Previous 30 Years Objective & Conventional solved questions of MADE ERSY Publications IAS St IFS Civil & Mechanical Engineering ST' NGTH of A RIALS Useful for Civil Services Examination Engineering Services Examination State Engineering Services Examinations Public Sector Examinations Dr. U.C. Jindal IAS & IFS IFS IAS & (Objective &&Conventional) (Objective & (Objective Conventional) Conventional) Previous Questions Previous Solved Previous Solved Solved Questions Questions Strength of of Materials Strength Strength ofMaterials Materials Previous Solved Questions of of Previous 30 Previous 30 30Years Years Years Solved Solved Questions Questions of Civil & Civil &Mechanical Mechanical MechanicalEngineering Engineering Engineering Useful for for ESE, CSE, State Engg. Services, PSUs Useful Useful for ESE, ESE, CSE, CSE, State State Engg. Engg. Services, Services, PSUs PSUs and Other Examinations and Other and OtherCompetitive Competitive CompetitiveExaminations Examinations Dr. Dr. U.C. U.C. Jindal Jindal Jindal M.Tech, M.Tech, Ph.D. Ph.D. Former Professor Former Professor & & Head Head of of the the Department Department Department ofofMechanical Engineering Department of Department Mechanical Mechanical Engineering Engineering Delhi College College College of of of Engineering, Engineering, Engineering,Delhi Delhi Delhi Delhi I I MADE EFISH Public:mime Public:mime Publications MADE EASY MADE EASY — Publications Publications Preface I am thankful to Mr. B. Singh, CMD of MADE EASY Group, who is ever ready to help the Student Community by providing them newest type of books, as in the present book with typical/thought provoking/mind racking questions asked in IFS and IAS. Prelims and Mains of UPSC, for the last 30 years for both Civil and Mechanical engineering, in the subject of Strength of Materials. For the solution of each question a student must be equipped with strong concepts in the subject, and the students are the beneficiaries of the latest and comprehensive knowledge of the subject of the qualified and dedicated faculty of MADE EASY. Further improvements in the text of the book will be made after getting the feedback from the students. Any error in printing or calculations pointed out by the reader will be acknowledged with thanks by the author. Dr. U.C. Jindal Author Contents Contents IAS Previous Solved IAS & & IFS IFS(Objective (Objective & & Conventional) Conventional) Previous Solved Questions Questions 1 Strength Strength of of Materials Materials 1 Sl. SI. Chapter Chapter Pages Pages 1. 1. Simple Stresses Stresses in Bars............................................................. 1-19 Simple inUniform Uniform and and Compound Compound Bars 1-19 2. 2. Principal Stresses........................................................................................................................20-38 Principal Stresses 20-38 3. 3. Thin Thin and and Thick Thick Shells Shells.................................................................................................................39-51 39-51 4. 4. Shear Force Diagrams..................................................................52-69 Shear Forceand and Bending Bending Moment Moment Diagrams 52-69 5. 5. Theory Simple Bending. Bending......................................................................................................70-88 Theory of of Simple 70-88 6. 6. Deflection of Beams Beams............................................................................................................... 88-107 Deflection of 88-107 7. 7. Torsion . ....................................................................................................................................108-123 Torsion 108-123 8. 8. Springs 124-130 Springs . ....................................................................................................................................124-130 9. 9. 131-137 Struts and Struts and Columns..............................................................................................................131-137 Columns 10. 10. Theories of of Failure Failure.................................................................................................................138-148 Theories 138-148 11. 11. 149-157 Strain Energy Strain Energy Methods........................................................................................................149-157 Methods 12. 12. Miscellaneous Questions Questions....................................................................................................158-165 Miscellaneous 158-165 13. 13. Rotational Stresses Stresses................................................................................................................166-169 Rotational 166-169 14. 14. Unsymmetrical Bending.....................................................................................................170-173 Bending 170-173 Unsymmetrical 15. 15. 174-179 General Objective Objective Type Type Questions Questions..................................................................................174-179 General 01 01 CHAPTER Simple Stresses in Uniform and Compound Bars Bars Q.1.1 AAsteel steelrod rodofoflength length300 300mm mmand anddiameter diameter30 30mm mm is is subjected subjected to to a pull pull P, and the temperature Q.1.1 rise is extension of the is 100°C. IfIf the total extension the rod rod is is 0.40 0.40 mm, mm, calculate the magnitude of P. Take a for steel= 1Cr61°C and E E = 0.215 x 106 106 N/mm2. N/mm2 . steel = 12 xx 10-6/°C [CSE-Mains, 2011, CE : 12 Marks] [CSE-Mains, Solution: P = Pull in N P= % 2 2 A= of cross-section cross-section == 4 x 30 2 = 706.86 mm mm2 = Area of Extension due to pull, pull, (assuming pull axial) &11 = P x 300 ~XL Px300 _ x -6 xL _ 1 974 x 10-6P mm 6 -= 1.974 · AE - 706.86 706.86x0.215x10 Pmm x0.215 x10- 612 extension due to 8/2 to temperature temperature change change a.UT = aLAT = 12 x 10-6 x 300 x 100 100 = =0.36 0.36 + 1.974 x 10-6 P = 0.4 1.974 xx 10-6 P = 0.04 1 6 4 x106 P = 0.04 0.0 x 0 = 20263 N = 20.263 kN P 1.974 metallicbar bar250 250mm mmxx100 100mm mmx x5050 mmisisloaded loadedasasshown shownininthe thefigure figure 1.1. 1.1.Work Work out out the the Q.1.2 AAmetallic mm change in volume. What should in the 4 MN load in in order that change in should be the change that should be made in should be no change in in the the volume of the bar. there should 4MN 4 MN z ,..x i 50 _ _ _ _ _ _ __ 50 400 kN mm T ,___ __,,,,,_____., 250 mm 2MN 2 MN Fig. 1.1 1.1 Fig. = 5 2 ratio = 0.25. 0.25. E = 2 x 106 10 N/mm2, N/mm , Poisson's Poisson's ratio Assume E= Marks] [IFS 2011, CE : 15 Marks] 22 ~= /AS & Previous Solved Solved Questions IAS & IFS IFS (Objective (Objective & & Conventional) Conventional) Previous MADE EASY EASY Solution: Stresses 0 x = + + 400,000 = +80 +80 MPa MPa 400,000 = 5000 2x10 2 x1066 a = +160 MPa CJy _ = + + 50x250 50x250 = +160 MPa 6 CJz = Volumetric strain, Et, - 4x10 4x1066 160MPa = 160 250x100 MPa 250 x100 = a -Fa +a 1a, +ay +az ) Y z 2v 80 = 40 40 = 80 80 -2 = - 2 x0.25x x 0.25 80 x- = E E E E E 3 Volume 250 xx 100 100 x 50 mm3 mm Volume == 250 40 xV = — x 250 x100 x 50 in volume= volume . s 6V = change in evv xV = Ex250x100x50 81/ 6 50 x1066 50 x106 50x10 = sax 10 = = 250 mm3 mm 3 E 2x10 2x1055 For ev = ;;;;; 0 0 Ey + ay + az 2v(a, + ay + az) = 0 (Jx (Jy + (Jz - 2v(ax (Jy az> = 0 + ay)(1 ay) (1-2v} - 2v) == 2v(az) (ax+ 2v(az>-az (80 + 160)(1-2v} 160) (1 - 2v) == (2v - 1) CJz az (2v-1} 240 CJz 240Xx0.5 0.5== (2v-1) (2v -1) = -0.5 az 120 = 120 ;;;;; -<Jz Xx 0.5 a az==-240 -240MPa MPa Py'' =; ; ; 240 x 250 x 100 100;;;;; where in load, P = 6 MN 4 MN load should be increased to 6 MN load in same same direction direction N/mm 2 . So that az becomes-240 becomes -240 N/mm2. (b) cranechain chainhaving having an an area 7.25 cm2 cm2 carries a load of 15 15 kN. kN. ItIt is is being being lowered lowered at at aa uniform uniform Q.1.3 AAcrane speed of 50 50 m/minute, the the chain gets gets jammed jammed suddenly, suddenly, at at that that time time the length of chain unwound speed of is 12 m. Estimate the stress stress induced induced in in the the chain chain due due to to sudden sudden stoppage. stoppage. Neglect Neglect weight of the chain. Assume E. E = 2.1 Hf N/mm2. N/mm2 . 2.1 x 105 Marks] [IFS 2012, CE : 10 Marks] Solution: (·: self weight of chain W = 15 kN = 15000 N (-; chain negligible) negligible) 2 725 mm2 mm A, chain area area of ofcross-section cross-section == 725 Length, L= ;;;;; 12000 mm Volumeof ofchain chain == 12000 Volume 12000 x 725 725 = 87 x 105 105 mm3 mm3 Say, stress Say, stressdeveloped developed== ai, instantaneous in N/mm N/mm22 Speed, 50 V= = 0.833 m/s 60 = 60 mV mV 22 15000 15000 0.8332= 0.833 2 Kinetic energy ;;;;; 530.5 Nm;;;;; energy absorbed absorbed by bychain chain ;;;;; = - - = 9.81 xx Nm = 530500 Nmm 9.81 2 2 Stresses in and Compound Compound Bars Bars Simple Stresses in Uniform Uniform and Strength of Materials Change in length, Change in length, 3 4 3 <1111 0 12 a.x 12000 6L = a a; xi= 1; x ooo SL= E E ltVlil Change in in potential potentialenergy energy == WSl x a; x12000 N = 15000 15000XO';X12000 -857143 N Nmm aiNmm mm = - 857.143 . a; mm 2.1x10s 2.1x 105 2 Gi Total strainenergy energyabsorbed absorbedby bychain chain == 530500 Total strain 530500 + 857.143 a.= cri = — ;~ volume 2E = 530500 + 857.14a; 857.14 a, Or or ai2 O'i x87x 10 5 == 20.7143 20.7143$52 crf x €3 7 x105 2x2.1x10 2 x2.1x10'5 cr~-41.38cr;-25610.3 O aF - 41.38a1 - 25610.3 == 0 - 41.38+.J41.38 + 44x25610.3 41.38 +V41.3822 + x 25610.3 2 41.38+J1712.30+102441.2 41.38 + ../1712.30+ 102441.2 2 41.38 322. 728 41.38++ 322.728 2 182.054 N/mm2 N/mm2 = 182.054 Provethat thatPoisson's Poisson's ratio ratio cannot cannot be greater than Q.1.4 Prove than 0.5. [CSE-Mains, 1988, ME ME:: 15 Marks] [CSE-Mains, Solution: Figure shows a sphere sphere under uniform uniform hydrostatic hydrostatic pressure p. E = Young's Young's modulus If E= modulus and vis v is Poisson's ratio, ratio, v, then Volumetric strain, Ev 3p (1-2v)=0 or v, Poisson's Poisson's ratio ratio == 0.5 0.5, then in place of decrease in in volume volume there will If vis v is greater than 0.5, will be increase in volume, volume, which is not possible. possible. p Fig. 1.2 Q.1.5 A A 1000 mm long bar subjectedtotoan anaxial axialpull pullPwhich Pwhichinduces inducesaamaximum maximum stress stress of 1500 kg/crn2. Q.1.5 1000 mm long bar isissubjected kg/cm2. The area of cross-section of the the bar is 2 cm cm22 over a length of 950 mm and for the central 50 mm length, the sectional cm2 . sectional area area is equal to 1 cm2. 2 2cm 2 cm2 P P P~-------------------~--------------------~P 2 1 an cm2 1- 47.5 cm -P-1l---47.5cm---l k 47.5 cm -id l---47.5cm---l 5.0cm 5.0 cm Fig. 1.3 Fig. Efor kgf/cm2 , calculate strain energy stored in Assuming that E for bar material material is is 20 20 xx 1a5 105 kgf/cm2, in bar [CSE-Mains, 1990, ME : 20 Marks] [CSE-Mains, 44 ~= /AS & Previous Solved Solved Questions IAS & IFS IFS (Objective (Objective & & Conventional) Conventional) Previous MADE EASY EASY Solution: 2 , Maximum stress will occur in central portion of area 1 cm cm2, 2 = 1500 1500 kgf/cm2 kgf/cm 1500 xx 1 1= -- 750 750 kgf/cm kgf/cm 22 Stress inother otherportion portion=- 1500 Stress in 2 2 2 1 5002 7502 1500 x 47.5 x 2 x 5 x1+ strainenergy energy =- ~x5x1+2Ex47.5x2 U, strain U, 2E 2E x 103 5625x 103 26718.75 26718.75x 103 5625 x103 = E + E E + E 32343750 32343750 32343750 32343750 = 20x10 5 20x105 E = 16.17 16.17kgf-cm kgf-cm (Strain (Strain energy stored) Q.1.6 AAsteel steelrod rodofofsquare squarecross-section cross-sectionisisloaded loadedas asshown shown in in the figure figure 1.4. B A T T E ~ 150 150 kN kN C 75kN 75 kN 4--------- 10 mm 100 kN -4- II) U) 10 mm k- D E E 25kN 25 kN N N N !k l----2m----1---2m----1---2m---l [4-2 m 2m >I< 2m>1 Fig. 1.4 Fig. Find the section which is subjected subjected to maximum stress, its its magnitude magnitude and nature. What will will be the change in its its length? length? Take E E == 200 GPa. change in Solution: A B B ....-----.---, 150 kN 150kN ► E E ~ 150 kN 4— c B 50kN 50 kN —110. cC C E E 0 ~ N 50kN 50 kN 4— 25 kN 4- D 25 kN I10 mm Hollow Fig.1.5 Considering compressive stress as (-ve) (-ve) and vice-versa dAB = 150 000 . 