000] = 401249.97 = 401250 mm4 Ans 80 × 53 5 × 403 + 400 (10 − 2.5)2 + + 200 (10 − 25)2 IY = 12 12 = [833.33 + 22500 + 26666.7 + 45000] = 950000 mm4 Ans E XAMPLE 8.4: Determine the moment of inertia of a plate with a circular hole (as shown in Fig. 8.15) about X-axis passing through its centroid. Y 1 G X 2 50 mm y 40 mm 80 mm Y Figure 8.15 @seismicisolation @seismicisolation X 180 mm 120 • Strength of Materials Let total area including hole is 1 and hole alone is 2 a1 = 180 × 80 = 14400, a2 = π 2 50 = 1962.5, 4 y1 = 90 y2 = 40 A = 14400 − 1962.5 = 12437.5 mm2 ȳ = 14400 × 90 − 1962.5 × 40 = 97.89 mm 12437.5 80 × 1803 I¯X1 = = 38880000 mm4 12 Moment of inertia of the area a1 about the centroidal x-axis of the shaded figure, G. IX = (I¯X )1 + a1 d 2 = 38880000 + 14400 (90 − 97.89)2 IX = 38880000 + 896430.24 = 39776430 mm4 π Moment of inertia of the area a2 about its own G, I¯X2 = (50)4 64 I¯X 2 = 306640.6 Moment of inertia of the area a2 about the centroidal axis (Ix )2 = (I¯X )2 + a2 d 2 = 306640.6 + 1962.5(97.89 − 40)2 = 306640.6 + 6576832.2 = 6883472.8 Moment of inertia of composite (shaded) figure, Ix = I X − (I X )2 = 39776430 − 6883472.8 = 32892957.2 mm4 @seismicisolation @seismicisolation Ans Moment of Inertia E XAMPLE 8.5: Determine IX and IY of the cross section of iron beam shown in Fig. 8.16. Y 12 cm A 5 cm X G G′ X B 15 cm Y Figure 8.16 IX = IX of rectangular 12 × 15 − IX of circle of radius 5 cm 12 × 15 π = − (10)4 12 64 = 3375 − 490.87 = 2884.13 cm4 Ans Similarly, Iy = Iy of rectangle − Iy of semicircular portions Iy of rectangle = 15 × 123 = 2160 cm4 12 For the semicircle part ACB, IAB = 1 π × (10)4 = 245.3 cm4 2 64 4r 4×5 = = 2.12 cm 3π 3π 1 1 Area A = π r2 = π × 52 = 39.27 cm2 2 2 Using IAB = IGG + Ah2 Distance of G of semicircle from AB = IGG = 245.3 − 39.27 (2.12)2 = 68.8 cm4 Again from theorem, IYY = IGG + Ah21 h1 = 6 − 2.12 = 3.88 cm ∴ IYY = 68.8 + 39.27 × 3.882 = 659.98 cm4 = 660 cm4 app. Because there are two semicircular cutouts, ∴ IYY for both = 2 × 660 = 1320 IY for section = 2160 − 1320 = 840 @seismicisolation @seismicisolation Ans • 121 122 • Strength of Materials Product of Inertia of Rectangle Y b Consider on elementary area dA = dx.dy dy dx Product of inertia of the element = xydA = xy.dx.dy x y For whole rectangle (b × h), the product of inertia, h b h = X 0 Figure 8.17 b xydxdy = 0 2 b x = 2 y2 × 2 0 0 h = 0 xdx × b y. dy 0 b2 h2 b2 h2 × = 2 2 4 Note: If a section in symmetrical about an axis, its product of inertia about that axis is zero. E XAMPLE 8.6: Find the product of inertia of the hatched area about xy shown in Fig 8.18. 60 mm 15 mm 55 mm X X With respect to x and y-axes, product of inertia b2 h2 of rectangle in 4 Now, product of inertia of the hatched area = Product of inertia of rectangle 60 × 55 − Product of inertia of rectangle 45 × 40 602 × 552 452 × 402 − = 2722500 − 810000 4 4 = 1912500 mm4 Ans 15 mm = Figure 8.18 Exercise 8.1 Determine the moment of inertia about a target X − X of semicircular of dia 4 cm as shown in Fig. 8.19. Y X X o 4 cm [Ans: 10.15 cm4 ] Figure 8.19 @seismicisolation @seismicisolation Moment of Inertia • 123 8.2 Find the moment of inertia of the section shown in Fig. 8.20, about the centroidal axis X − X perpendicular to web. 10 cm 2 cm 2 cm 10 cm 2 cm 20 cm [Ans: 2666.67 cm4 ] Figure 8.20 8.3 Find the moment of the area shown in Fig. 8.21 about edge AB. R=10 cm 25 cm A B 20 mm [Ans: 35359.7 cm4 ] Figure 8.21 8.4 Calculate the second moment of area of an equal sized trapezium as shown in Fig. 8.22 about the centroidal axis parallel to the base. C 40 mm D 30 mm A 80 mm B [Ans: 130909.1 cm4 ] Figure 8.22 @seismicisolation @seismicisolation • 124 Strength of Materials 8.5 Calculate second moment of area for the Fig. 8.23 about the centroidal axis parallel to the AB. 100 mm 30 mm 15 mm 70 mm 20 mm A B 50 mm [Ans: 802.885 × 104 mm4 ] Figure 8.23 8.6 Find the second moment of area of Fig. 8.24 about centroidal axis. 12 cm 5 cm 5 cm 20 cm 5 cm 24 cm [Ans: 70086 cm4 ] Figure 8.24 8.7 Determine moment of inertia IXX and IYY of section shown in Fig. 8.25. 4 cm 2 cm G 3 8 cm 3 cm 2 cm [Ans: 448 cm4 , 58 cm4 ] 4 cm Figure 8.25 @seismicisolation @seismicisolation C HAPTER 9 BENDING OF BEAMS The shearing forces and bending moments at the various sections of a loaded beam give rise to shearing and bending stresses in the beam and, by making a number of assumptions, the relationship existing between the stresses, the strains, the bending moment, the curvature as well as the elasticity of the beam will be established in the following sections of this chapter. If a beam ABCD of uniform section is subjected to bending moment M the beam will bend with radius R, subtending an angle θ at sector P OQ . After bending the beam, has taken shape as A BC D and centroidal axis PQ as P Q . Now upper longitudinal layer is subjected to compression and has become A B . The lowermost layer DC has become DC and is in tension. Since at top it is compression and at bottom it is tension, so obviously at centroidal axis, there will be no stress and is called neutral axis. Here P Q is neutral axis and layers of fibres here suffer no stress or strain due to bending. Consequently, its original length PQ remain, unchanged, i.e., PQ = P Q . Before we derive any relationship between stresses, we have to make a number of assumptions which are given below: i) ii) iii) iv) v) The material obeys Hooke’s law and is within elastic limit. The beam is initially straight and unstressed. The material is homogeneous and isotropic. The beam bends in the form of a circular arc. The value of modulus of elasticity E for the material of the beam has the same value in compression as in tension. vi) Plane transverse sections of the beam before bending remain plane after bending. Relationship Between Curvature and Strain After application of bending moment M, the radius of curvature R of the bent beam is measured from the centre O to the neutral surface. If θ is the angle is radius subtended by the arc A B at centre O, then since the neutral surface remains unchanged in length then, Line PQ = Arc P Q P Q = Rθ @seismicisolation @seismicisolation 126 • Strength of Materials Now let us consider the bottom layer of the beam distance y from the neutral surface. Length of DC before bending = PQ = P Q = Rθ Length of DC after bending = DC = (R + y)θ ∴ Extension of DC = (R + y)θ − Rθ = yθ yθ Extension = Original length Rθ y ε= R Strain in DC, ε = ∴ Also, strain in DC, ε = σ E ∴ y σ E σ = ; = E R y R (i) Since, Young’s modulus E is constant for the material of the beam, and R is also constant for the particular curvature considered, the stress σ varies across the depth of the beam from zero at the neutral axis XX to a maximum value at the outer layers of fibres. Since, we are only considering sections that are symmetrical about the neutral axis XX, the distance y from the neutral axis XX to the outer layers of fibres is equal to half the overall depth of the beam. The stress distribution is drawn in Fig. 9.1 where compressive stresses are plotted to the left of axis Y −Y , and tensile stresses to the right. Figure 9.1 Moment of Resistance The moment of resistance of a beam is the resultant moment about the neutral axis of the internal tensile and compressive forces resisting the applied bending moment M. @seismicisolation @seismicisolation Bending of Beams • 127 Referring again to Fig. 9.1, let us consider an elemental strip of the cross section of the beam at a distance x from the neutral axis XX and of thickness dx. If σ1 is the stress at the strip then, since the stresses at the various layers of the beam section are proportional to their distances from the neutral axis XX. σ1 x = ; σ y x σ1 = σ y ∴ σ is the stress at y. x Force (F) on strip = Stress × Area = σ × b dx y Moment of this force about the neutral axis XX = Force × Distance from the neutral axis XX x σ = σ × b dx x = bx2 dx y y E 2 bx dx R The total moment of resistance is the sum of the moments of all the elemental stripes above and below the neutral axis (NA), XX, i.e., from x = y = d/2 to x = y = −d/2 and is given by: From Eqn. (i), = +d/2 M= −d/2 M= i.e., M = The quantity E 2 bx dx R E bx R 3 ∴ E M= R +d/2 bx2 dx −d/2 3 +d/2 −d/2 E bd 3 × R 12 bd 3 is called the second moment of area of the section about NA and is denoted by I. 12 ∴ E ×I R M E = I R M= (ii) Combining Eqn. (i) and (ii) M E σ = = I R Y This is known as the simple bending equation. Note: If we had taken any other section, for example, circular, the result would have been same. Only I has to be determined for a particular section. @seismicisolation @seismicisolation 128 • Strength of Materials Modulus of Section M σ = I y I or M = σ y Since y represents the distance of the outer layer of fibres from NA, the ratio of the second moment of area (moment of inertia) I to y is called modulus of section and is denoted by τ . Thus, Z = I y ∴ M = Zσ (iii) Equation (iii) is very important since it enables us to determine readily the permissible or maximum bending moment on a beam. Since the beam of Fig. 9.1 is of rectangular section. ∴ bd 3 12 d 2 bd 3 2 bd 2 I × = Modulus of section, Z = = Y 12 d 6 I= y= and Note: EI is known as flexural rigidity. E XAMPLE 9.1: A mild steel bar of 35 mm × 15 mm section and 1.2 m length is simply supported at its ends with the 35 mm edge horizontal. If a load of 100 N is applied at the centre of the bar, determine the maximum stress, produced in the material due to bending. S OLUTION : 100 N 0.6 m 15 mm A 35 mm 50 N B 1.2 m 50 N Figure 9.2 bd 3 35 × 153 = = 9843.75 mm4 12 12 100 Reactions, RA = RB = = 50 N 2 I= M = 50 × 0.6 = 30 Nm = 30000 Nmm @seismicisolation @seismicisolation Bending of Beams • 129 Maximum stress will occur at 7.5 mm from NA. y = 7.5 mm Now, M M σ = ∴ σ= y I y I 30000 σ= × 7.5 = 22.86 N/mm2 = 22.86 MN/m2 9843.75 E XAMPLE 9.2: A tubular frame member of 85 mm external diameter is subjected to a bending moment of 600 Nm. If the stress set up in the material due to bending is 50 MN/m2 , find the internal diameter of the member. S OLUTION : Let D and d be the external and internal diameters of the tubular frame. Then, π (D4 − d 4 ) , σ = 50 MN/m2 = 50 N/mm2 64 M σ 85 = ; Here y = = 42.5 I y 2 I= M = 600 Nm = 600000 Nmm, I= π 2 (854 − d 4 ) π (52200625 − d 4 ) = 64 64 Substituting 600000 × 64 50 = π (52200625 − d 4 ) 42.5 π (52200625 − d 4 ) × 50 = 600000 × 64 × 42.5 52200625 − d 4 = 600000 × 64 × 42.5 π × 50 = 10385454.5 or d 4 = 52200625 − 10385454.5 ∴ ∴ d 4 = 41815170.5 d = 80.41 mm. Ans E XAMPLE 9.3: A beam, the cross-section of which is shown in Fig. 9.3, acts as a cantilever, which projects 1.8 m from the wall. The cantilever carries a load of 5 kN at the free end. Calculate the maximum bending stress. @seismicisolation @seismicisolation • 130 Strength of Materials 75 mm 10 mm 10 mm 85 mm 10 mm Figure 9.3 S OLUTION : Since the beam cross section is symmetrical about X and Y axis, therefore I can be obtained as follows: 5 kN 75 × (85 + 20)3 (75 − 10)(85)3 − 12 12 75 × 1157625 65 × 614125 − = 12 12 I= 1.8 m Figure 9.4 = 7235156.25 − 3326510.42 I = 3908645.83 mm4 M = 5 × 1000 × 1.8 × 1000 = 9000000 mm4 y = 10 + Now, 85 = 52.5 2 M σ = I y ∴ σ= M y I σ= 9000000 × 52.5 = 120.89 N/mm2 3908645.83 or σ = 120.89 MN/m2 . Ans E XAMPLE 9.4: A spring steel strip, 25 mm wide and 1.5 mm thick, is bent to an arc of a circle of 2 m radius. Calculate the bending moment necessary and the maximum stress set up. E for steel = 200 GN/m2 . S OLUTION : b = 25 mm; ∴ I= d = 1.5 mm bd 3 25 × 1.53 7.03 = = 7.03 mm4 = 12 m4 12 12 10 @seismicisolation @seismicisolation Bending of Beams Distance from N.A. to extreme layer of fibres } = y = • 131 1.5 d = mm 2 2 0.75 m 103 EI M E = OR M = I R R 9 m4 200 × 10 × 7.03 N × M= m 2 × 1012 m2 y = 0.75 mm = Now ∴ Bending moment = 0.703 Nm. For maximum stress Ey 200 × 109 × 0.75 N m = σ= × R 2 × 103 m2 m σ = 75 × 106 N/m2 = 75 MN/m2 . E XAMPLE 9.5: The beam cross section shown in Fig. 9.5 is subjected to a bending moment of 12 kNm. Calculate the maximum stress and its nature indicating its place. 110 mm C D 20 mm 20 mm 100 mm 25 mm A B 180 mm Figure 9.5 110 mm Y C 20 mm y D 20 mm 1 G 2 100 mm 3 A B Y 180 mm Figure 9.6 @seismicisolation @seismicisolation 25 mm 132 • Strength of Materials We see that the section is symmetrical about Y −Y axis. Let the centre of gravity G be at Y from AB. In order to find G, let us divide the section into three simple rectangles, i.e., 1, 2 & 3. a1 = 110 × 20 = 2200 mm2 ; y1 = 135, a2 = 100 × 20 = 2000 mm2 ; y2 = 75; a3 = 180 × 25 = 4500 mm2 ; y3 = 12.5 mm Total area, A = 2200 + 2000 + 4500 = 8700 mm2 y= ∴ 2200 × 135 + 2000 × 75 + 4500 × 12.5 503250 = 8700 8700 y = 57.84 mm Now to find I of the whole section, using parallel axis theorem: 110 × 203 2 + 2200 (135 − 57.84) I= 12 20 × 1003 2 + 2000 (75 − 57.84) + 12 180 × 253 2 + + 4500 (12.5 − 57.84) 12 = [73333.3 + 13098064.3] + [1666666.7 + 588931.2] + [234375 + 15054595.2] ∴ I = 30715965.7 mm4 M σ = I y yt = 57.84, yc = (100 + 25 + 20) − 57.84 = 87.16 mm t for tension, c for compression M y I 12000000 × 57.84 = 22.596 N/mm2 σt = 30715965.7 σ= = 22.6 MN/m2 (tensile at AB) σc = 12000000 × 87.16 = 34.05 N/mm2 30715965.7 = 34 MN/m2 (compressive at CD) Hence, maximum stress is 34 MN/m2 at CD of compressive nature. Ans @seismicisolation @seismicisolation Bending of Beams • 133 E XAMPLE 9.6: The horizontal beam of channel cross section as shown in Fig. 9.7 is 2.8 m long and is simply supported at the ends. Determine the maximum uniformly distributed load it can carry if the tensile and compressive stresses must not exceed 35 and 52 N/mm2 , respectively. Neglect the weight of channel. 2 1 y1 G 110 mm 20 20 mm 3 y y2 20 mm 110 mm Figure 9.7 In order to find position of G, let us divide the channel in three simple rectangles. Being symmetrical about Y axis, we have to only find y. a1 = 90 × 20 = 1800; y1 = 65 mm; a2 = 1800 mm2 , y2 = 65 mm, a3 = 150 × 20 = 3000 mm2 ; y3 = 10 mm. Total area, A = 1800 + 1800 + 3000 = 6600 mm2 y= 1800 × 65 + 1800 × 65 + 3000 × 10 = 40 mm 6600 y = Y2 = 40 mm, y1 = 110 − 40 = 70 mm 20 × 903 2 I= + 1800 (65 − 40) 12 150 × 203 2 + 3000 (10 − 40) ×2+ 12 = [1215000 + 1125000] × 2 + [100000 + 2700000] = 4680000 + 2800000 = 7480000 mm4 σ= M y I M = 1.4w × 1.4 2 = 0.98w Nm w/m 2.8 m 1.4 w 1.4 w Figure 9.8 @seismicisolation @seismicisolation 134 • Strength of Materials Since y1 > y2 , so stress will be greater at y1 of 52 = 0.98w × 1000 × 70 . . . (i) 7480000 ∴ w = 5669.97 N/m Now taking lower side, i.e., y2 = 40 35 = 0.98w × 1000 × 40 . . . (ii) 7480000 ∴ w = 6678.6 N/m Obviously, if we choose 6.678 kN/m uniformly distributed load then stress in equation (i) will exceed 52 N/mm2 which is not permissible. Hence, safe uniformly distributed load (w) = 5.67 kN/m Ans. E XAMPLE 9.7: A rectangular beam is to be cut from a circular log of wood of diameter D. Find the dimensions of the strongest section in bending. S OLUTION : M σ = I y or M = σ I = σZ y Let the strongest section cut out of the circular log of diameter D be of width b and depth d. The section modulus for it is: Z= Also b2 + d 2 = D2 d D bd 2 6 or b Figure 9.9 d 2 = D2 − b2 Z= b (D2 − b2 ) b D2 − b3 = 6 6 For strongest section Z should be maximum. Therefore, dZ = 0. db 1 2 (D − 3b2 ) = 0 6 D2 − 3b2 = 0 b2 = But 3b2 = D2 or D2 3 or D b= √ 3 d 2 = D2 − b2 = D2 − @seismicisolation @seismicisolation D2 2D2 = 3 3 Bending of Beams For strongest section d = D 2 3 • 135 Ans E XAMPLE 9.8: A horizontal cantilever 3.2 m long is having rectangular cross section 65 mm wide throughout its length. Its depth varies from 55 mm at free end to 200 mm at fixed end. A load of 6 kN at free end is applied, determine the position of maximum stress induced in the section. Also find the value of maximum bending stress. S OLUTION : 6 kN X 65 mm 65 mm x 200 mm 55 mm 200 mm X 3.2 m Fixed end 55 mm Free end Figure 9.10 Let us consider X − X section at a distance of x from the free end. dx, depth at section X − X = 55 + Sectional modulus Z = 145x 176000 + 145x = 3200 3200 1 bdx2 65 = (176000 + 145x)2 × 6 6 32002 Z = (176000 + 145x) × 1.058 × 10−6 σ= = M 6000x = Z (176000 + 145x)2 × 1.058 × 10−6 6000 × 0.945 × 106 x 5670 × 106 x = 2 (176000 + 145x) (176000 + 145x)2 For maximum stress dσ d = dx dx = 5670 × 106 x (176000 + 145x)2 5670 × 106 (176000 + 145x)2 + 2(176000 + 145x)145 × 5670 × 106 x (176000 + 145x)4 @seismicisolation @seismicisolation 136 • Strength of Materials 176000 + 145x = 290x x = 1213.8 mm The maximum stress will be at 1213.8 mm from the free end. Ans For maximum stress: σ= (Substituting for x) = 6000x (176000 + 145x)2 × 1.058 × 10−6 6000 × 1213.8 × 0.945 × 106 6882246 × 106 = (176000 + 145 × 1213.8)2 1.24 × 1011 = 55.5 N/mm2 = 55.5 MN/m2 Ans Beams of Uniform Strength In simply supported beams, carrying a uniformly distributed load, the maximum bending moment will occur at its centre. As we go near the supports, the bending moment reduces untill it become zero. Therefore, it is in interest to save the material near supports. Lot of material can be saved by designing beams of uniform strength. Naturally, we will have to reduce the section gradually untill supports. The section of a beam of uniform strength may be varied in the following ways: a) By keeping the width uniform and varying depth b) By keeping the depth uniform and varying width c) By varying both width and depth. Generally, uniform strength is maintained by keeping the width uniform and varying the depth. E XAMPLE 9.9: A simply supported beam of 2.6 m span has a constant width of 120 mm throughout its length with varying depth of 160 mm at the centre to minimum at the ends as shown in Figure 9.11. The beam is carrying a concentrated load P at its centre. P 0.5 m X A C dx X B 2.6 m Figure 9.11 Find the minimum depth of the beam at a section 0.5 m from the left hand support, such that the maximum bending stress at this section is equal to that at the mid-span of the beam. @seismicisolation @seismicisolation Bending of Beams • 137 S OLUTION : Width = 120 mm, span = 2.6 m = 2600 mm depth at centre dc = 160 mm. Let depth at 0.5 m from A be dx at X and σx be the bending stress at X. P Reaction RA = RB = 2 P Bending moment at C, Mc = × 1300 = 650 P 2 P and bending moment at X, Mx = × 500 = 250 P 2 Section modulus at centre of beam, Zc = b.dc2 120 (160)2 = = 512000 mm3 6 6 Zx = b.dx2 120 dx2 = = 20 dx2 mm3 6 6 Bending moment at C (Mc ), 650P = σc × Zc = σc × 512000 σc = 650P 512000 (i) Similarly, for stress at X, Bending moment at X, 250P = σx × Zx = σx × 20 dx2 250P 12.5P = 20 dx2 dx2 σy = for uniform strength, σc = σx Equating (i) and (ii) 12.5P 650P = 512000 dx2 dx2 = ∴ 12.5 × 512000 = 9846 650 dx = 99.23 mm Ans @seismicisolation @seismicisolation (ii) 138 • Strength of Materials Composite Beams or Flitched Beams A composite beam is one which consists of two or more materials rigidly fixed together throughout their length. Generally, wooden beams are reinforced with steel plates to make them stronger. The reinforcing material should have higher modulus of elasticity then the material of the beam. Some examples are given in Figure 9.12. Composite beam Figure 9.12 If M1 and M2 are the parts of the applied bending moment M carried by the two materials, then M1 + M2 = M (i) Here the radius of curvature of the two materials will be same as they are fixed together, R1 = R2 E1 I1 E2 I2 = M1 M2 or M1 E1 I1 = M2 E1 I2 (ii) M1 and M2 can be determined from Eqns. (i) and (ii) and the stresses in the two materials are then given by M1 M2 σ1 = and σ2 = Z1 Z2 Alternately, the composite section may be replaced by an equivalent homogeneous section. Thus, the section shown in Fig. 9.13(a) is equivalent to the section in Fig. 9.13(b) composed entirely of E2 = n. material (1) or to the section in Fig. 9.13(c) composed entirely of material (2). Where E1 (2) (1) (2) b b (a) nb (b) b n b (c) Figure 9.13 In this method first I of the equivalent section is obtained. Then stresses may be obtained from M σ = y, these being the stresses which would exist in the homogeneous section. In section (b), the I @seismicisolation @seismicisolation Bending of Beams • 139 actual stresses in material (2) would be n times those in the equivalent section and in section (c), the actual stresses in material (1) would be 1/n times those in the equivalent section. In this analysis, it has been assumed that the compound sections are all symmetrical about the plane of bending, otherwise twisting of the section would occur. E XAMPLE 9.10: A flitched beam consists of a wooden joist 110 mm wide and 220 mm deep reinforced by two steel plates 12 mm thick and 220 mm deep as shown in Fig. 9.14. If the maximum stress in wooden joist is 10 N/mm2 , determine the corresponding stress induced in steel. Also find the moment of resistance of composite section. E for steel = 200 GN/m2 and E for wood = 12 GN/m2 . S OLUTION : Modular ratio m = 200 Es = = 16.7 Ew 12 220 mm Maximum stress in wood = 10 N/mm2 Corresponding maximum stress in steel, 2 σs = mσw = 16.7 × 10 = 167 N/mm 12 mm Ans 110 mm 12 mm Figure 9.14 Total moment of resistance = Moment of resistance of wood + Moment of resistance by steel = σw Iw σs Is + y y Moment of inertia of wood = Iw = 110 × 2203 = 97606666.7 mm4 12 12 × 2203 = 21296000 mm4 12 σw 10 × Iw = × 97606666.7 Moment of resistance by wood = Mw = y 110 Moment of inertia of steel = Is = 2 × = 8873333.3 Nmm Moment of resistance by steel, Ms = = σs × Is y 167 × 21296000 = 32331200 Nmm 110 Total moment of resistance = Mw + Ms = 8873333.3 + 32331200 = 41204533.3 Nmm = 41.2 × 106 Nmm @seismicisolation @seismicisolation Ans 140 • Strength of Materials Alternative method Equivalent width in terms of steel 30.59 mm bw 110 = = 6.59 mm m 16.7 bs = 220 mm bd 3 30.59 × 2203 = 12 12 Moment of inertia, I = 12 6.59 Equivalent steel section 12 4 = 27143526.7 mm Figure 9.15 Moment of resistance of equivalent steel section M= σs 167 I= × 27143526.7 = 41208809 Nmm y 110 = 41.2 × 106 Nmm Ans Note: Same as before. Another alternative method Equivalent width of steel in terms of wood, 510.8 mm bw = mbs = 16.7 × 12 = 200.4 mm Moment of inertia I = 220 mm bd 3 510.8 × 2203 = 12 12 = 453249867 mm4 200.4 mm 110 mm 200.4 mm Equivalent wooden section Figure 9.16 Moment of resistance of equivalent wooden section, M= 10 × 453249867 σw ×I = = 41.2 × 106 Nmm Ans y 110 Note: Same as before. E XAMPLE 9.11: A wooden beam 120 mm wide and 180 mm deep is to reinforced by two plates 120 mm × 10 mm and 100 × 5 mm. The thicker plate is secured to the top and thinner one to the bottom surface as shown in Fig. 9.17. The permissible stress in steel is 140 MPa and the value of modular ratio is 18. Calculate moment of resistance of strengthened section. Also determine maximum stress in wood (timber). @seismicisolation @seismicisolation Bending of Beams • 141 1 10 mm N 10 mm A 180 mm G 180 mm 6.7 mm 3 2 5 mm B A 120 mm Flitched beam y = 114.46 mm 5 mm Equivalent steel section Figure 9.17 Let us find the equivalent steel section of flitched beam. Equivalent width of timber in steel = 120 bt = = 6.7 mm m 18 To find position of neutral axis, first find the centre of gravity of equivalent steel section from AB. For that we split equivalent steel section in three simple rectangles 1, 2 & 3. a1 = 120 × 10 = 1200 mm2 ; y1 = 190 mm; a2 = 180 × 6.7 = 1206 mm2 ; y2 = 95 mm; a3 = 120 × 5 = 600 mm2 ; y3 = 2.5 mm. Total area = 1200 + 1206 + 600 = 3006 mm2 1200 × 190 + 1206 × 95 + 600 × 2.5 = 114.46 mm 3006 120 × 103 2 + 1200(190 − 114.46) I= 12 6.7 × 1803 2 + + 1206(95 − 114.46) 12 120 × 53 2 + + 600(2.5 − 114.46) 12 Y (from AB) = = [10000 + 3423775] + [3256200 + 456702] + [1250 + 7521025] = 14668952 mm4 = 14.67 × 106 mm4 M = σs × I ymax = 140 × ; y from AB = 114.46 is max 14668952 = 17942104 Nmm 114.46 = 17.94 kNm Ans @seismicisolation @seismicisolation 142 • Strength of Materials Maximum stress in timber will exist at the top of bottom flange, so ytmax = 114.6 − 5 = 109.6 mm σtmax = M 17.94 × 106 × 109.6 ×Ytmax = I 14.67 × 106 = 134.03 N/mm2 Actual σtmax = 134.03 = 7.45 N/mm2 = 7.45 MPa. Ans 18 Combined Bending and Direct Stresses P In Fig. 9.18, a short column is shown with eccentric loading. In this case there will be two stresses (i) direct P.e P × distance of load stress and (ii) bending stress i.e., A I from the neutral axis. e e Figure 9.18 A combination of bending and direct stresses may occur in a variety of circumstances but in every case, the stresses due to bending moment and direct load may be calculated separately and the results combined to give the resultant stresses. Thus σ = σd ± σb where σd and σb are the direct and bending stresses. The shape of the resultant stress distribution diagram will depend on whether σb is greater or less than σd . Figure 9.18 shows a bar which is subjected to an axial load P and a bending moment due to load being eccentric. Figures 9.19(a), (b) and (c) show the possible forms of the resultant stress distribution. P a M P < Z a (a) M Z P a M Z M P = Z a (b) Figure 9.19 @seismicisolation @seismicisolation P a M Z M P > Z a (c) Bending of Beams • 143 E XAMPLE 9.12: A steel flat 180 mm wide and 30 mm thick is subjected to a pull of 180 kN, which is off the geometrical axis by 4 mm in the plane which bisects the thickness. Determine the maximum and minimum stresses induced in the section. S OLUTION : 4 mm P 30 mm 180 mm 28.89 MPa 37.77 MPa Figure 9.20 P = 180000 N A = 180 × 30 = 5400 mm2 , P = 180000 N M = P.e = 180000 × 4 = 720000 Nmm Z= bd 2 30 × 1802 = = 162000 mm3 6 6 σd = P 180000 = = 33.33 N/mm2 A 5400 σb = M 720000 = = 4.44 N/mm2 Z 162000 σmax = σd + σb = 33.33 + 4.44 = 37.77 N/mm2 = 37.77 MPa σmin = σd − σb = 33.33 − 4.44 = 28.89 N/mm2 = 28.89 MPa E XAMPLE 9.13: A mild steel T section as shown in Fig. 9.21 carries a load of 100 kN in the central plane bisecting the web at 45 mm from the base. Determine the maximum and minimum stresses induced in the section. @seismicisolation @seismicisolation 144 • Strength of Materials 180 mm A 10 mm 1 B y G 45 mm 120 mm 2 10 mm Figure 9.21 Let us first find the G of section from AB, a1 = 180 × 10 = 1800; y1 = 5 mm; a2 = 110 × 10 = 1100 mm2 ; y2 = 65 mm Total area = 1800 + 1100 = 2900 mm2 1800 × 5 + 1100 × 65 = 27.76 mm = yt ; 2900 180 × 103 2 I= + 1800(27.76 − 5) 12 10 × 1103 2 + + 1100(27.76 − 65) 12 y= yc = 92.24 = [15000 + 932432] + [1109167 + 1525499] = 3582098 mm4 Eccentricity of loading, e = 45 − 27.76 = 17.24 mm M = P.e = 100000 × 17.24 = 1724000 Nmm M.yc 1724000 × 92.24 = = 44.39 MPa I 3582098 1724000 × 27.76 M.yt = = 13.36 MPa σt = I 3582098 P 100000 = 34.48 MPa σd = = A 2900 σmax = σd + σc σmax = 34.48 + 44.39 = 78.87 MPa Ans σmin = σd − σt σc = σmin = 34.48 − 13.36 = 21.12 MPa Ans @seismicisolation @seismicisolation Bending of Beams • 145 Modulus of rupture: If a metallic beam, simply supported at its ends, is loaded by a gradually increasing transverse central load till rupture or breaking takes place, the actual stresses at the outer layers at rupture σ M = , since the condition or breaking, are not the ones calculated by the bending equation: I Y of elasticity assumed in this formula has ceased to exist. However, the stress calculated from the equation: M M y = = σ or say σr , is known as transverse rupture stress or modulus or rupture. Since, this I Z test is generally applied to different qualities of cast iron or timber beams, the modulus of rupture σr so calculated is called a guide to compare the strength of various qualities of these materials. This test is conducted on brittle materials. E XAMPLE 9.14: A 150 mm × 150 mm pine wood beam was supported at its ends on a 4.5 m span and loaded as shown in Fig. 9.22. The beam failed when a 8 kN load was placed at 1.5 m from each end. Find the modulus of rupture. 8 kN 1.5 m A 8 kN 1.5 m C 1.5 m B D RA RB Figure 9.22 Beam is of square section, side = 150 mm = b Section modulus = = b3 6 1503 = 56.25 × 10 mm4 6 Reactions RA = RB = 8 kN Portion CD of the beam is under pure bending, MC = MD = 8 × 1.5 = 12 kNm So maximum, M = 12 × 106 Nmm Modulus of rupture, σm = = M Z 12 × 106 56.25 × 104 = 21.33 N/mm2 @seismicisolation @seismicisolation Ans 146 • Strength of Materials Note: The theoretical value of the fracture stress obtained by flexure formula using ultimate bending moment is known as modulus of rupture. Conditions for no Tension in the Section P M > , then as per As we have seen in eccentric loading when load is of compressive nature, if Z A stress distribution diagram, stress changes sign, being somewhere tensile while compressive stress at other places. Now in masonry work, it is not desirable to have tensile stresses to avoid failure of structure. This limits eccentricity e to a certain value which we shall investigate for rectangular section and circular section. 1. Rectangular section: If a rectangular section ABCD is loaded at a point distant e along X − X axis and off the Y − Y axis, as shown in Fig. 9.23. Naturally, bending will take place about Y − Y axis. Here Compressive load P B A a Core of kernel of the section b 3 b e D C d d 3 Figure 9.23 bd 3 and Area, A = bd. 12 2 = AkYY Here IYY = Also IYY ∴ 2 = Equating the above two AkYY bd 3 12 2 = bd kYY bd 3 12 ∴ 2 kYY = d2 12 (i) Now, for no reverse stress, σd ≥ σb ≥ M Z P×e×d P P×e×d , we know for symmetrical section y = yt = yc = d/2. ≥ ≥ A 2I 2 A k2 or e ≤ 2k2 d where k is radius of gyration. @seismicisolation @seismicisolation (ii) Bending of Beams • 147 2k2 . Thus, for no tension in the section, the eccertricity must not exceed d Equation (ii) becomes with the help of (i) e≤ 2d 2 d ≤ d × 12 6 Therefore, the load should not be placed at a distance more than d/6 on either side of centroid on X − X axis to avoid any reverse stress occurs. d d d And therefore the limit of eccertricity (e) = + = . 6 6 3 Hence, the stress will be of the same sign throughout the section if the load line is within the middle third of the section. This is known as middle third rule for rectangular section. Likewise, if the load is placed on Y − Y axis, off the X-axis, middle on YY -axis which is b/3 is safe zone. If we join the four points of middle, third distances on XX and YY axis are joined, a rhombus of diamond shape is obtained as shown in Fig. 9.24. This portion is known as the core or kernel of the section. The reverse stress will not occur in any part of the entire rectangular section, if the load is placed anywhere inside he rhombus. 2. Circular section In this case IXX = IYY ∴ k2 = π d4 = Ak2 64 y π d 4 /64 d 2 = 16 π d 2 /4 2k2 e≤ d d 2 × d2 = e≤ d × 16 8 X d X We know that for no tension, d 4 ∴ e≤ d 8 y Figure 9.24 d d = . 8 4 Hence for load must fall within the middle fourth of the section. Hence, diameter of core or kernel = 2e = 2 × E XAMPLE 9.15: A short column of hollow cylindrical section 300 mm outside diameter and 150 mm inside diameter carries a vertical load of 500 kN at 80 mm away from the axis of the column. Determine maximum and minimum stresses and state their nature. @seismicisolation @seismicisolation 148 • Strength of Materials S OLUTION : Area, A = π 2 π (D − d 2 ) = (3002 − 1502 ) 4 4 80 mm 500 kN = 52987.5 mm2 P 500000 Direct stress, σd = = = 9.44 N/mm2 A 52987.5 P×e Bending stress, σ = Z Z= σb = π /64(3004 − 1504 ) I = = 2483789 mm3 y 300/2 500000 × 80 = 16.1 N/mm2 2483789 150 mm 300 mm Figure 9.25 Maximum stress = 9.44 + 16.1 = 25.54 N/mm2 = 25.54 MPa (Compressive) Ans Minimum stress = 9.44 − 16.1 = −6.66 N/mm2 = 6.66 MPa (Tensile) Ans E XAMPLE 9.16: A masonry column of 3.5 m × 4.3 m supports a load of 50 kN as shown in Fig. 9.26, i) determine the stresses developed at each corner of the column; ii) in order that there is no tension anywhere in the section of column, what additional load should be applied at the centre of column? iii) what are the stresses at the corners with the additional load in the centre? Here the load is applied Y A 1m B 0.5m X 3.5m D 4.3m Y C Figure 9.26 S OLUTION : i) Area, A = 4.3 × 3.5 = 15.05 m2 Load, P = 50 kN @seismicisolation @seismicisolation X Bending of Beams • 149 4.3 × 3.53 = 15.363 m4 12 3.5 × 4.33 = 23.2 m4 Moment of inertia about Y −Y , IY = 12 Moment of inertia about X − X, IX = eX = 1 m and eY = 0.5 m ∴ Moment Mx = P × 0.5 = 50 × 0.5 = 25 kNm My = P × 1 = 50 × 1 = 50 kNm 4.3 = 2.15 m 2 3.5 = 1.75 m Distance between XX axis and corners A & D = 2 Distance between YY axis and corners A & B = σA = P Mx × y My × x 50 25 × 1.75 50 × 2.15 − = + + − A IX IYY 15.05 15.363 23.2 = 3.32 + 2.848 − 4.63 = 1.538 kN/m2 σB = P Mx × y My × x 50 25 × 1.75 50 × 2.15 + = + + + A IX IY 15.05 15.363 23.2 = 3.32 + 2.848 + 4.63 = 10.798 kN/m2 σC = Ans P Mx × y My × x 50 25 × 1.75 50 × 2.15 + = − − + A IX IY 15.5 15.363 23.2 = 3.32 − 2.848 + 4.63 = 5.102 kN/m2 σD = Ans Ans 25 × 1.75 50 × 2.15 P Mx × y My × x 50 − − − − = A IX IY 15.5 15.363 23.2 = 3.32 − 2.848 − 4.63 = −4.158 kN/m2 = 4.158 kN/m2 (Tensile) Ans ii) Compressive stress due to additional load P , = P P = kN/m2 A 15.05 If there is to be no tension in the column section, the compressive stress due to P kN should be equal to the tensile stress, i.e., 4.158 kN/m2 . ∴ P = 4.158 Hence P = 62.58 kN 15.05 @seismicisolation @seismicisolation Ans 150 • Strength of Materials iii) Stresses at the corners with the additional load in the centre: Stress due to additional load = P 62.58 = = 4.158 kN/m2 = 4.16 kN/m2 app. A 15.05 σa = Stress at A = 1.538 + 4.16 = 5.698 kN/m2 Ans 2 σb = Stress at B = 10.798 + 4.16 = 14.958 kN/m 2 σc = Stress at C = 5.102 + 4.16 = 9.262 kN/m Ans Ans σd = Stress at D = −4.158 − 4.158 = 0 Ans E XAMPLE 9.17: Determine the maximum force that the screw on c-clamp can exert on the wooden block in Fig. 9.27, if the allowable stress at section A − A is not to exceed 83 MN/m2 . 250 mm A A 50 mm 12.5 mm Cross section A-A Figure 9.27 S OLUTION : Eccentricity of force, e = 250 mm Permissible stress, σt = 83 MN/m2 = 83 N/mm2 To find P exerted by the screw of c-clamp, Area of cross section at A − A, a = 50 × 12.5 = 625 mm2 Moment of inertia about N.A. at A − A, I = ∴ bh3 12.5 × 503 = 12 12 I = 13.0208 × 104 mm4 Bending moment due to P = P × 250 Nmm P P (i) Direct stress due to P, σd = = = 0.0016 P A 625 M P × 250 (ii) Bending stress, σb = y = × 25 = 0.048 P I 13.0208 × 104 @seismicisolation @seismicisolation • Bending of Beams 151 Total stress = σd + σb = 0.0016P + 0.048P = 0.0496P Total stress must not exceed allowable stress 83 N/mm2 ∴ 0.0496P = 83 ∴ P = 1673.38N = 1.673 kN Ans Exercise 9.1 A light wooden bridge is supported by six parallel timber beams, each 300 mm deep, and 200 mm wide. Each beam may be considered simply supported over a span on 4.5 m. If the allowable stress in the timber is 5.60 MN/m2 , calculate the greatest uniformly distributed load the bridge can support. [Ans 179.2 kN] 9.2 The crane beam shown in Fig. 9.28 is made up of two 25 mm thick steel plates each 400 mm deep at the middle section. Calculate the maximum allowable central load W if the span is 4.5 m and the ultimate tensile stress of the steel is 370 MN/m2 . Allow a factor of safety of 16. [Ans 73 kN] W Figure 9.28 9.3 Calculate the maximum bending moment which may be applied to the cast iron section shown in Fig. 9.29, if the permissible tensile stress of the material is 280 MN/m2 and a factor of safety of 10 is to be used. 100 mm 10 mm 120 mm 10 mm 10 mm 100 mm Figure 9.29 9.4 The bar of section shown in Fig. 9.30 is simply supported over a span of 1 m and carries a central load of 20 kN. Find the maximum bending moment and the maximum bending stress in the material. [I = 77.44 × 104 mm4 , 5 kNm, 194 MN/m2 ] @seismicisolation @seismicisolation 152 • Strength of Materials 20 mm rad 60 mm 50 mm Figure 9.30 9.5 A cast iron beam is in the shape of T section as shown in Fig. 9.31. The beam is simply supported on a span of 8 m. The beam carries a uniformly distributed load of 1.5 kN/m length on the entire beam span. Determine the maximum and minimum tensile and compressive stresses. 100 mm 20 mm 100 mm 20 mm [Ans Figure 9.31 123.1 MN/m2 ] 9.6 An I section beam as shown in Fig. 9.32, if the ratio of maximum compressive stress to tensile stress induced is 4:3, determine the dimension b assuming that the section is subjected to sagging moment. Show also the stress variation diagram across the section. Compressive side b 300 mm 20 mm 260 mm 200 mm 20 mm Tensile side Figure 9.32 [Ans b = 112.5 mm] 9.7 A composite beam is made by placing two steel plates, 12 mm thick and 240 mm deep, one each on both sides of a wooden section 90 mm wide and 240 mm deep. Determine the moment of resistance of the beam. Take Es /Ew = 15, the stress in the wood not to exceed 7 MPa. [Ans @seismicisolation @seismicisolation 30.24 Nm] Bending of Beams • 153 9.8 An uniformly tapering vertical post of height 10 m has a diameter of 150 mm at the top. A horizontal pull of 250 N is applied at the top of the post. Find the maximum bending stress for the post and state where it occurs. [Ans 1.12 N/mm2 , 5 m from top] 9.9 A timber beam 150 m wide is reinforced by a steel plate 100 mm wide and 10 mm deep attached at the lower face of the timber beam. Calculate the moment of resistance of the beam, if the allowable stresses in timber and steel are 6 MPa and 60 MPa respectively. Take Es = 16 Et . [Ans 9.45 kNm] 9.10 Prove that the moment of resistance of a beam of square section with diagonal in the plane of bending is increased by flattening the top and bottom corners, as shown in Fig. 9.33, and that 8 the moment of resistance in a maximum when, y = Y . 9 Y y Figure 9.33 9.11 A tie bar 75 mm wide and 25 mm thick sustains an axial load of 100 kN. What depth of metal may safely be removed from one of the narrow sides in order that the maximum stress over the reduced width may not exceed 100 MN/m2 . [Ans 12.05 mm] 9.12 A horizontal cantilever 3 m long is of rectangular cross section 60 mm wide throughout its length, the depth varying uniformly form 60 mm at the free end to 180 mm at the fixed end. A load of 4 kN acts at the free end. Determine the position of the most highly stressed section, and find the value of the maximum bending stress induced. Neglect the weight of the cantilever itself. [Ans 41.6 MN/m2 ] 9.13 A steel tube of 40 mm outside diameter and 30 mm inside diameter is used as a simply supported beam on a span of 1 m end. It is found that the maximum safe load it can carry at midpoint is 1.2 kN. Four of these tubes are placed parallel to another and firmly fixed together to form in effect a simple beam, the centres of the tubes forming a square of 40 mm side with one pair of centres vertically over the other pair. Find the maximum central load which this beam can carry if the maximum stress is not to exceed that of the single tube above. [Ans 8.64 kN] @seismicisolation @seismicisolation 154 • Strength of Materials 9.14 A short column has the cross section as shown in Fig. 9.34. An axial compressive force of 150 kN is applied at point F. Calculate the stress at each corner. 20 mm F 30 mm 200 mm [Ans σa = −7.5 N/mm2 (Compressive) σb = 23.2 N/mm2 (Tensile) σc = −9.8 N/mm2 (Compressive) 100 mm σd = 8.2 N/mm2 (Tensile)] Figure 9.34 9.15 A flitched beam consists of a wooden joist 120 mm wide and 200 mm deep, strengthened by a steel plates 10 mm thick and 180 mm deep on either side of the joist at centre. If the stresses in wood and steel are not to exceed 7.5 MN/m2 and 127.5 MN/m2 , find the moment of resistance of section of the beam. Take Es = 20 Ew . [Ans 19.436 kNm] 9.16 A steel bar 120 mm in diameter is completely encased in an aluminium tube of 180 mm outer diameter and 120 mm inner diameter so as to make a composite beam. The composite beam is subjected to a bending moment of 15 kNm. Determine the maximum stress due to bending in each material. Take Es = 3Eal . [Ans σs = 37.56 MN/m2 , σal = 18.78 MN/m2 ] 9.17 A cantilever of uniform strength and rectangular cross section and has the depth twice the width throughout. The beam is of 1.2 m long and to be loaded with a uniformly distributed load of 16 kN/m length. Allowable bending stress = 20 N/mm2 . Calculate the dimensions of the beam at the middle and at the fixed end. [Ans 120 mm at middle, 190.5 mm at fixed end] 9.18 Calculate the maximum tensile and compressive stresses induced in a box girded carrying a uniformly distributed load of 16 N/m on a simply supported length of 8 m. The box section is made of ISA 75 × 75 × 5 mm thick angles welded to four 360 mm × 6 mm structured steel plates as shown in Fig. 9.35. For each angle section area ‘A’= 7.27 × 102 mm2 , IXX IYY = 38.7 × 104 mm4 and the distance of centre of gravity from the outer face of each leg of angle Cxx = Cyy = 20.2 mm. [Ans σmax = 89.168 MPa both in tension and compression] @seismicisolation @seismicisolation Bending of Beams 6 mm • 155 16 kN/m 360 mm Angle 75 × 75 × 5 8m 5 75 75 6 mm 360 mm Figure 9.35 9.19 A tapering chimney of hollow circular section is 30 mm high. Its external diameters at the top end base are 1.5 m and 2 m respectively as shown in Fig. 9.36. There is wind pressure of 30 kN/m2 from right to left. If the weight of the chimney is 50,000 kN and the internal diameter at the base is 1 m, determine the maximum and minimum stresses at the base and show their variation. 1.5 m Wind force 30 m [Ans σmax = 18.552 N/m2 , 1m σmin = −5.62 N/m2 (Compressive)] 2.5 m Figure 9.36 9.20 A steel pipe of external diameter 1224 mm and internal dia 1200 mm is running full if the bending stress is not to exceed 56 N/mm2 , find the greatest span on which the pipe may be freely supported. Steel and water weight 76800 N/m3 and 10000 N/m3 respectively. [Ans 20.36 m] 9.21 A 4 m long horizontal beam is simply supported at the ends in a rectangular cross section 40 mm wide and 80 mm deep. It is loaded with centrally concentrated load of P. Find the value of P which produces a bending stress of 20 MPa at a point 1 m from the left support and 20 mm below the top surface of beam. What is the maximum bending tensile stress in the beam and where does it occur? [Ans 3413 N, 80 MPa at the bottom of beam at midspan] @seismicisolation @seismicisolation 156 • Strength of Materials 9.22 A beam of I section with flange t mm thick 3 t mm wide and web 2t deep ×t thick (making overall depth of beam section as 4t) is simply supported at ends. The beam carries uniformly distributed load over whole of its length in addition to a central concentrated load equal to the total uniformly distributed load. If the length of the beam is 3 m, find the relationship between t and uniformly distributed load for a maximum bending stresses of 120 MN/m2 . [Ans w = 261 × 106t 3 where w in N/m and t in m] 9.23 A rectangular beam, simply supported over a span of 4 m, is carrying a uniformly distributed load of 50 kN/m. Find the dimensions of the beam, its depth if the beam section is 2.5 times its width. Take maximum bending stress as 60 MPa. [Ans 125 mm, 300 mm] @seismicisolation @seismicisolation 10 C HAPTER SHEAR STRESSES IN BEAMS Normally, beams are designed for bending stresses and then checked for shear stresses, as in several situations arise in design in which mode of failure is likely to be shear rather than bending. For example, wooden beams which are weak in shear along the planes parallel to the grain of the wood, and thin webbed beams, where if the web is excessively thin, it would not have sufficient stiffness and stability to hold its shape and it would fail due to shearing stress. Therefore, it is necessary to study shear stress distribution as detailed under. On application of shear force, the shear stress on the cross section tends to slide the transverse elements of the beam and the complimentary shear stress of elements act on it (for example, wooden beams when tested to destruction). Though the mean shear stress is equal to shear force divided by cross-sectional area but the shear stress in fact is not uniform in the cross section. It is zero on top and bottom of section. In order to derive an expression for shear stress at any point in the cross section of beam, let us consider any normal section AB of a beam where bending moment and shear force be M and F act respectively. Now consider another section CD at a distance δ x from AB, where bending moment and shear force are M + δ M and F + δ F, respectively M δy A y2 N P y E M+δM C Q F y1 y G A B D δx Longitudinal section y1 b Cross section Figure 10.1 Now consider any elementary stripe PQ of thickness δ y at a distance Y from neutral axis. M y I M +δM y Bending stress at Q = I Bending stress at P = @seismicisolation @seismicisolation 158 • Strength of Materials where I is the moment of inertia of cross section about neutral axis ∴ Area at P or Q = bδy M Total force on face P = · y × b.δy I M +δM × y × b.δy Total force on face Q = I Hence, unbalanced force PQ, = M M +δM × y.b.δ y − · y × b · δy I I ∴ Total unbalanced force on portion AEFC, y2 = y1 δM δM .ybδ y = I I y2 b.y.δ y (i) y1 This force is balanced by the shear force across EF shear force across EF = τ × Area of EF where τ is the shear stress, Shear force at EF = τ × b.δx (ii) Equating (i) and (ii), δM I y2 b.y.δy = Zb.δx y1 δM 1 × τ= δ x Ib We know that y2 b.y.δy y1 δM = F, shear force at the section AB. δx y2 b.y.δy = Moment of area of part on force AB. (transverse) y1 between A and E about neutral axis Now if A= Area of the part on face AB (transverse) between A and E. y = distance of its centre of gravity from neutral axis @seismicisolation @seismicisolation (iii) Shear Stresses in Beams Then, y2 b.y.δy = Ay y1 δM and Substituting the values of δx y2 by · δy in Eqn. (iii), y1 we get, τ =F· 1 Ay Ib or τ= FAy , where I is moment of inertia of the whole section. Ib This is an important relation. Now we shall discuss shear stress distribution across few important cross sections: (A) Shear Stress Distribution for Beam of Rectangular Section Take a section ABCD as shown in Fig. 10.2. b A d/2 B E y τ F y A N D Parabola Section of beam C d Max shear stress τmax Shear stress distribution diagram Figure 10.2 y = distance of centre of gravity of shaded area from neutral axis d −y b A= 2 1 d −y y = y+ 2 2 1 d = +y 2 2 @seismicisolation @seismicisolation • 159 160 • Strength of Materials Moment of inertia of the shaded area about neutral axis bd 3 12 F.Ay Shear stress, τ = I.b I= Substituting for the values of A & y, we have 1 d d −y b +y 2 2 2 τ= I.b 2 b d 2 −y F× 2 4 = I.b 2 F d 2 −y ∴τ = 2I 4 F. (i) The shear stress distribution shown in Fig. 10.2 is parabolic. d When, y = , τ = 0 (by substituting in Eqn. (i)) 2 At N.A, y = 0 so τ will be maximum Fd 2 F d2 −0 = Hence, τmax = 2I 4 8I bd 3 12 2 Fd 3 F = . We get, τmax = 3 2 bd bd 8× 12 F = average shear stress at the section, and is denoted by τav But, bd Substituting for I = 3 ∴ τmax = τav 2 Therefore, maximum shear stress for a rectangular section is 1.5 times the average shear stress. (B) Shear Stress Distribution of a Solid Circular Section Let us consider a solid circular section of diameter D, radius R and Shear force F at that shaded section. @seismicisolation @seismicisolation Shear Stresses in Beams • 161 b Parabolic B A δy y y τ D τmax R Shear stress distribution diagram Beam section Figure 10.3 Consider any elementary strip of thickness δy over the layer AB at a distance y from NA. AB = b = 2 R2 − y2 ∴ Area of shaded strip = b.dy = 2 R2 − y2 × dy Moment of this area about neutral axis R Ay = R 2 2 2y R − y dy = by dy y y Because b = 2 R2 − y2 b2 = 4 R2 − y2 Differentiating both sides, 2bdb = 4(−2y)dy 1 or ydy = − b.db 4 Substituting in Eqn. (i), R Ay = y 1 b − bdb 4 When y = y, b = 0 and when y = R, b = 0 ∴ 1 Ay = − 4 0 1 −b3 b3 b db = − = 4 3 12 2 b @seismicisolation @seismicisolation (i) 162 • Strength of Materials Now shear stress at any section, τ= ∴ FAy Ib b3 2 12 = Fb Shear stress at section AB = I ×b 12I 2 2 F4(R − y ) = 12I F (R2 − y2 ) = 3 I F× (ii) we find from Eqn. (ii) that when y = R, τ = 0 (at the highest point of the section) and at y = 0, i.e., at neutral axis, shear stress is maximum. 2 D F 4 2 ∴ τmax (at y = 0) = R2 = F π 4 = 3 τav 3I 3× D 64 F Because average, τ = π D2 4 Hence, for solid circular beam section, 4 τmax = τav = 1.33 τav 3 (C) Shear Stress Distribution in an I Section F (D2–d2) 8I B y y d b A N Beam section Parabolic BF (D2–d2) 8Ib D Parabolic Shear stress distribution diagram Figure 10.4 i) Shear stress distribution in the flange D A = Shaded area = B −y 2 @seismicisolation @seismicisolation F [B(D2-d2)+bd2] 8Ib Shear Stresses in Beams • 163 y = Centroidal distance of this area from neutral axis 1 D 1 D 1 D −y = − y + 2y = +y = y+ 2 2 2 2 2 2 1 D D ∴ Ay = B −y × +y 2 2 2 FAy Ib Substituting values for A, y & b, we know, τ = 1 D D −y × +y F ×B F D2 2 2 2 2 = −y τ= IB 2I 4 d At the outer surface of the flange, where y = ; τ = 0 2 d At the inner surface of the flange, where y = 2 2 d2 F 2 F D − = τ= D − d2 2I 4 4 8I Obviously shear stress distribution in the flange is parabolic curve. ii) Shear stress distribution in the web: Width of the section in the web at a distance y from neutral axis = b Ay = Moment of flange area about neutral axis + Moment of web area about neutral axis d 1 D−d d 1 d D−d × + +b −y × y+ −y =B 2 2 2 2 2 2 2 D−d d D d d y =B × + +b −y + 2 2 4 2 4 2 b d2 B 2 2 2 −y = (D − d ) + 8 2 4 Substituting this value of A y in Eqn., τ = FAy Ib F B 2 b d2 2 2 (D − d ) + −y τ= Ib 8 2 4 F = B(D2 − d 2 ) + b(d 2 − 4y2 ) 8Ib @seismicisolation @seismicisolation (i) 164 • Strength of Materials Therefore, in the web, the shear stress τ varies with respect to y follows a parabolic curve. Also, τ increases as y decreases. At y = 0, τ is maximum. Substituting in Eqn. (i), τmax = F B(D2 − d 2 ) + b d 2 8Ib On the junction of the web and flange, y = d , 2 Substituting in Eqn. (i) F d2 2 2 2 B(D − d ) − b(d − ) τ= 8Ib 4 = FB 2 (D − d 2 ) 8Ib Of course in web also shear stress distribution is parabolic. It may be noted that abrupt change in width of flange at bottom surface of the flange, the numerical value of shearing stress F B F B suddenly changes from (D2 − d 2 ) to × (D2 − d 2 ) in the proportion (i.e., increases). 8I b 8I b E XAMPLE 10.1: An R.S.J. is of I section of overall height 200 mm and flange width 125 mm. The web thickness is 7 mm and the flange thickness 11 mm. The standard taper on the flange may be neglected and all corners may be assumed sharp. The beam is subjected to transverse loads acting parallel to the web, and at one section the shear force is 100 kN. Determine the maximum vertical shearing stress in the web at this section. Also determine the shear stress at top and bottom of the flange. S OLUTION : 125 mm 11 mm 7 X X 200 mm Shear force, F = 100 kN, Working in metre & N, 0.125 × 0.23 (0.125 − 0.007) × (0.2 − 0.022)3 − Ixx = 12 12 = 8.83 × 10−5 − 5.546 × 10−5 11 mm = 2.784 × 10−5 m4 Figure 10.5 @seismicisolation @seismicisolation Shear Stresses in Beams • 165 Maximum stress will be in the web at neutral axis F.A.y Ib 100000(0.125 × .011 × 0.945 + 0.007 × 0.089 × 0.0445) τmax = 2.784 × 10−5 = 80.78 MN/m2 Ans Now, τ= Shear stress at the top and bottom of the web, τ= 100000 × 0.125 × .011 × 0.0945 = 66.6 MN/m2 2.784 × 10−5 Ans E XAMPLE 10.2: A cast iron beam with flange and web section is 250 mm deep overall. The top flange is 125 mm × 50 mm deep, the bottom flange 200 mm × 50 mm deep and the web is 40 mm thick. If the transverse shearing force is 140 kN. Calculate the consequent shear stress is the web at the top and bottom junctions with flanges and maximum shear stress and sketch a diagram showing the variation of shear stresses over the depth of the beam. S OLUTION : Working in m and N throughout; 14.93 MN/m2 1 0.125 m 0.05 m 18.7 MN/m2 0.04 m N G Zmax 2 y 0.15 m A 3 0.05 m 17.1 MN/m2 A 0.20 m Shear stress distribution diagram Figure 10.6 @seismicisolation @seismicisolation B 166 • Strength of Materials For y from AB, a1 = 0.125 × 0.05 = 6.25 × 10−3 ; y1 = 0.225 m a2 = 0.04 × 0.15 = 6 × 10−3 ; y2 = 0.125 m a3 = 0.20 × 0.05 = 0.01; y3 = 0.025 m A = a1 + a2 + a3 = 6.25 × 10−3 + 6 × 10−3 + 0.01 = 0.02225 m2 6.25×10−3 ×0.225 + 6×10−3 ×0.125 + 0.01×0.025 0.02225 = 0.1083 m y= 0.125 × 0.053 2 + 0.125 × 0.05 × (0.1083 − 0.225) I= 12 0.04 × 0.153 2 + 0.15 × 0.04 × (0.1083 − 0.125) + 12 0.2 × 0.053 2 + 0.2 × 0.05 × (0.1083 − 0.025) = 0.000171 m4 + 12 Shear stress at top of web, τ = τ= FAy Ib 140000 × 0.125 × 0.05 × 0.1167 = 14.93 MN/m2 0.000171 × 0.04 Ans Shear stress at bottom of web, τ= 140000 × 0.2 × 0.05 × 0.0833 = 17.1 MN/m2 0.000171 × 0.04 Ans Max stress shear stress, τmax = 140000 × [0.125 × 0.05(0.1167) + 0.04 × 0.0917 × 0.04585 = 18.5 MN/m2 0.000171 × 0.04 @seismicisolation @seismicisolation Ans Shear Stresses in Beams • 167 Hint: 0.1167 0.0917 m N 0.1417 m A 0.04585 Figure 10.7 E XAMPLE 10.3: Figure 10.8 shows the section of a Tee-beam made of a uniform material, which is subjected to a shear force of 200 kN and a bending moment of 25 kNm. Calculate (a) the maximum bending stress and (b) the maximum shear giving sketches to show the form of stress distribution in each case. S OLUTION : 67.16 MN/m2 150 mm A 1 B 25 mm y 0.05625 150 mm 2 25 mm 65.3 MN/m2 = τmax 0.11875 σmax = 137.5 MN/m2 Figure 10.8 Let us first find position of C.G. from AB, a1 = 0.150 × 0.025 = 3.75 × 10−3 m2 ; y1 = 0.0125 m; a2 = 0.150 × 0.025 = 3.75 × 10−3 m2 y2 = 0.1 m; A = a1 + a2 = 0.0075 3.75 × 10−3 × 0.0125 + 3.75 × 10−3 × 0.1 = 0.05625 m 0.0075 0.15 × 0.0253 Ixx = + 3.75 × 10−3 (0.05625 − 0.0125)2 12 0.025 × 0.153 −3 2 + 3.75 × 10 (0.1 − 0.05625) + 12 y= @seismicisolation @seismicisolation 168 • Strength of Materials = 2158 × 10−8 m4 M 25000 ymax = × 0.11875 = 137.5 MN/m2 Ans I 2158 × 10−8 M 25000 σmin = ymin = × 0.05625 = 65.16 MN/m2 Ans I 2158 × 10−8 σmax = τ= F.Ay 200000 × [0.15 × .025 × 0.04375 + 0.025 × 0.03125 × 0.0156] ; τmax = Ib 2158 × 10−8 × 0.025 = 65.3 MN/m2 Ans E XAMPLE 10.4: A cantilever of I Section 200 mm × 100 mm has rectangular flanges 10 mm thick web and 7.5 mm thick. It carries a uniformly distributed load. Determine the length of the cantilever if the maximum bending stress is three times the maximum shearing stress. What is the ratio of the stresses halfway along the length of cantilever? 100 mm w N/m 10 mm l 7.5 mm X 200 mm (b) X 10 mm 100 mm (a) Figure 10.9 S OLUTION : 0.1 × 0.23 0.0925 × 0.183 − = 2.17 × 10−5 m4 Ixx = 12 12 Maximum BM = wl 2 M wl 2 × 0.1 = 2304.15 wl 2 , σmax = y = 2 I 2 × 2.17 × 10−5 FAy , Shear force(max) = wl Ib wl × (0.1 × 0.01 × 0.095 + 0.090 × .0075 × 0.045) τmax = 2.17 × 10−5 × 0.0075 τmax is at neutral axis & τ = @seismicisolation @seismicisolation • 169 wl(9.5 × 10−5 + 3.0375 × 10−5 ) wl × 10−5 (12.54) = 0.016275 × 10−5 0.016275 × 10−5 (i) Shear Stresses in Beams = = 770.5 wl Since σmax is three times τmax ; ∴ w l = 2304.15 w l2 3 × 770.5 l = 1 m Ans ∴ σmax = 2304.15 w(1)2 = 2304.15 w τmax = 770.5 w × 1 = 770.5 w Ratio of stresses half-way: 0.0125w w(0.5)2 0.125 w = 0.125 w ∴ σ = × 0.1 = 2 I I −5 0.5w × 10 (12.54) w = 10−5 × 836 F = w × 0.5 = 0.5w From Eq(i) τ = I × 0.0675 I 0.0125w/I σ = 1.5 Ans Required ratios = = τ 836 × 10−5 w/I M= E XAMPLE 10.5: Find the maximum shear stress in a hollow circular section of 100 mm external diameter and 75 mm internal when subjected to a total shearing force of 160 kN. S OLUTION : FAy Ib b = 2(R − r) π Shaded area = (R2 − r2 ) 2 π R2 4R π r2 4r − · · 2 3π y = 2 π3π (R2 − r2 ) 2 (R3 − r3 ) y= 3π (R2 − r2 ) τ= y b Figure 10.10 @seismicisolation @seismicisolation (i) 170 • Strength of Materials Substituting in Eqn. (i) π F × (R2 − r2 ) 4(R3 − r3 ) 2 τ= × I × 2(R − r) 3π (R2 − r2 ) = Substituting in Eqn. (ii); ∴ F R3 − r3 . 3I R − r (ii) I= π (0.14 − 0.0754 ) = 0.00000336 m4 64 R= 0.1 0.075 = 0.05; r = = 0.375 2 2 τ= 160000 0.053 − 0.03753 × 3 × 0.00000336 0.05 − 0.0375 = 91.7 MN/m2 Ans Exercise 10.1 Plot the distribution of shear stress over the section shown in Fig. 10.11, which is subjected to a shearing force of 300 kN, giving essential values. 300 mm 25 mm 300 mm 25 mm Figure 10.11 [Ans τmax = 51.5 MN/m2 ] 10.2 A transverse shear force F and a bending moment FL are applied to a uniform beam having the symmetrical cross section shown in Fig. 10.12. If the ratio of the transverse shear stress at the natural axis to the maximum direct stress due to bending is not to be less then 0.5, determine the maximum permissible value of l in terms of a. @seismicisolation @seismicisolation Shear Stresses in Beams • 171 0.8a 4a 2a [Ans Figure 10.12 l = 9.15a] 10.3 A rolled steel section is shown in Fig. 10.13, it is subjected to a vertical force of 20 kN. Determine shear stress at points A, B and C of the section. A 30 mm B R = 30 mm 60 mm dia C 30 mm 100 mm Figure 10.13 [Ans τA = 0, τB = 1.96 N/mm2 , τc = 5.88 N/mm2 ] 10.4 An extruded aluminium alloy section is of shape and dimensions as per Fig. 10.14. If a vertical force on the section is 4 kN. Find the dimension d, if the average shear stress in the section is 8 MN/m2 . What is the shear stress at neutral axis? 30 mm d/2 45° d 60 mm Figure 10.14 [Ans 20 mm; τ at neutral axis = 11.4 MN/m2 ] @seismicisolation @seismicisolation 172 • Strength of Materials 10.5 A T-section beam symmetrical about a vertical axis, is made with a top flange 100 mm wide and 14 mm thick to which a vertical web plate 150 mm deep and 10 mm wide is welded. At a certain point, the total shearing force is 40 kN. Calculate the percentage shear carried by the vertical web and the shearing force per metre run in the welded section. [Ans 94.2%; 308 kN] 10.6 An I Section shown in Fig. 10.15 is used as a beam. Determine the percentage of shear force resisted by web if the beam is subjected to a shear force F. 150 mm 20 mm 20 mm 250 mm 20 mm Figure 10.15 [Ans 92.53%] 10.7 A beam of triangular section having base width 200 mm and height of 300 mm is subjected to a shear force 3 kN. Determine the value of maximum shear stress and sketch the shear stress distribution diagram along the depth of beam. [Ans 3 Hint: τmax = τmean × ; 150 kN/m2 ] 2 10.8 A beam of channel section as shown in Fig. 10.16, is subjected at a vertical section, a shear force of 50 kN. Draw shear stress distribution diagram. Find the ratio of maximum and mean shear stresses. @seismicisolation @seismicisolation • 173 [Ans 2.2] Shear Stresses in Beams 60 mm 15 mm 120 mm 15 mm 15 mm 60 mm Figure 10.16 10.9 Find the maximum shear stress induced by a force of 4 kN is the vertical section of a hollow beam of a square section, if the outside width is 100 mm and the thickness is 20 mm. [Ans 1.35 MN/m2 ] 10.10 The beams of cross section (shown in Fig. 10.17) has shear force of 5.3 kN. Determine shear stress at points A and B. 325 mm 12 mm B A 200 mm 100 mm 100 mm 100 mm 12 mm Figure 10.17 [Ans τA = 287 kPa; τB = 396 kPa] @seismicisolation @seismicisolation 174 • Strength of Materials 10.11 The I section shown in Fig. 10.18 is used as a simply supported beam so that maximum shear stress developed in the beam is 16.8 N/mm2 . Determine the shear force to which the beams is subjected. 150 mm 20 mm 10 mm 30 mm 20 mm 150 mm Figure 10.18 [Ans 50 kN] @seismicisolation @seismicisolation 11 C HAPTER TORSION When a cylindrical shaft is subjected to equal and opposite couples at the end, shear stresses develop in the shaft whether it is rotating or in equilibrium. Our objective is to derive an equation connecting torque, modulus of rigidity, length, polar moment of inertia, length, shear strain, shear stress, radius and length. For that we have to make certain important assumptions before going ahead. Assumptions i) ii) iii) iv) v) vi) vii) The material is homogeneous and isotropic. Twist along the shaft is uniform. The shear stress is directly proportional to the radial distance from the axis of shaft. Transverse planes of the shaft remain plane. The modulus of rigidity is same throughout the material. All radii remain straight. The distortion along the shaft is uniform throughout. Now consider a circular shaft of outer radius R, fixed at one end. A torque T is applied at the force end as shown in Fig. 11.1. T φ B' B θ A 0 Radius = R Diameter = D l Figure 11.1 Let us take a straight line AB, which after application of torque T in clockwise direction takes the new position as AB so that shear strain φ as shown in Fig. 11.1 is set up. At the same time angle θ (in radians) is made at the free end as ∠BOB . Now shear strain = φ @seismicisolation @seismicisolation 176 • Strength of Materials ∴ BB = l φ , Also BB = Rθ Rθ Therefore, l φ = Rθ , or φ = l τ Shear stress at outer radius = Also we know shear strain φ = Modulus of rigidity C (i) (ii) Equating Eqns. (i) and (ii) Rθ τ = , l C We get Cθ τ = l R (iii) Torsional Moment of Resistance Take an elementary ring at radius r of thickness dr. Now as per assumption shear stress τ at r is proportional to shear stress τ at radius R. ∴ τ τ = R r ∴ τ = τ .r R Figure 11.2 Shear force acting on elementary ring = τ 2π rdr τ Substituting for τ , shear force acting on elementary ring = 2π r2 dr R τ Moment of this shear force at centre O = 2π r3 dr R R Total moment of resistance, T = τ 2πτ 2π r3 dr = R R 0 R r3 dr 0 = 2πτ r4 R 4 R 0 2πτ R4 πτ R4 × = ∴ T= R 4 2R D 4 4 4 π 2 πR πD = = Now 2 2 32 4 πD = J, Polar moment of inertia 32 IZZ [As we know from perpendicular axis theorem, IXX + IYY = IZZ is known as polar moment of inertia] @seismicisolation @seismicisolation ∴ π D4 π D4 π D4 + = 64 64 32 Torsion Hence, T = τJ ; R or • τ T = J R 177 (iv) Combining Eqns. (iii) and (iv), we get torsion equation: τ Cθ T = = J R l Power Transmitted: If T is torque and N is rotational speed in revolutions per minute, then, Power, P= T × 2π N kW, if T is in Nm. 60 × 1000 T = CJ , CJ is known as torsional rigidity, Torsional Rigidity: The stiffness or torsional rigidity = θr which can also be defined as the ratio of the torsion to angle of twist (in radians) per unit length of the shaft. E XAMPLE 11.1: A circular shaft of 60 mm diameter transmits torque from one shaft to another. Find the safe torque, which the shaft can transmit, if the shear stress is not to exceed 50 MPa. S OLUTION : T τ = J R ∴ Safe torque = or, T= J π d4 π d3 τ= τ= τ R 32.d/2 16 π (60)3 × 50 = 2119500 Nmm 16 = 2.1195 kNm Ans E XAMPLE 11.2: A solid circular shaft is to transmit a torque of 15 kNm. If the permissible shear stress is 55 MPa, find the diameter of the shaft. S OLUTION : τ π d3 J T = or, T = τ = τ J R R 16 T = 15000 Nm, τ = 55 MPa, d =? Substituting, π d3 × 55 × 106 16 15000 × 16 ∴ d3 = = 1.39 × 10−3 π × 55 × 106 or, d = 0.112 m = 111.6 mm Ans 15000 = @seismicisolation @seismicisolation 178 • Strength of Materials E XAMPLE 11.3: Torque is required to transmit through a hollow shaft of external diameter 100 mm and internal diameter 55 mm. If the permissible shear stress in 55 MPa, what is the safe torque that can be transmitted? S OLUTION : π (D4 − d 4 ) 32 τ π (D4 − d 4 ) T = J × ×2 = × 55 D 16 × D π (1004 − 554 ) π (100000000 − 9150625) × 55 = × 55 = 16 × 100 16 × 100 = 9806054 Nmm = 9.8 kNm Ans J= E XAMPLE 11.4: A solid steel shaft 60 mm diameter and 800 mm long transmits 35 kW at 200 r.p.m. Calculate: i) the maximum shear stress produced; ii) the angle of twist in degrees and iii) the shear stress at a radius of 25 mm. C = 80 GN/m2 S OLUTION : P= T= i) T τ = J r 2π NT 60 or, T = P × 60 2π N 35000 × 60 = 1671.97 Nm 2π × 200 ∴ τ= 1671.97 × 1000 × 30 × 32 T r= J π (60)4 = 39.44 N/mm2 = 39.44 MPa Ans 60 = 30 mm 2 Cθ 1671.97 × 1000 × 800 × 32 T.l T = ∴ θ= = = 0.01315 radians J l C.J 80000 × π (60)4 180 × 0.01315 = 0.754◦ Ans = π 39.44 × 25 τ τ1 39.44 τ1 = ; = ∴ τ1 = = 32.87 N/mm2 r r1 30 25 30 = 32.87 MPa Ans Because maximum stress will be at radius ii) iii) E XAMPLE 11.5: Compare the weights of hollow and solid shaft with equal lengths to transmit a 2 given torque for the same maximum shear stress if the inside diameter is of the outside. 3 @seismicisolation @seismicisolation Torsion S OLUTION : Whollow = Wsolid π /4 D2o − 2 Do 3 • 179 2 × l × density π /4 D4 × l × density Do = Outside diameter of hollow shaft D = Diameter of solid shaft 4 D2o − D2o 5D2 o 9 = = D2 9D2 Now, Thollow = Tsolid ; τ ×π Thollow = D4o − 2 Do 3 (i) 4 ×2 τπ = 32 Do 65 4 D 81 o 16Do = 0.1575D3o τ τ π Tsolid = × (D)4 × 2 = 0.19625D3 τ D 32 Equating, Tsolid = Thollow 0.19625D3 τ = 0.1575D3o τ D3o = 1.246 D3 Substituting in Equation (i) = ∴ Do = 1.076D Whollow 5(1.0760D)2 = = 0.643 Ans Wsolid 9D2 E XAMPLE 11.6: A hollow marine propeller shaft turning at 110 rev/min is required to propel a vessel at 47 km/h for the expenditure of 6.4 MW, the efficiency of the propeller being 68 per 2 cent. The diameter ratio of the shaft is to be and the direct stress due to thrust is not to exceed 3 8 MN/m2 . Calculate: (a) the shaft diameter and (b) the maximum shearing stress due to the torque. S OLUTION : Output power = 47 × 103 × P Watt 3600 where P is the propulsive force in N. 47 × 103 P = 0.68 × 6.4 × 106 3600 ∴ P = 334 kN π ∴ 334 × 103 = (D2 − d 2 ) × 8 × 106 4 ∴ @seismicisolation @seismicisolation 180 • Strength of Materials where D and d are outside and inside diameters, respectively 2 π 5 2 × D × 8 × 106 ; because d = D 4 9 3 D = 0.3093 m and d = 0.2062 m Ans 334 × 103 = ∴ 6.4 × 106 × 10 = 556000 Nm 2π × 110 J π 556000 = τ = τ × (D4 − d 4 ) R 16 π 0.30934 − 0.20624 =τ× 16 0.3093 T= ∴ Solving, we get τ = 128 MN/m2 ∴ Ans E XAMPLE 11.7: A solid shaft 220 mm diameter has the same cross-sectional area of that of a hollow shaft of the same material with inside diameter of 140 mm. Find the ratio of the power transmitted by the two shafts at the same speed. S OLUTION : Dh = External dia of the hollow shaft dn = 140 mm, the internal dia of the hollow shaft Because the two shafts have the same area, π 2 π (D − 1402 ) = × 2202 4 h 4 2 Dh − 19600 = 48400 ∴ D2h = 48400 + 19600 = 68000 Dh = 260.77 mm Ratio of power transmitted by the two shafts, Phollow Thollow Zhollow = = Psolid Tsolid Zsolid 4 4 π (Dh − dn ) π (260.774 − 1404 ) Zhollow = = = 3190915.75 mm3 16Dh 16 × 260.77 π (220)3 Zsolid = = 2089670 mm3 16 Phollow 3190915.75 = 1.527 Ans = Psolid 2089670 @seismicisolation @seismicisolation Torsion • 181 E XAMPLE 11.8: A hollow shaft is to transmit 320 kW at 100 r.p.m. If the shear stress is not to exceed 65 N/mm2 and the internal diameter is 0.5 of the external diameter, determine the external and internal diameters, assuming the maximum torque is 1.5 times the mean torque. S OLUTION : Dh = External diameter of the hollow shaft Di = 0.5 Dh , P = 320 kW 2 π N Tmean 2 × 100 × Tmean = 60 60 320000 × 60 × 1000 = 30573.2 Nm Tmean = 2π × 100 P= Tmax = 1.5 × 30573.2 = 45859.8 Nm = 45859.8 × 1000 = 45859800 Nmm Zn = π (D4h − D4i ) π (D4h − (0.5Dh )4 ) = 16Dh 16Dh = π × 0.9375D4h = 0.184D3h 16Dh Tmax = τ · Zh = 65 × 0.184D3h 45859800 = 65 × 0.184D3h ∴ ∴ Dh = (3834431.4)1/3 = 156.52 mm Internal diameter, Di = 156.52 × 0.5 = 78.26 mm E XAMPLE 11.9: A shaft shown in Fig. 11.3 rotates at 220 r.p.m. with 35 kW and 20 kW taken off at A and B, respectively and 50 kW applied at C. Find the maximum shear stress developed in the shaft and the angle of twist (degrees) of the gear A relative to gear C. Take C = 85 GN/m2 . Ans Ans B A 85 mm dia 50 mm dia 4.5 m 2m Figure 11.3 S OLUTION : Shaft between B and C: Power of the shaft = 50 kW, let the torque in this part be Tbc 2π NT 2π × 220Tbc ; 50000 = 60 60 Tbc = 2171.4 Nm = 2171400 Nmm Power = ∴ @seismicisolation @seismicisolation C 182 • Strength of Materials Let σs be the maximum shear stress in this part (between B and C) of the shaft. τs 16Tbc π d3 = Tbc or τs = 16 π d3 16 × 2171400 34742400 τs = = = 18 N/mm2 1928352.5 π 853 Shaft between B and A, Power of the shaft = 50 − 20 = 30 kW 2π × 220 × Tab For Tab , 30000 = 60 ∴ Tab = 1302.84 Nm Let τs be the maximum shear stress in this part of the shaft, π d3 T × 16 = Tab ∴ τ = ab 3 16 πd 16 × 1302.84 × 1000 ∴ τ = = 53.11 N/mm2 π (50)3 τs × Ans Therefore, maximum shear stress occurs at the 50 mm diameter shaft. Twist of the Shaft Let θbc be the twist of the shaft BC, l Tbc 2000 2171400 × 32 . = × C J 85000 π (50)4 = 0.0833 radians θbc = Now let θab be the twist of the shaft AB, l Tab 4500 1302.84 × 1000 × 32 × = × C J 85000 π (85)4 = 0.013466 radians θab = Hence, angle of twist of A with respect to C = θbc + θab = 0.0833 + 0.013466 = 0.096766 radians 0.096766 × 180 = = 5.55◦ = 5◦ 33 Ans π E XAMPLE 11.10: A maximum shear stress of 160 MN/m2 is induced in a hollow shaft of 120 mm and 70 mm external and internal diameters, respectively. What maximum shear stress will be developed in a solid shaft of the same weight, material and length, subjected to the same torque? @seismicisolation @seismicisolation Torsion • 183 S OLUTION : RH = 120 = 60 mm, 2 rH = 70 = 35 mm 2 Let τ be the maximum shear stress in the hollow shaft and τ1 be the maximum shear stress in the solid shaft TH = τ π τ1 × π Rs 4 × (R4H − rH ); Ts = ; Rs is the radius of solid shaft. RH 2 2 Since the torque is same, τ π 4 τ1 × π Rs 4 = × R − rH RH 2 H 2 (i) putting the values of τ , RH and rH in Eqn. (i) π 160 π 4 × (60 − 354 ) = τ1 × R3s 60 2 2 2.67(12960000 − 1500625) = τ1 R3s 30596531 = τ1 R3s (ii) Because the weight, length and material is same, Cross-sectional area of hollow shaft = Cross-sectional area of solid shaft π (RH − rH ) = π Rs 2 2 2 602 − 352 = R2s ∴ Rs = ∴ Rs = 48.73 mm √ 3600 − 1225 Substituting in Eqn. (ii), 30596531 = τ1 (48.73)3 ∴ τ1 = 264.41 N/mm2 264.41 Maximum shear stress in solid shaft = = 1.65 Ans Maximum shear stress in hollow shaft 160 Hence, hollow shaft is 1.65 stronger than solid shaft. E XAMPLE 11.11: The stepped steel shaft shown in Fig. 11.4 is subjected to a torque T at free end and a Torque (1.7T ) in the opposite direction at the junction of the two sizes. @seismicisolation @seismicisolation 184 • Strength of Materials 1.7 T 60 mm dia T 120 mm dia C A B 1.3 mm 2 mm Figure 11.4 What is the total angle of twist at the free end, if maximum shear stress in the shaft is limited to 80 MPa? Take C = 85 GN/m2 . S OLUTION : Now the torques at B and C are in opposite directions, therefore the effect of these two torques will be studied independently (sum of the two twists, one in clockwise direction and the other in anticlockwise direction). To find value of T at C: It must be noted that torque in AB will induce more stress in BC because of smaller diameter between B and C. Hence, let us first calculate the torque in BC because it is less stressed than permissible in AB. T= π π × 80 × 603 = 3391200 Nmm τ (DBC )3 = 16 16 Polar moment of inertia, π (DAB )4 = 32 π JBC = (DBC )4 = 32 JAB = π (120)4 = 20347200 mm4 32 π (60)4 = 1271700 mm4 32 For angle of twist due to T at C T.l T lAB lBC + = J.C C JAB JBC 3391200 1300 2000 = + 85000 20347200 1271700 θ= = 39.9 6.39 × 10−5 + 157.27 × 10−5 = 39.9 × 163.66 × 10−5 = 0.0653 radians Now angle of twist at C due to torque 1.7T at B, θ= T lAB 1.7T × 1300 = × C JAB 85000 × 20347200 Substituting for θ= 1.7 × 3391200 × 1300 = 0.00433 radians 85000 × 20347200 @seismicisolation @seismicisolation Torsion • 185 Hence, angle of twist = 0.0653 − 0.00433 = 0.06097 radians 0.06097 × 180 = = 3.495◦ Ans π Composite Shaft In a composite shaft, there are two or more shafts of different materials fixed together. The applied torque is shared by each material. 1 Total torque T is shared as torque T1 by material number 1 and T2 by material number 2. ∴ 2 T1 + T2 = T (i) Since the angle of twist of the two shafts is same, θ1 = θ2 T2 T1 = C1 J1 CJ2 Figure 11.5 or T1 C1 J1 = T1 C2 J2 (ii) Solving Eqns. (i) and (ii) T1 and T2 can be found out and the stresses in the two materials are given by: T1 T2 τ1 = ; τ2 = Z1 Z2 Twisting Beyond the Limit of Proportionality If the torque applied to a shaft is sufficient to cause yielding in the material, the relation between the shear stress and the angle of twist is assumed to be similar to that between the direct stress and angle of bending for an overstrained beam. Thus, the stress is proportional to the radians up to the limit of proportionality, after which it remains constant over the remainder of the section of the shaft. dx r R x τ Let us consider a shaft section of radius R Fig. (11.6), which is subjected to a torque sufficient to cause yielding to a radius r; let the shear stress at the limit of proportionality be τ . For the plastic part, then, π T = τ Z = τ × r3 2 For the plastic part, the torque on an elementary ring of radius x and thickness dx is z×2π x×dx×x. Figure 11.6 @seismicisolation @seismicisolation 186 • Strength of Materials R Total torque on plastic part = r τ × 2π x2 dx 2 = τ × π (R3 − r3 ) 3 The total torque carried by the shaft is then the sum of torques carried by the elastic and plastic parts. Note: T. J is called torsional rigidity. Polar Modulus: ZP = J = Polar modulus R E XAMPLE 11.12: A composite shaft consists of a steel rod of 70 mm diameter surrounded by a closely fitting tube of brass. Find the outside diameter of the tube so that when a torque of 1200 Nm is applied to the composite shaft, it will be shared equally by the two materials. C for steel = 85 GN/m2 , C for brass = 45 GN/m2 . Find also the maximum shear stress in each material and common angle of twist in a length of 3.8 m. S OLUTION : Total torque, T = Ts + Tb = Ts + Ts (∵ Ts = Tb given) T 1200000 = = 600000 Nmm 2 2 Tb = Ts = 600000 Nmm Cθ C.θ .J T = OR, T = J l l Cs θs Js For steel, Ts = ls Cb θb Jb For brass, Tb = lb Cs θs Js Cb θb Jb since Ts = Tb ; ∴ = ls lb Ts = As we know ls = lb ∴ Cs θs Js = Cb θb Jb But in composite shaft, the angle of twist in each shaft is same. ∴ θs = θb OR, Cs Js = Cb Jb π π 85000 × (70)4 = 45000 × [D4 − 704 ] 32 32 where D is the outside diameter of brass tube 2.04 × 1012 = 45000 D4 − 24010000 @seismicisolation @seismicisolation Torsion D4 − 24010000 = 4.533 × 109 D4 = 45333333 + 24010000 D4 = 69343333, OR, D = 91.25 mm Ans For maximum shear stresses: τ= T ×R ; J For steel, τs Ts × d/2 Js 70 600000 × 2 = 600000 × 16 = 8.913 N/mm2 Ans τs = π π (70)3 (70)4 32 D D Tb × 600000 × 300000D 2 2 τb = = π = Jb 0.1067(D4 − 24010000) (D4 − 704 ) 32 300000 × 91.25 273750000 = = = 5.66 N/mm2 4 4835819.3 0.1067 × (91.25 − 24010000) Common angle of twist, π 4 Cs × θs × Js 85000 × θs × 32 (70) = = 52699580.6 θs Ts = ls 3800 substituting for Ts , 600000 = 52699580.6θs ∴ θs = 0.0114 radians 0.0114 × 180 = = 0.6535◦ π Ans Torsion of a Tapering Shaft B x A R1 R x dx l Figure 11.7 @seismicisolation @seismicisolation R2 Ans • 187 188 • Strength of Materials Let the tapered shaft in Fig. 11.7 be subjected to a torque T . τ1 = Maximum shear stress at A τ2 = Maximum shear stress at B The shear stress at a distance x and section X = τ , π π π Now, T = τ1 × R31 = τ2 × R32 = τ × × R3 2 2 2 or, τ1 R31 = τ2 R32 = τ R3 Angle of twist of small length dx, R is the radius at section x. T dx 2T dx T dx = π 4 = CJ C× 2R Cπ R4 R2 − R1 R = R1 + × x = R1 + kx l dθ = Also, where k = (i) R2 − R1 , k is constant for this shaft. l dθ = 2T dx · Cπ (R1 + kx)4 (ii) ∴ Total angle of twist for length l of the shaft = θ d θ l 1 dx 2T 1 2T = . =− . Cπ (R1 + kx)4 Cπ 3k (R1 + kx)3 0 0 2T 1 1 1 =− . − Cπ 3k (R1 + kl)3 R31 l Now, k= R2 − R1 l or, kl = R2 − R1 ; OR ∴ R2 = R1 + kl 2 T 1 1 1 2 T 1 − − . . = 3k Cπ R32 R31 3k Cπ R31 R32 3 R2 − R31 2T l θ= 3Cπ R2 − R1 R31 R32 2 T l R21 + R1 R2 + R22 θ= . 3 Cπ R31 R32 θ =− In case of a shaft of uniform radius R, R1 = R2 = R @seismicisolation @seismicisolation (iii) Torsion • 189 2 T l 3R2 2 T l R2 + R2 + R2 = θ= . . 3 Cπ 3 Cπ R6 R6 2 Tl 3 Tl θ= . × = , 3 Cπ R4 CJ which is same as before. E XAMPLE 11.13: A 1.2 m long shaft tapers uniformly from a diameter of 100 mm to a diameter of 140 mm. If the shaft transmits a torque of 18 kNm, find: i) Angle of twist ii) Maximum shear stress developed. Take C = 85 GN/m2 S OLUTION : R1 R2 l Figure 11.8 140 100 = 70 mm, R1 = = 50 mm 2 2 T = 18000 Nm, l = 1.2 m, Now working in N and m; R2 = Angle of twist, θ : 2 T l R21 + R1 R2 + R22 ; working in N and m θ= . 3 Cπ R31 R32 2 18000 × 1.2 0.052 + 0.05 × 0.07 + 0.072 = × 3 85 × 109 π 0.053 × 0.073 −3 + 3.5 × 10−3 + 4.9 × 10−3 −8 2.5 × 10 = 5.3953 × 10 1.25 × 10−4 × 3.43 × 10−4 10.9 × 10−3 −8 = 5.3953 × 10 = 0.01372 radian 4.2875 × 10−8 @seismicisolation @seismicisolation 190 • Strength of Materials = 0.01372 × 180 = 0.786◦ π Ans Maximum shear stress developed: Tmax = τmax × π × D31 16 Because maximum shear stress occurs at the smallest diameter. π (0.1)3 16 τmax = 91.72 MN/m2 Ans 18 × 103 = τmax × ∴ Thin Circular Tube Subjected to Torsion Consider a thin circular tube of external diameter D and thickness t. Whereas t is very small compared to diameter D. t J = Area of the section × Square of radius D 2 π D3t = π Dt = 2 4 J π D3t × 2 =τ× R 4×D 2 τπ D t T= 2 T =τ× ∴ D Thin circular tube (i) Figure 11.9 and 4T l T.l T.l.4 = = 3 C.J C.π D t π D3tC 4T l θ= π D3tC θ= ∴ Strength weight ratio = T W W = π Dtlw T= ∴ Hence, (ii) where w is the weight density of the material. τπ D2t 2 T τπ D2t τD = = W 2π Dtlw 2lw τD T = W 2lw @seismicisolation @seismicisolation (iii) Torsion • 191 E XAMPLE 11.14: A thin steel tube of 90 mm diameter is 3 mm thick. If the allowable shear stress is 75 MN/m2 and C = 82 GN/m2 , find: i) Safe twisting moment that can be applied to the tube; ii) The twist in a length of 550 mm. S OLUTION : D = 90 mm = 0.09 m t = 3 mm = 0.003 m i) Safe twisting moment T : τπ D2t 75 × 106 × (0.09)2 (0.003) = 2 2 = 911.25 Nm Ans T= ii) Angular twist θ : 4 × 911.25 × 0.55 4T l = 3 π D tC π (0.09)3 (0.003) × 82 × 109 = 0.00356 radians 0.00356 × 180 = = 0.204◦ Ans π θ= E XAMPLE 11.15: A shaft LMN of 550 mm length and 45 mm external diameter is having a hole for a part of its length LM, of a 22 mm diameter and for the remaining length MN having a hole of 32 mm diameter. If the shear stress is not to exceed 80 N/mm2 , find the maximum power, the shaft can transmit at a speed of 250 r.p.m. If the angle of twist in the length of 22 mm diameter hole is equal to that in the 32 mm diameter hole, find the length of the shaft that has been bored to 22 mm and the length of the shaft that has been bored to 32 mm diameter. 