150,000 ; = 240 MPa suare = -240 MPa (q (squaresection section)) 6 5 625 aBc = 50 ,000 = -125 MPa 5 0,000 125 MPa 400 25 25000 Oen = + 000 = +83.33 MPa Gap — 400-100 400 -100 MPa (in amax (inportion portion AB) °max = = -240 MPa 240 125 x 2000 + 83.33 x 2000 240 xx2000 Change 2000 _ 125 x 2000 + 83.33 x 2000 Changeininlength length= = 200,000 200,000 200,000 200, 000 200, 000 200, 000 = -2.4 -1.25 ++0.833 -2.817 mm -2.4-1.25 0.833==-2.817 mm (contraction) (contraction) ► Simple Stresses in Uniform and Compound Bars Strength of Materials 4 Q.1.7 A rigid bar AD is pinned at A and attached to the bars BC and ED as shown in figure 1.5. The entire system is initially stress free and weight of all bars are negligible. The temperature of the bar BC is lowered by 25°C and that of bar ED is raised by 25°C. Neglecting any possibility of lateral buckling, find the normal stress in bars BC and ED. For BC, of brass, E = 90 GPa, a = 20 x 10-6/°C For ED, of steel, E = 200 GPa and a = 12 x 10-6/°C Cross-sectional area of BC is 500 mm2 and that of ED is 250 mm2. [IFS 2011, CE : 10 Marks] I E E 0 cv mm Solution: Free expansion in steel bar (if there were no resultant) 8/s = 250x 12x 10-6 x 25 = 0.075 mm Free contraction in brass bar, SlB = 300 x 20 x 10-6 x 25 = 0.15 mm If brass bar is free to move down, then end D would move by 0.15 x 5 350 mm-0-I Fig. 1.5 600 = 0.36 mm 250 (... ABD is rigid, vertical deflection in ABD will be linearly proportional to distance from A) But free expansion in steel bar is only 0.075 mm. Therefore steel bar will be pulled down and brass bar will be pulled up, or stretched to prevent, free contraction Say as = Stress in steel bar 6B = Stress in brass bar Ps = Force in steel bar = as x 250 N PB = Force in brass bar = 08 x 500 N Taking moments about B 600 x Ps = 250 PB 600 x as x 250 = 250 x 500 x as 150000 as = 1250000'B 6B = 0.833 GB or aB = 1.2 as Brass bar aB - x 300 = 81g' Final contraction = 0.15 - — Eg Steel bar as Final extension = 0.075+ Es x 250 = Sls' But Putting the value , = 600 RI X (N g' 250 R7 2.4 viB ' x 300 0.075 + Es x 250 = 2.4[0.15 EB ] s 1.2a x 720 x 250 = 0.36 0.075 + Gs 90,000 20Q 000 6 0- MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions as x 1.25 x 10-3 + as x 9.6 x 10-3 = 0.285 as [10.85] = 285 = 26.26 N/mm2 (tensile) aB = 31.52 Nimm2 (tensile) Checking Steel bar, total extension = 0.075 + 0.032825 = 0.107825 = 8/s 31.52x 300 = 0.15 — 0.105066 90,000 Brass bar, final contraction = 0.15 8/B = 0.04493 8/B x 2.4 = 0.10784 ^ 8/s Q.1.8 A plate is riveted to a channel section in a structure as shown in figure 1.6. An eccentric load of 12.5 kN acts as shown on the plate. Determine the rivet diameter so that the maximum shear stress in any rivet is not to exceed 40 MPa. Diameter of the rivet should be chosen from preferred series diameter (in mm) r = 141.4 mm, /2 = 20000 mm2 12, 14, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 42, 48 rivet diameter [CSE-Mains, 2009, ME : 40 Marks] 12.5 kN 'V Channel E E O c\I I1 II 200 mm 50 mm Solution: 400 mm Fig. 1.6 Number of rivets = 5 Load = 12.5 kN 12.5 = 2.5 kN 5 A = Area of rivet Direct shear load on each rivet — If i d— 2500 A N/mm2 Torsional shear stress, is « r Torsional shear force = krA 4 kA (r2 ) = P.e. = 12500 x (100 + 50 + 400) 4 kA (20000) = 12500 x 550 Nmm kA = k— 12500 x 550 80000 = 85.9375 85.9375 A Torsional shear stress "Cs , in outer rivets = kr — 'Cd = tir 85.9375 x 141.4 12151.5625 A A 2500 A = \('rs x sin45° + td)2 + (Ts x cos4512 Fig. 1.7 Simple Stresses in Uniform and Compound Bars Strength of Materials I(12151.5625 x 0.707+ 2500) \ A 2 + 0 2151.5625 A A 7 2 x0.707 ) 8591.15)2 (11091.15)2 A ) A 1000 x 14.03 1000 v -I23.01+ 73.80 = A A 14030 A = 40 MPa (permissible shear stress) A = 350.073 mm2 = 4 d 2 d = 21.11 mm Rivet diameter from preferred series = 22 mm. Q.1.9 A compound tube is made by shrinking a thin steel tube on a thin brass tube. The areas of crosssection of these tubes are As and Ab, while the Young's moduli are Es and Eb respectively. Show that for any tensile load, the extension of the compound tube is equal to that of a single tube of same length and total cross-section area, but having a Young's modulus of E s As + E b Ab AS + Ab [CSE-Mains, 2005, ME : 20 Marks] Solution: Say Steel tube P = axial load As, Es = asAs + abAb But 5Is = 81b a S1 Es Brass tube Ab, Eb = ab x L Es a, Es b Eb Fig.1.8 Es Es P = ab x — x As x ab x Ab = ab[Eb As + At) ] Eb [EsAs+AbEb] P = ob Eb 1 [Es As + AbEb ] Eb sib = sis ab xi klb = b 81b = a b 1 Eb P fit' = EsAs+EbAb (For single bar having modulus of elasticity E) P 1 x e= (As +Ab ) E Strain, (From eq. (i) 8 0. lAS & IFS (Objective & Conventional) Previous Solved Questions or Es As + Eb Ab E(A, + Ab ) E= or MADE EASY E s As + E b Ab , for a single bar As + Ab Q.1.10 A copper tube 22 mm internal diameter, 30 mm outer diameter and 150 mm long is compressed by a nut tightening over a steel bolt, 20 mm diameter and 1 mm pitch. (i) If the nut is tightened by a quarter of a turn beyond the just touching position, determine the stress in the bolt. (ii) what would be the final stress in the bolt if the temperature of the assembly is to increase by 10°C. = 18 x 10-6/°C. Assume E, = 2 x 106 kg/cm2, Ecu = 6 x 106 kg/cm2, as = 12 x 10-6/°C, [CSE-Mains, 1999, ME : 30 Marks] Solution: 2 As area of cross-section of bolt = 4 x 20 = 1 007t mm2 Acu, area of cross-section of copper tube = .(302 _222 )=1O47c (i) For equilibrium, compressive force in tube = Tensile force in bolt ac„ x ,4„ = as x As acu x 104n = as x 1007c as = 1.04o Due to tightening of nut as bolt Pitch = 1 mm 1 0.25 mm = 4 mm = Extension of bolt + Contraction in tube 0.25 mm = 6 sx 150 + a0U x 150 Es E„ 0.25 as a„ 1.04a„ a „ + , (here as and a are in kg/mm2) 150 E, E„ 2x10" 6 x10' 1.04a„ u„ 0.25 x103 150 — 20 6 1.6667 = 0.052 a„ + 0.1666 acu= 0.21866 a„ = —7.622 kg/mm2 (comp.) as = +7.927 kg/mm2 (tensile) = —762.2 kg/cm2 (comp.) = as +792.7 kg/cm2 (comp.) (ii) Increase in temperature a > as OT= 10°C Compressive stress will be developed in copper tube due to temperature rate and tensile stress will be developed in stress bolt Strain in Cu tube = Strain in steel bolt Hence, oc„,AT — E r = a. AT GsT s E Simple Stresses in Uniform and Compound Bars Strength of Materials 4 9 acur ▪ asr _ (a cu - as) AT E„ E, or acur asT 2x106 6x105 (here asT and Guff are in kg/cm2) - (18 -12) x 10-6 x 10 acur ▪ a6sT 6 x 10-6 x 105 x 10 = 6 20 acuT x 104n = sT x 10071 asT = 1.04 acuT But 1:5,,,T 1.040,,T " - 6 20 6 am[0.05 + 0.1733] = 6 -26.866 kg/cm2 (comp.) acuT a sT = +27.94 kg/cm2 (tensile) Final stress in steel bolt asT = 792.7 + 27.94 = 820.64 kg/cm2 (tensile) Objective Questions Q.1 In a semi-infinite plate shown in the figure 1.9. The theoretical stress concentration factor kt, for an elliptical hole of major axis 2a and minor axis 2b is given by (a) t S k A 2a (b) 1 Fig. 1.9 (a) kt = = 2b (c) kt Q.3 S 2a (d) kt = 1+ — [CSE-Prelims, ME : 1998 ] Q.2 0 (b) kt = 1+ -a- In a simple tensile test, Hooke's law is valid upto the (a) elastic limit (b) limit of proportionality (c) ultimate stress (d) breaking point [CSE-Prelims, ME: 1998] B (c) S (d) The stress strain curve for an ideal strain hardening material will be as in [CSE-Prelims, ME : 1998] 10 Po. lAS & IFS (Objective & Conventional) Previous Solved Questions Q.4 The percentage elongation of a material as obtained from static tension test, depends on (a) diameter of the test specimen (b) gauge length of the specimen (c) nature of end grips of the testing machine (d) geometry of the test specimen [CSE-Prelims, ME : 1998] Q.5 (c) Potential energy of strain : Body is in a state of elastic deformation (d) Hooke's law : Relation between stress and strain [CSE-Prelims, ME : 1999] 0.8 Match the List-I (material properties) with List-II (technical definition represents) and select the correct answer. List-I A. Hardness B. Toughness C. Malleability D. Ductility List-I I 1. Percentage elongation 2. Resistance to indentation 3. Ability to absorb energy during plastic deformation 4. Ability to be rolled into plates Codes: A BCD 1 4 (a) 3 2 3 1 (b) 2 4 (c) 2 3 4 1 4 2 (d) 1 3 [CSE-Prelims, ME : 1999] Q.9 A measure of Rockwell hardness is (a) depth of penetration of indentor (b) surface area of indentation (c) projected area of indentation (d) height of rebound [CSE-Prelims, ME : 1999] A horizontal force of 200 N is applied at 'A' to lift the load W at C, as shown in figure 1.10. Value of weight W is 200 N 4 A E 0.075 m (a) 200 N (c) 600 N Q.6 Fig. 1.10 (b) 400 N (d) 800 N [CSE-Prelims, ME : 1998] Which one of the following pairs is not correctly matched? If E. Young's modulus a = coefficient of linear expansion T= Temperature rise A= Area of cross-section L= Original length (a) Temperature strain with permitted expansion 8 is raTL-Sl L ) (b) Temperature stress — aTE (c) Temperature thrust — aTEA (d) Temperature stress with permitted expansion 8 — E(aTL - 8) Q.7 Which one of the following pairs is not correctly matched? (a) Uniformly distributed stress: Force passes through centroid of the cross-section (b) Elastic deformation : work done by external forces during elastic deformation is fully dissipated as heat MADE EASY Q.10 If a block of material of length 25 cm, breadth 10 cm and height 5 cm undergoes volumetric strain of 1 , then change in volume will be 5000 (a) 0.50 cm3 (c) 0.20 cm3 (b) 0.25 cm3 (d) 0.75 cm3 [CSE-Prelims, ME : 2000] Q.11 For an isotropic homogeneous and linearly elastic material, which obeys Hooke's law, The number of independent elastic constants are (a) 1 (b) 2 (c) 3 (d) 6 Simple Stresses in Uniform and Compound Bars Strength of Materials 1 1 1 Q.12 Assuming E= 180 GPa and G=100 GPa for a material a strain tensor is (0.002 0.004 0.006` 0.004 0.003 0 0.006 0 0 Shear stress xy is (a) 400 MPa (c) 800 MPa Dm- (b) 500 MPa (d) 1000 MPa [CSE-Prelims, ME : 2001] Q.13 With the increase of percentage of carbon In steel, which one of the following properties does increase (a) modulus of elasticity (b) ductility (c) toughness (d) hardness [CSE-Prelims, ME : 2001] 0.14 Match List-I (material) and List-II (Stress-strain curve) and select the correct answers. List-I A. Mild steel B. Pure copper C. Cast iron D. Aluminium List-II 1. a 2. a 3. 4. Codes: A (a) 3 (b) 3 (c) 2 (d) 4 BCD 1 4 1 2 4 1 1 4 3 1 3 2 [CSE-Prelims, ME : 2001] Q.15 Match List-I with List-I I and select the correct answers. List-I A. Ultimate strength B. Natural strain C. Conventional strain D. Stress List-II 1. Internal structure 2. Change in length per unit instantaneous length 3. Change in length per unit gauge length 4. Load per unit area Codes: A BCD (a) 1 2 3 4 (b) 1 3 2 4 2 1 (c) 4 3 2 3 1 (d) 4 [CSE-Prelims, ME : 2002] Q.16 A steel rod of diameter 1 cm and 1 m long is heated from 20°C to 120°C, its a = 12 x 10-6/K and E= 200 GN/m2. If the rod is free to expand, thermal stress developed in it is (a) 12 x 103 N/m2 (b) 240 kN/m2 (c) zero (d) infinite [CSE-Prelims, ME : 2002] Q.17 Consider the following statements: 1. There are only two independent elastic constants 2. Elastic constants are different in orthogonal directions 12 . MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions 3. Material properties are same everywhere 4. Elastic constants are same in all loading conditions 5. The material has ability to withstand shock loading Which of the above statements are true for a linearly elastic homogeneous and isotropic material? (b) 2, 3 and 4 (a) 1, 3, 4 and 5 (d) 2 and 5 (c) 1, 3 and 4 [CSE-Prelims, ME : 2002] Q.18 A cast iron specimen in a torsion test gives a (a) cup and cone fracture (b) fracture along a plane normal to the axis of the specimen (c) fracture along a helix of approximately 45° (d) fracture along a plane inclined at 60° to the axis [CSE-Prelims, ME : 2002] Q.19 The reactions at the rigid support A and B for the bar loaded as shown in figure 1.11 are respectively Fig. 1.11 (a) 0 kN 10 kN 3 3 (c) 5 kN, 5 kN (b) 10 kN 20 kN 3 3 (d) 6 kN, 4 kN Q.20 A steel rod 10 mm in diameter and 1 m long is heated from 20°C to 120°C, E = 200 GPa and a =12 x 10-6/°C. If the rod is not free to expand, the thermal stress developed is (a) 120 MPa (tensile) (b) 240 MPa (tensile) (c) 120 MPa (comp.) (d) 240 MPa (comp.) [CSE-Prelims, ME : 2003] Q.21 The force F is such that the bars AC and BC (AC and BC are equal to length) as shown in the figure are identically loaded is Fig. 1.12 (a) 70.7 N (c) 141.4 N (b) 100 N (d) 108 N [CSE-Prelims, ME : 2003] Q.22 A mild steel specimen is tested in tension up to fracture in a Universal Testing Machine. Which of the following mechanical properties of the material can be evaluated from such a test? 1. Modulus of elasticity 2. Yield stress 3. Ductility 4. Tensile strength 5. Hardness 6. Modulus of rigidly Select the correct answer using the code given below: (a) 1, 3, 5 and 6 (b) 2, 3, 4 and 6 (c) 1, 2, 5 and 6 (d) 1, 2, 3 and 4 [CSE-Prelims, ME : 2007] Q.23 A heavy uniform rod of length 'L' and material density '8' is hung vertically with its top end rigidly fixed. How is the total elongation of the bar under its own weight expressed? (a) 28L2g E 8L2g (c) (b) 8L2g E (d) 61_2g 2E [CSE-Prelims, ME : 2007] Q.24 A round bar length 1, elastic modulus E and Poisson's ratio ji is subjected to an axial pull 'P. What would be the change in volume of the bar? Simple Stresses in Uniform and Compound Bars Strength of Materials (a) (c) Pl (1— 21.0E P1µ (b) P1(1— 2µ) E (d) — P/ ptE [CSE-Prelims, ME : 2007] Q.25 Which one of the following material has the highest ultimate tensile strength (a) mild steel (b) cast iron (c) spring steel (d) wrought iron [CSE-Prelims, ME : 2007] Q.26 A straight uniform rod is subjected to axial load. Which one of the following is the correct statement? (a) It induces maximum shearing stress on a transverse plane (b) It induces maximum normal stress on a plane inclined at 45° to the axis of the rod (c) It induces maximum shear stress on a plane inclined at 45° to the axis of the rod (d) It induces zero shear stress on any inclined plane to the axis of the rod [CSE-Prelims, ME : 2006] Q.27 Which one of the following is rupture stress? (a) Breaking stress (b) Maximum load/original cross-sectional area (A) (c) Load at breaking point/A (d) Load at braking point/neck area [CSE-Prelims, ME : 2006] Q.28 A 2 m long rod of diameter 2 mm is subjected to an axial pull of 1 kN. The rod is extended by 0.5 cm. What is the approximate value of the modulus of elasticity of the material of the rod? (a) 127 G Pa (b) 180 G Pa (c) 125 G Pa (d) 200 G Pa Q.29 Which one of the following information cannot be obtained from the static tensile test of a mild steel specimen? (a) Modulus of elasticity (b) Qualitative determination of toughness (c) Ductility (d) Weldability [CSE-Prelims, ME : 2006] 41 13 Q.30 Consider the following statements relating to Rockwell Hardness Testing Method: 1 It is a quick method to determine hardness of industrial components. 2. Polishing of test sample is necessary. 3. Test load is applied in two stages. 4. Hardness of relatively softer metals cannot be determined by this method. 5. Inverted pyramid type of indenter is used. Which of the statements given above are correct? (a) 1, 2, 3 and 4 (b) 3, 4 and 5 (c) 1, 2 and 3 (d) 2, 4 and 5 [CSE-Prelims, ME : 2006] 0.31 If a material has numerically the same value for its modulus of rigidity and bulk modulus, then what is its Poisson's ratio? (a) 0.25 (b) 0.2 (c) 0.15 (d) 0.125 [CSE-Prelims, ME : 2009] Q.32 A free bar of length 1 is heated uniformly from 0°C to a temperature T°C. a is the coefficient of linear expansion and E is the modulus of elasticity. Which one of the following is the stress induced in the bar? (a) aTE 4 (c) aTE (b) aTE 2 (d) Zero [CSE-Prelims, ME : 2009] Q.33 What property of a material enables it to be drawn into wires with the application of tensile force? (a) Plasticity (b) Elasticity (c) Ductility (d) Malleability [CSE-Prelims, ME : 2009] Q.34 ABC is a rigid bar. It is hinged at A and suspended at B and C by two wires, BD and CE made of copper and steel respectively, as shown in the given figure. The bar carries a load of 10 kN at F, midway between Band C. Given that A,-= 4 cm2 Aa = 2 cm2 = 1 x 105 N/mm2 ES = 2 x 105 N/mm2 14 0. MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions Subscript c and s stands for copper and steel. If the extensions in the steel and copper wires are As and As respectively, the ratio As/Ac would be I I Q.38 A bar of uniform cross-section of 400 mm2 is loaded as shown in figure. The stress at section 1-1 is 71 40 kN -0. 20 kN 4E-- 10 kN ----0,- 30 kN 4- 1 400 mm 300 mm 400 mm m ,-1 Fig. 1.13 (a) 1/4 (c) 2 (b) 4 (d) 1/2 [CSE-Prelims, CE : 2000] Q.35 Match List-I (Material) with List-II (Modulus of elasticity N/mm2) and select the correct answer: List-II List-I A. Steel 1. 0.6 x 105 2. 1 x 105 B. Cast iron C. Aluminium 3. 2 x 105 D. Timber 4. 0.1 x 105 Codes: A B C D 2 1 4 (a) 3 4 (b) 2 3 1 (c) 3 2 4 1 (d) 2 3 4 1 [CSE-Prelims, CE : 2001] Q.36 The phenomenon of decreased resistance of a material due to reversal of stress is called (b) elasticity (a) resilience (d) fatigue (c) creep [CSE-Prelims, CE : 2002] Q.37 Total elongation of a prismatic bar of length L, weight W, cross-sectional area A and modulus of elasticity E, under its own weight while hanging vertically, is (a) WL AE (c) WL 2AE Fig. 1.14 1 . m---1 -- (b) 2WL AE WL 3AE [CSE-Prelims, CE : 2002] (d) (a) 50 N/mm2 (c) 25 N/mm2 (b) 100 N/mm2 (d) 200 N/mm2 [CSE-Prelims, CE : 2003] Q.39 As soon as the external forces causing deformation in a perfectly elastic body are withdrawn, the elastic deformation disappears (a) only partially (b) completely over a prolonged period of time (c) completely and instantaneously (d) completely after an initial period of rest [CSE-Prelims, CE : 2003] Q.40 Match List-I (Mechanical property) with List-II (Feature) and select the correct answer. List-I A. Creep B. Tenacity C. Ductility D. Brittleness List-II 1. Amenability to go through changes of shape without rupture 2. Susceptibility to deform with time under sustained loading 3. Ability to be drawn into wire 4. Susceptibility to fail suddenly without warning Codes: A BCD (a) 4 1 3 2 (b) 2 3 1 4 3 1 2 (c) 4 (d) 2 1 3 4 [CSE-Prelims, CE : 2003] Simple Stresses in Uniform and Compound Bars Strength of Materials 4 15 Q.41 If P is the direct load, L is the length of the member, A is the uniform area of cross-section and E is the Young's modulus; then strain energy due to direct stress caused by a gradually applied load is P 2L P 2L (a) 2AE PL (c) 2AE (b) 2 A2E PL (d) AE [CSE-Prelims, CE : 2003] Q.42 Consider the following statements regarding tensile test diagrams for carbon steels with varying carbon contents; As the carbon content increases 1. the ultimate strength of steel decreases 2. the elongation before fracture increases 3. the ductility of the metal decreases 4. the ultimate strength of steel increases Which of the statements given above are correct? (a) 3 and 4 (b) 1 and 3 (c) 1, 2 and 3 (d) 1 and 2 [CSE-Prelims, CE : 2005] Q.43 What is the maximum possible value of Poisson's ratio for a non-dilatant material? (a) 0.67 (b) 0.50 (c) 0.33 (d) 0.25 [CSE-Prelims, CE : 2005] Q.44 The shear modulus of a material is half of its Young's modulus. What is the value of its Poisson's ratio? (a) -1 (b) -0.5 (c) zero (d) 0.5 [CSE-Prelims, CE : 2006] Fig. 1.15 (a) 16.67 N/mm2 (c) 26.67 N/mm2 (b) 13.33 N/mm2 (d) 30 N/mm2 [CSE-Prelims, CE : 2007] Q.47 In a pure tensile member, the normal stress on a plane at right angles to the direction of load is 100 N/mrn2. What is the normal stress at a plane whose normal is inclined at 60° to the direction of the load? (a) 75 N/mm2 (b) 100 N/mrn2 (c) 125 N/mm2 (d) 150 N/mm2 [CSE-Prelims, CE : 2008] Q.48 How is the maximum strain energy stored per unit volume in a body without permanent distortion, termed as? (a) Modulus of resilience (b) Modulus of tenacity (c) Modulus of toughness (d) Proof resilience [CSE-Prelims, CE : 2008] Q.49 A cable is supported at P and Q in which P is higher than Q. At what point(s) among the following is the tension in the cable maximum? Q.45 What is range of values of Poisson's ratio for ductile materials? (a) 0.10 to 0.15 (b) 0.16 to 0.20 (c) 0.21 to 0.24 (d) 0.25 to 0.33 [CSE-Prelims, CE : 2006] Q.46 What is value of stress at the base CC for the bar shown in the figure given below? Fig. 1.16 16 0. lAS & IFS (Objective & Conventional) Previous Solved Questions (a) Ponly (c) R only (b) Q only (d) P and 0 [CSE-Prelims, CE : 2008] Q.50 In the following pin jointed truss, what is the displacement of support B due to the given load? 5005 kN MADE EASY Q.52 At a certain stage under elastic loading, the elongation observed was 0.03 mm, the gauge length was 150 mm and the modulus of elasticity was 2 x 105 N/mm2. What was the stress at that location? (a) 4 N/mm2 (b) 40 N/mm2 (c) 80 N/mm2 (d) 60 N/mm2 [CSE-Prelims, CE : 2009] Q.53 Modulus of elasticity and Poisson's ratio of a material are 2.1 x 105 N/mm2 and 0.25 respectively. What is the value of modulus of rigidity of the same material in 105 N/mm2? (a) 0.84 (b) 0.70 (c) 1.40 (d) 0.50 [CSE-Prelims, CE : 2010] Fig. 1.17 Answers (Cross-sectional area of each member = 500 mm2, modulus of elasticity E = 2 x 105 N/mm2) (a) 3.25 mm (b) 2.50 mm (c) 1.50 mm (d) 0.50 mm [CSE-Prelims, CE : 2009] Q.51 What is the normal stress on a plane inclined at 45° to the axis of a square rod of side a subjected to an axial tensile force of T? (a) (c) T a T (b) 2a 2 T (d) 4a2 2 1. (d) 2. (b) 3. (d) 4. (b) 6. (d) 7. (b) 8. (c) 9. (a) 10. (b) 11. (b) 12. (c) 13. (d) 14. (b) 15. (a) 16. (c) 17. (c) 18. (c) 19. (a) 20. (d) 21. (a) 22. (d) 23. (d) 24. (b) 25. (c) 26. (c) 27. (d) 28. (a) 29. (d) 30. (c) 31.(d) 32. (d) 33. (c) 34. (c) 35. (b) 36. (d) 37. (c) 38. (b) 39. (b) 40. (d) 41.(a) 42. (a) 43. (b) 44. (c) 45. (d) 46. (d) 47. (a) 48. (a) 49. (a) 50. (b) 51.(b) 52. (b) 53. (a) 5. (d) 2 8a [CSE-Prelims, CE : 2009] Explanations 1. without having holes, acts, shoulders or narrow passes. (d) 2a kt = 1+--b 2a is major axis perpendicular to axis of loading. Stress contraction factor (kt) is a dimensionless factor which is used to quantity how concentrated the stress is in a material. It is defined as the ratio of the highest stress in the element to the reference stress. 2. (b) but ae aP abreaking ame °max k= r aref Reference stress is the stress in the part within an element under the same loading conditions Simple tensile stress Fig. 1.18 Simple Stresses in Uniform and Compound Bars Strength of Materials Hooke's law valid only upto proportionality. 3. ' limit of (d) Eand [t are two independent elastic constants. 12. (c) shear strain = 0.004 0.008 G= 100 x 1000 N/mm2 ti~ = yxy x 4 = 0.008 x 100 x 1000 N/mm2 = 800 MPa a y = 13. (d) With the increase of carbon percentage in steel. Hardness of steel does increase. Fig. 1.19 OA = elastic AB= strain hardening portion (b) % elongation - 6L gauge length change in length gauge length 5. (d) 200 x 0.3 = 0.075 W W= 800 N 6. 17 11. (b) 2 4. 44 (moment about B) (d) is not correctly matched. Temperature stress with permitted explain is not E(aTL- 6). 14. (b) Mild steel - 3, Pure copper - 2, Cast iron - 4, Aluminium - 1. 15. (a) 1. A. internal structure 2. B. 6//instantaneous length 3. C. 6L/L 4. D. load per unit area 16. (c) Rod is free to expand, no stress will be developed. 17. (c) 1, 3 and 4 are correct statements. 7. (b) Elastic deformation work done by external forces during elastic deformation is not dissipated fully as heat. 18. (c) A cast iron specimen in a torsion test fails along a helix approx at 45° to the axis due to tensile principal stress p1 = +T(shear) 8. (c) 2. Resistance to inductation A. Hardness Toughness 3. Ability to absorb energy B. during plastic C. Malleability 4. Ability to be rolled into plates 1. Percentage of elongation D. Ductility 19. (a) 9. (a) A measure of Rockwell hardness is a depth of penetration of indentor. 10 kN 1 4- P2 2m k- 1m Fig. 1.20 P1 x1 =P2 x 2 AE AE P1 = 2P2 P1 + P2 = 1 0 10. (b) 6V- 25 x10x 5 - 0.25 cm3 5000 20 - 3 18 •• MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions 20. (d) TE = 12 x 10-6 x 100 x 200 x 103 = 240 MPa (comp.) Because expansion is prevented. 21. (a) Bars AC and BC are identically loaded. FAC = FBC Horizontal component of FAc and FBc are cancelled. 2FAB cos 30° = 100 cos45° Horizontal component of 100 N = 70.7 N is balanced by F So F= 70.7 N 22. (d) 1, 2, 3, 4 are evaluated by tension test on MS. 23. (d) 8L due to self weight = (pg) 2 2E 29. (d) Weldability cannot be determined by static tensile test. 30. (c) 1, 2, 3 are correct statements. 31. (d) 2(1+v) 3(1-2v) 2 + 2v= 3 - v 1 =8v v=0.125 32. (d) Bar is free to expand no stress is developed. 33. (c) Ductility enables the wire to the drawn. 34. (c) Since ABC is rigid, deflection will be linear. L A 8, 24. (b) 8V= evV = 1-1 m 2µ) x IA 1m-►I I Fig. 1.21 8s -=2, Similar triangles Sc 25. (c) Spring steel is a high carbon steel, it has highest out amongst the material gives. 