4 do − di4 π T= τ 16 do 22 mm 32 mm N L M l2 l1 500 mm Figure 11.10 @seismicisolation @seismicisolation 45 mm 192 • Strength of Materials For shaft LM, 4 π 45 − 224 = 0.00436 × 80 [4100625 − 234256] τ 16 45 = 1348589.5 Nmm T1 = For shaft MN, 4 4100625 − 1048576 45 − 324 π π = × 80 τ T2 = 16 45 16 45 = 1064826 Nmm So safe torque is minimum torque 1064826 Nmm or 1064.826 kNm. 2π × 250 × 1064.826 2π NT = = 27862.95 60 60 P = 27.86 kW Ans P= Now θ = T.l C.J T l1 C.J1 T l2 Angle of twist for MN, θ2 = C.J2 Angle of twist for LM, θ1 = But θ1 = θ2 T l1 T l2 l1 l2 = ∴ = CJ1 CJ2 J1 J2 π π J1 = [454 − 224 ] = [4100625 − 234256] 32 32 = 379387.46 mm4 π π J2 = [454 − 324 ] = [4100625 − 1048576] 32 32 = 299482.3 mm4 l1 l2 Substituting in Eqn. (i) = J1 J2 or, l1 J1 379387.46 = 1.267 = = l2 J2 299482.3 l1 = 1.267 l2 = 1.267(500 − l1 ) ∵ l1 + l2 = 500 mm ∴ @seismicisolation @seismicisolation (i) Torsion • 193 l1 = 633.5 − 1.267 l1 or l1 + 1.267 l1 = 633.5 633.5 = 279.4 mm Ans 2.267 32, l2 = 500 − 279.4 = 220.6 mm Ans 2.267 l1 = 633.5 of dia ∴ l1 = Exercise 11.1 A shaft is transmitting 100 kW at 160 r.p.m. Find a suitable diameter for the shaft if the maximum torque transmitted exceeds the mean by 25%. Take allowable shear stress as 70 N/mm2 . [Ans 80 mm] 11.2 A hollow shaft is to transmit 337.5 kW at 100 r.p.m. If the shear stress is not to exceed 65 N/mm2 and the internal diameter is 0.6 of the external diameter, find the external and internal diameters assuming that the maximum torque is 1.3 times the mean. [Ans 155.2 mm, 93.12 mm] 11.3 A hollow shaft of external diameter 120 mm transmits 300 kW power at 200 r.p.m. Determine the maximum internal diameter if the maximum stress in the shaft is not to exceed 60 N/mm2 . [Ans 88.5 mm] 11.4 A solid cylindrical shaft is to transmit 300 kW power at 100 r.p.m. a) If the shear stress is not to exceed 80 N/mm2 , find its diameter. b) What percent saving in weight would be obtained if this shaft is replaced by a hollow one, whose internal diameter equals to 0.6 of the external diameter, the length, the material and maximum shear stress being the same. [Ans a) 122 mm b) 29.55%] 11.5 A hollow shaft, having an inside diameter 50% of its outer diameter transmits 600 kW at 150 r.p.m. Determine the external diameter of the shaft if the shear stress is not to exceed 65 N/mm2 and the twist in a length of 3 m should not exceed 1.4 degrees. Assume maximum torque = 1.2 mean torque and C = 100 GMPa. [Ans 157.0 mm] 11.6 A solid shaft of 150 mm diameter is to be replaced by a hollow shaft of the same material with internal diameter equal to 60% of external diameter. Find the saving in material, if maximum allowable shear stress is the same for both the shafts. [Ans 30.9%] 11.7 A hollow shaft has an external diameter of 0.3 m and internal diameter of 0.15 m. Compare its strength with that of a solid shaft of the same weight per unit length. [Ans Hollow 44%, stronger than solid] @seismicisolation @seismicisolation 194 • Strength of Materials 11.8 Two shafts of the same material and same length are subjected to same torque. If the first shaft is of a solid circular section and second shaft is of hollow circular section whose internal diameter is 2/3 of the external diameter and the maximum shear stress developed in each is the same, compare the weights of two shafts. Wh [Ans = 0.643] Ws 11.9 A shaft running at 180 r.p.m. has to transmit 97.5 kW. The shaft must not be stressed beyond 70 N/mm2 and also it must not twist more than 1 degree in a length of 3 m. What diameter would you recommend? C = 90 GN/m2 . [Ans d = 87.82 mm] 11.10 A mild steel tube 80 mm external diameter has a wall thickness of 5 mm. Find the torque necessary to induce a maximum shear stress of 80 N/mm2 in a length of 1.8 m of the tube and the amount of twist in degrees. C = 90 GN/m2 [Ans T = 3.328 kNm, θ = 2.292◦ ] 11.11 A solid circular shaft tapers uniformly from D at one end to d at the other end. It D = 1.2d, determine the percentage error committed, if the angle of twist for a given length is calculated using the mean diameter. [Ans 2.727%] 11.12 A compound shaft is formed by surrounding a 50 mm diameter solid brass shaft by a steel tube having wall thickness of 5 mm. The metals are securely attached to each other at their ends. Determine the increase in torque carrying capacity of the composite shaft over that of brass shaft alone as above. Assuming C for brass = 50 GMPa; C for steel = 120 GMPa. The working stress for brass and steel may be taken as 75 N/mm2 and 120 N/mm2 respectively. [Ans 98.7%] 11.13 A composite shaft consists of a steel rod of 60 mm diameter surrounded by a close fitting tube of brass. Find the outside diameter of the brass tube, when a torque of 1 kNm is applied on the composite shaft end shared equally by the two materials. C for steel as 84 GPa and C for brass as 42 GPa. Also find the common angle of twist in 4 m. [Ans 79 mm, 1.07◦ ] 11.14 A solid shaft 3.6 m long and 75 mm in diameter is fixed at both its ends. A twisting moment of 22.8 kNm is applied at a distance of 15 mm from one end. Calculate i) twisting moment shared by each portion; ii) the angle of twist on both sides of the place of application of the twisting moment, and iii) the maximum shear stress at each side. C = 90 GN/m2 . [Ans 13.3 kNm, 9.5 kNm, 0.0715 radian, 114.7 MN/m2 , 160 MN/m2 ] 11.15 A steel shaft 1 m long, 30 mm diameter is rigidly fixed at the ends. A torque of 600 Nm is applied at a distance of 250 mm from one end. Calculate: i) Fixing couples at the ends, ii) maximum shearing stress and iii) angle of twist at the point of application of torque. C = 82 GN/m2 [Ans 150 Nm, 450 Nm, 84.8 MN/m2 ; 0.01725 radian] 11.16 Design a suitable diameter for a circular shaft required to transmit 117.6 kW at 100 r.p.m. The shear stress in the shaft is not to exceed 70 MPa and the maximum torque exceeds the mean by 40%. Also calculate the angle of twist in a length of 2 m. Take C = 90 GPa. [Ans 78 mm; 2.336◦ ] @seismicisolation @seismicisolation Torsion • 195 11.17 A hole equal to half the diameter of C shaft is drilled in a shaft throughout its length. By what percentage are its weight and torsional strength reduced? [Ans Weight = 25%; Strength = 6.3%] 11.18 A steel bar 19 mm diameter is encased in a closely fitting brass tube of 32 mm external diameter, securely fixed together at the ends. The compound bar is subjected to a torque of 520 Nm and the angle of twist measured on a gange length of 250 mm is found to be 1.8◦ . If Csteel = 80 GN/m2 and calculate C for the brass. Find also the maximum shearing stresses in the two materials and the proportions of the total strain energy taken up by each part. [Ans 34.6 GN/m2 ; 95.5 MN/m2 , 69.5 MN/m2 ; 1 0.248 : 0.752; Hint: Strain Energy = T θ ] 2 11.19 A hollow circular shaft 20 mm thick transmits 300 kW power at 200 r.p.m. Determine the external diameter of the shaft if the shear strain due to torsion is not to exceed 0.00086. Take C = 80 GPa. [Ans 107.5 mm] 11.20 A shaft is required to transmit a power of 300 kW running at a speed of 120 r.p.m. If the shear strength of the shaft material is 70 N/mm2 , design a hollow shaft with inner diameter equal to 0.75 times the outer diameter. [Ans Do = 136 mm Di = 102 mm] 11.21 A stepped shaft of 2 m length consists of three lengths of diameter of 90 mm, 70 mm and 50 mm in sequence. If the angle of twist is the same for each section, compute the length of each section and the total twist. The maximum shear stress in the entire shaft is not to exceed 50 N/mm2 . Take C = 80 GPa. [Ans l1 = 1368.2 mm, l2 = 501.7 mm, l3 = 130.3 mm] 11.22 Prove that a hollow shaft is always stronger than a solid shaft of the same material, weight and length when subjected to simple torque. @seismicisolation @seismicisolation C HAPTER 12 THIN CYLINDRICAL AND SPHERICAL SHELLS It is very important to study thin cylindrical and spherical shells as we generally come across in industries these containing fluids for example, boilers, tanks, compressed air receivers and in chemical industry particularly. The thickness of a thin shell is less than 1/10th to 1/15th of its diameter. Types of stresses in thin shells: Due to the pressure of internal fluid, the following types of stresses develop in thin cylinders: i) Hoop or circumferential stress ii) Longitudinal stress iii) Radial stress, but it is negligible as compared to hoop stresses or longitudinal stresses, so not considered. i) Hoop Stress or Circumferential Stress Thickness 't' d X P X (b) l (a) 't' (c) Figure 12.1 If pressure p succeeds in bursting the cylinder into two halves as shown in Fig. 12.1(c) [one half is not shown], it is evident that resisting area is tl + tl = 2tl. Let σh be the hoop stress or circumferential stress then, Bursting force = p d l (i) [Note: d is the internal dia and dl is the projected area where pressure p acts to split it into two halves.] @seismicisolation @seismicisolation Thin Cylindrical and Spherical Shells Resisting force = σh × 2tl 197 p d = dia (ii) Equating Eqns. (i) and (ii), • Figure 12.2 σh × 2tl = pdl pd ∴ hoop stress, σh = 2t (iii) ii) Longitudinal Stress As both ends of cylinder shown in Fig. 12.1(a) are closed with plates by rivets or welding, the force acting is on σh both ends. π So bursting force = p d 2 4 (iv) Let σl be the longitudinal stress developed in thin shell, then, Resisting force = σl × resisting area d = σl × 2π t 2 (v) Equating Eqns. (iv) and (v), d π σl × 2π t = p d 2 2 4 pd So longitudinal stress, σl = 4t If joint efficiency of longitudinal joint of the shell is ηl then Eqn. (iii) becomes: Hoop or circumferential stress, σh = ρd 2t ηl And if joint efficiency of circumferential joint in given as ηc then Eqn. (v) becomes: Longitudinal stress, σl = ρd 4t ηc Sometimes only one common efficiency of joint is given on set η , then, σh = ρd 2t η and σl = ρd 4t η Change in the dimensions of a thin cylindrical shell: Let μ be the Poisson’s ratio of the material of shell. pd pd and longitudinal stress σl = As we know, hoop stress σh = 2t 4t @seismicisolation @seismicisolation (vi) 198 • Strength of Materials σ σ Now, circumferential strain = h − μ l = εl E E Since the circumference is proportional to the diameter, so that the diameter strain is equal to circumferential strain. δd ∴ diameter strain, =ε d Using the above two equations, increase in diameter can be calculated. δh δl δl Again longitudinal strain = −μ = ε2 from which the increase in length can be deterl E E mined. Change in Volume Volume of the shell, v = π 2 d l 4 Taking log of both sides of above equation log v = log π + 2 log d + log l 4 Differentiating, δv δd δl =2 + v d l = 2ε1 + ε2 From this equation the increase in volume can be found out. It may be noted that in deriving above formulae, the following, assumptions have been made: i) The radial stresses in the cylinder wall are negligible. ii) The stresses are uniformly distributed throughout the wall of the cylinder. iii) There is no longitudinal stays in the cylinder. iv) Material of the cylinder obeys Hooke’s law and does not exceed elastic limit. v) Young’s modulus (E) is constant in material at all points. E XAMPLE 12.1: A cylindrical shell of 1.5 m diameter is made up of 20 mm thick plates. Determine the hoop and longitudinal stresses in the plates, if the shell is subjected to an internal pressure of 3 MPa. Take efficiency of the joints as 75%. S OLUTION : Hoop stress, Longitudinal stress, pd 2t η 3 × 1500 × 100 = 150 MN/m2 Ans = 2 × 20 × 75 pd σl = 4t η 3 × 1500 × 100 = 75 MN/m2 Ans = 4 × 20 × 75 σh = @seismicisolation @seismicisolation Thin Cylindrical and Spherical Shells • 199 E XAMPLE 12.2: A thin cylinder containing a fluid is of internal diameter 50 mm and of 5 mm thick sheet. If tensile stress in the material is not to exceed 38 MPa, find the maximum pressure to which a fluid can be subjected to. S OLUTION : Let p be the maximum pressure which can be allowed in the cylinder. Now, pd p × 50 σ= = =5p 2t 2×5 Now we know hoop stress is of tensile nature and maximum permissible tensile stress is 38 MPa. ∴ 38 = 5 p 38 = 7.6 MN/m2 Hence, maximum pressure, p = 5 Ans E XAMPLE 12.3: The internal pressure in a thin cylindrical shell of diameter 450 mm is 6 MPa. Determine the minimum thickness of the shell, if allowable tensile strength in the sheet material is 350 MPa and efficiency of the joints is 70 %. Take factor of safety as 4. S OLUTION : pd 2t η 6 × 450 × 100 350 = 2 × t × 70 6 × 450 × 100 = 5.51 mm ∴ t= 2 × 70 × 350 σh = Considering factor of safety as 4, Required thickness = 5.51 × 4 = 22.04 mm Ans E XAMPLE 12.4: A cylinder, 2.8 m long and 800 mm in diameter is subjected to an internal pressure of 2.5 N/mm2 . Determine the minimum thickness of the metal if the stress in the material of the shell is not to exceed 70 N/mm2 . Calculate: (i) the change in diameter (ii) the change in length (iii) the change in volume @seismicisolation @seismicisolation 200 • Strength of Materials Given E = 200 GN/m2 and μ = 0.25 S OLUTION : ∴ σh = pd 2t σh = 2.5 × 800 , 2×t but σh is given as 70 N/mm2 70 = 2.5 × 800 2×t ∴ σh = 2.5 × 800 = 70.03 MN/m2 2 × 14.28 σl = 2.5 × 800 pd = = 35.015 MN/m2 4t 4 × 14.28 Now, diametral strain, ε1 = t= 2.5 × 800 = 14.28 mm 2 × 70 Ans σ σh −μ l E E δd 1 = ( σh − μ σl ) d E = 1 (70.03 − 0.25 × 35.015) 200000 = 3.064 × 10−4 ∴ Change in diameter, δ d = 3.064 × 10−4 × 800 = 0.24512 mm (increase) Longitudinal strain, ε2 = = ε2 = = Ans σh σl −μ E E 1 (σl − μσh ) E δl 1 = (σl − μσh ) l E 1 (35.015 − 0.25 × 70.03) 200000 = 0.0000874 ∴ Change in length, δ l = 0.0000874 × 2800 = 0.24472 mm (increase) δV = 2ε1 + ε2 V @seismicisolation @seismicisolation Ans Thin Cylindrical and Spherical Shells • 201 = 2 × 3.064 × 10−4 + 8.75375 × 10−5 = 0.0003064 + 0.000874 = 0.0003939375 ∴ Change in volume, δ V = 0.0003938 ×V π 2 π d × l = (800)2 × 2800 4 4 = 1406720000 mm3 δ V = 0.0003938 × 1406720000 Volume, V = ∴ = 553966.3 mm3 (Increase) Ans E XAMPLE 12.5: A boiler is made of 12 mm thick plates and is subjected to a steam pressure of 2 N/mm2 . The efficiencies of the longitudinal and circumferential joints are 70% and 40%, respectively. If permissible tensile stress in boiler plates is 110 N/mm2 , determine the maximum allowable diameter. S OLUTION : σh = pd 2 × d × 100 = 0.119d = 2t ηl 2 × 12 × 70 Allowable σh = 110 N/mm2 ∴ or, 110 = 0.119d d = 924.37 mm Again, σl = pd 2 × d × 100 = = 0.1042d 4t ηc 4 × 12 × 40 Permissible tensile stress is 110 N/mm2 ∴ 110 = 0.1042 d Hence, d = 1055.6 mm For safety minimum of the two diameters is the required diameter Hence, d = 924 mm Ans E XAMPLE 12.6: A cylindrical water tank 6.5 m in diameter with its axis vertical is made from steel plates 3 mm thick. Determine the maximum height to which the tank may be filled if the hoop stress is permissible upto 65 MPa. @seismicisolation @seismicisolation 202 • Strength of Materials S OLUTION : p × 6.5 pd or, 65000000 = 2t 2 × 0.003 (Note: working in N and m) ∴ p = 60000 N/m2 σh = We know p = wh; w for water = 10 × 103 = 10000 N/m3 [Taking g = 10 m/s2 appr.] 60000 60000 = 10000 × h ∴ height of water = =6m 10000 Hence, h = 6 metre. Ans E XAMPLE 12.7: A thin cylinder of diameter 500 mm is made of 6 mm thick plate. The efficiencies of the longitudinal and circumferential joints are 70 and 40%, respectively. Determine the largest permissible pressure, if the tensile stress of the plate is limited to 100 MPa. pd 2t ηl p × 500 × 100 100 = 2 × 6 × 70 ∴ p = 1.68 MPa pd σl = 2t ηc p × 500 × 100 100 = 2 × 6 × 40 ∴ p = 0.96 MPa σh = Hence, permissible pressure, i.e., p = 0.96 MPa Ans Thin Spherical Shells: π 2 d 4 Bursting force = projected area × pressure π (i) = d2 × p 4 Projected area = Resisting force: let σ be the stress developed in the material of spherical shell. Resisting force = σ × π dt t d (ii) p Figure 12.3 @seismicisolation @seismicisolation Thin Cylindrical and Spherical Shells • 203 Equating (i) and (ii) π 2 d p 4 pd σ= 4t σ π dt = ∴ If η is the efficiency of the joint, then σ = pd 4t η Ans E XAMPLE 12.8: A spherical gas vessel of 1.4 m diameter is subjected to a pressure of 2.2 MPa. Determine the stress induced in the plate of vessel, if thickness of the vessel plate is 6 mm. S OLUTION : pd 4t 2.2 × 1400 = 128.3 MPa = 4×6 σ= Ans E XAMPLE 12.9: A spherical shell of 1.5 m diameter is subjected to an internal pressure of 3 MPa. Determine the minimum thickness of the plates required, if the permissible stress induced in the material of the shell is not to exceed 90 MPa. Take efficiency of joint as 75%. S OLUTION : pd 4t η 3 × 1500 × 100 90 = 4 × t × 75 3 × 1500 × 100 = 16.67 mm ∴ t= 90 × 4 × 75 σ= Ans Change in diameter and volume of a thin spherical shell due to an internal pressure: In case of stress everywhere in spherical shell is same try σ . Then, σ μσ − E E σ = (1 − μ ) E pd (1 − μ ) = 4tE Circumferential strain = @seismicisolation @seismicisolation 204 • Strength of Materials Because circumference is proportional to the diameter, the diametral strain is given by pd δd = (1 − μ ) d 4tE If V is the volume of the spherical shell, 4 4 V = π r3 = π 3 3 V= so, 3 d π = d3 2 6 π d3 6 Now, because the volumetric strain in a sphere is 3 times the linear strain, 3pd δV δd =3 = (1 − μ ) V d 4tE From the above equation change in volume, δ V can be calculated. E XAMPLE 12.10: Calculate the increase in diameter and volume of a spherical shell 1200 mm in diameter and 12 mm thick when it is subjected to an internal pressure of 2 N/mm2 . Take E = 200 GPa, μ = 0.3. S OLUTION : pd δd = (1 − μ ) d 4tE Change in diameter, δd = = pd 2 (1 − μ ) 4tE 2 × (1200)2 (1 − 0.3) 4 × 12 × 200000 = 0.21 mm increase δV 3δ d 3pd = = (1 − μ ) V d 4tE π π Volume, V = d 3 = (1200)3 6 6 = 904320000 mm3 Change in volume, δV = V × 3pd (1 − μ ) 4tE @seismicisolation @seismicisolation Thin Cylindrical and Spherical Shells = • 205 904320000 × 3 × 2 (1200) (1 − 0.3) 4 × 12 × 200000 = 678240 mm3 (increase) Ans Wire-Bound Thin Cylindrical Shells In order to strengthen a cylindrical shell against bursting in longitudinal section due to hoop or circumferential stress, a wire is bound tightly around the outer diameter of the shell. This wire will be under tension and is closely wound around the shell as shown in Fig. 12.4. This increases the strength of the thin cylinder to withstand high internal pressure without excessive increase in wall thickness. By winding wire, an initial tensile stress σw in wire is produced. This wire is of high tensile material. The bursting force due to internal pressure per mm length is equal to the resisting force due to pipe section plus wire section. The circumferential stress in the pipe is also equal to the strain in the steel wire. Wire Cylinder Wire wound around a pipe t 1R d σwσc σcσw l Figure 12.4 E XAMPLE 12.11: A cast iron pipe of 350 mm internal diameter and 15 mm thick is wound closely with a single layer of circular steel wire of 6 mm diameter under tension of 70 MPa. Determine the initial compressive stress in the pipe section. Also find stresses set up in the pipe and steel wire, when water under pressure of 7 MPa is admitted into the pipe. E for cast iron = 100 GPa, E for wire steel = 200 GPa and Poisson’s ratio = 0.3. S OLUTION : Initial compressive stress in the pipe section: Before admission of pressurised water, pipe is subjected to compression due to tension in the wire. Let us consider one mm length of the pipe. In 6 mm there are two sections (one at bottom and the other at top). Therefore 1 mm will have 2 = 0.333 wire sections. 6 @seismicisolation @seismicisolation 206 • Strength of Materials Initial compressive force in the wire, before the pressurised water is admitted into the pipe, π (6)2 × 70 = 658.7 N = 0.333 × 4 Hence, initial compressive stress in the pipe section, σc = 658.7 = 21.96 N/mm2 2 × 15 Stresses set up in the pipe and steel wire: Let σ p = Stress in the pipe section in N/mm2 and σw = Stress in the steel wire in N/mm2 Bursting force per mm length of the pipe, when water under pressure is admitted = p × d × l = 7 × 350 × 1 = 2450 N Total resisting force = Resisting force in pipe + Resisting force in wire π = [2σPtl] + σw × 0.333 × (6)2 4 = [2σP × 15 × 1] + [σw × 9.411] = 30 σP + 9.411 σw (i) (ii) Now the bursting force in pipe = total resisting force Therefore, equating Eqns. (i) and (ii), 2450 = 30 σP + 9.411 σw (iii) Circumferential strain in the pipe, σp 1 0.3 × 7 × 350 pd 1 = σp − − μ × E 4t Ec Ec 4 × 15 σ p − 12.25 = Ec σw Strain in the steel wire = Es = (iv) (v) Since circumferential strain in the pipe is equal to the strain in the steel wire, so equating Eqns. (iv) & (v) σ p − 12.25 σw = Ec Es Es σw = × (σ p − 12.25) Ec 200000 × (σ p − 12.25) = 2(σ p − 12.25) σw = 100000 @seismicisolation @seismicisolation (vi) Thin Cylindrical and Spherical Shells • 207 Substituting the value of σw in Eqn. (iii), 2450 = 30 σ p + 9.411 × 2(σ p − 12.25) 2450 = 30 σ p + = 18.822 σ p − 230.57 2450 = 48.822 σ p − 230.57 ∴ σ p = 54.9 N/mm2 σw = 2 (σ p − 12.25) {from Eqn. (vi) σw = 2 (54.9 − 12.25) = 85.3 N/mm2 Hence, final stress in the pipe section = 54.9 − 21.96 = 32.94 N/mm2 Final stress in steel wire = 70 + 85.3 = 155.3 N/mm2 Ans Ans E XAMPLE 12.12: A brass tube of internal diameter 160 mm and external diameter 180 mm, is wound closely with a steel wire of diameter 2.5 mm at an initial tension of 120 N/mm2 . If the tube is Es = 2, μ subjected to an internal pressure of 3 N/mm2 , find the stresses in the brass and the steel, Eb for brass = 0.25. S OLUTION : Let us consider 10 mm length (l) of the wire bound pipe, number of turns of wire in 10 mm length 10 = =4 2.5 Axial plane 180 mm dia 10 mm Figure 12.5 An axial plane will cut the wire at eight sections π Whose area = 8 × (2.5)2 = 39.2 mm2 4 Before internal pressure, Total tensile force in the wire = 120 × 39.2 = 4704 N 180 − 160 Thickness of tube = = 10 mm = t 2 @seismicisolation @seismicisolation 208 • Strength of Materials Area of brass section by the axial plane = 2 × l × t = 2 × 10 × 10 = 200 mm2 If the compressive stress in the tube is σbc , the total compressive resistance in the tube = σbc × 200 N Equating this to the tensile force in the wire, we get 200 σbc = 4704 ∴ σbc = 23.52 N/mm2 (Compressive) Therefore, before the tube is subjected to internal pressure, the compressive stress in it is 23.52 N/mm2 . Let the hoop stress in the tube and wire induced by the internal pressure be σb and σs , respectively. Bursting force in the tube = pdl = 3 × 160 × 10 = 4800 N Resistance offered by wire = σs × 39.2 N Resistance offered by tube = σb × 200 N Total resisting force = bursting force 39.2σs + 200σb = 4800 Hoop strain in the wire = (i) σs Es pd Hoop stress in the tube is σb which is not equal to due to force exerted by the wire. The 2t pd . longitudinal stress in the tube = 4t σb pd −μ Eb 4tEb 0.25 × 3 × 160 σ = b− Eb 4 × 10 × Eb 3 σb = − Eb Eb Hoop strain in the tube = Strain in wire and tube is equal ∴ σs σb 3 = − Es Eb Eb Es σs = (σb −3) Eb σs = 2 (σb −3) @seismicisolation @seismicisolation (ii) Thin Cylindrical and Spherical Shells • 209 Solving Eqns. (i) and (ii) Substituting for σs from Eqn. (ii) in Eqn. (i); 39.2 × 2 (σb −3) + 200σb = 4800 78.4 (σb −3) +200σ b = 4800 78.4σ b −235.2 + 200σ b = 4800 278.4σ b = 5035.2 σb = 18.086 N/mm2 (Tensile) From Eqn. (ii) & substituting for σb , σs = 2 (σb − 3) σs = 2 (18.06 − 3) = 30.12 N/mm2 tensile The initial stresses in the wire and tube are 120 N/mm2 (tensile) and 23.52 N/mm2 (compressive), respectively, Therefore, the final stresses are: Stress in the wire = 120 + 30.12 = 150.12 N/mm2 (tensile) Ans 2 Stress in the tube = 23.52 − 14.1 = 9.42 N/mm (compression) Ans E XAMPLE 12.13: A cast iron pipe of 220 mm inside diameter and 250 mm external diameter is closely wound with a layer of 6 mm diameter steel wire with initial tensile stress of 40 MPa. Find the stresses developed in the pipe and the steel wire when water is admitted into the pipe at a pressure of 4 MPa. Esteel = 200 GPa, Ecast iron = 100 GPa, Poisson’s ratio = 0.3. S OLUTION : Equating the circumferential strains of the wire and tube, Initial stresses Equivalent wire thickness,tw = Initial compressive hoop stress in the tube, σ = σ= π ·d π ×6 = = 4.712 mm 4 4 tw σw t 4.712 × 40 = 12.57 N/mm2 15 Stresses due to fluid pressure alone: When water is admitted at pressure p, let the stresses be σ tensile (hoop) in the tube and σw tensile in the wire due to pressure alone. @seismicisolation @seismicisolation 210 • Strength of Materials For equilibrium, σ .2t + σw 2tw = pd where d is internal diameter of pipe or σ ×2 × 15 + σ w ×2 × 4.712 = 4 × 220 or 30σ + 9.424σw = 880 σ = −0.314σw + 29.33 ∴ (i) 4 × 220 (σ − μσl ) pd σ = = w where σl = Eci Es 4t 4 × 15 2 σl = 14.67 N/mm Es (σ − 0.3 × 14.67) σw ∵ = =2 Eci 2Eci Eci 2σ − 8.802 = σw σ = 0.5 σw + 4.401 Equating Eqns. (i) and (ii) −0.314 σw + 29.33 = 0.5 σ w + 4.401 −0.814 σw = − 24.93 σw = 30.63 N/mm2 Substituting value of σw is Eqn. (ii) σ = 0.5 × 30.63 + 4.401 = 24.12 N/mm2 Final stresses: In the pipe = 24.12 − 12.57 = 11.55 N/mm2 2 In the wire = 30.63 + 40 = 70.63 N/mm Cylinder with hemispherical ends Hoop stress in the cylindrical part: σh c = pd 2tc σlc = pd 4tc Longitudinal stress in the cylindrical part @seismicisolation @seismicisolation Ans Ans (ii) Thin Cylindrical and Spherical Shells • 211 tc d ts p l tc = Thickness of cylinder ts = Thickness of hemisphere Figure 12.6 Hoop stress in the hemispherical part σh s = pd 4ts Circumferential strain in the cylindrical part = εc = σhc μσlc pd − = (2 − μ ) E E 4tc E Circumferential strain in hemispherical part = εh = σhs μσhs pd − = (1 − μ ) E E 4tc E For the condition of no distortion on the junction, εc = εh pd (2 − μ ) = 4tc E ts = tc pd (1 − μ ) 4ts E 1−μ 2−μ E XAMPLE 12.14: A thin cylindrical shell of diameter 1.6 m is provided with hemispherical ends. If it is subjected to an internal pressure of 3 N/mm2 , find the thickness of the cylindrical and hemispherical parts for a permissible stress of 120 N/mm2 . @seismicisolation @seismicisolation 212 • Strength of Materials S OLUTION : Cylindrical part: pd = 120 2tc ∴ tc = 3 × 1600 = 20 mm 2 × 120 Ans pd = 120 4ts ∴ ts = 3 × 1600 = 10 mm 4 × 120 Ans Hemispherical part: Exercise 12.1 A thin cylindrical shell of diameter 1200 mm is subjected to a fluid pressure of 4 MPa. What should be the thickness of the wall if the maximum stress is not to exceed 100 MPa. Hence, what will be the change in volume per metre length. E = 200 GPa; μ = 0.3. [Ans δ v = 1.074 × 106 mm3 ] 12.2 A cylinder of internal diameter 2.5 m and of thickness 50 mm contains a gas. If the tensile stress in the material is not to exceed 80 N/mm2 , determine the internal pressure of gas. [Ans 3.2 N/mm2 ] 12.3 A thin cylinder of internal diameter 1.25 m contains a fluid at an internal pressure of 2 N/mm2 . Determine the maximum thickness of the cylinder if : i) The longitudinal stress is not to exceed 30 N/mm2 . ii) The hoop stress is not to exceed 45 N/mm2 . [Ans 28 mm] 12.4 A cylindrical vessel of 3 m diameter is used for processing rubber and is 10 m long. If the steel plates have the thickness of 24 mm, and vessel operates at 800 kPa internal pressure. Determine: i) the change in length, ii) change in diameter and iii) change in volume. E = 200 GPa, μ = 0.3. [Ans δ l = 0.492 mm, δ d = 0.62 mm δ v = 32 × 106 mm3 ] 12.5 In a thin cylinder, the hoop strain is 3.5 times the longitudinal strain. How much is the Poisson’s ratio? [Ans μ = 0.25] 12.6 In a thin cylinder the longitudinal stress is 60 MPa. E = 200 GPa and μ = 0.25. How much is the volumetric strain. δV [Ans = 1.2 × 10−3 ] V 12.7 A steel pipe 900 mm diameter has to carry water under a head of 200 m. If the allowable tensile stress is 90 MPa, determine the minimum value of thickness of pipe. [Ans 9.8 mm] @seismicisolation @seismicisolation Thin Cylindrical and Spherical Shells • 213 12.8 A spherical shell of internal diameter 0.9 m and of thickness 10 mm is subjected to an internal pressure of 1.4 N/mm2 . Determine the increase in volume and increase in diameter. Take 1 E = 200 GPa, μ = . 3 [Ans δ V = 12028.5 mm3 , δ d = 0.0954 mm] 12.9 A boiler shell is to be made of 20 mm thick plate having a permissible tensile stress of 125 N/mm2 . If the efficiencies of the longitudinal and circumferential joints are 80% and 30%, respectively, determine: i) The maximum permissible diameter of the shell for an internal pressure of 2.5 N/mm2 , and ii) Permissible intensity of internal pressure when the shell diameter is 1.6 m. 12.10 12.11 12.12 12.13 12.14 12.15 [Ans i) 1200 mm ii) 1.875 N/mm2 ] A thin spherical shell of internal diameter 1.5 m and of thickness 8 mm is subjected to an internal pressure of 1.5 N/mm2 . Determine the increase in diameter and increase in volume. Take E = 2 × 105 N/mm2 and μ = 0.3. [Ans 0.369 mm, 1304 × 103 mm3 ] Find the increase in volume of a spherical shell of 900 mm diameter and 10 mm thick when subjected to an internal pressure of 1.5 N/mm2 . Take E = 205 GPa, μ = 0.3. [Ans 131900.1 mm3 ] A copper tube is closely wound with a steel wire of 1 mm diameter. If the internal and external diameters of the tube are 42 mm and 45 mm, find the required tension in the wire so that an internal pressure of 2000 kPa produces a circumferential tensile stress of 8000 kPa in the tube. Take Es = 1.65 Ec . [Ans 10.5 N] A cast iron thin cylindrical pipe of internal diameter 200 mm and 15 mm thick is closely wound by a simple layer of steel wire of 3mm diameter under a tension of 50 N/mm2 . Find the stressed set up in the pipe when the pipe is empty. Also determine the stresses set up in the pipe and steel wire, when water is admitted in the pipe under a pressure of 5 N/mm2 ECI = 100 GPa, Es = 200 GPa and μ = 03. [Ans 7.89 N/mm2 ; σp = 1865 N/mm2 ; σw = 93.08 N/mm2 ] A hollow cylindrical shell of diameter 1.5 m, thickness 10 mm and length 3 m is subjected to an internal pressure of 4 N/mm2 . Find the change in volume. Take E = 200 GPa, μ = 0.3. [Ans 15.1 × 106 mm3 ] An 1800 mm internal diameter cylindrical shell is provided with hemispherical ends, and subjected to internal fluid pressure of 5 N/mm2 . Determine the thickness of cylindrical and hemispherical portions if the maximum allowable stress s 90 N/mm2 . [Ans 50 mm, 25 mm] @seismicisolation @seismicisolation C HAPTER 13 THICK CYLINDERS AND SPHERES We shall now consider thick cylinders for which the thickness of the wall is not small as compared with the diameter of the cylinder. The thickness of a thick cylinder is more than 1/10 to 1/20 of the diameter. When the cylinder is subjected to internal pressure p, it will expand, resulting in increased length and diameter, i.e., axial and circumferential stresses develop in the walls of the cylinder and both these stresses are tensile stresses which in case of thick cylinders are analysed as per ‘Lame’s theory’. Assumptions made in Lame’s theory: i) The material is homogeneous and isotropic. ii) Expansion or contraction of all fibres are unstrained by adjacent fibres. iii) Increase in length is uniform throughout, irrespective of radius and thus the axial strain is the same at any radius of the cylinder. iv) The material is stressed within elastic limit. v) Plane sections, perpendicular to the longitudinal axis of the cylinder, remain plane even after applications of pressures. Let r1 and r2 be the external and internal radii of a thick cylinder of length l subjected to radial pressure. A cross section of a thick cylinder is shown in Fig. 13.1. dx dx r1 px x x x r2 x x px+dx Figure 13.1 Consider an elementary ring of the shell at radius x and thickness dx. Let px and (px + d px) be the intensities of radial pressure at the inner and outer sides of the elementary ring Bursting force, P = px × 2x × l − (px + dx).2(x + dx) × l @seismicisolation @seismicisolation Thick Cylinders and Spheres • 215 The resisting force = σx × 2dx × l Equating, σx 2dx l = px 2x l − (px + d px ).2(x + dx) × l Neglecting small quantities of the second order, σx = −px − x d px d = − (xpx ) dx dx (1) Another relation between px and σx will now be obtained from the consideration that longitudinal strain at any point in the section of the shell is constant. At any point in the section of the elementary ring, the principal stresses are: i) The radial pressure px ii) The hoop stress σx and iii) a longitudinal tensile stress σl equal to p r22 as explained under: r12 − r22 σl r1 p r2 σl Figure 13.2 p × π r22 = σl × π (r12 − r22 ) ∴ σl = p r22 r12 − r22 The longitudinal strain at any point in the ring of radius x and thickness dx is, obviously equal to σl μσx μ px − + E E E This is independent of x. ∴ σ l μσx μ px − + = a constant E E E ∴ (σx − px ) = a constant since σl , μ and E are constants. ∴ (σx − px ) = 2a, where a is a constant @seismicisolation @seismicisolation (2) 216 • Strength of Materials From the relations (1) and (2), we can solve for σx and px . Substituting (px + 2a) for σx in (1), we get d px dx d px 2dx ∴ =− (px + a) x 2dx d px =− or px + a x (px + 2a) = −px − x Integrating, loge (px + a) = −2 loge x + loge b where loge b is a constant of integration. ∴ b x2 b px = 2 − a x b σx = 2 + a x px + a = and from (2), (3) (4) Equations (3) and (4) are Leme’s formulae or equations for the radial pressure and hoop stress at any specified point in the section of a thick cylindrical shell. The constants a and b can be calculated for given conditions. E XAMPLE 13.1: A steel pipe of 350 mm internal diameter and 90 mm thickness carries water at a pressure of 10 N/mm2 . Determine the maximum and minimum intensities of hoop stress across the section. Show the radical pressure distribution and hoop stress distribution, across the section. S OLUTION : 350 = 175 mm, thickness = 90 mm 2 350 + 2 × 90 = 265 mm r1 = 2 r2 = r2 r1 Figure 13.3 px at radius 175 mm = 10 Using Lame’s equation, b −a x2 b − a OR 10 = 1752 b − 10 ∴ a= 30625 px = 10 = @seismicisolation @seismicisolation b −a 30625 (i) Thick Cylinders and Spheres • 217 px at radius 265 mm = 0 b −a 2652 b a= 70225 0= (ii) Equating Eqns. (i) and (ii), b b − 10 = 30625 70225 b b − = 10 30625 70225 3.2653 × 10−5 b − 1.42399 × 10−5 b = 10 1.84131 × 10−5 b = 10 ∴ b = 543091 Substituting b in Eqn. (ii), a= 543091 = 7.73 70225 Now, hoop stress at 175 mm radius, b b +a = + 7.73 2 x 1752 543091 σx = + 7.73 = 25.46 N/mm2 30625 σx = At 265 mm, 543091 + 7.73 2652 543091 + 7.73 = 70225 σx = = 7.81 N/mm2 mm 175 265 mm Ans p =10 N/mm2 7.81 N/mm2 sx=25.46 N/mm2 Radial pressure and hoop stress distribution Figure 13.4 @seismicisolation @seismicisolation Ans 218 • Strength of Materials E XAMPLE 13.2: A thick metallic cylindrical shell has 160 mm internal diameter. It is to withstand an internal pressure of 12 N/mm2 . Determine the necessary thickness of the shell, if the permissible tensile stress in the section is 25 N/mm2 . 160 = 80 mm. 2 b b −a 12 = 2 − a; 12 = 6400 80 b − 12 a= 6400 r2 = (i) From Lame’s equation the permissible stress (σx ), b +a r22 b 25 = 2 + a or, r2 σx = a = 25 − b 6400 Equating Eqns. (i) and (ii) b b − 12 = 25 − 6400 6400 b b b + = 25 + 12; = 37 6400 6400 3200 ∴ b = 118400 Substituting the value of b in Eqn. (i) a= 118400 − 12 = 6.5 6400 We know internal pressure at outer radius is zero, ∴ b 118400 −a = − 6.5 2 r1 r12 118400 118400 or r12 = = 18215.4 6.5 = 2 6.5 r1 √ r1 = 18215.4 = 134.96 mm 0= So required thickness = 134.96 − 80 = 54.96 mm Ans @seismicisolation @seismicisolation (ii) Thick Cylinders and Spheres • 219 Solid Circular Shaft Subjected to External Pressure p Let the radius of solid shaft be r, and external pressure applied on it be p. Applying Lame’s equations, px = b − a, x2 σx = b +a x2 p Figure 13.5 It is evident from the equation of radial pressure that px is infinite at radius = 0. Therefore, the value of constant b is zero, because it is not possible for pressure to be infinite at radius. Therefore, we get px = −a = p, Hence, pressure p is constant everywhere and its value is equal to external pressure p. Also, σx = 0 + a , where a = −p Therefore, σx = −p The minus sign shows that σx is compressive. Hence, the intensity of pressure is constant everywhere and is compressive, its value is equal to the external pressure p. Compound cylinders: A compound cylinder consists of two concentric cylinders as shown in Fig. 13.5, the outer cylinder being shrunk onto the inner cylinder so that the later is initially compression before the application of internal pressure. The final stresses are then the resultants of those due to pre-stressing and those due to the internal pressure. If the radius of common surface is r0 and the pressure at this surface before the application of the internal pressure is p0 , then the initial stresses are determined by considering the two cylinders separately, the boundary conditions for the outer cylinder being px = p0 when x = r0 and px = 0 when r = r1 and for the inner cylinder, px = p0 when r = r0 and pr = 0 when r = r2 . The stresses due to internal pressure are obtained by considering the cylinder to be homogeneous with pr = p at r = r2 and pr = 0 at r = r1 . The various stresses are then combined algebrically as shown in Fig. 13.7, from which it is clear that the maximum resultant stress is less than that for a homogeneous cylinder of the same cross section with same internal pressure. r1 r2 r0 Figure 13.6 px σc r2 r0 Initial stress r1 Stresses due to po Resultant stresses Figure 13.7 @seismicisolation @seismicisolation 220 • Strength of Materials E XAMPLE 13.3: A compound tube is composed of a tube 240 mm internal diameter and 21 mm thick shrunk on a tube of 240 mm external diameter and 23 mm thick. The radial pressure at the junction in 10 N/mm2 . The compound tube is subjected to an internal fluid pressure of 80 N/mm2 . Find the variation of hoop stress over the well of the compound tube. S OLUTION : Outer tube: px = b1 − a1 x2 and σx = b1 + a1 x2 r1 = r0 r2 r1 240 240 + 21 = 141 mm r0 = = 120 mm 2 2 r2 = 120 − 23 = 97 mm x = 141 mm, px = 10 N/mm2 b1 − a1 = 0 (141)2 b1 b1 − a1 = 0 ∴ a1 = 19881 19881 b1 − a1 = 10 Again (120)2 b1 b1 − a1 = 10 ∴ a1 = − 10 14400 14400 At Figure 13.8 Equating Eqns. (i) and (ii), b1 b1 = − 10 19881 14400 b1 b1 − = 10; 6.944 × 10−5 b1 − 5.0299 × 10−5 b1 = 10 14400 19881 1.9141 × 10−5 b1 = 10 b1 = 522439 From Eqn. (i) ∴ a1 = 522439 = 26.28 19881 Hoop stresses for the outer tube, 522439 + 26.28 = 52.56 N/mm2 (Tensile) 19881 522439 + 26.28 = 62.56 N/mm2 (Tensile) σ120 = 14400 b2 Inner tube : px = 2 −a2 x b2 σx = 2 + a2 x σ141 = @seismicisolation @seismicisolation (i) (ii) Thick Cylinders and Spheres px = 10 N/mm2 b2 b2 ∴ − a2 = 10 ∴ a2 = − 10 14400 14400 At x = 97 mm, px = 0 b2 b2 − a2 = 0 − a2 = 0 ∴ ∴ 2 9409 97 • 221 At x = 120, (iii) (iv) Substituting for a2 from Eqn. (iii) in Eqn. (iv), b2 b b2 b2 − − 10 = 0, − + 10 = 0 9409 14400 9409 14400 1.063 × 10−4 b2 − 0.694 × 10−4 b2 = −10 0.369 × 10−4 b2 = −10 ∴ b2 = −10 = −271003 0.369 × 10−4 Putting value of b2 in Eqn. (iii) a2 = −271003 − 10 = −28.82 14400 ∴ a2 = −28.82 Hence, the hoop stresses for the inner tube are calculated as follows: −271003 + (−28.82) 14400 = −47.64 N/mm2 (Compressive) −271003 σ97 = + (−28.82) 972 = −57.62 N/mm2 σ120 = Stresses due to internal fluid pressure alone: B −A x2 B σx = 2 +A x px = At x = 97 mm, px = 80 N/mm2 80 = B B −A − A, 80 = 2 9409 97 (v) At x = 141 mm; px = 0 0= B −A 1412 ∴ A= @seismicisolation @seismicisolation B 19881 (vi) 222 • Strength of Materials Substituting for A from Eqn. (vi) into Eqn. (v) B B − 9409 19881 80 = 1.0628 × 10−4 B − 0.503 × 10−4 B 80 = 80 = 0.5598 × 10−4 B, ∴ B = 1429082 Substituting for B in Eqn. (vi), A= 1429082 = 71.88 19881 Hence, the hoop stresses due to fluid pressure alone are given by, 1429082 + 71.88 = 223.76 N/mm2 (Tensile) 972 1429082 σ120 = + 71.88 = 171.12 N/mm2 (Tensile) 1202 1429082 σ141 = + 71.88 = 143.76 N/mm2 (Tensile) 1412 σ97 = Now due to the combined effect of shrinking tube (outer) on the inner tube and internal fluid pressure, the final hoop stresses are as follows: Outer tube: σ120 = 62.56 + 71.88 = 134.44 N/mm2 Ans σ141 = 52.56 + 143.76 = 196.32 N/mm (Tensile) Ans 2 Inner tube: σ120 = −47.64 + 171.12 = 123.48 N/mm2 (Tensile) Ans σ97 = −57.62 + 223.76 = 166.14 N/mm2 (Tensile) Ans Shrinkage Allowance: While making a compound cylinder, shrinkage allowance is kept, i.e., if it is necessary that the inner diameter of the outer cylinder to be slightly smaller than the outer diameter of the inner cylinder. Normally, in order to fit, outer cylinder is heated till it slides over the inner cylinder. Naturally, on cooling outer cylinder will shrink and exert the required pressure at the common surface. Interface pressure is p0 (at common surface). Outer cylinder: If E1 and μ1 are the modulus of material and Poisson’s ratio of material of outer cylinder, respectively. At outer surface of the outer cylinder, let the hoop stress be σc1 and radial stress being p0 = σr Then circumferential strain at outer cylinder: σc1 μ1 p0 + E1 E1 2r0 Increase in diameter = 2r0 εc1 = (σc1 + μ1 p0 ) E1 εc1 = @seismicisolation @seismicisolation Thick Cylinders and Spheres • 223 Inner cylinder: Let E2 and μ2 be the modulus of elasticity and Poisson’s ratio of the inner cylinder, respectively. At the outer surface let the hoop stress be σc2 and radial stress be p0 = σr σc2 μ2 p0 − E2 E2 2r0 Decrease in diameter: 2r0 εc2 = (σc2 − μ2 p0 ) E2 Therefore, shrinkage allowance = Increase in diameter of outer cylinder + Decrease in diameter of inner cylinder 2r0 2r0 = (σc1 + μ1 p0 ) + (σc2 − μ2 p0 ) E1 E2 σc1 + μ1 p0 σc2 − μ2 p0 = 2r0 + E1 E2 Circumferential strain: εc2 = If the two cylinders are made of same material, then E and μ will be common, so the above expression becomes, Shrinkage allowance = 2r0 (σc1 + σc2 ) E E XAMPLE 13.4: A steel tube of 120 mm inside diameter and 170 mm outside diameter is to have an outer tube of same material having 200 mm outside diameter shrunk on it, the shrinkage allowance being such that the radial pressure between the two tube is to be 25 N/mm2 . Take E = 200 GPa, μ = 0.25, calculate: i) ii) iii) iv) v) Hoop stress at the inner surface of outer tube Increase in internal diameter of outer tube Hoop stress at the outer surface of inner tube Reduction in external diameter of inner tube Shrinkage allowance S OLUTION : R1 = 100 mm R0 = ∴ 170 = 85 mm 2 R0 = 85 mm σr = p0 = 25 b b −a 25 = 2 − a, 25 = 7225 85 b b b −a ∴ a = = p = σr = 0 = 2 2 10000 100 100 Substituting the value of a in Eqn. (i) b b 1 1 25 = − =b − 7225 10000 7225 10000 @seismicisolation @seismicisolation (i) (ii) 224 • Strength of Materials 25 = b (0.0001384 − 0.0001) 25 ∴ b= = 651042 3.84 × 10−5 From Eqn. (ii) a= 651042 = 65.1 10000 b 651042 + 65.1 = 155.21 N/mm2 (Tensile) Ans +a = 2 7225 85 i) σc = ii) 155.21 0.25 × 25 155.21 + 6.25 σc μσr + = + = E E 200000 200000 200000 = 8.073 × 10−4 Increase in diameter of outer tube = 8.073 × 10−4 × 170 = 0.1372 mm. Ans iii) R0 = 85 mm σr = p0 = 25 = b b − a; 25 = −a 2 7225 85 (iii) 120 = 60 mm 2 b 6 σr = 0 = 2 − a a = 3600 60 when R2 = (iv) Putting value of a from Eqn. (iv) in Eqn. (iii), b 1 1 b − =b − 7225 3600 7225 3600 25 = −0001394b ∴ b = −179340 179340 = −49.82 ∴ a=− 3600 R0 = 85 mm b −179340 σ c1 = 2 + a = − 49.82 = −74.64 N/mm2 (Compressive) 7225 85 25 = iv) Circumferential strain of outer diameter of inner tube: 74.64 0.25 × 25 σc1 μ p0 + =− + E E 200000 200000 1 (−74.64 + 6.25) = −3.4195 × 10−4 (Compressive) = 200000 εc = @seismicisolation @seismicisolation Ans Thick Cylinders and Spheres ∴ • 225 Reduction in external diameter = 3.4195 × 10−4 × 170 = 581.3 × 10−4 = 0.05813 mm Shrinkage allowance = 0.1372 + 0.0581 = 0.1953 mm Ans E XAMPLE 13.5: A compound thick cylinder is formed by shrinking a tube of external diameter 320 mm over another tube of internal diameter 140 mm. After shrinking, the diameter at the junction of the tubes is found to be 240 mm and radial compression as 30 N/mm2 . Find the original difference in radii at the junction. Take E = 200 GPa. S OLUTION : 320 140 = 160 mm; Inner radius, r2 = = 70 mm 2 2 240 radius at junction, r0 = = 120 mm 2 b2 b2 30 = − a2 = − a2 1202 14400 b2 − a2 30 = 14400 r1 = (i) Also 0= b2 − a2 702 ∴ a2 = b2 4900 Substituting the value of a2 from Eqn. (ii) in Eqn. (i) b2 b2 − ; 30 = b2 6.944 × 10−5 − 2.041 × 10−4 14400 4900 30 30 =− b2 = (.00006944 − 0.0002041) 1.3466 × 10−4 b2 = −222783 30 = Substituting b2 value in Eqn. (i), −222783 − a2 14400 a2 = −15.47 − 30 = −45.47 30 = Considering the outer cylinder when r1 = 160 mm, px = 30 N/mm2 b 30 = 2 − a x @seismicisolation @seismicisolation (ii) 226 • Strength of Materials b2 b1 −a − a1 = 2 14400 (120) b1 − 30 a= 14400 30 = (iii) Similarly, b1 b1 − a1 = − a1 2 25600 160 b1 a2 = 25600 0= (iv) Substituting, value of a2 in Eqn. (iii) b1 b1 b1 b1 = − 30 OR, 30 = − 25600 14400 14400 25600 30 = b1 (6.944 × 10−5 − 3.906 × 10−5 ) 30 987492 = 987492, a2 = = 38.57 b1 = −5 25600 3.038 × 10 b1 σc1 = 2 + a1 x 987492 = + 38.57 = 88.24 N/mm2 1412 b2 σc2 = 2 + a2 x −222783 = + (−45.47) 1202 = −15.47 − 45.47 = −60.94 N/mm2 (Compressive) 1 2r0 (σc1 + σc2 ) Difference in radii at junction = 2 E r0 (σc1 + σc2 ) E 120 = {88.24 + (−60.94)} 200000 = 0.01638 mm Ans = E XAMPLE 13.6: A compound tube is made of a tube 270 mm internal diameter 30 mm thick shrunk on a tube of 220 mm internal diameter. At junction radial pressure is 10 N/mm2 . The compound tube is subjected to an internal pressure of 70 N/mm2 . Find the variation of hoop stress over the wall of the compound tube. @seismicisolation @seismicisolation Thick Cylinders and Spheres 270 + 60 = 165 mm 2 220 = 110 mm r2 = 2 270 r0 = = 135 mm 2 b b −a Outer tube : p = 10 = − a; 10 = 2 18225 135 b − 10 a= 18225 b b 0= −a = −a 27225 1652 b a= 27225 • 227 r1 = (i) (ii) Substituting for a from Eqn. (ii) into Eqn. (i) b b 1 1 = − 10 or, 10 = b − 27225 18225 18225 27225 10 = b (5.487 × 10−5 − 3.673 × 10−5 ) ∴ b= 10 1.814 × 10−5 b = 551268 551268 a= = 20.25 27225 b σx = 2 + a x 551268 At radius = 165, σc = + 20.25 = 40.50 N/mm2 1652 551268 At radius = 135, σc = + 20.25 = 50.50 N/mm2 1352 Ans Ans Inner tube At radius 135 mm, σr = pr = 10 = b −a 18225 b −a At radius 110 mm, σr = 1102 b −a 0= 12100 b a= 12100 10 = @seismicisolation @seismicisolation b −a 1352 (i) (ii) 228 • Strength of Materials Substituting for a from Eqn. (ii) into Eqn. (i), b b − 18225 12100 10 = b (5.487 × 10−5 − 8.264 × 10−5 ) 10 b=− 2.777 × 10−5 b = −360101 360101 a=− = −29.76 12100 10 = or At radius r0 = 135 mm σc = = −360101 + (−29.76) 1352 −360101 − 29.76 18225 = −49.52 N/mm2 (Compressive) Ans At radius r2 = 110 −360101 − 29.76 1102 = −59.52 N/mm2 (Compressive) Ans σc = Stresses due to internal fluid pressure: At radius r2 = 110 p110 = σ110 = b −a 1102 b −a 12100 b a= − 70 12100 70 = (i) At radius r1 = 165 σ165 = p165 = 0 = ∴ a= b −a 1652 b 27225 @seismicisolation @seismicisolation (ii) Thick Cylinders and Spheres • 229 Substituting the value of a from Eqn. (ii) into Eqn. (i), b b = − 70 27225 12100 1 1 70 = b − 12100 27225 70 = b(8.264 × 10−5 − 3.673 × 10−5 ) 70 = b(4.591 × 10−5 ) ∴ b = 1524722 1524722 = 56 and a = 27225 At radius r3 = 165 mm: 1524722 + 56 1652 = 112 N/mm2 (Tensile) Ans σc = At radius r0 = 135 mm: 1524722 + 56 1352 = 139.66 N/mm2 (Tensile) Ans σc = At radius r2 =110: 1524722 + 56 1102 = 182 N/mm2 (Tensile) Ans σc = Inner tube Radii σc due to shrinkage σc due to inner fluid pressure N/mm2 Resultant σc N/mm2 N/mm2 Outer tube 180 mm 135 mm 165 mm 135 mm −59.22 182 122.78 −49.52 139.66 90.14 40.50 112 152.5 50.50 139.66 190.16 Thick Spherical Shells The thick spherical shells are used to store fluids under high pressure. Equations for hoop stress and radial stresses are obtained in a similar manner as in thick cylindrical shells, by applying Lame’s theory. The equations for circumferential (hoop) stress and radial stresses are given below: b +a r3 2b σr = 3 − a r σc = @seismicisolation @seismicisolation 230 • Strength of Materials E XAMPLE 13.7: A spherical shell of 150 mm internal diameter has an internal pressure of 40 MN/m2 . If the permissible tensile stress in 85 MN/m2 . Determine thickness of the shell. 150 = 75 mm = 0.075 m 2 (p) = 40 MN/m2 ri = Internal pressure Permissible stress (σh ) = 85 MN/m2 2b σr = 3 − a r b σh = 3 + a r At r = 0.075 m, the radial stress = 40 MN/m2 2b − a = 40 (0.075)3 2b − a = 40 4.2187 × 10−4 4740.8b − a = 40 (i) At r = 0.075 m, σh or σc = 85 MN/m2 b + a = 85 (0.075)3 b + a = 85 4.2187 × 10−4 2370.4b + a = 85 a = 85 − 2370.4b Substituting the value of a in Eqn. (i) 4740.8b − (85 − 2370.4b) = 40 4740.8b − 85 + 2370.4b = 40 7111.2b = 125 ∴ b = 0.01758 from Eqn. (ii) a = 85 − 2370.4 × 0.01758 = 43.33 Let the external radius be r1 , σr = 0 2b 2b − a or, a = 3 3 r r 2b 2 × 0.01758 3 r = = = 8.114 × 10−4 a 43.33 0= @seismicisolation @seismicisolation (ii) Thick Cylinders and Spheres ∴ • 231 r = 0.0933 m = 93.3 mm thickness = 0.0933 − 0.075 = 0.0183 m = 18.3 mm Ans Principal and shear stresses: Because there is no torque acting on the cylinder, so σc , σr and σl are the principal stresses. σr is compressive and σc and σl are tensile. σc is maximum and σr is minimum, We know, τmax (maximum shear stress) = τmax = or = = τmax = σmax − σmin 2 σc − (−σr ) (∴ σr is compressive) 2 σc + σr 2 1 b b +a + 2 +a 2 r2 r b r2 Changes in cylinder dimensions: Circumferential strain = 1 [σc − μσr − μσl ] E Change in diameter at any radius r of the cylinder is given by δd = Change in length, δ l = 2r [σc − μσr − μσl ] E 1 [σl − μσr − μσc ] E Exercise 13.1 A thick pipe of steel is of 450 mm diameter and of thickness 100 mm. It is subjected to an internal fluid pressure of 60 bar. Find the hoop stress at internal and outer surfaces. [Ans σc = 15.6 MN/m2 and 9.6 MN/m2 at inner and outer surface respectively] 13.2 A cylinder 200 mm outside diameter and 100 mm inside diameter has an internal fluid pressure of 20 MN/m2 . Determine the longitudinal stress, assuming it to be uniform. [Ans 6.67 MN/m2 ] 13.3 Calculate the thickness of the metal necessary for a steel cylindrical shell of internal diameter 0.15 m to withstand an internal pressure of 50 MN/m2 , the permissible tensile stress is not to exceed 150 MN/m2 . [Ans 31 mm] @seismicisolation @seismicisolation 232 • Strength of Materials 13.4 A thick walled closed-end cylinder is made of an aluminium alloy having E = 72 GPa and μ = 0.33, has inside diameter of 200 mm and outside diameter of 800 mm. The cylinder is subjected to internal fluid pressure of 150 MPa. Determine the circumferential stresses at 0.1m radius and 0.4 m radius [Ans 170 MN/m2 and 20 MN/m2 ] 13.5 Calculate the ratio of thickness to internal diameter for a tube subjected to internal pressure when the pressure is 5/8 of the value of the maximum permissible circumferential stress. Find the increase in internal diameter of such a tube 100 mm internal diameter when the internal pressure is 80 MN/m2 . Also find the change in wall thickness. t = 0.54, δ d = 0.07306 mm, decrease in wall thickness = 0.01438 mm] [Ans d 13.6 A hub is shrunk onto a solid shaft and the final dimensions are: Diameter at the junction = 150 mm External diameter of hub = 200 mm Find the initial difference in the diameters of the hub and the shaft to produce a radial contact pressure of 30 MPa. E = 200 GPa. [Ans 0.1028 mm] 2 13.7 An external pressure of 10 MN/m is applied to a thick cylinder of internal diameter 150 mm and external diameter 300 mm. If the maximum hoop stress permitted on the inside wall is 35 MN/m2 . Calculate: i) The maximum internal pressure that can be applied ii) The change in outside diameter if the cylinder has the closed ends, E = 210 GPa, μ = 0.3 [Ans p = 37 MN/m2 , δ d = 0.03423 mm] 13.8 A compound cylinder formed by shrinking one tube onto another is subjected to an internal pressure of 60 N/m2 . Before the fluid is admitted, the internal and external diameters of the compound cylinder are 120 mm and 200 mm and the diameter at the junction is 160 mm. If, after shrinking on, the radial pressure at the common surface is 8 N/mm2 . Calculate the final stress set up by the section Ans [Ans See table below:] Hoop stress Inner tube Outer tube N/mm2 x = 60 mm x = 80 mm x = 80 mm x = 100 mm Initially Due to fluid pressure Resultant −36.58 +127.5 +90.92 −28.58 +86.48 +57.9 +36.44 +86.48 +122.92 +28.44 +67.5 +95.94 ‘+’ for tension ‘-’ for compression @seismicisolation @seismicisolation Thick Cylinders and Spheres • 233 13.9 A compound thick cylinder is formed by shrinking a tube of external diameter 300 mm over another tube of internal diameter 150 mm. After shrinking, the diameter at the junction of the tubes is found to be 250 mm and radial compression at 28 N/mm2 . Find the original difference in radii at the junction. Take E for the cylinder metal as 200 GPa. [Ans 0.13 mm] 13.10 A spherical shell of 80 mm internal diameter has to withstand an internal fluid pressure of 30 MN/m2 . Find the thickness of the shell if permissible stress in 80 MN/m2 . [Ans 7.7 mm] @seismicisolation @seismicisolation C HAPTER 14 DEFLECTION OF BEAMS Relation between Slope, Deflection and Radius of Curvature To find out relation between slope, deflection and radius of curvature, consider Fig. 13.1 where a small portion AB of a beam bent into an arc as shown in the figure. C = Centre of arc into which the beam has been bent α = Angle which the tangent at A makes with x − x axis ds = length of beam AB α + d α = Angle which the tangent at B makes with x − x axis. Y C da A a 0 B R ds dy D dx a+da X Figure 14.1 It is clear from the geometry of the figure that or ∠ACB = d α and ds = Rd α ds dx ∴ R= = (ds = dx, being very small) dα dα dα 1 = R dx (i) dy dy or α = , because tan θ = θ for very small angles. Now, tan α = dx dx Differentiating the above equation with respect to x, we have d2y dα = 2 dx dx 1 d2y = 2 R dx [because 1 dα = as shown in Eqn. (i)] dx R @seismicisolation @seismicisolation (ii) Deflection of Beams • 235 EI 1 M M E = or, M = or = I R R R EI 1 Substituting in Eqn. (ii) for , R From bending equation, M d2y = 2 EI dx ∴ M = EI d2y dx2 (iii) Equation (iii) is called the curvature-moment relation or the differential equation of flexure. The term EI is known as flexural rigidity. Deflection of beams: It can be found by the following methods: (A) (B) (C) (D) (E) (F) Double integration method Macaulay’s method Moment area method Conjugate beam method Superposition method Strain energy method. (A) Double Integration Method Cantilevers 1. Cantilever with a Concentrated Load at its Free End W l X x B A yB Figure 14.2 Consider section X at a distance of x from free end. Mx = −W x EI d2y dx2 (due to hogging minus sign is taken) = −W x (i) Integrating Eqn. (i) E W x2 dy =− +C1 dx 2 C1 is the constant of integration, we know that when x = l, (ii) dy =0 dx Eqn. (ii), O=− W l2 +C1 2 or, @seismicisolation @seismicisolation C1 = W l2 2 dy is slope . Substituting in dx 236 • Strength of Materials Now Eqn. (ii) becomes, EI dy W x2 W l 2 =− + dx 2 2 (iii) This is the required equation for the slope at any point on cantilever. The maximum slope occurs dy as slope i for at the free end. For maximum slope, substituting x = 0 in Eqn. (iii). (Let us denote dx very small angles). EI or W l2 2 W l2 iB = EI iB = Integrating the Eqn. (iii) once again, EIy = − W x3 W l 2 x + +C2 6 2 (iv) C2 is constant for integration. When x = 0, y = 0, substituting these values of x and y in the above equation, W l3 W l3 W l3 + +C2 = +C2 6 2 3 W l3 C2 = − 3 O=− ∴ Substituting for C2 in Eqn. (iv), EIy = − W x3 W l 2 x W l 3 + − 6 2 2 From Eqn. (v), deflection y can be calculated at any point of cantilever. For maximum deflection, x = 0 W l3 3 W l3 yB = − 3EI W l3 = 3EI EIyB = − or (minus sign is dropped because it is due to sign convention only). @seismicisolation @seismicisolation (v) Deflection of Beams • 237 2. Cantilever with a Concentrated Load not at the Free End W l l1 C yC A B C' yB B' Figure 14.3 It should be noted that portion AC will bend as AC ; while the portion CB remains straight. W l12 Slope, ic = 2EI Since portion C B is straight, W l12 iB = iC = 2EI and from previous article, y= W l13 3EI From the geometry of the figure, we find out that yB = yC = ic (l − l1 ) = W l13 W l12 + (l − l1 ) 3EI 2EI 3. Cantilever with Uniformly Distributed Load l A w/m x x x B iB B' yB Figure 14.4 wx2 2 2 d y wx2 EI 2 = − 2 dx Mx = − Integrating the above equation EI dy wx3 =− +C1 dx 3 @seismicisolation @seismicisolation (i) 238 • Strength of Materials When x = l, then slope dy = 0, substituting in the above equation; dx O=− wl 3 +C1 6 or C1 = wl 3 6 So, Eqn. (i) because, EI wx3 wl 3 dy =− + dx 6 6 (ii) wl 3 This is the equation for slope at any point on cantilever. Maximum slope at free end B, iB = 6EI Integrating Eqn. (ii), we get, EIy = − wx4 wl 3 x + +C2 24 6 (iii) When x = l, y = 0 substituting in Eqn. (iii) We get O= wl 4 wl 4 + +C2 24 6 ∵ C2 = wl 4 8 Substituting in Eqn. (iii) EIy = − wx4 wl 3 x wl 4 + + 24 6 8 This equation can give deflection at any point on cantilever. Maximum deflection is at free and B, putting x = 0. wl 4 8EI wl 4 = 8EI yB = − (Minus sign is removed. It depends on sign convention) or yB = W l3 ; where W = wl. 8EI 4. Cantilever Partially Loaded with Uniformly Distributed Load l1 A l C B iB yc C' B' Figure 14.5 @seismicisolation @seismicisolation yB Deflection of Beams • 239 Portion AC of the cantilever will bend into AC , while the portion CB will remain straight and will take new position as C as shown in Fig. 14.5, Cantilever with uniformly distributed load (partially) will have slope iB if the tangent is drawn from point B . ic = wl13 6EI (from previous article) Because the portion CB of the cantilever is straight, ∴ iB = ic = yc = wll3 6EI wll4 8EI (from previous article) From the geometry of the figure, we have yB = yc + ic (l − l1 ) = wl 3 wl14 + 1 (l − l1 ) 8EI 6EI 4 3 l l 7wl 4 w w l + × = 8EI 2 6EI 2 2 384EI 5. Cantilever Loaded from the Free End Cor. if l1 = l/2 then yB = A w/unit length B C l1 l (a) B A C l1 l (b) Figure 14.5a,b In each case redraw the figure with uniformly distributed load throughout the length l. And draw an upward distributed load (w/unit length from fixed end to point C for a length of (l − l1 ). Now obtain slope and reflections due to newly loaded beam as per Article 3 and Article 4. Then the slope at B is equal to the slope due to the total load minus the slope due to superimposed load upwards. Now, the deflection at B is equal to the deflection due to the total load minus the deflection due to superimposed load. @seismicisolation @seismicisolation 240 • Strength of Materials wl 3 w(l − l1 )3 − 6EI 6EI 4 wl w(l − l1 )4 w(l − l1 )3 yB = − + l1 8EI 8EI 6EI iB = and The slope and deflection at A due to superimposed uniformly distributed load from A to C is obtained by substituting (l − l1 ) for l 6. Cantilever with a Gradually Varying Load X x w per unit length B A iB l yB Figure 14.6 1 wx x wx3 Mx = − × ×x× = − 2 l 3 6l d2y wx3 EI 2 = − 6l dx Integrating, EI when x = l, then wx4 dy =− +C1 dx 24l (i) dy = 0, substituting in the equation above, dx wl 4 +C1 OR, 24l dy wx wl 3 EI =− + dx 24l 24 O= − ∴ C1 = + wl 3 24 (ii) This equation will give slope at any distance x from free end integrating Eqn. (ii) again, EI iB = wl 3 24 or, iB = wl 3 radian 24EI Integrating Eqn. (ii) again, EIy = − w wl 3 + + C2 120l 24 @seismicisolation @seismicisolation (iii) Deflection of Beams • 241 when x = l, y = 0, substituting in Eqn. (iii) wl 4 wl 4 wl 4 + + C2 , or C2 = − 120 24 30 wx5 wl 3 x wl 4 EIy = − + − 120l 24 30 O=− ∴ (iv) This is required equation for deflection at any distance x from free end. For maximum deflection, x=0 ∴ wl 4 30 EI wl 4 = 30 EI y=− (minus sign means downward deflection) E XAMPLE 14 A-1: A cantilever beam is 130 mm wide, 165 mm deep and 2 m long. Determine the slope and deflection at the free end of the beam, if it is loaded with a concentrated load of 25 kN at the free end. Take E = 200 Gpa S OLUTION : I= 130 × 1653 = 48664687.5 mm4 12 W l2 iB = 2EI iB = 25 kN A B (as derived before) 25000 (2000)2 2 × 200000 × 48664687.5 2m = 0.00514 radian Ans Figure 14.6a Deflection at free end B, yB = W l3 3EI yB = 25000 (2000)3 = 6.85 mm 3 × 200000 × 48664687.5 (as derived before) Ans E XAMPLE 14 A-2: A cantilever of length 2.5 m is carrying a point load of 60 kN at a distance of 1.8 m from the fixed end. If I = 9 × 107 mm4 and E = 200 GPa, find: (i) slope at the free end and (ii) deflection at the free end. @seismicisolation @seismicisolation 242 • Strength of Materials S OLUTION : 60 kN W l2 θB = 1 2EI 60000 × (1800)2 θC = θB = 2 × 200000 × 9 × 107 = 0.0054 radian C A B 1.8 m 2.5 m Figure 14.7 W (l1 )3 W (l1 )2 + (l − l1 ) 3EI 2EI 60000 (1800)3 60000 (1800)3 = + (2.5 − 1.8) 3 × 200000 × 9 × 107 2 × 200000 × 9 × 107 = 6.48 + 6.804 = 13.284 mm Ans yB = E XAMPLE 14 A-3: A cantilever of length 2.8 m carries a uniformly distributed load of 10 kN/m over a length of 2 m from the free end. If I = 8 × 109 mm4 and E = 200 GPa; determine: i) the slope at the free end, and (ii) deflection at the free end. S OLUTION : In such cases assume uniformly distributed load is extended up to fixed end A. And now assume 10 kN/m is acting upwards from A to C as show Fig. 14.9. Now using the results of derivations earlier in this section, we get θB = A 2m 2.8 m (l1) (l) Figure 14.8 10 kN/m B A wl 3 w(l − l1 )3 − 6EI 6EI 10 kN/m B C C 10 kN/m (l) Figure 14.9 10 × (2800)3 10 (2800 − 800)3 θB = − 6 × 200000 × 8 × 109 6 × 200000 × 8 × 109 = 0.0000229 − 0.00000833 = 0.00001457 radian Ans @seismicisolation @seismicisolation Note here 10000 = 10 N/mm 1000 Deflection of Beams • 243 wl 4 w(l − l1 )3 l1 − yB = 8EI 6EI (ii) 10 (2800 − 800)3 800 10 (2800)4 − = 8 × 200000 × 8 × 109 6 × 200000 × 8 × 109 = 0.04802 − [0.00667] = 0.04135 mm Ans E XAMPLE 14 A-4: A cantilever of length 3.5 m carries two point loads of 1.5 kN at the free end and 3 kN at a distance of 1.2 m from the free end. Find the deflection at free end. E = 200 GPa, I = 109 mm4 . S OLUTION : 3 kN Deflection at free end due to 1.5 kN, yB1 l1 = 2.5 m wl 3 = 3EI = C A 1500 × (3500)3 3 × 200000 × 109 1.5 kN B 1m 3.5 m Figure 14.10 = 0.1072 mm Deflection due to 3 kN at free end, yB2 = = wl13 W l12 (l − l1 ) + 3EI 2EI 3000 × (2500)2 (1000) 3000 (2500)3 + 3 × 200000 × 109 2 × 200000 × 109 = 0.078125 + 0.046875 = 0.125 mm Total deflection = yB1 + yB2 = 6.1072 + 0.125 = 0.2322 mm Ans E XAMPLE 14 A-5: A cantilever of length 2.5 m carries a uniformly distributed load of 2 kN/m run for a length of 1.4 m from the fixed end and a point load of 1 kN at the free end. Find the deflection at the free end if the section is rectangular: 120 mm wide 250 mm deep. E = 10000 N/mm2 @seismicisolation @seismicisolation 244 • Strength of Materials S OLUTION : I= 120 × 2503 = 156250000 mm4 12 2 kN/m A 1 kN B C 1.1 m 1.4 m = l1 2.5 m Figure 14.11 Deflection y1 due to 1 kN, y1 = W l3 1000 (2500)3 = = 3.33 mm 3EI 3 × 10000 × 156250000 y2 due to uniformly distributed load of 2 kN/m at free end y2 = = wl14 wl + (l − l1 ) 8EI 6EI 2 (1400)3 (2500 − 1400) 2 (1400)4 + 8 × 10000 × 156250000 6 × 10000 × 156250000 = 0.6147 + 0.6439 = 1.2586 mm Total deflection yB , yB = 3.33 + 1.2586 = 4.5886 mm Ans E XAMPLE 14 A-6: A cantilever of length 3.5 m carries a uniformly varying load of zero intensity at the free end and 20 kN at fixed end. E = 200 GPa, I = 0.9 × 108 mm4 . Find the slope and deflection at the free end. 20 kN/m B A 3.5 m Figure 14.12 @seismicisolation @seismicisolation Deflection of Beams • 245 S OLUTION : θB = wl 3 24EI (derived before) 20 (3500)3 = 0.001985 radian 24 × 200000 × 0.9 × 108 wl 4 (derived before) yB = 30EI 20 (3500)4 = = 5.56 mm Ans 30 × 200000 × 0.9 × 108 = 1. Simply Supported Beam with a Central Point Load X W x A C X l/2 W/2 yc B l/2 W/2 Figure 14.13 W Reactions: RA = RB = 2 Consider a section X at x from point A, Wx 2 d2y W x EI 2 = 2 dx Mx = (i) Integrating Eqn. (i) EI dy W x2 = × +C1 dx 2 2 (ii) l dy = 0, substituting When x = , 2 dx O= ∴ 2 W l 1 × × +C1 ; 2 2 2 C1 = − W l2 16 @seismicisolation @seismicisolation = W l2 +C1 = 0 16 246 • Strength of Materials Substituting for C1 in Eqn. (ii), EI This is required equation for slope dy W x2 W l 2 = − dx 4 16 (iii) dy (in radians). Integrating Eqn. (iii) again, dx W x3 W l 2 K − +C2 12 16 EIy = (iv) When x = 0; y = 0, substituting in Eqn. (iv), ∴ 0 = 0 − 0 +C2 C2 = 0 Now Eqn. (iv) becomes, EIy = W x3 W l 2 x − 12 16 This is the required equation for deflection. l For maximum deflection at midpoint where point load W is acting, substituting x = ; we get, 2 3 W EIy = = 12 y= − W l2 l 2 16 W l3 W l3 − 96 32 y=− So l 2 W l3 48EI (minus sign shows downward deflection) W l3 48EI 2. Simply Supported Beam with Uniform Distributed Load x X w unit of length A wl RA = 2 C yc l Figure 14.14 @seismicisolation @seismicisolation B wl RB = 2 Deflection of Beams • 247 wl Reactions = RA = RB = , c is the centre. 2 Consider a section X at a distance x from A, x2 2 wlx wx2 = − 2 2 d 2 y wlx wx2 EI 2 = − 2 2 dx Mx = RA x − w Integrating EI dy wlx2 wx3 = − +C1 dx 4 6 (i) l dy = o, substituting in Eqn. (i) When x = , 2 dx o= wl 3 wl 3 − +C1 16 48 ∴ C1 = − wl 3 24 Equation (i) becomes EI dy wlx2 wx3 wl 3 = − − dx 4 6 24 (ii) This is required equation for slope at any point. Integrating Eqn. (ii), EIy = wlx3 wl 4 wl 3 x − − +C2 12 24 24 when x = 0, y = 0 ∴ 0 = 0 − 0 − 0 +C2 ∴ EIy = wlx3 12 − wx4 24 − C2 = 0 wl 3 x 24 (iii) This is the required equation for deflection at one point. For maximum deflection yc at centre, x = l/2, substituting in Eqn. (iii), EIy = 5wl 4 wl 4 wl 4 wl 4 − − =− 96 384 48 384 ∴ Maximum deflection at mid span 5wl 4 384EI 5wl 4 yc = 384EI yc = − or (minus sign indicates downward deflection) @seismicisolation @seismicisolation 248 • Strength of Materials If we substitute wl = W , then yc = − 5W l 3 384EI Deflection due to impact: If a mass M drops through a height h produces a maximum instantaneous displacement y and let a load W gradually applied produces the same deflection y. Then the work done in the two cases is the same, 1 Mg (h + y) = wy 2 Y = kw ∴ where k depends on the nature of the beam and the position of load ∴ 1 Mg (h + kw) = kw2 2 This is a quadratic from which W , the equivalent static load, can be obtained. Then, y = kw 3. Simply Supported Beam with Point Load Eccentric Consider the beam shown in Fig. 14.15. W In this case point where slope is zero is not defined because of eccentric load. C A Wb RA = ; l Wa RB = l a B b l Wb l Wa l Figure 14.15 Let as consider portion AC, any point at a distance x from A, W bx l 2 dy W bx EI 2 = l dx M= Integrating, EI dy W bx2 = +C1 dx 2l Integrating again, EIy = W bx3 +C1 x +C2 6l @seismicisolation @seismicisolation (i) Deflection of Beams • 249 At A, x = 0, y = 0, substituting in the above equation, 0 = 0 + 0 +C2 ∴ EIy = W bx3 6l ∴ C2 = 0 +C1 x (ii) Now consider portion CB, at any point at a distance x from A, W bx −W (x − a) l d 2 y W bx EI 2 = −W (x − a) l dx Mx = Integrating, ET dy wbx2 W (x − a)2 = − +C3 dx 2l 2 (iii) EIy W bx3 (x − a)3 −W +C3 x +C4 6l 6 (iv) Integrating again, W bx2 +C1 2l W bx2 W (x − a)2 And slope at C from Eqn. (iii) = − +C3 2l 2 Therefore, equating the above two equations, Now slope at C from Eqn. (i) = W bx2 W (x − a)2 W bx2 +C1 = − +C3 2l 2l 2 When x = a Equation (v) becomes, W bx2 W bx2 +C1 = +C3 2l 2l or C1 = C3 Also deflection at C form Eqn. (ii) = deflection at C from Eqn. (iv) Equating, W bx3 W bx3 W (x − a)3 +C1 x = − +C3 x +C4 6l 6l 6 @seismicisolation @seismicisolation (v) 250 • Strength of Materials At C, x = a, C1 = C3 , substituting x = a, W ba3 W ba3 W (a − a)3 +C1 a = − +C3 a +C4 6l 6l 6 Since C1 = C3 ∴ At B, x = l, y = 0 From Eqn. (iv), C4 = 0 0= (l − a)3 W bl 3 −W +C3 l 6l 6 ∵ C4 = 0 W bl 3 W b3 − +C3 l = 0 6l 6 or, ∴ C3 = − W bl 3 W b3 + 6l 6l Wb 2 l − b2 6l From Eqn. (ii), if x ≤ a, then Since C1 = C3 ∴ C1 = − W bx3 W b(l 2 − b2 )3 x − 2l 6l W bx3 w(x − a)3 Wb 2 EIy = − − (l − b2 )x 6l 6 6l EIy = Also, This is for x ≥ a, ∴ yc = (from Eqn. iv) Wb 2 a + b2 − l 2 if x = a 6EIl Note: Thus from the above two equations, deflection at any point can be determined, provided right equation is used, depending upon value of x. If it is less or more than a. For maximum deflection we can work as follows: Let a > b, dy W bx2 W b 2 From Eqn. (i) EI = − (l − b2 ) dx 2l 6l dy =0 We know that at maximum deflection, slope, dx 0= Solving, x= W bx2 Wb 2 − (l − b2 ) 2l 6l l 2 − b2 3 @seismicisolation @seismicisolation Deflection of Beams Thus, maximum deflection, Wb EIy = 6l ymax = l 2 − b2 3 3 − Wb 2 (l − b2 ) 6l • 251 l 2 − b2 3 W b (l 2 − b2 )3/2 √ 9 3 EIl Distance of point of maximum deflection from the centre = l l l The maximum value of which can be = √ − = 3 2 13 Therefore it is to be noted that maximum deflection lies within l 2 − b2 l − 3 2 l of the centre. 13 (B) Macaulay’s Method As we have seen in the preceding article that we have to make separate expression for different sections in double integration method. This method is okay for simple cases, but becomes very tedious and complicated for complex problems. To make calculations easier Macaulay’s method is very convenient. In this method (x − a) is expressed in brackets and while solving the problems it is treated in a special manner which will be explained through problems. Let us take previous case again (Refer Fig. 14.16). W A a Wb Wa RA = ; RB = l l B b k l Wa l Wb l Figure 14.16. Consider a section X at a distance x from A, Mx = EI d 2 y W bx = −W [x − a] ; l dx2 The left of this line is for AC and the whole for CB. Integrating, EI X C W [x − a] 2 dy W bx2 = +C1 − dx 2l 2 @seismicisolation @seismicisolation 252 • Strength of Materials Again integrating, EIy = W [x − a]3 W bx3 +C1 x +C2 − 2l 6 [x − a]2 is integrated in special manner in, Macaulay’s method. Note: Now 2 At A, x = 0, y = 0, The equation for portion AC, W bx3 +C1 x +C2 2l 0 = 0 + 0 +C2 ∴ EIy = C2 = 0 At B, x = l, y = 0. Now as indicated before, the whole equation is for CB. ∴ ∴ 0= W b3 W bl 3 +C1 l − 2l 6 C1 = − or, 0= W bl 2 W b3 +C1 l − 2 6 Wb 2 (l − b2 ) 6l Therefore, the slope and deflection are given by equations: [x − a]2 dy W bx2 W b 2 = − (l − b2 )−W dx 2l 6l 2 [x − a]3 W bx3 W b 2 2 − (l −b )x−W EIy = 6l 6l 6 EI Deflection under load, Wab 2 2 2 W ba3 W b 2 2 − l −b a = − l −b −a 6l 6l 6l Wab (a + b)2 −b2 −a2 =− 6l Wab 2 2 a +b +2ab − b2 −a2 EIyc = − 6l EIyc = yc = − Wa2 b2 3l EI (minus sign indicates downward deflection) For maximum deflection we have to proceed as done already in double integration method. E XAMPLE 14 B-1: A simple supported beam has a span of 15 m and carries two point loads of 4 kN and 9 kN at 6 m and 10 m, respectively from one end. Find the deflection under each load and maximum deflection. E = 200 GPa, I = 400 × 106 mm4 . @seismicisolation @seismicisolation Deflection of Beams • 253 S OLUTION : 4 kN 9 kN A D 6m RA = 5.4 kN 5m 4m B 7.6 kN 10 m 15 m Figure 14.17 Taking moments about B, 15RA = 4 × 9 + 9 × 5 = 81 ∴ RA = 5.4) = 7.6 kN; Using Macaulay’s method, Mx = EI 81 = 5.4 kN RB = (4 + 9 − 15 d2y = 5.4x − 4 (x − 6) − 9 (x − 10) dx2 x2 (x − 6)2 (x − 10)2 dy = 5.4 −4 −9 +C1 dx 2 2 2 EI EIy = 5.4 (i) x3 4 (x − 6)3 9 (x − 10)3 − − +C1 x +C2 6 6 6 When x = 0, y = 0 0 = 0 − 0 − 0 + 0 +C2 ∴ C2 = 0, When x = 15, y = 0 0= 5.4 (15)3 4 (15 − 9)3 9 (15 − 10)3 − − +C1 x 6 6 6 0 = 3037.5 − 486 − 187.5 + 15C1 ∴ C1 = −157.6; Working in N and m, I = ∴ EIy = 400 × 106 (1000)4 5.4x3 4 (x − 6)3 9 (x − 10)3 − − −157.6x 6 6 6 (ii) = 4 × 10−4 m4 E = 2 × 1011 N/m2 Since we have been working in kN, so will multiply by 1000 the equation of y. For deflection at C, put x = 6 m in Eqn. (ii) @seismicisolation @seismicisolation 254 • Strength of Materials 1 5.4 × 63 yc = − 0 − 0 − 157.6 × 6 EI 6 = 1000 2 × 1011 × 4 × 10−4 =− [194.6 − 945.6] 751.2 × 1000 = −0.00939 m = −9.39 mm 2 × 1011 × 4 × 10−4 Ans Minus sign means downward deflection. ∴ deflection at c, yc = 9.39 mm Ans Deflection at D: Putting x = 10 m 1000 5.4 × 103 4 (10 − 6)3 yD = − −0 − 157.6 × 10 6 6 2 × 1011 ×4 × 10−4 = 1.25 × 10−5 [900 − 42.67 − 1576] =− 1.25 × 718.67 = −0.00898 m = 8.98 mm 105 Ans dy For maximum deflection =0 dx Using Eqn. (i) 0 = 2.7x2 − 2 x2 + 36 − 12x − 4.5 x2 + 100 − 20x − 157.6 0 = 3.8x2 − 114x + 679.6 0 = x2 − 30x + 178.84 √ +30 ± 900 − 715.36 30 ± 13.6 = x= 2 2 = 8.2 m Ans Maximum deflection will occur at 8.2 m. Substituting x = 8.2 in Eqn. (ii) EIy = 5.4 4(8.2 − 6)3 (8.2)3 − − 0 − 157.6 × 8.2 6 2 EIy = 496.23 − 7.09 − 1292.32 = −803.18 or y=− 803.18 × 1000 2 × 1011 × 4 × 10−4 @seismicisolation @seismicisolation Deflection of Beams • 255 y = 0.010 m y = 10 mm or Ans Note: If term is in bracket is −ve, then, it is regarded as zero in Macaulay’s method, for example, (8.2 − 10)3 in Eqn. (ii) will be regarded as zero as shown above. E XAMPLE 14 B-2: A horizontal beam AB a freely supported at A and B, 9 m apart and carries a uniformly distributed load of 12 kN/m run. A clockwise moment of 120 kNm is applied to the beam at a point C, 3.5 m from the left hand support A. Calculate the slope and defection of the beam at C. If flexural rigidity EI = 40 × 103 kNm2 . S OLUTION : 120 kN/m 12 kN/m A 3.5 m X x 40.7 kN B C 67.33 kN 9m Figure 14.18 Taking moments about A, 9 + 120 RB × 9 = 12 × 9 × 2 = 606 kN 606 = 67.33 kN ∴ RB = 9 RA = (12 × 9) − 67.33 = 40.7 kN Using Macaulay’s method, taking A as origin and considering section X at a distance of x metres from A, d2y x Mx = EI 2 = 40.7x − 12x × + 120 (x − 3.5)◦ 2 dx Note: (x − 3.5)◦ is taken when couple is applied. Integrating, x2 x3 dy = 40.7 − 12 + 120 (x − 3.5) +C1 dx 2 6 dy = 20.35x2 − 2x3 + 120 (x − 3.5) +C1 EI dx EI @seismicisolation @seismicisolation (i) 256 • Strength of Materials Integrating again, x3 2x4 120 (x − 3.5)2 − + +C1 x +C2 3 4 2 EIy = 6.78x3 − 0.5x4 + 60 (x − 3.5)2 +C1 x +C2 EIy = 20.35 (ii) Now when x = 0, y = 0 0 = 0 − 0 + 0 + 0 +C2 ∴ C2 = 0 Eqn. (ii) becomes, EIy = 6.78x3 − 0.5x4 + 60 (x − 3.5)2 +C1 x (iii) when x = 9, y = 0, substituting Eqn. (iii), ∴ 0 = 6.78 (9)3 − 0.5 × 94 + 60 (9 − 3.5)2 + 9C1 0 = 4942.62 − 3280.5 + 1815 + 9C1 C1 = −386.35 So Eqn. (ii) and Eqn. (iii) become, EI For slope dy = 20.35x2 − 2x3 + 120 (x − 3.5) − 386.35 dx EIy = 6.78x3 − 0.5x4 + 60 (x − 3.5)2 − 386.35x dy at C, substituting x = 3.5 in Eqn. (i) dx EI dy = 20.35 (3.5)2 − 2 (3.5)3 + 120 (3.5 − 3.5) − 386.39 dx = 249.3 − 85.75 + 0 − 386.35 1 [−222.45] = 40 × 103 = −0.00556 radian = 0.00556 radian. Ans For deflection at C, using Eqn. (v) and substituting x = 3.5; EIy = 6.78 (3.5)3 − 0.5 (3.5)4 + 60 (3.5 − 3.5)2 − 386.35 × 3.5 = 290.7 − 75 + 0 − 1352.225 1 (−1136.5) y= 40 × 103 = −0.0284 m (minus sign means deflection is downward) = 28.4 mm Ans @seismicisolation @seismicisolation (iv) (v) Deflection of Beams • 257 Beams of composite section (Flitched Beams): When a beam is made up of two or more materials joined together such that it behaves like a single piece and the deflection of each material at a certain point is same. It is possible to find slope and deflection of beam of composite section by algebraically adding the flexural rigidities (EI) of different materials. ∑ EI = E1 I1 + E2 I2 If should be noted that the moment of inertia of the composite section is to be found out about the centre of gravity of the section. E XAMPLE 14 B-3: A flitched beam of space 7 m consists of a timber section 170 mm wide and 250 mm deep. Two plates of 170 mm and 15 mm thick are fixed at the top and bottom of the timber section. The composite beam is subjected to a point load of 90 kN at the middle of the beam. Determine the deflection of the beam under the load. Take E for steel as 200 GPa and E for timber as 10 GPa, respectively. S OLUTION : In this case centre of gravity of steel plates and of timber is at the centre of the whole section because of symmetry. 15 mm 250 mm ∴ EItimber = 10000 15 mm 170 × 2503 12 = 2.21 × 1012 N mm2 170 mm Figure 14.19 170 (15)3 Esteel = 200000 2 × + 2 (170 × 15) × (132.5)2 12 = 200000 [95625 + 89536875] = 17.9 × 1012 Total flexural rigidity of the composite section about its centre of gravity, ΣEI = 2.21 × 1012 + 17.9 × 1012 = 20.11 × 1012 Deflection at the centre of the beam, yc = W l3 90000 × (7 × 1000)3 = 48ΣEI 48 × 20.11 × 1012 = 31.98 mm Ans @seismicisolation @seismicisolation 258 • Strength of Materials E XAMPLE 14 B-4: A composite beam consists of two timber joists 130 mm wide and 300 mm deep with a steel plate 240 mm deep and 15 mm thick fixed symmetrically between the timber joists. The beam carries a uniformly distributed load of 4 kN/m and is simply supported over a span of 5 metres. Find the slopes at supports and deflection of the beam at its centre. Take Es = 200 GPa and Et = 10 GPa. S OLUTION : It is clear from the figure that the centre of gravity of the beam section coincides with the centre of gravity of the steel plate. Therefore, flexural rigidity for the timber joists, 15 mm 300 mm 240 mm 130 mm 130 mm (a) 4 kN/m A B 5m (b) Figure 14.20 130 × 2803 EIt = 10 × 1000 2 × 12 = 4.756 × 1012 Nmm2 15 × 2403 EIs = 200 × 1000 12 = 3.456 × 1012 Nmm2 Total flexural rigidity of the composite beam about its centre of gravity, ΣEI = 4.756 × 1012 + 3.456 × 1012 = 8.212 × 1012 Nmm2 @seismicisolation @seismicisolation Deflection of Beams • 259 Slope at the supports, Because of symmetry θB = θA = wl 3 24ΣEI 3 4 × 5 × 103 = 24 × 8.212 × 1012 ∵ 5 × 1000 = 5 N/mm 5 kN/m 1000 = 2.537 × 10−3 = 0.00254 radian Deflection of the beam at its centre, Yc = = 5wl 4 384ΣEI 5 × 4 (5 × 1000)4 384 × 8.212 × 1012 = 3.96 mm Ans (C) Moment Area Method This method is based on the application of two theorems known as Mohr’s theorem. This method is partially convenient in case of beams acted upon with concentrated loads, in which case bending moment area consists of triangles and rectangles. However, this method may be conveniently used in certain cases of distributed load where the position of the centroid of the bending moment area is known. Theorem I: The change in slope between any two points on a beams is equal to the net area of the bending moment diagram between these points, divided by the flexural rigidity (EI) of the beam. Let AB [Fig. 14.21(a)] be part of C beam, which has been deflected to the position A B and let Fig. 14.21(b) represent the corresponding part of the BM diagram. x2 x1 A y1 A' B θ1 y2 B' (a) M dx (b) Figure 14.21 @seismicisolation @seismicisolation θ2 260 • Strength of Materials If the slopes at A and B, distances x1 and x2 from origin are θ1 and θ2 , respectively, We know EI d2y =M dx2 ∴ slope of beam at any point θ= dy = dx = M dx EI 1 EI M.dx, as E and I are constants. 1 × area of BM diagram between points A and B. Then θ2 − θ1 = EI If θ1 = 0, than θ2 the actual slope of the beam at B. M d2y = 2 EI dx Multiplying both sides by x, x d2y M = x 2 EI dx Integrating by parts between the limits of x, and x2 , x2 x2 dy 1 x −y = M.x.dx dx EI x1 x1 = 1 x EI moment of area of BM diagram about origin. dy By a suitable choice of origin, x can usually be made zero at both limits and y can be made zero dx at one limit leaving the value of y at other limit to represent the required deflection. In should be noted that moment-area method is usually convenient to use only when a point of zero-slope is known. For ready reference following table gives the value of area A, distance from origin x̄, and BM diagrams. @seismicisolation @seismicisolation Deflection of Beams BM Diagram x Area Remarks b h 1 bh 2 2 b 3 Concetrated Load for centilever 1 bh 3 3 b 4 Uniformly distributed load on centilever 1 bh 4 4 b 5 Uniformly varying load on a centilever 2 bh 3 5 b 8 Uniformly distributed load on single supported beam x b h Square parabola x b h Cubic parabola x h x b Figure 14.22 Let us take some standard cases: (a) Cantilever with concentrated end load; Fig. 14.23: θb − θa = 1 1 × W l.l EI 2 W A θb W l2 θb = since θa = 0 2EI Taking the origin at B, 1 1 2 dy − y]l = × W l.l × l dx EI 2 3 wl 3 [0 − 0] − [0 − yb ] = yb = 3EI [x wl yb G 2l 3 Figure 14.23 @seismicisolation @seismicisolation B BM Diagram • 261 262 • Strength of Materials (b) Cantilever with uniformly distributed load; Fig. 14.24: 1 1 wl 2 θb − θa = × .l EI 3 2 A wl 3 i.e. θb = since θa = 0 6EI w/unit length B θb yb l Taking the origin at B, l 1 1 wl 2 3 dy × l× l x −y = dx EI 3 2 4 6 i.e. (0 − 0) − (0 − yb ) = yb = wl2 2 wl 4 8EI 3 l 4 BM Diagram Figure 14.24 (c) Cantilever with end couple; Fig. 14.25: B A 1 θb − θa = ×Ml EI Ml since θa = 0 θb = EI Taking the origin at B, l l 1 dy ×Ml× x −y = dx EI 2 0 (0 − 0) − (0 − yb ) = yb = θb M yb l M l/2 BM Diagram Figure 14.25 Ml 2 2EI (d) Simply supported beam with central concentrated load Fig. 14.26 1 1 wl l . θb − θc = × elastic curve W EI 2 4 2 B y A 2 C Wl c since θc = 0 θB = qb W/2 W/2 16EI l Taking the origin at B and considering the part BC only, Wl l/2 4 dy 1 1 Wl l 2 l x −y = × . × × 2 l dx EI 2 4 2 3 2 0 3 2 BM Diagram W l3 (0 − yc ) − (0 − 0) = 48EI Figure 14.26 @seismicisolation @seismicisolation Deflection of Beams • 263 (e) Simply supported beam with uniformly distributed load Fig. 14.27: 1 2 wl 2 l × . . EI 3 8 2 θb = wl 3 since θc = 0. w/unit length B yc θb − θc = ∴ A l w l3 since θc = 0 24EI wl2 Taking the origin at B and considering BC 8 only, 5 l l/2 8 2 3 dy 1 2 wl l 5 l BM Diagram x −y = × . × × . dx EI 3 8 2 8 2 0 Figure 14.27 4 5 wl . yc (0 − yc ) − (0 − 0) = 384 EI 5wl 4 = 384 EI dy in radians. Note: θ is in radians and is slope dx Alternately Mohr’s theorem related to the moment-area method can be written as follows. i.e θb = Theorem I Change in slope between A and B = Net area of bending moment diagram A and B EI Theorem II Deflection of B with respect to A = Moment of area of bending moment diagram between A and B EI E XAMPLE 14 C-1: A cantilever of length 3 m carries a concentrated load of 50 kN at the free end and another 40 kN at 1 m from fixed end. Calculate the slope and deflection at the free end using moment-area method. E = 200 GPa and I = 107 mm4 . 40 kN 50 kN C A 1m B 2m Figure 14.28 @seismicisolation @seismicisolation 264 • Strength of Materials Bending moments: MB = 0, MC = −50 × 2 = −100 kNm MA = −50 × 3−40 × 1 = −190 kNm A B 2 3 1 100 kNm Bending moment diagram 190 kNm Figure 14.29 Let us divide the bending moment diagram in three parts as shown in Fig. 14.29. 