26. (c) Axial load on uniform straight bar induces maximum shear on a plane inclined at 45° to the axis of the rod. 35. (a) A. Steel B. Cast iron C. Aluminium D. Timber 3. 2 x 105 N/mm2 2. 1 x 105 N/mm2 1. 0.6 x 105 N/mm2 4. 0.1 x 105 N/mm2 36. (d) Fatigue 37. (c) 27. (d) Rupture load at breaking point/neck area. 28. (a) E= 1000 2000 x TC 0.5x10 =1.27 x 105 N/mm2 =127 GPa Fig. 1.22 Extension due to own weight 38. (b) 611 = Ebb = craa x (a)2 = 100 x (0.861)2 = 75 N/mm2 wL 2AE 8- 19 Simple Stresses in Uniform and Compound Bars Strength of Materials 48. (a) Maximum strain energy per unit volume without permanent distortion = Modulus of resilience. 40000 = 100 N/mm2 400 39. (b) Elastic deformation disappears completely over a prolonged period of time. 49. (a) P only, maximum slope of cable at P. 50. (b) 500 x,FakN 40. (d) A. Creep 2. Susceptibility to deform with time under sustained loading 1. Amenability to go through changes of shape without rupture 3. Ability to be drawn into wire 4. Susceptibility to fail without suddenly warning B. Tenacity C. Ductility D. Brittleness A 250 xsffkN 250 x5- kN Fig. 1.24 Extension in AB - Components of contraction in CA and CB 8B = 41. (a) 250 x 103 1000 500 x 103 x 1000 x 2 x x cos 60° 500 2 x 105 500 x2 x105 = +2.5 - 5 = -2.5 mm or 2.5 mm PL P2L Strain energy = -P x = 2 AE 2AE 1 51. (b) n \< 7 45° 42. (a) (3) and (4). n 43. (b) Maximum possible value of Poisson's ratio for a non-dilatant material is 0.5. Fig. 1.25 T 6a= a T 2 abb- -7---X(COS45) . a2 2a 2 44. (c) 1 E G= -E = , so, v = 0 2 2(1+v) 52. (b) 8L= 0.03 mm L= 150 mm E= 2 x 105 N/mm2 45. (d) v = 0.25 - 0.33, for ductile materials. 46. (d) 0cc - a= (20+25)1000 1500 = 30 N/mm2 LL xE _ 0.03 x2x105 150 = 40 N/mm2 53. (a) 47. (a) E= 2.1x 105 N/mm2, v = 0.25 b 60° E 3 100 b Fig. 1.23 2.1x105 2.5 = 0.84 x 105 N/mm2 G= 2(1+0.25) - 11111111• 02 Principal Stresses CHAPTER Q.2.1 At a section in a beam the tensile stress due to bending is 50 N/mm2 and there is shear stress of 20 N/mm2. Determine from first principles, the magnitude and direction of principal stress and calculate the maximum shear stress [IFS, 2012, ME : 10 Marks] Solution: 50 20 (F1 + Q2) F1 Q2 (b) (a) (c) Fig. 2.1 Figure shows an element ABC unit of thickness subjected to normal and shear stresses as given. On plane BC there is complementary shear stress of 20 N/mm2 as shown in figure 2.1(a). F1 = 50 x AC Normal force, Shear force, Q1 = 20 x AC Shear force, Q2 = 20 x BC Normal force on inclined plane AB Fr,_ (F1 + Q2 ) cos0 + Q1 sin° On x AB = (F1 + Q2) cos() + Q1 sine or on x AB = 50 AC cos0 + 20 BC cos° + 20 AC sine or an = 50 cos2O + 20 sine cos() + 20 cos() sin() 50 cos20 + 20 sin 20 tangential force on plane AB Ft x AB = (F1 + Q2 ) sin° - 1 cos° or To = sine 02 sine Q1 cos() AB AB AB Note from figures, that internal resistance is equal and opposite to the applied forces on inclined plane. Principal Stresses Strength of Materials 4 21 Putting the values of F1, Qi, 02 tie - 50 x ACsin0 20 xl3Csin0 20 x ACcos0 AB AB AB = 50 sin0 cos° + 20 sin20 - 20 cos20 50 = — x sin 20 - 20 cos 20 = 25 sin 20 - 20 cos 20 2 Principal planes On principal planes, to, shear stress is zero 25 sin20 - 20 cos20 = 0 So, tan20 = 0.8 201 = tan-10.8 = 38.66° 01 = 19.33° Another plane, 02 = 90 + 19.33 = 109.33° Principal stresses 01 = 19.33, cos01 = 0.943.6 cos201 = 0.890 Principal stress, sin201 = 0.6247 p1 = 50 x 0.8904 + 20 x 0.6247 44.52 + 12.494 = 57.016 MPa 02 = 109.33°, cos02 = -0.3310, cos202 = 0.10956 sin202 = -0.6247 Principal stress, p1 = 50 x 0.10956 + 20 x (-0.6247) = 5.478 - 12.494 = -7.016 MPa Principal stresses are + 57.016, - 7.016 MPa Principal directions are 19.33°, 109.33° For maximum shear stress, 0, de = 0 25 x 2 cos 20 + 20 x 2 sin 20 = 0 0 = -25.67° or 64.33° To = 25 x sin (-2 x 25.67°) - 20 x cos (-2 x 25.67°) = -19.521 - 12.495 = - 32.016 Mpa It9I = 32.016 MPa Q.2.2 Draw Mohr's circle for a 2-dimensional stress field subjected to (ii) Pure biaxial tension (i) Pure shear (iv) Pure uniaxial tension (iii) Pure uniaxial compression [CSE-Mains, 1997, ME : 15 Marks] Solution: = +T OB = -T ti B - ti 22 ► lAS & IFS (Objective & Conventional) Previous Solved Questions Air a MADE EASY 2 - Radius 6 0 OA = Pure Bi-axial tension OB= (72 oc = Cf + 6 ' ' 2 R= Pure uniaxial compression OA = OC = --cv/2 (iv) Pure uniaxial tension Fig. 2.2 Q.2.3 For the following state of stress, show the stresses on two given planes at right angle of an element. Find the magnitude and directions of principal stress and the maximum shear stresses in each case. (i) Simple uniaxial tension (ii) Pure equal normal stresses on given planes (iii) Pure shear stresses on given plane [CSE-Mains, 2002, ME : 20 Marks] Solution: (a) Fig. 2.3 (a) (i) Simple uniaxial tension. Principal stress, p1 = p2 = p3 = 0 p1 along x-axis. Maximum shear stress = /91 —P2 2 —P1 2 Principal Stresses i Strength of Materials 23 y x (b) Fig. 2.3 (b) (along x and y direction) (along z direction) Pure equal normal stresses , p1 = p2 = a p3 = 0 Maximum shear stress - P1 P2 2 =0 (iii) Mohr's circle diagram will be a point on +x axis with coordiantes (a, 0) (c) Fig. 2.3 (c) Pure shear stresses on given planes p1 = +T, P2 = 01 = 45°, 02 = —45° p3 = 0 Maximum shear stress = ti Note: In case (ii) Mohr circle diagram will be a point on (+ve) x-axis, having coordinate (a, 0) Q.2.4 At a point in an elastic material the stresses on three mutually perpendicular planes are : 50 MN/m2 tensile and 40 MN/m2 shear First Plane Second Plane : 30 MN/m2 compressive and 40 MN/m2 complementary shear Third Plane : No stress Find: (i) The position of principal planes and magnitude of principal stresses (ii) The position of the planes on which maximum shear stress act, calculate normal and shear stress on them. [CSE-Mains, 1992, ME : 20 Marks] Solution: Fig. 2.4 24 P. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions BC and AC are two mutually perpendicular planes. Reference plane BC 1 = +50 N/mm2 (i) c = +40 N/mm2 Plane AC 62 = -30 N/mm2 ti = +40 N/mm2 61 2 62 - 10, al 262 - - 40 Principal stresses P1 = 10-4402 +402 = 10 + 56.56 = 66.56 N/mm2 P2 = 10_J402 + 402 = 10 - 56.56 = -46.56 N/mm2 12 x 40 20 = tan-1 2T = tan= tan-1(1) = 45° 1 80 al a2 61 = 22.5°, 02 = 112.5° Tmax = V402 + 402 = 56.56 N/mm2 (ii) Angle of plane of Tam 03 = 0l+ 45° = 67.5° w.r.t plane BC 04 = 67.5 + 90° = 157.5° Normal stresses on planes of maximum shear stress, 'um, - 61 + az 2 (wrt plane BC) - 10 N/mm2 On both shear planes with +Tricia, and -Tma, Q.2.5 At a point in a material under stress, the intensity of the resultant stress on a certain plane is 50 MN/m2 (tensile) and inclined at 30° to the normal of that plane (Fig. 2.5). The stress on a plane at right angle to this plane has a normal tensile component of intensity 30 MN/m2. Find: (i) resultant stress on second plane (ii) principal planes and stresses (iii) maximum shear stress and the plane on which it occurs Solution: Shear stress on plane BC = 50 sin 30° = 25 MPa Normal stress on plane BC = 50 cos 30° a1= 43.3 MPa Complementary shear stress on plane AB ti = 25 N/mm2 (as shown) 50 MPa Resultant stress on plane AB = V302 + 252 = 39.05 N/mm2 = 39.05 MPa 62, normal stress on plane AB = 30 MPa (ii) al +62 2 433+ 3° = 36.65 MPa 2 Fig. 2.5 Principal Stresses Strength of Materials al — a2 ai — G2 )2 T_2 _ 652 252 25.87 MPa 2 Principal stresses 25 43.3 — 30 = 6.65 MPa 2 2 ( 4 (31 P1' 1-'2 2 + G2 + (G1 G2) +T2 2 2 p1 = 36.65 + 25.87 = 62.52 MPa p2 = 36.65 — 25.87 = 10.78 MPa Principal angle (wrt plane BC) 1 0 = — tan 1 2 2T cy1—a2 1 = 2 tan _12 x 25 13.3 = 37°33' 02 = 90 + 61 = 127°33' (iii) Maximum shear stress max = ±25.86 MPa 03 = 01 + 45° = 37°33' + 45° = 82°33' 04 = 03 + 90° = 172°33' 2 20 N/mm Q.2.6 The figure given below shows the state of stress at a point namely 30 N/mm2 tensile in x-direction, 20 N/mm2 tensile in y-direction and shear stress of 50 N/mm2. Find the position of principal planes, principal stresses and maximum shear stress graphically or other wise. [CSE-Mains, 2000, ME : 20 Marks] 50 2 30 N/mm 50 30 2 30 N/mm 50 Solution: Let us take plane AB as reference plane on which shear stress is negative shear stress cr / = 30 N/mm2, a2 = 20 N/mm2, ti = 50 N/mm2 D 50 4 20 — 25 Fig. 2.6 — 5 N/mm2 Principal stresses P1 ' P3 = + G2 (01 — 02 2 2 2 ) +ti2 = 25± V52 + 502 = 25± 50.25 p1 = +75.25 N/mm2 p2 = —25.25 N/mm2 Angle of principal plane w.r.t plane AB (in anti-clockwise direction) _1 2T = 1 1 2 x 50 2 tan= 42°8'41" 0 = -tan _1 1 10 0 G2 02 = 01 + 90° = 132°8'41" 2 l'Crinaxl maximum shear stress = ( al — (32 ) 2 Tina, will occur at angle, and T 2 = 50.25 N/mm2 03 = 01 + 45° = 87° 8' 41" 04 = 03 + 90° = 177° 8' 41" 26 . lAS & IFS (Objective & Conventional) Previous Solved Questions Q.2.7 A circle of 10 cm diameter is inscribed on a steel plate before if is stressed. Then the plate is loaded so as to produce stresses as shown in the figure 2.7 and circle is deformed into an ellipse. Determine the length of major and minor axes of the ellipse and their directions. Take modulus of elasticity of steel as 2.1 x 107 N/cm2 and Poisson's ratio = 0.3. [CSE-Mains, 2004, ME : 20 Marks] MADE EASY 2 12 kN/cm 17 6 P2 Solution: E = 21000 kN/cm2 Taking AB as reference plane 2 12 kN/cm = +12 kN/cm2 a2 = -6 kN/cm2 Fig. 2.7 ai +432 - 3 kN/cm2 2 al - 02 - 9 kN/cm2 2 ti = 4 kN/cm2 2 a2 2 2 2 +tie - V9 + 4 2 --9.849 kN/cm p1 = 3 + 9.849 = 12.849 kN/cm2 p2 = 3 - 9.849 = -6.849 kN/cm2 Principal angle (w.r.t. in plane CD in anti-clockwise direction) 1 1 = 12° = 1 01 = --tan2 ai --a2 2 tan-12184 02 = 102° Strains, E vp2 E 12.849 0.3 x 0.849 _ 14.9037 + 21000 21000 21000 10.7037 -6.849 0.3 x 12.849 21000 E2 - 21000 21000 = 0.70797 x 10-3 e2 = -0.5097 x 10-3 (In the direction of major principal axis) Maximum axis = 100 + 0.70797 x 10-3 x 100 = 100.0709 mm (In the direction of minor principal) Minimum axis = -100 - 5097 x 10-3 + 100 = 99.949 mm Q.2.8 At a point in a strained material, planes AB and AC pass through the point A as shown in the figure 2.8(a). Normal and shear stresses on plane AB are 30 MPa and 50 MPa respectively as shown. Normal and shear stress on plane AC are 60 MPa and 20 MPa respectively as shown. By graphical or analytical method, determine angle between the planes AB and AC. [CSE-Mains, 2001, ME : 30 Marks] Solution: Using sign convention and Principal Stresses Strength of Materials 1 27 D 60 (30, +50) (a) E OC = +30 CD = +50 OA = +60 AB = —20 Refer (60, — 20) Mohr's stress circle 29 = 90°, 0 = 45° (b) Fig. 2.8 Note: In shear stress produce clockwise couple at the center of element then it will be plotted above ax (+ve) and if shear stress causes anticlockwise couple at the center then it will be plotted below a axis. Q.2.9 (a) Under which condition of the state of stress at a point in the two dimensions, the Mohr's circle will be reduced to a point. (b) How much change in volume would a 100 mm side cube of steel will have when it is kept at a depth of 2 km in sea water? Assume specific gravity of sea water equal to 1.02, modulus of elasticity equal to 2.08 GPa and Poisson's ratio equal to 0.29. [CSE-Mains, 2007, ME : 20 Marks] Solution: (a) Mohr's stress circle will be reduced to a point (T = ay = If x or ax = 0y = Cr as shown fa Y X , 0) (—a. • rX (a, 0) Fig. 2.9 28 (b) 0. MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions h= 2 km y= 1.02 x 1 x 10-5 N/mm2 p, pressure = yh = 1.02 x 10-5 x 2000 x 1000 N/mm2 = 20.4 N/mm2 Depth, Weight density, K, bulk density = E 3(1-2v) 3(1-0.29 x 2) p Volumetric strain, Ev 2.08 x105 K= -1.65 x105 N/mm2 20.4 1.65x105 12.357x 10-5 V = 1003 = 106 mm3 8V = 12.357 x 10-5 x 106 = 123.57 mm3 Change in volume, Q.2.10 The normal stresses at a point in a strained material across two planes at right angles of each other are 120 N/mm2 tensile and 60 N/mm2 compressive. The shear stress on these planes is 40 N/mm2. Find Principal stresses and principal planes (i) The direct and shear stresses on a plane inclined at 30° to the vertical (ii) [CSE-Mains, 1998, ME : 30 Marks] Solution: Let us take BC as reference plane al = +120 MPa, a2 = -60 MPa, t = -40 MPa on BC a' 2 a2 (31 CF2 - 90 MPa - 30 MPa 2 ' 2 Pi, P2 = 30 ± V902 + 402 = 30 ± 98.49 p1 = 128.49 MPa p2 = -68.49 MPa 1 12T 9 ,_ -tan 1 2 al 0,2 (w.r.t palne BC in anticlockwise direction) 12 x 40 1 = 1tan= x 24° = 12° 2 180 2 02 = 90 + 12° = 102° Reference plane BC Inclined plane AB 0 = 60° (angle from plane BC) Normal, a o al 1- a2 a1 - a2 cos120°+ tsin120° 2 + 2 co 0_ m 0 co 1 = 30 + 90 cos120° + 40 sin120° = 30 - 45 + 34.64 = 19.64 MPa Shear stress, II T 40 MPa T = (a1 - a2 )sin120°-tcos120° o 2 IT' = 90 sin 120°- 40 x cos 120° = 90 x 0.866 - 40(-0.5) = 77.94 + 20 = 97.94 MPa = 97.94 MPa 0" T C al= + 120 MPa Fig. 2.10 Principal Stresses Strength of Materials 29 Q.2.11 Two planes AB and AC make an angle of 50° at point A. Plane AB is subjected to tensile stress of 3 kN/cm2 and a shear stress 3 kN/cm2 from B toward A. Plane AC neglected to a shear stress of magnitude AB — Ref. Plane 2 kN/mm2 (from C towards A) and an normal stress of unknown magnitude. Determine (i) Normal stress on plane AC (ii) Principal stresses. [IFS 2011, ME : 15 Marks] Solution: Stress on plane AB a = +30 N/mm2 = 3 kN/cm2 ti = -30 N/mm2 = -3 kN/cm2 AB is reference plane shear stress on reference plane is D negative. Consider a plane AD perpendicular to plane Fig. 2.11 AB, say normal stress on this plane is a2. Complementary shear stress on this plane is +30 N/mm2. Plane AC = 61 - 62 sin20 --tcos20 2 0 = -50° 20 - alsin(-100°)-30cos(-100°) 2 (30 2 a2)( 0 9848)- 30 (-0.17365) = -14.772 + 0.4924 a2 + 5.20945 62 - 20 +14.772 - 5.20945 0.4924 = 60.038 N/mm2 (i) Normal stress on plane BC a - 61 2 62 4. al a2 cos(-100°) + Tsin(-100°) 2 30 + 60.038 30 - 60.038 2 2 x cos(-100°)+ 30 x sin(-100°) = 45.019 - 15.019 (-0.173648) + 30 (-0.9848) = 45.019 + 2.6080 - 29.544 = 18.083 N/mm2 (Normal stress on plane AC) (ii) Principal stresses a1 = 30 N/mm2 a2 = 60.038 N/mm2 01 + 02 — 2 +45.019 N/mm2 61 2G2 - -15.019 N/mm2 T = 30 N/mm2 30 0. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions 2 (ai - az ) +,r2 = V(-15.019)2 + 302 = ±33.55 N/mm2 2 P1 — al + az +33.55 = 45.019 + 33.55 = 78.519 N/mm 2 2 P2 = 01+ 02 + 33.55 = 45.019 - 33.55 = 11.519 N/mm 2 2 Q.2.12 A1 m x 1m mild steel shell of 1 mm thickness is stretched in its own plane by stresses, ; = 20 MPa, ay = 30 MPa as shown in figure 2.12. Point 0 is centre of the plane. OC and OD are two mutually perpendicular lines inclined at 45° to x and y-direction respectively before application of stresses. Determine the change in angle (in degrees) between OC and OD after application of stresses. Take modulus of rigidity of plate material = 80 GPa. ax = 20 MPa 0 30 MPa ay = 30 MPa Fig. 2.12 [CSE-Mains, 2011 , ME : 20 Marks] Solution: t, shear stress on plane OC - 30 - 20 sin 90° = +5 MPa 2 Shear stress on plane OD = -5 MPa Change in angle of 90° between OD and OC 5 (I) xy - = G 5 rad=0.0036° 80000 30 kN 1r 500 mm 30 kN ;150 Q.2.13 A steel bracket is bolted a column by three similar bolts as shown in the figure 2.13. It is subjected to an eccentric load of 30 kN as indicated. Determine the suitable minimum diameter for the bolts using a factor of safety of 5.0. Assume the bracket and the attached plate to be rigid. Take the allowable stresses for the bolt material in tension as 600 MN/m2 and in shear as 350 MN/m2. [CSE-Mains, 1996, ME : 30 Marks] 300 mm 1 100 mm Fig. 2.13 Principal Stresses Strength of Materials 4 31 Solution: Load = 30 kN Load is eccentric about two axes, x and y Direct shear stress on each bolt — 10000 A Nimm2 (.0 Shear stress due to 30,000 x 150 Nmm moment will be maximum at bolt E, but ab due to moment 30,000 x 500 Nmm will be maximum at both D. Shear strain in bolts r1 = 180 mm G of the system bar as shown. r2 = 200 mm A = area of cross-section of bolt Shear stresses in bolts N/mm2 i d = direct shear stress — 10000 A 'Cs oc r G of system lies at G', 100 mm for upper bolts ri = 180 mm, r2 = 200 mm Ts = kr Force, fs = krA Torque, T = EWA k[2r12 A + r22A] = kA[2 x 1002 + 2002] kA x 10.40 [ 10.48] = 30,000 x 150 Ak — k— To = kri — 3 x15 x105 104 x 10.48 42.939 A 42.939 x 180 7729 A = A 150 10000 TA ,. — A 100 tanO = 0.667 0 = 33.7° i s cos 33.7 = 6430 i s sin 33.7 = 4288 Td + Ts cos 33.7 — Tr — T r' resultant shear stress — 16430 A 4288 12 11 6 430)2 10A00 270 16980 = + — 18.386 + A A 9 16980 A Nimm2 Bending stress KA [211 2 + 122] = Re KA [2 x 4002 + 1002] = P.e = 30000 x 500 = 15 x 106 Nmm KA [320,000 + 100,00] = 15 x 106 3 2 P. MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions k'A — 15 x 106 1500 . 33 x 104 33 k'A = 45.45, 45.45 a bi = 0,, — A A x 400 = 18180 A 9090 abi 2 A For upper bolt (max. principal stress) = (Gbi )2 9090 A — 2 + Tr 2 = 9090 + Pmax 45.45 K— + (9090)2 (16980 )2 A ) A ) 1000 A V82.628 + 288.32 = 9090 + 19260 A 28350 = 120 (since FOS = 5) A A= 236.25 mm2 = Ed 2 4 d = 17.344 mm -=-' 18 mm Bolt dia., For upper bolt (max. shear strain),FOS = 5 19260 'CMaX — A = 70 A = 275.143 = Ed2 4 d = 18.72 mm Bolt dia., Objective Questions Q.1 The complementary shear stresses of intensity T are induced at a point in the material, as shown in the figure 2.14. Which of the following is the correct set of orientation of principal planes with respect to AB. T D C T e I TI (a) 30° and 120° (c) 60° and 150° Q.2 A material element is subjected to a plane state of stress such that the maximum shear stress is equal to the maximum tensile stress would compound to T D C (a) 1 0 1 T A T< Flg. 2.14 B (b) 45° and 135° (d) 75° and 105° [CSE-Prelims, ME : 1998] A T<--- B Principal Stresses Strength of Materials 41 33 (d) (b) [CSE-Prelims, ME : 1999] Q.4 Biaxial stress system is correctly shown in 30 20 a (c) (a) 20... ►10 1 20 30 10 (d) 30 6 20 (b) 10 - 10 1 20 [CSE-Prelims, ME : 1998] Q.3 30 y For the given stress condition ax = 2 N/mm2, ay = 0, Txy =0, the correct Mohr's circle is 10 40 (a) 30 (C) 3 40 40 (b) 20 (d) 30 30 1 (c) 20 t 40 [CSE-Prelims, ME : 1999] 34 0. Q.5 Principal stresses at a point in a stressed solid are 400 MPa and 300 MPa respectively. The normal stress on plane inclined at ±45° to the principal planes will be (a) 200 MPa and 500 MPa (b) 350 MPa on both planes (c) 100 MPa and 600 MPa (d) 150 MPa and 550 MPa [CSE-Prelims, ME : 2000] (b) Q.6 A Mohr's stress circle is drawn for a body subjected to tensile stresses f, and fy in two mutually perpendicular directions such that fx > fy. Which one of the following statements in this regard is not correct? (a) Normal stress on a plane at 45° to fx is equal to fx tO fy 2 (d) C 0 2 (d) Maximum shear stress is equal to (0 = origin and C = centre of circle, OA = al , OB= 02 ) fy fx 2 [CSE-Prelims, ME : 2000] The resultant stress on a certain plane makes an angle of 20° with the normal to the plane. On the plane perpendicular to the above plane. The resultant stress makes an angle of 0 with the normal. The value of 0 can be (a) 0° to 20° (b) any value other than 0° to 90° (c) any value between 0° to 20° (d) 20° only [CSE-Prelims, ME : 2001] The correct Mohr's stress circle drawn for a point in a solid shaft compressed by a shrink fit hub is as (a) B A fy fx (c) Maximum normal stress is fx Q.8 0 (c) 0 (b) Shear stress an a plane at 45° to f is equal Q.7 MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions Q.9 When the two principal stresses are equal and like, the resultant stress on any plane is (a) equal to principal stress (b) zero (c) one half of the principal stress (d) one third of the principal stress [CSE-Prelims, ME : 2002] Q.10 On a plane, resultant stress is inclined at an angle of 45° to the plane. If the normal stress is 100 N/mm2, shear stress on plane will be (a) 71.5 N/mm2 (b) 100 N/mm2 (c) 86.6 N/mm2 (d) 120.8 N/mm2 [CSE-Prelims, ME : 2003] Q.11 The moduli of elasticity and rigidity of a material are 200 GPa and 80 GPa respectively. What is the value of Poisson's ratio of the material? (a) 0.30 (b) 0.26 (c) 0.25 (d) 0.24 [CSE-Prelims, ME : 2007] Q.12 State of stress at a point of a loaded component is ax = 30 MPa, ay = 18 MPa, 8 MPa. If the Txy = Principal Stresses Strength of Materials 4 35 larger principal stress at the point is 34 MPa, what is the value of smaller principal stress? (a) 12 MPa (b) 14 MPa (c) 16 MPa (d) 18 MPa [CSE-Prelims, ME : 2008] Select the correct answer using the codes given below: (a) 1 and 3 (b) 2 and 3 (c) 1 and 4 (d) 2 and 4 [CSE-Prelims, CE : 2004] Q.13 The maximum shear stress occurs on (a) principal planes (b) plane at 45° to the principal planes (c) planes at 90° to the principal planes (d) planes independent of the inclination to the principal planes [CSE-Prelims, CE : 2003] Q.17 If Ex and Ey are the maximum and minimum strains, respectively, in the neighbourhood of a point in a stressed material, then what is the expression for the maximum principal stress? (a) EE, (b) E(E, + µEy) Q.14 The ratio of the maximum shear stress to the difference of the two principal stresses is (a) 1/2 (b) 1/3 (c) 1/4 (d) 1/6 [CSE-Prelims, CE : 2002] [CSE-Prelims, CE : 2007] (c) E(E, + µEy ) 1— µ 2 E(Ey +µE,) (d) 1-112 Q.18 The principal stresses in N/mm2 on a square element are shown in the figure below. What is the intensity of tangential stress (pi) on the plane BD? 100 Q.15 For the two dimensional stresses shown in the figure, what is the normal stress on the 45° plane? oll llll 100 100 ctfffttB 100 2 10 N/mm Fig. 2.20 (a) 50 N/mm2 (c) 100 N/mm2 Fig. 2.19 (a) 20 N/mm2 (c) 4 N/mm2 (b) 12 N/mm2 (d) 8 N/mm2 [CSE-Prelims, CE : 2004] Q.16 A rectangular block is subjected to shear stress intensity of T. The stresses induced on a plane inclined at 135° to the horizontal axis shall be (b) 0 (d) 200 N/mm2 [CSE-Prelims, CE : 2006] Q.19 A point in an element is stressed as shown in the figure given below. 100 N/mm2 t E t 100 N/mm2 c„, O., 1. ax = 4-T (tensile) 2. crx = —T (compressive) 3. 4. atangential = -FT atangential = A 1 1 100 N/mm2 Fig. 2.21 —.-100 N/mm2 36 ► IAS & IFS (Objective & Conventional) Previous Solved Questions What is the value of normal stress on the oblique plane BE? (b) 175 N/mm2 (a) 200 N/mm2 (d) 100 N/mm2 (c) 150 N/mm2 [CSE-Prelims, CE : 2007] 400 N/mm2 400 N/mm2 Q.20 A specimen is subjected to pure shear, the shear stress being q. Tensile and compressive stresses of intensity a, occur on planes inclined at 45° to the shear stress. What is the value of the ratio a/q ? (b) 1.5 (a) 2 (d) 1 (c) 1.25 [CSE-Prelims, CE : 2008] Q.21 The square element in the figure is subjected to a biaxial stress of 400 N/mm2 as shown. What is the intensity of normal stress pn on the plane BD? MADE EASY 400 N/mm2 Fig. 2.22 (b) 400, (a) 200 (c) 400/V- 1. (b) 6. (d) 11. (c) 16. (d) 21. (d) 2. 7. 12. 17. (d) (b) (b) (c) (d) 0 [CSE-Prelims, CE : 2009] Answers 3. (d) 8. (d) 13. (b) 18. (b) 4. (d) 5. (b) 9. (a) 10. (b) 14. (a) 15. (d) 19. (d) 20. (d) Explanations 1. 3. (b) 45° and 135° (d) p1 = +r, P2 = Fig. 2.23 Cannot Mohr's stress circle. 5. (d) 40 20 2 Fig. 2.22 2. 30 30 20 (d) al -(-G1) max 2 = a = maximum tensile stress 40 Fig. 2.24 Principal Stresses Strength of Materials Complementary shear stresses are balanced Normal stresses is balanced in x and y-direction 5. 4 37 10. (b) 100 mpa (b) an — D1 + 10 , ,2 2 ± P1 — P-2 cos 90° 45° 2 400 + 300 = 350 MPa 2 On both planes. 6. (d) fx Amax 7. fy 2 Fig. 2.26 = (b) 100 MPa 11. (c) E = 2G(1 +v) 200 = 2 x 80 (1 + v) 1.25 = 1 + v v = 0.25 ar cos 20° 12. (b) ax + ay = p1 + P2 30 + 18 = 34 + p2 p2 = 14 MPa 13. (b) The maximum shear stress occurs on planes at 45° to the principal planes. Fig. 2.25 14. (a) 4) can be other than 0 or 90° where a = 0 8. 9. P1 — P2 (d) For a point in a solid shaft, compressed by a shrink fit hub is = ao = p' (junction pressure) OA = OB = OC = p', circle will be point (a) 2 P2 ti max — max =2 ti max — P2 = 0.5 15. (d) _ Pi 20 = +45°, sin29 = 1 ti 0 P2 sin 2 a Pi= P2 = P 2— — P2 2 = 10 N/mm2, p2= 6 N/mm2 P1 + P2 a on any plane = p. Pn — 2 + P1 P2 2 cos 20 38 ► MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions cos 20 = cos 90° = 0 18. (b) = p2 = —100 N/mm2 10+6 2 p_ = 8 N/mm n 2 pt _ Pi — P2 cos 90° = 0 2 16. (d) Mohr's stress circle is a point. 19. (d) P1 = P2 = +100 N/mm2 + P2 — 100 2 45° — P2 " —u 2 . 913° a= sin (-2 x 45°) = 100 = —t (compressive) 0 = 450 atangential = + P2 P1 P2 cos 60° 2 2 a _ e 20. (d) 't COS 90° = 0 —=1 Shear —t Fig. 2.27 Fig. 2.29 17. (c) 21. (d) p1 = +400 N/mm2 e _ Pi — 1-IP2 p2 = —400 N/mm2 E x v. Eex = P2 — 11,131 Pi + P2 E 2 — 1-1P2 ay = P2 — PP1 •• .