1 A1 = 100 × 2 × = 100 kNm 2 A2 = 100 × 1 = 100 kNm 1 A3 = (190 − 100) × 1× = 45 kNm 2 Now slope at B will be A1 + A2 + A3 A = EI EI 100 + 100 + 45 245 = = EI EI θC = 245 × 106 = 0.0001225 radians 200 × 103 ×107 180 = 0.0001225 × = 0.007◦ π = Deflection at B = Ax̄ ; EI Ans x̄ is measured from free end B, 2 2 100× ×2 + 100 × 2.5 + 45× 2 + ×1 3 3 = EI 133.33 + 250 + 120 503.33 = = EI EI = 503.33 × 109 = 0.25 mm 200 × 103 ×107 @seismicisolation @seismicisolation Ans Deflection of Beams • 265 E XAMPLE 14 C-2: A beam 7 m long is simply supported at its ends and carries point loads of 25 kN each at points 1.7 m from the ends. Calculate by moment-area method: i) the maximum slope and ii) deflection under each load. Take EI = 60000 kNm2 25 kN Due to symmetry RA = RB = 25 kN MC = 25 × 1.7 = 45.9 kNm A 25 kN 1.7 m Slope at the M is zero because of maximum deflection at M (the midpoint). Slope at A or B is maximum, slope at B with respect to M = Area of bending moment diagram between M and B EI 25 kN M C D B 1.7 m 25 kN 7m 45.9 kNm 1.7 m 1.8 m M 1.8 m 45.9 kNm 1.7 m Figure 14.30 1 45.9 × 1.7× +45.9 × 1.7 39.015 + 78.03 2 = θB = EI EI 117.045 = 0.001951 radian = 60000 180 = 0.001951× = 0.112◦ Ans π Moment of area bending moment diagram from M to B about B Deflection at M = ym = EI 1.8 45.9 2 1.8 × 45.9 × + 1.7 + × 1.7 × 1.7 × 2 2 3 = EI 214.812 + 44.217 = 60000 = 0.00432 m = 4.32 mm (maximum deflection) Deflection under load 25 kN, = ym − deflection of D with respect to M; Deflection of C with respect to M Moment of bending moment diagram area between M and D about D EI 1.8 45.9 × 1.8× 2 = 0.00124 m = 60000 = 1.24 mm = @seismicisolation @seismicisolation 266 • Strength of Materials Hence, deflection under load 25 kN at D = 4.32 − 1.24 = 3.08 mm Ans (D) Conjugate Beam Method The basis of this method depends upon the modification of moment-area method. The momentarea method is convenient for the beams of constant flexure rigidity, but conjugate beam method is particularly useful for the beams of different flexure rigidity. In fact conjugate beam is an imaginary M beam of length equal to that of the original beam but for which the load diagram is diagram. EI M M This diagram shows the variation of over the length of the beam. In other words, the load at EI EI any section on the conjugate beam is equal to the bending moment at that point divided by flexural rigidity (EI). In nutshell, the slopes and deflection at any section of a beam by conjugate beam method is given by: 1. The slope at any section of the given beam is equal to the shear force at the corresponding section of the Conjugate beam. 2. The deflection at any section for the given beam is equal to the bending moment at the corresponding section of the conjugate beam. Therefore, before applying the conjugate beam method, conjugate beam is constructed. The load on the conjugate beam at any section is equal to the bending moment at that section divided by EI. Hence, the loading on the conjugate beam is known. The shear force at any point on the conjugate beam gives the slope at the corresponding point of actual beam. And the bending moment at any section on the conjugate beam gives the deflection at the corresponding point of the actual beam. Simply supported beam with a concentrated load at the centre A simply supported beam AB of length l carrying a point load W at the centre C is shown in W Fig. 14.31. C A B l/2 (a) l RA = W 2 (b) A (c) A R'A = Wl2 16 EI D' Wl 4 C' D'' Wl EI Loaded c Conjugate Beam Figure 14.31 @seismicisolation @seismicisolation RB = W 2 Bending moment diagram B Load diagram B R'B = Wl2 16 EI Deflection of Beams • 267 (i) The bending moment diagram is drawn in usual way. For constructing a conjugate beam diagram, the load on the conjugate beam is obtained by dividing the bending moment at that point by EI. The shape of the loading on the conjugate beam is same as of bending moment diagram. It should be remembered that bending moment diagram (modified) as shown in Fig. 14.31 (c) is the loading diagram known as conjugate beam. ii) Bending moment diagram from loaded conjugate diagram gives the deflection at any section. iii) Shear force diagram drawn on the basis of conjugate diagram provides the slope at the desired section. Now, referring to Fig. 14.31, in case of simply supported beam with a central point load, because total load on the conjugate beam = Area of the load diagram Wl W l2 1 = = ×l× 2 4EI 8 EI ∴ Reaction RA = RB = dy Slope at A dx W l2 8 EI = θA And deflection at C = yc According to conjugate beam method, θA = Shear force at A for the conjugate beam = RA = W l2 , 16EI because shear force at A for conjugate beam = RA yc = Bending moment at C for the conjugate beam 1 l Wl 1 l W l2 l · − × × × × = 16EI 2 2 2 4EI 3 2 = W l2 3W l 3 −W l 3 W l3 − = 32EI 96EI 96EI = W l3 48EI @seismicisolation @seismicisolation 268 • Strength of Materials Simply supported beam carrying an eccentric concentrated load W C B A a (a) RA = Wb l (b) A (c) A' b l D Wab l BM Diagram C1 Load diagram D1 B Wab EIl R'A Conjugate beam C2 a RB = Wa l b B' R'B Figure 14.32 After constructing bending moment diagram, the conjugate beam shown in Fig. 14.32(C) is Wab . drawn. The vertical load on conjugate beam will be EIl To find RA and RB ; taking moments about A of the conjugate beam, we get RB × l 1 Wab 2 1 Wab b = ×a× × ×a+ ×b× × a+ 2 EIl 3 2 EIl 3 b Wa3 b Wab2 + a+ = 3EI l 2EI l 3 Wa3 b Wa2 b2 Wab3 + + 3EI l 2EI l 6EI l Wab = + 2a2 +3ab + b2 6EI l Wab 2 2 a +b +2ab + ab + a2 = 6EI l Wab = (a + b)2 + a(a + b) 6EI l Wab 2 l +a l = 6EI l Wa(l − a) . l (l + a) = 6EI l Wa 2 2 (l −a ) = 6EI = @seismicisolation @seismicisolation Deflection of Beams Similarly, RA = • 269 Wb 2 2 (l −b ) 6EI l dy at A dx yc = deflection at C for the given beam Using conjugate beam method, If θA = slope at A or θA = Shear force at A for the conjugate beam = RA = (because shear forece at A for conjugate beam = RA ) Wb 2 2 (l −b ) 6EI l yc = Bending moment at C for conjugate beam 1 Wab a ×a× × = RA ×a− 2 EI l 3 Wa3 b Wb 2 2 (l −b ).a − 6EI l 6EI l Wab 2 2 2 = l −b −a 6EI l = E XAMPLE 14 D-1: Using conjugate beam method, find the slope at the support points and deflection under the given load for the beam shown in Fig. 14.33(a). Take E = 200 GPa and I = 4 × 10−5 m4 . S OLUTION : 8 kN 2.5 I A RA = 3 kN I D C 4.8 m 6m 3.6 m D is the midpoint of beam AB. Figure 14.33(a) Taking moments about B, RA × 9.6 = 8 × 3.6 RA = 3 kN RB = 8 − 3 = 5 kN @seismicisolation @seismicisolation B RB = 5 kN 270 • Strength of Materials (b) 18 kNm 14.4 kNm A B BMD for actual beam A' (c) 5.76 EI 7.2 EI 18 EI B' R'A R'B Loaded conjugate beam Figure 14.33 (b, c) Bending moment diagram for actual beam: MC = 3 × 6 = 18 kNm MD = 3 × 4.8 = 14.4 kNm Conjugate beam (Fig. 14.33C) 7.2 1 = E (2.5I) EI 18 1 = Load intensity just to the right of C = 18 × EI EI 1 Load intensity at D = RA × 4.8 × E (2.5I) Load intensity just to the left of C = 18 × = 5.76 3 × 4.8 = 2.5 EI EI Reactions RA and RB . Taking moments about B, 7.2 1 6 18 1 3.6 RA × 8 = ×6× × + 3.6 + × 3.6 × EI 2 3 EI 2 3 120.96 38.88 = + EI EI = or 159.84 EI RA = 19.98/EI 1 7.2 1 18 RA + RB = × ×6 + × × 3.6 2 EI 2 EI @seismicisolation @seismicisolation Deflection of Beams • 21.6 32.4 + EI EI 54 = EI 19.98 54 + RB = EI EI 34.02 ∴ RB = EI = Now Slope at A, θA = shear force at A = RA = 19.98 × 1000 19.98 = EI 200 × 109 × 4 × 10−5 [Multiplying by 1000 to convert into newtons] = 0.002497 radian = 0.143◦ Slope at B, θB = shear force at B = RB 34.02 EI 34.02 × 1000 = 200 × 109 × 4 × 10−5 = = 0.00425 radian = 0.244◦ Deflection under the given load, Yc = Bending moment at C of the conjugate beam = 6 19.98 × 6 7.2 × 6 − × EI EI × 2 3 119.88 43.2 − EI EI 76.68 = EI 76.68 × 1000 = 200 × 109 × 4 × 10−5 = = 0.0096 m = 9.6 mm Ans @seismicisolation @seismicisolation 271 272 • Strength of Materials 5.76 1 4.8 19.98 × 4.8 − × 4.8 × × EI EI 2 3 22.12 95.9 − = EI EI 73.78 × 1000 = 200 × 109 × 4 × 10−5 Similarly, yD = = 0.00922 = 9.22 mm Ans E XAMPLE 14 D-2: Find the deflection at centre of the beam shown in the Fig. 14.34. Take E = 210 GPa and I = 280 × 10−5 m4 S OLUTION : 280 kN E C A RA = 140 kN (a) I 2I 1.8 m 3.6 m D B I RB = 140 kN 1.8 m (a) Actually loaded beam 504 kNm 252 kNm (b) 252 kNm (b) BMD for actual beam P 126 EI Q A′ C′ 1.8 m T 126 EI E′ 1.8 m 126 EI S 252 EI 126 EI (c) RA′ R D′ 1.8 m Loaded conjugate beam Figure 14.34 @seismicisolation @seismicisolation B′ 1.8 m RB′ = Deflection of Beams • 273 RB × 7.2 m = 280 × 3.6 RB = 140 kN RA + RB = 280 ∴ RA = 280 − 140 = 140 kN ME = 140 × 3.6 = 504 kNm MC = 140 × 1.8 = 252 kNm MD = 140 × 5.4 − 280 × 1.8 = 756 − 504 = 252 kNm Diagram (b) is drawn accordingly Conjugate Beam: Load at E = 504 2EI = 252 EI Load just to the left of C = 252 = PC EI Load just to the right of D = T D Load just to the right of C = (Due to symmetry) 252 126 = 2EI EI Load just to the right of D = SD Therefore, PQ = QC = and T S = SE = 126 EI 126 EI Because load on the conjugate beam is symmetrical, so the support reactions are equal each being equal to one-half of the total load on the conjugate beam. @seismicisolation @seismicisolation 274 • Strength of Materials RA = RB = Area of Δ A PC + area of trapezium C QRD 1 252 1 126 252 = ×1.8× + + ×1.8 2 EI 2 EI EI = 226.8 340.2 + EI EI = 567 EI Deflection of mid-point E of the actual beam: yD = Bending moment at E of the conjugate beam 1.8 yD = RA × 3.6 − area A PC × 1.8+ 3 1.8 126 1 1.8 126 ×1.8× + ×1.8× + − EI 2 EI 2 3 567 × 3.6 252 1 1.8 204.12 68.04 = − ×1.8× 1.8+ − + EI EI 2 3 EI EI = 2041.2 544.32 204.12 68.04 − − − EI EI EI EI = 1224.52 EI = 1224.52 × 1000 210 × 109 ×280 × 10−5 = 0.00208 m = 2.08 mm Ans E XAMPLE 14 D-3: Find the deflection at the free end of cantilever shown in Fig. 14.35. Take E = 200 GPa, I = 1.2 × 10−4 m4 Bending moment along cantilever: MB = 0 MC = 12 × 1.75 = 21 kNm MA = 12 × 3.5 + 22 × 1.75 = 42 + 38.5 = 80.5 kNm @seismicisolation @seismicisolation Deflection of Beams 22 kN A (a) Figure 14.35 12 kN C B 1.75 m 1.75 m 21 kNm BMD for actual beam 80.5 kNm (b) BMD for actual beam A′ C′ 80.5 EI B′ 21 EI 59.5 EI Loaded conjugate beam (c) Figure 14.35(a) Deflection at B, yB = Bending moment at B of conjugate beam 1.75 21 × 1.75 × +1.75 = EI 2 59.5 1 2 × 1.75 + × 1.75 × + 1.75 EI 2 3 + 1 2 21 × 1.75 × × ×1.75 EI 2 3 = 96.47 106.3 21.44 + + EI EI EI = 224.21 EI @seismicisolation @seismicisolation • 275 276 • Strength of Materials = 224.21 × 1000 200 × 109 ×1.2 × 10−4 = 0.00934 m = 9.34 mm Ans E XAMPLE 14 D-4: For the cantilever shown in Fig. 14.36 (a), find slope and deflection at free end. Take E = 200 GPa, I = 1.8 × 10−5 m4 . S OLUTION : 3 kN/m A B 2m (a) 6 kNm Bending moment diagram for actual beam 1.333 A′ 6 EI B′ G Loaded conjugate beam Figure 14.36 MA = 3×2×2 = 6 kNm 2 Variation of bending moment will be parabolic. Slope at B = θB = Shear force at B of conjugate beam 1 1.333 6 ×2× = = 3EI 3 EI 1.333 × 1000 = = 0.00037 radian 200 × 109 × 1.8 × 10−5 0.00037 × 180 = 0.0212◦ Ans = π @seismicisolation @seismicisolation Deflection of Beams • 277 Centroid of the loaded conjugate beam = 34 × 2 = 1.5 from B Deflection at B = yB = Bending moment at B of conjugate beam 1 2 6 ×3× × 1.5 = 3EI 3 EI 2 × 1000 = = 0.00055 m 200 × 109 ×1.8 × 10−5 = 0.55 mm Ans = (E) Superposition Method The resultant deflection at a point in a beam, in this method is obtained by adding the deflections at this point due to individual load on the beam. As long as stresses are within the elastic limit, the deflections at a point due to each individual load are superimposed or each other algebraically to give the resultant deflection due to the forces acting on the beam. E XAMPLE 14 E-1: Determine the deflection at centre C and point D, one meter from left hand support, for the simply supported beam shown in Fig. 14.37, EI = 1500 kN/m2 S OLUTION : D A 1m 12 kN C 3 kN/m B 2m 2m Figure 14.37 Let us split the problem in two parts as shown in Fig. 14.38(a) and (b) 12 kN C A B D 1m 2m 2m Figure 14.38a D A 3 kN/m B C 2m 2m 1m Figure 14.38b @seismicisolation @seismicisolation 278 • Strength of Materials i) Deflection due to concentrated load alone, yc = 12 × (4)3 W l3 = = 0.011 m = 11 mm 48EI 48 × 1500 For point D, using general equation, W x3 W l 2 x − 12EI 16EI 12(1)3 12(4)2 × 1 = − 12 × 1500 16 × 1500 = 0.000667 − 0.008 = 0.00733 = 7.33 mm (discarding minus sign) yD = ii) Deflection due to uniformly distributes load alone : 5wl 4 384EI 5 × 3(4)4 = 384 × 1500 = 0.00667 m = 6.67 mm yc = To find deflection at D due to general equation 1 wlx3 wx4 wl 3 x yD = − − EI 12 24 24 Substitution for x = 1, 1 3 × 4(1)3 3(1)4 3(4)3 1 − − 1500 12 24 24 1 = [1 − 0.125 − 8] 1500 = 0.00475 m (ignoring minus sign) = 4.75 mm YD = Total deflection at C = 11 + 6.67 = 17.67 mm Ans Total deflection at D = 7.33 + 4.75 = 12.08 mm Ans (F) Strain Energy Method Strain energy due to bending: Fig. 14.39 shows an element of a beam subjected to a bending moment which varies as shown from M at one end to M + dM at the other. The length of the element in dx, the mean radius, of curvature in R and the change of slope between the ends in d φ . @seismicisolation @seismicisolation Deflection of Beams • 279 dx dφ M+dM M R dφ Figure 14.39 Then the work done in bending moment = 1 × mean bending moment × angle of bending 2 (assuming that the moment is gradually applied) This work is stored in the element as strain energy. M + α dM Therefore, strain energy = × d φ where < α < 1 2 1 Md φ (to the first order of small quantities) 2 dx dx M M E dφ = = ∵ = R EI I R i.e. du = But (i) Substituting for d φ in Eqn. (i) ∴ ∴ du = M2 dx 2EI l total strain energy = M2 dx 2EI (ii) 0 where l is the length. When total strain energy is required M is terms of a to be substituted in the expression (ii) Castigliano’s Theorem: If a structure is subjected to a number of external loads (or couples) the partial derivation of the total strain energy with respect to any load or couple provides the deflection in the deflection of that load (couple). Figure 14.40 shows an elastic body which is subjected to forces W1 , W2, W3 , etc. At each load point, the deflection can be resolved into components in the direction of, and perpendicular to, the line of action of the force at that point. Let x1 be the deflection at A in the direction of W1 , @seismicisolation @seismicisolation 280 • Strength of Materials W3 W2 B C A D W4 W1 Figure 14.40 x2 be the deflection at B in the direction of W2 x3 be the deflection at C in the direction of W3 etc. 1 1 1 U = W1 x1 + W2 x2 + W3 x3 + ..... 2 2 2 or 2U = W1 x1 + W2 x2 + W3 x3 + ..... Work done, Differentiating with respect to W1 , remembering that the deflection at each point is a function of all the loads. ∂U ∂ x1 ∂ x2 ∂ k2 = x1 + W1 + W2 + W3 +... (iii) 2 ∂ w1 ∂ w1 ∂ w1 ∂ w2 If W increases gradually to W1 + σ W1 , the other forces remaining constant. This change will make x1 increase to x1 + δ x1 , x2 to x2 + δ x2 , etc. Then additional work done, 1 δ W1 ) δ x1 + W2 dx2 +W3 dx3 + .... 2 1 δU δ x1 δ x2 δ x3 = W1 + dx1 +W2 + W3 δ W1 δ W1 2 δ W1 δ W1 δ U = (W1 + ∴ (iv) Subtracting Eqn. (iv) from Eqn. (iii) x1 = ∂U ∂ W1 (v) That is, the movement of A in the direction of W1 is equal to the partial derivative of the total strain energy of the system with respect to W1 Similarly, x2 = ∂U ; ∂ W2 x3 = ∂U etc. ∂ W3 Castigliano theorem applies equally well to couples and to mixtures of forces and couples. The rotation at any point is the partial derivative of the total strain energy with respect to the couple at that point. @seismicisolation @seismicisolation Deflection of Beams • 281 Application of Castigliano’s theorem to deflection (including curved bars). From Eqn. (i), l U = M2 dx 2EI 0 If it is required to find the deflection δ at a point at which there is a load P in the direction of the required deflection, then 1 ∂U = δ= ∂P 2EI 1 = 2EI l 0 l 0 ∂ (M 2 ) dx assuming I is constant ∂P 1 ∂M dx = 2M ∂P EI l M 0 ∂M dx ∂P Note: If there is no force P at the required point and in required direction, much force can be applied and when an expression for the deflection is obtained, this force is then made zero. For the rotation in the direction of a couple C, 1 φ = EI l M 0 ∂M dx ∂C E XAMPLE 14 F-1: A simply supported beam of span l carries a concentrated load W at a distance a and b respectively from the two ends A and B. Obtain an expression for the deflection under the load. S OLUTION : W a b A x C dx Wb l l Figure 14.41 Wb l Wb x At any point between A and C, M = l ∴ UAC (strain energy between A and C) Reaction at A = @seismicisolation @seismicisolation B Wa l 282 • Strength of Materials 1 = 2EI 1 Wb x l 2 dx = 0 W 2 b2 a3 6EI l 2 Similarly, strain energy between B and C, UBC = W 2 a3 b2 6EI l 2 W 2 a2 b2 (a + b) 6EI l 2 (Because total strain energy = UAC + UBC ) Therefore, total strain energy = ∴ Total strain energy = W 2 a2 b2 6EI l 1 Also work down = W δ , which is equal to total strain energy. 2 Hence, 1 W 2 a2 b2 Wδ = 2 6EI l W a2 b2 δ= 3EI l Note: The sign (+ve or −ve) of M does not matter, because M is squared subsequently. E XAMPLE 14 F-2: Derive an expression for the maximum deflection (using Castigliano’s theorem) of a cantilever of length l carries a uniformly distributed load w/unit length over it entire span, when a point load P is applies at it free end. S OLUTION : P w unit length l x Figure 14.42 Consider a section ‘x’ from p; Bending moment due to P, M = Px Bending moment due to uniformly distributed load, Me = ∴ Mx = Px + wx2 2 @seismicisolation @seismicisolation wx2 2 Deflection of Beams • 283 Strain energy stored in the beam in given as e M 2 dx 2EI U = 0 As per Castigliano’s theorem, ∂U ∂⎡ P δ= l ∂⎣ 0 = ∂P l = 0 But ⎤ M 2 dx ⎦ 2EI 2M ∂ M . .dx 2EI ∂ P ∂M =x ∂P Therefore, δ= l M xdx EI 0 1 = EI l wx2 Px + 2 x.dx (subsituting for M) 0 1 = EI l Px2 + wx3 2 dx 0 l wx4 1 Px3 + = EI 3 8 0 3 1 Pl wl 4 = + Ans EI 3 8 when P = 0, then δ= wl 4 8EI E XAMPLE 14 F-3: A steel spring ABC of the dimensions shown in Fig. 4.43 in fully clamped at A. If a vertical force of 20 N in applied at C, find the vertical deflection of this point E = 200 GPa. @seismicisolation @seismicisolation 284 • Strength of Materials S OLUTION : For BC, taking origin at C P = 20 N 100 mm dx B C M = 20x ∴ x θ 0.10 1 UBC = 2EI (20x)2 dx dθ 0 60 mm R 0.2 = J 3EI 2 mm A 20 mm Figure 14.43 For AB, taking the origin at B, M = 20(0.10 + 0.06 sin θ ) dx = 0.06 d θ ∴ 1 UAB = 2EI π 20(0.10 + 0.06 sin θ )2 × 0.06 d θ 0 = 12 EI π (0.01 + 0.012 sin θ + 0.0036 sin2 θ )d θ 0 12 [0.01 × 0.024 + 0.0018π ] = EI 0.733 = EI Total strain energy 0.733 0.7996 0.2 + = 3EI EI EI 1 Work done = × 20 × δ 2 = Equating Eqns. (i) and (ii) 1 × 20 × δ = 2 0.7996 0.02 × 0.0023 200 × 109 × 12 δ = 0.03 m or 30 mm Ans @seismicisolation @seismicisolation (i) (ii) Deflection of Beams • 285 Props Props are sometimes needed in the loaded beams to counterbalance the deflection partially or completely. There are generally three types of props: i) Right props in which prop do not deflect at all (ii) Sinking props are those props which on loading sink into ground by some amount. iii) Elastic props are those which get compressed due to the elastic property of their material. On removal of load, these regain their original position. E XAMPLE 14.1: A simply supported beam carrying a uniformly distributed load over the whole span is propped to its with the help of a rigid prop. Calculate the load carried by the prop and each of the supports. S OLUTION : The downward deflection of point C due to uniformly distributes load. w/unit length B C A l P Figure 14.44 δ= 5wl 4 384 EI (i) Pl 3 48EI (ii) The upward deflection due to prop, δp= To make the downward deflection at C, disappear equation (i) should be equal to (ii) therefore equating 5wl 4 Pl 3 = 48EI 384EI 5wl ∴ P= 8 Also since RA = RB 1 RA = RB = (wl − P) 2 5 1 = (wl − wl) 2 8 3 = wl. Ans 16 @seismicisolation @seismicisolation 286 • Strength of Materials E XAMPLE 14.2: Solve example 14.1, assuming the prop to be elastic having the stiffness ‘S’ Find the Prop’s load and reactions. S OLUTION : P Downward deflection of the prop due to its elasticity = S Therefore, Pl 3 P 5wl 4 − = 384EI 48EI S P= 5wl 4 s 8 [48EI + Sl 3 ] Ans Then RA = RB = 1 (wl − P) 2 5Sl 3 1 = wl 2 8(48EI + Sl 3 ) 3 Sl + 128EI 3 Ans wl = 16 Sl 3 + 48EI Exercise 14.1 A cantilever beam of length 2 m carries a load of 50 kN at the free end. Find the maximum slope and deflection of the beam. Take E = 2 × 105 N/mm2 , I = 2 × 108 mm4 . [Ans: 0.0025 radian, 3.33 mm] 14.2 A cantilever beam of length 3 m having a rectangular cross section. Determine the slope and deflection at the force and of cantilever and under the load also when (i) point load of 20 kN acts at a distance of 1.5 m from fixed end and (ii) point load of 20 kN acts at a distance of 2 m from fixed end. Take EI = 8 × 1012 Nmm2 [Ans: 0.0028 radian, 0.878 mm downward, 0.005 radian, 11.67 mm, 6.67 mm] 14.3 A cantilever with a span of 4 m carries a point load at its free end. If the maximum slope is 1.5 degrees, calculate the deflection at the free end. [Ans: 69.81 mm] 14.4 A cantilever beam with a span of 3 m carries a point load 30 kN at a distance of 2 m from the fixed end. Determine the slope and deflection at the free end and at the point where load is applied. I = 11924 cm4 , E = 200 GPa. [Ans: 0.00256 radian, 3.35 mm] 14.5 A cantilever 3 m long is loaded with a uniformly distributed load of 15 kN/m over a length of 2 m from the fixed end. Determine the slope and deflection at the free end of the cantilever. Take E = 210 GPa and I = 0.000095 m4 . [Ans: 0.001 radian, 2.5 mm] @seismicisolation @seismicisolation Deflection of Beams • 287 14.6 A cantilever beam of 3 m span is 15 cm wide and 25 cm deep. If carries a uniformly distributed load of 20 kN/m over its whole span and 25 kN load at the free end. Calculate the maximum slope and deflection. E = 210 GN/m2 . [Ans: 0.004937 radian, 10.42 mm] 14.7 A cantilever of uniform section has a length ‘l’. It is propped at the free end and carries a point load W at a distance ‘a’ from the fixed end. (a) If the prop holds the free end at the level of the fixed end, find the prop reaction. (b) If now the prop is removed what will be the deflection at the free end. Wa2 Wa2 (3l − a); (3l − a)] [Ans: 3 6EI 2l 14.8 A cantilever beam is loaded by a moment M1 at the free end as well as with uniformly distributed load of w kN/m over half of its length at free end. Prove that the deflection at the free end in given by expression: M1 L2 7wl 4 + 2EI 384EI 14.9 A 2 m simply supported beam having cross section 150 mm × 500 mm carries a point load of 20 kN at a distance of 0.5 m from the left end. Find the slope at the two ends, deflection under the load and the maximum deflection. E = 2 × 104 mm2 [Ans : 8.02 × 10−3 degree 5.7 × 10−3 degree 0.06 mm 0.071 mm] 14.10 A wooden beam is 100 mm wide by 200 mm deep carries a load of 1.5 kN/m run over a simply supported span of 4 m. Take E = 200 GPa. Find the deflection at the centre of the span. [Ans: 8.33 mm] 14.11 A cast iron pipe 150 mm outer diameter and 120 mm inner diameter rests on simple supports 1m apart and is subjected to a central load. Determine the load necessary to cause a maximum deflection of 1.3 mm E = 105 N/mm2 . [Ans: 91.5 kN] 14.12 A horizontal cantilever AB, 1.5 m long is subjected to uniformly distributed to load 1.5 kN/mm to a length of 1 m from the fixed end A and a point load of 500 N at its free end B. The cantilever is of rectangular section of width 75 mm and depth 150 mm and its E = 1.25 × 105 N/mm2 . Determine the deflection at the free end B. [Ans: 0.146 mm] 14.13 Two point loads of 5 kN and 15 kN are acting on a 5 m simple beam AB, at 1 m and 2 m respectively from the left end. Determine: a) slopes at the two ends, b) deflections under each load c) position and magnitude of maximum deflection. E = 90 GPa. I = 18 × 106 mm4 . [Ans: y5 = 17.1 mm, y15 = 26.85 mm, Maximum deflection = 27.49 mm at 2.327 m from A.] @seismicisolation @seismicisolation θA = 1.06◦ , θB = 0.884◦ 288 • Strength of Materials 14.14 A uniform simply supported beam AB is of 6 m span. At a distance of 4 m from left hand support it carries a load of 10 kN on a bracket as shown in Fig. 14.45. Determine deflection at the point C. Take E = 200 kN/m2 and I = 2 × 107 mm4 . 10 kN 0.5 m B A C 4m 2m [Ans: Figure 14.45 7.8 mm] 14.15 A cantilever 150 mm wide, 200 mm deep and of 2 m span carries a uniformly varying load of 50 mm at the fixed and as shown in Fig. 14.46. Determine the slope and deflection at the free end. Take E = 100 GPa. 150 kN/m 50 kN/m A B 2m [Ans: Figure 14.46 0.01rad, 15.3 mm] 14.16 A cantilever 2 m long consists of a rectangular timber joint 150 mm × 240 mm deep. Two steel plates 150 mm × 10 mm thick are fixed at the top and bottom faces of the timber joist as shown in Fig. 14.47. 150 mm 10 mm 240 mm 10 mm Figure 14.47 @seismicisolation @seismicisolation Deflection of Beams • 289 Determine the slope and deflection of the cantilever at its free end, when it is carrying a uniformly distributed load of 10 kN/m. Take Es = 200 GPa and Et = 10 GPa. [Ans: 0.0012 rad, 1.8 mm] 14.17 A horizontal beam AB is freely supported at A and B, 8 m apart and carries a uniformly distributed load of 15 kN/m run. A clockwise moment of 160 kNm is applied to the beam at a point C, 3 m from the left hand support A. Calculate the slope and deflection of the beam at C, if EI = 40 × 103 kNm2 . [Ans: 0.0061 rad, 23.5 mm] 14.18 A simply supported beam of 2 m span carries a point load of 20 kN at its mid span. Determine the maximum slope and deflection of the beam. Take EI = 500 × 109 Nmm2 . Solve the problem by conjugate beam method. [Ans: 0.01 rad, 6.67 mm] 14.19 A simply supported beam is loaded as shown is Fig. 14.48. The beam has moment of inertia of its section as 2I for the middle left of its length, whereas it is only I at the quarter length on each end. Use conjugate beam method to find deflection at points C and E. Take E = 200 GPa and I = 4 × 10−6 m4 . 200 kN A C 1.5 m 1.5 m 0.48 mm, 0.67 mm] B D E 1.5 m [Ans: 1.5 m Figure 14.48 14.20 A beam of length 6 m is simply supported at its ends and carries two point loads of 48 kN and 40 kN at a distance of 1 m end 3 m respectively from the left support. Find: i) deflection under each load ii) maximum deflection, and iii) the point at which maximum deflection occurs. Take E = 200 GPa and I = 85 × 106 mm4 Use Macaulay’s method. [Ans: 9.019 mm, 16.7 mm, 16.745 mm, 2.87 m from left support] 14.21 A cantilever beam as shown in Fig. 14.49 is loaded by 500 N at the free end. Determine the value of W so that deflection at the free end is 10 mm. Take E = 15 GN/m2 , I = 10×106 mm4 W 1.5 m 500 N 1.5 m Figure 14.49 @seismicisolation @seismicisolation [Ans: 774.2 N] 290 • Strength of Materials 14.22 Using moment-area method, find the deflection at the centre of a simply supported beam carrying w N/m. The span of beam in l. 5wl 4 Ans : yc = 384 EI 14.23 A horizontal cantilever 2 m long has its free and attached to a vertical tie-rod 3 m long and 300 mm2 cross-section area, which is initially unstrained. If I of the cantilever cross-sectional is 6.5 × 106 m4 , determine the load taken by the tie-rod and the deflection of the cantilever when a uniformly distributed load of 30 kN/m is placed on the outer 1 m of the cantilever. Assume E for both cantilever and tie-rod to be 200 GPa. [Ans: 18.76 kN, 0.938 mm] 14.24 The loads acting on a simply supported beam are shown in Fig. 14.50. Use Macaulay’s method to determine deflection at points C, D and E. 26 kN C E 26 kN D 50 kN/m A B 1.5 m 1.5 m 1.5 m 1.5 m Figure 14.50 [Ans: 0.54 mm, 0.54 mm, 0.756 mm] 14.25 Use the method of superposition to find the slope at end A and the deflection at centre point D for the beam AB loaded as shown in Fig. 14.51. W W W 2.5Wa2 19 wa3 ; Ans : θA = EI 6 EI C D E B A a a a a Figure 14.51 14.26 Determine the deflections at the ends and centre of the beam shown in Fig. 14.52. Take EI = 5 × 106 N/m2 . Use superposition method. 40 kN 3m 20 kN 10 kN/m 40 kN E A C 2m B 6m D 2m Figure 14.52 @seismicisolation @seismicisolation [Ans: yc = 63.3 mm, yD = 63.3 mm, yE = 20.25 mm] C HAPTER 15 STRAIN ENERGY, IMPACT LOADING AND DEFLECTION DUE TO BENDING Whenever an elastic body is subjected to a force, it is stretched or compressed and on removal of that force, elastic body comes back to its original form. This happens because of strain energy or resilience stored in the body during compression or tension. For example, bow and arrow, the moment stretched bow is released, its tensile strain energy is taken by arrow which moves forward with great speed. This energy, which is absorbed in a body, when strained within elastic limit is known as strain energy. Strain energy is always capable of doing some work. The amount of strain energy in a body is found out by the principle of work. Therefore, Strain energy = Work done. Resilience: Many times total strain energy stored in a body is termed resilience. It is also defined as the capacity of a strained body for doing work when the strained body springs back to its original shape. Proof Resilience: Maximum strained energy which can be stored in a body is termed proof resilience. Modulus of Resilience: This is the proof resilience per unit volume of a material. This is an important property of the material. (A) Strain Energy Stored in a Body When the Load is Gradually Applied It an axial load P is gradually applied to a bar and producer extension x, then the work done or strain energy is represented by the area under load-extension diagram P U = Strain energy Load x Extension Figure 15.1 @seismicisolation @seismicisolation 292 • Strength of Materials 1 P.x, and P = stress × area = σ × a 2 σl and x = E σl 1 ∴ U = σ ·a× 2 E U= = The term 1 σ2 × Volume 2 2E σ2 × Volume represents strain energy. 2E (B) Suddenly Applied Load Sometimes if load P is suddenly applied, for example, load P is suddenly placed on the collar in Fig. 15.2, causing deflection x. (It should be noted that in previous case of gradually applied load, it was assumed that load is increased to P from zero value gradually.) l P Now work done = P.x But strain energy, U = σ2 σ2 × Volume = × Al 2E 2E where A is the area of cross section of the rod l is the length of rod supporting collar. Figure 15.2 Now because the energy stored is equal to the work done, therefore σ2 σ × A × l = P.x = P. l 2E E P or σ = 2 × A It proves that stress induced in this case is twice the stress induced when the load is applied gradually. After stress σ is found out, the corresponding instantaneous deformation x and the strain energy are obtained as usual. E XAMPLE 15.1: On a steel rod 3 m long, an axial pull of 25 kN is suddenly applied. If diameter of rod is 40 mm, calculate the strain energy which is absorbed in the rod. Take E = 200 GPa. S OLUTION : Cross-sectional area = π (40)2 = 1256 mm2 4 @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending • 293 P A 2 × 25000 = 39.81 N/mm2 = 1256 For suddenly applied load, σ = 2 × Volume of the rod, V = length × Area = 3000 × 1256 = 3768000 mm3 Hence, strain energy, U= = σ2 × Volume 2E (39.81)2 × 3768000 2 × 200000 = 14929 Nmm = 14.929 Nm Ans E XAMPLE 15.2: Find the maximum instantaneous stress and work done at maximum elongation, when an axial load of 80 kN is suddenly applied to the steel rod of 3.5 m length having a diameter of 35 mm. Also calculate the maximum instantaneous force in the rod. Take E = 200 Gpa. S OLUTION : π 2 π 2 d = 35 = 961.625 mm2 . 4 4 P σmax (instantaneous stress) = 2 × A 80000 = 166.38 N/mm2 Ans = 2× 961.625 A= Work done at maximum elongation, σmax × l E 166.38 × 3500 = 2.91 mm = 200000 Work done = P.x = 80000 × 2.91 = 232800 Nmm = 232.8 Nm Ans x= @seismicisolation @seismicisolation 294 • Strength of Materials Maximum instantaneous force developed in the rod = σmax × area = 166.38 × 961.625 = 159995.17 N = 159.995 kN Ans (C) Strain Energy Stored in a Body, When the Load is Applied with Impact Load P As we see in Fig. 15.3, if load P is suddenly released to fall through a height h, if will produce extension x in the rod. ∴ Work done = load × distance moved = P(h + x) (i) And strain energy stored = h σ2 × Al 2E (ii) Where A and l are area of rod and length of rod, respectively. Collar Figure 15.3 Equating Eqns. (i) and (ii), σ2 × Al = P(h + x) 2E σ = P h+ l E where σ is the instantaneous stress induced in the rod due to impact. Pσ l σ2 × Al = Ph + E 2E Al Pl ∴ σ2 −σ − Ph = 0 2E E E Simplifying by multiplying both sides by , Al σ2 PEh P −σ − =0 2 A Al @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending • 295 Solving this quadratic equation, PEh P 2 1 + 4× A 2 Al σ= 1 2× 2 P 2EAh 1± 1+ σ= A Pl P ± A ∴ After finding instantaneous stress, the corresponding deformation and strain energy can be found out as usual, 2P If h = 0, σ = that is the case of a suddenly applied load. A In case instantaneous extension x is very small as compared to h, then Work done = Ph σ2 × Al = Ph 2E 2EPh σ2 = Al 2EPh σ= Al But, of course, this will give approximate instantaneous stress. E XAMPLE 15.3: A steel bar 4 m long and 3000 mm2 in area hangs vertically and is secured to a fixed collar at its lower end. If a weight of 20 kN falls on the collar from a height of 12 mm, determine the stress developed in the bar. What will be the instantaneous strain energy stored in the bar? Take E = 200 GPa. S OLUTION : We know in this case, P σ= A 2EAh 1+ Pl 2 × 200000 × 3000 × 12 20000 1+ 3000 20000 × 4000 √ = 6.67 1 + 180 √ = 6.67 181 = = 89.73 N/mm2 Ans @seismicisolation @seismicisolation 296 • Strength of Materials Strain energy stored in the bar U= ∴ σ2 × Volume 2E (89.73)2 × 3000 × 4000 2 × 200000 = 241544.2 Nmm U= = 241.54 Nm Ans E XAMPLE 15.4: A steel bar 15 mm diameter gets stretched by 0.8 mm under a steady load of 8 kN. What stress would be produced in the bar by a weight 600 N, if this weight falls through 100 mm before striking the collar rigidly fixed to the lower end? Take E = 200 GPa. First we must find the length of bar. E= Pl Ax π EAx , A = (15)2 = 176.625 P 4 200000 × 176.625 × 0.8 = 3532.5 mm l= 8000 ∴ l= Now this is the case of impact loading, ∴ σ= = P A 1+ 1+ 600 176.625 1+ 2AEh P l 2 × 176.625 × 200000 × 100 600 × 3532.5 √ = 3.397(1 + 1 + 3333.3) = 3.397(1 + 57.74) = 199.55 N/mm2 Ans E XAMPLE 15.5: An unknown weight falls through 15 mm on a collar rigidly fixed to the lower end of a vertical bar of 3.5 m length and 750 mm2 area in section. If the extension is 2.5 mm, what is the corresponding stress and the value of the unknown weight? E = 200 GPa. Maximum stress = E × Maximum strain 2.5 = 200000 × 3500 = 142.86 N/mm2 If W is the unknown weight, then Loss of potential energy = Strain energy stored in the rod W (h + x) = σ2 × Volume 2E @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending • 297 (142.86)2 × 750 × 3500 2 × 200000 17.5W = 133934.1 133934.1 W= 17.5 = 7653.3 = 7.653 kN Ans W (15 + 2.5) = E XAMPLE 15.6: A 3.5 m long bar of 20 mm in diameter hangs vertically and has a collar attached at the lower end. Determine the maximum stress induced when a weight of 1000 N falls on the collar from a height of 40 mm. If the bar is turned down to half the diameter along half of its length, what will be the value of the maximum stress? E = 200 GPa. S OLUTION : π (20)2 = 314 mm2 4 P 2AEh σ1 1 = 1+ 1+ A Pl 2 × 314 × 200000 × 40 1000 1+ 1+ = 314 1000 × 3500 √ = 3.185[1 + 1 + 1435.4] = 3.185[1 + 37.89] A= = 123.86 N/mm2 Now the reduced diameter = Ans 20 = 10 mm, A1 = 314 mm2 2 A2 = 1750 mm A1 π (10)2 = 78.5 mm2 , l1 = l2 = 1750 mm 4 P = Falling load = 1000 N Pe = Equivalent gradually applied load which produces the same maximum stress and extension as is caused by the falling load, P. A2 1750 mm Figure 15.4 @seismicisolation @seismicisolation 298 • Strength of Materials Now the total extension, x = x1 + x2 = Pe l1 Pe le + A1 E A2 E = 1 Pe l1 1 + E A1 A2 = 1 Pe × 1750 1 + 200000 314 78.5 ∴ l1 = l2 x = 0.00875 Pe [0.00318 + 0.01274] x = 0.00875 Pe [0.01592] or x = 0.0001393 Pe Also, 1 P h + x = Pe × x 2 1 1000 (40 + 0.0001393 Pe ) = Pe × 0.0001393 Pe 2 40000 + 0.1393 Pe = 0.00006965 Pe2 0.00006965 Pe2 − 0.1393 Pe − 40000 = 0 Pe2 − 2000 Pe − 574300072 √ +2000 ± 4000000 + 4 × 574300072 Pe = 2 √ 2000 ± 4000000 + 2297200288 = 2 2000 + 47971 = 24985.5 = 2 Maximum stress is in smaller section, σ2 = 24985.5 = 318.3 N/mm2 78.5 Ans E XAMPLE 15.7: A steel rod is 2.5 m long and 60 mm in diameter. An axial pull of 130 kN is suddenly applied to the rod. Calculate the instantaneous stress induced and also the instantaneous elongation to the rod. Take E = 200 GN/m2 . @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending • 299 S OLUTION : π (60)2 = 2826 mm2 , l = 2500 mm 4 P = 130000 N, E = 200000 N/mm2 P 2 × 130000 σ = 2× = = 92 N/mm2 Ans A 2826 σl E= x σ l 92 × 2500 Instantaneous elongation = x = = E 200, 000 = 1.15 mm Ans Area = E XAMPLE 15.8: A tension bar 6 m long is 3.5 m long having cross-sectional area of 1200 mm2 while remaining 2.5 m long is having 2500 mm2 cross-sectional area, An axial load of 120 kN is gradually applied. Find the total strain energy produced in the bar and compare this value with that obtained in a uniform bar of the same length and having the same volume under the same load. E = 200 GN/m2 . S OLUTION : 1200 mm2 2 1 120 kN 2500 mm2 3.5 m 2.5 m Figure 15.5 120000 = 100 N/mm2 1200 120000 = 48 N/mm2 σ2 = 2500 σ1 = Strain energy in portion 1, U1 = σ2 × Volume 2E @seismicisolation @seismicisolation 120 kN 300 • Strength of Materials (100)2 × 1200 × 3500 2 × 200000 = 105000 Nmm = 105 Nm = Strain energy in portion 2, σ2 × Volume 2E (48)2 = × 2500 × 2500 2 × 200000 = 36000 Nmm U2 = = 36 Nm Total energy produced in the bar = 105 + 36 U1 = 141 Nm Total volume of the first bar, v = v1 + v2 v = 1200 × 3500 + 2500 × 2500 = 4200000 + 6250000 = 10450000 mm3 Length of uniform bar = 6000 mm v = A×l v 10450000 = = 1741.7 mm2 l 6000 120000 Stress in the uniform bar = = 68.9 N/mm2 1741.7 σ2 Strain energy in uniform bar = × Volume 2E (68.9)2 U2 = × 10450000 2 × 200000 = 124021 Nmm Area of uniform bar, A = = 124.02 Nm ∴ 141 Strain energy in the given bar, U1 = Strain energy in the uniform bar, U2 124.02 = 1.137 Ans @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending • 301 E XAMPLE 15.9: A bar of uniform cross section a and length l hangs vertically, subjected to its own weight. Prove that the strain energy stored in bar is given by U= A × ρ2 × l 6E where ρ is weight per unit volume. S OLUTION : ρ = Weight per unit volume. Consider section X − X which is acted upon by the weight of the bar of length X. Wx = (A × x) × ρ = ρ Ax dx x x Due to this weight the portion dx will have a small elongation d δ . Then l elongation in dx dx dδ = dx ρ Ax Stress in portion δ x = A = ρ ×x Strain in portion dx = x Figure 15.6 Since, ∴ Stress Strain ρx ρ xdx = = dδ dδ dx ρ × x × dx dδ = E E= Strain energy stored in portion dx is given by dU = Average weight × elongation of dx 1 W x dδ = 2 1 ρ × x × dx × ρ Ax × = 2 E 1 2 2 dx = ρ Ax × 2 E Total strain energy stored within the bar due to its own weight W is obtained by integrating from 0 to l. @seismicisolation @seismicisolation 302 • Strength of Materials l U= dU 0 l = 1 2 2 dx ρ Ax 2 E 0 = ρ 2A 2E l x2 dx 0 l3 1 × ρ 2A × 2E 3 2 3 Aρ l = 6E = E XAMPLE 15.10: The maximum stress produced by a pull in a bar of length 1.2 m is 180 N/mm2 . Figure 15.7 shows the cross section and length. Determine the strain energy stored in the bar if E = 200 GPa. 300 mm2 P A 150 mm2 B 0.4 mm 300 mm2 D C 0.4 mm P 0.4 mm Figure 15.7 S OLUTION : Stress will be maximum where cross-sectional area is minimum. Therefore, maximum stress, σ = 180 N/mm2 in area 150 mm2 . Now strain energy stored in part AB will be found out σBC × 150 = σAB × 300, because the load is equal in every portion. 180 × 150 = 90 N/mm2 300 σ AB2 × Volume UCD = UAB = 2E 902 = × 300 × 400 2 × 200000 = 2430 Nmm = 2.430 Nm σAB = @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending • 303 2 σbc × Volume 2E 902 × 150 × 400 = 2 × 200000 = 1215 Nmm = 1.215 Nm UBC = Total strain Energy, U = UBC +UAB +UCD U = 1.215 + 2 × 2.43 = 1.215 + 4.86 = 6.075 Nm Ans (∵ UAB = UCD ) E XAMPLE 15.11: A vertical compound tie member fixed rigidly at its upper end, consists of a steel rod 3 m long and 25 mm in diameter, placed within an equally long brass tube 27 mm is internal diameter and 35 mm external diameter. The rod and the tube are fixed together at the ends. The compound bar is then suddenly loaded in tension by a weight of 15 kN falling through a height of 5 mm on to a flange fixed to its lower end. Determine maximum stresses in steel and brass. Take Es = 200 GPa, Eb = 1.0 × 105 N/mm2 . S OLUTION : π (25)2 = 490.625 mm2 4 π π Ab = (352 − 272 ) = (1225 − 729) 4 4 2 = 389.36 mm Length = 3 mm = 3000 mm P = 15 kN = 15000 N 35 mm 27 mm As = Steel rod Brass tube 3m 25 mm 5 mm Flange Figure 15.8 Since both the ends are fixed together, Strain in steel rod = Strain in brass tube σs σb = Es Eb 20000 σ σs = b × Es = σb × Eb 100000 σs = 2σb @seismicisolation @seismicisolation (i) 304 • Strength of Materials Volume of steel rod, Vs = Area × length = 490.625 × 3000 = 1471875 mm3 Volume of brass tube = 389.36 × 3000 = 1168080 mm3 Strain energy in steel rod, σs2 × Volume 2Es (2σb )2 = × 1471875 2 × 200000 = 14.72 σb2 Us = Strain energy in brass tube, σb2 × Volume 2Eb σ 2 × 1168080 = b 2 × 100000 = 5.840 σb2 Ub = Total strain energy stored in compound bar, U = Us +Ub = 14.72σb2 + 5.840 σb2 = 20.56 σb2 (ii) Work done by falling load = P(h + x) = 15000 (3 + x) σ Strain in brass rod = b Eb x σb or = l 1 × 105 σ × 3000 x= b = 0.03σb 1 × 105 Substituting this value of x in Eqn. (iii), 15000 (3 + 0.03σb ) @seismicisolation @seismicisolation (iii) Strain Energy, Impact Loading and Deflection Due to Bending • Now equating the work done by the falling weight to the total strain energy. 15000 (3 + 0.03σb ) = 20.56 σb2 45000 + 450σb = 20.56 (from Eqn. (ii)) σb2 20.56 σb2 − 450σb = 45000 = 0 σb2 − 21.89σb − 2188.7 = 0 This is a quadratic equation. ∴ √ 21.89 ± 21.892 + 4 × 2188.7 σb = 2 √ 21.89 ± 479.17 + 8754.8 = 2 21.89 + 96.1 = 2 = 59 N/mm2 Ans σs = 2σb = 2 × 59 = 118 N/mm2 Ans Strain Energy in Pure Shearing Let a rectangular block (refer Fig. 15.9) of material be subjected to shearing forces F acting on the opposite faces as shown. 1 Now, the work done = F ×CC 2 1 = F ×CB × φ 2 MM [∵ = tan φ = φ for small value of φ ] MN D′ D F φ A C′ C φ F B Figure 15.9 Now F = τ × DC where τ is the shearing stress τ (φ = εs = shear strain) Also φ = C Taking unit depth normal to diagram, we get Strain energy = work done τ τ2 1 × DC ×CB Work done = τ × DC ×CB × = 2 C 2E Now DC ×CB is the volume of rectangular block, since it has unit depth normal to DCBA. τ2 × Volume of block. ∴ Strain energy = 2E @seismicisolation @seismicisolation 305 306 • Strength of Materials Strain Energy in Torsion a) Solid shaft T l φ θ R l Figure 15.10 Let a solid shaft be subjected to a torque T , which produces a twist θ in the length of shaft. Work done = But ∴ ∴ Cθ τ T = = J l R T= Work done = = Now J = ∴ 1 T.θ , which is stored as strain energy. 2 Work done = τ .J R and θ= τL CR 1 τJ τL × × 2 R CR 1 τ2 J × l × × 2 2 C R π R4 2 1 τ 2 π R4 × l × × 2 C 2R2 1 τ2 × × π R2 l 4 C τ2 × Volume ∵ or Strain energy = 4C = Volume = π R2 l b) Strain Energy for a Hollow Shaft 1 Work done = T θ 2 Work done = and θ= τl CR Jl τ2 × 2C R2 @seismicisolation @seismicisolation and T = τ J (as with solid shaft) R Strain Energy, Impact Loading and Deflection Due to Bending π 4 4 R −r 2 τ 2 π l R4 − r4 ∴ Work done = × 2C 2R2 2 τ 2 π l R + r2 R2 − r2 = × 2C 2R2 2 R + r2 τ2 × Thus, strain energy, V = × Volume ∵ 4C R2 • 307 J= π l R2 − r2 = Volume Strain Energy Due to Bending As we have already proved in chapter of Deflection of Beams that strain energy stored in beam l 2 M dx . in given by U = 0 2EI In case M is constant over the length l, then U= M2l 2EI Strain Energy and Deflection Due to Bending If y is the deflection under the load W , then 1 U = Wy 2 2U or y = W Strain Energy of a Simply Supported Beam having One Eccentric Load X W b a A C x RA=Wb l l X B RB=Wa l Figure 15.11 Refer Fig. 14.37, we have already proved that total strain energy = U = UAC +UCB = @seismicisolation @seismicisolation w2 a2 b2 6EIl 308 • Strength of Materials Strain Energy of a Simply Supported Beam having Uniformly Distributed Load on Whole Span X A RA= wl 2 w/unit length B RB= wl 2 x X Figure 15.12 Considering a section X − X, at x from A, wx2 wl x− 2 2 l 2 M U= dx 0 2EI 2 l wlx wx2 1 dx − = 2EI 0 2 2 w2 l 2 2 l x + x4 − 2lx3 dx = 8EI 0 3 l w2 l 2 x x5 2lx4 = + − 8EI 3 5 4 Mx = ∴ Strain energy, 0 = Hence, U= w2 l5 8EI 3 + l5 5 − l5 4 w2 l 5 240EI Strain Energy and Deflection in a Cantilever having Point Load at Free End W Y d A l X x Figure 15.13 @seismicisolation @seismicisolation B b Strain Energy, Impact Loading and Deflection Due to Bending Mx = −W x l Strain energy, U= M2 dx 2EI 0 l = W 2 x2 dx 2EI 0 W2 = 2EI l x2 .dx 0 = Hence, U= W2 2EI x3 3 l 0 W 2l3 6EI If y at free end is the deflection (maximum) 1 Then work done = W Ymax 2 Equating, work done and strain energy, 1 W 2l3 W ymax = 2 6EI W l3 (as before) ymax = 3EI Now if, σ = Maximum bending stress Wl Mmax 6W l = = 2 Z bd 2 bd 6 σ bd 2 W= 6l σ= ∴ Putting value of W in the expression for U, We have, σ 2 b2 d 4 l 3 bd 3 36l 2 × 6E × 12 σ2 = × bdl 18E σ2 × Volume of the beam U= 18E U= Hence, @seismicisolation @seismicisolation • 309 310 • Strength of Materials Strain Energy of a Simply Supported Beam with Point Load at Centre W X d A x C X l W/2 B l/2 b W/2 Figure 15.14 At x from A, Mx = W x 2 Strain energy for AC, = l/2 2 2 W x 4 0 × 1 dx 2EI l W2 W2 = x2 dx = 8EI = 8EI 0 2 3 W l x3 3 l/2 o 192EI Because of symmetry, For whole beam, strain energy U = 2 × W 2l3 Therefore, U = 96EI Now, ∴ W 2l3 192EI σ= W l/4 Mmax 3W l = 2 = Z bd /6 2bd 2 σ= 3W l 2bd 2 or W = 2σ bd 2 3l Substituting in total strain energy (U) formula deriver above, we have U= = 2σ bd 2 3l 2 × l3 96EI 4σ 2 b2 d 4 l3 × 2 96EI 9l @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending = 4σ 2 b2 d 4 × 9l 2 • 311 l3 96E × bd 3 12 σ2 (lbd) 18E σ2 U= × Volume of the beam 18E = Maxwell’s Reciprocal Theorem It states that “the work done by the first system of loads due to displacements caused by a second system of loads equals the work done by the second system of loads due to displacements caused by the first system of loads”. Or in simple words: “In any beam or truss the deflection at any point M due to load W at any other point C is the same as the deflection at C due to the same load W at D.” For example, the central deflection (at M) of a simply supported beam carrying an offset load, (See Fig. 15.15) is the same as the deflection at C if the load were moved to the centre M. W W A C M y B A C B M y (b) (a) Figure 15.15 Another example: If a cantilever carries a concentrated load not at the free end (Fig. 15.16), the deflection at C due to the load at B is same as the deflection at B if the load were moved to C. W B (a) W B y C C y (b) Figure 15.16 Proof: Let an elastic body be subjected to forces Wa and Wb at points A and B respectively (see Fig. 15.17). B A Wa Figure 15.17 @seismicisolation @seismicisolation Wb 312 • Strength of Materials Let δaa be the deflection at A in the direction of Wa due to Wa . Let δab be the deflection at A in the direction of Wa due to Wb . Let δbb be the deflection at B in the direction of Wb due to Wb . Let δba be the deflection at B in the direction of Wb due to Wa . 1 Let Wa be applied first then work done = Wa δaa , assuming the load is gradually applied. 2 1 1 If Wb is now applied, the additional work done = Wb δbb + Wa δab , the whole of Wa moving 2 2 through the additional distance δab . 1 1 Thus, the total work done = Wa δaa + Wb δbb +Wa δab . 2 2 If the loads are removed and then reapplied in reverse order (i.e., Wb is applied first), it will be 1 1 seen, by analogy, that the total work done = Wa δaa + Wb δbb +Wb δba . 2 2 Irrespective of the order in which the loads are applied, the body will assume the same strained position and have the work done by the loads will be the same, so that Wa δab = Wb δba . In the case where Wa = Wb , δab = δba , i.e., the deflection at A due to a load at B is the same as the deflection at B if the load is applied at A, the deflections being the movements of the points A and B in the directions shown in Fig. 15.17. Betti’s Theorem of Reciprocal Deflections It states “In an elastic system, the external work done by a force F acting at P during the deflections caused by another force at Q is equal to the external work done by the force at Q during the deflections caused by the force P.” Mathematically, Fp δ pq = Q p δqp E XAMPLE 15.12: A thin proving ring of radius r is subjected to a diametral tensile load W . Determine (a) the increase in diameter in the direction of W , (b) the decrease in diameter perpendicular to the direction of W . S OLUTION : W P A Q Mo r dθ θ B W (a) (b) Figure 15.18 @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending • 313 It is sufficient to consider one quarter of the ring AB (Fig. 15.18b) fixed at the midpoint of the W at the free end. Due to continuity of the ring at A, however sides and subjected to a force P = 2 the tangent there must remain horizontal and hence a moment M0 must be applied for this purpose, the moment corresponding to the bending moment in the actual ring at that point. The increase in the vertical diameter of the ring will be twice the deflection of A relative to B in the quadrant and the decrease in horizontal diameter will be twice the horizontal movement of A relative to B. For this purpose, it is necessary to add a horizontal force Q of zero magnitude. Taking the origin at A, M = M0 − Pr sin θ − Qr (1 − cos θ ) ∂M =1 ∂ M0 ∂M = −r sin θ ∂P ∂M = −r(1 − cos θ ) ∂Q and dx = r d θ There is no rotation in the direction of M0 , ∴ 1 ∂U = ∂ M0 EI π /2 i.e., 0 from which M0 = 2Pr π l M. 0 ∂M dx = 0 ∂ M0 {M0 − Pr sin θ } × 1 × r d θ = 0 ∂U 2 l ∂M M = dx δv = 2 ∂P EI 0 ∂P 2 π /2 2pr − Pr sin θ × {−r sin θ } × r d θ = EI 0 π 2Pr3 π 2 W r3 − = = 0.1484 EI 4 π EI 2 l ∂M ∂U M = dx ∂ Q EI 0 ∂ Q 2 π /2 2Pr = − Pr sin θ × {−r (1 − cos θ )} × r d θ EI 0 π 2Pr3 2 1 = − EI π 2 δh = 2 @seismicisolation @seismicisolation 314 • Strength of Materials = 0.137 W r3 EI Note: Since Q = 0, the term involving Q may be omitted from all the integrals but it cannot be ∂M will be obtained. omitted from the original expression for M, otherwise no value for ∂Q E XAMPLE 15.13: Find the strain energy stored by the structure shown in Fig. 15.19 and hence compute the vertical deflection of the end A. Assume that the section of the member is uniform. S OLUTION : a B At any section in AB distant x from A, the bending moment is given by x y M = Wx ∴ Strain energy stored by AB l M 2 dx 2EI a 2 2 W x .dx = 2EI 0 W 2 a2 UAB = 6EI UAB = C Figure 15.19 Any section is BC distant y from B the bending moment. is given by, M = Wa ∴ Strain energy stored by BC UBC = l 2 M dx 0 2EI = l 2 2 W a dy 0 2EI = W 2 a2 l 2EI ∴ Total strain energy stored U = UAB + UBC = W 2 a2 W 2 a2 l W 2 a2 + = (a + 3l) Ans 6EI 2EI 6EI @seismicisolation @seismicisolation A W Strain Energy, Impact Loading and Deflection Due to Bending • 315 Let δ be the vertical deflection at the end A, equating the work done to the strain energy stored, we have, 1 W 2 a2 Wδ = (a + 3l) 2 6EI Wa2 (a + 3l) ∴ δ= Ans EI Bending Under Impact Loads: If a weight W drops from a height h at the point C on simply supported beam as shown is Fig. 15.20, then in order to find instantaneous deflection under the load is sometimes required. W A h C y B Figure 15.20 Let We be the equivalent weight which when gradually applied at C will produce the same deflection y. As we know in both cases the deflection is same, then the strain energy in respective cases should also be same. And strain energy of the beam is equal to the work done by respective weights in both cases. 1 Therefore, W (h + y) = We y (i) 2 Depending upon type of beam (end conditions), the deflection y can be found out in terms of We using methods described in chapter ‘Deflection of Beams’. The deflection is directly proportional to the load applied on beam. So, or y ∝ We y = kWe (ii) where k is a constant which depends upon the type of beam and its end conditions. Substituting for y from Eqn. (ii) in Eqn. (i) 1 W (h + y) = We y 2 1 W (h + kWe ) = We2 k 2 (iii) Equation (iii) is a quadratic equation and its solution will give the value of We . Then deflection y can be calculated from Eqn. (ii) E XAMPLE 15.14: A concentrated load of 12 kN applied to a simply supported beam at midpoint of the beam, produces a deflection of 7 mm and a maximum bending stress of 25 MN/m2 . Calculate @seismicisolation @seismicisolation 316 • Strength of Materials the maximum value of the instantaneous stress produced when a weight of 6 kN is allowed to fall through a height of 20 mm on the beam at middle of the span. S OLUTION : Let We be the static load equivalent to the given impact or falling load. Since 12 kN static load produces a deflection of 7 mm, then We will produce a deflection y so that 7 ×We = 0.583We mm = 0.000583We m y= 12 1 Now using equation W (h + y) = We y 2 1 We × 0.000583 We 2 0.12 + 0.003498 W = 0.0002915 We2 6(0.020 + 0.000583 We ) = 0.0002915 We2 − 0.003498 W − 0.12 = 0 We2 − 12 W − 411.664 = 0 √ +12 ± 144 + 1646.66 We = 2 +12 ± 42.32 = 2 = 27.16 kN Now because a static load of 12 kN produces a maximum bending stress of 25 MN/m2 , in the impact case the maximum bending stress produced for which static load is 27.16 kN, will be 25 × 27.16 = 56.58 MN/m2 Ans σmax = 12 Load factor: In impact loading on a collar we have derived the instantaneous stress, σ= P A 1+ 1+ 2hAE Pl Static deflection δ l due to load P is given by δl = Pl AE ∴ σi = p A 1+ 1+ 2h δl [σ is replaced by σi for instantaneous stress] OR σi = σ σi = 1+ σ 1+ 1+ 1+ 2h δl 2h δl @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending The dimensionless ratio • 317 σi is usually called load factor and is denoted by n. σ ∴ n = 1+ 1+ 2h δl E XAMPLE 15.15: A 1.2 m long beam rectangular in section 40 mm wide × 50 mm deep is supported on rigid supports at its ends. If it is struck at the centre by a 15 kg mass falling through a height of 70 mm, find: i) The instantaneous stress developed; ii) The instantaneous strain energy is the beam. Take E = 200 GN/m2 15 kg h = 70 mm 40 mm 50 mm Beam A B y Beam Cross-section 1.2 m Figure 15.21 S OLUTION : I= 40 × 503 bd 3 = = 416666.7 mm4 12 12 Static deflection due to W W l3 48EI (15 × 9.81) (1.2)3 (1000)3 = 48 × 200000 × 416666.7 = 0.0636 mm δl = √ 2h 2 × 70 = 1+ = 1 + 2202.25 δl 0.0636 n = 46.93 (where n is load factor) n= 1+ @seismicisolation @seismicisolation 318 • Strength of Materials i) Instantaneous stress developed, σi : Static stress, ∴ M Wl 15 × 9.81 × 1200 = = 1 Z 4Z 4 × × 40 × 502 6 = 2.65 N/mm2 σ= σi = nσ = 46.93 × 2.65 = 124.36 MN/m2 Ans ii) Strain energy stored: Strain energy, 1 We δ li 2 1 U = (nW ) × (nδ l) [∵ We = nW, δ li = nδ l] 2 n2 (46.93)2 = W.δ l = × 15 × 9.81 × 0.0636 2 2 = 10306 Nmm = 10.31 Nm or J Ans U= Exercise 15.1 A steel rod at 100 mm in diameter is 5 m long. Find the maximum instantaneous stress induced when a pull of 200 kN is suddenly applied to it. Find also the instantaneous elongation. Take E = 200 GN/m2 . [Ans 1.273 mm] 15.2 Compare the strain energies of the two bars A and B of the same material and subjected to the same axial tensile loads. Bar A is of 50 mm diameter throughout while the bar B, though of the same total length as A, has diameter of 25 mm over the middle third of its length, the remainder being of 50 mm diameter. Also compare their proof resilience in simple tension. 1 [Ans , 8] 2 15.3 Two shafts A and B are of same length and made of steel and bronze, respectively. Both the shafts are subjected to equal torques. Find the ratio of the shaft diameters so that (i) strain energy stored per unit volume (ii) total strain energy stored, by each is the same. Take C for steel as 80 GPa and for bronze as 50 GPa. [Ans 0.924, 0.889] 15.4 A steel rod of cross-sectional area 1000 mm2 and 2 m long has a collar at its lower end, while its upper end is fixed. A weight of 200 N falling from height h and striking the collar produces an instantaneous maximum stress of 50 MPa in the rod. Assuming 5 per cent energy loss during impact, determine the value of h. Take E = 200 GPa for steel. 15.5 A cantilever of uniform section with length l and flexural rigidity EI carries a uniformly distributed load of w per unit length from its mid-span to free-end. Use Castiglieno’s theorem to find deflection at the free end of the cantilever. [Ans W l 4 /128EI] @seismicisolation @seismicisolation Strain Energy, Impact Loading and Deflection Due to Bending • 319 15.6 A hollow shaft subjected to a pure torque attains a maximum shearing stress τ . Given that the strain energy stored per unit volume is τ 2 /3C, calculate the ratio of shaft diameters. If such a shaft is required to transmit 3700 kW at 110 r.p.m. with uniform torque and the energy stored 2 is 20 kJ/m3 , determine the actual diameters. Take C = 80 GN/m√ . 3 : 1, 298.2 mm, 172 mm] [Ans 15.7 A weight of 2 kN falls from 24 mm on to a collar fixed to a steel bar which is 14 mm in diameter and 5.5 m long. Determine the maximum stress induced in the bar. E = 205 GPa. [Ans 166 MPa] 15.8 A steel bar of constant section, second moment of area I is bent as shown in Fig. 15.22 and fixed at one end. Find the horizontal and vertical deflections at the free end. P a l Ans Figure 15.22 a Pal 2 Pa2 (l + ); EI 3 2EI 15.9 Fig. 15.23 shows a flat ring made of steel 25 mm wide by 6 mm thick, loaded with a central load of 600 N. Calculate the maximum bending moment in the ring. Sketch the bending moment diagram and find the position of the point of inflexion. Taking E as 200 GN/m2 , find the vertical deflection produced by the load of 600 N. 600 N 150 mm 75 mm Mean radius 150 mm 6 mm 600 N Figure 15.23 [Ans 38.7 Nm (at load point); 129 mm from load point, 8.9 mm] 15.10 A simply supported rectangular beam has a width of 120 mm and depth of 240 mm covering a span of 3 m. A load of 12 kN is dropped at the midspan of the beam from a height of 12 mm. Find the maximum instantaneous deflection and the stress induced in the beam. E = 200 GN/m2 . [Ans 3.95 mm, 126 N/mm2 ] @seismicisolation @seismicisolation C HAPTER 16 THEORIES OF ELASTIC FAILURE A material is regarded as failed if it is loaded beyond the elastic limit and permanent deformation occurs when the particles of a material separate from each other (as in use of brittle material) accompanied by considerable plastic deformation. It is easy to guess failure of material when the material is subjected to simple stress followed by an axial loading. But when the material is subjected to complex stresses followed by biaxial or triaxial loading then it is difficult to predict the failure of material. For complex systems, we shall discuss five important theories of failure. In these five theories, the complex state has been related to the elastic limit in simple tension or compression. As we have studied that in any complex loading system, three principal stresses exist: σ1 , σ2 and σ3 such that σ1 > σ2 > σ3 . σ1 is the maximum principal stress, σ2 is the intermediate principal stress end σ3 is the minimum principal stress. In 2-D stress system we will discuss the following five theories of failure: 1. 2. 3. 4. 5. Maximum Principal Stress Theory (Rankine’s Theory) Maximum Shear Stress Theory (Guest’s or Tresca’s Theory) Maximum Principal Strain Theory (St. Venant’s Theory) Maximum Strain Energy Theory (Haigh’s Theory) Maximum Shear Strain Energy Theory (Von Mises’s Theory): Now we shall discuss these theories in details. 1. Maximum Principal Stress Theory (Rankine’s Theory) According to this theory, failure will occur when the maximum principal tensile stress (σ1 ) in the complex system reaches the value of the maximum stress at the elastic limit (σet ) in simple tension or the minimum principal stress (that is, the maximum principal compressive stress) reaches the elastic limit stress (σec ) in simple compression, i.e. σ1 = σet (in simple tension) = σo = σu (yield stress is simple tension or compression) σ3 = σec (in simple compression) σ3 means numerical value of σ3 . @seismicisolation @seismicisolation Theories of Elastic Failure • 321 Graphical representation is shown in Fig. 16.1. This theory is found to have good results when applied to brittle materials such as cast iron. σu σu and σ2 = ± Note: Working stress, σ1 = ± Factor of Safety F.O.S where σy , σx , τ direct and shear stresses on given planes is the complex system, σ1 = maximum principal stress σ2 σο Square (−) σο (+) σο σ1 (−) σο Figure 16.1 2. Maximum Shear Stress Theory (Guest or Teresa’s Theory) According to this theory the failure occurs when the maximum shear stress τmax reaches the value of the maximum shear stress in simple tension at the elastic limit, i.e., τmax = or σ1 − σ3 σet σu = = 2 2 2 in simple tension σ1 − σ3 = σet = σu While designing σet is replaced by the safe stress. This theory does not give accurate results for the state of stress of pure shear in which the maximum amount of shear in developed. This theory is preferred in case of ductile materials such as mild steel. In 3-D stress system, σ1 − σ3 τmax = 2 @seismicisolation @seismicisolation 322 • Strength of Materials σ2 B + σ1− σ2 = − σο A σ2 = σο σο σ1 = σο 45˚ (−) σο C O D + σο σ1 σ1− σ2 = σο σ2 = − σο F E (−)σο σ2 = − σο Figure 16.2 Graphical representation of maximum shear stress theory 3. Maximum Principal Strain Theory (St. Venant’s Theory) Failure occurs when the greatest principal strain reaches the strain at the elastic limit in a simple test. According to this theory, failure of material occurs when the maximum strain in the complex stress system equals the value of maximum strain at yield point in simple tension or compression test. εcomplex = εsimple 1 σ0 (σ1 − μσ2 ) = ± E E σ1 − μσ2 = ±σ0 (for tensile test) Also σ2 − μσ1 = ±σ0 (for compressive test) Except for brittle material, this theory does not match with experimental results, and so finds little general support. σ2 A σο σ2 − μσ1 = σο B Rhomboid σ1 − μσ2 = σο σο (−) σο C σ1 − μσ2 = − σο D σ2 − μσ1 = − σο σ1 Shear diagonal (−) σο Figure 16.3 Graphical representation of maximum principal strain theory @seismicisolation @seismicisolation Theories of Elastic Failure • 323 By plotting the above equation a rhomboid is obtained. 4. Maximum Strain Energy Theory (Haigh’s Theory) Failure occurs when the energy stored per unit volume in a strained material reaches the strain energy per unit volume at the elastic limit in a simple tension test, i.e., the maximum energy a body can store without permanent deformation in a fixed quantity, irrespective of the manner in which it is strained. σ2 1 2 σ1 + σ22 − 2μσ1 σ2 = 0 2E 2E σ12 + σ22 − 2μσ1 σ2 = σ02 2 2 σ1 σ2 σ1 σ2 · =1 + − 2μ σ0 σ0 σ0 σ0 or or σ0 σ0 This is the equation of an ellipse with semi-major and semi-minor axes √ ; √ respec1−μ 1+μ tively each at 45◦ . This theory gives good results for ductile materials to the coordinate axes as shown graphically in Fig. 16.4. σ2 Ellipse C B D A E F Figure 16.4 @seismicisolation @seismicisolation σ1 Shear diagonal 324 • Strength of Materials 5. Maximum Shear Strain Energy Theory (Von Mises and Henkey’s Theory) This is also known as distortion energy theory. Failure occurs when the shear strain energy per unit volume in a strained material reaches the shear strain energy per unit volume at the elastic limit in a simple tension test, this is similar to the preceding theory but it is assumed that the volumetric strain energy plays no part in producing elastic failures. (a) For 3D stress system: σ2 1 (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 0 12C 6C or (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 2σ 20 (b) For 2D stress system, putting σ3 = 0, we get σ12 + σ22 − σ1 σ2 = σ02 σ12 + σ22 − σ1 σ2 1 2 = σ0 σ2 45° 45° (–) σo σo (–) σo Figure 16.5 This theory is good for ductile materials. This is the best theory of the above five theories. @seismicisolation @seismicisolation σ1 Theories of Elastic Failure • 325 When σt = σc and εt = εc . σ2 (– , +) I II Maximum strain energy theory (+ , +) Maximum shear strain theory Maximum shear stress theory 45° 45° σ1 (+, –) III (– , –) Maximum principal strain theory Maximum principal stress theory IV Figure 16.6 Graphical representation of various theories of failure on the same diagrams E XAMPLE 16.1: The load on a bolt is an axial pull of 15 kN together with transverse shear force of 7 kN. Estimate the diameter of the bolt. Using all theories of failure. σut = 280 N/mm2 , factor of safety = 4. Poisson’s ratio (μ ) = 0.3 S OLUTION : 280 = 70 N/mm2 = σ0 Allowable simple tensile stress = 4 Let d = diameter of bolt (core dia). 15000 60000 = =σ Then, normal stress is = π π d2 d2 4 7000 28000 and shear stress (τ ) = π = π d2 d2 4 Therefore, principle stresses are σx σ1 , σ2 = ± 2 σ 2 x 2 + τ2 @seismicisolation @seismicisolation and σ3 = 0 326 • Strength of Materials σ1 σ2 = σ1 = 60000 ± 2π d 2 60000 2π d 2 2 + 2 30000 1 √ + 2 900000000 + 784000000 2 πd πd = 1 30000 + 2 × 41036.6 π d2 πd = 71036.6 π d2 σ2 = 28000 π d2 30000 41036.6 − π d2 π d2 =− 110.36 π d2 σ3 = 0 1. Applying Rankine Theory (Principal Stress Theory) Maximum principal stress in bolt σ1 = σ 1 + 2 2 σ 2 + 4τ 2 60000 π d2 2 60000 1 + 2 2π d 2 = 30000 1 √ + 3600000000 + 3136000000 π d2 2π d 2 = 1 30000 + × 82073.14 π d2 2π d 2 σ1 = +4 28000 π d2 2 = 71036.57 π d2 Maximum stream in simple tension = 70 N/mm2 = σ0 Now equating σ1 = σ0 71036.57 = 70 π d2 ∴ d 2 = 323.2 or d = 17.977 = 18 mm @seismicisolation @seismicisolation Theories of Elastic Failure • 327 2. Maximum Shear Stress Theory (Guest on Tresca’s Theory) τmax = 1 2 1 = 2 Now σ 2 + 4τ 2 60000 π d2 2 28000 +4 π d2 2 = 1 √ 3600000000 + 3136000000 2π d 2 = 1 × 82073.14 2π d 2 = 41036.6 π d2 τ= 10 σ2 − σ1 σ = = = 35 N/mm2 2 2 2 41036.6 = 35 π d2 ∴ d2 = ∴ 41036.6 = 373.4 35 × π d = 19.32 mm 3. Maximum Strain Energy Theory (Von Mises & Hencky Theory) or Distortion Theory σ12 + σ22 − 2μσ1 σ2 = σ02 71036.6 π d2 2 11036 11036 2 71036.6 = 702 + − + 2 × 0.3 π d2 π d2 π d2 5046198540 121793296 4703759506 + + = 4900 π 2d4 π 2d4 π 2d4 9871751392 = 4900 π 2d4 1001232443 = d4 4900 d 4 = 204333 d = 21.26 mm @seismicisolation @seismicisolation 328 • Strength of Materials 4. Maximum Strain Energy Theory (Haigh Theory) (σ1 − σ2 )2 + σ22 = σ02 71036.6 11036 + π d2 π d2 82072 π d2 1 π 2d4 2 + 2 11036 = 702 + − π d2 121793296 = 4900 π 2d4 (6735813184 + 121793296) = 4900 6747992480 = d4 π 2 × 4900 d 4 = 139675 d = 19.33 mm Maximum diameter d is 21.26 mm So the answer is 21.26 mm E XAMPLE 16.2: A subject is subjected to a maximum torque of 12 kNm and a maximum bending moment of 8.5 kNm at a particular section. If the allowable equivalent stress in simple tension is 170 MN/m2 , find the diameter of the shaft according to the maximum shear strain theory. Maximum torque, T = 12 kNm Maximum bending moment, M = 8.5 kNm Allowable equivalent stress in simple tension, σt = 170 MN/m2 ∴ M= π 3 d σb 32 σb = M × 32 π d3 τ= and 16T π d3 @seismicisolation @seismicisolation T =τ× π 3 d 16 Theories of Elastic Failure • 329 Principal stresses are given by: σ σ1 , σ2 = b ± 2 σ 2 b 2 + τ2 1 σb ± σb2 + 4τ 2 2 ⎡ ⎤ 1 ⎣ 32M 32M 2 32T 2 ⎦ = ± + 2 π d3 π d3 π d3 = 16 M+ π d3 16 σ1 = 3 M + πd = M2 + T 2 M2 + T 2 σ2 = 0 and σ3 = 16 M− π d3 M2 + T 2 According to the maximum shear stress theory, σt = σ1 − σ3 = = d3 = 16 M+ π d3 16 M2 + T 2 − 3 M − πd M2 + T 2 32 [M 2 + T 2 ] π d3 32 × 103 × π × 160 × 106 8.52 + 122 d 3 = 6.37 × 10−5 × 14.705 d 3 = 0.000937 d = 97.8 mm ∴ d = 0.0978 m Ans E XAMPLE 16.3: A steel shaft is subjected to an end thrust of 90 MPa and the maximum shearing stress on the surface arising from torsion is 68 MPa. The yield point of the material in simple tension was found to be 300 MPa. Calculate the factor of safety of the shaft according to the following theories: i) Maximum shear stress theory ii) Maximum distortion energy theory (AMIE Summer 2000) @seismicisolation @seismicisolation 330 • Strength of Materials S OLUTION : Given σ1 =? σ2 = 0, σ3 = −90 MN/m2 τmax = 60 MN/m2 ; σl t = 300 MN/m2 i) Maximum shear stress theory σ1 − σ2 = 60 2 or σ − (−90) = 120 or σ1 = 30 MN/m2 Also σ1 − σ3 = σt τmax = 30 − (−90) = σt ∴ σt = 120 MN/m2 300 σet F. O. S. = = = 2.5 Ans σt 120 ii) Maximum distortion energy theory σt2 = σl2 + σ32 − σ3 σ1 = 302 + (−90)2 − (−90)(30) = 11700 ∴ σt = 108.17 MN/m2 σet 300 F.O.S. = = = 2.77 Ans σt 108.17 E XAMPLE 16.4: In a steel member, at a point the major principal stress is 200 MN/m2 and the minor principal stress is compressive. If the tensile yield point of the steel is 250 MN/m2 , find the value of the minor principal stress at which yielding will commence, according to each of the following criteria of failure: (i) Maximum shearing stress, (ii) Maximum total strain energy, and (iii) Maximum shear strain energy Take, Poisson’s ratio = 0.3 S OLUTION : σ1 = 200 MN/m2 Yield point stress = σe = 250 MN/m2 Minor principal stress, σ2: @seismicisolation @seismicisolation Theories of Elastic Failure • 331 i) Maximum shearing stress criterion: σ1 − σ2 = σe σ2 = σ1 − σe = 200 − 250 = −50 MN/m2 σ2 = 50 MN/m2 (Compressive) Ans ii) Maximum total strain energy criterion: σ12 + σ22 − 2μσ1 σ2 = σe2 (or σ02 ) (200)2 + σ22 − 2 × 0.3 × 200 × σ2 = 2502 40000 + σ22 − 120 σ2 = 62500 σ22 − 120σ − 62500 + 40000 = 0 σ22 − 1206 − 22500 = 0 √ +120 − 14400 + 90000 σ2 = 2 √ +120 − 104400 σ2 = 2 = (only −ve sign is taken as σ2 is to be compressive) +120 − 323 2 = 101.5 MN/m2 (Compressive) Ans iii) Maximum shear strain energy criterion: (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = 2σe2 putting σ3 = 0, we get (σ1 − σ2 )2 + σ22 + σ12 = 2σe2 σ12 + σ22 − 2σ1 σ2 + σ22 + σ12 = 2σe2 σ12 + σ22 − σ1 σ2 = σ22 (200)2 + σ22 − 200 σ2 = (250)2 σ22 + 40000 − 62500 − 200σ2 = 0 σ22 − 200σ2 − 22500 = 0 @seismicisolation @seismicisolation 332 • Strength of Materials √ +200 − 40000 + 90000 σ2 = 2 = −80.27 MN/m2 (Compressive) Ans E XAMPLE 16.5: At a point two of the principal stresses are 150 N/mm2 and 100 N/mm2 . Determine the safe range of the third principal stress at the point by five different theories. Take E = 200 GPa and μ = 0.25, failure stress in tension test to be 220 N/mm2 . Failure stress in tension and compression is the same. S OLUTION : a) Maximum principal stress theory σ1 = σy = 220 N/mm2 σ3 = σy (σy = yield stress) 2 = −220 N/mm Hence, the range is −220 ≤ σ ≤ 220 b) Maximum strain theory σ1 − 0.3 (150 + 100) = σy = 220 Also, σ1 = 220 + 75 = 295 N/mm2 σ3 − 0.3 (150 + 100) = σ y = −220 σ3 = −220 + 75 = −145 N/mm2 Hence, the range is −145 ≤ σ ≤ 220 N/mm2 c) Maximum strain energy theory σ 2 + (100)2 + (150)2 − 2 × 0.3 (150σ + 100σ + 150 × 100) = 2202 σ 2 + 10000 + 22500 − 150σ − 9000 = 48400 σ 2 − 150σ − 24800 = 0 ∴ √ 150 ± 22500 + 99600 σ= 2 150 ± 349.4 σ= 2 σ1 = 249.7, σ2 = −99.7 Hence, the range is −99.7 ≤ σ ≤ 249.7. @seismicisolation @seismicisolation Theories of Elastic Failure • 333 d) Maximum shear stress theory σ1 − 100 = σy = 210 σ1 = 220 + 100 = 320 N/mm2 Also, 150 − σ3 = 220 σ3 = 150 − 220 = −70 N/mm2 Hence, the range is −70 ≤ σ ≤ 320 N/mm2 e) Maximum distortion theory or shear strain theory (σ − 100)2 + (σ − 150)2 + (150 − 100)2 = 2 (220)2 σ 2 + 10000 − 200σ + σ 2 + 22500 − 300σ + 2500 = 96800 2σ 2 − 500σ − 61800 = 0 σ 2 − 250σ − 30900 = 0 √ 250 ± 62500 + 123600 σ= 2 +250 ± 431.4 σ= 2 σ1 = 340.7, σ3 − 181.4 Hence, the range is −181.4 ≤ σ ≤ 340.7 N/mm2 . E XAMPLE 16.6: Principal stresses at a point in an elastic material are 120 MPa tensile, 60 MPa tensile and 20 MPa compressive. Determine the factor of safety against the failure based on various theories. The elastic limit in simple tension is 240 MPa and Poisson’s ratio 0.3. S OLUTION : i) Maximum principal stress theory Failure takes place when the maximum principal stress reaches the value of maximum stress at that limit. Thus, maximum principal stress, σ = 120 MPa 240 =2 So factor of safety = 120 ii) Maximum shear stress theory σ = 120 − (−20) = 140 MPa 240 = 1.71 Factor of safety = 140 @seismicisolation @seismicisolation 334 • Strength of Materials iii) Maximum principal strain theory σ1 − μσ2 − μσ3 = 120 − 0.3 × 60 − 0.3(−20) = 108 MPa Factor of safety = 240 = 2.03 108 iv) Maximum strain energy theory σ 2 = σ12 + σ22 + σ32 − 2 μ (σ1 σ2 + σ2 σ3 + σ3 σ1 ) = 1202 + 602 + (−20)2 − 2 × 0.3 [120 × 60 + 60 (−20) + (−20 × 120)] = 14400 + 3600 + 400 − 3600 = 14800 ∴ σ = 121.6 Factor of safety = 240 = 1.97 121.6 v) Maximum shear strain energy theory 2 σ 2 = (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 = (120 − 60)2 + [60 − (−20)]2 + [−20 − 120]2 = 3600 + 6400 + 19600 = 29600 σ2 = ∴ 29600 = 14800 2 σ = 121.65 MPa Factor of safety = 240 = 1.97 121.65 E XAMPLE 16.7: A solid shaft transmits 1000 kW at 300 r.p.m. Maximum torque is 2 times the mean. The shaft is subjects to a bending moment, which is 1.5 times the mean torque. The shaft is of ductile material for which the permissible tensile and shear stresses are 120 MPa and 60 MPa respectively. Determine the shaft diameter using suitable theory of failure. Give justification of the theory used. [AMI.E. Sec B Winter 94] @seismicisolation @seismicisolation Theories of Elastic Failure • 335 S OLUTION : Principal stresses: σb = Principal stresses = 32M ; π d3 1 σb ± 2 τ= 16T π d3 σb2 + 4τ 2 32 1 32M ± 3 M2 + T 2 3 2 πd πd 16 M ± M2 + T 2 = 3 πd 16 σ1 = 3 M+ M 2 + T 2 , σ2 = 0 πd = and σ3 = 16 M− π d3 M2 + T 2 Best theory is the distortion energy theory, which gives σ12 + σ32 − σ1 σ3 = σt2 or 16 π d3 2 M+ M2 + T 2 2 + M− M2 + T 2 2 × 16 2 16 2 2 2 2 − M −M +T = σt2 π d3 π d3 16 2 2 2 2 2 2M = σt2 or + 2M + 2T + T π d3 16 2 2 2 4M = σt2 or + 3T π d3 32 σt = 3 πd 3 M2 + T 2 4 2π NTmean = 1000 × 103 60 2π × 300 × Tmean = 1000 × 103 60 ∴ Tmean = @seismicisolation @seismicisolation 100 × 103 100 = kNm π π (i) 336 • Strength of Materials T = Max. torque = 200 kNm π M = Maximum bending moment = 150 kNm. π Putting in Eqn. (i), we get 150 2 200 2 3 × 103 + π π 4 32 × 103 3 3 d = (150)2 + (200)2 6 4 120 × 10 π 32 3 = 1502 + (200)2 4 120 π 2 × 100 3 −3 = d = 6.191 × 10 d = 0.1836 m. = 183.6 mm Ans 32 120 × 10 = π d3 6 Justification of using this theory: The theories used for ductile materials are: (i) Maximum shear stress theory and (ii) distortion energy theory. Out of these theories, distortion energy theory is best for ductile materials, as the experimental results for these materials fit very well in this theory. Maximum shear stress theory gives results on safer side, that is, a little more material is used than the actual required, as given by distortion energy theory. Maximum shear stress theory π 3 d σt 32 200 2 150 2 3 or 10 + π π π 3 or = d × 120 × 106 32 200 2 150 2 32 × + π π 3 d = × 103 6 π × 120 × 10 d 3 = 6.755 × 10−3 = 0.183 m ∴ d = 189 mm. M2 + T 2 = Because σt = 2τ (given), the equivalent torque criterion gives same results. @seismicisolation @seismicisolation Theories of Elastic Failure • 337 E XAMPLE 16.8: A mild steel shaft 100 mm diameter is subjected to a maximum torque of 15 kNm and a maximum bending moment of 10 kNm at a particular section. Find the factor of safety according to the maximum shear stress theory if the elastic limit in simple tension is 240 MN/m2 . [UPTU 2006–2007] S OLUTION : Tmax = 15 kNm = 15 × 106 Nmm Mmax = 10 kNm = 10 × 106 Nmm τ= = 16T π d3 16 × 15 × 106 π × [100]3 = 76.36 N/mm2 . σb = 10 × 105 M y= I I y = 10 × 106 Z 10 × 106 = π (100)3 32 107 × 32 = 101.81 N/mm2 π × 106 σb σb 2 σ1 , σ2 = + τ2 ± 2 2 = = 50.90 ± (50.90)2 + (76.36)2 = 50.90 ± 91.76 = 142.66 or τmax = = − 40.86 N/mm2 1 (σmax − σmin ) 2 1 (142.66 + 40.86) 2 @seismicisolation @seismicisolation 338 • Strength of Materials = 91.76 N/mm2 τmax = 240 × 106 2 × 106 = 120 N/mm2 Now, ∴ 91.76 = Factor of safety = 120 Factor of Safety 120 91.76 = 1.3 Ans E XAMPLE 16.9: Find the diameter of a shaft according to the maximum shear stress theory if the shaft is subjected to a maximum torque of 15 kNm and a maximum bending moment of 10 kNm at a particular section. Take allowable equivalent stress in simple tension as 200 MN/m2 . M= π 3 d σb 32 σb = 32M π d3 and T = π 3 d τ 16 16T π d3 1 2 2 σ= σb ± σb + 4τ 2 ⎡ ⎤ 32 2 32T ⎦ 1 ⎣ 32 M± + = 2 π d3 π d3 π d3 τ= 16 M± π d3 16 M+ σ1 = π d3 = ∴ M2 + T 2 M2 + T 2 According to the maximum shear stress theory, σ1 − σ3 = 32 π d3 M2 + T 2 d3 = 32 πσt M2 + T 2 @seismicisolation @seismicisolation Theories of Elastic Failure = 32 × 103 π × 200 × 106 • 339 102 + 152 = 5.095 × 10−5 (18.03) d = 0.0947 m = 94.7 mm Ans Exercise 16.1 A steel shaft is subjected to two principal stresses of +40 MN/m2 and −90 MN/m2 . If the elastic limit in simple tension or compression is 360 MN/m2 , find factor of safety according to each maximum principal stress, maximum shear stress and shear strain theories. [Ans 4,2.77 and 3.12] 16.2 A bar of circular section is subjected to an axial tensile load of 10 kN and a transverse sheer force of 5 kN. Find the diameter of the bar if the allowable stress in simple tension is 100 MN/m2 using each of the following theories of failure: i) Maximum principal stress theory ii) Maximum shear stress theory iii) Maximum Strain energy theory if μ = 0.25 [Ans i) 12.35 mm, ii) 13 mm and iii) 12.8 mm] 16.3 Determine the allowable shear stress for a circular solid shaft transmitting torque according to each of the following theories of failure: a) Maximum principal stress theory b) Maximum principal strain theory c) Maximum shear stress theory d) Maximum strain energy theory e) Distortion energy theory (shear strain energy theory) Allowable tensile stress in simple tension for the material is 160 MN/m2 . [Ans 160, 123.1, 80, 99.2 and 92.4 MN/m2 ] 16.4 A circular shaft of steel is subjected to combined bending and torsion, the bending moment being 20 kNm and torque 10 kNm. If safe equivalent stress in simple tension is 200 MN/m2 and μ = 0.25, find suitable diameter of the shaft based on the following theories. a) Maximum principal stress theory b) Maximum shear stress theory c) Maximum shear strain energy theory [Ans @seismicisolation @seismicisolation a) 102.56 mm; b) 104 mm; c) 103.5 mm] 340 • Strength of Materials 16.5 An axial pull of 20 kN along with a shear force of 15 kN is applied to a circular bar of 20 mm diameter. The elastic limit of the bar material is 230 MPa and the Poisson’s ratio, μ = 0.3. Determine the factor of safety against failure based on: a) maximum shear stress theory b) maximum strain energy theory c) maximum principal strain energy theory d) maximum shear strain energy theory [Ans 2; 2.3; 2.37; 2.2] 16.6 A thin cylindrical shell of diameter 200 mm and wall thickness t is subjected to material pressure of 3 N/mm2 . Yield strength of the material is 280 N/mm2 . Taking factor of safety of 3 and using the maximum shear stress theory, determine the wall thickness t of the cylinder. [Ans 3.45 mm minimum] 16.7 The load on a bolt consists of an axial thrust of 8 kN together with a transverse shear load of 4 kN. Calculate the diameter of the bolt using strain energy theory. Take factor of safety as 3. σy = 285 MN/m2 , μ = 0.3. [Ans 11.73 mm] 16.8 In a steel drum that is subjected to axial compressive force, the internal pressure is 10 N/mm2 . The maximum circumferential stress is 100 MPa and the longitudinal stress is 30 MPa. Find the equivalent tensile stress in a simple tensile test according to each of the five theories of failure. Take Poisson’s ratio, μ = 0.3. [Ans 96.43 MPa] 16.9 A mild steel shaft of 40 mm diameter when subjected to a pure torsion ceases to be elastic when torque reaches 2 kNm. A similar shaft is subjected to a torque of 1.2 kNm and a bending moment M kNm. If the maximum strain energy is the criterion for the elastic failure, find the value of M. Take Poisson’s ratio = 0.28. [Ans M = 1.28 kNm] 16.10 A shaft of 100 mm diameter is subjected to a bending moment of 5 kNm. Find the value of the maximum torque which can be applied to the shaft for each of the following conditions: a) maximum direct stress not to exceed 120 N/mm2 ; b) the maximum shearing stress not to exceed 60 N/mm2 ; c) maximum shear strain energy per unit volume not to exceed that induced by simple shear stress of 80 N/mm2 . [Ans a) 17.85 kNm; b) 10.65 kNm; c) 14.6 kNm] 16.11 A cast iron cylinder has outside and inside diameters of 200 mm and 125 mm. If the ultimate tensile strength of the cast iron is 150 MN/m2 , find according to each of the following theories, the internal pressure which would cause rupture: a) Maximum principal stress theory; b) Maximum strain theory and c) Maximum strain energy theory. Assume no longitudinal stress in the cylinder. μ = 0.25 [Ans a) 65.7 N/mm2 ; b) 59.2 N/mm2 ; c) 55.2 N/mm2 ] @seismicisolation @seismicisolation Theories of Elastic Failure • 341 16.12 A hollow mild steel shaft having 60 mm internal diameter and 120 mm as external diameter is subjected to a torque of 10 kNm and a bending moment of 3 kNm. Calculate the direct stress which acting alone would produce the same i) Maximum strain energy ii) Maximum shear strain energy iii) Maximum shear stress [Ans (i) −22.39 N/mm2 ; (ii) 56.78 N/mm2 and (iii) 64.642 N/ mm2 ] @seismicisolation @seismicisolation C HAPTER 17 COMBINED STRESSES (DIRECT, BENDING, TORSION) In engineering there are not always direct stresses, but combination of direct and bending or direct and torsion or combination of three is involved. Here, in this chapter we shall study such problems. E XAMPLE 17.1: A bending moment of 500 Nm and torque of 350 Nm is applied to a shaft of diameter 80 mm. Find: i) the maximum normal stress on a section perpendicular to the axis; b) the maximum shear stress on a section perpendicular to the axis; c) the principal stresses and d) the maximum shear stress: S OLUTION : a) Bending stress, σb = σb = b) τ = My I 500 × 0.080 × 64 2 × π (0.080)4 = 9952229.3 N/m2 = 9.95 MN/m2 T.r 350 × 0.040 × 32 = = 3483280 N/m2 = 3.48 MN/m2 J π × (0.08)4 Ans c) Principal stresses: σb 1 σb2 + 4τ 2 ± 2 2 9.95 1 ± = (9.95)2 + 4(3.48)2 2 2 1√ = 4.975 ± 99 + 48.44 2 σ1,2 = = 4.975 ± 6.07 = 11.045 N/mm2 , −1.095 N/mm2 @seismicisolation @seismicisolation Ans Ans Combined Stresses (Direct, Bending, Torsion) • 343 d) Maximum shear stress: σ1 − σ2 2 1 = (11.045 + 1.095) 2 = 6.07 N/mm2 Ans τmax = E XAMPLE 17.2: A cast iron column, hollow circular section, has a projecting bracket carrying a load of 120 kN. The load line is off axis of column by 0.3 m. The external diameter of the column is 0.35 m and thickness of metal 30 mm. Determine the extreme stresses in the section. S OLUTION : π × (0.352 − 0.292 ) = 0.03 m2 4 π I= (0.354 −0.294 ) = 0.0003891 m4 64 M = 120000 × 0.3 = 36000 Nm A= Direct stress, 120000 = 4000000 N/m2 0.03 = 4 MN/m2 (Compressive) σd = Bending stress, σb = 36000 × 0.35 My = = 16.19 MN/m2 (Tensile) also (Compressive) I 2 × 0.0003891 Extreme stresses: σ1 = −4 + 16.19 = +12.19 MN/m2 (Tensile) Ans σ2 = −4 − 16.19 = −20.19 MN/m2 = 20.19 MN/m2 (Compressive) Ans Combined Bending and Twisting 2τ σb A common application of combined stresses is that of a shaft subjected to bending and twisting and it is often convenient to express the resulting direct and shear stresses directly in terms of the applied moment and torque. We had already proved in the case of one direct stress only tan 2θ = @seismicisolation @seismicisolation 344 • Strength of Materials d σx T τ τ σx M (b) (a) Figure 17.1 If the bending moment is M and the torque is T (Fig. 17.1(a)) then the stresses acting on an element on the upper surface are shown in plan view (Fig. 17.1(b)) (those on the lower surface are the same, except that σx is compressive). M σx = π d3 32 and T τ= π d3 16 assuming solid shaft. Thus, the maximum principal stress, σ1 is given by σ1 = i.e., 1 σx + σx2 + 4τ 2 2 ⎫ ⎧ ⎛ ⎛ ⎞2 ⎞2 ⎪ ⎪ ⎬ ⎨ 1 M ⎜ T ⎟ ⎜ M ⎟ = π 3 ⎪ + ⎝ π 3 ⎠ + 4 ⎝ π 3 ⎠ 2⎪ ⎩ d ⎭ d d 32 32 16 π 3 d τmax = M 2 + T 2 16 π 2 d τmax is evidently the equivalent torque, i.e., that torque which, acting alone, 16 would produce the same maximum shear stress as M and T acting together. The quantity i.e., Te = M2 + T 2 E XAMPLE 17.3: A shaft of 100 mm diameter is subjected to at a certain section: a) a bending moment of 8 kNm and b) a twisting moment of 11 kNm. Compute the maximum direct stress induced in the section, indicating the position of plane on which it acts. Determine the stress, which acting alone, will produce the same maximum i) strain and ii) strain energy. μ = 0.3. S OLUTION : d = 100 mm; M = 8 kNm = 8 × 106 Nmm; T = 9 kNm = 9 × 106 Nmm. @seismicisolation @seismicisolation Combined Stresses (Direct, Bending, Torsion) Shear stress, τ = 16 T 16 × 11 × 106 = = 56 N/mm2 3 πd π (100)3 Bending stress, σb = 32 M 32 × 8 × 106 = = 81.53 N/mm2 3 π d3 π (100) tan 2 θ = ∴ 2τ 11 = 1.375 = σb 8 2 θ = 53.97 = 53◦ 58 ; θ1 = 26.5◦ 29 or with the axis of the shaft. θ2 = 116◦ 59 with the axis of the shaft. σb σb 2 σ1 = + τ2 + 2 4 81.53 81.53 2 = + 562 + 2 4 √ = 40.765 + 1662 + 3136 = 110 N/mm2 σb σb 2 σ2 = + τ2 − 2 4 81.53 2 81.53 − = + 562 2 4 √ = 40.765 − 1162 + 3136 = −24.79 N/mm2 i) For maximum strain, we have σ = E ε = σ1 − μσ2 = 110 − 0.3 × (−24.79) = 117.44 N/mm2 @seismicisolation @seismicisolation 63◦ 1 • 345 346 • Strength of Materials ii) For maximum strain energy, we have σ 2 = σ12 + σ22 − (μ × σ1 σ2 ) = (110)2 + (−24.79)2 − {0.3 × 110 × (−24.79)} = 12100 + 614.5 + 818; √ σ = 135325 σ = 116.33 N/mm2 . Ans E XAMPLE 17.4: A ship’s heavy duty propeller shaft is of 0.6 m diameter and supports a propeller weighing 170 kN. The propeller overhangs the supporting bracket by 2.5 m as shown in Fig. 17.2. The speed of the ship is 50 km/h. The engine develops 30 MW and the propeller receives it at 350 r.p.m. Assuming a propulsive efficiency of 72%, determine the maximum principal stresses and shear stress in the shaft. S OLUTION : T = Torque on the shaft P = Axial thrust 2π × 350 × T = 30 × 106 60 30 × 106 × 60 T= = 818926.3 Nm 2π × 350 Let P 2.5 m 170 kN Figure 17.2 Bending moment in shaft, Mmax = 170 × 2.5 = 425 kNm Work done by propeller = Axial thrust × Velocity P × 50 × 103 3600 30 × 106 × 0.72 × 3600 P= = 1555200 N = 1555.2 kN 50 × 103 30 × 106 × 0.72 = P −1555.2 × 103 × 4 = = −5.50 N/mm2 A π (0.6)2 32 × 425 × 1000 Bending stress, σb = π (0.6)3 Direct stress, σd = = 20.05 N/mm2 @seismicisolation @seismicisolation Combined Stresses (Direct, Bending, Torsion) • 347 Resultant maximum direct stress = −σd − σb = −5.50 − 20.05 = −25.55 N/mm2 16T 16 × 818926.3 = = 19.32 N/mm2 3 3 π (0.6) πd Principal stresses, Shear stress, τ = 1 1 σ1,2 = (σd + σb ) ± (σd + σb )2 + 4τ 2 2 2 1 1 = (−25.55) ± (−25.55)2 + 4 (19.32)2 2 2 1√ = −12.775 ± 652.8 + 1493.05 2 1√ = −12.775 ± 2145.85 2 = −12.755 ± 23.16 σ1 = 10.385 N/mm2 2 σ2 = −35.93 N/mm Ans Ans Maximum shear stress 1 2 σ + 4 τ2 2 1 = (25.55)2 + 4 (19.32)2 2 1√ = 2145.85 2 = 23.16 N/mm2 Ans = E XAMPLE 17.5: A solid shaft of diameter d is subjected to a bending moment M = 15 kNm and a twisting moment T = 25 kNm. What is the minimum diameter of the shaft if the shear stress in the shaft is not to exceed 160 N/mm2 and the direct stress is not to exceed 200 N/mm2 . S OLUTION : Bending moment = M = 15 kNm = 15 × 106 Nmm Torque T = 25 kNm = 25 × 106 Nmm Maximum permissible shear stress, τ = 160 N/mm2 Maximum permissible bending stress, σ = 200 N/mm2 @seismicisolation @seismicisolation 348 • Strength of Materials Minimum diameter of the shaft √ π d3 Equivalent torque, Te = M 2 + T 2 = τmax 16 √ √ π d3 M2 + T 2 M 2 + T 2 16 3 or d = × = ∴ 16 τmax τmax π 2 2 15 × 106 + 25 × 106 16 3 d = × 160 π d = 97.53 mm √ M + M2 + T 2 π d3 Similarly, equivalent bending moment = Me = = σmax 2 32 √ √ M+ M 2 + T 2 32 π d 3 M+ M 2 + T 2 = × ∴ ∴ d3 = 32 2 × σmax σmax × 2 π 6+ 6 2 + 25 × 106 2 15 × 10 15 × 10 d3 = × 32 2 × 200 × π d = 95.77 mm Naturally, between these two diameters the larger value is answer. Hence, d = 97.53 mm Ans. E XAMPLE 17.6: A circular shaft, transmitting 60 kW power at 150 rpm is supported in bearings that are 4.5 m apart. At 1.7 m from one bearing, it carries a pulley which exerts a transverse load of 18 kN on the shaft. Determine the suitable diameter of the shaft if: a) the maximum direct stress is not to exceed 120 N/mm2 , b) the maximum shear stress is not to exceed 70 N/mm2 and c) the stress which acting alone would produce the maximum strain energy is not to exceed 130 N/mm2 . Take μ = 0.3 S OLUTION : 60P 60 × 60000 = = 382.66 Nm = 382660 Nmm 2π N 2π × 150 Wab 18000 × 1.7 (4 − 1.7) 18000 × 1.7 × 2.3 = = = 17595 Nm M= l 4 4 = 17595000 Nmm T= i) Maximum principal stress criterion 16 σ = 3 M + M2 + T 2 πd 16 17595000 + 175950002 + 3826602 d3 = π × 120 d 3 = 0.0425 [17595000 + 17599161] d= 116.8 mm @seismicisolation @seismicisolation Combined Stresses (Direct, Bending, Torsion) • 349 ii) Maximum shear stress criterion 16 2 M +T2 π × d3 16 d3 = 175950002 + 3826602 π × 70 τmax = = 0.073 × 1.7599 × 107 = 1284727 ∴ d = 108.71 mm iii) Maximum strain energy criterion, σ= 16 4M 2 + π d3 2 1 +1 μ T2 1 μ 1 +1 2 16 0.3 2 d3 = 4 (17595000) + (382660)2 1 π × 140 0.3 = 0.0364 1.23834 × 1015 + 2.6 (1.4643 × 1011 ) = 0.0364 1.23834 × 1015 + 3.808 × 1011 = 0.0364 1.23834 × 1015 + 0.0003808 × 1015 = 1281280 d = 108.61 Hence, maximum diameter is chosen for safety, ∴ d = 117 mm Ans E XAMPLE 17.7: A solid circular shaft while transmitting power is subjected to a twisting moment of 10 kNm and a bending moment of 6 kNm at a particular location. If the shaft diameter is 80 mm, compute principal stresses and maximum shearing stress. S OLUTION : Diameter of shaft, d = 80 mm Torque, T = 10 kNm = 10 × 106 Nmm Bending moment, M = 6 kNm = 6 × 106 Nmm @seismicisolation @seismicisolation 350 • Strength of Materials Maximum principal stress 16 2 +T2 M + M π d3 ! 