(i) — P1 — P2 — 400 N/mm2 2 0 _ 450 EVEy = µP2—µp1 (1 —1,1,2) P1 = E(ex + 1.tev) o p — n p1 + P2 +Pi — P2 cos90° 2 2 = 0+400 x 0=0 P1 E — (1— µ2 ) [ex +"Y] 03 Thin and Thick Shells CHAPTER 0.3.1 Derive a formula for increase in volume of a thin metallic sphere when it is subjected to an internal pressure p. A thin spherical shell of copper has a diameter of 400 mm and a wall thickness of 2 mm and just full of water at atmospheric pressure. Calculate the volume of water pumped in to raise the inside pressure to 1.5 N/mm2. The modulus of elasticity of copper is 1 x 105 N/mm2. K (Bulk modulus) is 2.5 x 103 N/mm2 and Poisson's ratio v is 0.25) [CSE-Mains, 2000, ME : 20 Marks] Solution: (a) When a thin spherical shell is subjected to internal pressure p, circumferential stress, sac is developed in wall of the shell as shown in figure 3.1. pD a= 4t e where p = pressure, D = diameter, t = wall thickness, Ec = circumferential strain a, va, pD _ 4tE(1- v) ec - E E Volumetric strain, Ev = 3E, = 3pD (1 v) 4tE (Neglecting the effect of p) V, volume of shell - 703 6 3pD 703 8V, change in volume = evV = — (1 v) x 4tE 6 (b) D = 400 mm t = 2 mm p = 1.5 N/mm2 a_ EV pD 1.5 x 400 = = 75 N/mm 2 4x2 4t pD 3 x 75 (1 0.25) = (1-v) E = 3 -4 Fig. 3.1 40 0. IAS & IFS (Objective & Conventional) Previous Solved Questions 225 x 0.75 MADE EASY = 1.6875 x 10-3 1x105 V = Volume of shell - it133 TE X 40 03 = 33.51x 106 mm3 6 6 Expansion in volume of shell = ev x V= 1.6875 x 10-3 x 33.51 x 106 = 56.548 x 103 mm3 Decreases in volume of water = k xV = 1.5 , (33.51 x 106 ) 2.5 x 10' = 20.106 x 103 mm3 Volume of water pumped in = 56.548 x 103 + 20.106 x 103 mm3 = 76.654 x 103 mm3 = 76.654 cc Q.3.2 A thin cylinder 150 mm internal diameter and 2.5 mm thick has its ends closed by rigid plates and is then filled with water under pressure. When an axial pull of 37 kN is applied to the ends, water pressure is observed to fall by 0.1 N/mm2. Determine the value of Poisson's ratio. Assume E. 140000 N/mm2, K for water = 2200 N/mm2. [CSE-Mains, 2008, ME : 20 Marks] Solution: Op = 0.1 N/mm2 Reduction in volumetric strain of cylinder = Reduction in volumetric strain of water = x 4tE (5 4v) Op 0.1 K 2200 Axial force = 37000 N 37000 Axial stress - 37000 701- x 150 x 2.5 a (1- 2v) = Increase in volumetric strain due to a = T 31.406 E = 31.406 N/mm2 x (1- 2v) 0.1x or or 0.1 31.406 150(5-4v)+ (1 2v) 4 x 2.5E 2200 = E 1.5(5 - 4v) + 0.1x 140000 2200 31.406 (1 - 2v) 7.5 - 6v + 6.3636 = 31.406 - 62.812v 56.812v = 31.406 - 6.3636 - 7.5 = 17.5424 or Poisson's ratio, v - 17.5424 56.812 = 0.308 Q.3.3 A tyre is shrunk on a wheel of 12 meter diameter. Assuming the wheel to be rigid, calculate the internal diameter of the tyre if after shrinking, hoop stress in the tyre is 1200 kgf/cm2, a for the tyre = 11.7 x 10-6/°C and E = 2 x 106 kgf/cm2. Find the least temperature to which the tyre must be heated above that of the wheel before it could be fitted. [CSE-Mains, 1986, ME : 30 Marks] Thin and Thick Shells Strength of Materials Solution: Hoop stress, csc = 1200 kgf/cm2, E = 2 x 106 kgf/cm2, Hoop strain, E = 41 a = 11.7 x 10-6 /°C D= 12m 1200 _ a, = 600 x 10-6 = 0.6 x 10-3 E 2 x 106 = A Ta = AT x 11.7x 10-6 = 0.6 x 10-3 AT - 0.6 x 1000 = 51.28°C 11.7 Q.3.4 Distinguish between thin and thick cylinders. A thin cylindrical shell having hemispherical ends is subjected to internal pressure p. The internal diameter is 'd and thickness of cylinder and hemisphere are t1 and t2 respectively. Assuming Poisson's ratio, v = 0.3, prove that. 1. 2. t2 7 For no distortion at junction, -6- = 17 For equal maximum hoop stress in cylinder and hemisphere, ratio it = 0.5 . ti [CSE-Mains, 2002, ME : 20 marks] Solution: D In a thick shell or a thin shell, D is internal diameter and t is wall thickness, if — > 20 , then it is a thin shell. When the shell is subjected to internal pressure, the hoop stress developed in the shell does not vary much across the thickness. If — < 20 , i.e., thickness of shell is considerable in comparison to f diameter, then there is variation of hoop and radial stresses across the thickness of thick shell. Figure 3.2 shows a thin cylindrical shell with hemispherical ends, subjected to internal pressure p. In cylindrical portion: pD a hoop stress = 2t1 Fig. 3.2 pD Axial stress, as = 4t 1 c hoop strain = pD v pD (2 v) 1 E ptDE In hemispherical portion at junction pD a = hoop stress = 4t 2 E, c / hoop strain = PD (1 v) 4t 2E For no distortion Sc' = EC" 4 2 0. lAS & IFS (Objective & Conventional) Previous Solved Questions 4t0 (2 or v) = 4t2 (1— v) E t2 1—V 1— 0.3 t1 2—v 2 — .07 pD Maximum stress in cylindrical portion 0.7 = 7 1.7 17 = 2t1 Maximum stress in hemispherical portion = a MADE EASY, pD a = 4t2 = ac pD = pD 2t1 4t2 t2 = 0.5 tl Q.3.5 A steel sleeve is pressed onto a solid steel shaft which has 5 cm diameter. The radial pressure between shaft and sleeve is 1800 N/cm2 and hoop stress at the inner surface of sleeve is 4500 N/cm2. If an axial compressive load of 50 kN is applied to the shaft, determine change in radial pressure at the interface of shaft and sleeve. Assume, v = 0.3. [IFS 2011, ME : 15 Marks] Solution: Shaft diameter, Radial stress, Hoop stress, Poisson's ratio, Compressive load, Axial compressive stress, Say d = 50 mm a = 18 N/mm2 = p' at inner radius a = 45 N/mm2 v = 0.3 Pc = 50,000 N 50000 4 x , 25.46 N/mm2 lc 50' = additional radial stress developed in shaft due to axial compressive stress of 25.46 N/mm2 CYa = Additional stresses in shaft p,, = radial compressive stress ID" = hoop stress compressive aa" = axial compressive stress additional circumferential strain in shaft cs —p" vp" va," + + E E —p" + 0.3p" + 0.3 x 25.46 —0.7p" + 7.638 Sleeve Additional radial stress = p" at inner surface 45 additional hoop stress = — p" = 2.5pItensile) 18 Thin and Thick Shells Strength of Materials esi, strain (hoop direction) - 2.5p" + 0.3p" E E 4 43 2.8p" E Strain compatibility Es/ = Ecs 7.638 - 0.7p" 2.8p" E - E 7.638 = 2.18 N/mm2 3.5 = Increase in radial pressure p" = Q.3.6 (a) (b) (c) Solution: (a) (b) How will you distinguish between a thin walled and a thick walled pressure vessel. What advantages you obtain by wire winding a thin cylinder? What largest internal pressure can be applied to a cylindrical tank 1.8 meter in diameter and 14 mm wall thickness, if the ultimate tensile strength of steel is 467 MPa and factor of safety of 7 is desired? In a pressure vessel if D is diameter and t is wall thickness and internal pressure is p, then if D/t ratio is greater than 20 then it is termed as thin walled pressure vessel. Hoop stress and axial stresses developed in thin pressure vessel are assumed to be constant along radial thickness of pressure vessel. If Dlt ratio is much less then 20, then it is termed a thick walled pressure vessel. Hoop stress and radial stress developed in thick pressure vessel vary along its thickness. When a thin pressure vessel is subjected to internal pressure, then axial and hoop stresses developed in thin shell are as = - and ac= pD 2t ac = aya Material of the shell is not gainfully used in axial direction. By wire winding, a thin wire under tensile stress crw is wound tightly along the outer circumference of shell introducing initial compressive stress ac' in the hoop direction in shell. When the thin cylinder with wire wound on it is subjected to internal pressure then hoop stress developed in shell due to internal pressure is ac" resultant hoop stress will be ad, = 0:- ac'. Pressure bearing capacity of the cylinder is increased by wire winding. (c) aut = 467 MPa FOS = 7 Cr a I = allowable stress = = pD 2t 467 7 = 66.71 N/mm 2 , hoop stress D= 1800 mm, t = 14 mm P- 2taa 2x14 x66.71 . D 1800 = 1.038 MPa, maximum internal pressure 44 1. MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions Q.3.7 A compound cylinder is formed by shrinking a tube 16 cm external diameter and 12 cm internal diameter on to another which has an internal diameter of 8 cm. If after shrinking the radial compressive stress at common surface is 300 kg/cm2, find the circumferential stress at the inner and outer surfaces and at the common surface. [CSE-Mains, 1987, ME : 30 Marks] Solution: Inner cylinder p' = 300 kg/cm2 R2 = 8 cm Ri = 4 cm (junction pressure) R3 = 6 cm (junction) 2pRE acRi -Rf = —2 x 300 acR3 = 6 1080 kgf/cm2 x36 -16= 28 cm Compass cylinder RE +R-f P x R2 3 — R21 . -300 x Fig. 3.3 36 +16 = 780 kg/cm2 36 -16 + 1071.43 kgf/cm2 Outer cylinder + 771.43 kgf/cm2 RE +RE a cR3 +13' X 2 2 R2 - R3 . +300 x 64 + 36 = +1071.43 kgf/cm2 64 - 36 —1080 kgf/cm2 2R 6 cR2 = x De ,2 Dp2 112 - 13 Hoop Stress distribution Fig. 3.4 = +300 x 2 x 36 = 771.43 kgf/cm2 64 - 36 Q.3.8 On the outer surface of a closed thick cylinder of diameter ratio 2.5, were fixed strain gauges to measure the longitudinal and circumferential strains. At an internal pressure of 230 MN/m2 the strains were recorded as 91.8 x 10-6, and 369 x 10-6 respectively. Determine the value of Young's modulus and, modulus of rigidity and Poisson's ratio. [CSE-Mains, 1993, ME : 20 Marks] Solution: R2 Ri = 2.5 p = 230 N/mm2 Outer surface Hoop stress, ac 2/312 2 x 6.25 P X 2 - 2 = 230 x 6.25 -1 = +547.62 N/mm2 Thin and Thick Shells Strength of Materials Axial stress, aa = +PX 4 45 2 = 273.81 N/mm2 2 R2 — = 0, radial stress is zero at outer surface E strain = c axial strain — 547.62 v x 273.81 273.81 = (2 v) E E 273.81 v x 547.62 273.81 (1 2v) E E E 273.81 (2 v) = 369 x 10-6 E 273.81 (1 2v) = 91.8 x 10-6 E or 2—v 369 = 4.0196 1— 2v 91.8 4.0196-4.0196 x 2v = 2 —v 2.0196 = 7.0392v v = 0.287 E— Modulus of rigidity, 273.81(2 — 0.287) 369 E G= 2(1+v) x 10 6 = 1.271 x 106 N/mm2 1.271 X 1 06 4.94 x 105 N/mm2 2 x1.287 = Q.3.9 Find the ratio of thickness of internal diameter of a thick tube subjected to internal pressure when the pressure is 5/8 of the value of maximum permissible circumferential stress. Find the increase in internal diameter of such a tube of 100 mm internal diameter, when the internal pressure is 100 MN/m2, E. 200 x 109 N/m2, Poisson's ratio = 0.286. [CSE-Mains, 1992, ME : 20 Marks] Solution: Say, R1 = Internal radius, R2 = External radius, p= Internal pressure R2 R 22 + 1 ac max = P — 2 132 —1312 But, 5 P = 8 6 cmax R 2 +R12 5 —XpX 22 — p R2 —131 2 or 5F2 + 513 = 8/3 — 8R1 3F2 = 13131 R2 = 2.08 R1 R1 + t = 2.08 R1 t = 1.08 R1 = Di = 0.54D1 46 0. MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions = 0.54 = 50 mm Now, p = 100 N/mm2 R2 = 104 mm ac max = 50 x 13316 1042 + 502 _ 50 x 10816+2500 = 50x 10816 - 2500 8316 1042 - 50' = 1.6 x 50 = 80 N/mm2 Axial stress, as = p x R2 502 - 50 x 2 R2 1042 - 502 "2 "1 = 50 x 502 = 15.03 N/mm2 8.316 Principal stresses at inner radius ac max = +80, a,. = 15.03 N/mm2 p = -50 N/mm2 Strain, 80 Ec 0.286 x15.03 0.286 x 50 E 80 +10 E 90 = 4.5 x10-4 200,000 Change in internal diameter SDI = 4.5 x 10-4 x 100 = 0.045 mm Q.3.10 A compound cylinder is made by shrinking a jacket with outer diameter of 20 cm on a hollow cylinder with diameters 10 and 15 cm. When the compound cylinder is subjected to an internal pressure of 350 kgf/cm2, the maximum circumferential stress in both the cylinder is the same. Calculate the maximum stress developed and internal diameter of the jacket. Take the value of E = 2 x 106 kg/cm2. [CSE-Mains, 1988, ME : 30 Marks] Solution: Inner radius, Junction radius, Outer radius, Internal pressure, Inner cylinder, R1 = 5 cm R3 = 7.5 cm R2 = 10 CM p = 350 kgf/cm2 + acRi = 350 x R2 1=6 -13-F 100+25 100 - 25 = 583.33 kgf/cm2 = 350x Outer cylinder, acR3 = Px Fig. 3.5 FifF? +RfFIE .= 52 X 102 ± 52 X 7.52 350 x 02 (R 2 R2) 7.5 2 (10 2 -5 2 ) /13 2 - 1 / Thin and Thick Shells Strength of Materials 4 47 2500 +1406.25 3906.25 = 350 x 56.25(100 - 25) 75 x 56.25 = 324.074 kgf/cm2 = 350 x Junction pressure p', aCRi (inner cylinder) = -p' x 2R2 , 2 x7.5 2 3 = p x 56.25 - 25 RE -Fq acR, = -p' x 3.6 +Fq , 102 + 7.52 -+P x 2 acR3 = +13' X R22-R32 10 - 7.5 2 6CR3 = p' X 156.25 = +p' x 3.5714 43.75 Finally 583.33 - 3.6p' = 324.074 + 3.5714p' _ 259.256 amax = 36.15 kgf/cm2 7.1714 = 583.33 - 3.6 x 36.15 = 583.33 - 130.145 = 453.185 kgf/cm2 Shrinkage allowance D, -=[3.6p'+ 3.5714p1 8D3 = E 7.1714x 36.15 7.5x 2 x 10' = 1.944 x 10-3 cm D3, inner diameter of jacket = 15 -1.944 x 10-3 = 14.998 cm 0.3.11 A compound cylinder is made by shrinking an outer tube of outside diameter 200 mm and inside diameter 150 mm onto an inner tube, internal diameter 100 mm with a radial interference of 0.2 mm. Both the tubes are made of steel with elastic modulus E = 2 x 106 kg/cm2 and Poisson's ratio v = 0.3. Calculate the value of pressure at the interface and values of hoop stress in the two tubes at the interface. Work from the first principle assuming the basic Lame's equation. ar = A- r2 ae A+ 2 r [CSE-Mains, 1999, ME 30 Marks] Solution: = 5 cm, R2 = 10 CM, R3 = 7.5 cm, 8R3 = 0.02 cm 8R3 - p'R3 [R5 + R3 R3 +Rf] E + R2 -RE 14 -13? 48 ► IAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY 102 +7.52 7.52 +52 0.02 - P' x 7.5 + 52 2 x 106 102 - 7.52 7.52 5333.3 _ ,[156.25 81.25] + P 43.75 31.25 p'[3.571+ 2.6]= 6.171 p' p' = 864.2 kg/cm2 = 8.64 kg/mm2, Pressure at interface Fig. 3.6 Hoop stress at interface Inner tube Outer tube R3 +Ri2 a' CR3 = , 'IcR3 3 7 52 4. 52 = 864.2 x ' 2 ' 7.5 - 5 2 864 2 x +P X Fq+R = 2246.9 kg/cm2 10 2 +7.52 „ 10` -7.5' = 864.2 x 3.571 = 3086.06 kg/cm2 Objective Questions Q.1 Match List-I (Terms used in thin cylinder stress analysis) with List-II (Mathematical expression) and select the correct answer. A. B. C. D. Q.2 List-I Hoop stress, a Maximum in plane shear stress Longitudinal stress Cylinder thickness List-II (a) pd 2 pd 2t 3. pd 2o 4. (d) pd2 2tE (1+µ) pd2 (2+ it) 4tE Q.3 The percentage change in volume of a thin cylinder under internal pressure having hoop stress = 200 MPa, E. 200 GPa and Poisson's ratio = 0.25 (a) 0.40 (b) 0.30 (c) 0.25 (d) 0.20 [CSE-Prelims, ME : 2002] pd 8t Codes: A (a) 2 (b) 2 (c) 2 (d) 2 (b) [CSE-Prelims, ME :1998] 4t 2. pd2 (2 µ) 4tE (c) 7E-(2+11) pd 1. A thin cylinder of diameter d, thickness t is subjected to an internal pressure' p'. Change in diameter is (where E is the modulus of elasticity andµ is the Poisson's ratio.) BCD 3 1 4 4 3 1 4 1 3 4 1 3 [CSE-Prelims, ME : 1998] Q.4 A thin cylinder shell of mean diameter 750 mm and wall thickness 10 mm has its ends rigidly closed by flat steel plates. The shell is subjected to internal fluid pressure of 10 N/mm2 and an external pressure p1 . If the longitudinal stress in the shell is to be zero, what should be the approximate value of p1? Thin and Thick Shells Strength of Materials (a) 8 N/mm2 (c) 10 N/mm2 Q.5 Q.6 (b) 9 N/mm2 (d) 12 N/rnm2 [CSE-Prelims, ME : 2007] A thin walled water pipe carries water under a pressure of 2 N/mm2 and discharges water into a tank. Diameter of the pipe is 25 mm and thickness is 2.5 mm. What is the longitudinal stress induced in the pipe? (a) 0 (b) 2 N/mm2 (c) 5 N/mm2 (d) 10 N/mm2 [CSE-Prelims, ME : 2007] What is the value of volumetric strain in a thin walled cylindrical vessel with modulus of elasticity E, Poisson's ratio µ and of diameter 'd, wall thickness 't' and subjected to internal fluid pressure p? (a) pd (5 - 4µ) 4tE (c) pd (3 20 2tE (b) pd 3tE pd - 3µ) (d) — (2 tE ) [CSE-Prelims, ME : 2006] Q.7 Q.8 A thin cylindrical shell 500 mm in diameter and 25 mm thick having E. 2.0 x 105 N/mm2 and = 0.25, is subjected to internal fluid pressure of 20 N/mm2. What is unit change in volume of the shell? (a) 4 x 10-3 (b) 2.5 x 10-3 (c) 2 x 10-3 (d) 1.5 x 10-3 [CSE-Prelims, ME : 2008] Which one of the following is the correct expression for hoop stress, if p is internal pressure in a thin walled cylinder of diameter 'd and thickness t? (a) pd (c) pd — 4t Q.9 (b) 49 2. Circumferential stress is twice the magnitude of longitudinal stress. 3. Hoop stress and longitudinal stress remain constant along the vessel thickness. Which of these statements are correct? (a) 1 and 2 only (b) 2 and 3 only (c) 1, 2 and 3 (d) 2, 3 and 4 [CSE-Prelims, ME : 2009] Q.10 In a hollow thick cylinder the radial stress or under an internal pressure 'p' (a) increase from a minimum at the innermost surface to a maximum value at the outermost surface (b) decreases from a maximum at the innermost surface to a minimum value at the outermost surface (c) increase from zero at the innermost surface to a value or = p at the outermost surface (d) decreases from a value ar. = p at the innermost surface to zero at the outermost surface [CSE-Prelims, ME : 2001] Q.11 A seamless pipe with 80 cm diameter carries a fluid under a pressure of 2 N/mm2. If the permissible tensile stress is 100 N/mm2, the minimum required thickness of the pipe is (a) 2 mm (b) 4 mm (c) 8 mm (d) 10 mm [CSE-Prelims, ME : 2003] Q.12 Note: 'id is the internal fluid pressure A thin cylindrical shell is subjected to the loads as shown in figure. The element marked 'A' will be subjected to pd 2t pd 8t [CSE-Prelims, ME : 2009] (d) Consider the following statements in respect of thin cylindrical pressure vessel subjected to internal fluid pressure. 1. State of stress along the vessel thickness is biaxial except on the inside surface. 4 Fig. 3.7 (a) biaxial compressive stresses (b) biaxial tensile stresses 50 0. IAS & IFS (Objective & Conventional) Previous Solved Questions (c) uniaxial compressive stresses (d) tensile stress along the longitudinal direction only [CSE-Prelims, ME : 2004] Q.13 The maximum hoop stress in a thick cylinder under internal pressure would occur at which of the following locations? (a) At the inner surface (b) At the outer surface (c) At mid thickness (d) Between the inner surface and mid thickness [CSE-Prelims, ME : 2008] Q.14 A thin cylindrical pressure pipe with both ends closed has diameter 1000 mm. The pipe is subjected to an internal pressure of 4 N/mm2. The permissible tensile stress in the material is 100 N/mm2. What is the minimum required thickness of the pipe? (a) 5 mm (b) 10 mm (c) 40 mm (d) 20 mm [CSE-Prelims, ME : 2008] MADE EASY pD (b) tE-(1- ") pD2 (d) (c) 4tE (1 21.0 [CSE-Prelims, ME : 2010] Q.16 A thin cylindrical shell is subjected to internal pressure, such that hoop strain is approximately five times the axial strain, what is the material of the shell (depending on Poisson's ratio) (b) Cast iron (a) Steel (c) Aluminium (d) Wrought iron Q.17 A thin cylindrical shell made of steel is subjected to internal pressure such that hoop stress developed in shell is 120 MPa. This shell is subjected to axial compressive stress such that there is no change in the length of the shell. (There are end plates on the shell). If E= 200 GPa, Poisson's ratio is 0.3. What is axial compressive stress applied on cylinder (a) 90 MPa (b) 60 MPa (c) 36 MPa (d) 24 MPa Answers Q.15 What is the change in diameter D of a thin spherical shell of wall thickness t when subjected to an internal fluid pressure p? (E = Young's modulus of la = Poisson's ratio) 2. (a) 3. (d) 4. (c) 5. (c) 7. (c) 8. (b) 9. (c) 10. (d) 12. (b) 13. (a) 14. (d) 15. (c) 17. (d) Explanations 1. (d) od = pD al, Hoop stress -- 2 2t 3. Maximum shear stress, 13) - 4 8t Longitudinal Stress, ± 15 1 - 1 4t Cylinder thickness t , PD -3 2a 2. c xd = d (2 v) 4E (d) c = 200 as = 100 200 - 0.25 x 100 175 100 - 0.25 x 200 50 E° Ea = (a) pd - vpd _ pd E= (2-v) 2tE 4tE 4tE 2ec + ea = e=2x 400 E 0.2% 400 200 x 1000 Thin and Thick Shells Strength of Materials 4. (c) 10 x 750 pi x750 -0 4 x 10 4 x 10 p1 = 10 5. 14. (d) (c) p = 4 N/mm2 D = 1000 mm = 100 = (a) pd Volumetric strain = — (5-4g) 4tE 7. t (c) E - PD (5 - 4v) , as v = 0.025 v 4tE pD x 4 pD 4tE tE 20 x 500 9. pD 11) 4tE ( G ED=Ec xD pd pD 2 (1-v) 4tE (c) 1, 2, 3 are correct statements. 16. (c) = Poisson's ratio 10. (d) ar decreases from maximum value of ar = pat the inner most surface to zero at the outermost surface. 2hoop strain axial strain - 1- 2g or D = 800 m, p = 2 N/mm2 = 100 N/mm2 pD µ= 1 = 0.333 is for aluminium i.e., 0.33 2t pD 2 x 800 = 8 mm = t = 2a, 2 x 100 17. (d) a = 120 MPa as = 60 MPa 12. (b) as Via c E as 1' a, Fig. 3.8 5 2-g= 5- 10 p. 9g=3 11. (c) = - 4 x 1000 = .20 mm 2 x 100 6 (b) Hoop stress = D 2t 15. (c) Thin spherical shell t = wall thickness = Poisson's ratio p = internal pressure D = diameter 25 x 2 x105 = 200 x = 2 x 10-3 8. 51 13. (a) Maximum hoop stress in a thick cylinder under internal pressure would occur at the inner surface. pD 2 x 25 a_ = = 5 N/mm2 a 4t 2.5 x 4 6. 4 as' E 60 - 0.3 x 120 = cr; as = 24 MPa Axial compressive stress. 04 Shear Force and Bending Moment Diagrams CHAPTER Q.1 A beam ABCD, 10 m long is supported at B, 1 m from end A and at C, x is from end D. The beam carries a point load of 10 kN at end A and a UDL of 4 kN/m throughout its length fig. 4.1. Determine the value of x if centre of the beam is the point of contraflexure. Draw the BM diagram. 10 kN 4 kN/m Al B Rc R8 -'---1m (9—x) x -°1 Fig. 4.1 [CSE-Mains, 2004, ME : 30 Marks] Solution: Total load on beam = 10 + 40 = 50 kN Reaction Take moments about B (9 - x)Rc + 10 x 1 = 40 x 4 Rc - 150 9 -x RB = 50 150 450 - 50x -150 300 - 50x 9-x 9-x 9-x Moments about centre of the beam -10 x 5 + R8 x 4 - 4 x 5 x 2.5 = 0 -50+ 4 (300 - 50x) 9 -x 50 = 0 4 (300 - 50x) 9 -x -- 100 300 - 50x = 225 - 25x 75 = 25x x = 3m RB 300 - 50x 150 = 25 kN 9 -x 6 Shear Force and Bending Moment Diagrams Strength of Materials 11 53 BMD MA = 0 MB = -10 x 1 - 4 x 0.5 = -12 kNm M3 =-10 x 3 + 25 x 2-4 x 3 x 1.5 = -30 + 50 - 18 = +2 kNm + 2kN/m M5 =-10 x 5 +25 x 4-5 x 4 x 2.5 = -50 + 100 - 50 = 0 —12 kN/m Mc = M7 = -3 x 4 x 1.5 = -18 kNm — 18 kN/m BMD MD= 0 Fig. 4.2 Q.4.2 On a simply supported beam (10 m span), a concentrated load (10 kN) and a moment of 40 kNm act a section 7 m from one of the ends. Draw the shear force and bending moment diagrams. Indicate the points of contraflexure, if any 10 kN [ICSE-Mains, 2012, ME : 10 Marks] Solution: Fig. shows a beam AB, 10 m long, simply supported at ends carrying a load of 10 kN at Cand moment 40 kNm at RA = —1 kN RB = +11 kN C, 7 m from A Reaction 3 —0Taking moment about A 70 + 40 = 10 RB Reactions, RB = 11 kN — 11 kN (b) Reaction, Ftc = 10 - 11 = -1 kN SF diagrams 1 kN FAC FCB = 1 - 10 = -11 kN Fig. (b) shows SFD BM diagram A MA Mc = -1 x 7 = -7 kNm Mc = -7 + 40 = +33 kNm MB = 0, Fig. (C) shows BMD — 7 kNm BMD Fig. 4.3 Q.4.3 A beam ABCD is loaded as shown in figure 4.4. The beam is of rectangular section 50 mm x 100 mm. 600 N 60 N/m RA = 393.33 N t Rc = 386.67 N 5000 Nmm 1-*-3 m 3m 5m Fig. 4.4 (i) (ii) Sketch the SF and BM diagrams of the beam Determine the maximum bending stress at section B of the beam [CSE-Mains, 2011, ME : 25 Marks] 54 1. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions Solution: Reaction Moment about A 60 x 6 x 4 + 600 x 11 = 5000 + 6/3c (Note that CG of triangular load lies at 4 m from A) 2 720 + 6600 = 5000 + 6/3c 2320 = 6/qc = 386.67 N Total load on beam = 600 + 180 = 780 N RA = 780 - 388.67 = 393.33 N Reaction, SFD FA = +393.33 N F3 = 393.33 30 x 3 = 348.33 N 2 A 386.67 N 600N 393.33 213.33 N (Just left to C) F6 = 393.33-180=213.33N (Just right to C) F6 = 213.33 + 386.67 = 600.00 N • D A Fig. 4.5 BMD MA = 0 (Just left to B) MB = 393.33 x 3 30 x 3 x 1= 1180 - 45 = 1135 Nm 1135 Nm (Just right to B)M3'= 1135 - 5000 = -3865 Nm x M6 = 393.33x6-5000 6026 x2 = 2360 - 5000 - 360 = -3000 Nm MD = Jump moment at B, 5000 Nm bxd2 50x1002 = 8,33 x104 mm3 6 6 M 5000 x 10 3 = 60 N/mm2 = z 8.33 x 104 za max — 3000 Nm — 3865 Nm Fig. 4.6 Q.4.4 Draw the bending moment and shear force diagram of the beam as shown in fig. 4.7(a). 0= Constant. [CSE-Mains, 2012, CE : 15 Marks] Solution: BM diagram (Let us consider this as a combination of to cantilevers AEB and CFD and a simply supported beam BC) Portion BC Carries UDL = 4 kN/m Total load = 4 x 4 = 16 kN Reaction at end = 8 kN x 42 Maximum BM = w/2 = 4 8 = 8 kNm 8 Shear Force and Bending Moment Diagrams Strength of Materials 2 kN A 1 4 55 2 kN 4 kN/m Mt-VW\ 1 C F D --1-1 1 I 2 .(-- 4 m—•-1 2 I 1 [4-(a) 8kN E B (b) BMD —26 —26 kNm 10 kN — 10 kN Fig. 4.7 Cantilever AB MB = ME = -16 kNm MA = -8 x 3 - 2 = -26 kNm BM diagram for CD cantilever will be similar SF diagram, BC, FB = +8 kN, Fcenter = 0 FBE = -8 kN Fa4 = -10 kN SFD for portion CFD portion will be similar to that of portion BEA Figure shows BMD and SFD for the beam AB, BC, CD. Q.4.5 The bending moment diagram of a simply supported beam AB is shown in figure 4.8(a). Sketch the loading, diagram and SF diagram of the beam. [CSE-Mains, 2013, ME : 10 Marks] Solution: BM diagram is of a beam subjected to a building couple Mat C, at a distance of 3 m from A. Length of beam is 5 m. M = 12 + 8 = 20 kNm (anticlockwise) Moment at B, RB x5 = 20 56 P. lAS MADE EASY & IFS (Objective & Conventional) Previous Solved Questions RB = 4 kN1.. RA = 4 kNT (to balance RB) Mc = 4 x 3 = 12 kNm Mc = 12 - 20 = -8 kNm SFD Fig 4.8 SF diagram FAB = +4 kN (constant) Q.4.6 The shear force diagram of a beam is shown in the figure. Draw bending moment and load diagrams. 2P B + P/3 D E 2P/3 2a .1- a•-•-1 Fig. 4.9 [CSE-Mains, 2002, CE : 12 Marks] Solution: Beam is supported at B and D Reaction P RB = 2P — + 7P 3 3 2P 8P RD = 2P + — = 3 3 Shear Force and Bending Moment Diagrams Strength of Materials 1 57 7P 8P Total load on beam = — + — = 5P 3 3 Rate of loading between A and B = 2P sl, a P 2P , Vertical load at C = - + — = P 4, 3 3 Vertical load at E = 2P1 2P 1". a . A.,,rem.