16 6 6 2 + 10 × 106 2 6 × 10 σmax = + 6 × 10 π 803 = 9.95 × 10−6 6 × 166 + 36 × 1012 + 100 × 1012 = 9.95 × 10−6 6 × 106 + 11.66 × 106 σmax = = 174.83 MPa (tensile) Ans 16 6 6 6 × 10 σmin = − 11.66 × 10 π 803 = 9.95 × 10−6 −5.66 × 106 = −56.32 MPa = 56.32 MPa (Compressive) Ans Maximum shearing stress: 16 2 2 M + T π d3 16 6 2 + 10 × 106 2 = 6 × 10 π 803 = 9.95 × 10−6 (36 × 1012 + 100 × 1012 ) τmax = = 9.95 × 10−6 × 11.66 × 106 = 116.02 MPa Ans E XAMPLE 17.8: A propeller shaft 180 mm external diameter and 90 mm internal diameter transmits 1000 kW power at 110 r.p.m. It is also subjected to a bending moment of 10 kNm and an end thrust of 130 kN. Find: (i) the principal stresses and (ii) the stress which acting alone would produce the same maximum strain. Take μ = 0.3 S OLUTION : do = 180 mm di = 90 mm P = 1000 kW, N = 110 r.p.m, M = 10 kNm = 10 × 106 Nmm, A= End Thrust, E = 120 kN = 120 × 103 N π (1802 − 902 ) = 19075.5 mm2 4 @seismicisolation @seismicisolation Combined Stresses (Direct, Bending, Torsion) Z= π 32 do4 − di4 do π 32 = π Torsional sectional modulus = 16 1804 − 904 180 1804 − 904 180 = 536498.4 mm4 = 1072996.8 mm4 130000 = 6.82 N/mm2 19075.5 10 × 106 = 18.64 N/mm2 σb = 536498.4 σd o = Maximum compressive direct stress: σc = 6.82 + 18.64 = 25.46 N/mm2 60P 60 × 1000 T= = = 86.86 kNm 2π N 2π × 110 = 86.86 × 106 Nmm 86.86 × 106 = 80.95 N/mm2 Maximum shear stress = τ = 1072996.8 Principal stresses: σc σ1 = + 2 σ 2 c 25.46 + = 2 2 + τ2 25.46 2 2 + 80.952 √ = 12.73 + 671.5 = 12.73 + 81.94 = 94.67 N/mm2 σ2 = 12.73 − 81.94 (Tensile) Ans = −69.21 N/mm2 = 69.21 N/mm2 (Compressive) Ans Max shear stress: τmax = = 25.46 2 2 + 80.952 √ 6715 = 81.94 N/mm2 @seismicisolation @seismicisolation Ans • 351 352 • Strength of Materials Stress to give maximum strain σ = σ1 + μ σ2 = 94.67 − 0.3 (−69.21) = 94.67 + 20.76 = 115.43 N/mm2 Ans E XAMPLE 17.9: The maximum normal stress and the maximum shear stress, analysed for a shaft of 150 mm diameter under combined bending and torsion were found to be 120 MN/m2 and 80 MN/m2 , respectively. Find the bending moment and torque to which the shaft is subjected. If the maximum shear stress is limited to 100 MN/m2 , find by how much the torque can be increased if the bending moment is kept constant? (AMIE Summer 1995) S OLUTION : d = 150 mm = 0.15 m Maximum normal stress, σmax = 120 MN/m2 Maximum shear stress, σmax = 80 MN/m2 For combined bending and torsion, 16 2 +T2 M + M π d3 16 2 τmax = 3 M +T2 πd σmax = (i) (ii) Substituting the values in Eqns. (i) and (ii), we get 16 2 +T2 M + M π (0.15)2 16 2 +T2 80 × 106 = M π (0.15)2 80 × 106 × π (0.15)3 M2 + T 2 = 16 2 2 M + T = 53014.4 120 × 106 = ∴ Putting this value in Eqn. (iii), we get 16 [M + 53014.4] π (0.15)3 120 × 106 × π (0.15)3 M + 53014.4 = 16 M + 53014.4 = 79521.4 M= 26507 Nm Ans 120 × 106 = @seismicisolation @seismicisolation (iii) (iv) (v) Combined Stresses (Direct, Bending, Torsion) • 353 Substituting for M in Eqn. (v), we get (26507)2 + T 2 = 53014.4 (26507.2)2 + T 2 = (53014.4)2 70262 × 104 + T 2 = 281052.6 × 104 T 2 = 104 (281052.6 − 70262) = (210790.6)104 T = 459 × 102 = 45900 Nm Ans Now, τmax = 100 MN/m2 = 100 × 106 N/m2 Since the bending moment is kept constant, M = 26507 Nm (already found out) Substituting in Eqn. (ii), we have, 16 100 × 10 = π (0.15)3 6 ! (26507)2 + T 2 100 × 106 × π × (0.15)3 16 2 2 26507 + T = 66267.9 265072 + T 2 = Squaring both sides, 265072 + T 2 = 4.39 × 109 7.02 × 108 + T 2 = 4.39 × 109 T 2 = 108 (36.88) ∴ ∴ T = 104 × 6.07 = 60700 Nm Ans The increased torque = (60700 − 45900) = 14800 Nm Ans E XAMPLE 17.10: A shaft rotating at 220 r.p.m. is required to transmit 20 kW. It is simply supported between bearings 2.8 m apart. It carries at its centre a flywheel of mass 120 kg. Determine the suitable diameter of the shaft if neither the maximum shear stress nor the tensile stress increases above 50 MN/m2 and 70 MN/m2 respectively. @seismicisolation @seismicisolation 354 • Strength of Materials S OLUTION : Torque, T = 25000 × 60 = 1085.7 Nm 2π × 220 Maximum bending moment = Wl 120 × 9.81 × 2.8 = = 824.04 Nm 4 4 Equivalent torque, Te = M2 + T 2 = (824.04)2 + (1085.7)2 √ Te = 679041.9 + 1178744.5 = 1363 Nm Equivalent bending moment, 1 M + M2 + T 2 2 1 2 2 824.04 + (824.04) + (1363) = 2 √ 1 = 824.04 + 679041.9 + 1857769 2 1 = (824.04 + 1592.74) 2 = 1208.39 Nm Me = If maximum shear stress is not to exceed 40 MN/m2 , Te = 1363 Nm π 3 d τmax Also Te = 16 π 1363 = d 3 × 50 × 106 16 1363 × 16 d3 = = 0.0001389 π × 50 × 106 ∴ d = 0.0518 m = 51.8 mm @seismicisolation @seismicisolation Combined Stresses (Direct, Bending, Torsion) • 355 If maximum tensile stress or principal stress is not to exceed 70 MN/m2 . Me = 1208.39 Nm π Also Me = d 3 σmax 32 π ∴ 1208.39 = d 3 ×70 × 106 32 1208.39 × 32 3 d = = 0.000176 π × 70 × 106 ∴ d = 0.056 m = 56 mm Ans Hence, suitable diameter out of 51.8 and 56 mm is 56 mm. Ans Exercise 17.1 A shaft 100 mm diameter carries a flywheel of mass 2 tonne at the centre of a simply supported span of 2.4 m. The centre of gravity of flywheel is 2.5 mm from the axis of rotation. If the shaft transmits 750 kW at 300 r.p.m. estimate the maximum stress at the centre of gravity (a) when centre of gravity is vertically below the axis and b) when it is vertically above the axis. [Ans a) 10.6 N/mm2 Tensile b) 7.33 N/mm2 Compressive] 17.2 Find the principal stresses in a propeller shaft and the inclination to the axis of the planes over which they act. Given that: (a) the thrust results in an axis compressive stress of 15 MN/m2 and (b) the engine torque results in a shear stress at the shaft surface of 60 MN/m2 . [Ans σmax = 52.97 MPa Tensile at 41◦ 21 , σmin = 67.97 MPa Compressive at 31◦ 26 ] 17.3 A rolled steel of I section having flanges 100 mm × 20 mm and web 260 mm × 10 mm is used on a short column to carry a load of 100 kN. The load acts eccentrically 50 mm to the left of axis YY passing through the centre of web and 60 mm above the axis XX passing through the centroid of the section. Determine the maximum and minimum stress intensities induced in the section. [Ans 99.32 N/mm2 (Compressive), 69 N/mm2 (Tensile).] 17.4 A shaft of dia 100 mm is subjected to tensile load 157 kN, bending moment 6.28 Nm and torque 9.42 kNm simultaneously. Calculate maximum tensile, compressive and shearing stresses produced in the shaft material. [Ans 106 MPa, 74.8 MPa, 63.8 MPa] 17.5 A horizontal beam, 3 m long, 50 mm wide, 75 mm deep, is simply supported at its left and pin-jointed on its right end. It is subjected to a vertical downward concentrated load of 0.5 kN at its midspan and a horizontal thrust of 5 kN acting from left end to the right side, 20 mm below the longitudinal axis. Determine the stress at the extreme fibre at the midspan. [Ans 6.67 N/mm2 compressive (top fibre) 5.67 N/mm2 tensile (bottom fibre)] 17.6 A solid shaft 127 mm diameter transmits 630 kW at 300 r.p.m. It is also subjected to a bending moment of 9.1 kNm and an end thrust. If the maximum principal stress is limited to 77 N/mm2 , find the end thrust. [Ans 31.223 kN] @seismicisolation @seismicisolation 356 • Strength of Materials 17.7 A short column of rectangular cross section 80 mm by 60 mm carries a load of 40 kN at a point 20 mm from the longer side and 35 mm from the shorter side located in the first quadrant. Determine the maximum compressive and tensile stresses in the section. [Ans 19.785 N/mm2 , −3.125 N/mm2 ] 17.8 A hollow rectangular column is having external and internal dimensions as 2.4 m × 1.8 m and 1.2 m × 1.2 m respectively. Calculate the safe load, that can be placed at an eccentricity of 0.5 m on a plane bisecting the longer side, if the maximum compressive stress is not to exceed 5 MN/m2 . [Ans 720 kN] 17.9 A screw clamp is tightened on a workpiece exerting clamping force as 11 kN. Screw C clamp is shown in Fig. 17.3. Find the maximum tensile and compressive stresses in the material at section A − B due to bending and direct loading. Area of section = 480 mm2 , Ixx = 6.4 × 104 mm4 . Workpiece 15 mm 90 mm Centroid 40 mm B Section AB A Figure 17.3 [Ans 314.4 MN/m2 at top face, −414.4 MN/m2 (Compressive)] @seismicisolation @seismicisolation 18 C HAPTER FIXED BEAMS Fixed beams and continuous beams are also called indeterminate beams because reactions of these cannot be determined from equations of static equilibrium, [ΣFx = 0, ΣFy = 0 and ΣM = 0]. Fixed beams are also called built-in or encastre beams. Fixed Beam with a Point Load at the Centre A W C x B x MA l2 l D Wl 4 + A A C MA B B − MB E E Free bending moment diagram F + − −− A Fixed bending moment diagram F B Figure 18.1 Because of symmetry fixing moment at A and B are equal ∴ MA = MB @seismicisolation @seismicisolation 358 • Strength of Materials W Wl . (Because of symmetry, RA = RB = ; So maximum bending Maximum bending moment is 4 2 Wl W l × = ) moment = 2 2 4 Area of triangle ADB = Area of rectangle AEFB 1 Wl ×l× = MA × l 2 4 Wl ∴ MA = 8 Wl Due to symmetry, MA = MB = 8 Consider a point X, x from A, Mx = EI d2y = RA · x − MA dx2 d2y = RA · x − MA dx2 d2y W x W l − EI 2 = 2 8 dx EI Integrating, EI dy W x2 W lx = − +C1 dx 4 8 C1 is the constant of integration dy = 0, When x = 0, dx Therefore, C1 = 0 The equation becomes, EI W x2 Wl dy = − x dx 4 8 (i) Equation (i) given the slope at any point. Integrating Eqn. (i) again, EI y = W x2 W lx2 − + C2 12 16 where C2 is constant of integration, When x = 0, y = 0 ∴ C2 = 0 Therefore, EI y = W x3 W lx2 − 12 16 @seismicisolation @seismicisolation (ii) Fixed Beams • 359 Equation (ii) deflection given at any point For deflection at centre, substitute x = /2 in Eqn. (ii). Note that deflection at centre in this case will be maximum. W EI ymax = 12 = W l3 W l3 − 96 64 =− ∴ 3 Wl l 2 l − 2 16 2 ymax = W l3 192 (minus sign for downward deflection) W l3 192EI It is interesting to compare this deflection with similar case of simply supported beam having wl 3 which is four times more than beam with fixed ends. load at centre which is 48EI This shows that fixed beams are more stronger and stiffer. Fixed Beam with Uniformly Distributed Load Over the Span x w/m X A M RA A D wl2 12 A MB X RB l E + C − − wl . 2 Also due to symmetry, fixing moments, MA = MB . Maximum bending moment for simply supported beam, Due to symmetry RA = RB = B wl2 8 B Mmax = Figure 18.2 Consider a point at X, x from A, using Macaulay’s method, Mx = RA x − MA x0 − EI wx2 2 wx2 d2y 0 = R x − M x − A A 2 dx2 Integrating, EI dy wl x2 wx3 = − MA x − +C1 dx 2 2 6 C1 is the constant of integration. @seismicisolation @seismicisolation wl 2 8 360 • Strength of Materials At x = 0, dy =0 dx ∴ C1 = 0 ∴ EI wx3 dy wlx2 = − MA x − dx 4 6 (i) From Eqn. (i) slope at any point can be computed. Integrating Eqn. (i) again, EIy = At x = 0, y = 0 ∴ w l x3 x2 w x4 − MA − +C2 12 2 24 C2 = 0 EIy = w l x3 w x4 MA x2 − − 12 24 2 (ii) When x = l, y = 0 0= ∴ MA l2 w l4 w l4 = − 2 12 24 MA l2 w l4 = 2 24 ∴ Because of symmetry MA = MB = w l 4 w l 4 MA l 2 − − 12 24 2 MA = w l2 12 w l2 12 Substituting for MA in Eqn. (ii) w l x3 w x4 w l 4 − − 12 24 24 1 w l x3 w x4 w l 2 x2 − − y= EI 12 24 24 EIy = ∴ This equation will give deflection at any point. l For maximum deflection which will be at centre, i.e., x = , substituting for x in Eqn. (iii) 2 1 wl 4 wl 4 wl 4 ymax = − − EI 12(2)3 24(2)4 24(2)2 1 wl 4 wl 4 wl 4 ymax = − − EI 96 384 96 @seismicisolation @seismicisolation (iii) Fixed Beams ∴ Hence, ymax = − ymax = wl 4 384 EI • 361 (minus for downward deflection) wl 4 384 EI By Area Moment Method For fixed beam ΣA = 0 and ΣA x̄ = 0 2 w l2 × ×l +M×l = 0 3 8 ∴ M=− w l2 12 wm C l w −wl2 8 Free Moment Diagram A1 M = −wl 2 Fixed Moment Diagram A2 12 + − − Total BM Diagram wl2 24 −wl2 12 Figure 18.3 For maximum deflection at midpoint C, 1 [A1 x̄1 + A2 x̄2 ] EI 2 wl 2 l 3 l Ml l 1 × × + × = EI 3 8 2 8 2 2 4 1 3wl 4 wl 4 − = EI 384 96 yc = ymax = =− wl 4 384 EI ymax = yc = wl 4 384 EI Therefore, we find that the central deflection for a fixed beam having uniformly distributed load ‘w’ is one-fifth of the central deflection of the simply supported beam. @seismicisolation @seismicisolation • 362 Strength of Materials Deflection for a Fixed Beam with Concentrated Load Anywhere on the Span a A MA W b x x C MB x l RA Wb l Wab So free bending moment at C = l For free beam reaction at A = B RB Wab l + Free B.M.D − Ma − Mb + − Figure 18.4 At x from A, (using Macaulay’s method), bending moment, Mx = RA x − MA −W [x − a] EI d2y = RA x − MA −W [x − a] dx2 Integrating, EI x2 W dy = RA − MA x − [x − a]2 +C1 dx 2 2 Integrating again, EIy = RA When x = 0, y = 0, x3 x2 W − MA − [x − a]3 +C1 x +C2 6 2 6 dy =0 dx ∴ C1 = C2 = 0 ∴ 1 dy = dx EI and 1 y= EI x2 W RA − MA x − [x − a]2 2 2 x3 MA x2 W RA − − [x − a]3 6 2 6 @seismicisolation @seismicisolation (i) (ii) Fixed Beams • 363 dy =0 dx Substituting in Eqns. (i) and (ii) At x = l, y = 0, W l2 − MA l − b2 2 2 3 2 l MA l W 0 = RA − − b3 6 2 6 0 = RA and Simplifying the above two equations 0 = RA l 2 − 2MA l −W b2 and 0 = RA l 3 − 3MA l 2 −W b3 Solving further, RA = W b2 (l + 2a) l3 (iii) Wab2 l2 (iv) and MA = For static equilibrium, RA +RB −W = 0 (v) RA l −W (l − a) − MA +M B = 0 (vi) and From Eqns. (iv) and (vi), Wa2 b MB = l2 For deflection under load, etc., using Eqn. (ii), putting x = a a3 a2 1 RA − MA yc = EI 6 2 1 W b2 (l + 2a)a3 Wa3 b2 = − EI 6l 3 2l 2 Simplifying, yc = − or yc = Wa3 b3 3E I l Wa3 b3 3E I l @seismicisolation @seismicisolation 364 • Strength of Materials By Moment Area Methods: W a A b B l MA MB A1 Wab l A + + Free bending moment diagram + B 2a 3 l 2 (a + 3b ) A2 − MA MB Fixed bending moment diagram l 3 Figure 18.5 We know A1 + A2 = 0 MA +M B 1 W.a.b · ·l − ×l = 0 2 l 2 Wab or MA +M B = l Now A1 x̄1 + A2 x̄2 = 0 1 Wab 2a 1 Wab b ·a · + · b a+ 2 l 3 2 l 3 l l l − MB · l · − (MA − MB ) × =0 2 2 3 3 l2 l2 l2 Wa b Wa2 b2 Wab3 + + − MB · − MA · + MB · =0 or 3l 2l 6l 2 6 6 or or 2Wa3 b + 3Wa2 b2 +Wab3 − 2MB l 3 − MA l 3 = 0 1 2Wa3 b + 3Wa2 b2 +Wab3 l3 Wab = 3 2a2 +3ab + b2 l MA +2M b = @seismicisolation @seismicisolation (i) Fixed Beams • Wab 2 (a + b) +a (a + b) l3 Wab or MA + 2Mb = 3 (a + b) (a + b + a) l Wab = 3 × l × (l + a) l Wab or MA + 2Mb = 2 (l + a) l 365 = (ii) By solving Eqns. (i) and (ii) Wab2 l2 Wa2 b MB = 2 l MA = E XAMPLE 18.1: A fixed beam of 8 m span is loaded with point loads of 160 kN at a distance of 2.5 m from each support. Draw the bending moment and shear force diagrams. Also find the maximum deflection. Take E = 200 GN/m2 and I = 12 × 109 mm4 . S OLUTION : 160 kN 160 kN For simply supported beam C A RA = RB = 160 kN MA = 160 × 2.5 = 400 kNm D 3 2.5 m MA 8m 400 kNm B 2.5 m MB 400 kNm Free bending moment diagram + A B − MA MB = MA = 275 kNm Fixed bending moment diagram + 125 kNm − − 275 kNm 160 kN 160 kN SF Diagram Figure 18.6 @seismicisolation @seismicisolation 366 • Strength of Materials By equating the areas of free and fixed bending moment diagram 400 MA × 8 = (8 + 3) 2 2200 kNm = 275 kNm ∴ MA = 8 Bending moment at centre = 400 − 275 = 125 kNm Points of Contraflexure: Bending moment (actual) at any section in AC at a distant x from A, M = Free moment − Fixed moment = 160x − 275 At point of contraflexure M should be zero ∴ 160x − 275 = 0 275 = 1.72 m or x = 160 from either end due to symmetry. The bending moment at any section between A and D at a distance of x from A is given by, EI d2y = 160x − 275 − 160 (x − 2.5) dx2 Integrating, EI dy =0 ∴ dx Integrating again, When x = 0, dy x2 160 = 160 − 275x − (x − 2.5)2 +C1 dx 2 2 C1 = 0 EIy 160x3 x2 160 − 275 − (x − 2.5)3 +C2 6 2 6 When x = 0, y = 0, ∴ C2 = 0 For maximum deflection which will occur at the centre in this case, putting x = 4 m 160 (4)3 (4)2 160 − 275 − (4 − 2.5)3 6 2 6 = 1706.7 − 2200 − 90 = −5833 5833 5833 =− ymax = − EI 200 × 106 × 12 × 109 × 10−12 = −0.00243 m = −2.43 mm = −2.43 mm Ans EIymax = minus sign indicates downward deflection. @seismicisolation @seismicisolation Fixed Beams • 367 E XAMPLE 18.2: A fixed beam of 5 m span carries point loads of 120 kN and 85 kN as shown in Fig. 18.7. Find the following: 1. Fixing moments at the ends 2. Reactions at the supports Also draw bending moment and shear force diagrams. S OLUTION : 85 kN 120 kN C A D 2m 1.5 m MA B MB 1.5 m 5m 164.25 kNm 143.25 kNm + Free bending moment diagram − Fixed bending moment diagram 100.275 kNm 115 kNm + − − 7.55 Final bending moment diagram 92.5 kN 7.55 112.45 kN SFD Figure 18.7 For free reactions: Taking moments about B, RA × 5 = 120 × 3.5 + 85 × 1.5 RA = 109.5 RB = (120 + 85) − 109.5 = 95.5 kN @seismicisolation @seismicisolation 368 • Strength of Materials MC = 109.5 × 1.5 = 164.25 MD = 109.5 × 3.5 − 120 × 2 = 143.25 kNm Fixing moments: w a b2 l2 120 × 1.5 × 3.52 85 × 3.5 × 1.52 = + 52 52 = 88.2 + 26.8 = 115 kNm Ans MA = ∑ w a2 b l2 85 × 3.52 × 1.5 120 × 1.52 × 3.5 = + 52 52 = 62.475 + 37.8 = 100.275 kNm Ans MB = ∑ Reaction (R) at each support due to end moments alone, 115 − 100.275 5 = 2.945 R= Since MA > MB , The reaction R at A is upward and at B is downward, Final reaction at A = R f A = 109.5 + 2.945 = 112.45 kN R f B = 95.5 − 2.945 = 92.5 kN Ans Ans E XAMPLE 18.3: A fixed beam of 6.5 m span is subjected to a concentrated couple of 250 kNm applied at a section in from the left hand. Find the end moments by Macaulay’s method. Also draw bending moment and shear force diagrams. S OLUTION : MA 250 kNm B A C 4m Figure 18.8 @seismicisolation @seismicisolation 2.5 m Fixed Beams • 369 Let us assume directions of fixing moment and reaction as shown in Fig. 18.8. If after calculations, these turn out to be negative, then our assumed direction will be wrong. The bending at any section at a distance x from end A is given by d2y = RA x+ MA + 250 dx2 Now to facilitate application of Macaulay’s method is arranged as given below: EI d2y = RA x+ MA + 250 (x − 4)0 dx2 (i) x2 dy = RA + MA x + 250 (x − 4) +C1 dx 2 (ii) EI Integrating, EI dy = 0, ∴ C1 = 0 dx Integrating again, we here when x = 0, EIy = RA x3 x2 250 + MA + (x − 4)2 +C2 6 2 2 (iii) when x = 0, y = 0, ∴ C2 = 0 dy =0 when x = 6.5 m, dx Substituting these values in Eqn. (ii), (6.5)2 + 6.5 MA + 250 (6.5 − 4) 2 0 = 21.1 RA + 6.5 MA + 625 MA + 3.25 RA = −96.15 0 = RA (iv) At B, the deflection is zero ∴ When x = 6.5 m, y = 0 Substituting these values in Eqn. (iii), we get (6.5)3 (6.5)2 + MA × + 125 (65.4)2 6.5 2 0 = 42.25 RA + 21.12 MA + 781.25 42.25 RA + 21.12 MA = −781.25 0 = RA × Solving Eqns. (iv) and (v), 42.25 RA + 21.11 (−96.15 − 3.25 RA ) = −781.25 42.25 RA − 2030 − 68.61 RA = −781.25 −26.36 RA = 1249 or RA = −47.38 kN @seismicisolation @seismicisolation (v) 370 • Strength of Materials −ve sign shows our assumed direction was wrong. Substituting the value of RA in Eqn. (iv), 3.25 (−47.28) + MA = −96.15 ∴ MA = 57.85 kNm Ans Now for fixed beam, bending moment diagram and shear force diagrams are shown below. MA = 57.85 kNm 250 kNm A 4.0 m 47.38 kN B . C 2.5 m 47.38 kN Fixed beam 118.5 kNm 57.85 kNm + + Bending moment diagram − 131.7 kNm 47.38 kN 47.38 kN SFD Figure 18.9 Bending moment calculations: MA = 57.83 (sagging) just before C, Mc = 57.83 − 47.38 × 4 = −131.7 kNm (hogging) just after C, Mc = −131.7 + 250 = 118.5 (sagging) MB = 57.83 − 47.38 × 6.5 + 250 = 0 E XAMPLE 18.4: A built-up beam of span AB, 5 m carries a concentrated load of 70 kN at 1.3 m and 130 kN at 3.6 m from A. Determine the fixing end moments and draw shear force and bending moment diagrams. @seismicisolation @seismicisolation Fixed Beams S OLUTION : 70 kN A MA 130 kN D C B 1.3 m 2.3 m 1.4 m MB 3.6 m Figure 18.10 By principle of superposition, fixing end moments are: wab2 wab2 MA = + l2 l2 70 130 = 70 × 1.3 × 3.72 130 × 3.6 × 1.42 + 52 52 = 49.83 + 36.69 = 86.52 kNm Ans 2 2 wa b wa b MB = + l2 l2 70 130 = 70 × 1.32 × 3.7 130 × 3.62 × 1.4 + 52 52 = 17.51 + 94.35 = 111.86 kNm Ans For reactions at supports: Taking moments about B, 5 RA = 70 × 3.7 + 130 × 1.4 5 RA = 259 + 182 441 = 5 = 88.2 kN RA + RB = 70 + 130 = 200 kN ∴ RB = 200 − 88.2 = 111.8 kN @seismicisolation @seismicisolation • 371 372 • Strength of Materials 70 kN A C B D 1.3 m 86.52 130 kN 2.3 m RA 114.66 kNm 1.4 m 156.52 111.86 RB 112.86 kNm 86.52 kNm 83.13 kN BMD 13.13 kN SFD 116.87 kN Figure 18.11 Bending moment calculations: Bending moment at A and bending moment at B = 0 Mc , bending moment at C = 88.2 × 1.3 = 114.66 kNm MD , bending moment at D = 111.8 × 1.4 = 156.52 kNm Support reactions: Taking moments about B, RA × 5 + 111.86 = 70 × 3.7 + 130 × 1.4 + 86.52 = 527.52 RA = 83.13 kN RA + RB = 70 + 130 = 200 RB = (70 + 130) − 83.13 = 116.87 kN @seismicisolation @seismicisolation Fixed Beams • 373 Shear force calculations: Shear force at A = 0 rising to 83.13 kN Shear force at C = 83.13 falling to 83.13 − 70 = 13.13 kN Shear force at D = 13.13 falling to 13.13 − 130 = 116.87 kN Shear force at B = 116.87 rising to 116.87 − 116.87 = 0 E XAMPLE 18.5: Find reactions and fixing moments at fixed ends and draw shear force and bending moment diagrams for the fixed beam shown in Fig. 18.12. Use Macaulay’s method. 2 kN 2 kN 1.5 kN/m B A 1m 2m C 1m D Figure 18.12 For solving the problem extend uniformly distributed load up to D and to cancel the extended load (between C and D), let there be an upward uniformly distributed load from C to D. Now 2 kN 3 kN 1.5 kN/m D B A C MA MD 1.5 kN/m RA RD 1m 3.66 2m 1m Shear force diagram 1.66 kN E x' 1.34 kN 1.52 kN 4.34 kN 0.6 kNm 0.92 kNm Bending moment diagram 3.06 kNm 3.42 kNm Figure 18.13 @seismicisolation @seismicisolation 374 • Strength of Materials EI d2y 1.5(x − 1)2 1.5(x − 3)2 ◦ + − 3(x − 3) = R x − M x − 2(x − 1) − A A 2 2 dx2 Integrating twice, we get dy RA x2 1.5(x − 1)3 1.5(x − 3)3 3 = − MA x − (x − 1)2 − + − (x − 3)2 + A dx 2 6 6 2 3 2 3 4 4 MA x (x − 1) (x − 1) (x − 3) (x − 3)2 RA x − − − + − +C1 x +C2 EIy = 6 2 3 16 16 2 EI dy = 0 at x = 0 and at x = 4 dx We have C1 = 0 We know that, ∴ 0 = 8 RA − 4 MA − 9 − 27 1 3 + − 4 4 2 8 RA − 4 MA = 17 (i) y = 0 at x = 0 giving C2 = 0 and y = 0 at x = 4 giving 0= 64RA 1 1 81 − 8 MA − 9 − + − 6 16 16 2 Which comes to: 64RA − 48 MA = 87 Solving Eqns. (i) & (ii), RA = 3.66 kN Ans MA = 3.06 kNm Ans RB = 2 + 3 + 3 − RA = 8 − 3.66 = 4.34 kN Ans MB = −3.06 + 3.66 × 1 = +0.6 kNm Ans MC = −3.06 + 3.66 × 3 − 2 × 2 − 3 × 1 = 0.92 kNm MD = −3.06 + 3.66 × 4 − 2 × 3 − 3 × 2 − 3 × 1 = −3.42 kNm Bending moment where symmetry fixing is zero (ME ): Due to similar triangles x 2 − x = 1.66 1.34 On solving this equation, we have, @seismicisolation @seismicisolation (ii) • Fixed Beams 375 As shear force is zero, x = 1.11 m ∴ x 2 = −3.06 + 3.66 × 2.11 − 2 × 1.11 − 0.75(1.11)2 = 1.52 kNm Ans ME = −MA + RA (1 + x ) − 2 × x = 1.5x × Exercise 18.1 A fixed beam of 6 m span is loaded with point loads of 150 kN at a distance 2 m from each support. Draw bending moment and shear force diagrams. Find also the maximum deflection. Take E = 2 × 108 kN/m2 and I = 8 × 108 mm4 . [Ans Fixing moments: 200 kNm, 200 kNm, Ymax = 1.56 mm] 18.2 A fixed beam of 6 m span carries point loads of 100 kN and 75 kN as shown in Fig. 18.14. Find: i) Fixing moments at the ends ii) Reactions at the supports 100 kN C 75 kN D A B 2m 2m 2m Figure 18.14 18.3 18.4 18.5 18.6 Draw bending moment and shear force diagrams. [Ans MA = 122.22 kNm, MB = 111.11 kNm, RA = 93.52 kN, RB = 83.33 kN] A fixed-ended beam of span 9 m carries a uniformly distributed load of 15 kN/m and two equally concentrated loads, of 200 kN at the 3 m from left hand and another 3 m from right hand. Find the fixing moments and the deflection at the centre. EI = 210 MN/m2 . [Ans 392.75 kNm, yc = 6.835 mm] A horizontal I beam, rigidly built in at the ends, and 7.5 m long carries a total uniformly distributed load of 100 kN as well as a concentrated load of 40 kN. If the bending stress is limited to 75 MN/m2 and the deflection must not exceed 2.5 mm, find the depth of section required. E = 200 GN/m2 . [Ans 594 mm] A horizontal beam built-in each at end has a clear span of 4.5 m and carries loads of 50 kN at 1.5 m and 70 kN at 2.5 m from left hand and claculate the fixing moment and the position and amount of the maximum bending moment. [Ans 67.5 kNm, 60.5 kNm, 67.5 kNm at left hand end] A built-in beam of 5 m span carries a uniformly distributed load of 30 kN/m extending from one support to the centre of the span. If the moment of inertia of the section is 200 × 10−6 m4 , Calculate: (a) the end fixing moments; (b) the end reactions and (c) the position and magnitudes of the maximum deflection. Sketch the shear force and bending moment diagrams. E = 200 GN/m2 . [Ans 19.53 kNm; 43 kNm; 14.06 kNm; 60.94 kN, 3.15 mm at 2.8 m from unloaded end] @seismicisolation @seismicisolation 376 • Strength of Materials 18.7 A fixed beam of 6 m span carries uniformly distributed unloaded of 10 kN/m for a distance 4 m from the left hand end. Find the fixing moments at the supports. [Ans 26.66 kNm; 17.77 kNm] 18.8 A fixed beam of 6 m span supports two point loads of 300 kN each at 2 m from each end. Find the fixing moments at the ends and draw the shear force and bending moment diagrams. Also find the central deflection. Take I = 9 × 108 mm4 and E = 200 GPa. [Ans 400 kNm; yc = 2.78 mm] 18.9 A fixed beam of 6 m span carries a uniformly distributed load of 20 kN/m run over the right half and 30 kN/m over the right half and a concentrated load of 40 kN at the centre of the span. Calculate the fixed end moments. Assume uniform flexural rigidity. [Ans 93.375 kNm; 110.625 kNm] 18.10 A 250 mm × 112.5 mm steel beam, I = 47.6 × 10−6 m4 , is used as a horizontal beam with fixed ends and a clear span of 3 m. Calculate from the first principles the load which can be applied at one-third span if the bending stress is limited to 120 MN/m2 [Ans 103 kN] 18.11 A built-in beam of span 10 m carries a uniformly distributed load of 15 kN/m on the entire span along with two point loads each of 200 kN at a distance of 3 m and 7 m from left end. Find the fixing moments and the deflection at the centre. Take EI = 210 MN/m2 [Ans 545 kNm; 8.29 mm] 18.12 A girder of span 5 m is fixed at each end and carries two point loads of 90 kN and 120 kN placed at 1 m and 2 m from the left support respectively. Find the reactions and fixing moments at the ends. E = 200 GPa, I = 8 × 107 mm4 . Also find the deflections at the load points. [Ans 158.36 kN; 51.64 kN, 144 kNm, 72 kNm y under 90 kN = 2.84 mm y under 120 kN = 5.741 mm] @seismicisolation @seismicisolation 19 C HAPTER CONTINUOUS BEAMS When a beam rest on more than two supports, it is called continuous beams. Their applications are in long bridges, buildings, etc. The continuous beam is statically indeterminate and can be analysed by various methods, for example, by the theorem of three moments or the Clapeyron’s theorem method. Clapeyron’s Equation of Three Moments A C B l1 l2 x2 x1 a1 (a) A x D (b) MA A x′ (c) MA A MxC.G. dx Mx′ dx a2 C.G. a′1 Bending moment diagram due to vertical loads C B E a2′ MB F MC B C Bending moment diagram due to support moments MB1 MC B Resultant bending moment diagram C Figure 19.1 The beam ABC may be extended both ways but we are considering only portion ABC. Span AB: Now consider the span AB and let Mx be the bending moment due to vertical loads at a distance x as shown. Mx is positive, being sagging moment, and Mx is the bending moment due to support moments at a distance x from support A. This is negative because of hogging effect. @seismicisolation @seismicisolation 378 • Strength of Materials Hence, net bending moment at x is given by, EI d2y = Mx − Mx dx2 Integrating from zero to l1 after multiplying by x (both sides) EIx l EI 0 d2y = xMx − xMx dx2 d2y x 2 dx = dx l1 x Mx dx − 0 l1 xMx dx 0 l1 dy EI x − y = a1 x1 − a1 x1 dx 0 (i) Whereas a1 = area of BMD due to vertical loads Also Mx .dx = Area of bending moment diagram of length dx and, x Mx dx = moment of area of bending moment diagram of length dx about A. l1 x Mx dx = a1 x1 Hence, and similarly 0 l1 xMx dx = a1 x1 0 Equation (i) becomes EI dy dy l1 − yB − 0 × − yA dx atB dx atA = a1 x1 − a1 x1 EI [(l1 θB − yB ) − (0 − yA )] = a1 x1 − a1 x1 dy Because at B = θB . And deflections at A and B are zero, therefore yA = 0 and yB = 0. So the dx above equation becomes: @seismicisolation @seismicisolation Continuous Beams EI θB = a1 x1 − a1 x1 • 379 (ii) a1 = area of bending moment diagram due to supports moments = Area of trapezium ABED = 1 (MA + MB ) × l1 2 x = distance of centre of gravity of area ABED from A l1 1 2l1 + (MB − MA )l1 × 2 2 3 1 MA l1 + (MB − MA ) l1 2 MA l1 . = l1 l1 + (MB − MA ) × 2 3 1 MA + (MB − MA ) 2 MA . = 3MA l1 + 2l1 (MB − MA ) 6 = 2MA + MB − MA 2 l1 [3MA + 2MB − 2MA ] = 3 MB + MC MA + 2MB l1 = MA + MB 3 Substituting the values of a1 and x1 in Eqn. (ii), we have 1 MA + 2MB l1 EIl1 θB = a1 x1 − (MA + MB )l1 × 2 MA + MB 3 = a1 x1 − 6EIl 1 θB = l12 (MA + 2MB ) 6 6a1 x1 − l1 (MA +2M B ) l1 (iii) Span BC: Likewise considering the span BC with origin C and x positive to the left, it can be shown that 6a2 x2 6EI (−θB ) = − l2 (MC +2M B ) l2 @seismicisolation @seismicisolation 380 • Strength of Materials Slope at B in span BC will be obviously of opposite sign, i.e., negative because of opposite directions. Therefore, the above equation becomes as −6EI θ B = 6a2 x2 − l2 (MC +2M B ) l2 (iv) If we add Eqn. (iii) and (iv), then 0= 6a1 x1 6a2 x2 −l1 (MA +2M B ) + −l 2 (MC +2M B ) l1 l2 0= 6a1 x1 6a2 x2 + −l 1 MA +2l 1 MB −l 2 MC +2l 2 MB l1 l2 = l1 MA +l 2 MC +2M B (l1 +l 2 ) = MA l1 +2M B (l1 +l 2 ) +MC l2 = 6a1 x1 6a2 x2 + l1 l2 6a1 x1 6a2 x2 + l1 l2 This is known as Clapeyron’s equation of three moments. E XAMPLE 19.1: A continuous beam ABC is shown in Fig. 19.2, find the support moments at A, B and C if A and C are simply supported. Also draw bending moment and shear force diagrams. 12 kN/m 9 kN/m A 5m B C 7m Figure 19.2 S OLUTION : Because the ends A and C are simply supported, the support moments at A and C will be zero. ∴ MA = MC = 0 Ans. Using Clapeyron’s equation of three moments to find support moment at B, we get MA l1 + 2MB (l1 + l2 ) + MC l2 = 6a1 x1 6a2 x2 + l1 l2 0 + 2MB (5 + 7) + 0 = 6a1 x1 6a2 x2 + 5 7 @seismicisolation @seismicisolation (i) Continuous Beams For maximum bending moment between A and B, RA = RB = 9×5 2 = 45 = 22.5 kN 2 MABmax = 22.5 × 2.5 − 22.5 × 2.5 2.5 × 2 4 = 56.25 − 17.6 = 38.65 kNm For maximum bending moment between B and C, RB = RC = 12 × 7 2 = 84 = 42 kN 2 7 3.5 MBCmax = 42× −12 × 3.5× 2 2 = 147 − 73.5 = 73.5 kNm 12 kN/m 9 kN/m A C B 5m 7m 59.27 73.5 kNm A 38.65 Bending moment diagram C B 50.47 18.15 kN Shear force diagram 26.85 kN 53.53 kN Figure 19.3 @seismicisolation @seismicisolation • 381 382 • Strength of Materials Further calculations for shear force diagram: 18.15 − 9 × 5 = 26.85 kN, 50.47 = 77.32 − 26.85 The bending moment diagram on a simply supported beam carrying uniformly distributed load 2 will have area = × span × altitude. 3 So, now 2 a1 = ×5×38.65 = 129.48 3 5 x1 = =2.5 m 2 2 a2 = ×7×73.5=344.7 3 7 x2 = = 3.5 m 2 Substituting these values in Eqn. (i) 2M B (5 + 7) = 6 × 129.48 × 2.5 6 × 344.7 × 3.5 + 5 7 24 MB = 388.44 + 1034.1 MB = 59.27 kNm Ans Shear force diagram calculations: For span AB, taking moments about B, 5 RA ×5 − 12 × 5× = MB = −59.27 (MB is −ve because t is hogging) 2 5RA = −59.27+150 ∴ RA = 18.15 kN Similarly for span BC, taking moments about B, RC × 7 − 12 × 7 × 7 = −59.27 2 ∴ RC = 33.53 kN RB = Total load on ABC − (RA + RC ) = (9 × 5 + 12 × 7) − (18.5 + 33.53) = 129 − 51.68 = 77.32 kN @seismicisolation @seismicisolation Continuous Beams • 383 E XAMPLE 19.2: A continuous beam ABCD of length 12 m resting on four supports covering 3 equal spans carries a uniformly distributed load of 2 kN/m length. Calculate the moments and reactions at the supports. Draw the shear force and bending moment diagrams. S OLUTION : 2 kN/m A B 4m RA RB 4m RC RD 4 kNm 4 kNm B A D C 4m 4 kNm C 3.201 kNm D 3.201 kNm A B C A B C D Bending moment diagram due to vertical loads Bending moment diagram due to support moments Resultant bending moment diagram 4 kN 3.2 kN A 4.8 kN B C 4 kN 4.8 kN D SF diagram 3.2 kN Figure 19.4 Beam being simply supported, the moments at A and D are zero. MA = 0 By symmetry and MD = 0 MB = Mc For finding the support moments at B and D, Clapeyron’s equation of three moments is applied for ABC and BCD. For ABC: MA l1 + 2MB (l1 + l2 ) + MC .l2 = 6a1 x1 6a2 x2 + l1 l2 6a1 x1 6a2 x2 + 4 4 6a1 x1 6a2 x2 16MB + 4MC = + l1 l2 0 × 4 + 2MB (4 + 4) + MC × 4 = 16MB + 4MC = @seismicisolation @seismicisolation 6 (a1 x1 + a2 x2 ) 4 (i) 384 • Strength of Materials a1 = Area of bending moment diagrams due to uniformly distributed load on AB. 2 × AB × altitude of parabola 3 (4 × 2) 2 Altitude of parabola = × 2 − 2 × 2 × = 8 − 4 = 4 kNm 2 2 = Maximum bending moment of free bending moment diagram. = ∴ 2 × 4 × 4 = 10.67 3 4 x2 = = 2 m 2 a1 = Owing to symmetry, a2 = a1 = 10.67 and x2 = x1 = 2 m Putting the values is Eqn. (i), 16M B + 4MC = 6 (10.67 × 2 + 10.67 × 2) 4 Now MB = MC due to symmetry 16MB + 4MB = 1.5 (21.34 + 21.34) 20MB = 64.02 MB = 3.201 kNm = MC Ans Now, MA = 0, MD = 0, MB = MC = 3.201 kNm In bending moment diagram considering AB, BC and CD as simply supported, the parabolas have altitudes 4 kNm each. For support reactions: Due to symmetry RA = RD RB = Rc For the span AB, taking moments about B, we get MB = RA × 4 − 2 × 4 × 4 2 Since MB in hogging moment so giving it negative sign, −3.201 = 4RA − 16 @seismicisolation @seismicisolation Continuous Beams • 385 4RA = 16 − 3.201 ∴ RA = 3.2 kN Ans Due to symmetry, RD = RA = 3.2 kN Ans Now RA + RB + RC + RD = Total load on ABCD = 2 × 12 2(RA + RB ) = 24 RA + RB = 12 ∴ RB = 12 − RA = 12 − 3.2 = 8.8 kN RB = RC = 8.8 kN Ans Now bending moment and shear force diagrams are drawn as shown in Fig. 19.4. For shear force at B = 3.2 − 4 × 2 = −4.8 kN 8.8 − 4.8 = 4 kN and For shear force at C = 4 − 4 × 2 = −4 kN and 8.8 − 4 = 4.8 kN E XAMPLE 19.3: A continuous beam ABC consists of two spans AB of length 4 m and BC of length 3 m. The span AB carries a concentrated load of 100 kN at its midpoint. The span BC carries a point load of 120 kN at 1 m from C. Find: (a) moments at the supports and (b) reactions at the supports. (Poona University) S OLUTION : 100 kN A 120 kN B C 2m 3m 4m 1m 80 kNm 100 kNm O x1= 2 G x2= 1.33 O O Free bending moment diagram O 65.66 kNm Figure 19.5 @seismicisolation @seismicisolation 386 • Strength of Materials WP = 100 kNm 4 1 a) Area of bending moment diagram for span AB = a1 = × 100 × 4 = 200 kNm2 2 4 Due to symmetry x1 = = 2 2 1 Area of bending moment diagram for span BC = a2 = × 80 × 3 = 120 kNm2 2 The bending moment under the load 100 kN = x2 = l +b 3+1 = = 1.33 m 3 3 Let MA , MB & MC be the support moments at A, B and C, respectively. MA l1 +2M B (l 1 +l 2 ) + MC l2 = 6a1 x1 6a2 x2 + l1 l2 Being simply supported MA = MC = 0, we have 200×2 120×1.33 + 2MB (4 + 3) = 6 4 3 14MB = 6 (100 + 53.2) MB = 6×153.2 14 = 65.66 kNm Ans b) Considering the span AB simply supported and taking moments about B, RA × 4 − 100 × 2 + 65.66 = 6 200 − 65.66 4 RA = 35.58 kN Ans RA = ∴ Similarly, taking span BC as simply supported and taking moments about B, we get RC × 3 − 120 × 2 + 65.66 = 0 RC = ∴ Because 240 − 65.66 3 RC = 58.12 kN Ans RA + RB +RC = 100 + 120 ∴ RB = 220 − 35.58 − 58.12 = 126.3 kN Ans @seismicisolation @seismicisolation Continuous Beams • 387 Exercise 19.1 A continuous beam ABC 8 m long consists of two spans AB = 3 m and BC = 5 m. The span AB carries load of 50 kN/m while the span BC carries a load of 10 kN/m. Find: (i) Support moment at B (ii) Reactions at the supports [Ans MB = 179.7 kNm, RA = 48.4 kN, RB = 192.5 kN RC = 59.1 kN] 19.2 A continuous beam ABCD having three equal spans of length l each. It carries a uniformly distributed load w/unit length over its entire length. It is freely supported on all supports, which are at the same level. Draw the bending moment and shear force diagrams for this beam. Ans MB = MC = − wl 2 2wl 11wl , MA = 0, RD = = RA , RB = = RC 10 5 10 2 2 2 wl at l from A 25 5 wl 2 or D respectively. Maximum bending moment in span BC = 40 The maximum bending moment in span AB or CD = 19.3 A continuous beam 12 m long supported over spans AB = BC = CD = 4 m carries a uniformly distributed load of 3 kN/m run over span AB, a concentrated load of 4 kN at a distance of 1 m from point B on support BC and a load of 3 kN at the centre of the span CD. Find: i) support moments and ii) support reactions [Ans MA = 0, MB = −4.05 kNm, MC = −1.05 kNm RA = 4.99 kN, RB = 10.76 kN, RC = 2.01 kN, RD = 1.24 kN] 19.4 A continuous beam ABC 8 m long consists of two spans AB = 3 m and BC = 5 m. The span AB carries a load of 50 kN/m while the BC carries a load of 10 kN/m Find: i) Bending moment at B ii) Reactions at the supports [Ans MB = 179.7 kNm, RA = 48.4 kN, RB = 192.5 kN, RC = 59.1 kN] @seismicisolation @seismicisolation C HAPTER 20 SPRINGS Springs are of various types for serving different purposes. Some types are listed below: 1. Helical springs: a) Close-coiled helical springs b) Open-coiled helical springs c) Torsion helical springs d) Compression helical springs 2. 3. 4. 5. Flat springs Belleville springs Leaf or laminated springs Torsion springs Springs are usually made of high carbon steel (0.7 to 1.0% carbon) or medium carbon alloy steels. Sometimes phosphor–bronze or 18 Cr/ 8 Ni stainless steel are used for better corrosion resistance. These are used for railway carriages, cars, motorcycles, etc., for absorbing the shocks. Whereas sometimes they find use in brakes or clutches or spring balances, etc. Close-coiled helical spring with Axial load: If helix angle is 10◦ or less, the helical spring is called close-coiled helical spring. Fig. 20.1 shows a close-coiled spring under an axial load w. D = mean coil diameter d = wire diameter n = number of coils l = length of wire π Dn δ = axial deflection. Figure 20.1 @seismicisolation @seismicisolation Springs • 389 If may be noted that it is assumed to be subjected to torsion only. The effects of bending and direct shear are being negligible. D The torque on the wire of all sections = W × 2 T τ As we know = J R T.R ∴ Maximum shear stress, τ = J τ= W × D/2 × d/2 π 4 d 32 = 8W D 16W R = π d3 π d3 (i) The twist of one end of the wire relative to the other end is given by or ∴ θ= T.l CJ θ= W D/2 × π Dn π C × d4 32 = Cθ T = J l 16W D2 n Cd 4 Since the axial movement of the free end = θ × D 2 i.e., δ = 8W D3 n Cd 4 (ii) Also it is possible to derive this formula this way: Work done by the load = strain energy in the wire 1 1 Wδ = Tθ 2 2 1 T ·l = T× 2 C·J = T 2l 2CJ @seismicisolation @seismicisolation ∵ T Cθ = J l 390 • Strength of Materials Substituting for T and J, WD 2 ×l T 2l 2 = π 2CJ 2C d 4 32 4W 2 D3 n = Cd 4 1 2 As this is equal to 1 W δ so equating, 2 4W 2 D3 n 1 Wδ = 2 Cd 4 8W D3 n δ= same as proved in Eqn. (i). Cd 4 In terms of mean radius R which is D/2, 8W (2R)3 n Cd 4 64W R3 n = Cd 4 δ= Wahl’s Correction Factor While deriving the above equations, the effect of curvature of spring and direct shear stress is neglected. 8W D 16W R = is modified to include these effects by introducing a factor K called π d3 π d3 Wahl’s correction factor, so that 16 W R τ = K π d3 Equation τ = where K is found experimentally and is given by K= 4S − 1 0.615 + 4S − 4 S where S= D = Spring index. d Total strain energy of the spring: U= 1 Tθ 2 @seismicisolation @seismicisolation Springs • 391 τ Cθ T = = J r l π c τl 1 τ 32 d × U= 2 d/2 d/2C Using = π τ 2d2l 16 C 1 τ2 π = . × d2l 4 C 4 τ2 × volume of the spring wire 4C Stiffness of the spring is defined as force per unit deflection, So strain energy, U = Stiffness =k= W W Cd 4 = = δ 64 R3 n 64W R3 n 4 Cd Close-coiled Helical Spring with Axial Couple Figure 20.2 shows a close-coiled helical spring under an axial couple M. The bending moment on the wire at all sections = M M ∴ Maximum bending stress, σ = Z M σ= π d3 32 32 M = π d3 M M M M Figure 20.2 Owing to couple M, the radius of curvature of the coils changes from R to R and the number of coils changes fromn to n . 1 1 Then, M = EI approximately − R R The length of wire, l = 2π Rn = 2π R n ∴ ∴ 2π n 1 1 2π n = and = R l R l 2π (n − n) M = EI × l @seismicisolation @seismicisolation 392 • Strength of Materials 2π (n − n) is the angle of twist of the spring, φ , EI φ ∴ M= l Ml M × π Dn or φ = = π EI E d4 64 64MDn ∴ φ= E d4 Strain energy stored is the spring, M2l U= 2EI 2σ I M σ = OR M = and I d/2 d 2 2σ I l U= × d 2EI −4σ 2 I l × 2E d 2 π 4 d 4σ 2 l 64 × U= 2 2E d π σ 2d2l σ 2 π d2l = = 32 E 8E 4 U= or U= Stiffness for bending, k = (i) σ2 × volume of the spring wire 8E M φ M × Ed 4 64 MDn Ed 4 k= 64 Dn k= E XAMPLE 20.1: A closely-coiled helical spring is to carry a load of 750 N. Its mean coil diameter is to be 10 times that of the wire diameter. Calculate mean coil diameter and wire diameter if the maximum shear stress in the material of the spring is to be 100 MN/m2 . S OLUTION : W = 750 N D = 10d τ = 100 MN/m2 @seismicisolation @seismicisolation Springs • 393 We know, τ= 100 × 106 = 16W R π d3 16 × 750 × 10d 2 π d3 16 × 750 × 5 = π d2 16 × 750 × 5 d2 = π × 100 × 106 = 0.000191 d = 0.0138 = 13.8 mm ∴ Mean coil dia, D = 10 × 13.8 = 138 mm Ans E XAMPLE 20.2: A helical spring is made of 13 mm diameter steel wire wound on a 140 mm diameter mandrel. If there are 12 active coils, what is spring constant? Take C = 85 GN/m2 . What force must be applied to the spring to elongate 30 mm? S OLUTION : Diameter of steel wire, d = 13 mm = 0.013 m Diameter of mandrel, D = 140 mm = 0.14 m Number of active coils, m = 12 Modules of rigidity C = 85 GN/m2 Elongation of the spring, δ = 30 mm = 0.03 m Spring constant = stiffness of spring k W δ Cd 4 0.14 k= = 0.07 , R= 2 64R3 n 85 × 109 × (0.013)4 = 64(0.07)3 × 12 = 9215.9 N/m Ans W = 9215.9 δ W = 9215.9 0.03 W = 276.5 N Ans k= Now, or @seismicisolation @seismicisolation 394 • Strength of Materials E XAMPLE 20.3: A close-coiled helical spring has mean diameter of 75 mm and spring constant of 80 kN/m. It has 8 coils. What is the suitable diameter of the spring wire if maximum shear stress is not to exceed 250 MN/m2 ? Module of rigidity of the spring wire material is 80 GN/m2 . What is the maximum axial load the spring can carry? (AMIE Summer, 2000) S OLUTION : R= D 75 = = 37.5 mm = 0.0375 m 2 2 Spring constant, k = 80 kN/m Active coils = 8 Maximum shear stress = 250 MN/m2 Modulus of rigidity, C = 80 GN/m2 π Now, T = τ × d 3 16 W × 0.0375 = (250 × 106 ) × π (d)3 16 (i) Also we know w = k.δ W = 80000 × δ ∴ (ii) 64W R3 4 64W × (0.0375)3 × 8 = Cd 4 80 × 109 × d 4 W = 3.375 × 10−13 4 d δ= Putting value of δ in Eqn. (ii) W = 80000 × 3.375 × 10−13 d 4 = 80000 × 3.375 × 10−13 d = 0.0128 m = 12.8 mm Ans W d4 For maximum axial load the spring can carry, W : From Eqn. (i), π (0.0128)3 16 W = 2744 N Ans W × 0.0375 = (250 × 106 ) E XAMPLE 20.4: A helical compression spring has a coil diameter of 75 mm and must carry a maximum load of 900 N with a compression of 100 mm. Calculate the diameter of the wire and number of free turns required. Allow a maximum shearing stress of 400 MN/m2 and C = 80 GN/m2 . @seismicisolation @seismicisolation Springs • 395 S OLUTION : D = 0.075 m W = 900 N τ = 400 MN/m2 C = 80 GN/m2 , D =?, n =? δ = 100 mm = 0.1 m We know, τ= 8W D π d3 d3 = 8W D π .τ = 8 × 900 × 0.075 π × 400 × 106 = 0.42972 × 10−6 ∴ d = 10−2 (0.7546) = 0.007546 m = 7.55 mm Ans Also, δ= 8W D3 n Cd 4 h= 8Cd 4 8W D3 = 0.1 × 80 × 109 × 0.007554 8 × 900 × 0.0753 = 8.56 Ans E XAMPLE 20.5: A close-coiled helical spring is to have a stiffness of 80 kN/m and to exert a force of 2.7 kN. If the mean diameter of the coils is to be 75 mm and the maximum stress is not to exceed 250 MN/m2 , calculate the required number of coils and the diameter of the steel rod from which the spring should be made. The modulus of rigidity of the material is 80 GN/m2 . @seismicisolation @seismicisolation 396 • Strength of Materials S OLUTION : δ= 8W D34 Cd 4 n= δ Cd 4 8W D3 Since stiffness = 80 kN/m So for a force of 2.7 kN, deflection 2.7 80 = 0.03375 = Also, τ= 8W D π d3 d3 = 8W D πτ = 8 × 2700 × 0.075 π × 250 × 106 d 3 = 10−2 (1.273) ∴ d = 0.01273 m = 12.73 mm Now, n= = 8Cd 4 8W D3 0.03375 × 80 × 109 × (0.01273)4 8 × 2700× (0.075)3 = 7.81 = 8 Nos Ans Springs in Series and Parallel For springs in series, the load in the same for both springs (if two springs are attached in series). So, net deflation δ = δ1 + δ2 . @seismicisolation @seismicisolation Springs • 397 W W W + = where k1 and k2 are the stiffness of the respective springs. And k is the k1 k2 k equivalent stiffness. 1 1 k1 k2 1 = + or k = ∴ k k1 k2 k1 +k2 or δ = If two springs are of stiffness k1 and k2 and are joined together in parallel, they have the common deflection δ and the load is the mm of load taken by each W1 W2 W = = Thus, δ = k k1 k2 Since W = W1 +W2 or kδ = k1 δ + k2 δ or k = k1 + k2 where W1 and W2 are loads shared by each spring and k in the effective stiffness. Concentric (Cluster) Springs: When close coiled springs are placed inside the other, concentric (cluster) spring in formed. Do do di Di Figure 20.3 In case if the material and the free length of outside spring and inside springs are the same, then 8W i Di 8W o Do = maximum shear stress is equal, τ = π di π do Since deflection is same, so 8W 3 D3 n 8W 3 D3 n δ = i 4i = o 4o Cd i Cd o Suffix I is for inside spring and o for outside spring. In cluster springs, the solid length is the same, ni di = no do E XAMPLE 20.6: A composite spring has two close-coiled springs connected is series; one spring has coils of a mean diameter of 30 mm and wire diameter 3 mm. Find the wire diameter of the other spring, if it has 16 coils of mean diameter 42 mm. The stiffness of the composite spring is 1.8 kN/m. Determine the maximum load that can be carried by the composite spring and the corresponding extension of the maximum stream in 260 MN/m2 . Modulus of rigidity of the material of spring is 80 GN/m2 . @seismicisolation @seismicisolation 398 • Strength of Materials S OLUTION : W W W = + k k1 k2 where k is stiffness. 1 1 1 = + k k1 k2 Stiffness, k1 = = Cd 41 80 × 109 × (0.003)4 = 8 × 0.033 × 12 8D31 n1 6.48 2.592 × 10−3 = 2500 N/m k2 = 80 × 109 ×d 42 8 × (0.042)3 ×16 = 8.436 × 1012 d24 1 1 1 = + k 2500 8.436 × 1012 d24 8.436 × 1012 d24 +2500 1 = 1800 21090 × 1012 d24 21090 × 1012 d24 = 15184.8 d24 + 4500000 2.109 × 1016 d24 = 4500000 d24 = 4500000 2.109 × 1016 = 45 × 105 2.109 × 1016 = 21.34 × 10−11 ∴ dz = 3.82 mm Ans For maximum load we will take spring with minimum wire diameter τ= 8W G π d13 W= τπ d13 8D1 @seismicisolation @seismicisolation Springs = 399 260 × 106 × π × (0.003)3 8 × 0.03 = 91.85 N Total compression = • Ans 91.85 W = = 0.051 m k 1800 = 51 mm Ans E XAMPLE 20.7: In a compound helical spring the inner spring is arranged within and concentric with the outer one but is 10 mm shorter. The outer spring has 10 coils of mean diameter 25 mm and the diameter of the wire is 3 mm. Find the stiffness of the inner spring if an axial load of 150 N causes the outer one to compress 20 mm. If the radial clearance between the two springs is to be 1.6 mm, find the diameter of the wire of the inner spring when it has 8 coils C = 80 GN/m2 for both springs. n0 = 10 coils D0 = 25 mm d0 = 3 mm 0.010 m 8W D3 n Cd 4 δ Cd 4 W= 8D3 n 0.02 × 80 × 109 × (0.003)4 W= 8 (0.025)3 × 10 = 103.68 N δ= 0.025 Figure 20.4 Load shared by inner spring = 150 − 103.68 − 46.32 N Stiffness = 46.32 W = = 4.62 kN/m δ 0.01 Ans Diameter for inner spring = (0.025 − 0.0016 × 2 − 0.003) − d = 0.0188 − d δ Cd 4 8D3 n 0.01 × 86 × 109 × d 4 46.32 = 8(0.0188 − d)3 × 8 W= Solving the above equation, d = 2.03 mm Ans @seismicisolation @seismicisolation 400 • Strength of Materials E XAMPLE 20.8: Determine amount of compression and maximum shear stress produced when a load of 2200 N is dropped axially on a close-coiled helical spring from a height of 250 mm. The spring has 24 coils having mean diameter of 200 mm and wire diameter of 30 mm. Modulus of rigidity, C = 80 GPa. S OLUTION : Mean diameter of coil D = 200 mm Diameter of wire, d = 30 mm Number of coils, n = 24 Height of fall, h = 250 mm Falling load, w = 2200 N Modulus of rigidity, C = 80 GPa = 80000 N/mm2 For compression δ , Let We = Equivalent gradually applied load which shall produce the same deflection as by the falling load 2200 N. Work done by the falling load = work done by We 1 W (h + δ ) = We δ 2 Now δ= ∴ We = 8We D3 n Cd 4 δ Cd 4 8D3 n 2200(250 + δ ) = 1 δ Cd 4 ·δ × 2 8D3 n 2200(250 + δ ) = 1 80000(30)4 2 δ 2 8(200)3 24 2200(250 + δ ) = 21.1δ 2 550000 + 2200δ = 21.1δ 2 21.1δ 2 − 2200δ − 550000 = 0 δ 2 − 104.3δ − 26066.4 = 0 @seismicisolation @seismicisolation Springs • 401 √ 104.3 ± 104.32 + 4 × 26066.4 δ= 2 √ 104.3 ± 10878.5 + 104265.6 = 2 104.3 + 339.3 = 2 = 221.8 mm Ans For maximum stress: We = = δ Cd 4 8D3 n 221.8 × 80000(30)4 8(200)3 × 24 = 9357.2 N τ= = 8We D π d3 8 × 9357.2 × 200 π (30)3 = 176.6 N/mm2 Ans E XAMPLE 20.9: A close-coiled helical spring of steel wire of 7 mm diameter and having 12 complete turns is subjected to an axial couple M. The mean coil diameter is 85 mm. If the maximum bending stress in spring wire is not to exceed 250 MN/m2 , determine: i) the magnitude of axial couple M and ii) the angle through which one end of the spring is turned relative to the other end. Take modulus of elasticity for steel as 200 GN/m2 S OLUTION : Number of complete turns = 12 Diameter of steel wire = 7 mm Mean diameter of coil = 85 mm Maximum bending stress σb = 250 MN/m2 Esteel = 200 GN/m2 @seismicisolation @seismicisolation 402 • Strength of Materials i) Axial couple: 32M π d3 32M 250 × 106 = π (0.007)3 σb = ∴ 250 × 106 × π × (0.007)3 32 = 8.41 Nm Ans M= ii) 64MDn Ed 4 64 × 8.41 × 0.085 × 12 = 200 × 109 (0.007)4 = 1.1433 radians 180 = 1.1433 × π = 65.54◦ Ans φ= Open-Coiled Helical Spring In open-coiled helical spring, helix angle is more than 10◦ . Let the helix angle of the wire be α and mean radius of the coil be R. Then the length of wire, l = 2π Rn sec α n = no. of coils l α 2πR Y X α X Y W α WR Figure 20.5 @seismicisolation @seismicisolation Springs • 403 (a) Axial load Component in plane X − X = W R sin α , bending the wire Component in plane Y −Y = W R cos α , twisting the wire For axial deflection: Equating the external work to the strain energy in torsion and bending We get, 1 (W R sin α )2 l (W R cos α )2 l Wδ = + 2 2EI 2CJ 2 Sin α cos2 α + ∴ δ = W R2 l EI CJ 2π R 2π W R3 n sin2 α cos2 α + Because l = = cos α EI CJ cos α 64W R3 n 2 sin2 α cos2 α + δ= 4 E C d cos α 8W D3 n 2 sin2 α cos2 α δ= 4 + E C d cos α (i) (ii) Strain Energy: 1 U = Wδ 2 U= 4W 2 D3 n 2 sin2 α cos2 α + E C d 4 cos α (b) Axial couple: Y Xα X M < Y M Figure 20.6 @seismicisolation @seismicisolation < < M α (iii) 404 • Strength of Materials Figure 20.6 shows part of the spring, subjected to an axial couple M which is regarded positive if it increases the curvature of wire, i.e., winding up the spring. This moment may be resolved into components perpendicular and parallel to the wire. Component in plane X − X = M cos α , bending the wire Component in plane Y −Y = M sin α , twisting the wire Equating the work done by the couple to the strain energy in the wire, 1 (M cos α )2 l (Msinα )2 l Mφ = + 2 2EI 2CJ 2 cos α sin2 α + ∴ φ = Ml EI CJ M π Dn 32 2 cos2 α sin2 α + φ= cos α π d 4 E C 32MDn 2 cos2 α sin2 α = 4 + E C d cos α (iv) (v) Shear stress, τ : The torque component W R cos α causes the shear stress τ= = ∴ τ= W R cos α d π 4 ×2 d 32 16W R cos α π d3 8W D cos α π d3 Bending stress , σb The moment component W R sin α causes the bending stress σb = W R sin α d π 4 ×2 d 64 = 32W R sin α π d3 = 16W D sin α π d3 (vi) Composite Action of Axial Load and Couple When an axial load is applied to an open-coiled spring the spring winds up as well as extends. And similarly, when a winding-up couple is applied, the spring extends as well as twists. These movements and the formulae already derived can be obtained by the application of Castigliano’s theorem. @seismicisolation @seismicisolation Springs • 405 If the load W end couple M are applied simultaneously, total movement in plane X − X = −W R sin α + M cos α , bending the wire and total moment in plane Y − Y = W R cos α + M sin α , twisting the wire ∴ 1 1 1 1 (−W R sin α + M cos α )2 + (W R cos α + M sin α )2 2 EI 2 CJ 1 W 2 R2 sin2 α −W RM sin 2α + M 2 cos2 α W 2 R2 cos2 α −W RM sin 2α + M 2 sin2 α + = 2 EI CJ 1 2W R2 sin2 α − RM sin 2α 2W R2 cos2 α − RM sin 2α ∂U = + δ= ∂W 2 EI CJ 2 sin α cos2 α sin 2α 1 1 = W R2 l + + MRl − (vii) EI CJ 2 CJ EI l −W R sin 2α + 2M cos2 α W R sin 2α + 2M sin2 α ∂U = + φ= ∂M 2 EI CJ 2 sin 2α 1 1 cos α sin2 α + +W Rl − . (viii) = Ml EI CJ 2 CJ EI U= Additional terms obtained in Eqns. (vii) end (viii) represent the extension due to M and angle of twist due to W , respectively. If α → 0 these formulae reduce to those obtained for close-coiled springs. Note that Eqns. (vii) and (viii) can be used if only load or only couple is applied or both. Flat Spiral Spring It is made of uniform thin metallic stripe wound into a spiral is one plane. The inner end is fixed to a winding spindle C. R x b t B H A l y D V C E Flat spiral spring Figure 20.7 Refer Fig. 20.7, let a torque T be applied to the spindle C for winding. And let H and V be the horizontal and vertical reactions at D. Consider an element AB of length ol. Coordinates of element AB are xy as shown as Fig. 20.7. @seismicisolation @seismicisolation 406 • Strength of Materials AB = (V x − Hy) Bending moment on the element (i) If d θ is the change in angle between the tangents at A and B and if the length of the element is dl dl, then curvature of this element is r = dθ dl r M E but = I R 1 M = where R is the radius of curvature. R EI M.dl ∴ dθ = EI Vx −H y dl Substituting from Eqn. (i), d θ = EI The change in angle between targets at the extreme points of spring is: or dθ = θ= = Vx − Hy dl EI V EI xdl − H EI ydl Now xld is the moment of the whole length of the spring about the direction of Y whereas winding spindle being the centroid of the spring, we have xdl = moment of the profile of the spring about Y-axis = lR where l in the total length of spring. And ydl = the moment of the whole spring length about the axis joining the centres of the winding spindle and the pin D. So we have ydl = 0 V H V IR ∴ θ= lR − ×0 = EI EI EI But V R is equal to the torque T applied to the spindle θ= TP EI @seismicisolation @seismicisolation (ii) Springs • 407 The greatest bending moment occurs in the spring at E, which is M = V × 2R = 2T If σ is the bending stress at E, then σ = M Z where Z is the section modulus for the area of cross section of the spring about N.A. perpendicular to the plane of spring. ∴ 2T × 6 12T = 2 2 bt bt σ bt 2 or T = 12 Maximum bending stress, σ= (iii) The energy stored in winding the spring is 1 Tθ 2 1 TR T 2l = ×T × = 2 EI 2EI 2 1 σ bt 2 σ bt 2 1 12 × = × × 3 = 12 2EI 12 2E bt U= σ2 × bt l 24E σ2 = × volume of spring 24E = (iv) From the geometry of the circle into which the plates are initially formed. y l 2 l 2 2R-y Radius of circle = R @seismicisolation @seismicisolation 408 • Strength of Materials 2 l y (2R − y) = 2 or y = l2 8R Leaf, Laminated or Carriage Springs or Semi-elliptic Spring Laminated or leaf springs are made up of a number of stripes, made of 0.9% carbon (approximately) steel, these are also called carriage springs. The leaf spring is designed so that maximum stress is the same in all plates at all sections giving maximum utilisation of material. The arrangement of the spring is shown in Fig. 20.8, each plate free to slide relative to the adjacent plates as the spring deflects. The ends of each plate are tapered to provide a uniform change in effective breadth between the centre and the ends and if the plates are cut along their centre lines and placed side by side, they form a diamond-shaped plate. W W 2 x W nb x Figure 20.8 Let b = breadth of plate t = thickness of each plate n = number of plates @seismicisolation @seismicisolation 2 Springs • 409 Then effective width of plate at centre = nb Consider, section at a distance x from one end, x −W M 2 σ= = Z t2 x × nb × l/2 6 −3W l = 2nbt 3 (i) Note that this expression is independent of x. That means σ is constant at all sections. At any section M is proportional to x and I is proportional to x, so that M is proportional to I. E M = ; R is a constant. Thus, if the spring is to become flat when loaded, the plates Now since I R must initially be bent in the arc of a circle. The load which causes the plates to become flat is called the proof load. E d2y M −W x/2 −3W l = = = 2 3 I dx nbt 3 t x × nb × l/2 12 Integrating Eqn. (ii), E l When x = , 2 dy =0 dx ∴ A= dy 3W lx +A =− dx nbt 3 3W l l · nbt 3 2 3W l EIy = − 3 nbt ∴ x2 l − x +B 2 2 When x = 0, y = 0 ∴ B = 0 The maximum deflection occurs at the centre l where x = 2 i.e., ymax = 3W l 3 8E n bt 3 Strain energy, l M2 dx 2EI U= 0 l/2 =2 0 3W l W · × dx 2Enbt 3 2 @seismicisolation @seismicisolation (from Eqn. ii) (ii) 410 • Strength of Materials U= The strain energy 3 W 2l3 16 Enbt 3 3 W 2l2 may be written as 16 Enbt 3 3W l 2 nbtl U= 12E 2nbt 2 σ 2 nbtl × 6E 2 σ2 U= × volume 6E = ∴ Quarter-Elliptic Leaf Spring Figure 20.9 shows a quarter-elliptic leaf spring. In this case we substitute l = 2l and W = 2W in formulae of stress σ and deflection ymax of elliptical leaf spring discussed above. W l 3 2W × 2l 6W l × = 2 nbt 3 nbt 3 3 2W × (2l)3 ymax = × 8 Enbt 3 3 6W l = Enbt 3 σ= t Figure 20.9 E XAMPLE 20.10: Find the mean radius of an open-coiled spring having helix angle 30◦ to given vertical displacement of 25 mm and an angular rotation of the loaded and of 1.25◦ under an axial load of 40 N. The material available is steel rod of 6 mm diameter. E = 200 GN/m2 , C = 80 GN/m2 . S OLUTION : δ = WR l 2 So ∴ sin2 α cos2 α + EI CJ 200 E = = 2.5 C 80 E = 2.5 C J = 2I J EI = 2.5 C × 2 EI = 1.25 CJ @seismicisolation @seismicisolation (i) Springs • 411 Substituting in Eqn. (i) sin2 30◦ cos2 30◦ + 0.025 = W R l 1.25 CJ CJ 2 W R l 0.25 + 0.75 = CJ 1.25 2 0.025 = W R2 l (0.95) CJ or, W R2 l = 0.0263 CJ Also φ = Ml cos2 α sin2 α + EI CJ (i) sin 2α 1 1 +W Rl − 2 CJ EI Since, M = 0 So sin 2α 1 1 φ = W Rl − 2 CJ EI 1.25 × π W Rl sin 60 1 1 = − 180 2 CJ 1.25 CJ 1 0.433 W Rl 1− 0.0218 = CJ 1.25 0.0218 = 0.0866 W Rl CJ W Rl = 0.252 CJ (ii) From Eqns. (i) and (ii) 0.252R = 0.263 ∴ R = 0.1044 m = 104.4 mm Ans E XAMPLE 20.11: An open-coiled helical spring is made of 10 mm diameter steel wire, the coils having 13 complete turns and a mean diameter of 102 mm, the helix angle being 14◦ . Calculate the deflection under 250 N and intensities of direct and shearing stresses induced in the section wire. @seismicisolation @seismicisolation 412 • Strength of Materials If 250 N axial load is replaced by an axial torque of 9 Nm, calculate the angle of rotation of the coil and the axial deflection. E = 210 GPa, C = 85 GPa. S OLUTION : δ = WR l 2 sin2 α cos2 α + EI CJ sin 2α 1 1 + MRl − 2 CJ EI M = 0, W = 250 N, l = 2π Rn sec α l + 2π × ∴ 102 1 1 × 13 × = 4163.64 × 2 cos 14 cos 14 l = 4292.4 mm δ = 250 (51)2 × 4292.4 cos2 14 sin2 14 + 3 210 × 10 × I 85 × 103 J J = 2I I= ∴ π (10)4 = 490.6 mm4 64 J = 2 × 490.6 = 981.2 mm4 0.0585 09415 δ = 2791133100 + 210000 × 490.6 85000 × 981.2 = 2791133100 5.68 × 10−10 + 1.13 × 10−8 = 2791133100 5.68 × 10−10 + 113 × 10−10 = 33.12 mm Ans Stresses: Bending stress, σ = = 32W R sin α π d3 32 × 250 × 51 × sin 14 π (10)3 = 31.43 N/mm2 @seismicisolation @seismicisolation Ans Springs Shear stress, τ = = 16W R cos α π d3 16 × 250 × 51 × cos 14 π (10)3 = 63.04 N/mm2 When only torque of 9 Nm in applied: φ = Ml cos2 α sin2 α + EI CJ sin 2α +W Rl 2 1 1 − CJ EI W = 0, M = 9 Nm = 9000 Nmm 0.0585 0.9415 + φ = 9000 × 4292.4 210000 × 490.6 85000 × 981.2 φ = 38631600 9.14 × 10−3 + 7.014 × 10−10 +0 d = 0.380 radian = 0.38 × 180 π = 21.8◦ Ans For axial deflection; δ = WR l 2 sin2 α cos2 α + EI CJ sin 2α 1 1 + MRL − 2 CJ EI W = 0, M = 9 Nm = 9000 Nmm sin 2α 1 1 δ = MRl − 2 CJ EI 1 1 sin 2 × 14◦ − = 9000 × 51 × 4292.4 2 85000 × 981.2 210000 × 490.6 = 4624791594 12 × 10−9 − 9.7 × 10−9 = 10.6 mm Ans @seismicisolation @seismicisolation • 413 414 • Strength of Materials E XAMPLE 20.12: An open-coiled helical spring consists of 14 coils each of mean diameter 68 mm. If the spring wire of diameter 8 mm and helix angle 28◦ , determine: (i) The load required to elongate the spring by 23 mm and the bending and shear stresses caused by that load; (ii) The axial twist that would cause a bending stress of 60 MN/m2 in the coils. Take E = 200 GN/m2 and C = 85 GN/m2 S OLUTION : Wire diameter = 8 mm Coil mean diameter = 68 mm No. of turns = 14 Helix angle = 28◦ E = 200 GN/m2 C = 85 GN/m2 δ = W R2 l sin2 α cos2 α + EI CJ l = 2π Rn sec α ; I= π (8)4 = 200.96 mm4 , 64 J = 2 × 200.96 = 401.92 mm4 1 68 × 14 = 3385 mm 2 cos 28 cos2 28 sin2 28 2 + 23 = W (34) × 3385 200000 × 200.96 85000 × 401.92 5.878 × 10−6 = W 5.484 × 10−9 + 22.8 × 10−9 l = 2π × W= 5.878 × 10−6 = 207.8 N Ans 28.284 × 10−9 Bending stress,σb = = 16W D sin α π d3 16 × 207.8 × 68 × sin 28 π (8)3 = 66 MN/m2 @seismicisolation @seismicisolation Ans Springs Shear stress, τ = = • 415 8W D cos α π (8)3 8 × 207.8 × 68 cos 28 π (8)3 = 62.08 MN/m2 Ans (ii) Axial twist: T = Axial twist required to cause bending stress of 60 MN/m2 Component of axial torque causing bending = T cos α Now, σb = 32M 32T cos α = π d3 π d3 60 = 32T cos α π d3 T = 60 π d 3 32 cos α = 60 × π × 83 32 cos 28 = 3413.8 Nm = 3.1414 Nm Ans E XAMPLE 20.13: A flat spring is 6 mm wide, 0.3 mm thick and 3.5 m long. Calculate the torque, the work stored and the number of turns required to wind up the spring whereas the maximum bending stress does not exceed 520 N/mm2 , E = 200 GN/m2 . S OLUTION : b = 6 mm, t = 0.3 mm, l = 3.5 m = 3500 mm σ = 520 N/mm2 , E = 200 GN/m2 = 200000 N/mm2 σ bt 2 12 520 × 6 × (0.3)2 = 23.4 Nmm = 12 = 0.0234 Nm Ans Torque, T = @seismicisolation @seismicisolation 416 • Strength of Materials Work stored = Strain energy σ2 × volume of spring 24E 5202 U= × 6 × (0.3) × 3500 24 × 200000 = = 354.9 Nmm = 0.3549 Nm Ans Now θ= Tl , EI I= 6 × (0.3)3 12 = 23.4 × 3500 × 12 200000 × 6 × (0.3)3 = 30.33 radians = 30.33 2π = 4.8 turns Ans E XAMPLE 20.14: A flat spiral spring is having 10 mm broad and 0.6 mm thick steel stripe. The length of the steel stripe is 7 m. The end at the greatest radius is attached to a fixed point and the other end to a spindle. Find: (a) the maximum turning moment which can be applied to the spindle if the stress in the stripe is not to exceed 560 MN/m2 ; (b) the number of turns required to be given to the spindle and (c) the energy stored in the spring. Take E = 200 GN/m2 . S OLUTION : b = 10 mm = 0.01 m t = 0.6 mm = 0.0006 m l=7m E = 200 GN/m2 @seismicisolation @seismicisolation Springs • 417 a) We know, T= = σ bt 2 12 560 × 106 × 0.01 × (0.0006)2 12 = 0.168 Nm b) Ans 0.01 × (.0006)3 = 1.8 × 10−13 12 0.168 × 7 = 200 × 109 × 1.8 × 10−13 θ= Tl , EI I= = 32.67 radians = 32.67 2π = 5.2 turns Ans σ2 × volume of the spring 24E 2 560 × 106 × (0.01)(0.0006) × 7 = 24 × 200 × 109 Strain energy, U = = 313600 × 1012 × 0.01 × 0.0006 × 7 24 × 200 × 109 = 2.74 Nm Ans E XAMPLE 20.15: A laminated steel spring, simply supported at the ends and centrally loaded with a span of 0.75 m is repaired to carry a load of 7.5 kN and the central deflection is not to exceed 50 mm. The bending stress must not be greater than 400 MN/m2 . Plates are available in multiples of 4 mm for width. Determine the suitable value for the thickness, width and number of plates and the radius to which the plates should be formed. Assume the width to be twelve times the thickness. E = 200 GN/m2 . @seismicisolation @seismicisolation 418 • Strength of Materials S OLUTION : We know 3W l 3 8Enbt 3 W = 7.5 kN l = 0.75 m b = 12 t y= E = 200 GN/m2 Substituting in the formula, 3 × 7.5 × 103 ×0.753 8 × 200 × 109 ×n × 12t × t 3 4 nt = 9.89 × 10−9 0.05 = Which gives (i) Also, 3W l 2nbt 2 3 × 7.5 × 0.75 400 × 106 = 2 × n × 12t × t 2 σ= nt 3 = 1.758 × 10−6 or, (ii) From Eqns. (i) & (ii) nt 4 9.89 × 10−9 = nt 3 1.758 × 10−6 ∴ t= 0.00562 m Ans The nearest suitable value is 6 mm. b = 12 t (given) b = 12 × 6 = 72 mm (thickness is rounded to 6) Now nt 3 = 1.758 × 10−6 from Eqn. (ii) ∴ n= 1.758 × 10−6 (0.006)3 = 8.13 Hence, 9 plates are required @seismicisolation @seismicisolation Ans. Springs • 419 It must be noted that higher value to round off figure is taken so that laminated steel spring is safer. 3 × 7.5 × 103 × 0.753 3W l Actual deflection under load = ∴ y = 8 × 200 × 109 × 9 × 0.072 × 0.0063 8En bt 3 = 0.0424 m = 42.4 mm Now y= Ans l2 8R 0.0424 = 0.752 8R ∴ 0.752 8 × 0.0424 R= R = 1.66 m Ans E XAMPLE 20.16: A carriage spring has 12 plates each 65 mm wide by 6 mm thick and the longest plate is 0.8 m long. The greatest bending stress is not to exceed 185 MN/m2 and the central deflection when the spring is fully loaded is not to exceed 20 mm. Estimate the magnitude of the greatest central load that can be applied to the spring. Take E = 200 GN/m2 . S OLUTION : n = 12 plates b = 65 mm t = 6 mm l = 0.8 m δ = 185 MN/m2 Maximum deflection (central) = 20 mm. ymax = 3W l 3 8Enbt 3 W= ymax ×8Enbt 3 3W l 3 W= 0.02 × 8 × 200 × 109 × 12 × 0.065 × (0.006)3 3 × (0.8)3 = 3510 N = 3.51 kN Ans @seismicisolation @seismicisolation 420 • Strength of Materials E XAMPLE 20.17: A leaf spring is required to satisfy the following specifications: l = 0.75 m, w = 5 kN, b = 75 mm, maximum stress = 210 MN/m2 , maximum deflection = 25 mm and E = 200 GN/m2 . Find the number of leaves and their thickness. If the leaves become straight when this load in applied, find its radius of curvature. S OLUTION : We know σ= 3W l 2nbt 2 nt 2 = 3W l 2bσ = 3 × 5000 × 0.75 2 × (0.075) × 210 × 106 nt 2 =3.571 × 10−4 (i) Also, 3W l 3 8Enbt 3 3 × 5000 × (0.75)3 nt 3 = 8 × 200 × 109 × (0.075) × 0.025 ymax = nt 3 = 2.10 9 × 10−6 From Eqns. (i) and (ii) nt 3 2.1093 × 10−6 = nt 2 3.571 × 10−4 t = 5.906 × 10−3 m = 6 mm (say) Ans Using Eqn. (i) n= 3.571 × 10−4 (5.906 × 10−3 )2 3.571 × 10−4 (5.906)2 × 10−6 = 10 Nos Ans n= @seismicisolation @seismicisolation (ii) Springs • 421 We know, 3 × 5000 × 0.75 M 3W l = = 3 I nbt 10 × 0.075 × (0.006)3 = 69444444000 M E = I R E.I ∴ R= M = 200 × 109 69444444000 = 2.8 m Ans Exercise 20.1 A close-coiled helical spring is to have a stiffness of 70 kN/m and to exert a force of 2.25 kN. If the mean diameter of the coils is to be 90 mm, and the working stress 230 MN/m2 , find the required number of coils and the diameter of the steel rod from which the spring should be made. Take the modulus of rigidity as 80 GN/m2 . [Ans 6.58; 13.55 mm] 20.2 A close-coiled helical spring made of round steel wire is required just to fit a rod 30 mm diameter and to carry an axial load of 120 N without causing the deflection to exceed 20 mm. The maximum allowable shearing stress is 200 MN/m2 , and modulus of rigidity of steel is 80 GN/m2 . Find the diameter of wire, the mean diameter of coil and the number of turns. [Ans 3.72 mm; 33.72 mm; 8.52] 20.3 A close-coiled helical spring of circular section having a mean coil diameter of 60 mm a subjected to an axial load of 80 N applied at the end of spring producing a shear stress of 100 N/mm2 and a deflection of 50 mm. Find the diameter, the number of coils, the length of the spring wire and the strain energy stored in the spring. [Ans 4.96 mm; 17.5; 2 joules] 20.4 A close-coiled helical spring has a mean diameter of the coils 12 times the wire diameter. It is to be designed to absorb 300 J of energy with an extension of 150 mm. The maximum shear stress is not to exceed 140 MN/mm2 . Determine the mean diameter of the coil, diameter of the wire and number of turns. Also find the load with which an extension of 250 mm could be produced in the spring. Take modulus of rigidity, C = 80 GN/m2 . [Ans 30 mm; 360 mm, 1333.4 N] 20.5 Determine the maximum shear stress and the amount of compression produced when a mass of 200 kg is dropped axially on a close-coiled helical spring from a height of 250 mm. The spring has 20 coils each of mean diameter 200 mm and the wire diameter is 25 mm, C = 84 GN/m2 . [Ans 0.287 m, 239.6 kN/m2 ] @seismicisolation @seismicisolation 422 • Strength of Materials 20.6 Determine the mass of closely-coiled helical spring which would absorb the energy of a truck of mass 10 tonnes while moving with a velocity of 1 m/s. If the spring is compressed by the impact. Working bending stress is 300 N/mm2 ; C = 80 GN/m2 and specific gravity of the spring material = 7.8 20.7 Two springs A and B are connected is series Spring A consists of 12 coils of 6 mm steel wire and of 30 mm outside diameter. Spring B consists of 18 coils of 8 mm steel wire and 40 mm outside diameter. What is spring constant (stiffness) for the common system? What is the longest force that can be applied to these springs without exceeding a shear stress of 360 MPa? C = 82 GN/m2 . [Ans 19.44 N/mm, 1017.8 N] 20.8 One helical spring is placed inside the coil of the other helical spring, having same material, same number of coils and free axial lengths. The two springs are compressed by a load of 1.5 kN. The mean coil diameter and wire diameter of the outer spring and inner spring are 80 mm, 60 mm and 10 mm, 7 mm, respectively. Calculate the load taken and the maximum shear stress in each spring. [Ans 543.47 N; 956.53 N; 194.86 N/mm2 ; 242.09 N/mm2 ] 20.9 An open-coiled helical spring is made of 9.525 mm diameter steel wire, the coils having 14 complete turns and a mean diameter of 101.6 mm, the helix angle being 15◦ . Calculate the deflections under 225 N and intensity of direct and shearing stresses induced in the section of wire. If 225 N axial load is replaced by an axial torque of 8.5 Nm, calculate the angle of rotation of the coil and the axial deflection. E = 210 GPa, C = 84 GPa. [Ans 39.03 mm; 49.87 N/mm2 , 67.4 N/mm2 , 27◦ , 14.7 mm] 20.10 An open-coiled helical spring has 10 coils of a 12 mm diameter steel wire with a mean diameter of 150 mm. The helix angle of the coils is 32◦ . Find the axial extension produced by a load of 250 N. Any formulas used must be proved from fundamentals. E = 210 GN/m2 , C = 70 GN/m2 . [Ans 49.7 mm] ◦ 20.11 Find the mean radius of an open-coiled spring having helix angle of 30 to give a vertical displacement of 25 mm and an angular rotation of the loaded end of 1.25◦ under an axial load of 40 N. The material available is steel rod of 6 mm diameter. E = 200 GPa, C = 80 GPa. [Ans R = 0.1044 m] 20.12 An open-coiled helical spring made of 10 mm diameter rod has six free coils 100 mm mean diameter. The ends of the spring are fastened to two discs kept 0.75 m apart, which is the free length of the spring. Calculate the force or the discs, acting along the axis of the spring, when one disc is rotated through 10◦ to coil the spring. E = 200 GPa, C = 80 GPa. [Ans 11.5 N (Compressive)] 20.13 An open-coiled spring consists of 10 coils, each of mean diameter 50 mm, the wire forming the coil being 6 mm is diameter. Each coil makes an angle of 30◦ with the plane perpendicular to the axis of the spring. (i) Determine the load repaired to elongate the spring by 20 mm and the bending and shear stresses caused by that load. (ii) Calculate the axial twist that would cause a bending stress of 60 MN/m2 . @seismicisolation @seismicisolation Springs 20.14 20.15 20.16 20.17 20.18 20.19 • 423 Take E = 200 GN/m2 , C = 82 GN/m2 . [Ans 192.7 N; 113.6 N/mm2 , 98.37 N/mm2 , ii) 1.469 Nm] An open coiled helical spring is subjected to a combined axial load and torque. The spring has mean coil diameter of 120 mm, number of coils 10, wire diameter 12 mm and helix angle of 30◦ . Find the value of axial load and torque which would extend the spring by 5 mm with no rotation of coils, indicating if the torque tends to wind or unwind the spring. Find also the maximum normal and shear stresses in the spring. Take E = 210 GPa, C = 84 GPa. [Ans −0.355 Nm (trying to unwind), 58.667 N, 16.37 N/mm2 , 10.325 N/mm2 ] A carriage spring 600 mm long is constructed out of 60 mm wide steel plates. Find the thickness of leaves and the number of plates required, if the spring is subjected to a maximum load of 2680 N with a maximum central deflection of 12 mm and a maximum bending stress of 160 N/mm2 . Take E = 200 GPa. [Ans 6 mm; 7] A semi-elliptic leaf spring 800 mm long has 12 leaves. The cross section of leaves is 60 mm × 6 mm. For a central deflection of 15 mm and a bending stress not greater than 200 N/mm2 , find the value of the central load. E = 200 GPa. [Ans 3.24 kN] A steel carriage spring is 762 mm in span and carries a central load of 50 kN. The stress is to be limited to 188 MPa in plates which are 76.5 mm wide and 6.35 mm thick. What will be the stress in any plate and deflection of the spring at the centre? Also determine how much a plate will overhang the one immediately below it and to what radius of curvature should each plate be curve? Take E = 210 GPa. [Ans 185.3 MPa; 20.17 mm; 38.1 mm; R = 3.6 m] A 5 mm wide and 0.3 mm thick flat spiral spring is 2 m long. The maximum stress is 600 MPa at the greatest bending moment. Determine the torque the work stored and number of turns to wind up the spring. E = 208 GPa. [Ans T = 22.5 Nmm, 3.82 turns, 270 Nmm] A flat spiral spring is 5 mm wide, 0.25 mm thick and 3 m long. Assuming maximum stress of 1000 MN/m2 to occur at the point of greatest bending moment, calculate: i) the torque ii) the work that can be stored in the spring iii) the number of turns required to wind up the spring Take E = 200 GPa. [Ans 0.026 Nm, 0.781 Nm, 9.533 turns] 20.20 A quarter elliptic spring has a clear span of 800 mm and carries a load of 10 kN at the free end. The bending stress in stripe and deflection at free end are not to exceed 320 MPa and 80 mm, respectively. Find the number of plates if the width of a plate is 8 times the thickness. Take E = 210 GN/m2 . 3 W l3 3 Wl [Hint: δ = ] ; σ= 3 8 nbt E 2 nbt 2 [Ans t = 12.2 mm; 10] @seismicisolation @seismicisolation C HAPTER 21 COLUMNS AND STRUTS A strut usually means a compression member which is long in comparison with its cross-sectional area. If the strut is vertical, then it is known as column, pillar or stanchion. For very long column or strut, the effect of direct compression is negligible and they fail due to buckling load (or critical load or crippling load). Practically, many machine parts and members of structure behave as columns or struts. For example, connecting rod of an internal combustion engine is a strut with both ends hinged. Since, both the ends of a strut or column need be considered while analysing them. For columns and struts combination of end conditions are very important and are considered during derivation of formulae related to the same. The possible end conditions are: a) b) c) d) Both ends hinged Both ends fixed One end is fixed and the other hinged One end is fixed and the other free. It may be noted that if the member of the structure is not vertical and one or both of its ends are hinged or pin joined, the member is known as strut. As told before, connecting rods, piston rods, etc., are struts. Very long columns: Let us first consider the case of very long columns for which crippling load is loaded so that the effect of direct stress is negligible compared to bending stress. Long columns are those whose slenderness ratio is more than 120 or whose length is more than 30 times the least diameter. Long columns are analysed using Euler’s analysis. Leonhard Euler was Swiss mathematician, who was the first to analyse long columns mathematically. Following assumptions are made during derivation of Euler’s formulae: 1. 2. 3. 4. 5. The columns are made up of homogeneous material. The columns are initially straight. The cross section of column is constant throughout. The columns carry perfectly axial loads. Plane cross sections of column normal to centre line remain plane during buckling and the effect of shear stress is neglected. 6. Self-weight of columns is neglected. @seismicisolation @seismicisolation Columns and Struts • 425 7. The column is long compared to lateral dimensions. 8. The stresses do not exceed the limit of proportionality. 9. Longitudinal fibres of the column are free to expand or contract independently without any effect on adjoining fibres. 10. Shortening of the columns due to direct compression is negligible. Now we will derive formulae for different end conditions (as stated before) using Euler’s method. Before this let us first define important terms: Slenderness Ratio: It is the ratio of unsupported length of the column to the minimum radius of le is the slenderness ration where le is the gyration of the cross-sectional ends of the column. k equivalent length. Both ends hinged (a) One end fixed other free (b) Both ends fixed (c) l le= √2 le=l le=2l le=l/2 Equivalent Length (le ): It is the effective length of the column as shown in Fig. 21.1. One end fixed other hinged (d) Figure 21.1 Buckling factor = le , quad where k is the radius of gyration Minimum k Buckling load (or critical load or crippling load): The maximum limiting load at which the column tends to have lateral displacement or tends to buckle. The buckling takes place about the axis having minimum radius of gyration. It may be noted that columns are of the three types: i) Short columns: l = 8 times diameters or slenderness ratio is less than 32. ii) Medium columns: l = 8 to 30 times diameter or slenderness ratio is 30 to 100. iii) Long columns: l = more than 30 times diameter or slenderness ratio is more than 100. Safe load = Buckling load Factor of safety @seismicisolation @seismicisolation 426 • Strength of Materials (a) Both Ends Hinged P The column AB shown is Fig. 21.1 is hinged at both ends and let P be the critical load. Consider a section of column at a distance x from A, where the deflection is y. Moments due to critical load P, o B M = −Py ∴ EI dx2 X x d2y EI 2 = −Py dx d2y y l oA Figure 21.2 + Py = 0 d 2 y Py + =0 dx2 EI Solving the above differential equation, y = C1 cos x P EI +C2 sin To determine constants C1 and C2 , when x = 0, y = 0 ∴ x P EI C1 = 0 Now applying end condition at B. Here, x = l and y = 0; 0 = C2 sin If C2 is zero, then column would not bend at all therefore, sin l P = 0, π , 2π , 3π , etc. For which l EI Taking the least significant value, P EI P l EI =0 P =π EI π 2 EI π 2 EI P= 2 or = 2 l le l= or Note is this case equivalent length le = l. @seismicisolation @seismicisolation (i) Columns and Struts • 427 (b) Column with One End Fixed and the Other Free In this case, refer to Fig. 21.3, consider point x, P e B M = P(e − y) d2y = P(e − y) dx2 d 2 y P(e − y) = EI dx2 d2y P P + y= e EI dx2 EI EI or y l X x A Figure 21.3 Solution of this differential equation is y = e +C1 cos x P EI +C2 sin x P EI When x = 0, y = 0 ∴ Now Eqn. (i) becomes y = e − e cos x or, At A, or C1 = −e 0 = e +C1 ; P EI +C2 sin x P EI dy P P P P = +e sin x +C2 cos x dx EI EI EI EI dy = 0 and x = 0 dx ∴ Now since P cannot be zero Equation (i) becomes ∴ 0 = C2 P EI C2 = 0 y = e − e cos x @seismicisolation @seismicisolation P EI (ii) 428 • Strength of Materials At B, x = l, y = e ∴ Since, e is not zero, ∴ l = e − e cos l P EI p =0 cos EI P π 3 π 5π = , , or EI 2 2 2 Taking the least value, etc P π = EI 2 π 2 EI Hence P = 42 ∴ Note: In this case equivalent length le = 2l. (c) Both Ends Fixed P Since both end are fixed, so by symmetry end moments are equal (say M). At section x from end A, EI M d2y = −(Py − M) dx2 d2y P M or + y= 2 EI EI dx B l/4 y l/2 l x l/4 A M Figure 21.4 Solution of the above differential equation is P P M +C2 sin x y = +C1 cos x P EI EI P P P P dx = −C1 sin x +C2 cos x dy EI EI EI EI At A, x = 0 and dx =0 dy ∴ C2 P =0 EI @seismicisolation @seismicisolation Columns and Struts • 429 Since, P cannot be zero, therefore C2 = 0 Also at A, x = 0 and y = 0 M +C1 = 0 P M P M M P y = − cos x P P EI C1 = − ∴ But at B, x = l, y = 0 P M M − cos x =0 P P EI M P M cos x = ∴ P EI P P =1 Or cos l EI P = 0, 2π , 4π , . . . l EI Taking least significant value, we get P= 4π 2 EI l2 (iii) Note: In this case equivalent length le = l/2. (d) One End Fixed, Other End Hinged Let M be the fixing moment at A and for equilibrium F is the horizontal force as shown in Fig. 21.5. At x from A, d2y EI 2 = −Py + F(l − x) dx d 2 y Py F + = (l − x) 2 EI EI dx P B F y l x A M Figure 21.5 @seismicisolation @seismicisolation 430 • Strength of Materials Solution of the above differential equation is P P F x +C2 Sin x + (l − x) y = C1 Cos EI EI P At A, y = 0 when x = 0 Fl P Fl P P F ∴ y = − cos x +C2 sin x + (l − x) P EI EI P P P P P F dy Fl = sin x +C2 cos x − or dx P EI EI EI EI P C1 = − ∴ dy =0 dx Also, at A, x = 0 and ∴ C2 F P − =0 EI P F EI ∴ C2 = × P P F EI F Fl P P x + sin x + (l − x) ∴ y = − cos P EI P P EI P At B, x = l and y = 0 Fl 0 = − cos P F EI P P l + sin l EI P P EI Dividing equation stated above by cos or P l , EI P Fl F EI tan l 0=− + P P P EI P P Fl P tan l = × EI P F EI P P l = ×l tan EI EI @seismicisolation @seismicisolation Columns and Struts • 431 From mathematical tables, we find only two values for which tan θ = θ as in this case. Either θ = 0 or θ = 4.493 radians. Since θ = 0 is not admissible as then, P will be zero which is not possible. Therefore, we adopt θ = 4.493 radians. ∴ or l P = 4.493 EI 4.4932 EI P= l2 2π 2 EI P= l2 (iv) Note: In this case equivalent length le = √l . 2 Alternative methods: Equations (ii), (iii) and (iv) can be obtained by putting equivalent length le in Eqn. (i) in place of actual length l. For example, for case (b): le = 2l (Refer Fig. 21.1) Substituting in Eqn. (i) for equivalent length, P= π 2 EI l2 After substituting equivalent length 2l in place of l 2 to find critical load for one end fixed and the other free. π 2 EI π 2 EI = P= (2l)2 4l 2 which is same as before. Similarly, Eqn. (iii) and Eqn. (iv) can be derived by substituting in place l l of l; and √ . 2 2 For reference equivalent lengths are given again in the following table: Ends conditions Both ends hinged One end fixed and the other free Both ends fixed One end fixed and the other hinged Equivalent length le = l le = 2l le = l/2 l le = √ 2 E XAMPLE 21.1: A steel column is of length 7.5 m and diameter 550 mm with both ends hinged. Determine the crippling load by Euler’s formula. Take E = 210 GPa. S OLUTION : Actual length of the column l = 7.5 m = 7500 nmm Diameter of column, d = 550 mm Young’s modulus, E = 210 GPa = 210000 N/mm2 @seismicisolation @seismicisolation 432 • Strength of Materials Least moment of inertia of the column section, π d4 64 π (550)4 = 4489525391 mm4 = 64 I= Since, the column is hinged at both ends, ∴ Equivalent length = l π 2 EI l2 2 π × 210000 × 4498525391 = 75002 8 = 1.65 × 10 N Ans P= E XAMPLE 21.2: Compare the crippling load of solid circular column of diameter 250 mm and hollow circular column of same cross-sectional area and thickness 35 mm. Materials and lengths are same. Assume both are hinged at both ends. S OLUTION : π Sectional area of solid column = (250)2 4 Thickness of the hollow circular column = 35 mm Let the external diameter of hollow column = D mm ∴ Internal diameter of hollow circular column = D · 2t = D − 2 × 35 = (D − 70) mm ∴ Sectional area of hollow circular section π 2 = D − (D − 70)2 4 Since, sectional areas of both are same, π π 2 ∴ D − (D − 70)2 = (250)2 4 4 D2 − (D2 + 4900 − 140 D) = 62500 −4900 + 140 D = 62500 140 D = 62500 + 4900 D = 481.43 mm, Inner dia = 481.43 − 70 = 411.43 mm @seismicisolation @seismicisolation Columns and Struts • 433 Least moment of inertia: π 481.434 − 411.434 64 π 5.372 × 1010 − 2.865 × 1010 = 64 IH = = 0.123 × 1010 mm4 IS = π (250)4 64 = 191650391 mm4 π 2 E IH 2 PH IH 0.123 × 1010 = 2l = = PS IS 191650391 π E IS 2 l = 6.42 Ans E XAMPLE 21.3: A steel tube 4.5 m long, 40 mm internal diameter and 6 mm thick is used as a column: the crippling load if: i) both ends are hinged, ii) both ends are built-in, and iii) one end is built-in and other is free. Take E = 200 GPa S OLUTION : i) External diameter = 40 mm + 6 × 2 = 52 mm Internal diameter = 40 mm π 524 − 404 64 π (7311616 − 2560000) I= 64 = 233126.16 mm4 I= i) π 2 EI for both ends hinged l2 π 2 × 200000 × 233126.16 = 45002 = 22701.5 N Ans P= @seismicisolation @seismicisolation 434 • Strength of Materials ii) For both ends built-in (fixed): In this case effective length, l 2 4500 = 2250 mm = 2 π2 E I P= 2 le le = π 2 × 200000 × 233126.16 22502 = 90806 N Ans = iii) One end is built-in and other end free: Equivalent length le = 2l le = 2 × 4500 = 9000 mm π2 E I le2 2 π × 200000 × 233126.16 = (9000)2 = 5675.4 N Ans P= E XAMPLE 21.4: Determine the limiting length of a column with both ends hinged and having a cross-section 70 mm × 120 mm so that the critical stress is 300 N/mm2 . Take E = 200 GPa. Column section = 70 mm × 120 mm Critical stress = 300 N/mm2 E = 200000 N/mm2 Cross-sectional area = 70 × 120 = 8400 mm2 Critical load = critical stress × sectional area = 300 × 8400 = 2520 kN Imin = 120 × 703 = 3430000 m4 12 Using Euler’s formula, π 2 EI l2 π 2 × 200000 × 3430000 2520000 = l2 P= @seismicisolation @seismicisolation Columns and Struts • 435 π 2 × 200000 × 3430000 2520000 = 2684002 l = 1638.3 mm l2 = Limiting length of column =1.638 m Ans E XAMPLE 21.5: A column having a T section with a flange 150 mm × 15 mm and web 150 mm × 15 mm is 3.5 m long. Assuming the column to be hinged at both ends, find the buckling load by using Euler’s formula. Take E = 200 GPa. S OLUTION : 1 150 mm A B 15 mm Y X X G 2 150 mm 15 mm Y Figure 21.6 Let us first find y form AB a1 = 150 × 15 = 2250 mm2 , y1 = a2 = 150 × 15 = 2250 mm2 y2 = 15 = 7.5 mm 2 150 + 15 = 90 mm 2 A = a1 + a2 = 2250 + 2250 = 4500 ȳ = = 2250 × 7.5 + 2250 × 90 4500 16875 + 202500 4500 = 48.75 mm. IXX = 150 × 153 15 × 1503 + 2250 (48.75 − 7.5)2 + + 2250 (48.75 − 90)2 12 12 @seismicisolation @seismicisolation 436 • Strength of Materials = [42187.5 + 3828.5] + [4218750 + 1701.6] = [46016] + [4220452] = 4266468 mm4 IYY = 15 × 1503 150 × 153 + 12 12 = 4218750 + 42187.5 = 4260937.5 mm4 IYY < IXX , so we will take IYY in the formula of Euler. π 2 EI Buckling load = P = 2 (for both ends hinged) l = π 2 × 200000 × 4260937.5 (3500)2 = 685896 N = 685.9 kN Ans E XAMPLE 21.6: Determine the Euler’s crippling load for a strut, square in section and 4.5 m long with one end hinged and the other end free. The same bar is found to deflect 4 mm at the centre when a load of 90 N is placed at midspan with the bar simply supported at the ends. S OLUTION : Length of strut = 4.5 m = 4500 mm δ= W l3 48EI 4= 90 × 45003 48EI EI = 90 × 45003 48 × 4 = 4.27 × 1010 N mm2 Equivalent length for one end hinged and the other free, le = 2l = 2 × 4500 = 9000 mm P= = π 2 EI le2 π 2 × 4.27 × 1010 (9000)2 @seismicisolation @seismicisolation Columns and Struts = 5197.6 N = 5.197 kN • 437 Ans Limitation for the Use of Euler’s Theory In case of short column, Euler’s theory may give a critical load which is greater than that required to produce failure due to direct compression. The limiting case occurs when Pc = σc A where σc is the compression stress at the yield point and A is the cross-sectional area. π 2 EI l2 2 nπ E Ak2 = l2 P=n But where n is a constant depending on the end fixing conditions and k is radius of gyration (least) of the cross section ∴ n π 2 E A k2 = σc l2 l nπ 2 E = or k σc The quantity l/k as we know is called the slenderness ratio and its value at this point is called validity limit for Euler’s theory. l If the slenderness ration is small, the crippling stress will be high. But for the column material, k the crippling stress cannot be greater than the crushing stress. Hence, when the slenderness ratio is less than a certain limit, Euler’s formula gives a value of crippling stress greater than the crushing l stress which is wrong. In the limiting case, we can find the value of for which crippling stress is k equal to crushing stress. For example, for a mild steel column with both ends hinged. Crushing stress = 335 N/mm2 Young’s modulus, E = 200 GN/m2 @seismicisolation @seismicisolation 438 • Strength of Materials Equating the crippling stress to the crushing stress corresponding to the minimum value of slenderness ratio, we have Crippling stress = Crushing stress Crippling stress = = Crippling load P = Area A π 2E × A 2 le A k π 2E = 2 le k ∴ π 2E (For a column with both ends hinged 2 = 335 le k 2 π 2 × 200000 l = k 335 = 5886 l = 76.7, say 77 ∴ k le = l) Therefore, if the slenderness ratio is less than 80 for mild steel column (with both ends hinged), then Euler’s formula is not valid. Rankine’s Formula As we have discussed earlier that Euler’s formula gives correct results for only long columns, which fail mainly due to buckling. Short columns fail mainly due to directly crushing whereas medium columns which are neither long nor short fail by both buckling and direct crushing. Rankine derived an empirical formula based on practical experiments for determining the crippling load which is applicable to all columns irrespective of whether they are short or long. Let PC = Crushing load PR = Actual crippling load for column PE = Crippling load per Euler’s formula The Rankine hypothesis is 1 1 1 = + PR PC PE or 1 PE + PC = PR PC × PE @seismicisolation @seismicisolation Columns and Struts PC × PE PE + PC PC or PR = PC 1+ PE • 439 or PR = Substituting for PC = σC A and for PE = PR = or PR = ∴ PR = where a = σC A σC × A 1+ 2 π EI le2 π 2 EI in Eqn. (i) le2 σc A σc × A × le2 1+ 2 π × E × Ak2 σc A 2 le 1+a k σc ; π 2E (i) (This is known as Rankine formula) (Also known as Rankine-Gordon formula) a is Rankine constant σc and a both are constants in Rankine formula. The following table shows the values of σc and a for different materials. Material σc (MN/m2 ) Mild steel 320 Cast iron 550 Wrought iron 250 Timber 50 Aluminium 120 a= σc for both ends hinged π 2E 1 7500 1 1600 1 9000 1 750 1 5000 E XAMPLE 21.7: A rolled steel joist ISMB 300 is to be used as a column of 3.5 m length with both ends fixed. Find the safe load on the column taking factor of safety as 4, σc = 330 N/mm2 , 1 . Properties of the column section: a= 7500 Area = 5626 mm2 , Ixx = 8.603 × 107 mm4 , Iyy = 4.539 × 107 mm4 @seismicisolation @seismicisolation 440 • Strength of Materials S OLUTION : Since, IYY is less than IXX therefore, the column will tend to buckle about y-y axis Least moment of inertia of column section, I = 4.539 × 107 mm4 ∴ I = Ak2 k= k= I A 4.539 × 107 5626 = 89.82 mm Crippling load as given by Rankine formula, Pcr = For both fixed ends, le = σc A 2 le 1+a k l 3500 = = 1750 mm 2 2 Pcr = = 330 × 5626 1 1750 2 1+ 7500 89.82 1856580 1856580 = 1 + 0.0506 1.0506 = 1767161.6 N Allowing factor of safety 4, Safe load = 1767161.6 4 = 441790 N = 441.79 kN Ans E XAMPLE 21.8: A hollow cast iron pipe of length 4 m is used as a column. The external diameter is 60 mm and internal diameter is 35 mm. Determine the crippling load if both of its ends are fixed. 1 . Take σc = 540 N/mm2 and a = 1600 @seismicisolation @seismicisolation Columns and Struts S OLUTION : D, external diameter = 60 mm d, internal diameter = 35 mm π 2 (D − d 2 ) 4 π = (662 − 352 ) 4 Area = = 1864.375 mm2 Moment of inertia, π (604 − 354 ) 64 π = (12960000 − 1500625) 64 I= = 562225.6 mm4 I k= A 562225.6 = = 17.36 mm 1864.375 Crushing stress σc = 540 N/mm2 1 Rankine constant, a = 1600 Using Rankine formula, PR = For both ends fixed le = σc A 2 le 1+a k 4000 l = = 2000 mm 2 2 540 × 1864.375 P= 2000 2 1 1+ 1600 17.36 = 1006762.5 1 + 8.295 = 550293.8 = 550.29 kN Ans @seismicisolation @seismicisolation • 441 442 • Strength of Materials E XAMPLE 21.9: A hollow cylindrical cast iron column is 4.5 m long with both ends fixed. Determine the minimum diameter of the column if it has to carry a safe load of 350 kN with a factor of safety of 4. Take the internal diameter as 0.7 times the external diameter. Take σc = 540 N/mm2 and 1 in Rankine’s formula. a= 1600 S OLUTION : Length of column = 4.5 m = 4500 mm Equivalent length le for both ends fixed = ∴ 4500 l = 2 2 le = 2250 mm Safe load = 350 kN If external diameter = D Then internal diameter = 0.7D Crushing stress σc = 540 N/mm2 a= 1 1600 for cast iron Taking factor of safety into consideration Crippling load Area, P = 540 × 4 = 2160 kN π 2 A= D − (0.7 D)2 4 π 2 = D − 0.49 D2 4 = 0.1275 π D2 π 4 I= D − (0.7D)4 64 π 4 = D − 0.2401 D4 64 = 0.0119 π D4 But, I = A × k2 I k= A 0.0119 π D4 = 0.1275 π D2 = 0.3055 D @seismicisolation @seismicisolation Columns and Struts • 443 Now, using Rankine’s formula, P= 2160000 = 2160000 = σc A 2 le 1+a k 540 × 0.1275 π D2 2 1 2250 1+ 1600 0.3055D 216.189D2 33901.8 1+ D2 2160000 D2 = 2 216.189 D + 33901.8 D2 2160000D2 + 7.323 × 1010 = 216.189 D4 D4 − 9991.25D2 − 338731388 = 0 √ 9991.25 + 99825076 + 1354925552 D = 2 2 D2 = 9991.25 + 38141.2 2 D2 = 155.13 mm Ans Internal diameter, d = 155.13 × 0.7 = 108.59 mm Ans E XAMPLE 21.10: A column for a crane gantery consists of two ISMB 400 × 140 mm connected by two plates 10 mm thick as shown in Fig. 21.7. The length of column is 8 m and both ends are hinged. Taking a factor of safety of 6, calculate the safe load for the column using Rankine’s formula 1 . For ISMB 400 × 140 : Iyy = 422.11 cm4 area with failure stress σc = 320 MN/m2 and a = 7500 = 78.46 cm2 and Ixx = 20458.4 cm4 . @seismicisolation @seismicisolation 444 • Strength of Materials Y 320 mm 10 mm 400 mm X X 10 mm 480 mm Y Figure 21.7 S OLUTION : Iyy = 2 (IC.G + A x̄2 ) + 1 (48)3 × 1 12 = 2 422.11 + 78.46 × (16)2 + 48 × 48 × 48 12 = 59240 cm4 Ixx = 2 2058.4 + 48 (1)3 + (20.5)2 × 1 × 48 12 = 91124.8 cm4 Area of cross section = 78.46 × 2 + 48 × 1 × 2 = 252.92 cm2 Taking the least moment of inertia k= 59240 258.92 = 15.3 cm le = l = 8 m = 800 cm le 800 = = 52.29 k 15.3 @seismicisolation @seismicisolation Columns and Struts • 445 252.92 320 × × 106 A σc 100 × 100 6 P= 2 = 1 le (52.29)2 1+ 1+a 7500 k P= 1348907 = 988500 N 1.3646 = 988.5 kN Ans The following formulae are in addition to above, but are rarely used, and are mentioned for their academic importance only. 1. Johnson’s Parabolic formula P = σc A 1 − b le k 2 For mild steel, σc = 300 MPa and b = 0.00003 for pinned ends. 2. Straight line formula P l = σc − Z A k where Z is an empirical constant and where the range of application is strictly defined. Some of the approximate empirical formulae used in practical designing are given below: i) Stress at critical load for cast iron P = 23.8 − 0.6 A l N/mm2 k ii) Stress at critical load for structural steel P = 367.5 − 2 A l N/mm2 k Gordon’s formula P= σc A 2 l 1+a b where b is the least diameter or the width of the strut and a is the Gordon constant and is given by σc Ab2 a = . EI π 2 @seismicisolation @seismicisolation 446 • Strength of Materials Columns Subjected to Eccentric Loading (Secant Formula) (A) Euler’s Formula: AB is a column of length l subjected to an eccentric load P at eccentricity e. Here the top of column is free and bottom is fixed. a P e B y l X x A Figure 21.8 The bending moment at the section x from fixed end is given by d2y = P(a + e − y) dx2 P(a + e) d2y P + y= EI dx2 EI EI ∴ This is differentiation equation and the solution is given by Y = C1 cos x P +C2 sin x EI P + (a + e) EI The slope at any section is given by, dy = −C1 dx At A, x = 0 and y = 0 and also P sin x EI P +C2 EI P cos x EI dy =0 dx ∴ 0 = C1 + (a + e) P and 0 = C2 EI ∴ C2 = 0 and C1 = −(a + e) @seismicisolation @seismicisolation P EI Columns and Struts At B, x = l, y = a • 447 P + (a + e) EI P ∴ a = (a + e) 1 − cos l EI P ∴ (a + e) cos l =e EI P a + e = e sec l EI ∴ a = −(a + e) cos l The maximum bending moment for the column occurs at A and is equal to P (a + e) P ∴ Maximum bending moment = M = P.e sec l EI Therefore, the maximum compressive stress for the column section at A, P P.e sec l P EI σmax = σd + σb = + (i) A Z Z is section modulus. If both the ends are hinged Eqn. (i) can be modified as P le P.e sec P 2 EI σmax = σd + σb = + (ii) A Z because for a column with one end fixed and the other free, equivalent length le = 2l (as one end is fixed and the other is free). In general, for any end condition, formula can be written as P le Pe sec P 2 EI σmax = + (iii) A Z where le is equivalent length depending upon the end condition and is given in the earlier table before for ready reference. It may be noted that in the case of short columns (with no buckling) maximum moment is P.e le P which is increased to Pe sec in case of long columns. 2 EI Formula of Eqn. (iii) is also known as Secant formula for eccentric loads. @seismicisolation @seismicisolation 448 • Strength of Materials E XAMPLE 21.11: A circular hollow column having external diameter of 250 mm and internal diameter of 170 mm is having a length of 4.2 m with both ends fixed. The column is eccentrically loaded by a load of 220 kN with eccentricity 30 mm. Determine the extreme stresses on the column section. Also find the maximum eccentricity in order that there may be no tension anywhere in the section. Take E = 94 GN/m2 S OLUTION : π (0.252 − 0.172 ) 4 = 0.026376 m2 Area = Moment of inertia of the section, π (0.254 − 0.174 ) 64 = 1.507 × 10−4 m4 I= Equivalent length (both ends fixed) le = l 2 le = 4.2 = 2.1 m 2 Maximum bending moment le M = P.e. sec 2 P EI le P To calculate the angle 2 EI √ 2.1 220000 = 1.05 0.0155 2 94 × 109 × 1.507 × 10−4 = 0.131 radians 0.131 × 180 = 7.51◦ = π 1 sec 7.51◦ = cos 7.51 1 = 1.0087 = 0.9914 ∴ Maximum bending moment, Mmax = P · e · ×1.0087 = 220 × (0.03) × 1.0087 = 6.66 kNm @seismicisolation @seismicisolation Columns and Struts • 449 Maximum compressive stress, P M + ; P.e = 220 × 0.03 = 6.66 kNm A Z 6.66 × (0.125) 200 0.25 + = Because y = 0.026376 2 1.507 × 10−4 = 7582.6 + 5524.2 σmax = and Z= I y = 13106.8 kN/m2 = 13.11 MN/m2 Ans For no tension corresponding to the maximum eccentricity: P M = A Z P le P · e sec P 2 EI = A Z 220 × e × 1.0087 × 0.125 220 = 0.026376 1.507 × 10−4 1.507 × 10−4 e= 0.026376 × 1.0087 × 0.125 = 0.0453 m = 45.3 mm Ans Perry’s Formula for Eccentrically Loaded Column Prof. Perry has given a formula to determine the safe load of an eccentrically loaded column. This is known as Perry’s formula. Before deriving this formula which is limited, following assumptions are made: i) The formula gives the average value of stress in a column at failure. ii) The formula is applicable only when the column is subjected to both axial as well as eccentric loading. l yc iii) Prof. Robertson also performed experiments and found that when the value of 2 = 0.003 , k k Perry’s formula is valid for initial curvature as well as eccentric loading. ( yc = distance of extreme fibre from neutral axis.) In a column of effective length le subjected to a load P at an eccentricity of e, them bending moment = P · e σd = stream due to direct load P So σd = A @seismicisolation @seismicisolation 450 • Strength of Materials σmax = Maximum permissible stress le = Effective or equivalent length of the column. σb = Maximum compressive stress due to bending moment M Myc = = Z Ak2 P P · e sec l2e EI = · yc Ak2 π P P · e · yc = sec 2 2 PEuler Ak where PEuler = π 2 EI le2 P P · e · yc π σmax = + sec 2 A 2 Ak ∴ Substituting P = σd A σmax = σd As per Prof. Perry, sec ∴ ∴ π 2 π e yc 1 + 2 sec 2 k P PEuler P PEuler 1.2PEuler P = PEuler PEuler − P approximately PEuler Let σEuler = A 1.2 PEuler π P 1.2 σEuler sec = = 2 PEuler PEuler − P σEuler − σd σmax = σd 1 + eyc 1.2 PEuler · k2 PEuler − P σmax = σd 1 + eyc 1.2 σEuler · k2 σEuler − σd σmax −1 = σd σmax σEuler − σd −1 = σd σEuler σmax σ = −1 1− d σd σEuler eyc 1.2 σEuler · k2 σEuler − σd 1.2 e · yc k2 1.2 e · yc k2 This is known as Prof Perry’s formula. @seismicisolation @seismicisolation Columns and Struts • 451 E XAMPLE 21.12: A hollow mild steel column has an external diameter 35 cm and internal diameter 32 cm and is 4.5 m long. It is subjected to a vertical load P acting at an eccentricity of 5.5 cm when both the ends are fixed. The maximum compressive stress is limited to 3800 N/cm2 . Find the maximum value of P it can carry. Use Perry’s formula and take E = 2 × 107 N/cm2 . S OLUTION : l 450 = 225 cm. Effective length le = = 2 2 Cross-sectional area, π 2 35 − 322 4 = 157.785 cm2 A= Eccentricity of vertical load, e = 5.5 cm. Maximum compressive stream, σmax = 3800 N/cm2 Young’s modulus, E = 2 × 107 N/cm2 Moment of inertia, π 4 35 − 324 I= 64 = 22178.6 cm4 I = Ak2 k2 = ∴ k2 = I A 22178.6 157.785 = 140.56 cm2 PEuler = π 2 EI π 2 × 2 × 107 × 22178.6 = = 86388987.5 N le2 2252 σEuler = 86388987 157.785 = 547510.8 N/cm2 yc = D 35 = = 17.5 cm 2 2 Applying Perry’s formula, σmax σ 1.2 eyc −1 1− d = σd σEuler k2 σd 1.2 × 5.5 × 17.5 3800 −1 1− = σd 547510.8 140.56 @seismicisolation @seismicisolation 452 • Strength of Materials 3800 − σd σd 547510.8 − σd 547510.8 = 0.822 (3800 − σd ) (547510.8 − σd ) = 0.822 × σd × 547510.8 2080541040 − 547510.8 σd − 3800 σd + σd2 = 450 σd σd2 − 551760.8σd + 2080541040 = 0 √ 551760.8 ± 3.044 × 1011 − 8.32 × 109 σd = 2 551760.8 − 544132.3 σd = 2 = 3814.25 N/cm2 Hence, maximum value of P = 3814.25 × Area = 3814.25 × 157.785 = 601831 N = 601.8 kN Ans Exercise 21.1 A built-up beam shown in Fig. 21.9 is simply supported at its ends. Calculate its length, given that when it is subjected to a load of 40 kN/m length, it deflects by 10 mm. Find out the safe load, if this beam is used as a column with both ends fixed. Assume a factor of safety as 4. Use Euler’s formula, take E = 210 GN/m2 . 30 cm 5 cm 2 cm 100 cm 5 cm 30 cm Figure 21.9 [Ans @seismicisolation @seismicisolation l = 14.15 m, 2332.5 kN] Columns and Struts • 453 21.2 A hollow cylindrical cast iron column is 4 m long with both ends fixed. Determine the minimum diameter of the column, if it has to carry a safe load of 250 kN with a factor of safety of 1 in Rankine’s 5. Take the internal diameter as 0.3 times the external diameter. Take a = 1600 formula and σc = 550 N/mm2 . [Ans 108.8 mm] 21.3 A straight cylindrical bar has 16 mm diameter and is 1.2 m long. It is freely supported at its two ends in a horizontal position and loaded at the centre with a concentrated load of 90 N. The central deflection is found to be 5 mm. If placed vertical and loaded along its axis, what load would cause it to buckle? What is the ratio of the maximum stresses in the two cases? [Ans 4.444 kN, 3.04] 21.4 Calculate the diameter of the piston rod from the following data: Diameter of engine cylinder = 350 mm Maximum effective stress pressure in the cylinder = 700 kN/m2 Distance from piston to cross-head centre = 1.5 m Factor of safety = 4. σc = 330 MN/m2 1 a= for both ends fixed. 30, 000 [Ans 41.9 mm] 21.5 A strut 3 m long is constructed of steel tube 75 mm outside diameter and 3 mm thick. The ends are pin-jointed, but the end load of 50 kN is applied eccentrically through a line parallel to and 2.5 mm away from the axis of the strut, which is initially straight. Find the deflection and the maximum stress at the centre of length, E = 200 GN/m2 . [Ans 3.38 mm, 98.7 MN/m2 ] 21.6 A column having a section as shown in Fig. 21.10 is 3 m long. If the column is hinged at both ends, find the crippling load by using Euler’s formula. E = 200 GPa. Y 120 mm A B 16 mm y = 54.1 mm X X G 150 mm 16 mm Y Figure 21.10 [Ans 5.16.5 kN] @seismicisolation @seismicisolation 454 • Strength of Materials 21.7 A hollow column of external and internal diameters 150 mm and 110 mm, respectively and of length 2.5 m is hinged at top and bottom ends. A load of 80 kN is applied at an eccentricity of 50 mm from the geometrical axis of the column. Determine the maximum stress in the column section and also calculate the maximum eccentricity so that no tensile stress develops in the column. Take E = 200 GPa. [Ans 26.78 N/mm2 , 28.81 mm] 21.8 Find (Euler’s critical load) for a hollow cylindrical cast iron column 200 mm external diameter and 25 mm thick, 5 m long. The column is hinged at both ends. Take E = 80 GPa. For what length the critical load by Euler’s and Rankine’s formula be equal? Take 1 . σc = 600 N/mm2 and a = 1600 [Ans 1692.8 kN, 5.3 m] 21.9 A circular rod of diameter 40 mm and length 800 mm is subjected to an axial compressive load of 100 kN at an eccentric distance of e from the axis of the rod, Assuming the lateral deflection at the midpoint of the rod to be 0.8 mm, find the distance of eccentricity, e and the maximum stress in the rod. Take E = 200 GPa. [Ans 0.59 mm, 92.33 N/mm2 ] 21.10 A 2 m long column has a circular cross section of 60 mm diameter. One of the ends of the column is fixed in direction end position and the end is free. Taking a factor of safety as 3, calculate the safe load using: i) Rankine’s formula. Take yield stress = 550 N/mm2 and a = ii) Euler’s formula, E for material of column = 130 GPa 1 for pinned ends. 1600 [Ans 11.4 kN, 170 kN] 21.11 A hollow cast iron column with fixed ends supports an axial load of 1 MN. If the column is 4.5 m long and has an external diameter of 250 mm, find the thickness of metal required. Use 1 the Rankine’s formula, taking a constant of for pinned ends and a working stress of 80 6400 2 MN/m . [Ans 28 mm] 21.12 A pin-ended square cross-section column of length 3 m is subjected to a compressive stress of 10 MPa. Using a factor of 2.5, find the cross section if the column is to safely support (a) a 100 kN load and (b) a 200 kN load. Take E = 15 GPa. [Ans 117 × 117 mm cross-section] 21.13 A 1.2 m long column has a circular cross section of 5 mm diameter. One of the ends of the column is fixed in the direction and position and other end is free. Taking factor of safety 1 as 3, calculate the safe load using (i) Rankine’s formula with σC = 560 N/mm2 and a = 1600 for pinned ends and (ii) Euler’s formula with E for C.I = 120 GPa. [Ans i) 9.9 kN, ii) 13.45 kN] 21.14 A straight bar, 1 m long, 25 mm wide and 25 mm thick is bent in the shape of a bow, the elastic deflection at the middle being 50 mm. Its ends are joined by a bowstring. Assuming the modulus of elasticity, E = 200 GPa, determine the force in bow string and the maximum stress in the bar. [Ans P = 0.645 N, σmax = 125 N/mm2 ] @seismicisolation @seismicisolation Columns and Struts • 455 21.15 An I-joist ISMB 250 @ 37.3 kg/m has an effective length of 5 m. It is used as a stanchion with two plates 250 mm × 10 mm welded to it sides, as shown in Fig. 21.11. Compare the load carrying capacity. What will be its load capacity if one plate is attached to each flange? For ISMB 250 @ 37.5 kg/m, a = 47.55 cm2 , Ixx = 5137.6 mm4 , Iyy = 334.5 × 104 mm4 10 mm 10 mm 250 mm 125 mm Figure 21.11 [Ans @seismicisolation @seismicisolation 814.25 kN, 876.2 kN] 22 C HAPTER BENDING OF CURVED BARS M σ E So far we have dealt with the bending of straight bars using equation = = . But for bending I Y R curved bars like crane hooks, chain links, rings, etc., simple bending equation cannot be used. Actually curved bars mean the bars with small or large initial curvature. We have to make certain assumptions while deriving formulas for curved bars, these are stated as follows: 1. Each layer of the beam is free to expand or contract independently of the layer above or below it. 2. The bar material obeys Hooke’s law and is loaded within elastic limit. 3. The longitudinal fibres of the bar, parallel to centroidal axis exert no pressure on each other. In other words, the distance between longitudinal fibre from centroidal axis is same before and after bending. 4. The transverse sections which are plane before bending, remain plane after bending. 5. The value of Young’s modulus E is the same in tension and compression. 6. Radial strain is neglected. Beams with Large Radius of Curvature (or Small Curvature) Consider a curved bar as shown in Fig. 22.1(a) having Ri as its initial radius of curvature and R1 as radius of curvature after application of end moments, M. y C N D A Ri θ+δθ (b) o D' A' Ri M θ (a) y C' N' Figure 22.1 Let Ri = Initial radius of curvature of the bar R1 = Final radius of curvature @seismicisolation @seismicisolation o M Bending of Curved Bars • 457 θ = Initial angle subtended at the centre of the bar θ + δ θ = Final angle after bending. Change in length After bending strain, ε = Original length C D −CD CD (R +y)(θ + δ θ ) − (Ri +y)θ = 1 (Ri −y)θ = ε= R1 (θ + δ θ ) + yδ θ − Ri θ (Ri +y)θ (i) As shown in Fig. 21.1(b), NA = Ri θ and N A = R1 (θ + δ θ ) and, N A = NA As we know the length of neutral axis NA will remain same, so NA = N A Ri θ = R1 (θ + δ θ ) = R1 θ + R1 δ θ Ri δ θ = θ (Ri −R1 ) (R −R1 ) δθ = i θ R1 (ii) Substituting for Ri θ in Eqn. (i) ε= = Now put the value of R1 (θ + δ θ ) + yδ θ − R1 (θ + δ θ ) (Ri − y)θ yδ θ (Ri − y)θ (iii) δθ from Eqn. (ii) in Eqn. (iii) θ ε= = Ri −R1 y × Ri + y R1 y(Ri −R1 ) R1 Ri Since, y is very small compared Ri and R1 , so negligible 1 1 − Hence, ε = y R1 Ri 1 σ 1 − =y E R1 Ri @seismicisolation @seismicisolation 458 • Strength of Materials Therefore, Hence, 1 1 − R1 Ri M 1 1 σ = =E − y I R1 Ri σ =E y E XAMPLE 22.1: A round steel bar of 55 mm in diameter is bent into a circular arc of 3.8 m radius and the subtended angle is 85◦ . A couple is applied at each end so that the slope is changed to 90◦ at one end relative to other. If E is 200 GN/m2 , determine the maximum stress due to couple. S OLUTION : Diameter of bar = 55 mm Initial radius of curvature = 3.8 m = 3800 mm Initial angle, θ = 85◦ Final angle (θ + δ θ ) = 90◦ E = 200 GN/m2 = 200 × 103 N/mm2 55 Distance between outermost layer of bar and neutral axis = = 27.5 mm and δ θ = 90 − 85 = 5◦ 2 R1 = Final radius of arc Now δθ (Ri − R1 ) = θ R1 3800 − R1 5 = 85 R1 5R1 = 323000 − 85R1 90R1 = 323000 323000 = 3588.8 mm 90 1 1 σ = E ·y − R1 Ri R1 = Therefore, σmax = 200000 × 27.5 1 1 − 3588.8 3800 = 5500000 2.79 × 10−4 − 2.63 × 10−4 = 5500000 0.16 × 10−4 = 88 N/mm2 Ans @seismicisolation @seismicisolation Bending of Curved Bars • 459 Beam with Small Initial Radius of Curvature (or Large Curvature) If the radius of curvature is more than 5 times the depth of the beam, a simple flexure formula can be used. But analysis for beams having small initial radius of curvature was first dealt with by Winkler and later by Andrews and Pearson. The following derivation is known as Winkler-Bach theory: The beam ABCD is having an initial radius R. EF is its centroidal axis and NA is neutral axis on NA, there will be no change in length when moment M is applied. After application of moment M, A and B are new positions of points A and B such that A O B subtends an angle φ . Here, M is considered positive which tends to increase the curvature of the beam. EF is centroidal axis. Let PQ be any fibre at a distance of y from centroidal axis. From Fig. 22.2 it is understood that after moment is applied ABCD takes the form A B C D . Note: Same assumptions as stated in beam with small curvature are applicable here. A A' C' M G H y F E E' N C D' A F' C' φ o' B R' B' H' M θ R o Figure 22.2 Now strain in fibre GH, G H − GH GH (R + y) φ (R + y) φ − (R + y) θ = −1 = (R + y) θ (R + y) θ ε= It can be written as Strain in fibre EF = ε = R +y φ = (1 + ε ) θ R+y (i) E F − EF R φ R φ − Rθ = −1 = EF Rθ Rθ or R φ = 1+ε θ R @seismicisolation @seismicisolation (ii) 460 • Strength of Materials From Eqns. (i) and (ii) (1 + ε ) R+y R = 1 + ε R + y R R R + y (1 + ε ) = 1 + ε × R R+y R + y/R ε = 1+ε −1 R + y/R 1 + y/R ε = 1+ε −1 1 + y/R or Adding and subtracting ε on right-hand side, ε = (1 + ε ) 1 + y/R −1−ε +ε 1 + y/R 1 + y/R − 1(1 + ε ) + ε 1 + y/R 1 + y/R −1 +ε = (1 + ε ) 1 + y/R 1 + y/R − 1 − y/R +ε = (1 + ε ) 1 + y/R ⎤ ⎡ 1 1 ⎢ y R − R ⎥ ⎥ = ε + (1 + ε ) ⎢ ⎣ 1 + y/R ⎦ = (1 + ε ) 1 1 − = ε + (1 + ε ) R R σ = E ε = E ε + (1 + ε ) y 1 + y/R 1 y 1 + 1 + y/R R R (iii) Total force on cross section C A or D B in the circumferential direction, σ dA y = E ε dA (∵ σ = ε E) y 1 1 =E ε dA + (1 + ε ) + dA 1 + y/R R R 1 1 y F = E ε A + E(1 + ε ) + dA R R 1 + y/R F= @seismicisolation @seismicisolation (iv) Bending of Curved Bars • 461 The moment of resistance about an axis through the centroid, M= σ dA · y = E ε y dA + =E ε y dA (1 + ε ) y2 1 + y/R 1 1 − R R dA Since y = 0 at the centroidal axis end y dA = 0 ∴ and 1 1 + M = E(1 + ε ) R R y2 dA = 1 + y/R y2 dA 1 + y/R (v) yR + y2 − yR dA 1 + y/R yR (1 + y/R) − yR yR dA = yR − dA 1 + y/R 1 + y/R y y dA = R ydA − R dA y− =R 1 + y/R 1 + y/R = = 0−R yR dA = −R 1 + y/R let Now let us find out the value of As we assumed yR dA 1 + y/R y2 dA = Ah2 (1 + y/R) y dA (1 + y/R) y2 dA = Ah2 (1 + y/R) Now, y2 2 y + − y /R y R dA = dA (1 + y/R) (1 + y/R) y y2 − y 1 + R R dA = (1 + y/R) y2 1 = dA y− × R (1 + y/R) @seismicisolation @seismicisolation (vi) 462 • Strength of Materials 1 y2 × dA R (1 + y/R) 1 y2 = 0− y dA = 0 dA ∵ R (1 + y/R) 1 y2 2 = − Ah dA = Ah2 ∵ R (1 + y/R) = y dA − Now Eqn. (iv) becomes, 1 1 h2 F = EA ε − (1 − ε ) − R R R As net force F = 0 ∴ ε = (1 + ε ) 1 1 − R R h2 R Putting this value of ε in Eqn. (iv) or 1 1 h2 y 1 1 σ = E (1 + ε ) − − + (1 + ε ) R R R 1 + y R R 1 1 1 2 yR − h + = E(1 + ε ) R R R 1 + y/R y2 1 1 dA − Also we know, M = E(1 + ε ) R R (1 + y/R) 1 1 ∴ M = 1−ε Ah2 − R R 1 M 1 = 2 E 1 + ε − R R Ah M 1 2 yR2 σ = 2. h + R+y Ah R M R2 y = 1+ 2 AR h R +Y (vii) At inside of the centroidal axis, y is negative thus, σ= M R2 y 1− 2 AR h R−y @seismicisolation @seismicisolation (viii) Bending of Curved Bars • 463 The position of neutral axis can be found out because we know σ = 0 at NA. M R2 y 1+ 2 =0 ∴ σ= AR h R+y yR2 = −Rh2 − yh2 or or y= −Rh2 R2 + h2 (ix) Note: 1. If bending moment M tends to decrease the curvature, then Eqn. (vi) will be compressive stress and Eqn. (vii) will be tensile stress. 2. The bending moment is positive when it decreases the radius of curvature and negative when it increases the radius of curvature. Therefore, σ should be positive if it is tensile in nature. 3. y is positive if it is measured away from the centre of curvature and negative when measured towards the centre of curvature. Crane Hooks: (Inside) (Outside) W M R2 y1 σx1 = + 1− 2 A AR h (R − y1 ) W M R2 y2 σx2 = − 1+ 2 A AR h (R + y2 ) Centre of curvature X2 G X1 O y2 W y1 x R Figure 22.3 Values of h2 for Various Sections: h is known as link radius. h2 = we know, 1 A y2 dx 1 + y/R R y2 dx A R+y 2 R R dA = ydA− RdA + A R+y 2 R dA R 0 − RA + = A R+y = ∴ h2 = R2 A dA − R2 R+y @seismicisolation @seismicisolation 464 • Strength of Materials Y Rectangular Section A rectangular section with centre of curvature 0 lying on Y -Y axis is shown in Fig. 22.4. Consider an elementary stripe of width b and depth dy at a distance y from the centroidal axis. Area of the strip, dA = b dy Area of section, A = b d R3 h = bd +d/2 2 −d/2 dA dy y X b b.dy − R2 R+y R3 R Axis of curvature O Centre of curvature +d/2 [loge (R + y)]−d/2 − R2 d 2R + d R3 2 loge − R2 h = d 2R − d 2R + d 2 2 R or h = R loge −1 d 2R − d = X d G Figure 22.4 Rectangular section b Circular Section dy π A = d2 4 Considering a stripe of width b and the depth dy at a distance y from the centroidal layer as shown in Fig. 22.5 2 d b=2 − y2 2 Area of strip, dA = b.dy 2 d =2 − y2 · dy 2 R3 h = bd +d/2 2 −d/2 8R3 = π d2 d2 R O Axis of curvature Centre of curvature Figure 22.5 Circular section d 2 2× − y2 2 .dy − R2 R+y d +d/2 −d/2 y 2 − y2 4 dy − R2 R+y @seismicisolation @seismicisolation Bending of Curved Bars • 465 Expanding binomial expression and then integrating, we have h2 = d2 d4 +... + 16 128 R2 Triangular Section b' R+y = a dy = da Width of elementary strip b = R2 −a b d G b Area of elementary strip, dy d R1 R R2 R3 = A R1 O dA −R2 R+y R2 b da R1 a Centre of curvature Figure 22.6 Triangular section −R2 da R2 −a b. −R2 d a R1 ⎡ ⎤ R2 R2 3 R ⎣ R2 b da b − = da⎦ −R2 A d a d R1 R1 3 R R2 b R2 b loge − (R2 −R1 ) −R2 = A d R d 3 R 2h b 3R − 2d = d+ loge −b −R2 A 3 d 3R − d R3 = A 3 Axis of curvature Now, R3 h = A d R2 dA = b .dy = b .da 2 y R2 Because, d 3 d R1 = R− 3 R2 − R1 = d R2 = R+ @seismicisolation @seismicisolation 466 • Strength of Materials 3R + 2d b 2R3 (3d + 2d) loge − b −R2 bd 3d 3R − d 1 Because A = bd 2 3 3R + 2d 2R 3R + 2d 2 loge −1 −R2 h = d 3d 3R − d ∴ or h2 = Trapezoidal Section Figure 22.7 shows a trapezoidal section. Consider an elementary strip of width b and depth dy at distance y from centroidal axis. r = R + y ∴ dr = 0 + dy OR dr = dy dy b1 +b2 y A= ×d 2 R2 Area of stripe, dA = b × dy b1 +b2 (R2 −r) where b = b2 + r R1 d b1 −b2 , (R2 −r) dy dA = b2 + d b1 −b2 (R2 −r) dr {∴ = b2 + d Now b2 d b G b1 R Axis of curvature O dy = dr} Centre of curvature Figure 22.7 R3 dA −R2 A (R + y) b1 −b2 (R2 −r) dr b + 2 R3 d = −R2 A (R + y) h2 = R3 b2 + (b1 −b2 ) (R2 −r) dr −R2 A r R3 b2 (b1 −b2 ) ×R2 (b1 −b2 ) ×r = + − dr − R2 A r d ×r d ×r ⎡ ⎤ R2 3 R ⎣ b1 −b2 (b1 −b2 ) R2 ⎦ 2 b2 + ×R2 (log e r)RR21 − (r)R1 −R = A d d R1 3 R b1 −b2 R2 b1 −b2 = b2 + ×R2 log e r (R2 −R1 ) −R2 − A d R1 d R3 (b1 −b2 ) R2 2 Hence, h = b2 + ×R2 loge − (b1 −b2 ) −R2 A d R1 = [∴ R2 −R1 = d] h2 = @seismicisolation @seismicisolation Bending of Curved Bars T-Section • 467 b2 Let its initial radius of curvature be r Then, r = R + y ∴ dy t2 dr = 0 + dy R3 h = A 2 dA −R2 R+y y G t1 Let T section be made of two rectangles b1 × t1 R3 Axis of and b2 ×t2 . curvature Therefore, ⎡ ⎤ R2 R2 R3 ⎣ b1 ×dr b2 ×dr ⎦ 2 2 −R h = + A r r R1 R1 R2 R2 R3 = × b1 loge +b2 loge −R2 A R1 R2 where, A = b1t1 +b2t2 r b1 R2 R O R1 Centre of curvature Figure 22.8 I-Section t3 b3 r = R+y dr = 0 + dy dr = dy dy R3 1 dA − R2 h = A R+y R3 dA = −R2 A r 2 y G b2 r R4 R3 t1 b1 R2 R1 R O Axis of curvature Centre of curvature Figure 22.9 I section is made up of three rectangles: b1t1 , b2t2 and b3t3 ⎡ ⎤ R R R 3 2 dA 3 dA 4 dA R ⎣ ⎦ −R2 Now, h2 = + + A r r r R1 R2 R3 ⎡ ⎤ R R 3 2 R4 R3 ⎣ b1 dr b2 dr b3 dr ⎦ 2 = + + −R A r r r R1 R2 R3 3 R2 R3 R4 l + b2 loge + b3 loge − R2 = b1 loge A R1 R2 R3 @seismicisolation @seismicisolation t2 468 • Strength of Materials Total area of cross section of the beam is A = b1 t1 + b2 t2 + b3 t3 E XAMPLE 22.2: A curved bar of square section 85 mm sides and mean radius of curvature 125 mm is initially unstressed if a bending moment of 800 Nm is applied to straighten it then determine the stresses that develop at the inner and the outer faces of the bar. S OLUTION : 85 d = 85 mm Area of cross section, A = 852 = 7225 mm2 Radius of curvature R = 125 mm Applied bending moment = −800 Nm 85 mm Axis of curvature 125 mm = −800 × 103 Nmm (negative sign is because curvature is being decreased) R3 2R + d 2 Now, h = loge − R2 d 2R − d 1253 2 × 125 + 85 = loge − 1252 85 2 × 125 − 85 Figure 22.10 325 − 15625 165 = 22978 × −15625 = 22978 loge = 643.4 mm2 M y R2 Now σ = 1+ 2 × AR R+y h For outer surface, 85 = 42.5 mm 2 800 × 103 1252 42.5 σ =− 1+ × 7225 × 125 643.4 125 + 42.5 y= = −0.886 [1 + 6.16] = −6.34 MN/m2 Stress at inner surface: 85 = 42.5 mm 2 M y R2 σ= 1− 2 × AR R−y h y= @seismicisolation @seismicisolation (Compressive) Ans Bending of Curved Bars =− • 469 1252 42.5 800 × 103 1− × 7225 × 125 643.4 125 − 42.5 = −0.886 [1 − 12.51] = +10.19 MN/m2 (Tensile) Ans E XAMPLE 22.3: A beam of circular section of diameter 18 mm has its centre line curved to a radius of 44 mm. Find the maximum stresses in the beam when subjected to 4.5 kNmm moment. S OLUTION : Diameter, d = 18 mm Radius of curvature, R = 44 mm Bending moment = 4.5 kNmm = 4500 Nmm π Area = (18)2 = 254.34 mm2 4 Distance between centre line and extreme fibre, 18 = 9 mm 2 d2 d2 h2 = + 16 128R2 182 182 = + 16 128(44)2 y= = 20.25 + 0.00131 = 20.251 mm2 Maximum stress at bottom surface R2 y M 1+ 2 σ= AR h (R + y) 4500 442 × 9 = 1+ 254.34 × 44 20.251(44 + 9) = 0.4021 (1 + 16.23) = 6.93 MN/m2 (Tensile) Ans Maximum stress at top surface, M R2 y σ= 1− 2 AR h (R − y) @seismicisolation @seismicisolation 470 • Strength of Materials = 4500 (1 − 11.31) 254.34 × 44 = 0.4021 (−10.31) = −4.14 MN/m2 (Compressive) Ans E XAMPLE 22.4: A curved bar with a mean radius of curvature of 140 mm is initially unstressed and is 70 mm wide by 86 mm deep rectangular section in the plane of bending. To straighten it a bending moment of 2.4 kNm is applied. Determine the position of neutral axis and also the maximum bending stresses. S OLUTION : b = 70 mm; d = 86 mm R = 140 mm; M = 2.4 kNm = −2.4 × 106 Nmm (negative sign is taken as bending moment tends to decrease curvature) Area of cross-section, A = bd = 70 × 86 = 6020 mm2 2R + d R loge −1 h2 = R2 d 2R − d 2 × 140 + 86 140 = 1402 loge −1 86 2 × 140 − 86 = 1402 [1.628 loge 1.89 − 1] = 19600 [0.0363] = 711.48 mm2 Position of neutral axis: y=− = Rh2 R2 + h2 −140 × 711.48 (140)2 + 711.48 =− 99607.2 20311.5 = −4.9 mm Ans (−ve sign means the neutral axis is on compression zone (i.e., concave side) of the centroidal axis) maximum bending stresses: @seismicisolation @seismicisolation Bending of Curved Bars • 471 86 = 43 mm 2 R2 M y 1− 2 · σ max = AR h R−y 2.4 × 106 1402 43 =− 1− × 6020 × 140 711.48 140 − 43 Inner side of curve concave side y = = −2.848 (1 − 12.21) = +31.93 MN/m2 (Tensile) Ans Outer side (convex side) of curve: R2 y H 1+ 2 σ =− AR h R+y 2.4 × 106 1402 43 =− 1+ × 6020 × 140 711.48 140 + 43 = −2.848 (1 + 6.47) = −21.27 MN/m2 (Compressive) Ans E XAMPLE 22.5: Figure 22.11 shows a frame of rectangular cross section subjected to a load of 3 kN. Find: i) the resultant stresses at points X and Y and ii) position of neutral axis. 3 kN 130 mm X Y 52 mm 3 kN 20 G 50 mm Figure 22.11 @seismicisolation @seismicisolation 472 • Strength of Materials S OLUTION : Area cross section at X-Y A = 20 × 50 = 1000 mm2 = 1 × 10−3 m2 (m) Bending moment = −3 × 103 (130 + 50) × 10−3 = −540 Nm (Negative sign because moment tends to decrease the curvature.) (i) Resultant stresses at points X and Y : 3 × 103 = 3 MN/m2 (Tensile) 1 × 10−3 2R + d 2 2 R h =R loge −1 d 2R − d Direct stress = R = 52 mm = 0.052 m, d = 50 mm = 0.05 m 0.052 2 × 0.052 + 0.05 h2 = (0.052)2 loge −1 0.05 2 × 0.052 − 0.05 0.154 −3 = 2.7 × 10 1.04 loge −1 0.054 = 2.7 × 10−3 [1.04 × 1.048 − 1] = 2.43 × 10−4 m2 Bending stress due to bending moment at Y : y = 0.05 = 0.025 2 M R2 y σby = 1− 2 . AR h R−y (0.052)2 0.025 −540 1− × = 1 × 103 × 0.052 2.43 × 10−4 0.052 − 0.025 = −10.385 [1 − 10.303] = +96.61 MN/m2 (Tensile) Bending stress due to bending moment at X : y = Ans 0.05 = 0.025 2 M y R2 1+ 2 · AR h R+y (0.052)2 0.025 −540 1 + × = 3 −4 0.052 + 0.025 1 × 10 × 0.052 2.43 × 10 = −10.385[1 + 3.61] σbx = = −47.87 MN/m2 (Compressive) @seismicisolation @seismicisolation Ans Bending of Curved Bars • 473 Resultant stress at point Y , σY = 3 + 96.61 = 99.61 MN/m2 (Tensile) Ans Resultant stress at point X, σX = 3 − 47.87 = 44.87 MN/m2 (Compressive) Ans (ii) Position of neutral axis: Y =− Rh2 R2 + h2 =− 0.052 × 2.43 × 10−4 (0.052)2 + 2.43 × 10−4 =− 0.12636 × 10−4 27.04 × 10−4 + 2.43 × 10−4 = −0.00429 m = −4.29 mm Negative sign means neutral axis is at 4.29 mm away from centroidal axis towards the centre of curvature. Ans E XAMPLE 22.6: Figure 22.12 shows a circular ring of rectangular section, with a slit and subjected to load P. Calculate the magnitude of the force P if the maximum stress along the section 1-2 is not to exceed 250 MN/m2 . 55 mm P G 85 2 G 1 m 10 m R 1 1 m 55 m 200 A Figure 22.12 Area at section 1 − 2, A = 0.055 × 0.085 = 4.675 × 10−3 m2 Allowable stress, σ = 250 MN/m2 @seismicisolation @seismicisolation 474 • Strength of Materials Bending moment M = P(0.155) = 0.155 PNm (M is taken +ve because it tends to increase the curvature) Magnitude of the force P: P P = = 213.9 PN/m2 A 4.675 × 10−3 R 2R + d loge −1 h2 = R2 d 2R − d Direct stress = R = 0.155 m, d = 0.085 m 2 × 0.155 + 0.085 2 0.155 2 loge −1 h = (0.155) 0.085 2 × 0.155 − 0.085 0.395 −1 = 0.024 1.8235 loge 0.225 = 0.024 [1.0262 − 1] = 6.28 × 10−4 m2 Resultant stress at point 2, y = 0.085 = 0.0425 2 M y R2 P σmax = 1− 2 · + (Compressive) AR A h R−y 0.1552 P 0.155P 0.0425 1 − 250 × 106 = + × −3 −4 0.155 − 0.0425 4.675 × 10 × 0.155 6.28 × 10 4.675 × 10−3 250 × 106 = 213.9P [1 − 14.45] + 0.214 × 103 P = −2876.9P − 214P = 3090.9P ∴ P= 250 × 106 3090.9 = 80882.6 N = 80.88 kN Ans E XAMPLE 22.7: A crane hook is shown in Fig 22.13, lifting a load of 160 kN. Determine the maximum compressive and tensile stresses in the critical section A-B of the crane hook. @seismicisolation @seismicisolation Bending of Curved Bars • 475 B O G b1 = 130 A b2 = 40 d = 175 d2 d1 10 0 Section along AB All dimensions are in mm 160 kN Figure 22.13 S OLUTION : b1 = 130 mm = 0.13 m b2 = 40 mm = 0.04 m d = 175 mm = 0.175 m d b1 + 2b2 for locating centre of gravity, d1 = 3 b1 + b2 Centre of gravity may be found out by splitting the section as usual or use the above formula. 0.175 0.13 + 2 × 0.04 d1 = 3 0.13 + 0.04 = 0.072 m d2 = 0.175 − 0.072 = 0.103 m, A= 0.175 (0.13 + 0.04) 2 = 0.014875 m2 = 0.0149 R = 0.1 + 0.072 m = 0.172 mm, R1 = 0.1 m, R2 = 0.1 + 0.175 = 0.275 m R2 R3 b1 − b2 2 b2 + × R2 loge h = − (b1 − b2 ) − R2 A d R1 (0.13 − 0.04) 0.275 0.1723 2 0.04 + × 0.275 loge − (0.13 − 0.04) h = 0.0149 0.175 0.1 − 0.1722 = 0.3415 [{0.04 + 0.514 × 0.275} loge 2.75 − (0.09)] − 0.0296 = 0.3415 [0.18135 × 1.012 − 0.09] − 0.0296 = 0.00234 m2 Bending moment = −160 × 103 × 0.172 = −27520 Nm (Negative sign is taken because bending moment is tending to decrease the curvature.) @seismicisolation @seismicisolation 476 • Strength of Materials 160 × 103 × 10−6 = 10.74 MN/m2 (Tensile) 0.0149 Bending stress at A, R2 M y 1+ 2 × σbA = AR h R+y 27520 0.1722 0.103 =− 1+ × 0.0149 × 0.172 0.00234 0.172 + 0.103 Direct stress, σbA = = −10738255 [1 + 4.735] ×10−6 = −61.85 MN/m2 (Compressive) Ans. Bending stress at B, M y R2 σbB = 1+ 2 × AR h R−y 27520 0.1722 0.172 =− 1+ × × 10−6 0.0149 × 0.172 0.00234 0.172 − 0.072 = −10738255 [1 − 9.1] ×10−6 = 86.98 MN/m2 (Tensile) Ans E XAMPLE 22.8: When a curved beam of trapezoidal section of bottom width 35 mm, top width 24 mm and height 44 mm is subjected to pure bending moment of +650 Nm. The bottom width is towards the centre of curvature. The radius of curvature is 54 mm and beam is curved in a plane parallel to depth. Determine i) location of neutral axis and ii) Maximum and minimum stresses. Also plot the variation of stresses across the section. S OLUTION : 55.53 24 d2= 23.36 mm Centroidal Axis N d1= 20.64 mm G 44 mm A 35 mm R = 54 mm 15 10 5 42.58 33.21 22.26 A 9.27 N 6.37 5 25.55 10 49.66 15 85.47 Stresses are in MPa Figure 22.14 @seismicisolation @seismicisolation Bending of Curved Bars • b1 = 35 mm, b2 = 24 mm, d = 44 mm M = +650 Nm = 650000 Nmm; R = 54 mm b1 + 2b2 d 35 + 2 × 24 44 d1 = = × = 1.407 × 14.67 b1 + b2 3 35 + 24 3 = 20.64 mm R1 = R − d1 = 54 − 20.64 = 33.36 mm R2 = R + d2 = 54 + 23.36 = 77.36 mm Area, 44 = 1298 mm2 A = (35 + 24) 2 R3 b1 − b2 R2 2 h = b2 + × R2 loge − (b1 − b2 ) − R2 A d R1 35 − 24 77.36 543 24 + × 77.36 loge − (35 − 24) − 542 = 1298 44 33.36 = 121.31 [{24 + 0.25 × 77.36} 0.842 − 11] − 2916 = 121.31 [{43.34} 0.842 − 11] − 2916 = 176.47 mm2 i) Location of neutral axis: Let y = distance of neutral axis from centroidal axis. The location of neutral axis is given by, y=− =− R × h2 54 × 176.47 =− 2 2 2 R h 54 + 176.47 9.529.4 = −3.08 mm 3092.47 Ans −ve sign means the neutral axis is below the centroidal axis at a distance of 3.08 mm. ii) Stresses: y M R2 σ= 1− 2 RA R−y h For max stress, y = 20.64, h2 = 176.77 mm2 M = +650 Nm = 650000 Nmm R = 54 mm, A = 1298 mm2 650000 542 20.64 = 1− 54 × 1298 176.47 54 − 20.64 @seismicisolation @seismicisolation 477 478 • Strength of Materials = 9.27 [1 − 10.22] = −85.47 MN/m2 = 85.47 MN/m2 (Compressive) Ans Minimum stress occur at y = 23.36 mm 542 23.36 650000 1+ 54 × 1298 176.47 54 + 23.36 = 9.27 [1 + 4.99] σmin = = 55.53 MN/m2 (Tensile) Ans iii) Plotting the stresses across the section General formula: R2 y M 1+ 2 R×A R+y h 650000 542 y = 1+ 54 × 1298 176.47 54 + y y = 9.27 1 + 16.53 54 + y σ= At y = 23.36, σ = +55.53 MN/m2 At y = 15 mm, σ = 9.27 1 + 16.53 15 = 42.58 MN/m2 54 + 15 10 At y = 10 mm, σ = 9.27 1 + 16.53 = 33.21 MN/m2 54 + 10 5 At y = 5 mm, σ = 9.27 1 + 16.53 = 22.26 MN/m2 54 + 5 0 = 9.27 MN/m2 At y = 0 mm, σ = 9.27 1 + 16.53 54 + 0 −5 At y = −5 mm, σ = 9.27 1 + 16.53 = −6.37 MN/m2 54 − 5 −10 = −25.55 MN/m2 At y = −10 mm, σ = 9.27 1 + 16.53 54 − 10 −15 = −49.66 MN/m2 At y = −15 mm, σ = 9.27 1 + 16.53 54 − 15 y = −20.64 mm σ = −85.47 MN/mm2 (already calculated) @seismicisolation @seismicisolation Bending of Curved Bars • 479 E XAMPLE 22.9: Fig. 22.15 shows an open ring having T -section. Determine the stresses at the points X and Y . If the ring in subjected to a load of 160 kN. 160 kN t2=220 mm A R b1 = 160 mm X b2 = 35 mm y2 5 27 Y y1 R1 B t = 32 mm 1 R2 R3 160 kN Figure 22.15 Area of T section = b1 t1 + b2 t2 = A A = 160 × 32 + 35 × 220 A = 5120 + 7700 A = 12820 mm2 = 0.01282 m2 Let position of centre of gravity from AB = y1 y1 = 160 × 32 × 16 + 35 × 220 × 142 = 91.7 mm = (0.0917 m) 12820 y2 = (220 + 32) − 91.7 = 160.3 mm = (0.1603 m) R3 R2 R3 2 h = + b2 loge b1 loge − R2 A R1 R2 R1 = 275 mm = 0.275 m R2 = 275 + 32 = 307 mm = (0.307 m) R3 = 275 + 32 + 220 = 527 mm = (0.527 m) R = 275 + 91.7 = 366.7 mm = (0.3667 m) (0.3667)2 0.307 0.527 2 h = 0.160 loge + 0.035 loge − (0.3667)2 0.01282 0.275 0.307 = 3.846 [0.0181 + 0.0019] − 0.1344 = 0.00829 m2 @seismicisolation @seismicisolation 480 • Strength of Materials 160000 × 10−6 = 12.48 MN/m2 (Compressive) 0.01282 Bending moment = M = +P × R (+ve sign is taken because bending moment is tending to increase the curvature) Bending stress at X, M R2 y2 (σb )X = 1+ 2 AR R + y2 h (0.3667)2 0.1603 160000 × 0.3667 1+ = 0.01282 × 0.3667 0.00829 (0.3667 − 0.1603) Direct stress = = 12480499.2 [1 + 12.6] = 169.7 MN/m2 tensile M R2 y1 (σb )Y = 1+ 2 AR R − y1 h 0.0917 160000 × 0.3667 (0.3667)2 = 1− 0.01282 × 0.3667 0.00829 (0.3667 − 0.0917) = 12480499.2 [1 − 5.41] = −55.04 MN/m2 (Compressive) Hence, resultant stress at x = σd + (σb )X = −12.48 + 169 = 156.52 MN/m2 (Tensile) Ans resultant stress at y = σd + (σb )Y = −12.48 − 55.04 = 67.52 MN/m2 (Compressive) Stresses in circular ring: P R M R θ d d1 d2 O P Figure 22.16 In such case there will be two stresses: P × sin θ (i) Direct stress, σ0 = 2A @seismicisolation @seismicisolation θ Pcos θ P 2 Ans Bending of Curved Bars (ii) Bending stress due to bending moment M, where M = bending moment = P × R M R2 y Bending stress, σb = 1+ 2 A×R R+y h 1 sin θ − π 2 • 481 Bending stress will be maximum at the outer edge and inner edge of the cross section Where y = +d2 for extreme outer edge from centre line and y = −d1 for extreme inner edge from centre line Resultant stress, σ = σ0 + σb E XAMPLE 22.10: A ring of mean radius of curvature 100 mm is subjected to a pull of 4 kN. The line of section of load passes through the centre of ring. Determine the maximum tensile and compressive stresses in the material of the ring if the diameter of cross section of the ring is 20 mm. S OLUTION : R = 100 mm Pull , P = 4 kN, diameter of cross section = 20 mm. π Area of cross section, A = (20)2 = 314 mm2 4 h2 = = 1 d4 d2 + 16 128 R2 1 204 202 + × 16 128 1002 = 25 + 0.125 = 25.125 mm2 (i) direct stress, σ0 = P x sin θ 2A Direct stress at θ = 0 σ0 = Bending moment at θ = 0 P × sin θ = 0 2A 1 1 M = P×R − sin θ π 2 1 1 − sin 0 = P×R× π 2 = 4000 × 100 = 127388.5 Nmm π @seismicisolation @seismicisolation 482 • Strength of Materials Bending stress at θ = 0, at outer edge: R2 M y 1+ 2 × σb o = A×R R+y h 127388.5 1002 10 = 1+ × 314 × 100 25.12 100 + 10 20 + sign because y = = 10 mm at outer edge 2 = 4.06 [1 + 36.19] = 146.93 N/mm2 tensile 20 Bending stress at the innermost of the cross section when θ = 0 and y = − = −10 mm. 2 [−ve sign because y is towards centre of curvature.] 1002 10 (σb )i = 4.06 1 − × 25.12 100 − 10 = 4.06 [1 − 44.23] = −175.51 N/mm2 (Compressive) ∴ Ans Resultant stress at the outermost edge, (σr )0 = σ0 + (σb0 ) = 0 + 146.93 = 146.93 (Tensile) Ans And resultant stress if the innermost edge (σr )i = σ0 + (σbi ) = 0 − 175.51 = −175.51 = 175.51 N/mm2 (Compressive) Ans (ii) Stresses when θ = 90◦ = π 2 P p × sin 90◦ = 2A 2A 4000 ×1 = 4 × 314 Direct stress, σ0 = = 3.185 N/mm2 (Tensile) @seismicisolation @seismicisolation Bending of Curved Bars Bending moment 1 1 − × sin 90◦ π 2 1 1 = 4000 × 100 − π 2 M = P×R = 400000 (0.3185 − 0.5) = −72600 Nmm The bending stress at outermost edge where y = 10 mm R2 M y 1+ 2 × (σb )o = AR R+y h 1002 10 −72600 1+ × = 314 × 100 25.12 100 + 25.12 = −2.312 [1 + 31.82] = −73.57 N/mm2 (Compressive) The bending stress at innermost edge where y = −10 mm R2 M y 1− 2 × A×R R −y h 1002 10 −72600 1− × = 314 × 100 25.12 100 − 25.12 (σb )i = = −2.312 [1 − 53.16] = +120.6 N/mm2 (Tensile) Resultant stress at outer most edge, (σr )o = σo + (σb )o = 3.185 − 73.57 = −70.385 (Compressive) Ans @seismicisolation @seismicisolation • 483 484 • Strength of Materials Resultant stress at innermost edge, (σr )i = σo + (σb )i = 3.185 + 120.6 = 123.78 N/mm2 Hence maximum tensile stress = 146.93 N/mm2 Maximum compressive stress = 175.51 N/mm2 Ans Ans Ans Stress in a Chain Link P A A R Direct stress σo = X P sin θ 2A Bending moment, M in curved portion P × R 2R + l = − sin θ 2 2R − l θ B l Bending moment, M in straight portion P × R 2R + l −1 = 2 2R + l P Figure 22.17 E XAMPLE 22.11: On a simple chain link as shown in Fig. 22.18, a pull of 30 kN is applied. The radius of curvature of the semicircular ends is 43 mm. The circular cross section of the link has diameter of 46 mm. If the length of straight portion is 48 mm, determine the stresses in curved portion and straight portion of link. @seismicisolation @seismicisolation Bending of Curved Bars • 485 S OLUTION : 30 kN P = 30 kN = 30000 N R = 45 mm θ 45 mm Cross-sectional diameter d = 46 mm l = 43 mm l = 43 mm Area of the cross-section = π (46)2 4 = 1661.1 mm2 h2 = = 30 kN 462 1 464 + × 4 16 128 45 = 132.25 + 8.5 × 10−3 Figure 22.18 = 132.258 mm2 (a) Stress in curved portion: P × sin θ 2A R2 M y 1+ 2 × Bending stress, σb = A×R R+y h Direct stress, σo = (Sign of y depends on its position) Where M is the bending moment in curved portion, P × R 2R + l M= − sin θ 2 2R + l When θ = 0 1 d4 d2 + × 4 16 128 R P × R 2R + l = − sin 0 2 πR + l P × R 2R + l (as sin 0 = 0) = 2 πR + l 30000 × 45 2 × 45 + 43 = 2 π × 45 + 43 @seismicisolation @seismicisolation 486 • Strength of Materials = 675000 [0.722] = 487350 Nmm 46 487350 452 23 At outer fibre, y = = 23, (σb )o = 1+ 2 1661.1 × 45 132.258 45 + 23 = 6.52[1 + 5.18] = 40.29 N/mm2 (Tensile) ∴ Resultant stream at the extreme outer edge, Since, θ = 0, σ0 = So P × sin 0 = 0 A×R (σr )0 = r0 + (σb )0 = 0 + 40.29 = 40.29 N/mm2 (Tensile) Ans Bending stress at the extreme inner edge where θ = 0 and y = r = −23 mm 23 (45)2 (σb )i = 6.52 1− × 132.26 45.23 = 6.52[1 − 16] = −97.8 N/mm2 (Compressive) Ans Resultant stress at the extreme inner edge, ( σ r )i = σ 0 + ( σ b )i = 0 − 97.8 N/m2 = −97.8 N/mm2 (Compressive) Ans (b) Stresses at θ = 900 : P × sin θ 2A 30000 = × sin 90◦ 2 × 1661.1 Direct stress σ0 = = 9.03 N/mm2 P × R 2R + l M= −1 2 πR + l @seismicisolation @seismicisolation Bending of Curved Bars 30000 × 45 2 × 45 + 43 −1 = 2 π × 45 + 43 133 = 675000 −1 1843 = −187886.6 Nmm Bending stress at outer edge y = 23 mm (σb )0 = 452 23 −187886.6 1+ × 1661.1 × 45 132.26 45 + 23 = −2.513(1 + 5.18) = −15.53 N/mm2 (Compressive) Resultant stress = σ0 + (σb0 ) = 9.03 − 15.53 = −6.5 N/mm2 (Compressive) Ans (c) Stresses in straight portion: σ0 = direct stress = 30000 P = = 9.03 N/mm2 2A 2 × 1661.1 σb = bending stress M ×y where M is bending moment I P × R 2R + l σb = −1 2 2R + l = = −187886.6 Nmm I= (Refer (b)) π d4 64 π (46)4 = 219675.2 mm4 64 −187886.6 M ×23 at outer edge, y = 23 mm (σb )0 = ×y = I 219675.2 = = −19.67 N/mm2 (Compressive) @seismicisolation @seismicisolation • 487 488 • Strength of Materials Resultant stress at outer edge, = σ0 + (σb )0 = 9.03 − 19.67 = −10.64 N/mm2 (Compresive) Ans (σb )i = bending stress at inner edge of straight position where y = −23 mm (σb )i = −187886.6 × (−23) 219675.6 = 19.67 N/mm2 (Tensile) Resultant stress at inner edge = σ0 + (σb )i = 9.03 + 19.67 = 25.7 N/mm2 (Tensile) Ans Exercise 22.1 Determine the ratio of maximum and minimum value of the stresses for a curved bar of rectangular section in pure bending. Radius of curvature is 80 mm and depth of the beam is 60 mm. Locate neutral axis also. [Ans 1.682,4 mm] 22.2 A crane hook whose horizontal cross section is trapezoidal, 50 mm wide at the inside and 25 mm wide at the outside, thickness 50 mm carries a vertical load of 9810 N whose line of action is 38 mm from the inside edge of this section. The centre of curvature is 50 mm from the inside edge. Calculate the maximum tensile stress and compressive stress set up. [Ans 49.89 N/mm2 (Tensile), 30.92 N/mm2 (Compressive)] 22.3 A central horizontal section of a hook is a symmetrical trapezium of inner width = 67.5 mm and outer width = 22.5 mm. The depth of section is 90 mm. The load line passes through the centre of curvature. The radius of the hook is 52.5 mm. Determine the maximum compressive and tensile stresses in the section of the hook. [Ans 83.7 MN/m2 (Tensile), 43.2 MN/m2 (Compression)] 22.4 A round bar of steel 38 mm diameter is bent into a curve of mean radius 31.7 mm. If a bending moment of 4.6 Nm tending to increase the curvature acts on the bar find the maximum tensile and compressive stresses. [Ans 0.564 MN/m2 (Tensile), 1.605 MN/m2 (Compressive)] 22.5 A ring of mean diameter 250 mm is made of round steel of 20 mm diameter. A pull of 4 kN acts on the ring. Calculate the greatest tensile and compressive stresses. [Ans 33.6 MN/m2 (Tensile), 30.9 MN/m2 (Compressive)] 22.6 The links of a chain are made of 12.5 mm diameter round steel and have semicircular ends, the mean radius of which is 37.5 mm. The ends are connected by straight pieces 37.5 mm @seismicisolation @seismicisolation Bending of Curved Bars • 489 long. Find the tensile and compressive stresses at various points of the link under a load of 10 kN. [Ans Stress in straight position: 310.2 MPa (Tensile) 228.8 MPa (Compressive). At junction of semicircle and straight position: 349.6 MPa (Tensile) 200.6 MPa (Compressive). Curved portion: 633.44 MPa (Tensile), 810 MPa (Compressive))] 22.7 A crane hook whose horizontal cross section is trapezoidal 50 mm wide at the inner side, 25 mm wide at the outer side and 50 mm thick carries a load of 10 kN whose line of action is 60 mm from the inner edge of the section. The centre of curvature is 50 mm from the inside edge. Determine the maximum tensile and compressive stresses set up in the material. [Ans 67.68 MPa (Tensile), 43.52 MPa (Compressive)] 22.8 Figure 22.19 shows a frame subjected to a load of 2.4 kN. Find (i) The resultant stress at points 1&2; (ii) Position of neutral axis. 2.4 kN 1 2 [Ans 91.72 MN/m2 (Tensile) 39.84 MN/m2 (Compressive) 4.35 mm] 120 48 2.4 kN Dimensions are in mm. 18 48 Figure 22.19 22.9 A circular ring made of a steel bar of 20 mm diameter is subjected to a pull of 15 kN along its vertical diameter. The outer diameter of the ring is 140 mm. Determine the maximum tensile and compressive stresses in the ring. [Ans 321.3 MPa (Tensile), 413.5 MPa (Compressive)] 22.10 A curved beam of T section as shown is Fig. 22.20 is subjected to pure bending moment of 900 Nm which try to decrease the curvature. Find the position of neutral axis and the bending stresses at the outer most and inner most fibres. 20 mm C. G. 60 mm [Ans σouter = −29.45 N/mm2 (Compressive), σinner = 17.54 N/mm2 (Tensile) −1.562 mm] R 20 mm 80 mm Figure 22.20 300 mm Axis of curvature @seismicisolation @seismicisolation 490 • Strength of Materials 22.11 A circular ring of 60 mm wide and 90 mm deep rectangular cross section is subjected to an axial vertical diameter load W . Calculate the magnitude of W , if the maximum stress at the section cut by horizontal diametral plane does not exceed 225 MN/m2 . [Ans 83.19 kN] 22.12 A chain link made of round bar of 12 mm diameter has a straight length of 60 mm and mean radius of curvature of 40 mm. The ring is subjected to an axial pull of 1.5 kN. Determine the maximum stress of the intrados of the link. [Ans 149.7 MN/m2 ] @seismicisolation @seismicisolation 23 C HAPTER UNSYMMETRICAL BENDING Unsymmetrical bending results if the section is unsymmetrical but the load line is inclined to both the principal axes or if the section itself is unsymmetrical, for example, angle section of channel section (with vertical web), and the load line is along any centroidal axis. For studying unsymmetrical bending let us revise the product of inertia for an angle section as discussed below. In order to determine the product of inertia of the angle section shown in Fig. 23.1, we proceed with the formula xy da: The quantity xy da is called product of inertia. Let us find the product of inertia with respect to X and Y axes. Y 15 mm A1 45 mm A2 15 mm 45 mm X Figure 23.1 Using parallel axis theorem, Rect. 1. IXY = IXY 45 + A1 (7.5) 2 A1 = 45 × 15 = 675 mm2 A2 = (45 − 15) × 15 = 450 mm2 @seismicisolation @seismicisolation and xy da = 0 492 • Strength of Materials IXY = 0 + 675 × 7.5 × 22.5 (because xy da = 0) = 113906.2 mm4 Rect. 2. IXY = IXY + A2 30 15 + 2 15 2 = 0 + 450 × 30 × 7.5 = 101250 mm4 Product of inertia of the total area, = 113906.2 + 101250 = 215156.2 mm4 In unsymmetrical bending the principal axes and principal moments of inertia are important to be found before proceeding for further calculations. To understand principal axis, it may be noted that if a figure has an axis of symmetry, that axis is a principal axis. During unsymmetrical bending, the principal axes has to be found as follows. Determination of Principal Axes and Principal Moments of Inertia Figure 23.2 shows a figure of axis A in which the principal axes through centroid G are UU and VV . Axes XX and YY are another pair of perpendicular axes passing through G making an angle α with principal axes. Axes XX and YY are through centre of gravity of section which is unsymmetrical or on where inclined load is applied. After application of load on unsymmetrical bending new axes UV and UV are inclined at α with XX and YY and about which the bending takes place. Let us consider any figure where da in positive quadrant has co-ordinates u and v relative to VV and UU and x and y relative to YY and XX respectively. V If IXX , IYY and IXY are given about XX and Y − Y axes, d is required to find the principal moments of inertia and direction of the principal axes. Y x da u y V X U α X G u = x cos α + y sin α v = y cos α + x sin α U Y V Figure 23.2 IVV = 2 v da = y2 cos2 α − 2xy sin α cos α + x2 sin2 α da = IXX cos2 α − IXY sin 2α + IYY sin2 α @seismicisolation @seismicisolation (i) Unsymmetrical Bending IVV = IUV u2 da = 493 x2 cos2 α + 2xy sin α cos α + y2 sin2 α da = IXX sin2 α + IXY sin 2α + IYY cos2 α xy(cos2 α − sin2 α ) + (y2 − x2 sin α cos α ) da = uvda = = IXY cos 2α + (IXX − IYY ) ∴ • tan 2α = (ii) sin 2α =0 2 2IXY IYY − IXX (iii) Equation (iii) gives two values of 2α differing by π , i.e., two values of α differing by π /2. Given the principal moments of inertia, it is required to find the moments of inertia about XX and YY x = u cos α − v sin α IXX = IYY y = v cos α + u sin α v2 cos2 α + uv sin 2α + u2 sin2 α da y2 da = = IUV cos2 α + IVV sin2 α (because IUV = 0) = x2 da = u2 cos2 α − uv sin 2α + v2 sin2 α da (iv) = IVV cos2 α + IUU sin2 α (v) ∴ (vi) IXX + IYY = IUU + IVV (= J) Beam with Unsymmetrical Bending Moment Figure 23.3 shows the cross section of a beam in which the applied bending moment act in the plane YY , inclined at an angle α to the principal axis VV . Let the applied bending moment is M. Y Component in plane VV , V N = M cos α u M u da v U α Component in plane UU = M sin α X A U Y Figure 23.3 @seismicisolation @seismicisolation X β G V 494 • Strength of Materials Therefore, stress on element da, of coordinates u and v relative to the principal axis, M cos α M sin α ×v+ ×u IVV IVV v cos α u sin α σ =M + IUU IVV = (i) Equation (i) applies to all points in the cross section provided that the appropriate signs are given to the coordinates u and v. For points on the neutral axis, σ = 0 i.e., u cos α u sin α =− IUU IVV IUU or v = − tan α xu IVV (ii) IUU tan α . IVV If β is the inclination of this axis to UU, than This is a straight line of slope − −1 β = − tan IUU tan α IVV (iii) The most highly stressed point is that which is the farthest from the neutral axis at an angle of β as shown in Fig. 23.4. The stress at this point is then obtained by substituting the appropriate values of u and v in Eqn. (i). To Find I UU and I VV using Mohr’s Circle As we determine principal stresses σ1 and σ2 using Mohr’s circle, it is also possible to find IUU and IVV , as explained in Fig. 23.3A. Diagram is self-explanatory. C is the centre of AB. And as a result OA = IVV and OB = IUU . It is understood that perpendicular DG and EF are equal to IXY . Hence, if IXX , IYY and IXY are known, Mohr’s circle can help to determine IUU and IVV . G IYY IXY IVV O C A D α E 2α IXY F IXX IUU Figure 23.3A @seismicisolation @seismicisolation Unsymmetrical Bending • 495 Radius of the circle = IXX cosec 2α = IXX − IYY 2 2 2 + IXY Momental Ellipse Figure 23.3B 1 1 and Gb = in Fig. 23.3B and draw an ellipse with Ga and Gb as the major kUU kVV 1 IUU 1 . Then Gc = and minor semi-axes kUU = and Gd = . A kXX kYY Let Ga = Proof. We know from the equation of the ellipse, u2 1 kUU 2 + v2 1 2 = 1 kVV But we know u = Gc cos α and v = Gc sin α Therefore, 2 2 cos2 α + Gc2 kVV sin2 α = 1 Gc2 kUU ∴ 2 2 kUU cos2 α + A kVV sin2 α = @seismicisolation @seismicisolation A Gc2 496 • Strength of Materials We have already proved that, IXX = IUU cos2 α + IVV sin2 α A Then, IUU cos2 α + IVV sin2 α = = IXX Gc2 1 Hence, Gc = kXX For figures such as circle, square, equilateral triangle or other regular polygon having more than two axes of symmetry, the momental ellipse becomes a circle and therefore the moment of inertia about any axis through G is the same as about the principal axes. Deflection of Beams due to Unsymmetrical Bending Consider Fig. 23.4 where components of load W are shown acting at G. The load W is resolved in to i) W sin α along UG ii) W cos α along V G Y V U N G Let X δu δ δu = deflection due to W sin α as shown δv = deflection due to W cos α as shown α β δv X β A U Y V Figure 23.4 Then depending upon the end conditions of the beam, values of δu and δv are given by: δu = K (W sin α ) l 3 EIVV (i) δv = K (W cos α ) l 3 EIUU (ii) 1 K = A constant depending on end conditions of the beam. For example, K = for a beam 48 simply supported beam with a point load at centre. l = length of the beam Resultant deflection δ = (δ u)2 + (δ v)2 @seismicisolation @seismicisolation Unsymmetrical Bending • 497 Hence, Kl 3 δ= E or δ = W sin α 2 W cos α 2 + IVV IUU sin2 α cos2 α + 2 2 IVV IUU KW l 3 E (iii) The inclination β of the deflection δ , with the line GV is given by: tan β = δu IUU = tan α δv IVV (iv) Method for Finding Bending Stream is Unsymmetrical Bending i) Find centre of gravity of the given section. ii) Determine IXX , IYY and IXY of the given section. 2IXY . If the value of α is positive, then IYY − IXX the principal axis UU will be in the anticlockwise direction with X axis. And the location of VV axis will be at right angle to UU axis. iv) Find the values of IUU and IVV . If IXX = IYY , then value of IUU will be obtained from equation: iii) Calculate the value of α from relation tan 2α = IUU = IXX cos α 2 − IXY sin 2α + IYY sin2 α v) Find M and its components along principal axes GU and GV . vi) Find the resultant bending stress using relation: δb = (M sin α ) × u (M cos α ) ×v + IVV IUU =M u sin α u cos α + IVV IUU E XAMPLE 23.1: An unequal angle of dimensions 120 mm by 75 m and 12 mm thick is shown in Figure 23.5. Determine. i) Position of the principal axes end ii) Magnitude of principal moment of inertia for the given angle. @seismicisolation @seismicisolation 498 • Strength of Materials S OLUTION : Y 75 mm A C 1 15 mm y x X 120 mm X C.G. 2 Y B 15 mm Figure 23.5 a1 = 75 × 15 = 1125 mm2 ; y1 = 7.5, x1 = a2 = 105 × 15 = 1575 mm2 y2 = 75 = 37.5 mm 2 105 15 + 15 = 67.5; x2 = = 7.5 mm 2 2 Total area = a1 + a2 = 1125 + 1575 = 2700 mm2 ȳ = = x̄ = = IXX = 1125 × 7.5 + 1575 × 67.5 1575 8437.5 + 106312.5 = 72.86 mm 1575 1125 × 37.5 + 1575 × 7.5 1575 42187.5 + 11812.5 = 34.29 mm 1575 75 × 153 15 × 1053 + 1125 × (65.36)2 + + 1575 (5.36)2 12 12 ∵ (72.86 − 7.5 = 65.36 and 72.86 − 67.5 = 5.36) @seismicisolation @seismicisolation Unsymmetrical Bending • = [21093.75 + 4805.92] + [1447031.25 + 45249.12] = 1518180 mm4 IYY = 105 × 153 15 × 753 + 1125 (34.29 − 37.5)2 + + 1575 (34.29 − 7.5)2 12 12 = [527343.75 + 11592.1] + [29531.25 + 1130384] = 1698851.1 mm4 IXY = A1 h1 k1 + A2 h2 k2 h1 = (72.86 − 7.5) = 65.36 k1 = (34.29 − 37.5) = +3.21 h2 = (72.86 − 67.5) = 5.36 It is below X − X & hence − ve = −5.36 k2 = (34.29 − 7.5) = 26.79 = It is towards left of Y −Y. Hence, − ve = −26.79 IXY = 1125 × 65.36 × 3.21 × +1575 × (−5.36) × (−26.79) = 236031.3 + 226161.2 = 462192.5 mm4 i) Position of principal axes tan 2α = 2IXY IYY − IXX = 2 × 462192.5 1658851.1 − 1518180 = 924385 140671 = 6.571 2α = 81.35 ∴ α = 40.67◦ Axis UU will be 40.67◦ anticlockwise. @seismicisolation @seismicisolation 499 500 • Strength of Materials ii) Value of principal of moment of inertia IUU = IXX cos2 α − IXY sin 2α + IYY sin2 α = 1518180 cos2 40.67 − 462192.5 sin 2 × 40.67 + 1658851.7 sin2 40.67 = 873386.7 − 456923 + 704538.6 = 1121002.3 mm4 IVV = IXX + IYY − IUU = 1518180 + 1658851.1 − 1121002.3 = 2056028.8 mm4 E XAMPLE 23.2: A 80 mm × 80 mm angle as shown in Fig. 23.6 is used as a freely supported beam with one leg vertical. IXX = IYY = 0.8736 × 10−6 m4 . When a bending moment is applied in the vertical plane YY , the mid-section of the beam deflects in the direction AA at 30◦ 15 to the vertical. Calculate the second moments of area of the section about its principal axes. Find also the bending stress at the corner B, if the bending moment is 2 kNm. B Y Y A 10 mm V U 30˚15' 80 mm α = 45˚ 23.4 mm X X X β X 10 mm 80 mm Y U A Y V Figure 23.7 Figure 23.6 Because the angle has an axis of symmetry, axis (UU in Fig. 23.7) is a principal axis and α = 45◦ . From equations: IXX + IYY = IUU + IVV and the inclination of the resultant deflection to VV = tan −1 δUU δVV −1 = tan IUU + IVV = IXX + IYY = 2 × 0.8736 × 10−6 m4 @seismicisolation @seismicisolation IUU tan α IVV Unsymmetrical Bending = 1.7472 × 10−6 m4 IUU IUU tan α = (∵ tan α = 1) tan β = IVV IVV IUU ∴ = tan 45◦ + 30◦ 15 = 3.79 IVV • 501 (i) (ii) From Eqns. (i) and (ii) IUU = 1.382 × 10−6 m4 IVV = 0.365 × 10−6 m4 and By drawing or calculation, u = 23.5 mm, v = 56.56 mm v cos α u sin α ∴ σ =M + IUU IVV ◦ 0.0235 sin 45◦ 3 0.05656 cos 45 N/m2 + = 2 × 10 1.382 × 10−6 0.365 × 10−6 = 149 MN/m2 Ans E XAMPLE 23.3: A beam of T section having flange 120 mm × 20 mm and web 150 mm × 10 mm is 2.5 m in length and is simply supported at its ends. It carries a load of 3.5 kN inclined at 20◦ to the vertical and passing through the centroid of the section. If E = 200 GN/m2 , calculate: i) ii) iii) iv) Maximum tensile stress Maximum compressive stress Deflection due to the load Position of neutral axis. S OLUTION : Y,V 20˚ 3.5 kN A Wv B 120 mm 1 3.5 kN 20˚ 70˚ Wu 20 mm 42.7 = y G X,U 2 150 mm 127.3 C X,U D 10 mm Y,V Figure 23.8 @seismicisolation @seismicisolation 1.25 m 1.25 m 502 • Strength of Materials a1 = 120 × 20 = 2400 mm2 , y1 = 10 mm a2 = 150 × 10 = 1500 mm2 , y2 = 150 + 20 = 95 mm 2 A = 2400 + 1500 = 3900 ȳ = 2400 × 10 + 1500 × 95 3900 = 42.7 mm (150 + 20) − 42.7 = 127.3 mm Because the section is symmetrical about the vertical axis, therefore the principal axes pass through the centroid G and are along UU and VV axis shown. IXX = IUU = 10 × 1503 120 × 203 + 2400 (42.7 − 10)2 + + 1500 (42.7 − 95)2 12 12 = [80000 + 2566296] + [2812500 + 4102935] = 9561731 mm4 = 9.562 × 10−6 m4 IYY = IVV = 20 × 1203 150 × 103 + 12 12 = [2880000 + 12500] = 2892500 mm4 = 2.892 × 10−6 m4 Components of W Wu = W sin 20 = 3.5 sin 20◦ = 1.197 kN Wv = W cos 20 = 3.5 cos 20◦ = 3.29 kN Bending moments: Mu = 1.197 × 2.5 Wu × l = 4 4 = 0.748 kNm @seismicisolation @seismicisolation Unsymmetrical Bending Mv = • 503 3.29 × 2.5 Wv × l = 4 4 = 2.056 kNm Mu will cause maximum compressive stresses at B and D and maximum tensile stress at A and C. Mv will cause maximum compressive stresses at A and B and maximum tensile stress at C and D. i) Maximum tensile stress: Maximum tensile stress at C, Mu × (5 × 10−3 ) Mv × (127.3 × 10−3 ) + IVV IUU σc = 0.748 × 5 × 10−3 2.056 × 127.3 × 10−3 −3 × 10 + × 10−3 MN/m2 9.562 × 10−3 2.892 × 10−6 = = 1.293 + 27.37 = 28.663 MN/m2 Ans ii) Maximum compressive stress: Maximum compressive stress at B, σB = Mu × (60 × 10−3 ) Mv × (42.7 × 10−3 ) + IVV IUU 0.748 × (60 × 10−3 ) 2.056 × 42.7 × 10−3 × 10−3 + × 10−3 MN/m2 −6 2.892 × 10 9.562 × 10−6 = = 15.52 + 9.181 = 24.7 MN/m2 Ans iii) Deflection due to load, δ= Here K = ∴ KW l 3 E sin2 α cos2 α + 2 IUU IVV 1 for a beam simply supported and carrying a load at the centre. 48 KW l 3 δ= EIVV δ= sin α 2 IUU IVV 2 + cos2 α 3.5 × 103 × (2.5)3 48 × 200 × 109 × 9.562 × 10−6 sin2 20◦ × @seismicisolation @seismicisolation 9.562 × 10−6 + cos2 20◦ 2.892 × 10−6 504 • Strength of Materials √ = 0.000596 ± 0.117 × 10.93 + 0.883 = 0.000596 × 1.47 = 8.7 × 10−4 m = 0.876 mm Ans iv) Position of neutral axis: tan β = IUU tan α IVV 9.562 × 10−6 tan 20◦ 2.892 × 10−6 = 1.20 = β = 50.19◦ Ans E XAMPLE 23.4: The section of a beam is rectangular 100 mm deep and 75 mm wide. It is subjected to a bending moment of 10 kNm. The loading plane is inclined at 45◦ to the Y axis of the section. Locate the neutral axis of section. Also find the maximum bending stress anywhere in section. [A.M.I.E Summer 2000] S OLUTION : Figure 23.9 shows a rectangular section with the given dimensions. This section is symmetrical about XX and YY axis, therefore, XX and YY are the principal axes UU and VV . 1 × 75 × 1003 = 6250000 mm4 12 1 × 100 × 753 = 3515625 mm4 = 12 IUU = IXX = IVV = IYY Bending moment M = 10 kNm = 10 × 103 kNmm Component of bending moment, M1 = M sin 45◦ = 10 × 103 × 0.707 = 7070 kNm M2 = M cos 45◦ = 10 × 103 × 0.707 = 7070 kNm Position of neutral axis (NA) Let β be the angle which the neutral axis makes with the principal axis UU. IUU tan α IVV 6250, 000 = tan 45◦ = 1.7777 3515625 tan β = @seismicisolation @seismicisolation Unsymmetrical Bending ∴ β = tan−1 1.7777 = 60.64◦ = 60◦ 38 • 505 Ans Maximum bending stress anywhere in section. E XAMPLE 23.5: The section of a beam is rectangular 100 mm deep and 75 mm wide. It is subjected to a bending moment of 10 kNm as shown is Fig. 23.9. The loading plane is inclined at 45◦ to the y-axis. Find the maximum stress in the section. S OLUTION : On examining Fig. 23.9 we see that the maximum bending stress in the section may occur either at the point B or D. Let us determine the coordinates x, y for the point B x= 75 = +37.5 mm; 2 y=− 100 = −50 mm 2 y is minus because it is downward. Then u = x cos α + y sin α , α = 45◦ = 37.5 × cos 45◦ − 50 × sin 45◦ = 37.5 × 0.707 − 50 × 0.707 = 26.5125 − 35.35 = −8.8375 mm v = y cos θ − x sin θ = −50 × 0.707 × −37.5 × 0.707 = −35.35 − 26.5125 = −61.8625 mm σB = Y,V M = 10 kNm A M2 D E 8' 45˚ 60˚3 β= XU XU M1 G A N B = 100 mm M1 u M2 v + IVV IUU 7070 × (−8.8375) 7070 − (−61.8625) + 3515625 6250000 = −0.0177 − 0.0699 = −0.0876 kN/mm2 = −87.6 N/mm2 (Tensile) Ans Maximum bending stress at the point D will be 87.6 kN/mm2 (Compressive). Ans Y,V 75 mm Figure 23.9 @seismicisolation @seismicisolation 506 • Strength of Materials E XAMPLE 23.6: Figure 23.10 shows an unequal angle section, for which IXX = 0.8 × 10−6 m4 and IYY = 0.382 × 10−6 m4 . Find the moment of inertia about the principal axes UU and VV , given 1◦ that the angle between the axes UU and X − X is 28 . 2 If the angle with the 80 mm leg vertical, is used as a beam, freely supported on a span of 2 m carrying a vertical load of 2 kN at the centre, find: (a) the maximum stress at the point A and (b) the direction and magnitude of the maximum deflection. Neglect the weight of the beam and take E = 200 GN/m2 S OLUTION : V V 10 mm Y 80 mm G 28.5˚ X X X 26.4 mm 60 mm U Y V X δ VV δ UU 10 mm U U Y U β δ (i) Y 16.4 mm V Figure 23.11 Figure 23.10 IUU + IVV = IXX + IYY = (0.8 + 0.382) × 10−6 = 1.132 × 10−6 m4 As we know IXX = IUU cos2 α + IVV sin2 α and ∴ IYY = IVV cos2 α + IUU sin2 α IXX − IYY = (IUU − IVV ) cos2 α IUU − IYY = = IXX − IYY cos2 α (0.8 − 0.382) × 10−6 cos 57◦ = 0.2072 × 10−6 m4 @seismicisolation @seismicisolation (i) Unsymmetrical Bending • 507 (a) Maximum bending moment 2 × 103 × 2 Wl = 4 4 = 1000 Nm = By drawing or calculation, u = 20 mm, v = 50 mm (See Fig. 23.11) ∴ v cos α u sin α σ =M + IUU IVV 0.05 cos 28.5◦ 0.02 × sin 28.5◦ N/m2 + = 1000 0.9748 × 10−6 0.2072 × 10−6 = 91.2 MN/m2 Ans (b) W l3 δ= 48E cos α IVV 2 2 × 103 × 23 = 48 × 200 × 109 sin α + IVV 2 cos 28.5◦ 0.9748 × 10−6 = 0.004125 m = 4.125 mm 2 sin 28.5◦ + 0.2072 × 10−6 2 Ans Now β = tan = tan −1 −1 IUU tan α IVV 0.9748 × 10−6 tan 28.5◦ 0.2072 × 10−6 = 68◦ 48 ∴ Angle to vertical = 68◦ 48 − 28◦ 30 = 40◦ 18 . Ans Shear Centre In case of symmetrical section such as rectangular and I sections, the applied shear force F is balanced by the set of shear forces summed over the rectangular section as over the flanges and the web of I section and shear centre coincides with the centriod of the section. If the applied load is not @seismicisolation @seismicisolation 508 • Strength of Materials placed at the shear centre, the sections twists about this point and this point is also known as centre of twist. Therefore, the shear centre of a section can be defined as a point about which the applied shear force in balanced by the set of shear forces obtained by summing the shear stresses over the section. The shear centre discussed above is a point in or outside a section through which the shear force applied produces no torsion or twist of the member. Shear Centre for Channel Section Figure 23.12 shows a channel section. Let e be the distance of the shear centre from the web. B dA F1 = t1 X F.AȲ dA It1 A = t1 x F1 F τ dA = dA = t1 dx t1 x d o X C.G. S.C. e h 2 F b h t1 x · t1 dx F1 = It1 o 2 Ft1 h 2 b = 4I 2 h 1 I = t1 h3 + 2Bt1 12 2 ȳ = h F1 t1 B Figure 23.12 (i) 1 t1 h 1 = Bt1 h2 1 + 2 6 Bt1 Substituting value of I in Eqn. (i) F1 = = Ft1 h B2 1 l t1 h 4 × Bt1 h2 1 + · 2 6 Bt1 = Fb 1 t1 h 2h 1 + 6 t1 B Fb 1 ht1 2h 1 + 6 Bt1 Taking moments about 0, F × e = F1 × h h + h1 × = F1 h 2 2 @seismicisolation @seismicisolation (ii) Unsymmetrical Bending • 509 Substituting for F1 from Eqn. (ii), FB h 1 t1 h 2h 1 + · 6 t1 B 1 B 2 e= 1 Aw 1+ 6 Af F.e = ∴ Aw = area of web A f = area of flange. E XAMPLE 23.7: Find the position of shear centre for a channel section 50 mm × 50 mm × 5 mm. 47.5 mm F1 F 5 mm 50 × 503 45 × 403 − 12 12 = 520833.3 − 240000 I= 22.5 mm e N A S.C. 45 mm = 280833.3 mm4 5 mm 5 mm F1 50 mm Figure 23.13 Horizontal maximum shear stress in the flange, F.AȲ I.b F (50 − 2.5) (25 − 2.5) × 5 = I ×5 F = 1068.75 I Zmax = Shear force in flange: F × 1068.75 × (50 − 2.5) × 2.5 I F = 126914 I F1 = @seismicisolation @seismicisolation 510 • Strength of Materials Clockwise moment due to F1 , = F1 × (50 − 5) = 126914 × 45 × = 5711130 F I F I Equating this to anticlockwise moment, F × e, F × e = 5711130 × F I 5711130 280833.3 = 20.34 mm e= Ans Alternatively, 1 B 2 e= 1 Aw 1+ . 6 Af 1 × 47.5 2 = 1 45 × 5 1+ × 6 47.5 × 5 23.75 = 1 × 0.158 = 20.51 mm Ans Shear centre for unequal I section F b2 b1 F2 F1 x dx h X F3 O t1 F e X S.C. t2 F2 F1 b2 Figure 23.14 @seismicisolation @seismicisolation b1 t1 Unsymmetrical Bending Shear stress is any layer, τ = FAȲ It I = IXX t3 = 2 (b1 + b2 ) 1 + (b1 + b2 )t1 × 12 2 h h3 + t2 . 2 12 For shear force F1 dA = t1 x dx Aȳ = t1 x b1 F1 = h 2 τ dA = 0 b1 = 0 F × h.t1 .x dx 2IXX Fht1 x2 = 2IXX 2 = F.x.t1 h . .t1 dx IXX t1 2 b1 0 Fht1 b21 4IXX Also likewise, F2 = Fht1 b22 4IXX Taking moment of the shear forces about the centre of the web 0, we get. F2 h = F1 h + F.e (F3 = F (F2 − F1 ) h = F.e Fh2t1 2 b2 − b21 = F.e 4IXX t1 h2 b22 − b21 or e = 4IXX @seismicisolation @seismicisolation for equilibrium) • 511 512 • Strength of Materials E XAMPLE 23.8: Determine the position of the shear centre of a section of a beam shown in Fig. 23.15. 70 mm 55 mm 35 mm 15 mm X X 250 mm 35 mm 70 55 Figure 23.15 t1 = 35 mm b1 = 55 mm b2 = 70 mm h = 250 − 35 = 215 mm IXX = 2 15 × 1803 125 × 253 + 125 × 35 × 107.52 + 12 12 = 2 [446614.6 + 50558.6] + 7290000 = 2 [497173.2] + 7290000 = 8284346.4 mm4 We know, ∴ t1 h2 b22 − b21 e= 4IXX 35 × 2152 702 − 552 e= 4 × 8284346.4 1617875(1875) 33137385.6 = 91.54 mm Ans = @seismicisolation @seismicisolation Unsymmetrical Bending • E XAMPLE 23.9: Determine the position of shear centre for the Fig. 23.16. 60 mm F F5 F4 3 mm F3 N S.C. 3 e F1 F2 25 mm 35 mm A 35 mm 25 mm 60 mm Figure 23.16 F1 , F2 , F3 , F4 , F5 = shear forces in different positions. It may be noted that F3 is shear force in upward direction. FAȳ We know that shear stress, τ = shear force F1 will be equal to F5 IXX × b 60 mm dx x N 35 mm C A B dy 25 mm 60 mm y 3 mm A Figure 23.17 Consider an element of length y, width 3 mm and thickness dy. A = area of element = 3 × y = 3y y ȳ = distance of centre of gravity of area A from NA = 35 + ; width of element = 3 mm 2 IXX = moment of inertia of whole section about x axis. y F × 3y × 35 + 2 = F y 35 + y ∴ Z= IXX × 3 IXX 2 dA = area of thickness dy = 3dy @seismicisolation @seismicisolation 513 514 • Strength of Materials Shear force F1 (= F5 ) = τ dA 25 = 0 y y 35 + × 3dy IXX 2 3F = IXX F y2 dy 35y + 2 25 0 25 3F 35y2 y3 = + IXX 2 6 0 3F (25)3 2 = 17.5 (25) + IXX 6 ∴ = 3F [10937.5 + 2604.2] IXX = 3F (13541.7) IXX F1 = F5 = 40625.1 F IXX Shear force F2 (= F4 ) in top and bottom flanges. Shear stress, τ= FAȳ IXX × b Note that Aȳ is made up of two parts here. i) Moment of flange area upto length x about neutral axis. ii) Moment of shaded area of vertical projection about neutral axis. dx x 25 35 mm A N Figure 23.18 @seismicisolation @seismicisolation Unsymmetrical Bending ∴ Aȳ = (3x) × (35 + 25) + (3 × 25) × 35 + 25 2 = 180x + 3562.5 b = 3 mm τ= ∴ F× (180x + 3562.5) IXX ×3 Area of thickness dx, dA = 3dx ∴ Shear force, F4 (= F2 ) = τ .dA 60 = 0 = F (180x + 3562.5) × 3dx IXX × 3 60 F(180x + 3562.5) IXX 0 = 180x2 + 3562.5x 2 F IXX = = ∴ F IXX dx 60 0 180 (60)2 + 3562.5 (60) 2 F [324000 + 213750] IXX F4 = F = F IXX [537750] Taking the moments of all forces about the centre of web, we get F × e = F4 × 60 + F2 × 60 + (F5 + F1 ) × 50 = 120 F4 + (F5 + F1 ) × 50 (∵ F4 = F2 & F5 = F1 ) = 120F4 + 100F1 = 120 × = or e= F IXX F IXX F (537750) + 100 × 40625.1 IXX [64530000 + 4062510] 68592510 IXX @seismicisolation @seismicisolation • 515 516 • Strength of Materials IXX = Moment of inertia of the given section about x − x 60 × 33 3 × 1203 2 +2 + 60 × 3 × 60 = 12 12 +2 3 × 253 + 3 × 25 × (60 − 12.5)2 12 = 432000 + 2 [135 + 648000] + 2 [3906.25 + 169218.7] = 432000 + 1296270 + 173124.95 = 1901394.95 ∴ e= 68592510 = 36 mm 1901394.95 Ans Exercise A 23.1 A steel beam 50 mm × 25 mm in cross section is supported over a span of 1 m with one 50 mm face inclined at 30◦ to the vertical. A load of 600 N acts vertically at the centre of the span. Neglecting the weight of the beam, and assuming that the ordinary beam theory applies, find: (a) the maximum stress in the beam due to bending and (b) the magnitude and direction of the maximum deflection. E = 200 GN/m2 . [Ans 26.9 MN/m2 , 0.523 mm at 36◦ 35 to vertical] 23.2 A 8 cm × 8 cm angle as shown in Fig. 23.19 is used as a freely supported beam with one leg vertical IXX = IYY = 0.8736 × 10−6 m4 . When a bending moment is applied in the vertical plane YY , the mid section of the beam deflects in the direction AA at 30◦ 15 of the vertical. Calculate the second moments of the area of the section about its principal axis. Also find the bending stress at the corner B, if the bending moment is 2 kNm. 1 cm B 8 cm 30˚15' Y X X 1 cm A 8 cm Y Figure 23.19 [Ans 1.382 × 10−6 m4 ; 0.365 × 10−6 m4 ; 149 MPa] @seismicisolation @seismicisolation Unsymmetrical Bending • 517 23.3 Determine the centroidal principal moments of an equal angle 30 mm × 30 mm × 10 mm. [Ans 84166.67 mm4 , 12166.67 mm4 ] 23.4 The tension flange of a girder of I section is 240 mm × 40 mm whereas the compressive flange is 120 mm × 20 mm. The web is 300 mm deep and 20 mm thick. If the girder is used as a simply supported beam of 8 m span, determine the load per metre run if the allowable stress is 90 MPa in compression and 30 MPa in tension. Hint: This question is not of unsymmetrical bending. [Ans 9,277 kN/m] 23.5 A 1200 mm long cantilever of I section 50×30 mm having horizontal flange each of thickness 2.5 mm vertical web of thickness 2 mm is subjected to a load of 40 N at its free end. The load is inclined at an angle of 15◦ to the vertical. Determine the resultant bending stress at top corners of the upper flange. [Ans 4.917 N/mm2 (Compressive); 28.12 N/mm2 (Tensile)] 23.6 A I section beam shown is Fig. 23.20, 2.4 metre is used as cantilever beam to support the load of 200 N at the free end which makes 30◦ with the vertical. Determine the resulting bending stresses at corners A and B. 200 N 30° Y A B 2.5 mm 45 mm X X 2 mm C 2.5 mm 30 mm Y D [Ans Figure 23.20 11985 × 105 N/m2 Tensile at A, 8799 × 105 N/m2 Tensile at B] 23.7 An equal angle 100 mm × 100 mm × 20 mm has its one horizontal leg on top. As a cantilever of 1.5 m span, it carries, at it free end, a point load of 6 kN inclined at an angle of 30◦ to the vertical and its line of action passing through the centroid of the section (Fig. 23.21). Determine stresses at points A, B and C. Also locate neutral axis. @seismicisolation @seismicisolation 518 • Strength of Materials B C 20 mm 100 mm 100 mm 20 mm A Figure 23.21 [Ans σA = 193.63 N/mm2 comp, σB = 296.183 N/mm2 tension, σC = 126.61 MPa compression.] 23.8 Determine the position of shear centre for a channel having dimensions: flanges 120 mm × 20 mm and web 160 mm × 10 mm. [Ans e = 50.84 mm] 23.9 Locate the shear centre of section shown is Fig. 23.22. t1 b1 t2 t3 X X h t2 b1 b2 t1 Ans Figure 23.22 e= h2 b22t2 b1 b2t1 h2 2b21 − + + IXX 2 3 4IXX 23.10 Find the shear centre of the section shown in Fig. 23.23. 10 mm 25 mm 10 kN 210 mm e 10 mm 25 mm 110 mm Figure 23.23 [Ans @seismicisolation @seismicisolation e = 44.7 mm to the left of web corner] Unsymmetrical Bending • 23.11 Determine the position of shear centre for the unequal I section shown in Fig. 23.24. 60 mm 80 mm 40 mm 220 mm 40 mm 60 mm 80 mm Figure 23.24 [Ans e = 9.08 mm] 23.12 Find the shear centre of an arc of a circle as shown in Fig. 23.25. F θ θ e R t Figure 23.25 Ans @seismicisolation @seismicisolation e= 2R(sin θ − θ cos θ ) (θ − sin θ cos θ ) 519 C HAPTER 24 ROTATING DISCS AND CYLINDERS ω Fc Rotating Ring Consider a thin ring which is rotating about its centre of gravity at an angular velocity of ‘ω ’ radians per second. Fig. 24.1 shows the ring of thickness ‘t’. Let r be its mean radius and ρ the density of the material of the ring. dθ θ r σc t σc Figure 24.1 Let us consider an element of the ring subtending an angle d θ at θ from horizontal axis. Centrifugal force on the element per unit length = mrw2 = [ρ (rd θ )t]rw2 Total vertical force per unit length, π = ρ r2 d θ t ω 2 sin θ 0 = ρr t ω 2 π 2 sin θ d θ 0 = ρ r t ω (− cos θ )π0 2 2 = 2ρ r2t ω 2 If σc is the hoop stress induced in the ring. Then for equilibrium, σc (2t)1 = 2ρ r2t ω 2 @seismicisolation @seismicisolation Rotating Discs and Cylinders 2 V σc = ρ r ω = ρ r r ∴ 2 2 [ω = 2 • 521 V ] r σc = ρ r 2 where V is the linear velocity. E XAMPLE 24.1: A steel ring of cross-sectional area 900 mm2 rotating about its axis passing through its centre of gravity at an angular speed of 2500 r.p.m. The mean diameter of the ring is 275 mm. Determine the hoop stress produced in the material of the ring. Mass density of the steel is 7800 kg/m3 . If the rotational speed is increased by 25%, what will be then the increase in the centrifugal stress? S OLUTION : σc = ρω 2 r2 = 7800 2π × 2500 60 2 × 0.275 2 2 = 7800 × 68469.4 × 0.0189 = 10093758.9 = 10.09 MN/m2 Ans Increased speed = 1.25 × 2500 = 3125 r.p.m. 2π × 3125 Stress at increased speed = 7800 60 2 0.275 × 2 2 = 7800 × 106983.5 × 0.0189 = 15771507.6 = 15.7 MN/m2 % increase in stress = 15.7 − 10.09 × 100 10.09 = 55.6 % Ans E XAMPLE 24.2: A flywheel with a moment of inertia of 350 kgm2 rotates at 275 r.p.m. If the maximum stress is not to exceed 8 MPa, find the thickness of the rim. Take the width of the rim as 160 mm and mass density of the material 7600 kg/m2 . Neglect the effect of inertia of spokes. S OLUTION : σc = 8 N/mm2 = 8 MN/m2 , ρ = 7600 kg/m3 @seismicisolation @seismicisolation 522 • Strength of Materials Width = 0.16 m, let t be the thickness of rim. ω= 2π × 275 = 28.78 radians/s 60 At the outer radius, stress will be maximum, σc = ρ .ro2 .ω 2 8 × 106 = 7600 × ro2 (28.78)2 ro2 = 1.271 ro = 1.127 m, say 1 m (as mean radius) Let us assume radius of gyration be 0.9 m Moment of inertia = mk2 350 = (7600 × 2π × 1 × 0.16 × t)0.92 t = 0.0566 m t = 56.6 mm Ans E XAMPLE 24.3: A composite ring is made of two materials. The outer ring is of steel and the inner ring is of copper. The diameter of the common surface is 850 mm. Each ring has a width of 35 mm and a thickness of 25 mm in radial direction. The ring rotates at 1850 rpm. Find the stresses set up in the steel and copper. Take Es = 2 Ec . Density of steel = 7600 kg/m3 k, Density of copper is 9000 kg/m3 . S OLUTION : Rc d = 850 mm r = 425 mm = 0.425 m t = 25 mm = 0.025 m 2π × 1850 ω= 60 = 193.63 rad/s Rs Copper Steel Figure 24.2 850 25 + = 437.5 mm 2 2 850 25 Rc = − = 412.5 mm 2 2 Rs = Let p be the pressure due to shrinkage at the common surface when disc is not rotating. @seismicisolation @seismicisolation Rotating Discs and Cylinders Then hoop stress due to shrinkage in the steel ring = Hoop stress in copper ring = • 523 p × 850 pd = = 17 p (Tensile) 2t 2 × 25 pd p × 800 = = 17 p (Compressive) 2t 2 × 25 Hoop stress due to rotation, In steel σc = ρ r2 ω 2 = 7600 × (0.4375)2 × 193.632 = 54.54 MN/m2 (Tensile) In copper σc = ρ r2 ω 2 = 9000 (0.4125)2 (193.63)2 = 57.42 MN/m2 (Tensile) As the strain in steel = strain in copper 17p + 54.54 −17p + 57.42 = Es Ec 17p + 54.54 −17p + 57.42 = 2Ec Ec 17p + 54.54 = 2(−17p + 57.42) 17p + 54.54 = −34p + 114.82 51p = 60.28 p = 1.18 MN/m2 Total stress in steel = 20 × 1.18 + 54.54 = 23.6 + 54.54 = 78.14 MPa Ans Total stress in copper = −20 × 1.18 + 57.42 = −23.6 + 57.42 = 33.82 MPa Ans E XAMPLE 24.4: A wheel 850 mm in diameter has a thin rim. If density is 7500 kg/m3 and E = 200 GPa, calculate: (i) How many revolutions per minute may it make without the hoop stress exceeding 150 MN/m2 ? (ii) Change in diameter Neglect the effect of spokes. @seismicisolation @seismicisolation 524 • Strength of Materials S OLUTION : r= 850 = 425 mm = 0.425 m 2 ρ = 7500 kg/m3 E = 200 GN/m2 σc = 150 MN/m2 (i) σc = ρω 2 r2 2π N 2 150 × 106 = 7500 0.4252 60 N2 = 10107304.5 N = 3179.2 r.p.m. Ans (ii) σc E δ d 150 × 106 = d 200 × 109 150 × 106 δd = × 0.85 200 × 109 = 0.00064 m = 0.64 mm Ans Circumferential strain εc = or Rotating Disc of Constant Thickness In engineering, the design of steam and gas turbines and other discs require analysis of stresses and deformation, as they have to rotate at high speeds. The constant thickness discs are often called flat discs. It is important to study radial and hoop stresses developed in them. Interestingly the theory related to these is quite similar to thick cylinders where we used Lame’s theory. ρr2ω2δrδθ Derivations of equations for hoop and radial stresses: Figure 24.3 shows an element of rotating disc, extending angle θ at the centre. Let σr and σc be the radial and hoop stresses respectively developed in the rotating disc. Both stresses are tensile in nature. (σr+δσr) (r+δr)δθ σcδr σrrδθ dθ Figure 24.3 @seismicisolation @seismicisolation σc δr Rotating Discs and Cylinders • 525 If at a radius r from axis of rotation σr and σc are developed when disc rotates uniformly with speed ω . Then if u is the radial shift, the stress-strain equations are: du = σr − μσc dr u E · = σc − μσr r E· Obtaining from E · (i) (ii) du Eqn. (ii) and equating to Eqn. (i), we get dr (σc − σr ) (1 + μ ) + r d σc /dr − μ r(d σr /dr) = 0 (iii) If the density of the material of disc is ρ , then for the element shown in Fig. 24.3, the centrifugal force = (ρ r δ θ .δ r) rω 2 = ρ r2 ω 2 δ r.δ θ for unit thickness In the radial direction equilibrium equation is 1 2σc dr sin δ θ + σr rδ θ − (σr + δ σr )(r + δ r)δ θ = ρ r2 ω 2 δ r δ θ 2 Dividing the whole equation by δ rδ θ , 1 2σc dr sin δ θ σ rδ θ (σ + δ σ )(r + δ r)δ θ r r r 2 + − = ρ r2 ω 2 δrδθ δrδθ δrδθ δθ (r + δ r) dr sin 2 r + σ r − ( σ r + δ σr ) = σc · = ρ r2 ω 2 δθ δr δr δr 2 δθ r r (rδ δ r) dr sin 2 − σ2 · − σr − − δ σr = ρ r 2 ω 2 = σc · + σ2 · δθ δθ δr δr δr 2 Neglecting δ σr being too small And taking limits, = σc − σr − r d σr = ρ r2 ω 2 dr Substitute for σc − σr from Eqn. (iv) in Eqn. (iii) d σr d σc d σr 2 2 r ) + ρr ω − μr =0 (1 + μ ) + r dr dr dr @seismicisolation @seismicisolation (iv) 526 • Strength of Materials Rearranging, d σc d σr + = − ρ r ω 2 (1 + μ ) dr dr Integrating, σc + σr = −(ρ r2 ω 2 /2) (1 + μ ) + 2A (v) 2A is constant of integration Subtracting Eqn. (iv) 2 d σr 2ω = − ρr (3 + μ ) + 2A 2σr + r dr 2 1 d(σr r2 ) ρ r2 ω 2 (3 + μ ) =− + 2A r dr 2 or, Integrating, σr r2 = −(ρ r4 ω 2 /8) (3 + μ ) + Ar2 − B or σr = A − From (v), σc = A + B − (3 + μ ) + (ρ r2 ω 2 /8) r2 (vi) B − (1 + 3μ ) + (ρ r2 ω 2 /8) r2 (vii) Solid Disc The stress cannot be infinite at the centre in a solid disc ∴ σr = 0 = A − B 0 or B = 0 If R is the outside radius, then at the outer surface, σR = 0 3+μ 3+μ ρω 2 R2 or A = ρ R2 ω 2 8 8 3+μ therefore σr = ρω 2 R2 − r2 8 ρω 2 (3 + μ )R2 − (1 + 3μ )r2 and σc = 8 ∴ 0 = A− (viii) (ix) At the centre, r = 0, then, σr = σc = 3+μ ρω 2 R2 8 (maximum value of stress) @seismicisolation @seismicisolation (x) Rotating Discs and Cylinders • 527 And the outer surface, σc = 1−μ ρω 2 R2 4 and σr = 0 3 + 0.3 If μ = 0.3, σr = σc = ρω 2 R2 = 0.413 ρω 2 R2 at the centre (maximum) 8 and at the end surface, 1 − 0.3 ρω 2 R2 = 0.175ω 2 R2 4 0.175 σc = σc (max) = 0.424σc (max) 0.413 σc = σe Stress σr Radius Figure 24.4 Hollow Disc In this case the boundary conditions are From Eqn. (vii), At r = ri , σr = 0 At r = r0 , σr = 0 2 B 2ω σc = A + 2 − (3 + μ ) ρ r 8 r At inner radius, 0 = A− B (3 + μ ) 2 2 − ρ ri ω 8 r2 0 = A− B (3 + μ ) − ρω 2 ro2 8 r2 and outer radius, @seismicisolation @seismicisolation (xi) 528 • Strength of Materials Solving above two equations B= A= and 3+μ 8 3+μ 8 ρω 2 .ri2 . ro2 ρω 2 ri2 + ro2 If we now put the value of A and B in Eqns. (vi) and (vii), we get the values of σr and σc respectively as given below: σr = and σc = 3+μ 8 3+μ 8 ρω 2 ri2 + ro2 − ri2 ro2 − r2 r2 (xii) ρω 2 ri2 + ro2 − ri2 ro2 1 + 3μ 2 − r 3+μ r2 (xiii) Let the value of r at which σr is maximum be R1 . Now differentiating Eqn. (xii) and equating the same to zero, we get, d σr = dr ∴ 3+μ 8 ρ w2 2ri2 ro2 =0 R31 R41 = ri2 ro2 √ R1 = ri r0 σr is maximum at the value of radius which is the geometric mean of the inner and outer radius of the disc √ Substituting r = ri r0 in Eqn. (xii), σr(max) = = r2 r2 3+μ ρω 2 ri2 + ro2 − i o − ri r0 8 ri r0 3+μ ρω 2 (ro − ri )2 8 It is clear from Eqn. (xiii) that σc is maximum when r is minimum, i.e., when r = ri ∴ σc(max) = = r2 r2 1 + 3μ 3+μ ρω 2 ri2 + ro2 − i 2o − ri 8 3+μ ri 3+μ 1−μ 2 ρω 2 ro2 + .r 8 3+μ i @seismicisolation @seismicisolation (xiv) Rotating Discs and Cylinders • 529 The stress distribution, i.e., σr and σc as given by Eqns. (xii) and (xiii) are shown in figure below. ro σc σr ri Figure 24.5 E XAMPLE 24.5: A disc of uniform thickness and 650 mm diameter rotates at 1850 r.p.m. Find the maximum stress developed in the disc. If a hole of 95 mm diameter is made at the centre of the disc, find the maximum values of radial and hoop stresses. Density of the material of the disc = 7800 kg/m3 and μ = 0.3. S OLUTION : R= 650 = 0.325 m, 2 ρ = 7800 kg/m3 ω= and 2π × 1850 = 193.63 rad/s 60 μ = 0.3 Maximum radial stress and hoop stress are at the centre are equal. 3+μ ρω 2 r2 8 3 + 0.3 × 7800 (193.63)2 (0.325)2 σr = σc = 8 σr = σc = = 0.4125 × 7800 × 37492.6 × 0.105625 = 12741801.5 N/m2 = 12.74 MN/m2 With hole of radius ri = r0 = 0.325 m 0.095 = 0.0475 m 2 Maximum radial stress is at i.e., Ans ri2 ro2 radius √ 0.0475 × 0.325 = 0.124 m @seismicisolation @seismicisolation 530 • Strength of Materials 3+μ ρω 2 (ro + ri )2 8 3 + 0.3 = × 7800(193.63)2 (0.325 − 0.0475)2 8 σr = = 0.4125 × 7800(37492.6)(0.2775)2 = 9289452 N/m2 = 9.28 MN/m2 Ans Maximum hoop stress is at the inner radius, σc = = 3+μ 1−μ 2 ρω 2 ro2 + r 8 3+μ i 3 + 0.3 1 + 0.3 × 7800 (193.63)2 0.3252 − (0.0475)2 8 3 + 0.3 = 0.4125 × 292442099.8 [0.106 + 0.212 × 0.00226] = 120632366 (0.1065) = 12847347 = 12.85 MPa Ans E XAMPLE 24.6: A thin steel disc of uniform thickness and of 240 mm diameter with a central hole of 50 mm diameter rotates at 9500 r.p.m. calculate the maximum principal stress and the maximum shear stress in the disc. ρ = 7000 kg/m3 , μ = 0.3. S OLUTION : B 3+μ We know, radical stress, σr = A − 2 − ρω 2 r2 8 r B 1 + 3μ and hoop stress, σc = A + 2 − ρω 2 r2 8 r 9500 × 2π = 994.3 rad/s 60 B 3 + 0.3 7000 × 994.32 r2 σr = A − 2 − 8 r ω= B − 2854676315 r2 r2 B 1 + 3 × 0.3 σc = A + 2 − × 7000 × 994.32 r2 8 r = A− = A+ B − 1643601515 r2 r2 @seismicisolation @seismicisolation Rotating Discs and Cylinders • 531 Now σr = 0 when r = 0.025 m and also when r = 0.120 m 0 = A− B − 2854676315 (0.025)2 (0.025)2 B − 1784173 0.000625 B 0 = A− − 2854676315 (0.12)2 (0.120)2 0 = A− Also = A− B − 41107339 0.0144 (i) (ii) From Eqn. (ii) A= B + 41107339 0.0144 Equating (i) & (ii), B B + 1784173 = + 41107339 0.000625 0.0144 B B − = 41107339 − 1784173 0.000625 0.0144 1600B − 69.44B = 39323166 ∴ Using (iii) ∴ ∴ ∴ 39323166 = 25692 1530.56 25692 + 41107339 A= 0.0144 B= = 1784166.7 + 41107339 = 42891506 B 1 + 3μ σc(max) = A + 2 − ρω 2 r2 [∵ r = 0.025 m] 8 r σc(max) = 42891506 + 42506080 − 1027251 = 84370335 = 84.37 MPa Ans σmax 2 84.37 = 2 Max shear stress τmax = = 42.185 MPa Ans @seismicisolation @seismicisolation (iii) 532 • Strength of Materials E XAMPLE 24.7: Calculate to the largest value of radial and hoop stress for a rotating disc of internal diameter 180 mm and external diameter 360 mm. The disc is rotating at 1700 r.p.m. ρ = 7200 kg/m3 S OLUTION : and μ = 0.3 B 3+μ σr = A − 2 − ρω 2 r2 8 r B 1 + 3μ and σc = A + 2 − ρω 2 r2 8 r ω= Now also at ∴ 2π × 1700 = 177.93 rad/s 60 σr = 0 when r = 0.09 m r = 0.180 m 3 + 0.3 0 = A− − × 7200 × 177.932 × 0.092 2 8 (0.09) B 0 = A − 123.4B − 761622.6 A = 123.4B + 761623 B 3 + 0.3 Also 0 = A − − × 7200 × 177.932 × 0.0182 8 (0.18)2 (i) 0 = A − 30.86B − 3046490 A = 30.86B + 3046490 Equating (i) and (ii) 123.4B + 761623 = 30.86B + 3046490 92.54B = 2284867 B = 24690 A = 30.86 × 24690 + 3046490 = 3808423 ∴ 24690 3+μ − ρω 2 r2 8 r2 24690 1 + 3μ σc = 3808423 + 2 − ρω 2 r2 8 r σr = 3808423 − @seismicisolation @seismicisolation (ii) Rotating Discs and Cylinders σr is maximum when • 533 r2 = ri ro = 0.09 × 0.18 = 0.0162 ∴ σr( max) = 3808423 − 24690 − 1523245 0.0162 = 3808423 − 1524074 − 1523245 = 761104 N/m2 = 0.76 MN/m2 Ans σc in maximum when, r = 0.09 m ∴ σc max = 3808423 + 24690 (0.09)2 − 1 + 3 × 0.3 × 7200 × 177.932 × (0.09)2 8 = 3808423 + 3048148 − 438510 = 6418061 = 6.42 MPa Ans E XAMPLE 24.8: A solid disc of uniform thickness and having a diameter of 500 mm rotates at 7800 r.p.m. Determine the radial and the hoop stresses at radii of 0, 100 mm, 150 mm and 250 mm. ρ = 7500 kg/m3 , μ = 0.25. S OLUTION : 2π × 7800 = 816.4 rad/s; R = 0.25 m 60 3+μ σr = ρω 2 R2 − r2 8 3 + 0.25 = ρω 2 R2 − r2 8 = 0.406 × 7800 × 816.42 0.252 − r2 = 2110700575 0.0625 − r2 = 2110.7 0.0625 − r2 MPa ω= At r = 0 @seismicisolation @seismicisolation 534 • Strength of Materials σr = 2110.7 × 0.0625 = 131.92 MPa At r = 100 mm = 0.1 m σr = 2110.7 0.0625 − 0.12 = 2110.7 (0.0625 − 0.01) = 110.81 MPa At r = 150 mm = 0.15 m σr = 2110.7 0.0625 − 0.152 = 2110.7 (0.04) = 84.43 MPa At r = 250 mm = 0.25 m σr = 2110.7 0.0625 − 0.252 =0 Now ω2 (3 + μ ) R2 − (1 + 3μ ) r2 8 7800 × 816.42 3.25 × 0.252 − (1 + 3 × 0.3) r2 = 8 = 649846236 0.2031 − 1.9r2 σc = ρ = 649.85[0.2031 − 1.9r2 ] At r = 0 σθ = 649.85 [0.2031 − 0] = 131.98 MPa At r = 150 mm = 0.15 m σθ = 649.85 0.2031 − 1.9 × 0.152 = 649.85 [0.16035] = 104.20 MPa At r = 250 mm = 0.25 m @seismicisolation @seismicisolation Rotating Discs and Cylinders • 535 σθ = 649.85 0.2031 − 1.9 × 0.252 = 649.85 [0.08435] = 54.81 MPa 130 σθ 110 Stress (MPa) σr 90 0 100 mm 150 mm 250 mm r Figure 24.6 Stress distribution curves E XAMPLE 24.9: A thin disc of uniform thickness is of 900 mm outer diameter and 60 mm inner diameter. It rotates at 3500 r.p.m. Determine the radial and hoop stresses at radii 0, 25 mm, 50 mm, 100 mm, 150 mm, 200 mm, 300 mm and 450 mm. ρ = 7800 kg/m3 . μ = 0.25 S OLUTION : Ri = 60 = 30 mm 2 = 0.03 m Ro = 900 = 450 mm 2 = 0.45 m, ρ = 7800 kg/m3 2π × 3500 = 366.33 rad/s 60 r2 r2 3+μ σr = ρω 2 ri2 + r2o − 1 2o − r2 , σro = 0 8 r 3 + 0.25 .032 × 0.452 7800 (366.33)2 (0.03)2 + (.45)2 − σ0.025 = − 0.0252 8 0.0252 ω= = 425238863 [0.0009 + 0.2025 − 0.00292 − 0.000625] = 425238863 × 0.199855 = 84986113 = 84.99 MPa @seismicisolation @seismicisolation (i) 536 • Strength of Materials From Eqn. (i) σr = 425238863 0.2034 − σr0.05 = 425238863 0.2034 − 0.0001822 − r2 r2 0.0001822 2 (0.05) − (0.05)2 = 425238863 [0.2034 − 0.073 − 0.0025] = 425238863 [0.1279] = 54388051 = 54.39 MPa σr0.1 = 425238863 0.2034 − 0.0001822 − 0.12 0.12 = 425238863 [0.2034 − 0.01822 − 0.01] = 425238863 × 0.17518 = 74493344 = 74.49 MPa σr0.15 = 425238863 0.2034 − 0.0001822 − 0.152 0.152 = 425238863 [0.2034 − 0.0081 − 0.0225] = 425238863 [0.1728] = 73.48 MPa σr0.2 = 425238863 0.2034 − 0.0001822 − 0.22 0.22 = 425238863 [0.2034 − 0.00455 − 0.04] = 425238863 [0.15885] = 67.55 MPa σr0.3 = 425238863 0.2034 − 0.0001822 − 0.32 0.32 = 425238863 [0.2034 − 0.002 − 0.09] = 425238863 [0.1114] = 47.37 MPa @seismicisolation @seismicisolation Rotating Discs and Cylinders σr0.45 = 425238863 0.2034 − • 537 0.0001822 − 0.452 0.452 = 425238863 [0.2034 − 0.0008997 − 0.2025] = 425238863 [0.0000003] Now, = 0 approx. r2 r2 1 + 3μ 2 3+μ σc = ρ w2 ri2 + ro2 + i 2o − r 8 3+μ r 3 + 0.25 (0.03)2 (0.45)2 1 − 3 × 0.25 2 = − × 7800 × (366.33)2 (0.03)2 + (0.45)2 + r 8 3 + 0.25 r2 = 425238863 0.0009 + 0.2025 + 0.0001822 − 0.077r2 r2 σco = 0 σc0.025 = 425238863 0.0009 + 0.2025+ 0.0001822 −0.077 (0.025)2 0.0252 = 425238863 [0.0009 + 0.2025 + 0.29125 − 0.0000481] = 425238863 [0.4946] = 210.32 MPa σc0.05 = 425238863 0.0009 + 0.2025+ 0.0001822 −0.077 (0.05)2 0.052 = 425238863 [0.0009 + 0.2025 + 0.07288 − 0.0001925] = 425238863 [0.2761] = 117.41 MPa σc0.1 = 425238863 0.0009 + 0.2025+ 0.0001822 −0.077 (0.1)2 0.12 = 425238863 [0.0009 + 0.2025 + 0.01822 − 0.00077] = 425238863 [0.22085] = 93.91 MPa σc0.15 = 425238863 0.0009 + 0.2025+ 0.0001822 −0.077 (0.15)2 0.0225 = 425238863 [0.0009 + 0.2025 + 0.0081 − 0.00173] @seismicisolation @seismicisolation 538 • Strength of Materials = 425238863 [0.20977] = 89.20 MPa σc0.2 = 425238863 0.0009 + 0.2025+ 0.0001822 −0.077 (0.2)2 2 0.2 = 425238863 [0.0009 + 0.2025 + 0.00455 − 0.00308] = 425238863 [0.20487] = 87.11 MPa σc0.30 = 425238863 0.0009 + 0.2025+ 0.0001822 −0.077 (0.3)2 2 0.3 = 425238863 [0.0009 + 0.2025 + 0.00202 − 0.00693] = 425238863 [0.19849] = 84.4 MPa σc0.45 = 425238863 0.0009 + 0.2025+ 0.0001822 2 (0.45) −0.077 (0.45)2 = 425238863 [0.0009 + 0.2025 + 0.000899 − 0.0156] = 425238863 [0.188699] = 80.24 MPa 200 100 σc 80 σr 60 Stress 40 20 0 25 50 100 200 300 450 r (mm) Figure 24.7 E XAMPLE 24.10: Calculate the largest value of radial and hoop stresses for a rotating disc of internal dia 160 mm and external diameter 340 mm. The disc is rotating at 1600 r.p.m. ρ = 7100 kg/m3 , μ = 0.3. @seismicisolation @seismicisolation Rotating Discs and Cylinders • 539 B 3+μ σr = A − 2 − ρω 2 r2 8 r B 1 + 3μ σc = A + 2 − ρω 2 r2 8 r S OLUTION : ω= 2π × 1600 = 167.47 rad/s 60 σr = 0 when r = 0.08 m 3.3 B 7100× (167.47)2 (0.08)2 − o = A− 8 (0.08)2 o = A − 156.25 B − 525698 Also, σr = 0 when (i) r = 0.170 m 3.3 B − ×7100 (167.47)2 (0.170)2 o = A− 8 0.1702 = A − 34.6 B − 2373855 From Eqns. (i) and (ii) A = 156.25B + 525698 Also ∴ A = 34.6B + 2373855 156.25B + 525698 = 34.6B + 2373855 121.65B = 1848157 B = 15192 A = 156.25 × 15192 + 525698 A = 2899448 15192 3.3 − × 7100× (167.47)2 r2 8 r2 15192 σr = 2899448 − 2 −47292906r2 r σr = 2899448 − σ is maximum when r2 = ri ro = 0.080 × 0.170 @seismicisolation @seismicisolation (ii) 540 • Strength of Materials = 0.0136 m2 15192 = 2899448 − σr − 47292906(0.0136) (max) 0.0136 = 2899448 − 1117059 − 643183.5 = 1139205.5 = 1.14 MPa Ans σc is max when r = 0.08 σc(max) = 2899448 − 15192 − 47292926 (0.08)2 r2 = 2899448 − 2373750 − 302675 = 223023 = 0.22 MPa Ans E XAMPLE 24.11: Determine the largest values of radial and hoop stresses for a rotating disc of internal diameter 160 mm and external diameter 340 mm. The disc is rotating at 1650 r.p.m. For the disc material, density = 7100 kg/m3 , μ = 0.3. S OLUTION : B 3+μ σr = A − 2 − ρω 2 r2 8 r B 1 + 3μ σc = A + 2 − ρω 2 r2 8 r ω= 2π × 1650 = 172.7 rad/s 60 Now σr = 0 when when r = 80 mm = 0.08 m r = 170 mm = 0.17 m 3.3 B − 7100 × (172.7)2 (0.08)2 o = A− 8 (0.08)2 = A − 156.25B − 559045 (i) σr = 0 when r = 0.17 m B 3.3 ∴ o = A− 7100 × (172.7)2 (0.17)2 − 8 (0.17)2 Also = A − 34.6 B − 2524439 @seismicisolation @seismicisolation (ii) Rotating Discs and Cylinders From Eqn. (i), A = 156.25 B + 559045 From Eqn. (ii), A = 34.6 B + 2524439 Equating, 156.25 B + 559045 = 34.6 B + 2524439 121.65 B = 1965394 B = 16156 A = 156.25 × 16156 + 559045 = 30832420 ∴ 16156 3.3 7100 × 172.72 r2 σr = 30832420 − 2 − 8 r 16156 − 87350818 r2 r2 16156 1.9 × 172.72 r2 σc = 30832420 + 2 − 8 r 16156 = 30832420 + 2 − 21171 r2 r = 30832420 − and Now σr is maximum when r2 = ri ro = 0.08 × 0.17 = 0.0136 σr(max) = 30832420 − 16156 − 87356818 × 0.0136 0.0136 = 30832420 − 1187941 − 1187971 = 28456508 = 28.46 MPa Ans σc is maximum when r = 0.08 m σc(max) = 30832420 + 16156 − 21171(0.08)2 0.082 = 30832420 + 2524375 − 135.4944 = 30832420 = 30.83 MPa @seismicisolation @seismicisolation • 541 542 • Strength of Materials Rotating Long Cylinder Centrifugal force σr+δσr σz δθ 2 r0 ri σc dθ 2 w δr σc L σz σr r δθ ω Figure 24.8 Let the density of the cylinder is ρ and length L, rotating about its axis with an angular velocity ω . and, σr = radial stress produced at radius r σr + δ σc = radial stress produced at radius (r + δ r) σc = hoop stress at radius r assumed constant over δ r. σz = Axial stress constant over cylinder cross section. u = radial shift displacement at radius r u + δ u = radial shift (displacement) at radius (r + δ r) 2π (r + u) − 2π r u Circumferential strain, εc = = 2π r r u + δu Applying Hooke’s law, δr Unstrained εc = u Strained u 1 [σc − μ (σr + σz )] = E r Eu = r [σc − μ (σr + σz )] ∴ r ∴ Figure 24.9 1 [σc − μ (σr + σz )] E From Fig. 24.9, radial shift at radius r is u, whereas of radius (r + δr ), it is (u + δu ) δu ∴ Radial strain = δr @seismicisolation @seismicisolation Rotating Discs and Cylinders Also, radial strain = • 543 1 [σr − μ (σc + σz )] E By equating, ∴ E 1 u = [σr − μ (σc + σz )] r E δr → 0 when E du = σ r − μ (σ c + σ z ) dr (i) Differentiating Eqn. (i), we get E d σc dy d σr d σz =r + σ c − μ (σ r + σ z ) − μ r + dr dr dr dr = σ r − μ (σ c + σ z ) ∴ (σc − σr ) (1 + μ ) + r d σc −μ dr from Eqn. (i) d σr d σz + =0 dr dr (ii) Now the centrifugal force on the element (Fig. 24.6) is Centrifugal force = ρ (rδθ .δr .L) ω 2 r = ρω 2 r2 Lδθ δr Now equating forces on the element in the direction of centrifugal force, we have ρ w2 r2 δθ δr L + (σr + δ σr ) [(r + δ r) δθ L] = σr .rδθ L + 2σc .δr . δθ δ δ L (Because sin θ = θ ) 2 2 2 After simplifying, σc − σr = ρ w2 r2 + r d σr dr (iii) Substituting value of (σc − σr ) from Eqn. (iii) to Eqn. (ii), we get d σr d σz d σr d σc (1 + μ ) + r −μ + =0 ρ w2 r2 + r. dr dr dr dr ρ w2 r2 (1 + μ ) + r d σr d σr σc d σr d σz + μr + r − rμ − rμ =0 dr dr dr dr dr d σc d σr d σz + = −ρ w2 r (1 + μ ) + μ dr dr dr 1 [σz − μ (σc + σr )] E ∴ E εz = σz − μ (σc + σr ) = a constant say C ∴ σz = C + μ (σc + σr ) Now, εz = @seismicisolation @seismicisolation (iv) 544 • Strength of Materials or d σc d σr d σz =μ + dr dr dr (v) Substituting this in Eqn. (iv) d σc d σz d σc d σr + = −ρ ω 2 r (1 + μ ) + μ 2 + dr dr dr dr ∴ ρ ω 2r −ρ ω 2 r (1 + μ ) d σc d σr = − + = dr dr 1−μ 1 − μ2 Integrating, we get σc + σr = ρ ω 2 r2 + 2A 2 (1 − μ ) (vi) where 2A is a constant of integration. From Eqns. (iii) and (vi), 2σr = d σr ρ ω 2 r2 + 2A − ρ ω 2 r2 − r 2(1 − μ ) dr 3 − 2μ d σr = −ρω 2 r2 + 2A 2 σr + r dr 2 (1 − μ ) 1 d 3 − 2μ + 2A σr .r2 = −ρω 2 r2 r dr 2 (1 − μ ) d σr r 2 3 − 2μ = −ρω 2 r3 + 2Ar or dr 2 (1 − μ ) Integrating, σr .r2 = − ρω 2 r4 + Ar2 − B 8 Where −B is a constant of integration. σr = A − B 3 − 2μ − ρ ω 2 r2 2 8(1 − μ ) r Substituting this value of σr in Eqn. (vi) σc = − B 3 − 2μ ρ ω 2 r2 + 2A − A + 2 − ρω 2 r2 2 (1 + μ ) 8 (1 − μ ) r = A+ 3 − 2μ ρ ω 2 r2 B 1− − 4 r2 2 (1 − μ ) @seismicisolation @seismicisolation (vii) Rotating Discs and Cylinders • 545 or σc = A + B 1 + 2μ − ρ ω 2 r2 8 (1 − μ ) r2 (viii) Now to evaluate σz , from Eqn. (v) d σc d σr d σz =μ + dr dr dr 2B 1 + 2μ 2B 3 + 2μ 2 2 ρω r + − 3 − ρω r =μ − 3 − 4 (1 − μ ) 4 (1 − μ ) r r μ ρ ω 2r =− 1−μ Integrating, σz = − μ ρ ω 2 r2 +C 1−μ 2 where C is a constant of integration. To determine C, consider the equilibrium of cylinder in the axial direction ro σz (2π rdr) = 0 ri or ro σz rdr = 0 or ri − μ 1−μ 8 ρω 2 ro4 − ri4 + c 2 ro ri μ ρω 2 r3 +Cr = 0 − (1 − μ ) 2 ro2 − ri2 = 0 c= μ ρω 2 2 2 ro + ri 1−μ 4 μ ρω 2 r2 μ ρω 2 2 2 . + ro − ri 1−μ 2 1−μ 4 μ ρω 2 ri2 + ro2 − 2r2 or σz = 4 (1 − μ ) ∴ σz = (ix) Solid Cylinder If we consider Eqns. (vii) and (viii) carefully, we find that the constant B becomes zero because stresses become infinite at r = 0 Now for solid cylinder boundary conditions is r = ro ; σr = 0 @seismicisolation @seismicisolation 546 • Strength of Materials Therefore, Eqns. σr = A − Becomes B 3 − 2μ − ρω 2 r2 2 8 (1 − μ ) r 3 − 2μ ρω 2 ro2 8 (1 − μ ) 3 − 2μ or, A = ρω 2 ro2 8 (1 − μ ) 0 = A− [since constant B = 0] Substituting this value of A in Eqns. (vii) and (viii), 3 − 2μ σr = ρω 2 ro2 − r2 8 (1 − μ ) ρω 2 σc = (3 − 2μ ) ro2 − (1 + 2μ ) r2 and 8 (1 − μ ) (x) (xi) Now for σz , putting ri = 0 in Eqn. (viii), maximum values of σr , σc and σz when r = 0 at the centre. 3 − 2μ σr(max) = ρω 2 ro2 8 (1 − μ ) 3 − 2μ σc(max) = ρω 2 ro2 8 (1 − μ ) i.e., σr(max) = σc(max) μ and σz(max) = − ρω 2 ro2 (Compressive) 4(1 − μ ) Also σz = 0 when 2r2 = ro2 √ or when r = ro 2 σc σz σr ro ro √2 Figure 24.10 Stress distribution of σr , σc and σz in a rotating cylinder @seismicisolation @seismicisolation Rotating Discs and Cylinders • 547 Hollow cylinder: In case of hollow cylinder constants A and B can be determined from Eqns. (vii) and (viii) taking boundary conditions, σr = 0 when r = ri and r = ro . From Eqn. (vii) and 0 = A− 3 − 2μ B − ρω 2 ri2 2 8 (1 − μ ) ri 0 = A− B 3 − 2μ − ρω 2 ro2 2 ro 8 (1 − μ ) Solving above two equations, B= σr = ∴ and σc = 3 − 2μ ρω 2 ri2 ro2 8 (1 − μ ) and, A= 3 − 2μ ρω 2 ri2 + ro2 8 (1 − μ ) r2 r2 3 − 2μ ρω 2 ri2 + ro2 − i 2 o − r2 8 (1 − μ ) r (i) r2 r2 1 + 2μ 2 3 − 2μ ρω 2 ri2 + ro2 − i 2 o − r 8 (1 − μ ) 3 − 2μ r (ii) Also it was derived earlier that σz = Now σr is maximum where μ ρω 2 ri2 + ro2 − 2r2 4 (1 − μ ) d σr =0 dr ∴ i.e., r = σr(max) = (iii) √ ri ro 3 − 2μ ρω 2 (ro − ri )2 8 (1 − μ ) (iv) Also σc is maximum where r is minimum, i.e., r = ri ∴ σc(max) = σz is maximum at r = ri and r = ro ∴ σz(max) = when μ ρω 2 ro2 − ri2 4(1 − μ ) (Tensile) r = ri and when r = r0 σz(max) = ∴ 1 − 2μ ri2 3 − 2μ ρω 2 ro2 1 + . 4 (1 − μ ) 3 − 2μ ro2 μ ρω 2 ro2 − ri2 4(1 − μ ) Also σz = 0 ri2 + ro2 r= 2 when 2r2 = ri2 + ro2 @seismicisolation @seismicisolation (Compressive) (v) 548 • Strength of Materials σc σz σr ri ro Figure 24.11 Variation of stresses in rotating hollow cylinder E XAMPLE 24.12: A shaft of 250 mm diameter is rotating at 3200 r.p.m. Determine the maximum radial and hoop stresses in the shaft. For the shaft material take ρ = 7800 kg/m3 and μ = 0.3. S OLUTION : The shaft may be taken as a solid cylinder. ∴ σr = A − ω= Now, ∴ 3 + 2μ ρω 2 r2 8 (1 + μ ) 3200 × 2π = 334.9 rad/s 60 σr = 0 when r = 125 mm = 0.125 m 0 = A− 3 − 2 × 0.3 × 7800 × 334.92 × (0.125)2 8 (1 + 0.3) 0 = A− 2.4 × 13669257 10.4 A = 3154444 σr = 3154444 − 2.4 × 7800 × 334.92 × r2 10.4 σr = 3154444 − 201884418r2 Now σr and σc are maximum at the centre of the cylinder (shaft), where r = 0 σr(max) = σc(max) = 3154444 = 3.15 MPa Ans @seismicisolation @seismicisolation Rotating Discs and Cylinders • 549 E XAMPLE 24.13: Also determine radial stress and hoop stress at radius 20 mm, 40 mm, 100 mm and 250 mm and show stress distribution for the example 24.11. σr = A − As already found 3 − 2μ ρω 2 r2 8 (1 − μ ) A = 3154444 σr = 3154444 − 2.4 × 7800 × 334.92 r2 10.4 σr = 3154444 − 201884418 r2 μ ρω 2 2 2 σc = ro − r 1−μ 4 0.3 7800 × 334.92 × 0.1252 − r2 = 0.7 4 σc = 93732051 0.015625 − r2 At r = 0.02 m Using Eqns. (i) and (ii) σr = 3154444 − 201884418(0.02)2 = 3073690 = 3.07 MPa Ans σc = 93732051(0.015625 − 0.022 ) = 93732051(0.015225) = 1427070 = 1.4 MPa Ans At r = 0.04 m σr = 3154444 − 201884418(0.04)2 = 3154444 − 323015 = 2831429 = 2.8 MPa Ans σc = 93732051(0.015625 − 0.042 ) = 93732051 × 0.014025 = 1314592 = 1.31 MPa Ans @seismicisolation @seismicisolation (i) (ii) 550 • Strength of Materials Now σr and σc are maximum at the centre of the cylinder (shaft) where r = 0 σr(max) = σc(max) = 3154444 = 3.15 MPa Ans E XAMPLE 24.14: For example 24.11, determine the radial and hoop stresses at radii 20 mm, 40 mm, 100 mm and 125 mm and show stress distribution in a diagram. S OLUTION : σr = A − 3 − 2μ ρω 2 r2 8 (1 − μ ) A = 31544444 Now, σr = 3154444 − (already calculated) 2.4 × 7800 × 334.92 r2 10.4 = 3154444 − 201884418 r2 σc = A − 1 + 2μ ρω 2 r2 8 (1 − μ ) = 3154444 − 1.6 × 7800 × 334.92 r2 5.6 = 3154444 − 185861845 r2 Using Eqns. (i) and (ii) At r = 20 mm = 0.02 m σr = 3154444 − 201884418 (0.02)2 = 3073690 = 3.07 MPa Ans σc = 3154444 − 185861845 (0.02)2 = 3154444 − 74345 = 3080099 At (i) = 3.08 MPa Ans r = 40 mm = 0.04 m σr = 3154444 − 201884418 r2 = 3154444 − 201884418 (0.04)2 = 3154444 − 323015 @seismicisolation @seismicisolation (ii) Rotating Discs and Cylinders = 2831429 = 2.8 MPa Ans σc = 3154444 − 185861845 r2 = 3154444 − 185861845 (0.04)2 = 3154444 − 297379 = 2857065 = 2.86 MPa Ans At r = 100 mm = 0.1 m σr = 3154444 − 201884418 (0.1)2 = 3154444 − 2018844 = 2952599 = 1.13 MPa Ans σc = 3154444 − 185861845 r2 = 3154444 − 185861845 (0.1)2 = 3154444 − 1858618 = 1295825 = 1.29 MPa At r = 125 mm = 0.125 m Ans σr = 3154444 − 201884418 (0.125)2 =0 σc Figure 24.12 @seismicisolation @seismicisolation σr • 551 552 • Strength of Materials E XAMPLE 24.15: A steel cylinder of 350 internal diameter and 700 mm external diameter is rotating at 2200 r.p.m. Calculate the maximum value of radial, circumferential and longitudinal stresses. Also determine the maximum shear stress in the cylinder. Take ρ = 7700 kg/m3 and μ = 0.3 S OLUTION : σr = A − B 3 − 2μ − ρ ω 2 r2 r2 8 (1 − μ ) B 1 + 2μ − ρ ω 2 r2 2 8 (1 − μ ) r μ σz = − ρω 2 (ri2 + r02 − 2r2 ) 4 (1 − μ ) σc = A + 2π × 2200 = 230.3 rad/s 60 B 3 × 2 × 0.3 σr = A − 2 − × 7700 × 230.32 r2 8 (1 − 0.3) r ω= B 2.4 × 7700 × 230.32 r2 − r2 5.6 B = A − 2 − 175 × 106 r2 r B σc = A + 2 − 116.83 × 106 r2 r 0.3 σz = × 7700 × 230.32 0.1752 + 0.352 − 2r2 4 (1 − 0.3) = 43.76 × 106 0.153 − 2r2 σr = A − Similarly, (i) (ii) (iii) In order to find out constants A and B, From Eqn. (i) σr = 0 when r = 0.175 m and r = 0.35 m B − 175 × 106 (0.175)2 0 = A− (0.175)2 0 = A − 32.65 B − 5359375 Also 0 = A− B 2 (0.35) − 175 × 106 (0.35)2 0 = A − 8.16 B − 21.4 × 106 32.65 B + 5359375 = 8.16 B + 21.4 × 106 24.49 B = 21400000 − 5359375 B = 654987 @seismicisolation @seismicisolation (iv) Rotating Discs and Cylinders Putting the value of B in Eqn. (iv) 0 = A − 32.65 × 654987 − 5359375 A = 21385325 − 5359375 = 16025950 ∴ Equations (i) and (ii) become, 654987 − 175 × 106 r2 r2 654987 σc = 16025950 − − 11683 × 106 r2 r2 σr = 16025950 − Now σr is maximum when r2 = ri ro = (0.175) (0.35) = 0.06125 ∴ σrmax = 16025950 − 654987 − 1.75 × 106 × 0.06125 0.06125 = 16025950 − 10693665 − 107187 = 5225098 = 5.22 MPa Ans σc is maximum at r = 0.175 m σc(max) = 16025950 + 654987 (0.175)2 − 116.83 × 106 × (0.175)2 = 16025950 + 21387331 − 3577918 = 33835363 = 33.83 MPa Ans σz is maximum r = ri = 0.175 m (Tensile) and r = ro = 0.35 (Compressive) From Eqn. (iii), σzmax = ± 43.76 × 106 0.153 − 2r2 = ± 43.76 × 106 0.153 − 2 (0.175)2 = ± 43.76 × 106 (0.153 − 0.06125) = ± 4.01 MPa Ans @seismicisolation @seismicisolation • 553 554 • Strength of Materials For shear stress, σc = 33.83 MPa, σr = 0, σz = 4.01 MPa σmax − σmin ∴ 2 = 16.92 MPa Ans τmax = Zmax = 33.83 − 0 2 Disc of Uniform Strength As we have seen that radial and hoop stresses in a rotating disc of uniform thickness keep on changing with the change in radius. Since hoop stress is maximum at the centre and decreases towards the periphery. And even the radial stress is zero at the periphery. Therefore, it is possible to design a disc of uniform strength and vary the thickness to make it economical. Let us consider a flat rotating disc of uniform strength (refer it Figure 24.13). Centrifugal force t+dt dr r Element dθ 2 A σ B σ D C σ dθ 2 t dθ t0 Axis of rotation (a) o (b) Figure 24.13 Figure (a) shows the elevation of half such a disc and (b) shows the free body diagram of an element ABCD of the disc which subtends an angle d θ at centre o. Let the uniform stress in the radial and circumferential directions be σ as shown in figure (b) Volume of the element = r.d θ .t.dr Centrifugal force acting on the element ABCD due to rotation = ρ .r.d θ .t.dr.ω 2 r = ρ d θ .t.dr.ω 2 r2 Radial force on the face DC = r.d θ .t.σ Radial force on the face AB = (r + dr)d θ .(t + dt)σ Circumferential force on faces BC and DA = t.dr.σ Resolving all the forces along the radial direction and considering equilibrium, we get dθ ρ .d θ .t.dr.ω 2 .r2 + (r + dr)d θ (t + dt)σ = rd θ .t.σ + 2tdr sin .σ 2 @seismicisolation @seismicisolation Rotating Discs and Cylinders • 555 dθ dθ = because d θ is very small. 2 2 Cancelling d θ on both sides, the above equation is simplied as follows: It may be noted that sin p.t.dr.ω 2 r2 + r.dt.σ = 0 dt σ = −ρω 2 r dr t or Integrating both sides, we get loge t = ρω 2 r2 + loge 2σ (loge A is a constant of integration) loge t ρω 2 r2 =− A 2σ −ρω 2 r2 t = e 2σ A At r = 0, t = to to = A ∴ (Note this point) ρ ω 2 r2 2σ Thickness at any radius, t = t0 e − E XAMPLE 24.16: The rotor disc of a turbine is of 900 mm diameter at the blade ring and is fixed to a 76 mm diameter shaft. If the minimum thickness of the disc is to be 10 mm, find the thickness at the shaft for a uniform stress of 200 MPa at 8000 r.p.m. Density ρ of the disc material is 7800 kg/m3 . S OLUTION : −ρω 2 r2 We know t = t0 e 2σ At At r = 0.45 m, r = 0.038 m, −ρω 2 (0.45)2 2σ 10 = t0 e −ρω 2 (0.038)2 2σ t = t0 e Dividing Eqn. (ii) by (i) −ρω 2 (0.038)2 ρω 2 (0.2025 − 0.00144) /2σ 2σ = 8e t = 10 × −ρω 2 (0.45)2 t0 e 2σ t0 e @seismicisolation @seismicisolation (i) (ii) • 556 Strength of Materials Where 2π × 8000 2 0.20106 ρω 2 × 0.20106 = 7800 × × 2σ 60 2 × 200 × 106 = 2.749 ∴ t = 8e−2.749 = 8 × 1.5627 = 125 mm Ans E XAMPLE 24.17: A grinding wheel is 320 mm diameter with the bore at the centre 30 mm diameter. If the thickness of the wheel at the outer radius is 25 mm, what should be the thickness at the bore diameter for a uniform allowable stress of 12 MN/m2 at 3000 r. p. m. Take density of the wheel material as 2750 kg/m3 . S OLUTION : r1 = 0.16 m, t1 = 0.025 m. r2 = 0.015 m, t2 =? 3000 × 2π = 314 rad/s 60 σ = 12 × 106 N/m2 and ρ = 2750 kg/m3 ω= Now, −ρω 2 r2 t = t0 e 2σ ρω 2 r2 2750 × (314 × 0.16)2 = = 0.2892 2σ 2 × 12 × 106 0.025 = t0 e−0.2892 t0 = 0.025 e−0.2892 m 0.025 m 1 e0.2892 0.025 m t0 = 0.749 t0 = 0.0333 m t0 = = 33.3 mm Again, −2750×(314×0.0125)2 2×12×106 t2 = 33.3e t2 = 33.3e −1765.2 106 @seismicisolation @seismicisolation Rotating Discs and Cylinders • 557 = 33.3e−0.001765 = 33.3 × 1 e0.001765 1 1.00177 = 33.24 mm Ans = 33.3 × Exercise 24.1 A flywheel rim with a mean diameter of 6 m rotates at a speed such that the hoop stress in the material is 10 MPa. The density of the material of the rim is 7000 kg/m3 . Determine the speed ignoring the effect of arms. [Ans 120.3 r.p.m.] 24.2 Calculate the hoop stress in a thin rim, 0.6 m mean diameter revolving about its axis at 800 r.p.m. Steel weighs 7700 kg/m3 . [Ans 47.7 N/m2 .] 24.3 Determine the greatest values of radial and hoop stresses for a rotating disc in which the outer and inner radii are 0.3 m and 0.15 m. The angular speed is 150 rad/s. Take μ = 0.309 and ρ = 7700 kg/m3 . [Ans 1.6 MN/m2 ; 13.6 MN/m2 ] 24.4 A steel cylinder of 300 mm internal and 600 mm external diameters is rotating at 2000 r.p.m. Calculate the maximum value of radial, circumferential and longitudinal stresses. Also find the maximum shear stress is the cylinder. Take ρ = 7800 kg/m3 and μ = 0.3 [Ans σr(max) = 3.3 MPa, σcmax = 27.48 MPa, τmax = 13.74 MPa] 24.5 A disc of uniform thickness having inner and outer diameter 100 mm and 400 mm respectively is rotating at 5000 r.p.m. about its axis. Take ρ = 7800 kg/m3 , and μ = 0.28. Determine the stress distribution along the radius of disc [Ans σr(0.15) = 0; σr(0.1 m) = 19.74 MPa σr(0.15) = 13.64 MPa σr(0.2) = 0 σr(max) = 19.74 MPa σc0.05 m = 71.09 MPa; σc0.1 m = 41.11 MPa σc0.15 = 30.09 MPa σc(0.2) = 19.78 MPa; σc(max) = 71.09 MPa, τmax = 35.54 MPa] 24.6 A disc having inner and outer diameters 150 mm and 300 mm respectively is rotating at an angular speed of 150 rad/s. Calculate the greatest values of radial and circumferential stresses. Take ρ = 7700 kg/m3 and μ = 0.304. [Ans 16 MN/m2 , 13.6 MN/m2 ] 24.7 A stress turbine rotor running at 3500 r.p.m. is to be designed so that the radial and circumferential stresses are to be same and constant throughout and equal to 80 MPa. If the axial thickness at the centre is 15 mm, what is the thickness at a radius of 500 mm? Density = 7800 kg/m3 . [Ans @seismicisolation @seismicisolation 2.92 mm] 558 • Strength of Materials 24.8 A disc of turbine rotor is 0.5 m diameter. At the blade ring its thickness is 55 mm. It is keyed to a shaft of 50 mm diameter. If the uniform stress in the rotor disc is limited to 200 MN/m2 at 9000 r.p.m. Determine the thickness of the disc at the start. Take ρ = 7700 kg/m3 . [Ans 158.43 mm] 24.9 A solid cylinder, 250 mm diameter is rotating at 1500 r.p.m. Determine the maximum hoop stress and the radial stress produced in the cylinder if the density of material is 7800 kg/m3 and μ = 0.28. Also determine the places where these stresses are maximum. What is the radial stress at the outer radius? [Ans σc = 1.273 MPa at the centre, σr = 1.273 MPa at the centre, zero] @seismicisolation @seismicisolation C HAPTER 25 FRAMEWORKS A framework is an assembly of bars connected by hinged or pinned joints and intended to carry loads at the joints only. Each hinge joint is assumed to rotate freely without friction, hence all the bars in the frame exert direct forces only and are therefore in tension or compression. A tensile load is taken as positive and a member carrying tension is called a tie. A compressive load is negative and a member is compression is called a strut. The bars are usually assumed to be light compared with the applied loads. In practice the joints of a framework may be riveted on welded but the direct forces are often calculated assuming pin joints. This assumption gives values of tension or compression which are on the safe side. Frames are sometimes classified as 1. Perfect frame and 2. Imperfect frame. Perfect frame: In a perfect frame, this is made up of members sufficient to keep it in equilibrium without any change in its frame. The simplest perfect frame is a triangle which has three members. The number of members, in a perfect frame, may also be expressed by the relation: W A n = (2 j − 3) n = no. of members j = no. of joints C B Figure 25.1 Deficient frame: This is an imperfect frame which does not satisfy the equation n = (2 j − 3). In this case members are more or less than required number n. Redundant frame: This is also an imperfect frame in which the number of members is more than (2 j − 3). @seismicisolation @seismicisolation 560 • Strength of Materials To solve the frames to determine if the members are in tension or in compression, there are three methods, named i) Graphical method (Bow’s notation) ii) Analytical method or method of joints iii) Method of sections Bow’s Notation for Graphical Solution E XAMPLE 25.1: The framework shown in Fig. 25.2 is composed of equilateral triangles. Determine the nature and magnitude of the forces in each member of the framework. 800 N 400 N B 3 4 C A F E 1 5 G 2 D R1 = 700 N R2 = 500 N Figure 25.2 S OLUTION : Taking moments about the extreme left hand pin-joint to find R2 : Clockwise moments = Anticlockwise moments 1 3 800 span + 400 span = CD(span) 4 4 800 1200 + = CD 4 4 ∴ CD = 200 N + 300 N = 500 N Now, downward forces = Upward forces 800 + 400 = R1 + 500 ∴ R1 = 1200 − 500 = 700 N Note: It may be noted that reaction R2 is called CD, reading clockwise around this joint. Framework Remember if we read member 1 − 3 around joint A, it will be read clockwise as AE (i.e., A to E near joint A and EA near joint 3). In vector diagram we use small letters. Now we proceed to draw the vector diagram to a suitable scale (for example, 1 kN = 1.5 cm). The vector diagram is shown in Fig. 25.3. @seismicisolation @seismicisolation Frameworks • 561 The sequence used in constructing the diagram is given below: The equilibrium of external loading is a vertical line of loads: ab = 800 units downward bc = 400 units downward cd = 500 units upward da = 700 units upward. a g e f d b c Figure 25.3 Vector diagram Proceeding to the extreme left hand pin-joint, a line from a is drawn parallel to member AE, and a line from d is drawn parallel to member ED. The point e is the intersection of these lines. The same procedure is applied to the extreme right-hand pin-joint permitting the positioning of point g. The next joint chosen is the pin-joint at which 800 N load in acting. Starting from member EA and proceeding clockwise round the joint, vectors ea and ab already appear on the vector diagram. From b a line is drawn parallel of FE, this positions the point f . The closing line g f is obtained by considering the forces acting at the pin-joint under the load of 400 N. This makes the diagram complete. And it should be observed that the diagram is a set of closed figures, there being no lines which do not form a part of a polygon. We can now return to the force diagram and place arrows near the pin-joints. For instance, for the extreme left-hand pin-joint, the arrow on member AE points towards the joint, since the sense of e to d on the vector diagram is ‘horizontally to the right’. Arrows at the other ends of AE and ED are in opposing direction to those near the extreme left-hand pin-joint. @seismicisolation @seismicisolation 562 • Strength of Materials B A F E C G D Figure 25.4 The same procedure is applied to the remaining joints, producing arrows shown in Fig. 25.4. We are now in position to state the nature of the forces in the various members, since if arrows point together or pointing towards each other, the member is in tension while if they point apart (away) the member is in compression. The magnitude of forces could have been obtained by scaling the diagram, but the vectors of ef and f g are relatively small, and inaccuracy in the graphical work could cause significant errors. Now db represents 100 N hence e f = f g = 100 N × (cosec 60◦ ) = 115.5 N. It would be advisable in many cases to obtain results by scaling the diagram. The solution is completed by using trigonometry to obtain length of vectors and by tabulating the answers to a degree of accuracy of three significant figures. Answers: Member Load(N) Nature AE 808 C BF 346 C CG 577 C DG 289 T DE 404 T EF 115 C FG 115 T C = Compression, T = Tension E XAMPLE 25.2: The framed structure shown in Figure 25.5 is maintained in the position by a force P. Determine: (a) the magnitude of the force P; (b) the reaction R at the hinge in magnitude and direction; (c) the magnitude and nature of the force in each member. 400 N C P B D G F H 1m R Hinge 2m A 400 N 2m E Figure 25.5 @seismicisolation @seismicisolation 400 N Frameworks • 563 Taking moments about the hinge: Clockwise moments = Anticlockwise moments 400 × 2 + 400 × 3 + 400 × 4 = P × 1 800 + 1200 + 1600 = P ∴ P = 3600 N The vector diagram is shown in Fig. 25.6 (a) and the structure with arrows on the members is shown in Fig. 25.6 (b). g c b d e h a f Figure 25.6a B n= ctio Rea 400 N C P R D G H F A E 400 N 400 N Figure 25.6b Answers: P = 3600 N Hinge reaction, R = 3790 at 18◦ 26 Member Load(N) Nature BF 1700 C CG 1200 T DH 566 T with horizontal EH 400 C C = Compression, T = Tension @seismicisolation @seismicisolation AF 2400 C FG 1700 T GH 1130 C 564 • Strength of Materials E XAMPLE 25.3: The wall crane shown in Fig. 25.7 is loaded by a vertical force of 10 kN at the joint ABD. Determine graphically the magnitude and nature of the force in each member and the vertical support reactions at the pin-joints ADC and BCD. 1 kN A B D 30° 60° C RCA RBC Figure 25.7 S OLUTION : Let DC be 5 units, then joint ABD is a right angle, it can be worked out A = 4.33 and BD = 2.5 units P s nit 3u 3 . 4 30° 2.5 units 60° M 5 units O N Figure 25.8 In ONP, ∴ ON = cos 60◦ NP ON 1 = 2.5 2 2.5 = 1.25 units ON = 2 Taking moments about joint DBC (Fig. 25.7) RCA × 5 = 1 × 1.5 1.25 RCA = = 0.25 kN 5 RBC = 1 − 0.25 = 0.75 kN @seismicisolation @seismicisolation Frameworks • 565 Now we proceed to draw vector diagram: a c d b Figure 25.9 Vector diagram 1 kN A B D C 0.25k N 0.75 kN Figure 25.10 Note: According to Bow’s notation, force 1 kN will be read as ABD (reading clockwise around joint). After measurement from vector diagram, results are tabulated below: Member Load (kN) Nature AD 0.5 C BD 0.867 C CD 0.433 T Reactions : RCA = 0.25 kN RBC = 0.75 kN E XAMPLE 25.4: The wall crane shown in Fig. 25.11(a) is loaded by a vertical force of 10 kN at joint ABD. The bar AD is horizontal and the framework is pinned to the vertical wall at joints ADC and BC. Determine the magnitude and nature of the forces in the framework. What are the magnitude and direction at the top wall joint? Note: The arrowheads on members and reactions are shown in Fig. 25.12(a) as answer are not in question. @seismicisolation @seismicisolation 566 • Strength of Materials A 45 45° D 45 10 kN C C d a B 30° b (a) joint ABD (b) Figure 25.11 S OLUTION : It may be noted that vertical line of the wall is not a load or bar of the frame. Space C separates the reaction BC at the lower wall joint and the reaction CA at the upper wall joint. The reaction CA is unknown in magnitude and direction. The reaction of the wall at the lower joint is equal and opposite to the force in bar BC. Joint ABD Fig. 25.11(b) Draw ab vertically downwards from a to represent the 10 kN load AB. Draw bd parallel to bar BD, and ad parallel to bar DA. The intersection d of bd and ad complete the force diagram for the joint. The directions of forces at the joint are fixed by the known downward vector ab. Following the vectors in order round triangle abd determines the direction bd (towards the joint) and the direction da (away from the joint). These directions are shown in Fig. 25.11(b). Joint DBC (Fig. 25.12 a): The known force at this joint is db equal and opposite to bd. From b draw force bc parallel to BC. From d draw force dc parallel to CD. Thus joint c is determined. The force directions at the joint are fixed by the vector arrow from d to e as shown. Joint ADC (Fig. 25.12(b)): The known force is ad in AD, equal and opposite to da, already found. Join ca. This, then represents the force in CA, the reaction at the joint. It acts away from the joint. The polygon abdc in complete force diagram for the framework. By construction, the forces are found as follows: Reaction at top well joint =7.8 kN (CA) at 20◦ . Reaction at bottom = 14.6 as shown in Fig. 25.11(a) Member Load (kN) Nature BC 14.6 C CD 3.8 C @seismicisolation @seismicisolation DB 14.14 C DA 10 T • Frameworks 567 c c d a Joint DBC a d b (b) Joint ADC b (a) Figure 25.12 E XAMPLE 25.5: The Warren girder shown in Fig. 25.13(a) is loaded vertically by forces of 20 and 30 kN at the lower panel pin-joints. It is simply supported at the joints ABG and AED. Determine the magnitude and nature of forces in each bar and the magnitudes of the support reactions. Note: Arrowheads shown in Fig. 25.13(a) are not in question but meant as answers. S OLUTION : A E A G F D b A B C 20 kN 30 kN b (a) b e g a c f a g Joint ABG (b) a g c f Joint GBCF (c) Figure 25.13 @seismicisolation @seismicisolation d (d) 568 • Strength of Materials There are more than two unknown forces at every joint; hence we cannot begin the force diagram immediately. It is necessary first to calculate the vertical reactions at the supports by taking moments about each support in turn. let L = vertical reaction DA R = vertical reaction AB Assume each horizontal bar of the frame to be of length 2 units. Taking moments about joint AED, 4 × R = (30 × 3) + (20 × 1) R = 27.5 kN Taking moments about joint ABG, 4 × L = (30 × 1) + (20 × 3) L = 22.5 kN Check: L + R = 22.5 + 27.5 = 50 kN = total load. Joint ABG (Fig. 25.13(b)) Draw ab vertically upwards to represent the 27.5 kN reaction AB. Draw bg through b parallel to BG. Draw ga through a parallel to GA. The force directions at the joints are fixed by the upward vector ab. The direction of the force in BG is away from the joint. The direction of the force GA is towards the joint. The direction arrows are now added to Fig. 25.13(a). Joint GBCF (Fig 25.13(c)): Draw bc to represent the downward load BC of 30 kN. Draw f g through g parallel to FG. Draw e f through e parallel to CF. The intersection of f g and e f given point f and complete the diagram for the joint. The directions of the forces at the joint are fixed by the known direction bc of the load BC. Tranversing the polygon be f g in order determines the force directions. Joint CDEF (Fig 25.13 d): Draw cd vertically downward from c to represent the known load CD of 20 kN. Draw de parallel to DE. Draw e f parallel to EF. If correctly drawn, the point e lies on the intersection of de and e f , and also on the horizontal line ae, parallel to AE. @seismicisolation @seismicisolation Frameworks • 569 The directions of the forces at the joint are fixed by the known direction of cd. The directions are found by following the arrows round the polygon cde f in order. Results: Reaction AB = 27.5 kN Reaction DA = 22.5 kN Member Load (kN) Nature BG 39 T GA 27.5 C CF 25 T FG 3.5 T DE 3.2 T EF 3.5 C EA 22.5 C T for tension and C for compression. E XAMPLE 25.6: Figure 25.14 shows a truss of total span 30 m. Determine the magnitude and nature of forces in all members of the framework. P P P C K P/2 B A D E M L N O 2m 2m R1 = 3P 2m P F P J 2m P Q R S 2m R2 = 3P 6P = 3P 2 b p c r j,s R,g n,o k d i e f m Vector diagram Figure 25.14 @seismicisolation @seismicisolation H 2m Framework Due to symmetry reaction R1 = R2 = P/2 G g 570 • Strength of Materials Answers: Member Magnitude Nature T for Tension C for Compression BJ, GS JI, IS JK, RS CK, RF KL, QR LI, IQ LM, PQ DM, EP MN, OP NI, IO NO 6.73 P 6.25 P 1.00 P 6.73 P 1.60 P 1.00 P 1.50 P 5.40 P 1.95 P 4.75 P 0 C T C C T T C C T T − Method of joints: In this method, every joint is considered separately and conditions of equilibrium are applied. Of course reactions in some cases are found out beforehand. E XAMPLE 25.7: The framework shown in Fig. 25.15 has a span of 6 m and a load of 10 kN is applied at top. Find the magnitude of nature of forces in the members AB, AC and BC. S OLUTION : 10 kN A 60° B RB 30° D 6m Figure 25.15 For taking moments we will need BD, so BD = cos 60 AB ∴ BD = 0.56 AB In a right angled triangle BAC cos 60◦ = AB AB = BC 5 AB = 6 cos 60◦ @seismicisolation @seismicisolation C RC Frameworks ∴ AB = 6 × 571 F1 1 2 60° =3m Now • F2 7.5 kN BD = 0.5 AB Figure 25.16. ∴ BD = 0.5 × 3 = 1.5 m Now taking moments about B, 10 × 1.5 = RC × 6 = 0 ∴ RC = 15 = 2.5 kN 6 RB = 10 − 2.5 and = 7.5 kN Joint B: Let force F1 act towards the joint B and F2 acting away from the joint B. Resolving the force at B, vertically, F1 sin 60 = 7.5 7.5 sin 60 7.5 = 0.866 F1 = F1 = 8.66 kN As F1 is pushing the joint B, therefore this force will make member BA compressive. Now resolving horizontally, F1 cos 60 = F2 1 8.66 × = F2 2 ∴ F2 = 4.33 kN (member BC will be in tension) @seismicisolation @seismicisolation 572 • Strength of Materials Joint C: F2 F3 30° F2 is already known equal to 4.33 kN. Now resolving vertically, F3 sin 30 = 2.5 2.5 F3 = 0.5 = 5 kN 2.5 kN Figure 25.17 (It will make the member AC in compression). Ans: Results : RB = 7.5 kN RC = 2.5 kN Member Force (kN) Nature BC 4.33 T AB 8.66 C AC 5 C E XAMPLE 25.8: A truss as shown in Fig. 25.18 in loaded. Determine the magnitude and nature of force in each member. S OLUTION : 2000 N 4000 N C B A 60° 60° 60° 60° E 4m RA = 2500 N D 4m RD = 3500 N Figure 25.18 Taking moments about D, RA × B = 2000 × 6 + 4000 × 2 = 12000 + 8000 20000 = 2500 N RA = 8 @seismicisolation @seismicisolation Frameworks RA + RD = 2000 + 4000 = 6000 2500 + RD = 6000 RD = 6000 − 2500 = 3500 N Joint A: Resolving vertically B F2 sin 60 = 2500 F2 2500 sin 60 2500 = 0.866 F2 = 60° A E F1 2500 N Figure 25.19 = 2887 N Resolving horizontally F2 cos 60 = F1 ∴ F1 = 2887 × 0.5 = 1443.5 N (∴ member AE will be in tension) Joint D Resolving vertically C F4 sin 60 = 3500 3500 F4 = 0.866 = 4041.6 N ∴ 60° E Member CD will be under compression F4 D F3 3500 N Figure 25.20 @seismicisolation @seismicisolation • 573 574 • Strength of Materials Resolving horizontally, F3 = F4 cos 60◦ F3 = 4041.6 × 0.5 = 2020.8 N ∴ Member ED will be under tension. Joint B: We already know from study of joint A that F2 = 2887 N 2000 N Resolving vertically 2000 + F6 sin 60 = 2887 sin 60 0.866F6 = 2500 − 2000 500 F6 = 0.866 = 577 N F5 B 60° 887 N 2 C 60° F6 A E Figure 25.21 Resolving horizontally, 2887 cos 60 + 577 cos 60◦ = F5 (∵ F6 = 577 N) 1443.5 + 288.5 = F5 F5 = 1732 N Therefore, BC will be under compression. Joint C: 4000 N We have already found out found in CD = 4042 N and force in BC = 1732 N B 1732 N E Resolving vertically, 4000 = F6 sin 60 + 4042 sin 60 4000 = 0.866 F6 + 3500 F6 = 4000 − 3500 0.866 = 577.4 N @seismicisolation @seismicisolation C 60° F6 60° 4042 N A Frameworks • 575 Results: RA = 2500, RB = 3500 N Member Force N Nature AB 2887 N Comp BC 1732 N Comp CD 4041.6 Comp BE 577 Tension CE 577 N Comp AE 1443.5 Tension ED 2021 Tension E XAMPLE 25.9: For the truss shown in Fig. 25.22, determine the force in all the members in magnitude and nature. 5 kN F G H J K 5m θ A C 3.75 m RA = 7.5 kN D 3.75 m E 3.75 m B 3.75 m RB = 7.5 kN 5 kN 5 kN Figure 25.22 S OLUTION : Due to symmetry of truss and loading, RA = RB = 5+5+5 = 7.5 kN 2 √ In right angled triangle ACF, CF = 3.752 + 52 CF = √ 14 + 25 CF = 6.245 m ∴ sin θ = 5 = 0.8 6.245 θ = 53.13◦ cos θ = 3.75 = 0.6 6.245 @seismicisolation @seismicisolation 576 • Strength of Materials Joint A: Resolving horizontally F1 F2 = 0 A F2 Resolving vertically F1 = 7.5 kN RA = 7.5 kN ∴ Member AF is under compression. Figure 25.23 Joint F: Resolving horizontally, F F4 = F3 cos θ = F3 × 0.6 F4 θ F3 F1 = 7.5 kN Figure 25.24 Resolving vertically, 7.5 = F3 sin θ 7.5 = F3 × 0.8 7.5 F3 = = 9.375 kN (Member FC is under tension.) 0.8 F4 = 0.6 F3 F4 = 0.6 × 9.375 = 5.625 kN ∴ Member FG under compression. Joint C: As we found out earlier F2 = 0 Resolving horizontally, F5 F3 = 9.375 F6 = 9.375 cos θ = 9.375 × 0.6 = 5.625 kN F6 θ F2 = 0 C 5 kN ∴ Member CD is under tension. Figure 25.25 @seismicisolation @seismicisolation Frameworks • 577 Resolving vertically, F5 + 9.375 sin θ = 5 F5 + 9.375 × 0.8 = 5 F5 = 5 − 7.5 = −2.5 kN The negative sign shows that the choice in assuming direction of F5 was wrong. Therefore, the assumed direction of F5 is member CG need to be reversed. Hence, CG is under compression. Joint G: Resolving horizontally, 5.625 + F7 cos θ = F8 5.625 + F7 × 0.6 = F8 Resolving vertically, G FA = 5.625 kN 2.5 = F7 sin θ 2.5 = F7 × 0.8 2.5 = 3.125 kN F7 = 0.8 ∴ ∴ F7 F5 = 2.5 kN Figure 25.26 GD is under tension ∴ F8 θ 5.625 + 3.125 × 0.6 = F8 ∴ F8 = 5.625 + 1.875 = 7.5 kN GH is under compression Joint H: Resolving horizontally, 5 kN 7.5 = F10 ∴ F10 member HJ is under compression. Resolving vertically, 5 = F9 ∴ F8 = 7.5 kN H F9 F5 = 2.5 kN member HD is under compression. @seismicisolation @seismicisolation Figure 25.27 578 • Strength of Materials Since the the truss is symmetrical, there is no need to consider remaining joints. Results: RA = RB = 7.5 kN. Member Force (kN) Nature AF, BK 7.5 C AC, BE 0 - CF, EK 9.375 T FG, JK 5.625 C GC, JE 2.5 C CD, DE 5.625 T DG, DJ 3.125 T GH, HJ 7.5 C HD 5 C E XAMPLE 25.10: A cantilever truss is shown in Fig. 25.28. Determine the magnitude and nature of the force in each member. 50 kN F 60° A 37.5 kN 50 kN E 60° 60° B 12.5 m D 60° 37.5 kN C 12.5 m Figure 25.28 Joint A: Resolving horizontally, F1 F2 = F1 cos 60 F2 = 0.5 F1 60 A F2 37.5 kN Figure 25.29 Resolving vertically, 37.5 = F1 sin 60 37.5 = F1 × 0.866 F1 = 37.5 = 43.3 kN 0.866 ∴ F2 = 0.5 × 43.3 = 21.65 kN Therefore, the member AF is under tension and the member AB is under compression. @seismicisolation @seismicisolation Frameworks Joint F: Resolving horizontally, 50 kN 60° F 60° F4 B F3 F 1= 43 .3 F4 = 43.3 cos 60 + F3 cos 60◦ F4 = 21.65 + 0.5F3 A Figure 25.30 Resolving vertically, F1 sin 60◦ + 50 = F3 sin 60 43.3 × 0.866 + 50 = 0.866F3 37.5 + 50 = 0.866 F3 F3 = 87.5 0.866 = 101.04 kN Therefore, member FB is under compression F4 = 21.65 + 0.5 × 101.04 = 21.65 + 50.52 = 72.17 kN Member FE is under tension Joint B: E F5 Resolving horizontally, 21.65 + 101.04 cos 60 + F5 cos 60 = F6 21.65 + 50.52 + 0.5F5 = F6 101.04 60° 60° C A F6 21.65 kN B Resolving vertically, 37.5 + 101.04 sin 60 = F5 × sin 60 37.5 + 101.04 × 0.866 = 0.866F5 37.5 + 87.5 = 0.866F5 125 = 144.34 kN F5 = 0.866 @seismicisolation @seismicisolation 37.5 kN Figure 25.31 • 579 580 • Strength of Materials Member BF is under tension Now, 21.65 + 50.52 + 0.5 × 144.34 = F6 F6 = 144.34 kN Therefore, member BC is under compression. Joint E: Resolving horizontally, 50 kN 72.17 + 144.34 cos 60 + F7 cos 60◦ = F8 72.17 + 72.17 + 0.5F7 = F8 72.17 kN F 60° 144.34 kN Resolving vertically, E F8 60° ◦ D F7 50 + 144.34 sin 60 = F7 sin 60 50 + 144.34 × 0.866 = 0.866F7 Figure 25.32 175 = 0.866F7 F7 = 175 = 202.08 kN 0.866 72.17 + 72.17 + 0.5 × 202.08 = F8 F8 = 245.38 N Therefore member EC is under compression. And member ED is under tension Answer: Member Force Nature AF 43.3 T AB 21.65 C BF 101.04 C FE 72.17 T BE 144.34 T BC 144.34 C CE 202.08 C CD 245.38 T The Method of Sections The general conditions of equilibrium can be developed into a method of finding the magnitudes of forces in frame structures; it is particularly useful when one or few of the members is being considered. The method simply consists of finding reactions, and there making a section cut the member under consideration, finally applying the general condition of equilibrium: Leftward forces = Rightward forces Upward forces = Downward forces Clockwise moments = Anticlockwise moments. @seismicisolation @seismicisolation Frameworks • 581 In certain cases, a simple solution is effected by arranging for the section to cut the member under consideration and two other members. Moments are taken about the intersection of the other two members. We will now solve the following example by this method. E XAMPLE 25.11: Figure 25.33 shows a loaded framework. Determine the magnitude of the force in the horizontal member CF, and state whether this member is in tension or compression. 2000 N 4000 N C F B E R1 D G 4m A R2 Figure 25.33 We will first find R1 by taking moments about the right handed support. Clockwise moments = Anticlockwise moments R1 × 4 m = 4000 N × 1 m + 2000 N × 3 m = 400 Nm + 6000 Nm = 10000 Nm 10000 = 2500 N, ∴ R2 = 2000 + 4000 − 2500 ∴ R1 = 4 = 3500 N Although it is not completely necessary in the solution, let us find R2 to check the value of R1 , Moments about the left hand support: Clockwise moments = Anticlockwise moments 2000 N × 1 m = 4000 N × 3 m = R2 × 4 m 2000 Nm + 1200 Nm = R2 × 4 m 1400 = 3500 N. ∴ R2 = 4m Upward forces = Downward forces 2500 × 3500 N = 2000 N + 4000 N 6000 N = 6000 N @seismicisolation @seismicisolation 582 • Strength of Materials The reactions therefore appears to be correct. We now make a section, if we can, to cut the members under consideration (the force in that member being designates as P) and two other members. This gives us Fig. 25.34. It does not matter, for the moment, in which directions we assume the force P to be acting. 2000 N Section line P O 2500 N Figure 25.34 We now consider the loadings on the portion cut off by the section. We will take the left-hand side but it does not matter which. The two other members intersect at 0. Taking moments about 0: Clockwise moments = Anticlockwise moments P × 1.732 m + 2500 N × 2 m = 2000 N × 1 m P × 1.732 m = 2000 Nm − 5000 Nm = −3000 Nm −3000 Nm ∴ P= 1.732 Nm = −1732 N The negative result tells us the direction we assumed P to be acting was incorrect. P in fact pushes on the joint BCFE. The arrow on the other end of members CF pushes on the opposing joint CDCF. The arrows on members CF therefore point away from each other, and members CF is in compression. Answer: Force in member CF is a compressive force of 1732 N (Note : Had the value of P been positive it would have indicated that the direction in which the force P was assumed to act was correct.) The question has now been solved, but to show the validity of the method, let us repeat the solution with another section, with the members to the right of this section line as shown in Fig. 25.34. This time we will let P push on the joint. @seismicisolation @seismicisolation Frameworks P • 583 4000 N O 3500 N Figure 25.35 Taking moments about 0: Clockwise moments = Anticlockwise moments P × 1.732 m + 4000 N × 1 m = 3500 N × 2 m P × 1.732 m = 7000 Nm − 4000 Nm = 3000 Nm 3000 Nm ∴ P= = 1732 N 1.732 m P is positive, hence the assumption for the direction was correct. The arrow on the other end of member CF points in the opposite direction. The arrows point away from each other. Member CF is therefore in compression. Hence, CF is loaded is compression by a force of 1730 N, as before. In introducing this method, it was suggested that the section cuts the member under consideration and two other members, moments then being taken about the point of intersection of the other two members. This can not apply on every section and in cases it may be necessary to adopt aspects of the general conditions of equilibrium other than moments. To demonstrate this we will follow another question. E XAMPLE 25.12: The framework in Fig. 25.36 is an equilibrium triangle, and the member AD is kept horizontal by a force P. Determine, either by calculation or graphically: a) the magnitude of the force P b) the direction and magnitude of the higher reaction R, and c) the force in member AD, stating it is in tension or in compression. P B C D R A 500 N Figure 25.36 @seismicisolation @seismicisolation 584 • Strength of Materials S OLUTION : Moments about hinge, taking the length of side of the equilateral triangle as L: P × L = 500 N × L ∴ P = 500 N Leftward forces = Rightward forces Let the hinge reaction = R at θ . Horizontal component of P = Horizontal component of R P cos 30◦ = R cos θ 500 × 0.866 = R cos θ = 433 N Upward forces = Downward forces Vertical component of P+ vertical component of R = 500 N P sin 30◦ + R sin θ = 500 N 500 N × 0.5 = R sin θ = 500 N ∴ R sin θ = 500 N − 250 N = 250 N R sin θ 250 = R cos θ 433 or tan θ = 0.5774 ∴ θ = 30◦ R sin θ = 250 N ∴ R= = 250 N 250 = sin θ sin 30 250 0.5 = 500 N ∴ R = 500 N at 30◦ We, now take a section to cut the member under consideration and at least one other member. For simplicity, the section is an indicated in Fig. 25.37. @seismicisolation @seismicisolation Frameworks • 585 Section line Q P 500 N Figure 25.37 Although we only require the value of the force in the member AD, it is necessary to include one other member. Further more, we cannot take moments, so we must use other aspects of the general conditions of equilibrium. Upward forces = Downward forces Q sin 60 = 500 Q × 0.866 = 500 Q= 500 = 577.4 N 0.866 Leftward forces = Rightward forces P = Q cos 66◦ = 577.4 × 0.5 = 288.7 N A positive value of P shows that the assumption made for its direction is correct. The arrow on the other end of member AD points in the opposite direction The arrows point away from each other, hence member AD is in compression. Answers: a) P = 500 N b) R = 500 N at 30◦ c) Force in D = 289 N compressive. E XAMPLE 25.13: The framework shown in Fig. 25.38 carries a load of 19.6 kN at the lower middle point. Find the forces in bars 1, 2, 3, and 4 and state whether the state whether the bars are in tension or compression. @seismicisolation @seismicisolation 586 • Strength of Materials A 1 4 2 60º 30º 60º 30º 3 19.6 kN A Figure 25.38 S OLUTION : √ 3 m, bar 4 = 1 m and the height Let the horizontal members each of length 2 m, then bar 2 = √ 3 of the frame is m. The weight of 19.6 kN is acting as shown. From symmetry the reaction 2 19.6 R1 = R2 = = 9.8 kN 2 A 22.63 kN 19 .6 11.32 kN 19.6 kN 60º 30º 30º 5.66 kN 19.6 kN 9.8 kN 11.32 kN kN 60º 5.66 kN 9.8 kN A Figure 25.39 Get the frame cut by an imaginary plane AA, shown in Fig. 25.38 and let X,Y, Z be the internal forces in the cut bars 1, 2, 3 respectively. If we assume that the cut bars are all in tension then the forces X,Y, Z will act as shown in Fig. 25.40, i.e. forces X and Y are pulling away from the top right-hand joint and force Z is pulling away from the lower right hand support. Q cos 60 1 X 2 4 Q Y 3 Z Q sin 60 Z R = 9.8 kN R = 9.8 kN (a) (b) Figure 25.40 Consider the equilibrium of the frame section to the right of the cutting plane AA. To find force Y resolve external and internal forces on the frame section, thus in the vertical direction: Y sin 30◦ = R = 9.8 Y = +19.6 kN @seismicisolation @seismicisolation Frameworks • 587 Y is positive therefore the assumed direction is correct and bar 2 is in tension. To find force X we may take moments about the joint carrying weight 19.6 kN, then eliminating the forces Y and Z. √ 3 +R×2 = 0 X 2 9.8 × 4 X =− √ = −22.63 kN 3 X is negative (therefore the direction of the force must be reversed) end bar 1 is in compression. To find force Z, resolve forces horizontally for the frame section, thus: X +Y cos 30◦ + Z = 0 −22.63 + 19.6 × 0.866 + Z = 0 Z = +5.66 kN Z is positive, therefore bar 3 is in tension. To find the force in bar 4, we may consider the forces acting at the lower right hand support joint. There are three forces R, Z and the force in bar 4 say Q. Resolving forces horizontally, Q × cos 60◦ = Z = 5.66 Q = 11.32 kN The forces in the remaining bars can be found from symmetry and the complete solution is shown in Fig. 25.39. In practice the correct directions of the forces in the bars can often be obtained by inspection. Exercise 25.1 Fig 25.41 shows a simple roof truss. A wind load normal to the longer sloping side is assumed to be equivalent to a 10 kN load at each pin joint. The reaction at the right hand joint may be taken vertical. Determine the reactions and the nature and magnitude of the force in each member. 10 kN B A 10 kN E 60º 30º D [Ans Members: AE, 10 comp; BE, 0; DE −5 Comp. Reactions CD: 8.66, DA, 13.25 at 41◦ to and above horizontal. Note:- −ve sign denotes compression otherwise member is in tension.] C Figure 25.41 @seismicisolation @seismicisolation 588 • Strength of Materials 25.2 Figure 25.42 shows a loaded framework, the triangle being equilateral. Find by the method of sections, the magnitude of the force in member FG stating whether this is tensile on compressive. 1000 N 3000 N C B D F E G 60º 60º 60º 60º A R1 R2 [Ans Figure 25.42 577 N in compression] 25.3 Determine by the method of sections, the magnitude and nature of forces in members AB, BC and CD only, of the pin-jointed framework shown in Fig. 25.43. (All sloping members lie at 45◦ to the horizontal). R1 1 kN A B C D 2 kN [Ans AB = 5 kN (Compressive) BC = 0 CD = 5 kN (Tensile)] 5 kN R2 Figure 25.43 25.4 In the warren girder shown in Fig. 25.44 all bars are of equal length and the loads are vertical. Find the magnitude and nature of the forces in the members A, B and C. Both reactions are vertical. 20 kN 60° 80 kN 60° 60° B C 60° 60° 60° A 60° 60° 60° [Ans Figure 25.44 @seismicisolation @seismicisolation A: 46.2, B: −23.1 C: −69.3 kN] Frameworks • 589 25.5 For the framework shown in Fig. 25.45, find analytically the magnitude and nature of the forces in all the members of the frame and state the reactions at the supports. F B A J 45° C H E 45° 45° X Y D G 100 kN [Ans AB − 66.7; DA = vertical AJ = Horizontal AJ = 0, BC, 94.3; BE = HG = FH = 33.3 HJ = EF = −47.1, EG, 66.7, AF, −33.3, AC = −66.7 kN; reaction at X = 66.7 kN vertical, reaction at Y = 33.3 kN vertical.] Figure 25.45 25.6 Determine the magnitude and nature of the forces in the framework shown in Fig. 25.46 by any method and state magnitude and direction of the reactions at the supports. [Ans AD, 11.54; BD, −5.77; DC, −11.54; CA, 11.54 kN Reaction at X = 11.54 kN horizontal; Reaction at Y = 15.27 at 48◦ 8 to vertical.] A X C D B Y 10 kN Figure 25.46 25.7 For the framework shown in Fig. 25.47, find by any method magnitude and nature of force in each member. Also find magnitude and direction at hinge X, and force in cable at Y . Cable Y A 30° 30° E 90° C D B 10 kN [Ans AE, 17.32, EB, −20, BD, −17.32; CD = 5; DE, −10 kN Reaction at X = 13.23 kN at 40◦ 54 to vertical. Reaction at Y = pull in cable = 8.66 kN] 30° X Figure 25.47 @seismicisolation @seismicisolation 590 • Strength of Materials 25.8 For the framework shown in Fig. 25.48, find by any method, the force and its nature of each member. Also find reactions with directions. Note: 10 tonnes should be converted into kN. A 10 tonne C 75° B [Ans AC = 98.1, BC = −170, CD = −25.4, AD = 94.8 kN Reaction at X = 94.8 kN parallel to bar AD Reaction at Y = 178 kN at 76◦ 12 to horizontal] D 45° 60° 45° X Y Figure 25.48 25.9 Find graphically the force and its nature in each member of the framework (Fig. 25.49). Also find reaction at X and Y with direction. A 5m 1.5 m E D C X 30° 5m B 20 kN 30° Figure 25.49 @seismicisolation @seismicisolation Y [Ans AD = AE = −33.3 kN, DC = BE = 29.6, DE = 33.3 kN Reaction at X & Y are vertical = 10 kN] C HAPTER 26 DAMS Throughout the year large quantity of water is required for irrigation and power generation. In order to meet this requirement, construction of dams is important to store the water. A retaining wall is constructed to retain the earth in hilly areas to construct a dam. There are many types of dams for example, rectangular dams, trapezoidal dams and so on. Trapezoidal dams are more popular because of being economical and easier to construct. Rectangular Dams Figure 26.1 shows a rectangular dam. For study we will consider a unit length of this dam which is retaining water on one of its vertical sides. b Water Dam G h O F h/ 3 N H C θ M K A J W R Water pressure diagram x Figure 26.1 w = Specific weight of water h = Height of water F = Force exerted by water on the side of the dam H = Height of dam W = Weight of dam per metre length of dam b = Width of dam wd = Weight density of dam masonry Weight of dam per unit length, W = wd b.H 591 @seismicisolation @seismicisolation 592 • Strength of Materials This weight acts through centre of gravity, G of the dam. Intensity of water pressure will be zero at water surface and will increase linearly to wh at the bottom surface. wh The average intensity of water pressure on the face of the dam = 2 The force F is given by F = wA h = w (h × 1) × = h 2 h= h 2 w × h2 2 h This force F will act horizontally at a height of above the base (called centre of pressure). 3 Thus, there are only two forces acting on the dam as shown in Fig. 26.1. The resultant force R may be found out by parallelogram of forces as shown in figure. Take OC = F and ON = W to some suitable scale. On completing rectangle, diagonal OD will represent the resultant force R. √ ∴ Resultant R = F 2 +W 2 F MN = tan θ = ON W so θ can be calculated. The diagonal OM represents the resultant of F and W . If we extend the diagonal OM so that it cuts the base of the dam at point K. And if line of weight W is extended to meet the base at J, Let x = distance JK The distance x is obtained from similar triangles ONM and OJK as explained below: ∴ NM JK = h/ 3 ON x F = h/ 3 W F h x= × W 3 E XAMPLE 26.1: A masonry dam of rectangular section, 22 m high and 12 m wide has water up to a height of 18 m on its one side. Determine: i) pressure force due to water on one metre length of the dam; ii) Position of centre of pressure; and iii) the point at which the resultant cuts the base. Take the weight density of masonary = 20 kN/m3 and weight density of water = 9.81 kN/m3 . @seismicisolation @seismicisolation Dams • 593 S OLUTION : (i) F = pressure force due to water on one metre length of dam F = wAh h 2 18 = 9.81 × 1000 × (18 × 1) × 2 = 9.81 × 1000 × (h × 1) × = 1589220 N ∴ h A = h×1 & h = 2 Ans (ii) Position of centre of pressure: The force F is acting horizontally at a height of h/3 above the base. ∴ 18 3 = 6 m Ans Position of centre of pressure = (iii) Let x be horizontal distance between the line of action of W and the point through which the resultant cuts the base (refer Fig. 26.1) W = weight of dam per length of dam = weight density of masonry × Area of dam × 1 = Wd × b × H × 1 = 20000 × 12 × 22 × 1 = 5280000 N We know, F h × W 3 1589220 18 = × 5280000 3 = 1.806 m Ans x= E XAMPLE 26.2: A concrete dam of rectangular section 18 m high and 7 m wide contain water up to a height of 14 m. Determine: (a) Total pressure per metre length of the dam. (b) Point where the resultant cuts the base and (c) Maximum and minimum intensities of stress at the base. Assume weight of water and concrete as 10000 N/m3 and 25000 N/m3 respectively. @seismicisolation @seismicisolation 594 • Strength of Materials S OLUTION : Total pressure per metre length of the dam: F= = wh2 2 10000 (14)2 = 980000 = 980 kN 2 Ans 7m Water 18 m 14 m O F 14 m 3 A J K W B R σmin σmax Figure 26.2 Point where the resultant cuts the base: Let the resultant R cut the base at K as shown in Fig. 26.2. The weight of the concrete per metre length, W = ρ ×b×H = 25 × 7 × 18 = 3150 kN Horizontal distance between the centre of gravity of the dam section and the point where the resultant cuts the base (i.e., distance JK), F h × W 3 14 980 × = 3150 3 = 1.452 m Ans x= Maximum and minimum stress at the base: eccentricity of the resultant e = x = 1.452 m @seismicisolation @seismicisolation Dams • 595 b F h b × d = AJ + JK = + x = + 2 2 W 3 and the eccentricity of the resultant, e=d− b 2 [d = AK] It is understood that as a result of eccentricity e, some moment will come into play causing some bending stress at the base section of the dams, let this moment be M, M = weight of the dam × eccentricity = W.e For unit length, moment of inertia of the base section about its centre of gravity, l × b3 1 × b3 = 12 12 b3 = 12 I= Let y be the distance between the centre of gravity of the base section and extreme fibre of the b base and σb be the bending stress in the fibre at a distance y from the centre of gravity of the 2 base section. M σ Also, = I y M.y ∴ σb = I b W.e × 2 = b3 12 σ We = 2 b Now direct stress at the base Weight of dam Width of dam W = b σd = Now the stress across the base at B will be maximum and at A stress will be minimum. W 6We W 6e ∴ σmax = σd + σb = + 2 = 1+ b b b b @seismicisolation @seismicisolation 596 • Strength of Materials W 6We W σmin = σd − σb = − 2 = b b b 6e 1− b In this example, ∴ e = x = 1.452 m W 6e σmax = 1+ b b 3150 6 × 1.452 = 1+ 7 7 = 450 (1 + 1.245) = 1010.25 kN/m2 (Compressive) Ans W 6e σmin = 1− b b 3150 6 × 1.452 = 1− 7 7 = 450 (1 − 1.245) = −110.25 kN/m2 = 110.25 kN/m2 (Tensile) Ans Trapezoidal Dams with Water Face Vertical Figure 26.3 shows trapezoidal dam having its water face vertical. a Water h G L W H M N h/ 3 A K J B R W b σmin Figure 26.3 @seismicisolation @seismicisolation σmax Dams • 597 Let a be top width of the dam and b be bottom width of the dam. Weight of the dam per unit length, W = w× (a + b) ×H 2 wh2 2 Horizontal distance between the centre of gravity of the dam and the point where the resultant R cuts the base, F h x= × W 3 Total pressure on a unit length of the trapezoidal dam = F = The distance between the toe of the dam A and the point where the resultant R cuts the base which is distance AK, F h + d = AJ + JK = AJ + W 3 AJ may be found either by splitting the dam section into a rectangle and a triangle or a2 + b2 + ab AJ = . 3(a + b) Now taking the moments about A and equating the same with the moment of the dam section about A, ∴ Eccentricity, e = d − AJ The stress at A will be minimum and at B will be maximum, such that: 6e W 1+ σmax = b b W 6e and σmin = 1− b b E XAMPLE 26.3: A concrete dam of trapezoid section having water on vertical face is 17 m high. The base of the dam is 8.5 m and top 4 m wide. Determine: (a) resultant thrust on the base per water length of dam; (b) point, where the resultant thrust cuts the base; and (c) maximum and minimum stresses across the base. Take weight of the concrete as 25 kN/m3 and water level upto top of the dam. @seismicisolation @seismicisolation 598 • Strength of Materials S OLUTION : Water 4m D 17 m F C L J A K B R W 8.5 m σmin σmax Figure 26.4 (a) Resultant thrust on the base per metre length: wh2 2 9.81 × 172 F= 2 F = 1417.5 kN F= Weight of concrete per metre length a+b H W =ρ 2 4 + 8.5 × 17 = 25 2 = 2656.25 kN √ Resultant thrust per metre length = R = F 2 +W 2 √ = 1417.52 + 2656.252 = 3010.8 kN Ans (b) Point, where the resultant cuts the base: Let the resultant cut the base at K (refer Fig. 26.4). To find the centre of gravity of the dam section, taking moments of the area about and equating the same, 17 × 4.5 4 4.5 4.5 (17 × 4) + AJ = 17 × 4 × + 17 × (4 + 2 2 2 3 (68 + 38.25) AJ = [136] + [210.375] @seismicisolation @seismicisolation Dams • 599 346.375 106.25 = 3.26 m AJ = Now F h × W 3 17 1417.5 × = 2656.25 3 = 3.06 m JK = x = ∴ horizontal distance AK, d = AJ + x = 3.26 + 3.06 = 6.32 m Ans (c) Maximum and minimum stresses: Eccentricity, e=d− b 2 = 6.32 − ∴ 8.5 2 = 2.07 m 6e W 1+ σmax = b b 2656.25 6 × 2.07 = 1+ 8.5 8.5 = 312.5 (1 + 1.461) = 769.06 kN/m2 W 6e σmin = 1− b b (Tensile) Ans = 312.5(1 − 1.461) = −144.06 kN/m2 (Compressive) Ans Rule of Middle Third As far as possible, there should not be any tension in the base of a dam wall as masonry work is weak in tension. We have seen that at A (which is called the heel of the wall), the direct compressive stress is W/b and the bending stress is tensile as determined below: @seismicisolation @seismicisolation 600 • Strength of Materials Modulus of section of the base AB, Z= 1 b2 × 1 × b2 = 6 6 Maximum bending stress on the section σb = M 6We = 2 Z b This is tensile at A and compressive at B (refer Fig. 26.4) Now there will be no tension at A, if 6We W > 2 b b b e< 6 If K lies at a distance not greater than b/6 from J. In other words, if the resultant of F and W cuts the base width within middle third, there would be no tension anywhere in the base. This is called the rule of the middle third. b 6 b b x + JK − = 2 6 If e = or or Ak = x + JK = 2b 3 Note: e is the distance between the point where the resultant R cuts the base and the centre point of the dam. E XAMPLE 26.4: A trapezoidal masonry dam is of 20 m height. The dam is having water upto a depth of 16 m on its vertical side. The top and bottom width of the dam are 4 m and 9 m respectively. The weight density of the masonry is given as 20 kN/m3 . Determine: a) the resultant force on the dam per metre length; b) the point where the resultant cuts the base; and c) the maximum and minimum stress intensities cut the base. S OLUTION : Height of dam, H = 20 m Depth of water, h = 16 m Top width of dam, a = 4 m Bottom width of dam, b = 9 m Weight density of masonry, wd = 20 kN/m3 = 20000 N/m3 . @seismicisolation @seismicisolation Dams C 16 m = h a 20 m = H M N h 3 A 601 D G L F • J K B R W b Figure 26.5 a) Resultant force on dam Force, F = w A h 4 2 16 = 9810 × (16 × 1) × 2 = 1255680 N = 9810 × (h × 1) × 16 h = 5.33 m above the base. It is acting at a distance of , i.e., 3 3 Weight of dam is given by w = weight density of masonry × area of dam × 1 a+b = wd × ×H ×1 2 4+9 = 20000 × × 20 × 1 2 = 2600000 N The distance of the line of action of W from the line AC is obtained by splitting the dam into rectangle and triangle. Taking the moments of their areas about the line AC and equating to the moment of the area of the trapezoidal about the line AC. A × 20 × 2 + 4 × 20 4 4+9 4+ = × 20 AJ 2 3 2 160 + 213.3 = 130 AJ AJ = 2.87 m @seismicisolation @seismicisolation 602 • Strength of Materials Resultant force R is given by F 2 +W 2 = 12556802 + 26000002 R= = 2887340 = 2.887 MN Ans b) The point where the resultant cuts the base. x = JK = the horizontal distance between the line of action of W and the point at which the resultant cuts the base We know, F h × ω 3 1255680 16 × = 2600000 3 = 2.57 m x= The distance AK = AJ + JK = d = 2.87 + 2.57 Eccentricity, = 5.44 m b e=d− 2 = 5.44 − = 0.94 m 9 2 Ans c) The maximum and minimum stress intensities W 6e σmax = 1+ b b 2600000 6 × 0.94 = 1+ 9 9 = 288889 [1 + 0.627] = 470022 N/m2 Ans 6e W 1− σmin = b b 2600000 6 × 0.94 = 1− 9 9 @seismicisolation @seismicisolation Dams = 288889 [1 − 0.627] = 107756 N/m2 Ans Trapezoidal Dams with Water Face Inclined E h F a D I C H G θ h 3 θ J K B A R W b Figure 26.6 Let us consider a trapezoidal dam with water face inclined. Let a = Top width of the dam b = Bottom width of the dam H = Height of the dam wd = Specific weight of the dam masonry h = Height of water retained by the dam w = Specific weight of the water θ = Inclination of the water face with the vertical. AI = l Length of sloping side AI, which is subjected to water surface. h = cos θ l h ∴ l= cos θ The weight of the dam per unit length, W = ωd × (a + b) ×H 2 Total pressure on a unit length of the dam, F= wh whl ×l = 2 2 @seismicisolation @seismicisolation • 603 604 • Strength of Materials Refer Fig. 26.6, the water pressure F will act at a height of h/3 from the bottom of the dam. Horizontal component at this water pressure F, FH = F cos θ = whl h wh2 × = 2 l 2 And vertical component of this water pressure, Fr = F sin θ = whl EI × 2 l w × EI × h 2 = Weight of the wedge AEI of water. = w2 h2 on the It is evident that such a dam may be taken to have a horizontal pressure equal to 2 imaginary vertical face AE. The weight of wedge AFE of water may be considered a part of the weight of dam when determining the centre of gravity of the dam section. The distance between the centre of gravity of the dam section and the point, where the resultant R cuts the base is given by the relation, x= ∴ F h × W 3 (x = JK) Total stress across the base at B, σmax = W b 6e 1+ b σmax = W b 6e 1− b And total stress across the base at A, E XAMPLE 26.5: A masonry dam of trapezoidal section is 12 m high. The top width of dam is 1.5 m and bottom width is 8 m. The face exposed to water has a slope of 1 horizontal to 10 vertical as shown in Fig. 26.7. Calculate the maximum and minimum stresses on the base, when the water level coincides with the top of dam. Take weight of masonry as 21 kN/m3 and that of water as 10 kN/m3 . S OLUTION : H = 10 m a = 1.5 m @seismicisolation @seismicisolation Dams b=8m h = 12 m wd = 21 kN/m3 w = 10 kN/m3 1.5 m C θ E D 12 m F θ J K W R B A 1.2 m 1.5 m S 8m Figure 26.7 10 1 θ = 84.29◦ 12 tan θ = AS 12 AS = tan θ 12 AS = 10 = 1.2 m tan θ = ∴ Total water pressure F per meter length of dam, wh2 2 10 × (12)2 = 2 = 720 kN F= @seismicisolation @seismicisolation • 605 606 • Strength of Materials Weight of the dam per metre length (including wedge AEC of water) (a + b) h + wd × H W = w× 2 2 (1.5 + 8) 12 + 21 × × 12 = 10 × 2 2 = 60 + 1197 = 1257 kN In order to find out the centre of gravity of the dam section (including wedge AEC of water). Taking moments about A and equating the same, 12 2 12 1 × + 21 × × W × AJ = 12 × 2 3 2 3 1.5 + 21 × 12 × 1.2 + 2 12 × (8 − 1.2 − 1.5) (8 − 1.2 − 1.5) + 21 × × + (1.2 + 1.5) 2 3 1257 × AJ = 24 + 84 + 491.4 + (21 × 31.8 × 4.47) = 599.4 + 2985 AJ = 2.85 m F h JK = x = × W 3 720 12 = × = 2.291 m 1257 3 ∴ AK = d = AJ + x = 2.85 + 2.291 = 5.141 m b and eccentricity, e = d − 2 8 = 5.141 − = 1.141 m 2 W 6e 1257 6 × 1.141 ∴ σmax = 1+ = 1+ b b 8 8 = 157.125 (1.856) = 291.6 kPa Ans 6e 1257 6 × 1.141 W 1− = 1− σmin = b b 8 8 = 157.125 × 0.144 = 22.63 kPa Ans Horizontal distance @seismicisolation @seismicisolation Dams • 607 Stability of dam depends upon several factors and conditions as described below. Condition A: Resultant thrust cuts the base within the middle third of the base width to avoid tensile stress at the base. When bending stress is greater than the direct stress, tension is developed at the base of the dam. To avoid the tension at the base of the dam, bending stress must be equal to or less than the direct stress. σb = bending stress and d = direct stress. σb ≤ σd 6We ω 6We or ≤ when σb = 2 b b2 b 6e or ≤1 b b or e ≤ 6 b Obviously, it means eccentricity must not be greater than at either side from the centre of the 6 base width. Hence, the resultant thrust R must interact with the base within the middle-third of the base width to avoid tension at the base, which is known as middle-third-rule. Let d = AJ + JK 4 F × = AJ + W 3 Eccentricity, e F A J K W b = d − AJ (as before) R Figure 26.8 Condition B: To avoid crushing of masonry at the base of the dam (or wall) σmax should not be more than the permissible or safe stress of the dam material. Condition C: The total horizontal thrust F causes sliding of the dam. To counteract sliding of the dam, the maximum μ w is set up at the base of the dam, where μ is the coefficient of friction between the masonry of the dam (or wall) and the soil on which it rests. Hence, condition to prevent sliding of the dam (or wall) at the base, μ w > F is to be satisfied. In other words, μw >1 F μw is taken at least 1.5. For designing purposes F Condition D: A dam (or wall) can be overturned tilted about the toe B as shown in Fig. 26.9 as the water almost F causes an overturning moment. To maintain the equilibrium condition, the weight of dam W provides a restoring moment about B. @seismicisolation @seismicisolation 608 • Strength of Materials h H F h 3 G J K A B W b R Figure 26.9 h 3 Restoring moment abut B = W × JB Overturning moment about B = F × Condition to prevent overturning about the toe B is: Restoring moment > Overturning moment h 3 h or W (b − AJ) > F 3 JK F F h = or (b − AJ) > . W 3 h/3 W or b > AJ + JK ∴ W × JB > F. ∴ AB > AK as shown in Fig. 26.9. This condition will satisfy if K lies between AB. For no tension condition, we have such that point E must lie within the middle third of the base width AB. So, if the a dam is checked against no tension condition at the base, it will be automatically checked for overturning of the dam (or wall). Exercise 26.1 A rectangular masonry dam 6 m high and 3 m wide has water level up to its top. Find: (i) total pressure per metre length of the dam; (ii) point at which the resultant cuts the base and (iii) maximum and minimum intersection of stresses at the bottom of the dam. Assume the weight of water and masonry as 10 kN/m3 and 20 kN/m3 respectively. [Ans 180 kN, 1.0 m, 360 kPa, −120 kPa Tensile] 26.2 A masonry dam 12 m traperzoidal in section has top width 1 m and bottom width 7.2 m. The face exposed to water has a slope of 1 horizontal to 10 vertical. Check the stability of the @seismicisolation @seismicisolation Dams • 609 dam, when the water level rises 10 m high. The coefficient of friction between the bottom of the dam and the soil is 0.6. Take the weight of masonry as 22 kN/m3 . [Ans Safe against tension, safe against or returning, safe against sliding] 26.3 A masonry dam of trapezoidal section is 6 m high, 2 m wide at the top and 5 m wide at the bottom. The water face of the dam has a slope 1 in 6. Calculate the maximum and minimum stresses at the base of the dam, when the height of water is same as the height of dam. Take unit weight of masonry and water as 20 kN/m3 and 9.81 kN/m3 respectively. [Ans 100133 N/m2 ; 79638.9 N/m2 ] 26.4 A masonry dam of trapezoidal section is 10 m high and retains water up to the top. The width at the top is 3 m and at the bottom 8 m. Water face has a slope of 1 in 10. Find the maximum and minimum intersities of stress at the base. Density of water = 10000 N/m3 and density of masonary = 24000 N/m3 . [Ans σmax = 231600 N/m2 Comp, σmin = 110900 N/m2 Comp] 26.5 A concrete dam of trapezoidal section has a vertical face on the water side. Its height is 5 m, top width 2 m and bottom width 3 m. Determine the maximum and minimum stresses at its base when the reservoir is full. Take weight of water 9810 N/m3 and weight of masonry 20000 N/m3 . [Ans 178335 N/m2 , −11667 N/m2 Tensile] @seismicisolation @seismicisolation C HAPTER 27 RIVETED JOINTS Riveted joints are permanent joints used to fasten one plate with another to counteract tensile forces, shear forces, etc. These are used in pressure vessels such as boiler, at structure works such bridge truss, roof truss, etc. In ship building also they are extensively used. We shall now study and investigate the strength of riveted joints. Rivet A rivet is a cylindrical solid bar with a load integral to it. The middle portion of the rivet is called ‘body’ or ‘shank’ and the bottom portion is called tail. According to IS: 2998-1982 (Reaffirmed 1992), the material of a rivet must have tensile strength not less than 40 N/mm2 and it should not crack when flattened. Elongations should not be less than 25 per cent. In cold condition or when heated to 650◦ C and quenched, the shank shall be bent on itself through 180◦ C without cracking. The rivet when hot must flatten without cracking to a diameter 2.5 times the diameter of shank. Head Tail Body or shank Figure 27.1 Rivet These are manufactured either by cold heating or by hot forging. In case if rivets are cold forged, they shall be properly heat-treated to remove stresses. Types of Rivet Heads According to Indian standard specifications, the rivet heads are available into many types. Rivet heads for general purposes. According to IS: 2155 − 1982 (Reaffirmed 1996) some types are shown in Fig. 27.2 (having diameter below 12 mm). @seismicisolation @seismicisolation Riveted Joints d 1.6 d 0.7 d 0.7 d d d 1.6 d Pan Head Snap Head 2d 0.25 d 1.516 d 0.5 d d d 2.25 d Flat Head Mush Room Head 2d 90° d Flat Countersunk Head 90° Figure 27.2 @seismicisolation @seismicisolation 0.5 d • 611 612 • Strength of Materials Material of Rivets It should be tough and ductile. They are usually made of steel (low carbon). Sometimes brass, aluminium or copper is used for special applications. The rivets for general purposes are manufactured from steel conforming to the following Indian standards: (a) IS: 1148–1982 (Reaffirmed 1992): Specification for hot rolled rivet bars upto 40 mm diameter for structural purposes. (b) IS: 1149:1982 (Reaffirmed 1992): Specification for high tensile steel rivet bars for structural purposes. For boiler work rivets are made conforming to IS: 1990–1973 (Reaffirmed 1992). Types of Riveted Joints 1. Lap joint 2. Butt joint. 1. Lap Joints 10° d M = 1.5 d p Single riveted lap joint Figure 27.3 @seismicisolation @seismicisolation Riveted Joints • 613 d d p F F pb Double rivets lap joint with chain riveting M = 1.5 d d F F p pb Double rivets lap joint with jig-jag riveting Figure 27.4 2 Butt Joints: In this type of joint, the plate to be joined and placed end to end as shown in Fig. 27.5. The covering plates are called cover straps. @seismicisolation @seismicisolation 614 • Strength of Materials M=1.5 d F p F Single riveted butt joint with two cover plates with chain riveting M = 1.5 d Section A-A F p p/2 F A A Double riveted butt joint with two cover plates with zig-zag riveting Figure 27.5 Some Definitions Related to Riveted Joints 1. Pitch of rivets: It is the distance from the centre of any rivet to the centre of next rivet in the same row. It is denoted by p. 2. Margin: It is the distance between the edge of the plate and centre of the nearest hole. It is denoted by M. In all types of riveted joint, margin (M) is taken as equal or greater than 1.5 d where d is the diameter of hole. 3. Back pitch: Back pitch is the distance between any two consecutive rows of rivets in the same plate. It is denoted by pb (Fig. 27.4). 4. Diagonal pitch: It is the diagonal distance between two adjacent rivets in two consecutive rows in the same plate. It is denoted by pd (Fig. 27.6). pb pd Figure 27.6 @seismicisolation @seismicisolation p Riveted Joints • 615 Failure of a Riveted Joint Possible ways of failure of riveted joint are described below. 1. Tearing of the plate at an edge: A joint may fail due to tearing of the plate at an edge as shown in Figs. 27.7 and 27.8. This can be prevented by keeping the margin M = 1.5d, d is the diameter of hole. M F d F F p p-d F d Tearing of the plate at an edge Tearing of the plate across the rows of rivets Figure 27.7 Figure 27.8 2. Tearing of the plate across a row of rivets: Because of tensile stresses in the main plates, the main plate or cover plates may tear off across a row of rivets as shown in Fig. 27.8. For such cases, only one pitch length is considered. The resistance offered by the plates against tearing is known as tearing resistance or tearing strength of the plate. Let p = pitch of the rivets. d = diameter of rivets hole t = thickness of the plate σt = permissible tensile stress for the plate material Tearing area per pitch length, At = (p − d)t Therefore, tearing resistance or pull required to tear off the plate per pitch length, Ft = At σt = (p − d)t σt If the tearing resistance Ft is greater then the applied load F, then this type of failure will not occur. 3. Shearing of the rivets: The rivets are being exerted by tensile forces and thus inducing shearing stresses in the rivets as is evident from Fig. 27.4 and Fig. 27.5 and if not properly designed, can be sheared off as shown in Fig. 27.9. @seismicisolation @seismicisolation 616 • Strength of Materials F F Shearing off a rivet in a lap joint F F Shearing off a rivet in a single cover butt joint F F Shearing off a rivet in a double cover butt joint Figure 27.9 Let d be the diameter of the rivet hole. τ = safe permissible shear stress for the rivet material, and π n = number of rivets per pitch length. Shearing area is As = × d 2 (in single shear) 4 π 2 As = 2 × × d (in double shear) 4 Therefore, shearing resistance or pull required to shear off the rivet per pitch length, π × d 2 × τ (in single shear) 4 π Fs = n × 2 × × d 2 × τ (in double shear) 4 Fs = n × 4. Crushing of the plate or rivets: Instead of shearing, sometimes, the rivets are crushed as shown in Fig. 27.10. This results in oval shape of rivet hole and hence the joint becomes loose. Resisting area against crushing in the projected area of the hole or rivet. The resistance offered by a rivet to be crushed is known as crushing resistance or crushing strength of the rivet. F F Crushins of a rivet Figure 27.10 @seismicisolation @seismicisolation Riveted Joints • 617 d = diameter of the rivet hole t = thickness of the plate σc = safe permissible crushing stress for the rivet or plate material n = number of rivets per pitch length under crushing. Crushing area = projected area per rivet Ac = dt Total crushing area = n.d.t Therefore, crushing resistance or pull required to crush the rivet per pitch length, Fc = n · d · t · σc If Fc is greater than applied load F then the joint will fail due to crushing. Efficiency of a Riveted Joint It is given by Least of Ft , Fs and Fc p × t × σt p = pitch of the rivets t = thickness of plate σt = Allowable tensile stress of the plate material. η= E XAMPLE 27.1: Determine the efficiency of the following rivets joints: (i) Simple riveted lap joint of 8 mm plates with 24 mm diameter rivets having a pitch of 55 mm. (ii) Double riveted lap joint of 8 mm plates with 24 mm diameter rivets having a pitch of 70 mm. Assume Permissible tensile stress in plate = 130 GPa Permissible shearing stress in rivets = 100 MPa Permissible crushing stress in rivets = 185 MPa. S OLUTION : t = 8 mm, d = 24 mm, σt = 130 MPa = 130 N/mm2 , τ = 100 MPa = 100 N/mm2 , σc = 185 MPa = 185 N/mm2 (i) Efficiency of the first joint: Pitch = 55 mm Let us find the tearing resistance of plate, shearing and crushing resistance of the rivets. @seismicisolation @seismicisolation 618 • Strength of Materials a) Tearing resistance of the rivet: Ft = (p − d)t σt = (55 − 24) × 8 × 130 = 32240 N b) Shearing resistance of the rivet: Because the joint is a single lap joint, the strength of one rivet in single stress is taken. The shearing resistance of one rivet, π 2 d ×τ 4 π = (24)2 × 100 4 = 45216 N Fs = c) Crushing resistance of the rivet: Again as the joint in a single riveted, therefore strength of one rivet is taken. The crushing resistance of one rivet, Fc = d × t × σc = 24 × 8 × 185 = 35520 N. Hence strength of joint = least of Ft , Fs and Fc = 32240 N Now strength of unriveted or solid plate, F = p × t × σt = 55 × 8 × 130 = 57200 N ∴ Efficiency of joint, η= = Least of Ft , Fs and Fc F 32240 = 0.563 or 56.3 % 57200 Ans (ii) Efficiency of the second joint: As we know pitch = 70 mm a) Tearing resistance of the plate, The tearing resistance of the plate per pitch length, Ft = (p − d) × t × σt = (70 − 24) × 8 × 120 = 44160 N @seismicisolation @seismicisolation Riveted Joints • 619 b) Shearing resistance of the rivets: Because the joint is double riveted lap joint, strength of two rivets in single layer is taken. The shearing resistance of the rivets, π × d2 × Z 4 π = 2 × (24)2 × 100 4 = 90432 N Fs = n × c) Crushing resistance of the rivet Because the joint is double riveted, strength of two rivets must be taken. The crushing resistance of rivets, Fc = n × d × t × σc = 2 × 24 × 8 × 185 = 71040 N Therefore strength of the joint, = Least of Ft , Fs and Fc = 44160 N Now the strength of the unriveted or solid plate, F = p × t × σt = 70 × 8 × 130 = 72800 N ∴ Efficiency of the joint, least of Ft , Fs , and Fc F 44160 = 72800 = 0.606 or 60.6% Ans η= E XAMPLE 27.2: A double riveted double cover butt joint in plates 25 mm thick is made with 30 mm diameter rivets at 110 mm pitch. The allowable stresses are: σt = 135 MPa, τ = 95 MPa, σc = 140 MPa. Find the efficiency of joint, taking the strength of the rivet in double shear as twice that of single shear. @seismicisolation @seismicisolation 620 • Strength of Materials S OLUTION : t = 25 mm, d = 30 mm, p = 110 mm, σt = 135 MPa, τ = 95 MPa, σc = 140 MPa. Let us first find the tearing resistance of the plate, shearing resistance and crushing resistance of the rivet. (i) Tearing resistance of the plate: Tearing resistance of the plate per pitch length, Ft = (p − d) × t × σt = (110 − 30) × 25 × 135 = 270000 N (ii) Shearing resistance of the rivets Because the joint is double riveted butt joint, the strength of two rivets in double shear is taken. Shearing resistance of the rivets, π × d2 × 7 4 π = 2 × 2 × × (30)2 × 95 4 = 268470 N Fs = n × 2 × (iii) Crushing resistance of the rivets Because the joint is double riveted, the strength of two rivets is taken. Crushing resistance of the rivets, Fc = n × d × t × σc = 2 × 30 × 25 × 140 = 210000 N Therefore strength of the joint = Least of Ft , Fs and Fc = 210000 N Strength of the unriveted or solid plate, F = P × t × σt = 110 × 25 × 135 = 371250 N Therefore efficiency of the joint Least of Ft , Fs and Fc F 210000 = 371250 = 0.566 or 56.6% Ans = @seismicisolation @seismicisolation Riveted Joints • 621 Note: When thickness of the plate is greater than 8 mm, Unwin’s empirical formula to determine the rivet hole is quite useful: √ d=6 t Diamond Riveting In many structural constructions, for example, bridge truss members, it becomes necessary to provide a large number of rivets to connect a member to the gusset plate. If F is the force to be transF . Obviously the mitted, the number of rivets necessary will be given by n = strength of one rivet strength of the rivet will be taken least of Ft , Fs and Fc . In the case of tension members of a roof truss, the number of rivets required for the joint is small and the section is usually weakened by one rivet hole only. But in bridge truss, the number of rivets required is large and the section will depend upon how these rivets are arranged. For example, suppose a bridge diagonal is a flat of thickness t requiring 9 rivets to connect it by a butt joint to a gusset plate, we may arrange the rivets in three rows of three each as diamond riveting shown in Figure 27.11. Such type of arrangement is also known as Lozenge joint. F 1 2 3 4 1 2 3 4 Figure 27.11 For the joint shown in Fig 27.11, assuming the section 1-1 through the first rivet hole to be the weakest, the width of flat will be given by, F = σt (b − d)t where σt is tensile stress b is the width of flat and d is the rivet hole. In diamond riveting section 1-1 which is weakened by one rivet hole only is usually the weakest, sections 2-2, 3-3 and 4-4 successively getting stronger. If σt is the permissible tensile stress for the plate, the pull required to tear the plate at section 1-1 is given by, F1 = σt (b − d)t To fail at section 2–2 by tearing of the plate, the rivet in front must first give way. The pull required for the failure of the plate at section 2–2 is given by, @seismicisolation @seismicisolation 622 • Strength of Materials F2 = σt (b − 2d)t + strength of one rivet in front. Similarly, for failure at section 3–3 by tearing of the plate, the three rivets in front must give way and therefore the pull required for tearing at section 3–3 is given by, F3 = σt (b − 3d)t + strength of three rivets in front. Similarly, F4 = σt (b − 3d)t+ strength of six rivets in front, for failure at section 4-4. Out of forces F1 , F2 , F3 and F4 , usually F1 is the least, which means that the weakest section is 1-1. The efficiency of such a joint is the greatest, and is given by η= F1 (b − d)t σt b−d = = F b t σt b Such a joint thus provides the most economical arrangement of rivets. E XAMPLE 27.3: Two lengths of a mild steel tie rod, having width 200 mm and thickeness 12.5 mm are to be connected by means of a butt joint with double cover straps. Design the riveted joint, if the allowable stresses are 100 N/mm2 in tension, 70 N/mm2 in shear and 180 N/mm2 in crushing. Also determine the efficiency of joint. S OLUTION : b = 200 mm; t = 12.5 mm; σt = 100 N/mm ; 2 τ = 70 N/mm2 , σc = 180 N/mm2 . Diameter of rivet (d) For t ≥ 8 mm the diameter of rivet hole √ is d = 6 t (using Unwin’s empirical formula) √ d = 6 12.5 = 21.21 mm. The standard diameter of rivet hole in 21.5 mm and the corresponding diameter of rivet is 20 mm. ∴ d = 21.5 mm Number of rivets: The pull acting on the joint is Ft = (b − d)t σt = (200 − 21.5) × 12.5 × 100 = 223125 N @seismicisolation @seismicisolation Riveted Joints The shearing strength of one rivet is π Fs = 1.75 d 2 × τ 4 π = 1.75 × (21.5)2 × 70 4 = 44451 N Note: Instead of 2, value of 1.75 is taken per Indian standard for safety. The crushing strength of one rivet is Fc = d × t × σc = 21.5 × 12.5 × 180 = 48375 N The number of rivets for joint is Ft minimum of ps MPa 223125 = 44451 = 5 rivets n= Number of rows: Five rivets are arranged as shown in Fig. 27.12. 1 2 3 1 2 3 Figure 27.12 Thickness of butt cover: t1 = 0.75t = 0.75 × 12.5 = 9.375 = 9.4 mm Pitch of rivet (p) p = 3d + 5 = 3 × 21.5 + 5 = 69.5 mm. @seismicisolation @seismicisolation • 623 624 • Strength of Materials Spacing of rows of rivets (Pb ) and margin (M) Pb = 3 d = 3 × 21.5 = 64.5 mm M = 1.5 d = 1.5 × 12.5 = 32.25 mm. Efficiency of joint (η ): The tearing resistance of joint along section 1–1 is Ft1 = (b − d)t σt . = (200 − 21.5) × 12.5 × 100 = 223125 N The tearing resistance of joint along 2–2 as Ft2 = (b − 2d) × t × σt + strength of one rivet in front = (b − 2d) × t × σt + Fs = (200 − 2 × 21.5) × 12.5 × 100 + 44451 = 240701 N The tearing resistance of joint along section 3–3 is, Ft3 = (b − 2d) × t × σ t + strength of three rivets in front. = (b − 2d) × t × σ t + 3Fs = (200 − 2 × 21.5) × 12.5 × 100 + 3 × 44451 = 196250 + 133353 = 329603 N The shearing resistance of five rivets is n × Fs = 5 × 44451 = 222255 N Crushing resistance of five rivets is n × Fc = 5 × 48375 = 241875 N @seismicisolation @seismicisolation Riveted Joints • 625 Or efficiency of joint η Minimum of Ft1 , Ft2 , Ft3 , nFs , nFc Strength of solid plate 223125 = 260 × 12.5 × 100 = 0.8925 or 89.25% Ans = Exercise 27.1 Find the efficiency of the following joints: a) Single riveted lap joint of 6 mm plates with 20 mm diameter rivets having a pitch of 50 mm. b) Double riveted lap joint of 6 mm plates with 20 mm diameter rivets having a pitch of 60 mm. Assume: Permissible tensile stress in plate = 100 MPa Permissible shearing stress in rivets = 80 MPa Permissible crushing stress in rivets = 160 MPa [Ans 60%, 66.6%] 27.2 Two mild steel bars of 350 mm width and 15 mm thickness are to be connected by means of Lozenge joint with two straps. The tensile load acting on the joint is 15 × 104 N. Design and sketch the joint, if the permissible stresses are: 80 MPa in tension, 120 MPa in crushing and 60 MPa in shear. [Ans Total number of rivets = 9; Thickness of butt straps = 12 mm, efficiency of joint = 93%, Pitch of rivets = 72 mm margin at the ends = 36 mm] 27.3 A double riveted double cover butt joint with plates 20 mm thick made with 20 mm diameter rivets at 110 mm pitch, the permissible stresses are σt = 120 MPa, τ = 100 MPa and σc = 150 MPa. Find the efficiency of joint taking strength of rivet in double shear is twice than that of single shear. [Ans 56.82%] 27.4 A double riveted butt joint, in which the pitch of the rivets in outer rows is twice that in the inner rows, connects two 16 mm thick plates with two cover plates each 12 mm thick. The diameter of rivets is 22 mm. Determine the pitches of the rivets in the two rows if the working stresses are not to exceed in the following limits: Tensile stress in plates = 100 MPa, shear stress in rivets = 75 MPa and bearing stress in rivets and plates = 150 MPa. Make a fully dimensioned sketch of the joint by showing at least two views. [Ans 107 mm, 53.5 mm] 27.5 A diamond riveted double strap butt joint for two plates each 400 mm wide and 20 mm thick. The covers are 15 mm thick each while the rivets used are of 25 mm diameter. Determine the strength and efficiency of the joint. The maximum allowable stresses in tearing, shearing and bearing (crushing) are 120 MPa, 100 MPa and 180 MPa respectively. [Ans 900 kN, 93.75%] @seismicisolation @seismicisolation C HAPTER 28 WELDED JOINTS Welding is a process of joining metals in which parent metals are fused together to form a single piece. The difference between welding and soldering or brazing is that in later filler material is used whose melting temperature is lower than that of the workpieces. Welding is used wherever strength is required and soldering and brazing are primarily employed as simple joints to take up only light loads. In forge welding pressure is used to join parent metals whereas if two parts are joined without any pressure but with a separate weld material, the process is known as fusion welding. The joints so made are called welded joints. Classification of Welding Broadly various welding processes are classified as given below: Gas Welding i) ii) iii) iv) Air-acetylene welding Oxy acetylene welding Oxy-hydrogen welding Pressure gas welding Arc Welding i) ii) iii) iv) v) vi) vii) viii) ix) Carbon arc welding Flux coated arc welding Shielded metal arc welding Submerged arc welding Tungsten Inert Gas (TIG) welding Metal Inert Gas (MIG) welding Plasma Arc welding Electroslag welding Stud arc welding Resistance Welding i) ii) iii) iv) Spot welding Seam welding Projection welding Flash butt welding @seismicisolation @seismicisolation Welded Joints • 627 v) High frequency resistance welding vi) Percussion welding vii) Resistance butt welding. Solid State Welding i) Cold welding welding ii) Explosion welding iii) Forge welding iv) Roll welding v) Ultrasonic welding vi) Hot pressure welding vii) Friction welding viii) Diffusion welding. Thermo-chemical Welding Processes i) Thermit welding ii) Atomic hydrogen welding Radiant Energy Welding Processes i) Electron beam welding ii) Laser beam welding Advantages and Disadvantages of Welded Joints over Riveted Joints Advantages: i) The welded joints provide maximum efficiency (sometimes upto 100%) which is not possible in riveted joints. ii) Changes, alterations and additions can be easily made in the welded structures. iii) The welded structures are usually lighter than riveted structures. iv) As the welded structure is smooth in appearance, so it is better in looks. v) Sometimes certain structures are difficult to be riveted, for example, circular steel pipe, but they can be easily welded. vi) Welding provides rigid joints. This is in line with the modern trend of providing rigid frames. vii) The process of welding is less time-consuming compared to riveting. viii) It is possible to weld any part of structure at any point. But riveting requires enough room. Disadvantages: i) Due to uneven heating and cooling, undesirable stresses develop in structures. ii) Welding requires highly skilled labour and supervision. iii) As in welding, there is no provision of expansion and contraction is provided, therefore there is a possibility of cracks developing in it. iv) The inspection of welding work is more difficult than riveting work. Types of welds: Welds may be classified mainly into two types: i) Fillet weld: This type of weld is used when the members to be connected overlap each other as shown in Fig. 28.1. The section of the fillet weld for design purposes will be taken as an isosceles right-angle triangle. The length of either of the equal sides of the triangle is called size of the weld. @seismicisolation @seismicisolation 628 • Strength of Materials let fil fillet Figure 28.1 t = Throat thickness Reinforcement C A 45° D s B s s = Leg as size of weld Figure 28.2 Enlarged view of circled portion For determining the strength of the fillet joint, it is assumed that the section of fillet as a right angled triangle as discussed before. BD is hypotenuse making equal angles with other two sides CB and AB (refer Fig. 28.2). The length of each side is called leg or size of the weld. Perpendicular distance BD is known as throat thickness. For calculation and design purposes, the minimum area of the weld is obtained at the throat BD. This minimum area is the product of the throat thickness and length of weld. It is evident from Fig. 28.2 that, Throat thickness, t = 5 × sin 45◦ = 0.707 s As described, minimum area of the weld as Throat area, A = Throat thickness × Length of weld = t × p = 0.707 s × l In order to determine tensile strength F, F = Throat area × Permissible tensile stress = 0.707 s × l × σt (for, single fillet weld as shown in Fig. 28.2) Tensile strength of the welded joint for double fillet weld shown in Fig. 28.1, F = 2 × 0.707 s × l × σt Many times weld is weaker than the plate owing to slag and blow holes. That is why weld is given a reinforcement shown is Fig. 28.1 to make weld stronger. ii) Butt weld: Figures 28.3 and 28.6 shows butt welds. @seismicisolation @seismicisolation Welded Joints • 629 Reinforcement a) Reinforcement Single V - butt weld Figure 28.3 b) Double V - butt weld Figure 28.4 c) Single U - butt weld Figure 28.5 d) Double U - butt weld Figure 28.6 E XAMPLE 28.1: A plate 90 mm wide and 12 mm thick is to be welded to another plate by means of double parallel fillets. The plates are subjected to a load of 90 kN. Find the length of the weld if the allowable shear stress in the weld does not exceed 65 MPa. S OLUTION : Width = 90 mm Thickness = 12 mm Load = 90 kN = 90 × 103 N Zmax = 65 MPa = 65 N/mm2 Note: Size of weld is generally equal to thickness of plate. ∴ s = 12 mm @seismicisolation @seismicisolation 630 • Strength of Materials So maximum load (F) which the plates can carry for double parallel fillet joint weld is given by the following equation: 90000 = 2 × 0.707 × 5 × l × τ 90000 = 1.414 × 12 × l × 65 90000 l= 1.414 × 12 × 65 = 81.6 mm, say 82 mm Ans E XAMPLE 28.2: A tie bar 120 mm × 15 mm thick is to be welded to another plate as per Fig. 28.7. Find the minimum overlap required if 8 mm fillet welds are used. σt =120 N/mm2 , τ = 80 N/mm2 . 120 mm x Figure 28.7 Total length of weld, taking tensile load, = 120 × 2 = 240 mm Maximum tensile force in the tie bar = 120 × 15 × 125 = 225000 N (i) Throat thickness, t = 0.707 × s Total shear load taken by joint, = 2x × 0.707 × 15 × 80 = 1697 x (ii) Total tensile load taken by weld joint, = 120 × 0.707 × 15 × σt = 120 × 0.707 × 15 × 125 = 159075 N Adding loads (ii) and (iii) and equating with total 225000 N @seismicisolation @seismicisolation (iii) Welded Joints 1697x + 159075 = 225000 225000 − 159075 x= 1697 65925 = 1697 = 38.85, say 40 mm • 631 Ans E XAMPLE 28.3: A steel plate 120 mm wide 15 mm thick is required to join with another plate by a single lap weld and parallel fillet as shown in Fig. 28.8. The maximum tensile and shear stresses are 100 N/mm2 and 60 N/mm2 respectively. Find the length of each parallel fillet if the joint is subjected to a total load of 180 kN. S OLUTION : l2 F 120 mm l1 F l2 Figure 28.8 Width of plate = 120 mm Thickness of plate = 15 mm Maximum tensile stress = 100 N/mm2 Maximum shear stress = 60 N/mm2 Now, length of single fillet parallel lab weld, l1 = width of plate = 120 mm l2 = length of each parallel fillet weld. Let F1 be the load carried by single fillet lap weld and F2 be the load carried b parallel fillet weld Total load carried by the plates, = 180 kN = 180000 N F1 = 0.707 × l1 × t × σt = 0.707 × 120 × 15 × 100 = 127260 N @seismicisolation @seismicisolation 632 • Strength of Materials F2 = 2 × 0.707 × l2 × t × τ = 2 × 0.707 × l2 × 15 × 60 = 1272.6 l2 F1 + F2 = 180000 127260 + 1272.6 l2 = 180000 − 127260 180000 − 127260 l2 = 1272.6 = 41.4 mm, say 42 mm Ans Unsymmetrical Welded Section l1 F1 a G F b F2 l2 Figure 28.9 For equilibrium ∑ H = 0 and ∑ M = 0 F = F1 + F2 F1 × a = F2 × b F1 = 0.707 s × L1 × σ F1 = 0.707 s × L2 × σ We know Solving eqns. F1 L1 b = = F2 L2 a F ×b F1 = a+b F ×a F2 = a+b Axial load in terms of shear stress: Axial load, F = Shear stress × Area of weld = τ × area of well But area of weld = (Total length of weld) × Throat thickness = (l1 + l2 ) × 0.707 × s Hence, F = τ × (l1 + l2 ) 0.707 s @seismicisolation @seismicisolation (i) (ii) (iii) Welded Joints • 633 E XAMPLE 28.4: A 220 mm × 150 mm × 15 mm angle carries a load of 250 kN. The angle is welded to a steel plate by fillet welds or shown in Fig. 28.10(a). Find the lengths of the weld at the top and bottom. Permissible shear stress is 100 N/mm2 . l1 15 mm 220 mm 148.98 mm G 71.02 mm 15 mm l2 150 mm To find C. G. of angle Figure 28.10a 15 mm 2 220 mm G A 150 mm 1 B 15 mm Figure 28.10b a1 = 150 × 15 = 2250 mm2 , a2 = 205 × 15 = 3075 mm , 2 y1 = 7.5 mm y1 = 117.5 mm 2 A = 2250 + 3075 = 5325 mm . Let bottom line AB be reference line, 2250 × 7.5 + 3075 × 117.5 5325 = 71.02 mm ȳ = Thickness of angle = 15 mm Size of weld = 15 mm Allowable shear stress = 100 N/mm2 Distance of the top edge of the angle section from neutral axis = 148.98 mm Distance of the bottom edge of the angle section from neutral axis = 71.02 mm l1 = length of weld at the top l2 = length of weld at the bottom @seismicisolation @seismicisolation 634 • Strength of Materials Now , F = τ (l1 + l2 ) × 0.707 × t 250000 = 100(l1 + l2 ) × 0.707 × 15 250000 100 × 0.707 × 15 Total length, l = 235.74 mm (l1 + l2 ) = let l1 + l2 = l b×l a+b 71.02 × 235.74 = 148.98 + 71.02 16742.3 = 220 = 76.1 mm Ans l1 = Now l = l1 + l2 l2 = 235.74 − 76.1 = 159.64 mm Ans E XAMPLE 28.5: A welded lap joint is to be provided to connect two tie bars 200 mm × 15 mm a shown in Fig. 28.11. The working stress in the bar is 180 N/mm2 . Investigate the design of the size of the fillet welds be 14 mm. Safe stress for the weld may be taken as 110 N/mm2 . 60 mm 60 mm 60 mm 90 mm 120 mm 90 mm F F Figure 28.11 S OLUTION : Safe load in the tie bar = 200 × 15 × 180 = 540000 N Total length of weld = 2 × 60 + 4 602 + 902 √ = 120 + 4 × 11700 @seismicisolation @seismicisolation Welded Joints • 635 = 120 + 4 × 108.17 = 552.7 mm Strength of weld = 0.707 × 14 × 552.7 × 110 = 601768.7 N Because the strength of the weld is greater than even the maximum tension in the tie bar, the design is safe. T Fillet welds under torsion A. Circular fillet weld: Figure 28.12 shows a circular shaft welded to a plate. Let this be subjected to torque T . In a horizontal plane, the shear stress in the weld, t Figure 28.12 T × d/2 J 2 d and J = π td 2 τ= because τ T = J R T × d/2 2T 2 = π td 2 d π td 2 t Throat thickness = √ 2 √ 2T 2 Maximum shear stress, τmax = π td 2 2.83T = π td 2 Therefore, τ = ∴ (i) z B. Long adjacent fillet welds: A vertical plate is welded to a horizontal plate by two identical fillet welds as shown in Fig. 28.13. Let torque T be applied on vertical plate as shown. Now t = leg length of weld l = length of the joint. T b t t Figure 28.13 @seismicisolation @seismicisolation l 636 • Strength of Materials Torque T will induce shear stress varying from zero at the axis and maximum at the plate ends. T × l/2 3T = 2. Induced shear stress, τ = 1 3 tl 2 × tl 12 The maximum value of the shear stress in the throat of weld is given by √ 3T 2 4.24T τmax = = (ii) tl 2 tl 2 Fillet Welds Under Bending Moment Let a circular rod connected to a rigid plate by a fillet weld. d = Diameter of rod M = Bending moment acting on the rod s = Size or leg of the weld t = Throat thickness Z = Section modulus of the weld section = d M s t s Figure 28.14 π td 2 4 Therefore, bending stress, M Z M = π td 2 4 4M = π td 2 σb = The maximum bending stress occurs on the throat of the weld which is inclined at 45◦ to the horizontal plane, ∴ length of throat t = s sin 45◦ = 0.707 s 4m σ b(max) = π × 0.707 s × d 2 5.66 m = π sd 2 For ready reference the table is given below showing polar moment of inertia J and section modulus Z for different type of weld. @seismicisolation @seismicisolation Welded Joints • Note: s = Size of weld Type of weld b Polar moment of inertia, J Section modulus, Z sl 3b2 + l 2 8.49 sbl √ 2 sl b2 + 3l 2 8.49 sb2 4.242 l b l l y G b s (b + l)4 − 6b2 l 2 16.97 (l + b) 4lb + b2 at bottom 8.49 2 b (4lb + b) at top s 8.49 (2l + b) s x l2 2l+b b2 y= 2(l+b) x= s b2 √ bl + 3 2 s (b + l)3 8.49 b l l G b s √ 2 (b + 2l)3 l 2 (b + l)2 − 12 b + 2l s b2 √ lb + 6 2 x x= l2 2(l+b) d π sd 3 5.66 s @seismicisolation @seismicisolation π sd 2 5.66 637 638 • Strength of Materials E XAMPLE 28.6: Two steel shafts of 55 mm diameter are connected by means of a flange coupling. The flanges are welded on the shaft ends. Determine the sign of the welds required on the surface of each shaft, both on the inner and outer faces of the flange, to transmit full torque capacity of the shafts. Assume the permissible shear stress in the shaft in 70 MPa and that in the weld is 100 MPa. S OLUTION : d = 55 mm, τshaft = 70 MPa, τweld = 100 MPa Maximum torque, Ts transmitted is given by π d3 × τshaft 16 π Ts = × 553 × 70 16 = 2285576.6 Nmm Ts = ∴ (i) Since the welds are provided at two places on each flange, the torque capacity, Tw of the welds, is given by, Tw = 2 × π sd 2 τweld 2.83 (as proved earlier) where s is the size of weld 2 × π s × 552 × 100 2.83 = 671272.1 s Nmm Tw = (ii) Equatity Eqns. (i) and (ii) 2285576.6 = 671272.1 s 2285576.6 ∴ s= 671272.1 = 3.4 mm Ans E XAMPLE 28.7: A circular shaft of dia 60 mm and length 220 mm is welded to a flat plate. A force of 15 kN is acting at 220 mm long shaft at the free end. The shaft is horizontal and the size of weld is 22 mm. Determine the maximum normal stress in the weld. @seismicisolation @seismicisolation Welded Joints S OLUTION : d = 60 mm, F = 15 kN, l = 220 mm, s, size of weld = 22 mm. 22 Throat thickness, t = √ = 15.5 mm 2 Throat area, A = π d = π × 60 × 15.5 = 2920.2 mm2 F Transverse shear stress = A 15000 = 2920.2 = 5.14 N/mm2 Bending moment = F × l = 15000 × 220 = 3300000 Nmm For a circular filled weld section modulus Z is given by π sd 2 5.66 π × 22 × 602 or Z = 5.66 = 43937.8 mm3 M Bending stress, σb = Z 3300000 = 43937.8 = 75.1 N/mm2 Z= Resultant normal stress, σmax = τ 2 + σb2 = 5.142 + 75.12 √ = 26.4 + 5640 = 75.27 N/mm2 @seismicisolation @seismicisolation Ans • 639 640 • Strength of Materials E XAMPLE 28.8: A rectangular cross-section bar is welded to support by means of fillet welds as shown in Fig 28.15. Determine the size of the welds, if the permissible shear stress in the weld is limited to 90 MPa. 30 kN 160 450 mm 120 Figure 28.15 S OLUTION : F = 30, kN = 30 × 103 N, τmax = 90 MPa, l = 120, b = 160 mm, L = 450 mm. The throat area for a rectangular fillet weld, A = t(2b + 2l) = 0.707 s (2b + 2l) = 0.707 s (2 × 160 × 2 × 120) = 0.707 s (320 + 240) = 396 s F 30 × 103 = A 396 s 75.75 = s Direct shear stream, τ = b2 lb + 3 1602 = 0.707 s 120 × 160 + 3 = 0.707 s (19200 + 8533.3) 5 From table, Z section modulus = √ 2 Z = 19607.5 s mm3 Bending stress, M Z M = P×L σb = = 30000 × 450 = 13500000 @seismicisolation @seismicisolation Welded Joints • 641 13500000 19607.5 s 688.5 N/mm2 = s ∴ σb = 1 2 (σb )2 + 4τ 2 Maximum shear stress (τmax ), 90 = 1 = 2 688.5 s 2 75.75 +4 s2 2 1√ 474632.25 + 22952 2s 352.5 90 = s 352.5 or s = 90 = 3.92 mm = = Say 4 mm Ans E XAMPLE 28.9: The bracket shown in Fig. 28.16 is designed to carry a dead weight of 15 kN. What sizes of fillets are required at the top and bottom of the bracket? Assume that the forces act through A and B. 25 mm Fillet weld A 75 mm B 15 kN 50 mm Figure 28.16 The welds are produced through shielded arc welding process with a permissible strength of 150 N/mm2 . S OLUTION : Shearing stress = Shearing force Area of weld group @seismicisolation @seismicisolation 642 • Strength of Materials Let t be the throat thickness and s be the size of weld. t = 0.707 s Area of weld group, A = 2 × 25 t = 50 t 15000 τ= 50t 300 = N/mm2 t Therefore, (i) To determine bending stress σ , moment of inertia of weld group is important to calculate. Ixx = 2 25t 3 25t × 752 + 12 4 Because t is a small quantities as compared to other quantities like ∴ 25t × (75)2 2 = 70312.5t mm4 M = 15000 × 50 = 750000 Nmm M 75 σ= × Ixx 2 75 750000 × = 70312.5t 2 400 = t Ixx = Bending moment, ∴ The maximum stress (shearing) on the plane of throat, τmax = σ 2 = 25t 3 can be neglected. 12 2 200 t + τ2 2 + 300 t 1√ 40000 + 90000 t 360.6 N/mm2 = t = @seismicisolation @seismicisolation 2 (ii) Welded Joints • 643 As stated in the problem, permissible stress is 150 N/mm2 ∴ 360.6 t 360.6 or t = 150 = 2.4 mm 2.4 = 3.39 ∴ s= 0.707 Say 3.4 mm Ans 150 = Eccentrically Loaded Welded Joint F B R b A θ θ τ1 G τ2 a e dA A B r R G Figure 28.17 Let us choose an elemental area dA at a distance r from G and assume that stress acting on dA is τ , so that τ τ2 = r R r or τ = τ2 (i) R The force acting on elemental area dA, dF = τ dA The moment due to dF about G dM = τ r dA From Eqns. (i) and (ii) dM = τ2 r2 dA R @seismicisolation @seismicisolation (ii) 644 • Strength of Materials Integrating, dM = M = F.e τ2 r2 dA R τ2 F.e R Fe = JG ∴ τ2 = R JG or F.e = JG is the polar moment of inertia of a fillet weld about G. Now τ , and τ2 act at an angle θ . The resultant stress, τA = τ12 + τ22 + 2 τ1 τ2 cos θ To determine JG , the following procedure is adopted. i) Find the C.G. of weld group with reference to inside edges of vertical and horizontal fillets. ii) Determine the second moment of inertia IXX1 and IYY 1 respectively with respect to horizontal and vertical axes passing through C.G. And then transfer them to axes of C.G. of weld group. iii) Similarly find Ixx2 and IYY 2 and then transfer them to horizontal and vertical axes passing through C.G. of weld group. iv) In order to find JG, add the above four moment of inertia: JG = I XX1 + I YY 1 + I XX2 + I YY 2 , etc. neglect terms like a t 2 , etc. Remember since t is too small while computing IXX1 , IYY 1 E XAMPLE 28.10: A bracket is welded to its support as shown in Fig. 28.18. All welds are fillet welds of equal thickness. Determine the fillet size if the permissible stress in the welds is 80 N/mm2 . [U.P.S.C. Engg. Services 1979] 50 mm 60 mm 1 s E Y θ A θ τ2 τ1 150 3 OG D 13.33 mm B C 2 16.7 mm Figure 28.18 @seismicisolation @seismicisolation X 25 kN Welded Joints • 645 To find C.G: A1 = 60 s, A2 = 60 s, A3 = 150 s As the weld group is symmetrical about horizontal axis, C.G. will be on OX-axis, let its distance from DE be x. A1 (30) + A2 (30) + A3 (0) A1 + A2 + A3 60 s × 30 + 60 s × 30 x= 60 s + 60 s + 150 s 2 × 60 × 30 = 270 = 13.33 mm x= To find polar moment of inertia of weld group about C.G. Let Ixx and Iyy be the second moments of inertia of plane rectangular about horizontal and and I be the moments of vertical axes passing through C.G. of respective rectangles. And Ixx yy inertia about horizontal and vertical axes passing through C.G. of weld group, G. The suffixes 1, 2, 3 are for areas 1, 2 and 3. Since s is much smaller, it can be neglected. Also value of s is not known initially. 60 s3 = 5 s2 12 60 s3 = 5 s3 = 12 t (150)3 = 281.25 × 103 s = 12 = 5 s2 + 60 s (75)2 Ixx1 = Ixx2 Ixx3 Ixx 1 = 5 s2 + 337.5 × 103 s Ixx = 5 s2 + 60 s (75)2 2 = 5 s2 + 337.5 × 103 s = Ixx3 = 281.25 × 103 s Ixx 3 s (60)3 = 18 × 103 s 12 s (60)3 = = 18 × 103 s 12 150 (s)3 = = 12.5 s3 12 = 18 × 103 s + 60 s (16.7)2 Iyy = 1 Iyy 2 Iyy3 Iyy 1 = 34.7 × 103 s @seismicisolation @seismicisolation 646 • Strength of Materials = 18 × 103 t + 60 s (16.7)2 Iyy 2 = 34.7 × 103 s Iyy = 12.5 s3 + 120 s (13.33)2 3 = 12.5 s2 + 21.32 × 103 s The polar moment of inertia of the weld group about C.G. JG = Ixx + Ixx + Ixx + Iyy + Iyy + Iyy 1 2 3 1 2 3 = 5 s3 + 337.5 × 103 s + 5 s3 + 337.5 × 103 s + 281.25 × 103 s + 34.7 × 103 s + 34.7 × 103 s + 12.5 s3 + 21.32 × 103 s = 22.5 s3 + 1047 × 103 s mm4 Because s3 is a small quantity, s3 may be neglected. Hence, JG = 1047 × 103 t mm4 . The secondary shearing stress at any point on the fillet at a distance R from G is given by τ = FeR JG This stress will act perpendicular to R, its value will be higher if R is larger R at A and C is greatest. From triangle ABG, RA = (AB)2 + (BG)2 = (75)2 + (46.7)2 = 88.35 mm e = 50 + BG = 50 + 46.7 = 96.7 mm 25000 × 96.7 × 88.35 1047 × 103 s 204 = MPa s τ2 = For resultant shearing stress, τ1 and τ2 are to be added vectorially, τs = τs = τ12 + τ22 + 2 τ1 τ2 cos θ 1 s (132.3)2 + (204)2 + 2 × 132.3 × 204 × 0.53 @seismicisolation @seismicisolation Welded Joints • 647 1√ 17503.3 + 41616 + 28608.55 s 296.2 MPa = s = Permissible stress in the weld is 80 MPa or 80 N/mm2 . 296.2 s 296.2 s= 80 = 3.70 mm 80 = Say 4 mm Ans Exercise 28.1 A steel plate 100 mm wide and 12.5 mm thick is welded by a parallel fillet welds to another plate. The plates are subjected to a load of 60 kN. Find the length of the weld if the maximum shear stress is not to exceed 56 N/mm2 . [Ans 60.6 mm] 28.2 A 12 mm thick bracket plate is connected to column flange as shown as Fig. 28.19 and transmits a load of 180 kN. Design a suitable weld. Permissible stresses in the weld are: in bending = 15.45 kN/cm2 , in shear = 10.05 kN/cm2 . 180 kN 150 mm 150 mm 600 mm [Ans Figure 28.19 Size of weld = 3.68 mm] 28.3 A 100 mm wide and 12.5 mm thick plate is welded to another plate by two parallel fillet welds. A load of 50 kN acts along the axis of this plate. Find the length of each fillet if the permissible shear stress is 55 MPa. [Ans 38.5 mm] 28.4 A plate is attached to a frame by two side fillet welds as shown in Fig. 28.20. Determine the size of the welds to resist vertical load of 50000 N. The permissible stress is 10250 N/cm2 . @seismicisolation @seismicisolation 648 • Strength of Materials 50 kN D 100 mm E 100 mm B 150 mm A [Ans Size of weld DE = 12.3 mm, size of weld AB = 11 mm] Figure 28.20 28.5 A rectangular cross-section bar is welded to a support by means of fillet welds as shown in Fig. 28.21. Determine the size of welds, if the permissible shear stress in the weld is limited to 75 MPa. 25 kN 150 500 mm 100 mm [Ans Figure 28.21 @seismicisolation @seismicisolation 5.32 mm] C HAPTER 29 MECHANICAL TESTING OF MATERIALS 1. Tensile Test a) To investigate for a tensile specimen, the relationship between gauge length and percentage elongation. b) To find percentage reduction in area after fracture. Apparatus: Denison Universal testing machine (Fig. 29.2) or equivalent machine, mild steel specimen (dimensions shown in Fig. 29.1) centre punch, hammer, rule, divider, vernier calliper. 26 mm dia Gauge length = 5.65 √A A = cross-sectional area of specimen 20 Rad 20 mm 14 mm 70 mm Gauge length 90 mm 55 mm 240 ± 1 mm Figure 29.1 Procedure: 1. With specimen aid of divider and punch mark two points 70 mm apart (gauge length) with centre punch, as shown in Fig. 29.1. 2. The specimen should be in annealed condition 3. Mount the specimen symmetrically. Fix Lindley extensometer on gauge length tightly. 4. Start apply load after bringing the load printer to zero. And take 5 − 6 readings of load and extension. @seismicisolation @seismicisolation 650 • Strength of Materials 5. Before yield point remove the extensometer after taking reading of extension. This is very important otherwise extensometer will get damaged. 6. Now if automatic graph is fitted, keep an eye on it and see how two yield points on mild steel appears. 7. Apply load until fracture takes place. It will be observed that before breaking, necking shape takes place. 8. Observe the fractured piece after taking them out from U.T.M. It will be cup and cone shaped fracture. Results Place the broken pieces together so that no gap appears between two pieces. Now measure the gauge length which will obviously be more than before let original gauge length is equal to L1 and after breaking and putting the two pieces to make them straight without any gap and measure the new gauge length. Let new gauge length after fracture = l2 . Then, l2 − l1 × 100 = % elongation l1 Maximum load indicated by pointer on dial B. Ultimate tensile strength = Original area of cross section C. Percentage reduction in area: A. Let initial area = A1 Now measure the diameter of neck where after necking fracture has taken place and calculate this area. Let it be A2 . Now, Percentage reduction in area = A1 − A2 × 100 A1 Conclusion: Note down carefully the sources of error and draw sketch of portion of both broken pieces (near portion of fracture) Cup Cone @seismicisolation @seismicisolation Mechanical Testing of Materials • 651 Plot stress-strain curve, if automatic graph is not available on U.T.M. and mark all important points. E B A C F D Figure 29.2 Universal testing machine @seismicisolation @seismicisolation 652 • Strength of Materials Figure 29.3 Use of Lindley extensometer Brittle material Ductile material Figure 29.4 Fractured ends of specimens. Note: For brittle material graph will be almost straight line without showing yield point. For future reference the following table is quite useful, showing various properties: @seismicisolation @seismicisolation Mechanical Testing of Materials • 653 Table 29.1 Typical Mechanical Properties of Some Metals & Alloys Metal or Alloy Lead Aluminium Duralumin Magnesium with 10% Al Copper 70 − 30 brass cartridge metal Phosphorbronze 95% Cu, 5% Sn Wrought iron Mild steel Nickel-chromiummolybdenum steel 2.5% Ni, 0.75% Cr, 0.5% Mo Stainless steel Cast iron (grey) Condition Soft sheet Wrought and annealed Extruded and heattreated Cast and heat-treated Wrought and annealed Annealed sheet, deep drawn Hard rolled Hot-rolled bar Hot rolled sheet Hardened and tempered 0.1% Proof stress MN/m2 − Ultimate tensile strength MN/m2 17.7 % elongation on 70 mm gauge length 64 Brinell hardness number 4 Fatigue limit MN/m2 2.78 − 58.7 60 15 30.9 278 432 15 115 170 116 247 2 80 61.8 46.3 216 60 42 66.4 84.9, 371 324, 463 67.5, 19.5 62, 132 114, 151 649 711 5.5 188 185 201 0.5% proof stress 232 309 30 100 185 309 28 100 185 1310 1540 10 444 432 185 − 463 309 30 0.0 170 250 263 137 Softened cast Fatigue Limit: The fatigue limit of a metal is the maximum range of stress which the test piece will endure without failure after an infinite number of reversals usually 107 cycles about a mean stress of zero. 2. Compression Test To investigate the behaviour of the following materials when subjected to compressive loading: i) cast iron, ii) mild steel, iii) aluminium, iv) copper and v) concrete @seismicisolation @seismicisolation 654 • Strength of Materials Apparatus: Denison universal testing machine or equivalent machine with attachments for compression tests, a cylindrical specimen of suitable length / diameter ratio of each of the above materials, micrometer, dial test-indicator with supporting stand. The compression testing attachment consists of upper and lower hardened steel compression plates. The upper plate is located in a recess in the lower wedge-box of the machine and is retained by a headed peg which forms a baynot fastening. The lower plate is located in a reverse on the transverse beam of the machine. Procedure: Measure the length and diameter of the specimen and check that the ends are machined perfectly flat and perpendicular to the longitudinal axis. Locate the cast-iron specimen vertically on the lower compression plate and lower the straining wedge-box until the upper compression plate just touches the specimen. Position the dial test indicator so that it is bearing on the top of the lower wedge-box and zero the pointer. Gradually, apply the axial compressive load to the specimen and record corresponding readings of contractor, as given by the dial test indicator reading, and load until fracture occurs. Examine and sketch the fractured specimen and note particularly the approximate angular position of the plane of fracture. Test the specimens of the other materials using the same procedure and if, after a sufficient period of testing, failure seems unlikely to occur for any specimen, discontinue the test. Plot graphs for each test, using the readings recorded, of load against contraction. ∴ σθ = σ cos θ · AB = σ cos2 θ . AC Resolving parallel to plane AC, τθ AC = σ sin θ · AB AB ∴ τθ = σ sin θ = σ sin θ cos θ AC Due to friction between the surfaces the resistance to sliding will be increased by an amount μσθ where μ may be regarded as the coefficient of friction. Let τ be the ultimate shear stress in cast-iron. Then if fracture is to occur, τθ = τ + μσθ ∴ τ = τθ − μσθ = σ sin θ cos θ − μσ cos2 θ σ μσ (cos 2θ + 1) = sin 2θ − 2 2 @seismicisolation @seismicisolation Mechanical Testing of Materials • 655 For a maximum value of τ , dτ =0 dθ σ cos 2θ + μσ sin 2θ = 0 ∴ tan 2θ = − 1 μ For cast-iron μ is assumed to be 0.364 ∴ 1 0.364 = −2.747 2θ = 110◦ θ = 55◦ tan 2θ = − That is, fracture occurs on a plane making an angle of 55◦ with the ends of the specimen. Typical forms of specimens after testing. Copper Aluminium Mild steal Cast iron Concrete Figure 29.5 Theory: Compression tests are normally carried out on specimens of circular cross section which, to eliminate bending, are normally such that the length /diameter ratio is less than 2. It is extremely important that the compressive load should be axial, and for this reason the ends of the specimen must be perfectly flat and perpendicular to the axis of the specimen. Even if the applied load is absolutely axial and the surfaces are in perfect contact the stress distribution is not uniform over the cross sections of the specimen, due to friction, and for this reason the specimens tend to become barrel shaped. @seismicisolation @seismicisolation 656 • Strength of Materials The behaviour of the specimens when subjected to loading depends largely on the particular material. For ductile materials their behaviour is similar to that when they are subjected to tensile loading. During the elastic stage the load is proportional to the contraction produced, and after the yield point, which is not so pronounced as when tested in tension, the deformation is plastic. Fracture of ductile materials is seldom obtained, and the compressed cylinder ultimately becomes a flat disc. Brittle materials when tested do not exhibit a yield point, and the plastic deformation is much less noticeable than for ductile materials. Since the shear resistance is respectively small, failure occurs due to shear on internal planes, which make an angle of approximately 55◦ , with the crosssection of the specimen. Theoretically, the plane on which the maximum shear stress occurs is at 45◦ , but due to friction between the sliding surfaces this angle is always somewhat greater. For example, consider a cast-iron specimen subjected to a compressive stress σ . Let the direct and shear components on a plane make the equilibrium of the portion ABC and assume unit thickness. σ σθ C σ θ σ cosθ τθ A θ B σ sin σ Figure 29.6 Resolving perpendicular to plane AC. σθ AC = σ cos θ · AB Results: i) Cast-iron Original diameter of specimen, d = mm Original length of specimen, l = mm Load W (kN) Contraction (mm) @seismicisolation @seismicisolation θ Mechanical Testing of Materials • 657 Record similar results for each different materials. Conclusions: Draw up separate conclusions for each specimen with regard to the following: i) Whether fracture occured or not, ii) Nature of fracture, if any. Compare the graphs of load against contraction obtained for each material. What do you infer from the tests on the cast-iron and concrete specimens with regard to the ratio between their ultimate shear and ultimate compressive strengths? 3. Shear Tests Object: To determine experimentally, the ultimate shear strength, in single and double shear of the following materials: i) mild steel, ii) cast-iron, iii) copper, iv) brass and v) aluminium. Apparatus: Denison universal testing machine together with attachments for single and double shear tests, round specimens of the above materials, micrometre, etc. The shear attachments are shown in Fig. 29.7. The shear die engages with the lower straining wedge-box, which is provided with a spigot and headed pegs to form a bayonet fastening. The shear box, which locates the specimen horizontally, rests on the transverse beam of the testing machine which locates the specimen horizontally, rests on the transverse beam of the testing machine. Figure 29.7 Procedure: Attach the shear die to the wedge-box, and suitably position the shear box on the transverse beam. Check the diameters of the specimens using a micrometre. Insert a single shear specimen of mild steel in the hole in one side of the shear box so that it projects two-thirds of the distance across the gap. Apply the load gradually to the specimens by means of the shear die, and record the ultimate (maximum) load required before fracture occurs. Test a second mild steel specimen in single shear as before, and record the average value of the ultimate load. Repeat the procedure for the single shear specimens of cast-iron, copper, brass and aluminium and in each case record the average value required for fracture in double shear. @seismicisolation @seismicisolation 658 • Strength of Materials Insert a specimen of mild steel symmetrically through both holes in the shear box. Apply the load to the specimen and record the average ultimate load required for fracture in double shear. Test a second mild steel specimen in double shear an before and record the average value of the ultimate load. Repeat the procedure for the double shear specimens of cast-iron, copper, brass and aluminum, and record for each material the average value of the ultimate load. Examine the form of the fractured specimen for each test. Results: Average ultimate load kN Material Diameter of specimen (mm) Cross-sectional area mm2 Single shear Double shear Mild steel Cast iron Copper Brass Aluminium Calculations: Ultimate shear strength (N/mm2 ) Material Single shear Double shear Mild steel Cart iron Copper Brass Aluminium Conclusions: Compare for each material the values obtained for the ultimate shear stress in single and double shear and account for any discrepancies. Compare also the experimental values with those normally accepted for the particular materials. Deduce from the form of the facture obtained, what effects other than pure shear will have influenced the results given. Compare the values of ultimate shear strengths of the materials with ultimate tensile strength of respective material. @seismicisolation @seismicisolation Mechanical Testing of Materials • 659 Table 29.2 Mechanical properties of metals Material Young’s modulus GN/m2 Tensile strength MN/m2 % Elongation 0.2% C steel 0.4% C steel 0.8% C steel 18/8 Stainless steel Grey cast iron White cast iron Brass (70 Cu 30 Zn) Bronze 95 Cu 5 Sn 210 210 210 210 210 210 110 110 350 600 800 700 250 − 460 600 30 20 8 65 − − 50 30 4. Modulus of Rupture Object: To determine experimentally, by commercial transverse tests on stand cast-iron specimens i) The transverse breaking load ii) The deflection at fracture iii) Modulus of rupture. Apparatus: Denison universal machine or equivalent machine with transverse test attachments, standard cast-iron specimens, dial test-indicator with supporting stand, steel rule. The transverse test attachment are a shown in Fig. 29.8 and consist of a loading knee together with two supporting dogs. Figure 29.8 The loading knee is located in a recess in the lower wedge-box of the machine and is retained by a headed peg which forms a bayonet fastening. The support dogs are located in recesses on the transverse beam of the machine to give the required span. @seismicisolation @seismicisolation 660 • Strength of Materials Procedure: Measure the diameter and length of the specimens and check that they conform to the standard requirements. Locate the transverse attachments in the testing machine ensuring that the support dogs are at the required distance apart. Place a specimen symmetrically on the supports and then lower the loading knee until it just touches the surface of the specimen. Position the dial test indicator so that it is bearing on the underside of the lower wedge-box and zero the pointer. Apply the load gradually to the specimen, through the loading knee, and record corresponding values of load and dial test indicator reading until fracture occurs. Repeat this procedure for several specimens, in each case recording corresponding readings of load, and the dial test-indicator reading until fracture occurs. Note in particular, for each specimen, the breaking load and maximum deflection. Plot graphs of load against deflection from the readings recorded. Determine the modulus of rupture using average breaking load by substitution in the equation given in the theory. Theory: The transverse test is universally recognized as the standard test for cast-iron. The strength of the material is indicated by the maximum central load the simply supported standard bar carries, and the maximum deflection at the point of application of the load gives and approximate measure of the toughness of the material. The transverse strength of cast-iron is commonly expressed by a figure known as either modulus of rupture or transverse rupture stress, this being the maximum stress that would here existed if the material conformed to the theory of bending. However, this is not so, since the material is stressed beyond the elastic limit. However, the modulus, of rupture is normally accepted as being a convenient way of expressing the results of a transverse test. Consider a circular specimen of diameter d simply supported over a span L. Let the maximum load required for fracture be W . From the theory of bending, modulus of rupture or transverse rupture stress, σ= and My I where Simply supported span, L = WL 4 d π d4 and I = 2 64 WL d/2 8W L 4 σ= = 4 πd π d3 64 y= Results: Diameter of specimen, d = M= mm mm @seismicisolation @seismicisolation Mechanical Testing of Materials • 661 Specimen Nos Load (kN) Deflection (mm) Draw up similar tables for the other specimens. Calculations: Average breaking load = Modulus of rupture, σ = kN kN/mm2 σ= 8W L π d3 Conclusion: Compare the values found for the breaking load and maximum deflection with the minimum standard requirements for the particular grade of cast-iron. Comment on the form of the graphs obtained of load against deflection. 5. Impact Test Object: To investigate the effect upon the Izod impact value for an alloy steel. Apparatus: Avevy Izod impact testing machine or equivalent machine, selection of about eight rounds or square Izod specimens being hardened with manufacturer’s instructions, but one only being correctly tempered. Of the remainder, one is left in the dead hard condition, and the others are tempered at various temperature, differing from the prescribed temperature, quenched in various media, e.g., air, oil, water. The dimensions of the specimen are given below: 28 mm 28 mm 28 mm 28 mm Hammer Strikes here 22 10 mm 45˚ 130 mm 2 mm 10 mm Izod specimen Hammer Strikes here 60 mm 10 mm square 2 mm deep, 1 mm Radius Notch 30 mm 40 mm No Charpy specimen Figure 29.9 @seismicisolation @seismicisolation 662 • Strength of Materials Here specimen’s dimensions for charpy test are also shown. The machine consists of a rigid construction as shown in Fig. 29.10 which can carry Izod test and Charpy test both. The specimens are held in gripping dies mounted in a vice on the machine base, the dies being clamped by means of a screw and spanner. A position gauge is used to ensure proper mounting of specimen. (a) Impact testing machine (b) Izod specimen and striker (c) Charpy support block. Figure 29.10 A trigger retains the pendulum in the raised position while the specimen is set up in the vice. The scale is an inverted quadrant graduated in joules. Procedure: Before beginning of the experiment, the pendulum is released to see that the pointer comes to zero. Because there in no specimen mounted, so the energy consumed is nil. Now fix the pendulum at the required height and hold the hammer (pendulum) in position with the help of ratchet or another mechanism. After fixing Izod specimen in the vice, release the hammer, so that in breaking the specimen, some energy will be spent and the pointer will give a reading in joules. This energy is the indicator of toughness of material. More the energy spent, more is strong the specimen in toughness. @seismicisolation @seismicisolation Mechanical Testing of Materials • 663 Striker 5˚ 22 mm 10˚ 1 mm 4 rad 2 to 3 mm radius specimen 45˚ Gripping dies Figure 29.11 Izod test After breaking the specimen, the hammer will swing backwards which is either stopped by a handle actuating a mechanism or if mechanism is not working, it should be carefully caught by the hand and lowered onto the rest. Now remove the broken piece (both) from gripping dies and raise the hammer to the right height held by a mechanism provided with a trigger. Repeat the test in respect of each specimen and record the Izod impact value in each case. Compare the impact value obtained for each specimen with that obtained for the specimen heat-treated in accordance with manufacturer’s instructions. Results: Write the exact composition of the steel used, the recommended heat-treatment and the minimum Izod value specified by the manufacturers for this condition. Specimen No Hardening temp(◦ C) Tempering temperature (◦ C) 1 2 3 4 5 6 @seismicisolation @seismicisolation Quenching media Izod impact value (J) 664 • Strength of Materials Conclusion: From the tabulated results deduce the effects of variation of tempering temperature and punching media upon the impact value of the meterial. Compare the relative severities of air, oil and water punching and deduce evidence to support your comparison from results. 6. Cupping Test This test is used to determine the ductility and drawing properties of sheet metal. Pressure ring d 20 mm dia Material under test Crack Die Figure 29.12 In Erichsen test, the material The sheet is kept between the die and pressure ring as shown in Fig. 29.12. The force on the ball is applied gradually over a steel (hardened) ball of 20 mm diameter. The sheet is then checked if after how much depth (d) the crack develops in specimen sheet. This test is very useful to test ductility of sheets, drawn sheets or deep-drawn sheets. Figure 29.13 Erichsen test machine @seismicisolation @seismicisolation Mechanical Testing of Materials • 665 7. Modulus of Rigidity of Rubber To determine the relationship between shear stress and sheet strain for rubber, and also to find modulus of rigidity of the material, a wall mounted rubber specimen fitted with dial-indicator and hanger for weights is needed. Set the dial test indicator to zero with the hanger in position. Now apply a load to the hanger. The rubber block in new under sheet (refer Fig. 29.14) and the loading plate will move downwards relative to the fixed plate, the relative displacement can be read from the dial indicator. Figure 29.14 This experiment should be repeated with increasing loads and record the varied displacement of the loading plate in each case. Unload and note the corresponding displacements with the decreasing loads. All results (readings) should be tabulated. Calculate the shear stress and corresponding shear strain and from the graph, determine the modulus of rigidity of rubber. @seismicisolation @seismicisolation 666 • Strength of Materials Theory (Refer Fig. 29.15) P = load on hanger l = length of rubber block h = width of rubber block t = thickness of rubber block d = vertical displacement of loading plate h P Area under stress = l × t P ∴ Shear stress = A P = l ×t l P Figure 29.15 Neglecting any displacement due to bending, shear strain, φ = d h Results: Displacement Load P(N) Load increasing d1 Load decreasing d2 Calculations: Load P(N) Mean displacement (d + d2 ) mm d= 1 2 t Shear stress P σ = (N/mm2 ) lt @seismicisolation @seismicisolation Shear strain d φ = (rad) h Mechanical Testing of Materials B Shear stress N/mm2 O • 667 From the graph (Fig. 29.16) plotted of shear stress against shear stain the modulus of rigidity C for Shear stress Shear strain = Slope of graph AB or C = N/mm2 OA rubber = A Shear strain φ While performing the experiment be careful in taking reading, from dial indicator. Quote the value obtained for modulus of rigidity and list probable causes of inaccuracy in the result. 8. Torsion of a Round Bar Using the torsion apparatus shown in Fig. 29.16 we have to determine: (i) the relationship between the angle of twist and the torque applied for a round bar and (ii) the modulus of rigidity of the material of the bar. Effective length of specimen Pointer Graduated circular dial for angle of twist Light pulleys set up to form a couple Pointer Scale x F Figure 29.16 @seismicisolation @seismicisolation F 668 • Strength of Materials We shall require metal bar, load-hangers, steel rule and weights in addition to torsion-testing apparatus. Clamp the specimen securely at its ends, and adjust the position of the upper protractor scale to give a convenient gauge length. Pass the draw cords over the pulleys and attach load-hangers to the free-ends. Zero the protractor scales with the hangers in position. Attach equal loads to the hangers, thus applying a couple to the bar, and record the angle of twist over the gauge length by observing the difference of the protractor-scale readings. Repeat this procedure for increasing values of load, and in each case, record the corresponding angle of twist. Measure the diameter of the bar, the diameter of the pulley, and the gauge length. Calculate the torque applied for each load and compile a table of readings of load, torque and angle of twist. Draw a graph against angle of twist and by selecting two points on the graph, find the law relating the two variables in the form T = K θ . Using the slope of the graph, also determine the modulus of rigidity. The two forces F constitute a couple Fx, i.e., applied torque, T = Fx (Nmm) Using Formula, ∴ F Cθ T = J l Tl N/mm2 C= θ x/2 x/2 F where J = Figure 29.17 π d4 mm4 32 C = modulus of rigidity (N/mm4 ) θ = angle of twist (radians) l = effective length of specimen (mm) ∴ C= T l × (N/mm2 ) θ J l Therefore C = slope of graph × N/mm2 J Results and calculations: Diameter of bar, d = mm ∴ J= π d4 = 32 @seismicisolation @seismicisolation mm4 Mechanical Testing of Materials Effective length of rod, l = 669 mm Effective diameter of pulley, x = mm Angle of twist Load F(N) • Angle of twist π (rad) θ =θ× 180 θ (degrees) T = Fx (Nmm) After the experiment, compare value C obtained with normally accepted for the material of the bar. If there is some error, state the sources of error. 9. Verification of Macaulay’s Method for Beam Deflection To verify Macaulay’s method, we need a beam deflection apparatus, steel beam, two dial test indicators and stands, micrometer, steel rule, two hangers, and weights. w1 w2 C A a x RA D X b RB l Figure 29.18 Set the beam on apparatus. Take convenient points C and D. Make sure that beam is simply supported and are free at A and B. Calculate reactions RA and RB after loading W1 and W2 at distance a and b from A respectively. Take point X at x mm from point A such that both loads are covered by x (loads W1 & W2 ) Now measure the vertical deflection at C and D with dial indicators, so that to verify Macaulay’s method we can compare these deflections with theoretical values. Now, EI Mx = RA x −W1 [x − a] −W2 [x − b] (i) d2y = RA x −W1 [x − a] −W2 [x − b] dx2 (ii) As we know in Macaulay’s method the bracketed terms are integrated as a whole. @seismicisolation @seismicisolation 670 • Strength of Materials Integrating Eqn. (ii) EI x2 W1 W2 dy = RA − [x − a]2 − [x − b]2 +C1 dx 2 2 2 EIy = RA (iii) W2 x3 W1 − [x − a]3 − [x − b]2 +C1 x +C2 6 6 2 (iv) When x = 0, y = 0 Substituting in Eqn. (iv) C2 = 0 Now, C1 can be found out by substituting x = l, because at B, deflection is zero so finally we have, W1 W2 x3 − [x − a]3 − [x − b]3 + C1 x EI y = RA 6 6 6 Since C1 is known, so find out deflection at C and D. If breadth of beam b = mm and bd 3 depth = mm, then I = . E is already known for the material of the beam. 12 While substituting to find yc and yD , remember any term in bracket becomes –ve, then it is to be ignored. Finally, to compare the deflection use following table: Experimental deflections At C At D Theoretical values or Calculated values using Macaulay’s method At C At D If results are somewhat differing, state the sources of error. 10. Verification of Maxwell’s Reciprocal Theorem The reciprocal theorem asserts that the deflection at any point A due to load at any point B is equal to the deflection at point B when the same load is placed at A. Let us use a small cantilever apparatus. We will need in addition hanger, weights and a dial indicator. First place hanger at B and apply load on the hanger. Using dial indicator, find the deflection at A. Let it be YBA . Now remove hanger and load. And place the same load at A and observe the deflection at B with dial indicator. Let this deflection be YAB . W A B l L Figure 29.19 @seismicisolation @seismicisolation Mechanical Testing of Materials • 671 Now you will notice that YBA = YAB Hence, Maxwell’s theorem is verified. If there is some error, find out the sources of error. May be due to vibration because of some compressor or some other machine, dial indicator is not giving correct readings. So try again at some stable place. 11. Hardness Testing Machines The most common hardness testing machines are: i) ii) iii) iv) Brinell Hardness Testing Machine Rockwell Hardness Testing Machine Vickers hardness Testing Machine Shore Scleroscope (i) Brinell Hardness Testing Machine BN (Brinell number) P is the load (kgf). An alternative and more precise equation than given is BN = P √ πD (D − D2 − d 2 ) 2 (i) d = diameter of ball impression and D = diameter of ball. It is apparent that the harder the metal, the smaller the indentation diameter, and consequently the higher the Brinell number. For thin sheets, however the load of 300 kgf is too heavy, assuming the relation of indentation depth is D2 is constant. The ball and load are therefore variable and the load is applied slowly for 30 seconds. So here also some Eqn. (i) is applicable. When the load is varied in standard machines, the variation increases by 500 kgf each. After application of the load to the softer metals for 30 s minimum, the pressure is released and the test piece removed by lowering the table or platform. The impression diameter is measured by a low power microscope with a graduated scale in the eyepiece. The middle of the indentation should be more than 2.5× diameter from the testpiece edge and the thickness more than 7× indentation depth. The indentation must not be visible on the other side of the thin testpiece. Uses of the Brinell test: Following are the main purposes for which the test is useful. (a) As a test of hardness: It is possible to find the amount of hardness induced by cold working. (b) To explore machineability. Brinell number gives a guide to the machining possibilities. If To explore BN a material is more 280–320, machining becomes difficult. But if Brinell number is below 120–100 the material may be too soft and may tear under the cutting edge. (c) To estimate the tensile strength (of steel): For annealed (normal steel): tensile strength (N/mm2 ) = 3.6 × B.N For hardened and tempered steel: tensile strength (N/mm2 ) = 3.2 × B.N (d) The shape of ball impression gives an indication of the work-hardening capacity. @seismicisolation @seismicisolation 672 • Strength of Materials Description: P (a) Metal being tested. (b) Piling up. (c) Sinking. (W&T. Avery Ltd. Brinell test. Figure 29.20 In this machine a small hardened and tempered chromium steel or tungsten carbide ball of 10 mm diameter is forced by pressure into the metal to be tested. The test-piece is placed on the platform of the testing machine either universal or specially designed, the platform being elevated by screw to make contact with the ball. This achieved, an oil pump applies increasing pressure, which is read from a manometer or pressure gauge fixed to the machine. A regulated balance is also fitted, the centre pin of which rests on the balls in the cylinders. These act as a piston and carry weights, rendering pressure variable with hardness. The load is known in advance as is usually 3000 kgf, under which load the weights float. The pressure is applied for 10 seconds and then released, the machined and fine emery polished test piece being removed and taken to a microscope whose inner lens has a millimetre scale divided into tenths, so that the diameter of the indentation made by the ball can be measured with accuracy. Non-ferrous metals are mostly subjected to a load of 500 kgf for 30 seconds. The Brinell number, corresponding to the hardness of the metal is obtained from the equation: BN = P Spherical area of impression (mm2 ) Brinell hardener values Material Brinell Number Soft brass 60 Mild steel 130 235 Annealed chisel steel 415 White cast-iron Nitride surface 750 @seismicisolation @seismicisolation Mechanical Testing of Materials • 673 (ii) Rockwell Hardness Testing Machine Figure 29.21 shows a Rockwell hardness testing machine like the Vickers machine, this employs a diamond but one conical in form, though both Vickers and Rockwell machines balls can be used if desired. The ball in these instances is 1.588 mm diameter. However, the modern tendency is to use the diamond cone (angle 120◦ ) and though machines are graduated in two scales, C and B, most values are given on the C scale for the harder metals. There are other scales, namely A, D, E and F, each representing a different combination of indenter and load. A recent machine of British manufacture based on the same principles in the Avery hardness tester, which is an improvement on the Rockwell as regards the means of measurement and in certain other respects. The hardness numeral is obtained by measuring the depth of the impressions superimposed on one another. The first impression made is produced by a another. The first impression made is produced by a load of 10 kgf, termed the preliminary, which penetrates the superficial layers of the steel or other metal. The minor load is applied and is indicated on a small subsidiary dial at a point marked ‘set’. The primary pointer is then set at zero by revolving the inclined edge of the did until it shows zero. The load is now raised to 150 kgf (140 kgf is added to the initial load 10 kgf) by button pressure. The 140 kgf load is then removed by a side lever and the hardness number is shown on the instrument dial, which is marked off in graduations from 0 to 100, representing a single rotation of the pointer and a penetrator movement limited to 0.02 mm. In the Rockwell testing machine the diamond scale load limit is 10 to 140 kgf. The test takes only about 10 seconds which means that it is almost invisible to the naked eye, no particular harm is done to the finish of the work. The surface concerned needs no other preparation than the elimination of oxide. On the other hand, the machine’s scale is not nearly so open, so that errors may occur in some instances from misreading of the scale rebound, while should there be any modification in the form of the penetrator face, this may also impair the accuracy of the readings. To ensure that the maximum accuracy is obtained, therefore certain precautions are essential. The machine should be so positioned that vibration has no serious disturbing effect on the dial group readings. When the minor load is being used, the large scale needles on the dial range must be ‘set’ within + or −5 scale divisions. It may happen that as the platform of the machine is raised, the Figure 29.21 Rockwell testing machine @seismicisolation @seismicisolation 674 • Strength of Materials needle stops, though it does not do so on ‘set’. It is then necessary to adjust the lifting screw again, the dial being rotated so that ‘set’ comes under the needle. Assuming the machine is of orthodox pattern, the dashpot speed in giving if the major load should allow the crack to finish its traverse in 5 seconds when no test piece is in the machine, the machine being set-up to give 100 kgf load. Similarly, the surface testing machine is set for seconds to give a major lord of 30 kgf. Calibrating the Rockwell Hardness Testing Machine The makers of the Rockwell machine suggest that the user should check it each time against the test blocks to ensure that the penetrator is not damaged and the machine is not out of calibration. The practice they advise is to check at the high, low and middle ranges of the given scale. Thus, to check the full C scale, the machine should be checked at C 63, 45 and 25. If only a single range or two ranges are employed, blocks should fall within ±5 hardness numbers on the C scale or any scale using diamond penetrator and within ±10 numbers on the B scale on induced, any scale using ball penetrators. (iii) The Vicker’s Diamond Test Machine Because of the limitations of the Brinell test, it was found desirable to replace ball by a pyramidal from of diamond. When the Brinell machine is put to work on soft metal, the surface of the metal is evaluated by plastic flow about its initial height an shown in Fig. 29.22. This means that indentation diameter d is greater than it should be, so that the hardness numeral is lower than the true value. Deformation of the half by extremely hard metals in another source of error. The square pyramidal diamond in studied under a microscope and to ensure that one corner just makes contact with a fixed knife-edge that can be seen through microscope eyepiece. d' Figure 29.22 The basic principle underlying the diamond pyramid test (also called Vicker’s hardness test) is similar to that of the Brinell hardness number, designated as HV or sometime also as DPN representing diamond pyramid number is determind as the ratio of the impressed load in kg to the area of the impression in square mm made by a pyramid with an accurately ground square base. The indenter is a diamond in the form of a square based pyramid with an included angles between opposite faces equal to 136◦ (Refer Fig. 29.23. Vicker’s hardness testing machine and impression as seen through microscope (vertical lines are edges of microscope measuring shutters). On examing Fig. 29.24 it reveals that there is a geometric similarity for all values of load divided by areas ratio. Apex angle is 136◦ is obtained by keeping the value of d/D = 0.375. Load (kg) should remain the same It is quite clear that the hardness value equal to Impressed area (mm2 ) under conditions, for a specimen of uniform hardness irrespective of the impressed load on the @seismicisolation @seismicisolation Mechanical Testing of Materials • 675 (Vickers-Armstrong, Ltd. -Impression as seen thorugh Microscope (Vertical lines are edges of Microscope Measuring Shutters). (Vickers-Armstrong, Ltd. Figure 29.23 x = 0.375D Diamond pyramid D x d d2 1 Test specimen Figure 29.24 indenting pyramid. The Vicker’s hardness number is defined as the load per unit surface of contact, i.e., impressed, in kg per square mm, and can be calculated from the average diagonal as follows: 2P sin θ2 d2 p = 1.854 2 d HV = θ = 136◦ d1 + d2 where, d = length of average diagonal in mm 2 since @seismicisolation @seismicisolation 676 • Strength of Materials θ = 136◦ apex angle P = impressed load in kg. The main advantages in this form of indentor are: i) The impressions irrespective of size are geometrically similar. ii) The deformation of the diamond is practically nil owing to its enormous hardness value. iii) The surface of impression is extremely well defined and being square can be measured with great accuracy across the diagonal corners. iv) The hardness numbers found on homogeneous materials are unaffected by variation of load. (iv) Shore Scleroscope This is shown in Fig. 29.25. The share scleroscope employs the principle of the drop and rebound of diamond-tipped hammer inside a glass tube. The hammer which is raised to the top of the tube by vacuum, is allowed to fall freely from a fixed height onto the test specimen. The height of the resulting rebound of the hammer is read at the graduated scale and indicates the degree of hardness of the material. The scale is graduated in units, which are obtained by dividing the average rebound from quenched pure high carbon steel into 100 part and then extended proportionally for metals having exceptional hardness. A lens and pointer are provided for accurate scale reading. These are principally used in connections with the soft materials. There is one important advantage of scleroscope that it can be taken out from laboratory and test may be carried out in field on heavy machinery. Figure 29.25 @seismicisolation @seismicisolation Mechanical Testing of Materials • 677 Table I Approximate Equivalent Hardness Number Conversion Table Rockwell Hardness C Scale No. Vicker’s Diamond Pyramid hardness No. Brinell Hardness No 10 mm ball, 3000 kg load standard ball Rockwell Hardness B Scale No Shore Scleroscope No 68 940 – – 97 67 900 – – 95 66 865 – – 92 65 832 – – 91 64 800 – – 88 63 772 – – 87 62 746 – – 85 61 720 – – 83 60 697 – – 81 59 674 – – 80 58 653 – – 78 57 633 – – 76 56 613 – – 75 55 595 – – 74 54 577 – – 72 53 560 – – 71 52 544 500 – 69 51 528 487 – 68 50 513 475 – 67 49 498 464 – 66 48 484 451 – 64 47 471 442 – 63 46 458 432 – 62 45 446 421 – 60 44 434 409 – 58 43 423 400 – 57 42 412 390 – 56 41 402 381 – 55 40 392 371 – 54 39 382 362 – @seismicisolation @seismicisolation 52 (Continued ) 678 • Strength of Materials Table I (Continued) Rockwell Hardness C Scale No. Vicker’s Diamond Pyramid hardness No. Brinell Hardness No 10 mm ball, 3000 kg load standard ball Rockwell Hardness B Scale No Shore Scleroscope No 38 372 353 – 51 37 363 344 – 50 36 354 336 (109) 49 35 345 327 (108.5) 48 34 336 319 (108.0) 47 33 327 311 (107.5) 46 32 318 301 (107.0) 44 31 310 294 (106.0) 43 30 302 286 (105.5) 42 29 294 279 (104.5) 41 28 286 271 (104.0) 41 27 279 264 (103.0) 40 26 272 258 (102.5) 38 25 266 253 (101.5) 38 24 260 247 (101.0) 37 23 254 243 100.0 36 22 248 237 99.0 35 21 243 231 98.5 35 20 238 226 97.8 34 (18) 230 219 96.7 33 (16) 222 212 95.5 32 (14) 213 203 93.9 31 (12) 204 194 92.3 29 (10) 196 187 90.7 28 (8) 188 179 89.7 27 (6) 180 171 87.1 26 (4) 173 165 85.5 25 (2) 166 158 83.5 24 (0) 160 152 81.7 24 Note: The Brinell hardness No. can be used to estimate the approximate tensile strength (N/mm2 ) of steels from the relationship: T S = 3.4 HB @seismicisolation @seismicisolation Mechanical Testing of Materials Table II Comparative Hardness Scaler for steels (un herdened) and non-ferrous Alloys Rockwell B scale 1/16” ball Penetrator 100 kg load Brinell scale 10 mm standard Ball 3000 kg load 100 240 99 234 98 228 97 222 96 216 95 210 94 205 93 200 92 195 91 190 90 185 89 180 88 176 87 172 86 169 85 165 84 162 83 159 82 156 81 153 80 150 79 147 78 144 77 141 76 139 75 137 74 135 73 132 72 130 71 127 70 125 (Continued ) @seismicisolation @seismicisolation • 679 680 • Strength of Materials Table II (Continued) Rockwell B scale 1/16” ball Penetrator 100 kg load Brinell scale 10 mm standard Ball 3000 kg load 69 123 68 121 67 119 66 117 65 116 64 114 63 112 62 110 61 108 60 107 59 106 Some Other Hardness Tests Knoop Hardness Number: The Knoop hardness test is applied to extremely thin metals, plated surfaces, exceptionally hard and brittle materials, very shallow carburized or nitrided surfaces, or whenever the applied load must be kept below 3600 g. The knoop indentor is a diamond ground to an elongated pyamidal form and it produces an inclination having long and short diagonals with a ratio of approximately 7 to 1. The longitudinal angle of the indentor is 172 degrees 30 minutes and tranverse angle 130 degrees. The tukon tester in which the Knoop indentor is used is fully automatic under electronic control. The Knoop hardness number equals load in kilogram divided by the projected area of indentation in square millimeters. The indentation number corresponding to the long diagonal and for a given load, may be determined from a table computed for a theoretically perfect indentor. The load, which may be varied from 25 to 3600 grams is applied for a definite period and always normal to the surface tested. Lapped place surfaces free from scratches are required. Monotron Hardness Number: With this apparatus a standard steel drill is caused to make a definite number of revolutions, while it is pressed with standard force against the specimen to be tested. The hardness is automatically recorded on a diagram on which deed soft material given a horizontal line, while a material as hard as the drill itself given a vertical line, intermediate hardness being represented by the corresponding angle between 0 to 90◦ . Turner’s Sclerometer: In making this test a weighted diamond point is drawn, once forward and once backward, over the smooth surface of the material to be tested. The hardness number is the weight required to produce a standard scratch. Mohr’s Hardness Scale: Hardness in general is determined by what is known as Mohr’s scale, a standard for hardness which is mainly applied to non-metallic elements and minerals. In this hardness scale there are ten degrees or steps each designated by a mineral, the difference in the series will scratch any of the preceding members. @seismicisolation @seismicisolation Mechanical Testing of Materials • 681 This scale is as follows: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Talc Gypsum Calcite Fluor spar Apatite Orthoclase Quartz Topaz Sapphire or corundum Diamond. These minerals, arbitrarily selected as standards, are successively harder, from talc, the softest of all minerals to diamond, the hardest. This scale, which is now universally used for non-metallic minerals, is however, not applied to metals. 12. Experiment 1) Close-coiled Helical Springs Object: To determine experimentally, for close-coiled helical springs, the relationship between deflection and i) applied load; ii) number of free coils; iii) mean coil diameter and iv) wire diameter. And to compare these with those indicated by the theory. Apparatus: Spring deflection apparature as shown in Fig. 29.26 close coils springs, steel rule, calipers, micrometer, hanger and weights. Figure 29.26 Procedure: i) Set up any spring in the deflection apparatus and apply initial load zero the vernier scale. Apply increasing loads to the spring in suitable increments and in each case record the corresponding deflection produced as indicated by the vernier scale. Plot using the readings, a graph of deflection against load. ii) Select a suitable number of springs having the same wire end mean coil diameters but a different number of ‘free’ coils. Place each spring, in turn, in the deflection apparatus and apply the same load to each. Record corresponding readings and deflection and number of ‘free’ coils for each spring. Plot a graph of deflection against number of ‘free’ coils for each spring. iii) Select a suitable number of springs which have the same wire diameter and number of ‘free’ coils but which differ in their mean coil diameter. @seismicisolation @seismicisolation 682 • Strength of Materials Apply the same load to each spring, in turn, and record corresponding readings of mean coil diameter and deffection produced. Plot a logarithmic graph, using the readings which have the same mean coils diameter and number of coil but differ in their wire diameter. Apply the same load to each spring, in turn, and record corresponding readings of wire diameter and deflection produced. Plot a logarithmic graph, using the readings obtained, of deflection against wire diameter. Determine from the graphs plotted the four required relationships. Theory: In a close-coiled helical spring the helix angle is very small, and it may be assumed that each coil is in a plane perpendicular to the spring axis. When a load is applied to a close-coiled spring stream are set up due to torsion, direct share and bending. The stresses due to direct share and bending are very small and may be nested in comparison with those set up due to torsion. Consider a spring of mean coil diameter D, wire diameter, d, and number of ‘free’ coils n, objected to an axial load W which produces deflection S (see Fig. 29.27). The work done by the load in deflecting spring, that is, W T 1 D δ = θ = W. θ 2 2 2 2 wire diameter d where T = applied torque = W.D 2 θ = angle of twist of wire in radius. Therefore deflection, D/2 δ= Dθ 2 From the torsion equation, Cθ T = J l W Figure 29.27 ∴ θ= T.l C.J where l = length of wire = π Dn J = polar second moment of area of cross section of wire, π d4 = 32 WD π Dn ∴ θ = 2π C d4 32 θ= 16W D2 n Cd 4 δ= 8w D3 n C d4 Therefore deflection @seismicisolation @seismicisolation Mechanical Testing of Materials • 683 Results: (i) (ii) (iii) (iv) Load (N) Deflection δ (mm) Number of free coils n Deflection δ (mm) Mean coil Diameter D (mm) Wire diameter d (mm) Deflection δ Deflection δ (mm) Calculations: From the graphs of deflection against load and deflection against number of ‘free’ coils show that δ α W and δ α n By determining the slopes of the logarithmic graphs show that δ α D3 and 1 δα 4 d Conclusions: Comment on the form of the graphs obtained and compare the relationship found with those indicated by the theoretical equation for the deflection. 13. Experiment To determine, experimentally, the value of the modulus of rigidity of C material of a close-coiled helical spring. Apparatus: Spring deflection apparatus as used (shown in Fig. 29.25) in previous experiment. Close coiled spring, steel rules, calipers, micrometer, load hanger and weights. @seismicisolation @seismicisolation 684 • Strength of Materials Procedure: Measure the wire diameter, mean coil diameter and count the number of ‘from’ coils. Set up the spring in the deflection apparatus and apply a suitable initial load to open the coils, and zero the vernier scale. Apply increasing loads to the spring in suitable increments and in each case record the corresponding deflection. Using the slope of this graph, in conjunction with spring data, determine the modulus of rigidity of the material of the spring. Theory: As proved for the previous experiment, when an axial load, W, is applied to a close-coiled spring the deflection, δ is given by, δ= 8W D3 n Cd 4 D = mean coil diameter d = diameter of wire n = number of ‘free’ coils Therefore, c = 8W D3 n δ d4 = 8D3 n d4 = 8D3 n × slope of graph d4 W δ Results: Wire diameter, d = mm Mean coil diameter, D = mm Number of ‘free’ coils, n = Spring load W (N) Spring Deflection δ (mm) Calculations: Slope of graph = W = δ Modulus of rigidity = @seismicisolation @seismicisolation N/mm N/mm2 Mechanical Testing of Materials • 685 Conclusions: Comment on the form of the graph obtained and compare the value found for modulus of rigidity with that normally accepted for the material. 14. Experiment To find Young’s modulus of elasticity of a material using simply supported beam. Apparatus: A simply supported beam, hanger, weights, dial test indicator with its stand. Dial test indicator, steel rule d l Weights Apparatus Figure 29.28 Theory: If a central load W is applied at the centre of a simply supported beam, then deflection of the centre is given by relation: W l3 48EI W l3 E= 48δ I δ= or Procedure: 1. Measure breadth and depth of steel beam (it can be any other material also), let it be b and d respectively. 2. At the centre of beam fix a book as shown in diagram (see Fig. 29.27). Also measure length l of the beam between supports. 3. Place the dial test indicator on the top of the hanger (placed at the centre of beam). Make the dial read zero. 4. Place the hanger whose weight is known on the hanger and take the reading of deflection with the help of dial test indicator. 5. Remove the weight and repeat the experiment two times and note deflection. @seismicisolation @seismicisolation 686 • Strength of Materials Readings: Weight (N) + Weight of Hanger δ (mm) 1. W δ1 2. 3. W W δ2 δ3 Sr. No. Average deflection δ1 + δ2 + δ3 3 =δ Calculations: I= bd 3 = . . . . . . mm4 12 Average deflection from above table, δ = . . . . . . mm L = . . .. . . mm E= Now W l3 (N/mm2 ) 48δ I Conclusion: If the value of E found from the experiment are different from standard value of the material, then find out the sources of error, e.g., length (l) measured is wrong or hook (weight) is not at exact centre or dial indicator is sticky etc. Note: This experiment can also be used to find deflection at the centre using moment area method. Deflection at centre C, = Moment of area bending moment diagram between left hand support and centre of span EI Of course, material is to be taken whose modulus of elasticity, E is already known. δ= W l3 48EI Then the deflection by experiment can be compared by theoretical value. 15. Fatigue Testing It has been observed during that if a component is subjected to repeated stresses, it fails stresses below the yield point stresses. Particularly, this kind of failure happens if stresses are due to reserved bending or reversed torque. Such kind of failure is known as endurance or fatigue limit. Experiment: To find the endurance limit of a metal specimen. @seismicisolation @seismicisolation Mechanical Testing of Materials • 687 Theory: In order to study the effect of fatigue of a material, a rotating fine standard specimen as shown in Fig. 29.29 is rotated in a fatigue testing machine while the specimen is loaded in bending. As the specimen rotates, the bending stream at the upper fibres varies from maximum tensile to maximum compressive as shown in the Fig. 29.30. In other words, the specimen is subjected to a completely reversed stress cycle. 7.675 mm R = 251 mm 875 mm Figure 29.29 Standard specimen Standard specimen: In the market, variety of fatigue testing instruments are available. Basically, it should consist of some method to produce alternating loads on the specimen. Also a counter should be incorporated in it which can show number of cycles before the specimen fails. Figure 29.30 shows a rotary bending fatigue machine. Bearing Bearing Motor COUNTER Shaft from motor fixed specimen by fastening system W Figure 29.30 Procedure: 1. 2. 3. 4. 5. 6. 7. 8. 9. The standard specimen is fitted in the bearings as shown in Fig. 29.30. Care should be taken that specimen in concentric with bearing within 0.04 mm. Apply suitable load by adjusting the arrangement of supporting the load. Set the revolution counter to zero. Start the motor of the machine. It is understood from the set up shown that the upper layer is under tension and when the same layer comes to bottom, it is under compression, thus stresses are completely reversed. Keep motor running until the specimen fails due to fatigue count the number of revolutions from the counter. Test is repeated with increased load on a new specimen and the procedure is repeated. For every test calculate the stress (σ ) applied. Plot a curve between σ and log N. Where N denotes cycles before failure. @seismicisolation @seismicisolation 688 • Strength of Materials Diameter of the specimen d (mm) Load W (N) Number of cycles (N) Stress (σ ) in N/mm2 Observation: The experiment is rotated at least four times with different loads on say stresses induced at the centre of the specimen in case of material of one kind. while testing new specimens. Endurance limit (106 cycles) = . . .. . . It may be noted that the term endurance limit is used for reversed bending only while for other types of loading, the term endurance strength may be used when referring to the fatigue strength of the material. It may be defined as the maximum stress which can be applied to the machine part working under actual conditions. Endurance limit or fatigue limit (σc ) a the maximum value of the completely reversed bending stress which a polished standard specimen can withstand without failure, for infinite number of cycles (usually 107 cycles). Exercise 29.1 29.2 29.3 29.4 29.5 29.6 29.7 29.8 29.9 Discuss the term ‘fatigue’ and ‘fatigue stress’. What is the necessity of a bend test? Explain how would you determine the modulus of rupture of a timber specimen. Draw stress-strain cure of tensile test on mild steel specimen and show important points on it. Explain the procedure of fatigue testing. State the names of tests which can be carried out on Universal Testing Machine. Describe briefly how a torsion test on a mild steel specimen is performed. What is ‘necking’ in tensile test and why it occurs? Explain each of following property: (i) Malleability (ii) Ductility (iii) Hardness (iv) Elasticity (v) Plasticity (vi) Brittleness. 29.10 How do you find the yield point in tensile test on mild steel? 29.11 Giving dimensions, draw sketch of a standard tensile test specimen of mild steel. @seismicisolation @seismicisolation 30 C HAPTER Miscellaneous Solved Problems: Stresses and Strains E XAMPLE 30.1: A tapered rod shown in Fig. 30.1 carries a tensile load of 18 kN at the free end. Calculate the total extension of the rod E = 200 GPa. Neglect self-weight. 65 mm dia A 480 mm B 35 mm dia 480 mm C 12 mm dia 18 kN Figure 30.1 S OLUTION : For portion BC, Extension δ = 4Pl π Ed2 d1 689 @seismicisolation @seismicisolation 690 • Strength of Materials Extension δBC = 4 × 18000 × 480 π × 200000 × 35 × 12 = 0.131 mm For portion AB, Extension δAB = 4 × 18000 × 480 π × 200000 × 65 × 35 = 0.0242 mm Total extenstoin = δBC + δAB = 0.131 + 0.242 = 0.373 mm Ans E XAMPLE 30.2: A bar of length 25 cm has varying cross section. It carries a load of 15 kN. Find x2 cm2 where x is the distance from one end the extension if the cross section is given by 6 + 100 in cm. Take E = 200 GPa. Neglect weight of the bar. S OLUTION : Stress at distance x from the given end, = = Load = Area 15000 N/m2 x2 10−4 6+ 100 1.5 × 108 x2 6+ 100 If da is the extension of an element of length dx at distance x from the given end, then strain at the section, = da σ = dx E da 0.075 1.5 × 108 = = 2 dx 600 + x2 x × 200 × 109 6+ 100 @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains da = ∴ a= • 691 0.075 dx (if x is in cm, a will also be in cm) 600 + x2 25 0.075 0 600 + x2 dx 25 k 0.075 −1 Total extension, a = √ ∴ tan √ 600 600 25 0.075 tan−1 √ ∴ a= 24.475 600 dx x 1 = tan−1 2 2 a a x +a = 3.062 × 10−3 tan−1 1.02 = 3.062 × 10−3 × 45.57◦ = 3.062 × 10−3 × 0.795 = 2.434 × 10−3 cm = 0.02434 mm [∴ degrees has been changed to radians] Ans E XAMPLE 30.3: A rigid beam is loaded as shown in Fig. 30.2. Determine the stresses in the steel wire and in the aluminium member. What is the reaction at the hinge A? Take E for aluminium as 60 GPa and E for steel as 200 GPa. Steel wire Area = 0.6 cm2 Hinge B 20 cm 100 cm Al bar Area = 50 cm2 Hinge A 40 cm 100 cm Rigid support 40 cm 50 kN Figure 30.2 S OLUTION : For static equilibrium sum of vertical forces should be equal to zero. Let the forces in wire and steel wire be Ps and Pa respectively as shown in Fig. 30.3. Then ΣV = 0 Pa + 50 = Ps + R And sum of moments about the hinge A should be zero. @seismicisolation @seismicisolation (i) 692 • Strength of Materials Pa Ps Hinge A 40 cm 100 cm R 40 cm 50 kN Figure 30.3 Then, ΣM=0 Pa × 40 + Ps × 140 = 50 × 100 140 50 × 100 Pa + Ps × = 40 40 ∴ Pa + 3.5 Ps = 125 (ii) Pa and Ps are in kN. If we consider the deformations of the aluminium bar and the wire. The deformed condition of the system is shown in Fig. 30.4. Pa 40 cm 100 cm 40 cm δa δs R Ps Figure 30.4 From Fig. 30.4, we get δs δa = 140 40 140 δa ∴ δs = 40 ∴ or, δ s = 3.5 δ a Pa la Ps ls = 3.5 As Es Aa Ea @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains • 693 Substituting, 20 Pa × × 103 Ps × 1 × 103 100 = 3.5 0.6 50 × 200 × 109 × 60 × 109 100 × 100 100 × 100 ∴ Pa = 35.7 Ps (iii) From Eqns. (i), (ii) and (iii), the values of Ps , Pa and R can be found out. From Eqns. (ii) and (iii), we get 35.7 Ps + 3.5 Ps = 125 39.2 Ps = 125 ∴ Ps = 125 39 = 3.205 kN Hence, Pa = 35.7 × 3.205 = 114.42 kN Stress in bar = Pa 114.42 × 103 = 22.88 MN/m2 = 50 Aa 100 × 100 Stress in wire = = Ans Ps As 3.205 × 103 60 100 × 100 = 0.534 MN/m2 Ans From the Eqn. (i) Pa + 50 = Ps + R 114.42 + 50 = 3.205 + R ∴ R = 161.215 kN Ans E XAMPLE 30.4: A rigid steel plate is supported by three vertical posts of 2 m height. But the middle post is 0.5 mm less as shown in Fig. 30.5. The cross-sectional area of each post is 180 mm × @seismicisolation @seismicisolation 694 • Strength of Materials P 0.5 mm 2.0 m A B C Figure 30.5 180 mm. Determine the safe value of the load P if the permissible stress for concrete in compression is 16 N/mm2 . Take E for concrete as 12 kN/mm2 S OLUTION : Allowable stress in concrete, σ = 16 N/mm2 . Due to the action of load P, there will be less strain or less stress in the middle post B. Therefore, stress in the outer posts A and C = 16 N/mm2 . Now, 16 σ × 2000 − × 2000 = 0.5 Econ Econ 32000 − 2000 σ = 12000 × 0.5 = 6000 2000 σ = 26000 ∴ σ = 13 N/mm2 Safe load, P = 2 × 180 × 180 + 180 × 180 × 13 = 1036800 + 421200 = 1458000 N = 1458 kN Ans E XAMPLE 30.5: A composite bar of steel and copper is shown in Fig. 30.6. Determine the compressive force developed in the bars after the rise in temperature by 90◦ C. Also find the change of length in the copper bar. The area of cross section of copper bar is 650 mm2 and that of steel bar is 1100 mm2 Ecu = 105 GPa, Es = 200 GPa αcu = 18 × 10−6 /◦ C, σs = 11 × 10−6 /◦ C Clearance between support and composite bar is 0.6 mm. @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains 0.6 mm Copper Steel 0.5 m 0.5 m Figure 30.6 S OLUTION : Free expansion in bars due to temperature rise, = 18 × 10−6 × 90 × 500 + 11 × 10−6 × 90 × 500 = 0.81 + 0.495 = 1.305 mm Temperature rise = 90◦ C Clearance = 0.6 mm Contraction in length of bars = 1.305 − 0.6 = 0.705 Compressive force in the copper bar, 0.705 = 500 P 500 P × + × 650 105000 1100 200000 P = 73445 N = 73.445 kN (Compressive) Ans Change in length of the copper bar, = 18 × 10−6 × 90 × 500 − 73445 500 × 650 105000 = 0.81 − 0.538 = 0.272 mm Ans @seismicisolation @seismicisolation • 695 696 • Strength of Materials E XAMPLE 30.6: A rod having constant cross section is tightly clamped at both ends as shown in Fig. 30.7. If the temperature is raised by 65◦ C, determine the stress in the rod. Es = 200 GPa, Ec = 100 GPa αs = 10 × 10−6 /◦ C, αc = 15 × 10−6 /◦ C Copper Steel 1.6 m 0.9 m Figure 30.7 S OLUTION : For free extensions, δcu = αcn lcu × t = 15 × 10−6 × 900 × 65 = 0.8775 mm δs = αs × ls × t = 10 × 10−6 × 1600 × 65 = 10.4 mm Total extension = 0.8775 + 1.04 = 1.9175 mm If P is the compressive force in bar Acu = As = A P lcu ls δ= + A Ecu Es 1600 900 1.9175 = σ + 100000 200000 1.9175 = σ 9 × 10−3 + 8 × 10−3 1.9175 = 17 × 10−3 σ ∴ σ = 112.79 MN/mm2 @seismicisolation @seismicisolation Ans Miscellaneous Solved Problems: Stresses and Strains • 697 Complex Stress and Strain E XAMPLE 30.7: A point in a strained component has the following stresses as shown in Fig. 30.8. Determine: a) Principal stresses and principal planes b) Maximum shear stress with its orientation. 140 MPa 70 MPa 80 MPa P Q 80 MPa 70 MPa R S 140 MPa Figure 30.8 S OLUTION : σx = −80 MPa, σy = 140 MPa, τ = 70 MPa Principal stress = σ1,2 = σx + σy ± 2 −80 + 140 ± 2 √ = 30 ± 6700 = σx + σy 2 + τ2 2 (−80 + 140)2 + 702 2 = 30 ± 81.85 = −51.85 = 51.85 MPa (Compressive) Location of principal plane, tan 2θ = = 2τ σx − σy 140 2 × 70 =− −80 − 140 320 = −0.636 tan(180 − 2θ ) = −(−0.636) tan(180 − 2θ ) = 0.636 @seismicisolation @seismicisolation Ans 698 • Strength of Materials 180 − 2θ = 32.46◦ 2θ = 147.54 θ1 = 73.77◦ Ans θ2 = θ1 + 90 = 73.77 + 90◦ = 163.77 Ans σ1 − σ2 2 111.85 − (−51.85) = 2 = 81.85 MPa Ans Maximum shear stress, τmax = Location of planes having maximum shear stress, θ3 = 73.77 + 45 = 118.77◦ Ans θ4 = 163.77 + 45 = 208.77◦ Ans Shear Force and Bending Moment Diagram E XAMPLE 30.8: Draw shear force and bending diagrams for the beam shown in Fig. 30.9 and indicate the values at important points. 6 kN B A 2m 7 kN C 2m E D 2m 1m Figure 30.9 Taking moments about A, 6 × 4 + 7 × 7 = 14 + RD × 6 24 + 49 = 14 + 6RD ∴ RD = 9.8 kN RA = 6 + 7 − RD = 13 − 9.8 = 3.2 kN ∴ RA = 3.2 kN SF Calculations SF at A = 0 rising to 3.2 kN SF at B = 3.2 kN SF at C = 3.2 kN falling to 3.2 − 6 = −2.8 kN SF at D = −2.8 kN rising to − 2.8 + 2.8 = 7.0 kN SF at E = 7 kN finally to zero @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains • 699 BM Calculations at A = 0 at B = 3.2 × 2 = 6.4 at left 6.4 − 14 = 7.6 kNm at right BM at C = 3.2 × 4 − 14 = −1.2 kNm BM at D = 3.2 × 6 − 14 − 6 × 2 = −6.8 BM at E = 3.2 × 7 − 14 − 6 × 3 + 9.8 × 1 = 0.2 0 BM BM 6 kN B A 3.8 kN 7 kN C D E 9.8 kN 2m 2m 2m (a) 1m 7 kN 3.2 o o 2.8 kNm (b) 2.8 kNm SFD 6.4 kNm o o 1.2 kNm 7.6 kNm (c) 6.8 kNm BMD Figure 30.10 Bending Stresses E XAMPLE 30.9: A timber beam 160 mm wide and 200 mm deep is reinforced by bolting on two steel flitches each 160 mm × 12 mm in section. Find the moment of resistance when (a) flitches are attached symmetrically at the sides and (b) the flitches are attached symmetrically at the top and bottom. Permissible stress in timber is 8 N/mm2 . What is the maximum stress in steel in each case? Take Es = 20Et @seismicisolation @seismicisolation 700 • Strength of Materials S OLUTION : 160 mm 160 mm 200 mm 12 mm 160 mm Figure 30.11 a) Let stress in timber σt = 8 N/mm2 at 100 mm from N.A. Stress in timber at 80 mm from N.A, 80 × 8 = 6.4 MPa 100 Maximum stress in steel, σs = mσw = 20 × 6.4 σt1 = = 128 MPa ∴ Moment of inertia, It = = bd 3 13 160 × 200 3 = 106.7 × 106 mm4 12 Is = 2 td 3 2 × 12 × 1603 = = 8.2 × 106 mm4 12 12 Moment of resistance = Ms + Mt = σs σt × Is + × It y y = 128 8 × 8.20 × 106 + × 106.7 × 106 100 80 = 10.5 × 106 + 10.67 × 106 = 21.17 × 106 mm (b) If the stress in timber, σt = 8 MPa Then stress in steel σs = 20 × 8 = 160 MPa @seismicisolation @seismicisolation Ans Miscellaneous Solved Problems: Stresses and Strains 160 mm 12 mm 200 mm 12 mm 160 mm Figure 30.12 For maximum stress in steel, 160 σs = 112 100 σs = 179.2 MPa Moment of Inertia: MOI of timber, It = MOI of steel, Is , 160 × 2003 = 106.7 × 106 mm4 12 =2 12 160 × 123 + (160 × 12)(100 + )2 12 2 = 2 [276480 + 21573] = 43699200 mm4 = 43.7 × 106 mm4 Moment of resistance = Mt + Ms = σt σs × It + × Is yt ys = 8 179.2 × 106.7 × 106 + × 43.7 × 106 100 112 = 8.536 × 106 + 69.92 × 106 = 78.456 × 106 Nmm Ans @seismicisolation @seismicisolation • 701 702 • Strength of Materials Shear Stresses E XAMPLE 30.10: For Fig. 30.13, evaluate the maximum shear stress if it is subjected to a shear force of 120 kN. A B 110 mm 25 mm D 14.88 N/mm2 T K x 40.56 N/mm2 M P 25 mm L 90 mm C Figure 30.13 S OLUTION : For this section it is obvious that maximum shear stress will develop at the centre, i.e., at M and P 90 × 1103 π (50)4 − 12 64 = 9.98 × 106 − 0.307 × 106 I= = 9.673 × 106 mm4 Now, Aȳ = Moment of area above N.A. FAȳ Shear stress, τ = bI 110 − 50 Area A between KT and BA = 90 × 2 90 × 30 = 2700 mm2 110 30 − = 40 mm 2 2 120000 × 2700 × 40 At T, τ = 90 × 9.673 × 106 C.G. of this area from N.A; ȳ = = 14.88 N/mm2 Aȳ = Moment of area above N.A. = (Moment of area of rectangle 90 × 55 above N.A.) − (Moment of area of circular portion between T.P. about N.A.) @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains Aȳ = 90 × 55 × = 136125 − = 136125 − 55 − 2 25 25 0 703 25 2xdy.y 0 2× 0 • 625 − y2 × y × dy ∵ x= 252 − y2 − 625 − y2 (−2y).dy 25 = 136125 + 0 625 − y2 (−2y).dy 625 − y2 = 136125 + 3/2 3/2 25 2 (625 − 252 )3/2 − (625 − 02 )3/2 3 2 = 136125 + −6251.5 3 2 = 136125 − × 15625 3 = 136125 + = 136125 − 10416.6 = 125708.4 mm3 Now, τmax = F.A.ȳ bI = 120000 × 125708.4 [90 − (2 × 25)] × 9.673 × 106 = 120000 × 125708.4 40 × 9.673 × 106 = 38.63 N/mm2 Ans Torsion E XAMPLE 30.11: A steel bar of 2.4 cm diameter is subjected to a torque of 310 Nm produces an angle of 1.4◦ in a length of 22 cm. The same bar when subjected to tension elongates 0.012 cm in length of 16 cm under a load of 75 kN. Determine the Poisson’s ratio for the material. @seismicisolation @seismicisolation 704 • Strength of Materials S OLUTION : Bar diameter, d = 2.4 cm Length, l = 22 cm Torque, T = 310 Nm Angle of twist θ = 1.4◦ = 1.4 × π 180 = 0.0244 radians J= π d 4 π (2.4)4 = 32 32 = 3.255 cm4 = 3.255 × 10−8 m4 We know, T Cθ = J l T.l C= Jθ = 310 × 22 100 × 3.255 × 10−8 × 0.0244 = 8.587 × 109 N/m2 = 85.87 GPa Cross-sectional area A = (i) π × 2.42 4 = 4.5216 cm2 = 4.5216 × 10−4 m2 E= = P.l A dl 75000 × 16 × 100 4.5216 × 10−4 × 0.012 × 100 = 2.211 × 1011 = 221 GPa @seismicisolation @seismicisolation (ii) Miscellaneous Solved Problems: Stresses and Strains • 705 E 2(1 + μ ) Substituting for C and E from Eqns. (i) and (ii), We know, C = 85.87 = 221 ; 2(1 + μ ) 1+μ = 221 = 1.287 85.87 × 2 1 + μ = 1.287 ∴ μ = 0.287 Ans E XAMPLE 30.12: A shaft of diameter D is subjected to a combined bending moment M and torsion T . The allowable stress in shear is 72% of the allowable stress in tension. If the factor of safety for failure due to shear and failure due to maximum tensile stress are the same, determine the ratio of M/T. S OLUTION : Let σb be the stress in bending and τs the maximum shear stress due to torsion. M M(D/2) 32 M = = I π D3 (π D4 ) 64 τ T = R J τs 16T = (D/2) π D4 ) ( 32 16T τs = π D3 σb + σb2 + 4τs2 The maximum tensile stress is σmax = 2 32M 16T 32M 2 +4 + π D3 π D3 π D3 = 2 16 = (M + M 2 + T 2 ) π D3 σb = And or But τmax = 0.72σmax @seismicisolation @seismicisolation 706 • Strength of Materials ∴ 16T 16 = 0.72 × (M + 3 πD π D3 M2 + T 2) T = 0.72 M + 0.72 M 2 + T 2 T − 0.72M = 0.72 M 2 + T 2 or Squaring both sides T 2 + 0.52M 2 − 1.44T M = 0.52M 2 + 0.52T 2 T 2 − 0.52T 2 = 1.44T M 0.48T 2 = 1.44T M M = 0.333 Ans T or Springs E XAMPLE 30.13: Calculate the thickness and number of leaves of a laminated spring which is required to support a central load of 3 kN on a span of 1.0 m, if the maximum stress is limited to 235 MN/m2 and central deflection to 82 mm. Breadth of each leaf can be assumed to be 110 mm. Take E for spring material = 200 GN/m2 S OLUTION : W = 3 × 103 N span, l = 1.0 m Maximum stress σmax = 235 × 106 N/m2 E = 200 × 109 N/m2 , b = 110 mm = 110 × 10−3 m Wl 3 Because, σmax = × 2 nbt 2 235 × 106 = 3000 × 1 3 × 2 n × (110 × 10−3 )t 2 where t is thickness of a leaf. nt 2 = 3000 × 1 3 1 × × 2 110 × 10−3 235 × 106 nt 2 = 1.741 × 10−4 @seismicisolation @seismicisolation (i) Miscellaneous Solved Problems: Stresses and Strains 3 Now central deflection is yc = 8 82 × 10−3 = nt 3 = W l3 nEbt 3 • 707 3 3000 × 13 × 8 n(200 × 109 )(110 × 10−3 )t 3 3000 3 × 8 200 × 109 × 82 × 10−3 × 110 × 10−3 nt 3 = 6.236 × 10−7 (ii) Dividing (ii) by (i) t= ∴ 6.236 × 10−7 1.741 × 10−4 t = 3.582 × 10−3 m = 3.582 mm = 3.6 mm Ans Substituting value of t in Eqn. (i), n= 1.741 × 10−4 = 13.56 (3.582 × 10−3 )2 The nearest whole number is 14 Ans E XAMPLE 30.14: An open-coiled helical spring made of 12 mm diameter rod has seven free coils each of 110 mm mean diameter. The ends of the spring are fastened to two discs kept 0.8 m apart, which is the free length of the spring. Calculate the force on the discs, acting along the axis of the spring, when one disc is rotated through 12◦ to coil. Take the spring E = 200 GN/m2 and C = 80 GN/m2 . S OLUTION : 0.8 = 0.114 m 7 0.114 tan α = = 0.33 0.11π Pitch of coils = ∴ ∴ α = 18.26◦ l = 2π Rn sec α = 2π × 0.055 × 7 × 1 cos 18.26◦ = 2π × 0.055 × 7 × 0.9496 = 2.29 m @seismicisolation @seismicisolation 708 • Strength of Materials 200 E = = 2.5C C 80 And J = 2I ∴ We know EI = 1.25 CJ sin 2α 1 1 sin2 α cos2 α + + MRl − =0 EI CJ 2 CJ EI 2 sin2 ×18.26 1 1 sin 18.26 cos2 18.26 + +M× − =0 W × 0.055 1.25CJ CJ 2 CJ 1.25CJ 0.3132 1 0.5951 W × 0.055 + 0.94962 + M × 1− =0 1.25 2 1.25 δ = W R3 l or or 0.055 W {0.0784 + 0.9} + M × 0.298 × 0.2 = 0 4.312 × 10−3 W + 0.0495 W + 0.0595 M = 0 0.0538 W = −0.0596 M 0.0538 W 0.0596 or M=− or M = −0.903 W Now, φ = Ml = 12 × or cos2 α sin2 α + EI CJ sin2 α +W Rl 2 1 1 − CJ EI π 180 2 Wl cos 18.26 2 −0.903 + sin 18.26 CJ 1.25 1 sin 2 × 18.26 1− = 0.2093 +0.055 × 2 2.5 W × 2.29 × [−0.903 {0.721 + 0.0982} π 80 × 109 × × 0.0124 32 + {0.055 × 0.0298 × 0.6}] = 0.2093 0.122 W × [−0.7397 + 0.009834] = 0.2093 − 0.122 W × 0.073 = 0.2093 W = −23.5 N (Compressive) Ans @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains • 709 E XAMPLE 30.15: A composite spring has two close-coiled helical steel springs in series; each spring has a mean diameter of 8 times the diameter of its wire. One spring has 20 coils and wire diameter of 2.5 mm. Find the diameter of the wire of the other spring, if it has 15 coils and the stiffness of the composite spring is 1.5 kN/m. Find the greatest axial load that can be applied to the spring and the corresponding extension for a maximum shearing stress of 310 MN/m2 C = 80 GN/m2 . S OLUTION : Figure 30.14 shows the arrangement of the two springs, which are subjected to an axial load W . d1 = 0.0025 m, ∴ S1 = = Total extension, δ = ∴ Effective stiffness = W W W + S1 S2 1.5 = = ∴ D1 = 0.02 m Cd14 W = d1 8D31 n1 80 × 109 × 0.00254 = 2.44 kN/m 8 × 0.02 × 20 W W + S1 S2 S1 S2 S1 + S2 2.44 S2 2.44 + S2 3.66 + 1.5 S2 = 2.44 S2 S2 = ∴ 3.66 = 3.894 kN/m 0.94 2.57 × 103 = 80 × 109 × d24 8 × (8d2 )3 × 15 from which, d2 = 0.001975 m 8W D 64W Also, τ = = Since D = 4d π d3 π d2 The maximum shear stress therefore occurs in spring (2) 310 × 106 = δ= 64W ; π × 0.0019752 ∴ W = 59.33 N W 59.33 = 0.0395 m = S 1.5 × 103 @seismicisolation @seismicisolation Ans Ans 710 • Strength of Materials W 1 2 W Figure 30.14 Application of Castigliano’s Theorem E XAMPLE 30.16: The ring shown in Fig. 30.15 is made of a flat steel strip 20 mm × 3 mm and is shaped in the form of a circle of mean diameter 0.2 m. The ends at B are cut square and not joined. A pull P is applied along the diameter CD which is at right angles to the diameter AB. If the maximum tensile stress due to P is 125 MN/m2 , find the increase in the opening at B due to P. E = 200 GN/m2 . 3 mm P P m 0m 20 A C C dθ dθ B A θ Q θ B D P Figure 30.15 Figure 30.16 S OLUTION : The increase in opening at B is twice the movement of B relation to A in the half ring shown in Fig. 30.16. Since there is no force at B in the direction of the required deflection, a force Q, of zero magnitude, must be applied there. @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains • 711 The required deflection is given by δ =2 2 ∂ UAB = ∂Q EI M ∂M dx ∂Q For BC, taking the origin at B, M = Qr(1 − cos θ ) ∂M = r(1 − cos θ ) and dx = rd θ ∂Q For AC, taking the origin at A, M = Qr(1 + cos θ ) + Pr cos θ ∂M = r(1 + cos θ ) and dx = rd θ ∂Q π /2 2 ∴ δ= Qr(1 − cos θ ) × r(1 − cos θ ) × rd θ EI 0 + π /2 0 2Pr3 = EI = {Qr(1 + cos θ ) + Pr cos θ } × r(1 + cos θ ) × rd θ π /2 0 cos θ (1 + cos θ )d θ · · · since Q = 0 2Pr3 π (1 + ) EI 4 The maximum stress occurs at A, where M = Pr, ∴ Pr = σ Z = 125 × 106 × ∴ ∴ 0.02 × 0.0032 6 = 3.75 Nm 3.75 P= = 37.5 N 0.10 2 × 37.5 × 0.103 π 1+ δ= 3 4 0.02 × 0.003 200 × 109 × 12 = 0.0149 m or 14.9 mm Ans E XAMPLE 30.17: Obtain the deflection under a concentrated load of 65 kN applied to a simply supported beam has shown in Fig. 30.17. Use Castigliano’s theorem. EI = 2.4 MN/m2 . @seismicisolation @seismicisolation 712 • Strength of Materials 65 kN X B A 1m 3m x X Figure 30.17 For purpose of calculation let us replace load 65 kN at C by load W . RA × 4 = W × 1 W RA = 4 Let there be a section X − X at x m beyond load W . Then, Mx = RA × x −W × (x − 3) Wx −W (x − 3) = 4 ∂M x = − (x − 3) ∂W 4 M ∂M · · dx EI ∂ W 1 3 Wx x 1 4 Wx x = × dx + −W (x − 3) × − (x − 3) dx EI 0 4 4 EI 3 4 4 3 4 2 W W x − (x − 3) dx = x2 dx + 16EI 0 EI 3 4 3 W 4 x − 4x + 12 2 W 2 x dx + dx = 16EI 0 EI 3 4 Again δ = = W 16EI 3 0 x2 dx + 9W 16EI 4 3 x2 − 8x + 16 dx 3 3 9W x W + [12.33 − 28 + 16] = 16EI 3 0 16EI 0.75W 9 (1 + 0.33) = , Now substituting for W & EI 16EI EI 0.7 × 65 × 103 δ= = 0.0203 m = 20.3 mm Ans 2.4 × 106 = E XAMPLE 30.18: A steel bar of constant section moment of area I is bent as shown in Fig. 30.18 and fixed at one end. Find the horizontal and vertical deflections at the free end. @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains P C a l A B Figure 30.18 Horizontal δ : ∂M =x M = Px, ∂P 1 a 2 δbc = Px dx EI 0 = δab = = Pa3 3EI 1 EI l Pa2 dx 0 Q Pa2 l P EI Q is introduced as zero force. Figure 30.19 Total horizontal deflection = δbc + δab = Pa2 Pa2 l + 3EI EI = Pa2 a +l EI 3 Ans Vertical deflection δ , BC : it is zero AB : M = Pa + Qx, δab = 1 EI δab = 1 EI = l 0 dM =x dQ (Pa + Qx) x · dx l Pal 2 2EI Pa xdx 0 Ans @seismicisolation @seismicisolation • 713 714 • Strength of Materials Moment of Inertia E XAMPLE 30.19: Derive expression for moment of inertia of a cirle about X − X axis passing through centroid. A B dy C y o X θ θ X R Figure 30.20 S OLUTION : Consider an elementary strip of thickness dy as shown in Fig. 30.20. This strip is at a distance y from X − X axis. We have: BC = OC cos θ = R cos θ ∴ Length of strip = AC = 2BC = 2R cos θ Area of strip = 2R cos θ × dy But y = r sin θ Differentiating both sides, dy = R cos θ d θ ∴ Area of strip = 2R cos θ × R cos θ d θ = 2R2 cos2 θ · d θ Moment of inertia of strip about X-axis, = 2R2 cos2 θ d θ y2 = 2R2 cos2 θ × R2 sin2 θ = 2R4 sin2 θ cos2 θ d θ The moment of inertia of the whole circle about X-X axis is IXX = +π /2 −π /2 2R4 sin2 θ cos2 θ d θ @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains Now sin 2θ = 2 sin θ cos θ 4 IXX = R ∴ sin2 2θ = 2 sin2 θ cos2 θ 2 +π /2 sin2 2θ −π /2 2 dθ +π /2 1 − cos 4θ 1 − cos 4θ dθ = sin2 2θ ∵ =R 4 2 −π /2 R4 π sin 2π π sin(−2π ) = − − − − 4 2 4 2 4 4 = R4 π π + 4 2 2 Because sin 2π = sin(−2π ) = 0 = Now radius R = π R4 4 D 2 4 D π 2 = 4 ∴ IXX = π D4 64 Ans or Alternate Method Y dr Z X r X R Z Y Figure 30.21 Consider an elementary ring at radius r and of thickness dr. Area of elementary ring = 2π r · dr @seismicisolation @seismicisolation • 715 716 • Strength of Materials Moment of inertia of this elementary ring about central axis, i.e., Z-Z axis. = 2π r dr · r2 = 2π r3 dr Total moment of inertia about Z axis R = 0 = 2π 2π r3 dr r4 4 R 0 2π R4 π R4 = = 4 2 (i) Substituting R = D/2 in Eqn. (i) 4 D 2 = 2 π D4 = 32 π IZZ IZZ Also IXX + IYY = IZZ Since IXX = IYY being symmetrical area 2IXX = π D4 32 or IXX = π D4 64 Ans Strain Energy E XAMPLE 30.20: A bar of 120 cm in length is subjected to an axial pull, such that the maximum stress is equal to 160 MN/m2 . Its area of cross section is 2.5 cm2 over a length of 120 cm except for the middle 6 cm length, where the area of cross section is 1.2 cm2 . If E = 200 GN/m2 , calculate the strain energy stored in the bar. S OLUTION : 2 1.2 cm2 1 P 2.5 cm2 2 P 6 cm 120 cm Figure 30.22 2 and central portion of length as 1. Portions on left and right are marked as @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains • 717 Maximum stress will be at lowest are, i.e., 1.2 cm2 σ1 = 160 MN/m2 Because σ1 A1 = σ2 A2 160 × 1.2 × 10−4 = σ2 × 2.5 × 10−4 ∴ σ2 = 76.8 MN/m2 Strain energy stored in the bar, σ12 A1 l1 σ22 A2 l2 + 2E 2E 2 2 160 × 106 × 1.2 × 10−4 × 6 × 10−2 76.8 × 106 × 2.5 × 10−4 × 114 × 10−2 = + 2 × 200 × 109 2 × 200 × 109 U= 2 = 120 − 6 = 114 cm] [∵ Total portion of = 0.4608 + 4.2025 = 4.6633 Nm or J Ans E XAMPLE 30.21: A vertical composite tie bar rigidity fixed at the upper end consists of a steel rod of 15 mm diameter enclosed in a brass tube of 15 mm internal diameter and 25 mm external diameter, each being 1.8 m long. Both are fixed together at the ends. The tie bar is suddenly loaded by a weight of 7.5 kN falling through a distance of 5 mm. Determine the maximum stresses in the steel rod and the brass tube. Take Es = 200 GPa & Eb = 100 GPa. S OLUTION : Refer Fig. 30.22. π 2 15 = 176.625 mm2 4 π Ab = (252 − 152 ) = 314 mm2 4 As = Let x = Extension in mm σs = Es x l and σb = @seismicisolation @seismicisolation Eb x l 718 • Strength of Materials Steel rod Brass tube 1.8 m 7.5 kN 5 mm 15 mm 25 mm Figure 30.23 Strain energy of the bar, σ2 σs2 As l + b Ab l 2Es 2Eb 2 2 E2 E x = 2s As l + 2 b Ab l l · 2Es l · 2Eb Eb x2 Es x2 = As + Ab 2l 2l x2 = (Es As + Eb Ab ) 2l x2 = (200000 × 176.625 + 100000 × 314) 2 × 1800 x2 (35325000 + 31400000) = 3600 = 18534.7x2 Nmm = (i) Potential energy lost by weight, = W (h + x) = 7500(5 + x) Nmm Equating Eqns. (i) and (ii) 18534.7x2 = 7500(5 + x) 18534.7x2 − 7500x − 37500 = 0 x2 − 0.405x − 2.02 = 0 x= +(0.405) ± @seismicisolation @seismicisolation (−0.4052 − 4 × 1 × (−2.02)) 2 (ii) Miscellaneous Solved Problems: Stresses and Strains • 719 √ +0.405 ± 0.164 + 8.08 = 2 = +0.405 ± 2.87 2 = 1.6375 mm σs = Es x 200000 × 1.6375 = = 181.91 MPa l 1800 σb = Eb x 100000 × 1.6375 = = 90.97 MPa l 1800 Ans Ans Theories of Failure E XAMPLE 30.22: A bolt has an axial pull of 15 kN together with a transverse shear force of 8 kN. Find the diameter of the bolt required according to: i) ii) iii) iv) v) Maximum principal stress theory Maximum shear stress theory Maximum principal strain theory Maximum strain energy theory Maximum distortion energy theory. Take allowable tensile stress at elastic limit = 120 MPa and Poisson’s ratio = 0.3 S OLUTION : Pt = 15 kN, Ps = 8 kN, σt at elastic limit = 120 MPa, μ = 0.3 Let d be the diameter of bolt in mm. Then cross-sectional area of the bolt, A = = 0.785d 2 mm2 Axial tensile stress, σ1 = Pt 19.1 15 = 2 kN/mm2 = A 0.785d 2 d and transverse shear stress, τ= 8 Ps 10.19 = = kN/mm2 A 0.785d 2 d2 (i) Maximum principal stress theory: σ1 + σ2 1 2 2 + σt1 = (σ1 − σ2 ) + 4τ 2 2 σ1 1 2 2 (σ1 ) + 4τ + = 2 2 @seismicisolation @seismicisolation π d2 4 720 • Strength of Materials Because σ2 = 0 ⎡ 2 = 19.1 1 ⎣ + 2d 2 2 = 1 √ 9.55 + 2 364.81 + 415.34 2 d 2d = 9.55 13.97 23.52 + 2 = kN/mm2 d2 d d2 = 23520 N/mm2 d2 19.1 d2 +4 10.19 d2 2 According to the maximum principal stress theory, σt at elastic limit 2 23520 120 = 2 d2 d = 19.8 mm Ans τmax = or or (ii) According to the maximum shear stress theory, 1 2 2 τmax = (σ1 − σ2 ) + 4τ 2 1 2 2 (σ1 ) + 4τ = 2 Because σ2 = 0 ⎡ ⎤ 10.19 2 ⎦ 19.1 2 1⎣ +4 = 2 d2 d2 1 = 2 2d 19.12 + 4(10.19)2 = 1 √ 364.81 + 415.34 2d 2 = 13.97 kN/mm2 d2 = 13970 N/mm2 d2 @seismicisolation @seismicisolation ⎤ ⎦ Miscellaneous Solved Problems: Stresses and Strains According to the maximum shear stress theory, σt at elastic limit 2 13970 120 = 2 d2 ∴ d = 15.26 mm Ans τmax = (iii) According to the maximum principal strain theory, σ1 1 + σt1 = (σ )2 + 4τ 2 Maximum 2 2 = and minimum σ1 1 − σt2 = 2 2 23520 d2 ( σ1 ⎡ = 19.1 1 ⎣ − 2d 2 2 = 1 9.55 − 2 2 d 2d [already calculated in (i)] )2 + 4τ 2 19.1 d2 2 +4 10.19 d2 2 (19.1)2 + 4(10.19)2 1 √ 9.55 − 364.81 + 415.34 d2 2d 2 9.55 13.97 = 2 − 2 d 4420 4.42 = − 2 kN/mm2 = − 2 N/mm2 d d = We know that according to the maximum principal strain theory, σt1 μσ2 σt at elastic limit − = E E E or σt1 − μσ2 = σt at elastic limit ∵ or d = 14.38 mm 23520 (−4420) × 0.3 − = 120 d2 d2 23520 1326 + 2 = 120 d2 d 24846 = 120 d2 Ans @seismicisolation @seismicisolation ⎤ ⎦ • 721 722 • Strength of Materials (iv) According to the maximum strain energy theory, (σt1 )2 + (σt2 )2 − 2μσt1 σt2 = (σt at elastic limit)2 23520 2 −4420 2 23520 −4420 = 1202 + − 2 × 0.3 × × d2 d2 d2 d2 553190400 19536400 62375040 + + = 14400 d4 d4 d4 635101840 = 14400 d4 ∴ d = 14.49 mm Ans (v) According to the maximum distortion theory, (σt1 )2 + (σt2 )2 − 2σt1 σt2 = (σt at elastic limit)2 23520 2 −4420 2 23520 −4420 = 1202 + − 2 × × d2 d2 d2 d2 553190400 19536400 207916800 + + = 14400 d4 d4 d4 780643600 = 14400 d4 ∴ d = 15.26 mm Ans E XAMPLE 30.23: A compound tube is made of a tube 300 mm internal diameter of 30 mm thick shrunk on a tube of 250 mm internal diameter. The interface radial pressure at the junction is 12 N/mm2 . The compound tube is subjected to an internal pressure of 75 N/m2 . Determine the variation of hoop stress over the wall of the compound tube. S OLUTION : r2 = 125 r0 = 150 r1 =1 80 mm Figure 30.24 @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains • 723 Interface radial pressure p0 = 12 N/mm2 Stress due to shrinkage: Outer tube: At radius r0 = 150 σr = 12 = b −a 1502 (i) σr = 0 = b −a 1802 (ii) At radius r1 = 180 mm Subtracting Eqn. (ii) from Eqn. (i) 12 = b b − 1502 1802 12 = b b − 22500 32400 12 = b(32400 − 22500) 22500 × 32400 {∴ a is cancelled} 12 = 1.358 × 10−5 b ∴ b = 883652.4 For A, from Eqn. (ii) a= b 1802 a= 883652.4 32400 = 27.27 At radius r1 = 180 mm, σc = = b +a 1802 883652.4 + 27.27 32400 = 54.54 N/mm2 At radius r0 = 150 mm, σc = b +a 1502 @seismicisolation @seismicisolation (tensile) 724 • Strength of Materials = 883652.4 + 27.27 22500 = 66.54 N/mm2 (Tensile) Inner tube: At radius r0 = 150 mm σr = 12 = b −a 1502 (iii) σr = 0 = b −a 1252 (iv) At radius r2 = 125 mm Subtracting Eqn. (iv) from Eqn. (iii), 12 = = 12 = b b − 2 150 1252 b b − 22500 15625 b(15625 − 22500) 22500 × 15625 12 = − 6875b 351562500 ∴ b = −613636.4 From Eqn. (iv) −613636.4 −a 15625 a = −39.27 0= And at radius r0 = 150, σc = = b +a 1502 −613636.4 − 39.27 22500 = −66.54 N/mm2 (Compressive) At radius r2 = 125 mm, σc = −613636.4 − 39.27 15625 = −78.54 N/mm2 (Compressive) @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains • 725 Stresses due to internal fluid pressure: At radius r2 = 125 σr = 75 = b −a 1252 (v) σr = 0 = b −a 1802 (vi) At radius r1 = 180 Subtracting Eqn. (vi) from Eqn. (v), b b − 1252 1802 b b − 75 = 15625 32400 b(32400 − 15625) 75 = 15625 × 32400 75 ∴ b= 3.3136 × 10−5 75 = b = 2263399.3 Using Eqn. (vi) a= 2263399.3 = 69.86 32400 ∴ a = 69.86 At radius r1 = 180 mm b +a 1802 2263399.3 + 69.86 = 32400 σc = = 139.72 N/mm2 (Tensile) At radius r0 = 150 mm, σc = 2263399.3 + 69.86 1502 = 170.45 N/mm2 (Tensile) At radius r2 = 125 mm, σc = 2263399.3 + 69.86 1252 = 214.72 N/mm2 @seismicisolation @seismicisolation (Tensile) 726 • Strength of Materials + for tensile and − for compressive Inner tube Radii 125 mm 150 mm σc due to shrinkage (N/mm2 ) −78.54 −66.54 σc due to inner pressure (N/mm2 ) +214.72 +170.45 RESULTANT σc N/mm2 +136.18 +103.91 Outer tube 180 mm 150 mm +54.54 +66.54 +139.72 +170.45 +194.26 +236.99 Rotating Discs and Cylinder E XAMPLE 30.24: A disc of uniform thickness and 625 mm diameter rotates at 2000 r.p.m. Determine the maximum stress developed in the disc. If a hole of 100 mm diameter is made at the centre of the disc. Find the maximum values of radial and hoop stresses. Density of the material = 7700 kg/m3 and μ = 0.27 S OLUTION : 625 2π × 2000 R= = 372.5, w = = 209.33 rad/s, ρ = 7700 kg/m3 and μ = 0.27. 2 60 Maximum radial stress and hoop stress are at the centre and equal. 3+μ 2 2 ρw r 8 3 + 0.27 = × 7700 × (209.33)2 (0.3125)2 8 = 13.4682597 N/m2 σr = σc = = 13.47 MN/m2 100 = 50 mm = 0.05 m r0 = 0.3125 m 2 √ Maximum radial stress is at ri2 · ro2 radius, i.e. 0.052 × 0.31252 = 0.015625 m With the hole of radius ri = 3+μ 2 ρ w (r0 + ri )2 8 3 + 0.27 × 7700 × 209.332 (0.3125 + 0.05) = 8 σr = = 137914979 × 0.3625 = 49994179.9 N/m2 = 49.99 MN/m2 Ans Maximum hoop stress is at the inner radius, 3+μ 2 2 1−μ 2 σc = ρ w r0 + r 8 3+μ i @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains • 727 1 − 0.27 3 + 0.27 2 2 2 × 7700 × 209.33 0.3125 + × 0.05 = 8 3 + 0.27 = 137914979 0.0977 + 5.58 × 10−4 = 13551250 N/m2 = 13.55 MN/m2 Ans Bending of Curved Bars E XAMPLE 30.25: A curved bar of rectangular section of 25 mm width and 42 mm depth and mean radius of curvature of 65 mm is initially unstressed. If a bending moment of 480 Nm is applied to the bar which tends to straighten it, determine the stresses at the inner and outer surfaces. Also find the position of neutral axis. S OLUTION : 2R + d R3 − R2 h = ln d 2R − d 2 × 65 + 42 623 ln − 652 = 42 2 × 65 − 42 172 = 6538.7ln − 4225 88 2 = 6538.7 × 0.67 − 4225 = 155.93 mm3 Bending moment is negative as it tends to straighten the bar. y is negative at inside and positive at outside. Stress at outside force M y R2 1+ 2 · AR h R+y 652 21 −480 × 1000 1+ × = 65 × (25 × 42) 155.93 65 + 21 569 = −7.033 1 + 86 = −53.56 MPa (Compressive) σ0 = @seismicisolation @seismicisolation 728 • Strength of Materials Stress at inside force R2 M y 1− 2 · AR h R−y 652 21 = −7.033 1 − × 155.93 65 − 21 569 = −7.033 1 − 44 σi = = 83.92 MPa (Tensile) Position of neutral axis y= −Rh2 R2 + h2 = −65 × 155.93 652 + 155.93 = −10135.45 4069.07 = −2.491 mm Fixed Beams E XAMPLE 30.26: A fixed beam of 7 m span is loaded with concentrated loads of 200 kN at distance 2.5 m from each support. Draw the BM and SF diagrams. Find also the maximum deflection. Take E = 200 GPa and I = 9 × 108 mm4 S OLUTION : By equating the areas of free and fixed BM diagrams, 1 MA × 7 = (7 + 2) × 500 2 MA = 321.43 kNm MB = 321.43 kNm Position of contraflexure: BM (actual) at any section in AC at a distance x from A is given by M = Free moment − fixed moment = 200x − 321.43 @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains 200 kN (a) Fixed Beam 200 kN C A 2m 2.5 m MA B D 2.5 m 7m fixing moments MA = MB due to symmetry RA = RB = 200 kN MB BMD + C = 200 × 2.5 = 500 kNm 500 kNm 500 kNm + (b) Free BMD A (c) Fixed BMD B 321.43 kNm 321.43 kNm 200 kN 200 kN C A MA 2.5 m B D 2m 2.5 m 7m MB 178.57 kNm (d) Final BMD 321.43 kNm 500−321.43 = 178.57 200 kN (e) SFD 200 kN Figure 30.25 For point of contra-flexure equate M = 0 200x − 32143 = 0 x = 1.607 m from either end @seismicisolation @seismicisolation • 729 730 • Strength of Materials Slope and Deflection: Take BM at x between A and D, then M = EI d2y = 200x − 321.57 − 200(x − 2) dx2 Integrating, EI x2 dy 200 = 200 − 321.57x +C1 − (x − 2)2 dx 2 2 dy When x = 0, = 0 ∴ C1 = 0 dx Integrating again, EIy = 200 x2 200 x3 − 321.57 − (x − 2)3 +C2 6 2 6 When x = 0, y = 0 C2 = 0 Hence, EIy = 33.33x3 − 160.785x2 − 33.33(x − 2)3 To get maximum deflection which occurs at the centre. Putting x = 7 = 3.5 m 2 EIymax = 33.33(3.5)3 − 160.785(3.5)2 − 33.33(3.5 − 2)3 = 1429 − 1969.6 − 112.49 = −653.1 653.1 ymax = − m 2 × 108 × 9 × 108 × 10−12 Note: Since we are working in kN so to convert 200 × 109 N/m2 into kN/m2 , divide it by 103 to make it 2 × 108 ∴ ymax = −3.63 × 10−3 m = −3.63 mm Ans Columns and Struts E XAMPLE 30.27: A T -section 150 mm × 120 mm × 20 mm is used as a column of 4.5 m long with hinged at both ends. Determine the crippling load. Take E = 200 GPa. S OLUTION : E = 200 GPa @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains = 200 kN/mm2 = 200000 N/mm2 1 and 2 Let AB be the reference line and divide the area into two simple sections a1 = 100 × 20 = 2000 mm2 100 = 50 mm y1 = 20 a2 = 150 × 20 = 3000 mm2 20 y2 = 120 − 2 = 110 mm, Total Area, A = 2000 + 3000 = 5000 mm2 2000 × 50 + 3000 × 110 ȳ = 5000 = 86 mm 150 mm Y 20 mm 2 X X G 1 120 mm 20 mm A B Y Figure 30.26 IXX = 20 × (100)3 150 × (20)3 + 2000(36)2 + + 3000 × (24)2 12 12 = 4.26 × 106 + 1.83 × 106 = 6.09 × 106 mm4 100 × (20)3 20 × (150)3 + 12 12 = 66666.7 + 5625000 IYY = = 5691666.7 mm4 @seismicisolation @seismicisolation • 731 732 • Strength of Materials Now because 5691666.7 (IYY ) is less than 6090000 mm4 (IXX ). So we shall take IYY as the value of I in formula (Euler’s). Also as the column is hinged at its both end, so length of columns, Le = l = 4500 mm Now, Euler’s crippling load, π 2 EI Le2 π 2 × 200000 × 5691666.7 = 45002 = 554247 N = 554.25 kN Ans ρE = Combined Loading E XAMPLE 30.28: For the inclined cantilever shown in Fig. 30.26, find the dimensions of the rectangular section of the beam if the permissible maximum direct stress is 120 MN/m2 . Check for the maximum shear stress which in this case should not exceed the value of 75 MN/m2 . Neglect the additional bending moment due to deflection of end of cantilever. Take value of P = 130 kN. S OLUTION : Refer to Figs. 30.27 and 30.28. √ Load P is resolved along the beam axis and perpendicular to it. 3 The resolved component P produces bending and shear force, while the component 0.5 P 2 produces direct tension. A B 1m b √3 P 2 P P 2 30° P Se at Action B Figure 30.27 @seismicisolation @seismicisolation 2b Figure 30.28 Miscellaneous Solved Problems: Stresses and Strains • 733 Depth of beam = 2b (given). P M y+ I A √ 3 P×1 P = 2 ×b+ 3 b × 2b b(2b) 12 Maximum tensile stress in section = 120 × 106 = = 0.433 × 130000 × 1 130000 + 0.667b3 2b2 0.433 × 130000 + 130000 × 0.3335b 0.667b3 80040000b3 = 56290 + 43355b 80040000b3 − 43355b − 56290 = 0 By trial b = 0.101 m ∴ b = 101 mm depth = 2b = 202 mm Ans Check for shear stress: 0.101 × 0.2023 = 6.94 × 10−5 m4 12 0.101 130000 × 0.866 × 0.101 × 0.101 × 2 = 0.101 × 6.94 × 10−5 = 8273.98 N = 8.27 kN which is quite safe Ans τmax = FAȳ , bI I= E XAMPLE 30.29: A shaft is subjected to torque T and a load F as shown in Fig. 30.29. Determine the maximum shear stress on a cross section of the shaft. Diameter of shaft is 5 cm, torque T = 1.2 kNm and force F = 22 kN. S OLUTION : In Fig. 30.30, point A has maximum shear stress because both the shearing stresses are directed downward. Resultant shear stress at A = τ1 + τ2 Resultant shear stress at B = τ1 − τ2 T 4F ∴ Maximum stress = π + d 3 3A 16 4F ∵ τr = 3A @seismicisolation @seismicisolation τr = modulus of rupture 734 • Strength of Materials A T τ1 τ1 τ1 τ1 A 2 cm Torsional shear stress A 8 cm Figure 30.29 ∴ τmax = B τ2 τ 2 τ2 Direct shear stress Figure 30.30 16 × 1000 × 1.2 4 × 22000 + π π × (0.05)3 3 × (0.05)2 4 = 48917197.4 + 14946921.4 = 63864118.8 = 63.86 MN/m2 Ans E XAMPLE 30.30: A shaft of 12 cm diameter is subjected to a bending moment of 1200 Nm and a torque of 600 Nm. Find (a) the maximum normal stress on a section perpendicular to the axis (b) the maximum shear stress on a section perpendicular to the axis and (c) the principal stresses. S OLUTION : (a) The maximum normal stress on a section perpendicular to the axis. Let σb be the maximum normal stress Now, M y I 1200 × 64 0.12 since × = 2 π (0.12)4 σb = π (0.12)4 m4 64 0.12 = 7077140.835 N/mm2 , y = 2 = 7.08 MN/m2 Ans I= (b) The maximum shear stress on a section perpendicular to the axis. Let τ be the maximum shear stress We know, T ·r J 600 × 0.06 = π (0.12)4 32 τ= @seismicisolation @seismicisolation Miscellaneous Solved Problems: Stresses and Strains = 1769285.2 N/mm2 = 1.77 MN/mm2 Ans (c) σb 2 1 σ1,2 = σ1,2 − σb ± + τ2 2 2 7.08 2 1 = × 7.08 ± + 1.772 2 2 √ = 3.54 ± 12.53 + 3.133 = 3.54 ± 3.96 = 7.5 MPa, −0.42 MPa @seismicisolation @seismicisolation Ans • 735 Index A Arc welding 626 Area moment method 361 B Beams of composite section 257 of uniform strength 136 with large radius of curvature 456 with unsymmetrical bending moment 493 deflection of 496 Bending moment 80 Bending stream 497 Bending under impact loads 315 Betti’s theorem of reciprocal deflections 312 Bow’s notation for graphical solution 560 Brinell hardener values 672 Brinell hardness testing machine 671 Brinell test 671 Buckling factor 425 Buckling load 425 Bulk modulus 43, 44 Butt joints 613 C Cantilever with a, concentrated load at its free end 235 concentrated load not at the free end 237 loaded from the free end 239 partially loaded with uniformly distributed load 238 uniformly distributed load 237 gradually varying load 240 of concentrated end load 261 end couple 262 uniformly distributed load 262 Carriage springs 408 Centre of gravity 97 for solid bodies 99 Centroid 97 of different sections 101 Circumferential stress 196 Clapeyron’s equation of three moments 377 Close-coiled helical spring with, axial couple 391 axial load 388 Columns and struts 424 Combined bending 142 Combined bending and twisting 343 Combined stresses 342 Composite action of axial load and couple 404 Composite beams 138 Composite shaft 185 Compound cylinders 219 Compression test 653 Compressive stress 4 Concentric (cluster) springs 397 Conjugate beam method 266 Continuous beams 377 Crippling load 425 Critical load 425 Cupping test 664 Curvature and strain, relationship 125 Curved bars, bending 456 Cylinders 520 @seismicisolation @seismicisolation 738 • Index D Dams 591 Deficient frame 559 Deflection 234 due to bending 307 due to impact 248 Diamond riveting 621 Direct stresses 142 Disc of uniform strength 554 Distortion theory 327 Double integration method 235, 251 Hardness testing machines 671 Hollow cylinder 547 Hollow disc 527 Hoop stress 196 E Eccentrically loaded welded joint 643 Elastic constants 40 Equivalent hardness number conversion table 677 Equivalent length 425 Erichsen test 664 Euler’s formula 446 Euler’s theory 437 L Laminated spring 408 Lap joints 612 Large curvature 459 Leaf spring 408 Limit of proportionality 185 Lindley extensometer 652 Longitudinal stress 197 Longitudinal tensile stress 215 I Impact test 661 J Johnson’s parabolic formula 445 K Knoop hardness number 680 F Fatigue limit 653 Fillet welds, under bending moment 636 under torsion 635 Fixed beam, deflection for 362 uniformly distributed load over the span 359 with a point load at the centre 357 Flat spiral spring 405 Flexural rigidity 235 Flitched beams 138, 257 Fractured ends of specimens 652 Frameworks 559 G Gas welding 626 Gordon’s formula 445 Graphical (Mohr’s) method 69 Guest or Teresa’s theory 321, 327 H Haigh’s theory 323 Hardness scaler for steels 679 M Macaulay’s method 251, 669 Material subjected to direct and shear stresses 55 shear stresses 54 two perpendicular stresses 50 Materials, mechanical testing of 649 Maximum principal strain theory 322 Maximum principal stress theory 320 Maximum shear strain energy theory 324 Maximum shear stress theory 321, 327 Maximum strain energy theory 323, 327 Maxwell’s reciprocal theorem 311, 670 Metals and alloys, mechanical properties of 653 Metals, mechanical properties of 653, 659 Method of sections 580 Middle third rule for rectangular section 147 Modulus of elasticity 43, 44, 45 Modulus of resilience 291 Modulus of rigidity 43, 665 Modulus of rupture 659 Modulus of section 128 @seismicisolation @seismicisolation Index • Mohr’s circle 494 Mohr’s circle of stresses 59 Mohr’s hardness scale 680 Mohr’s theorem 263 Moment area method 259 Moment of inertia 111 Moment of inertia of a, circular section 114 hollow rectangular section 114 triangle 115 semi-circular lamina 116 rectangular section 113 Moment of resistance 126 Momental ellipse 495 Monotron hardness number 680 N Non-ferrous alloys 679 O Offset stress 6 Open-coiled helical spring 402 Over riveted joints 627 P Perfect frame 559 Permanent set 6 Perry’s formula 449 Principal stress 47, 231 Principal strains 47 Product of inertia of rectangle 122 Proof resilience 291 Proof stress 6 Props 285 Q Quarter-elliptic leaf spring 410 R Radiant energy welding processes 627 Radius of curvature 234 Radius of gyration 112 Rankine’s formula 438 Rankine’s theory 320 Rectangular dams 591 Rectangular section 111 Resilience 291 Resistance welding 626 Rivet 610 efficiency of a 617 failure of a 615 material of 612 Rivet heads, types of 610 Riveted joints 610, 612 Rockwell hardness testing machine 673 Rotating disc of constant thickness 524 Rotating discs 520 Rotating long cylinder 542 Rotating ring 520 Rule of middle third 599 S Secant formula 446 Semi-elliptic spring 408 Shear centre 507 for channel section 508 for unequal I section 510 Shear force diagram calculations 382 Shear stress 4, 231 in beams 157 principle of 43 Shear stress distribution, for beam of rectangular section 159 in an I section 162 Shear tests 657 Shearing force 79 and bending moment diagrams 82 Shore scleroscope 676 Shrinkage allowance 222 Simply supported beam with, a central point load 245 a concentrated load at the centre 266 a uniform distributed load 246 point load eccentric 248 with central concentrated load 262 Slenderness ratio 425 Slope 234 Small curvature 456 Small initial radius of curvature 459 @seismicisolation @seismicisolation 739 740 • Index Solid cylinder 545 Solid disc 526 Solid state welding 627 Springs 388 in series and parallel 396 St. Venant’s theory 322 Straight line formula 445 Strain 29 Strain energy method 278 Strain energy 291 due to bending 307 in pure shearing 305, 307 in torsion 306 Stress 4 at critical load for cast iron 445 at critical load for structural steel 445 on a oblique section 47 Stress-strain curve 5 Suddenly applied load 292 Superposition method 277 Thick spherical shells 229 Thin circular tube subjected to torsion 190 Torsion 175 of a round bar 667 of a tapering shaft 187 Torsional moment of resistance 176 Trapezoidal dams, with water face inclined 603 with water face vertical 596 Turner’s sclerometer 680 U Universal testing machine 651 Unsymmetrical bending 491 Unsymmetrical welded section 632 V Vicker’s diamond test machine 674 Von mises and Henkey’s theory 324, 327 T Temperature stress 29 Tensile stress 4 Theorem of parallel axis 113 Theorem of the perpendicular axis 113 Theories of elastic failure 320 Thermal stresses in a bar of tapering section 32 Thermo-chemical welding processes 627 Thick cylinders and spheres 214 W Wahl’s correction factor 390 Welded joints 627 Welded joints 626 Welding, classification of 626 Wire-bound thin cylindrical shells 205 Y Young’s modulus 5 @seismicisolation @seismicisolation Strength ofofMaterials Strength Materials 978-93-89307-19-1 978-93-89307-31-3 Kaushik Strength ofof Materials Strength Materials Kaushik This book comprehensively covers all the major topics involving application of concepts of strength materials which covers a mechanical engineer encounters day. of Structural This bookofcomprehensively all the major topics involving every application conceptsand of machine elements which come under the purview of this subject and covered in the book and are strength of materials which a mechanical engineer encounters day to day. Structural beams ofelements all kinds, which thin and thickunder cylinders and spherical columns and struts, machine come the purview of this cells, subject and covered in thebars, bookdiscs are: and cylinders, springs, dams, trusses so on. Solid parameters covered beams of all kinds; thinframes, and thick cylinders andand spherical cells;mechanics columns and struts; bars; discs in the book aresprings; all typesframes; of stresses and strains, torsion, moments,parameters moments ofcovered inertia, and cylinders; dams; trusses and so on.bending Solid mechanics centre of gravity, constants, and and so on. Various theories of elastic failuremoments have been in the book are allelastic the types of stresses strains, torsion, bending moments, of coveredcentre in a focused chapter. dedicated chapter on mechanical testing materials will aid inertia, of gravity, elasticA constants, and so on. Various theories of of elastic failure have students in understanding strength of materials is determined in laboratory been covered in a focusedhow chapter. A dedicated chapter on mechanical testing conditions. of materials will aid students in understanding how strength of materials is determined in laboratory conditions. Learning Aids: Aids: ŸLearning At the end of the chapters, a large number (more than 300) of questions have been given. • At the end of the chapters a large number (more than 300) of questions have been Ÿ The text is supplemented with ample line diagrams/pictures so that students can given. understand the concepts easily. • The text is supplemented with line diagrams/pictures profusely so that the students can Ÿ The topics are discussed in a easily. sequence of gradually increasing level of difficulty, making it understand the concepts easier for students to follow the after building appropriate background strongly. so that • The topics are discussed intopics a sequence of gradually increasing level of difficulty, Ÿ Relatively difficult like principal stresses, theory failure, bending of curved bars, it is easier fortopics the students to follow the topics afterofbuilding appropriate background thick strongly. cylinders, strain energy, continuous beams, fixed beams, etc., have been treated with • Relatively difficult topics like is principal stresses, theory of failure, bending of curved special care to see that the material accessible to all students. bars, thick cylinders, strain energy, continuous beams, fixed beams, etc., have been withand special care tothis see book that the is accessible to all With its treated coverage approach, willmaterial be immensely useful to students. students of B. Tech (Mechanical Engineering) as well as that of A.M.I.E. (India). It will also be useful for competitive With its coverage and approach, this book will be immensely useful to the students of B. Tech examinations like GATE, IES and Civil Services. (Mechanical Engineering) as well as that of A.M.I.E. (India). It will also be useful for competitive examinations like[ GATE, and Civil Services. R. K. Kaushik D.M.E.,IES Gr.I.E (I), M.Tech (Mech.), M.I.E, Chartered Engineer, Ph.D.] is Professor, Department of Mechanical Engineering, Ganga Institute of Technology & R. K. Kaushik [ D.M.E., Gr.I.EHe (I),was M.Tech (Mech.), Chartered Engineer,atPh.D.] is Management, Kablana - Jhajjar. formerly head ofM.I.E, department (Mechanical) Haryana Professor, Department of Mechanical Engineering, Ganga Institute of Technology & Institute of Technology. Dr Kaushik has overall 47 years of experience, out of which 28 years Management, Kablana Jhajjar. He was formerly head of department (Mechanical) at Haryana were in different departments of ABB Ltd., earlier known as Taylor Instrument Ltd (both MNCs), Institute Technology. Dr Kaushik 47 in years of experience, which years where heofmostly discharged the rolehas of aoverall manager its R&D cell. He hasout 19of years of 28 teaching were in different departments of ABB Ltd., earlier known as Taylor Instrument Ltd (both MNCs), experience, teaching subjects like Strength of Materials, Mechanical Design and Engineering where he mostly the roleengineering) of a managerstudents. in its R&D hasworked 19 years teaching Mechanics to B. discharged Tech (mechanical Hecell. hasHe also asoflecturer in experience, teaching subjects like Strength of Materials, Mechanical Design and Engineering Kenya and Botswana for 12 years, teaching at the degree level. He has also been to Mechanics to B. Tech (mechanical engineering) students. He has also worked as lecturer in Switzerland in 2007 briefly on a promotion assignment for an India-based company. Kenya and Botswana for 12 years, teaching at the degree level. He has also been to Switzerland in 2007 briefly on a promotion assignment for an India-based company. Strength of Materials Strength of R.K. Kaushik Materials R.K. Kaushik Distributed by: 38 89 9 3 99 778899 3 30 07 7311931 @seismicisolation @seismicisolation TM TM