„, B H'— a BMD RB P 2P 1, C 11 1 E a -044- 2 a RD (a) A4A = 0 MB= 2P a -- x ax- = -P.a 2 a — Pa — 2Pa 3 — 2Pa ME= ° MD = -2Pa BMD Fig. 4.10 8P 2Pa Mc= -2P x 3a + — x 2a = 3 3 Verification of reaction Moment about B 2P a ax3RD+- xax- = Pa + 8Pa 2 a 3RD = 8P 8 RD = iP Q.4.7 A beam ABCD, 8 m long hinged at end A and roller supported at end D carries transverse loads as shown in the figure 4.11(a). Determine support reactions and draw BM diagram for the beam. Solution: A load at end of the lever equal to 20 kN can be replaced by a load of 20 kN and a CWmoment of 40 kNm at point C. Reaction Taking moments at A 2 x 20 + 20 x 4 + 40 = 8 RD RD = 20 kN RA = 20 + 20 - 20 = 20 kN BMD MA =O MB = +20 x 2 = +40 kN Mc = 20 x 4 - 20 x 2 = 40 kNm Mc = 40 + 40 = 80 kNm 20 kN 2m (a) 20 kN 20 kN Fig. 4.11 58 0. IAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY Q.4.8 Construct the SF and BM diagrams for the beam shown in the figure 4.12. 100 kNm 20 kNm A 50 kN D -id 0.5 1-4-- 3 m 40 kN E F G 1-1.4 1.5 m--1.-1 0.5 1 0.5 RB =45kN RE = 55 kN Fig. 4.12 [CSE-Mains, 2006, ME : 20 Marks] Solution: Taking moments about B 20 x0.5 x 0.25+100+4.5 xRE = 50 x 3 +40 x5 2.5+ 100+4.5 RE = 150+200 0, 350 -102.5 = 55 kN RE 4.5 RB = 20 x 0.5 + 50 + 40 - 55 = 45 kN SF diagram AB Portion = x x20 = 0 at 0 = -10 kN, at x = 0.5 BCD Portion DE Portion EF Portion Fx = -10 + 45 = +35 kN (remains constant) = -10 + 45 - 50 = -15 kN (remains constant are DE) = -10 + 45 - 50 + 55 = +40 kNm + 40 kN Fig. 4.13 + 32.5 kNm BMD MA MB = -20 x 0.5 x 0.25 = -2.5 kNm Mc = -20 x 0.5 x 1.25 + 45 x 1 = +32.5 kNm Mc = 32.5 - 100 = - 67.5 kNm MF = ME = -4 x 0.5 = -20 kNm MD =-40x 2+55 x 1.5 = -80 + 82.5 = +2.5 kNm - 67.5 kNm BMD Fig. 4.14 Shear Force and Bending Moment Diagrams Strength of Materials Q.4.9 A beam ABCD, 6 m long hinged at both the ends A and D, is subjected to moment 6 kNm (CW) at B and 3 kNm (CCW) at C as: shown in figure 4.15(a). Sketch the bending moment diagram of the beam. How many point of contraflexure are then in the beam) [CSE-Mains, 2009, ME : 5 Marks] Solution: Reaction: Moment about A, 6 kNm 4 59 3 kNm AD 0.5 kN t ho— 2m 2m RA = 0.5 kN RD x 6 + 3 = 6 kNm 2m —0-1 (a) +5 kNm RD = 0.5 kN T +4 kNm RA = 0.5 kN BMD MA = + 1 kNm MB = -0.5 x 2 = -1 kNm MB = -1 + 6 = 5 kNm Mc = -0.5 x 4 + 6 = +4 kNm — 1 kNm Mc = 4 - 3 = +1 kNm Mo = There is only one point of contraflexure in the beam. BMD (b) Fig. 4.15 Q.4.10 For the semicircular simply supported member shown in the figure 4.16, find the value of bending moment at a function of 0 and also draw its bending moment diagram. [CSE-Mains, 2013, CE : 15 Marks] Solution: Consider a section xx at angular displacement 8 as shown Taking moment about A, for reaction Px 1.5R = 2R x RB RB = 0.75P and so RA = 0.25P Bending moment, Me = (R - Rcos0) = 0 at x = 0 = 0.0335 PR, at 0 = 30° = 0.07325 PR, at 8 = 45° = 0.125 PR, at 0= 60° = 0.25 PR, at 0 = 90° Portion BC (0 = 0 - 60°) Fig. 4.16 0.375 PR 3P Me = — (R - R cos0)= 0.75PR (1- cos()) At 90° = 0 at 0 = 0 = 0.1005 PR, at 0= 30° = 0.21975 PR, at 0 = 45° = 0.375 PR (at point C), at 0 = 60° M90 = 0.75 P(R)- Px 0.5 R = 0.25 PR 1.5 R 0.5 R-0-1 BMD Fig. 4.17 60 0. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions Objective Questions Q.1 Which one of the following figure represents the correct shear force diagram for the loaded beam shown in the figure 4.18. a I.- 3a a Q.3 The beam is loaded as shown in the fig. 4.19. Select the correct BM diagram 1.4 Fig. 4.19 Fig. 4.18 (a) (a) (b) (b) (c) (d) (c) [CSE-Prelims, ME : 1998] Q.2 A cantilever carrying an uniformly distributed load throughout its length. Select the correct bending moment diagram of the cantilever (d) [CSE-Prelims, ME : 1999] (a) Q.4 If the beam shown in the fig. 4.20 is to have zero bending moment at its middle point, the overhang x should be r (b) w (c) Fig. 4.20 (a) wL2 4P (c) wL2 8P (d) [CSE-Prelims, ME : 1999] (b) wL2 6P wL2 12P [CSE-Prelims, ME : 2000] (d) Shear Force and Bending Moment Diagrams Strength of Materials Q.5 Match List-I with List-II and select the correct answer. List-I A. f Q If the SF diagram for a beam is a triangle with length of the beam as its base, the beam is (a) A cantilever with a concentrated load at its free end (b) A cantilever with UDL over its whole span (c) Simply supported with a concentrated load at its mid point (d) Simply supported with a UDL over its whole span [CSE-Prelims, ME : 2000] Q.7 The bending moment diagram for a simply supported beam is a rectangle over a larger portion of the span except near the supports. What type of load does the beam carry? (a) A uniformly distributed symmetrical load over a larger portion of the span except near the supports (b) A concentrated load at mid span (c) Two identical concentrated loads equidistant from the supports and close to midpoint of the beam (d) Two identical concentrated loads equidistant from the mid span and close to supports [CSE-Prelims, ME : 2007] Q.8 Which one of the following is the correct statement? B. P Q Q t D. F F F F List-II P T 1. dM If for a beam, — dx R 4. p S T T R 5. P Codes: A (a) 4 (b) 1 (c) 1 (d) 4 Q BCD 2 5 3 4 5 3 4 3 5 2 3 5 [CSE-Prelims, ME : 2001] 0 for its whole length, the beam is a cantilever (a) free from any load (b) subjected to a concentrated load at its free end (c) subjected to an end moment (d) subjected to a UDL over its whole span [CSE-Prelims, ME : 2007] Q S 61 Q.6 tT SI 41 Q.9 Which one of the following is the correct answer? The point of contra flexure in a beam is a point on its length where (a) the shear force is zero (b) the bending moment is maximum (c) the bending moment changes its algebraic sign and is zero at that point (d) the shear force changes its algebraic sign [CSE-Prelims, ME : 2006] 62 I,. lAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY Q.10 A overhanging beam AB has simple supports at D and E as shown in the figure below. It carries concentrated loads P at its free ends and a uniformly distributed load of intensity w (N/m) over its length DE. If bending moment at the middle point C of the beam is to be zero, then what is the value of P? P (N) P (N) Fig. 4.22 (b) 4(2W+ w) (a) 8(W + 2 J (c) 212W + — w 2 11 [CSE-Prelims, ME : 2008] Fig. 4.21 (b) (d) 4(W + ;) w/ Q.14 A cantilever beam is subjected to moments as shown in the given figure. The BM diagram for the beam will be w/ (d) 16 30 kNm 50 kNm [CSE-Prelims, ME : 2006] Q.11 For a cantilever dM = a , constant for its whole dx length. What is the shape of the SF diagram for the beam (a) rectangle (b) triangle (c) a parabola (d) a hyperbola [CSE-Prelims, ME : 2008] Q.12 SF diagram for a simply supported beam is a rectangle with its longer side equal to beam length. What type of load is acting on the beam? (a) Concentrated load at its mid span (b) UDL over its whole span (c) Concentrated load alongwith a couple at a point on beam length (d) Couple at a point on the beam length [CSE-Prelims, ME : 2008] Q.13 A beam AB of length 1 and weight wsupports a load Wat its free end Band is hinged at its end A. A wire CD supports the beam at the mid point C making an angle of 30° with the horizontal. What is the value of tension T in the wire CD? 2m 2m Fig. 4.23 A (a) 20 kNm 50 kNm 50 kNm (b) 80 kNm A (c) (d) 20 kNm 50 kNm 50 kNm 80 kNm [CSE-Prelims, CE : 2001] Q.15 A cantilever is loaded as shown in the below figure. The bending moment along the length is p Q.18 Consider the following statements in regard to the shear force diagram for an overhanging beam supported at A and C a P Fig. 4.24 (a) (b) (c) (d) 6 kN 2 kN uniform uniformly varying zero concentrated at the free end A 1-4-- 2m Q.17 A cantilever AB carries loading as shown in the figure. The BM diagram for the beam is 4 kNm 1m M Fig. 4.25 6kN-m (a) 4kN-rn 6kN-m (-) (b) B 2 kN 2m I.- .1 C 2 kN m 3 kN Q.16 Consider the following statements: 1. Point of contraflexure is the point where the bending moment is maximum 2. Point of contraflexure is the point where the bending moment changes sign 3. Point of contraflexure is the point where the shear force is zero Which of these statements is/are correct? (a) 1, 2 and 3 (b) 2 and 3 (c) 2 only (d) 1 only [CSE-Prelims, CE : 2002] F- 63 Shear Force and Bending Moment Diagrams Strength of Materials 4kN-m 7 kN Fig. 4.26 1. The beam is carrying a uniformly distributed load of 2 kN/m throughout 2. The beam is carrying a uniformly distributed load of 2 kN/m over the supported span AC and concentrated load of 2 kN at the free end B 3. The beam is carrying a uniformly distributed load of 2 kN/m over the supported span AC, and concentrated load of 5 kN at the centre of supported span BC and also a point load of 2 kN at the free end B. 4. The points of contraflexure occurs between the supported region AC and nearer to support C. Which of the above statements is/are correct? (a) 1, 2, 3 and 4 (b) 4 only (c) 2 and 3 (d) 3 and 4 [CSE-Prelims, CE : 2004] Q.19 A propped cantilever beam shown in the figure given above is having internal hinge at its mid span. Which one of the following is the shape of bending moment diagram for the given loading? eWslY=NPWY-1e- 6kN-m l (c) Fig. 4.27 4kN-m (d) 4kN-m 6kN-m 4kN-m [CSE-Prelims, CE : 2003] Parabolic (a) A B 64 ► IAS & IFS (Objective & Conventional) Previous Solved Questions Q.21 A cantilever beam of span 1 carries a uniformly varying load of zero intensity at the free end and w per metre length at the fixed end. What does the integration of the ordinate of the load diagram between the limits of free and fixed ends of the beam give? (a) Bending moment at the fixed end (b) Shear force at the fixed end (c) Bending moment at the free end (d) Shear force at the free end [CSE-Prelims, CE : 2006] Straight line (b) A Parabolic (c) — A (d) MADE EASY A [CSE-Prelims, CE : 2005] Q.20 A simply supported beam AB is loaded as shown in the figure given below. CDE is a rigid member. A load 4 kN is applied at E. Which one of the following is the SFD for the beam? Q.22 In the cantilever beam shown below, what is the percentage of bending moment at 1/2 with respect to the maximum bending moment at the fixed support? w/unit length C1.4— 1/2 4 kN D .1 E 0.5m C (a) 15% (c) 25% t k0.5 m 2m —141 2m Fig. 4.28 2 kN (a) 2 kN (b) 1.5 kN 2.5 kN (c) 2.5 kN 1.5 kN 1 kN (d) 3 kN [CSE-Prelims, CE : 2005] Fig. 4.29 (b) 20% (d) 30% [CSE-Prelims, CE : 2006] Q.23 Match List-I with List-II and select the correct answer using the code given below the lists: List-I A. Cantilever beam B. Overhanging beam C. Fixed beam D. Simply supported beam List-II 1. At the supports, shear force exists, BM is zero 2. Deflection is zero, but BM exists at the supports 3. Shear force and BM exist at the supports 4. BM exists, deflection is zero at the supports Codes: A BCD (a) 4 1 2 3 (b) 2 3 4 1 (c) 4 3 2 1 (d) 2 1 4 3 [CSE-Prelims, CE : 2006] Shear Force and Bending Moment Diagrams Strength of Materials Q.24 What is the maximum bending moment at mid span of the beam given below? r W (Rate of loading) Codes: A (a) 3 (b) 4 (c) 3 (d) 4 B 2 1 1 2 65 C 4 3 4 3 [CSE-Prelims, CE : 2007] Q.26 The figure shows the shear force diagram for an overhanging beam ACDB. 55 kN Fig. 4.30 (a) wi2 2 (b) w/2 (c) (d) 12 1/4/12 25 kN 3 m —0,40— 3 m 6 Wi t Fig. 4.31 [CSE-Prelims, CE : 2007] Q.25 Match List-I (Loaded Beam) with List-II (Maximum bending moment) and select the correct answer using the code given below the lists: List-I r w/m A. L H w/m Consider the following statements with respect to the above beam: 1. The beam has supports at A and D 2. The beam carries a concentrated load at C of 25 kN 3. Bending moment at D is 15 kNm 4. The beam carries a uniformly distributed load of 80 kN over the portion AC. Which of the statements given above is/are are correct? (a) 1, 2 and 4 (b) 1 only (c) 2, 3 and 4 (d) 1 and 4 only [CSE-Prelims, CE : 2009] Q.27 A simple supported beam AB of span 2 m is loaded as shown in the figure. Which one of the following pairs corresponds to SFD and BMD for the beam? B. f L 1 kN-m 3 kN-m G C. E) 2m 'Alm Fig. 4.32 List-II 1. wL2 8 2. wL2 12 3. wL2 2 4. wL2 6 2 kN (+) 2 kN 2 kN-m (—) 3 kN-m (a) 66 0. MADE EASY 1AS & IFS (Objective & Conventional) Previous Solved Questions 1 kN (-) 1 kN 1 kN-m (-) 2 kN-m 3 kN-m 2 kN (b) (d) 2 kN-m (-) 1 kN-m [CSE-Prelims, CE : 2010] (-) 1 kN 1 kN Answers 1. (a) (c) 3 kN-m (-) 1 kN-rn 2. (c) 3. (d) 4. (c) 5. (d) 6. (b) 7. (d) 8. (c) 9. (c) 10. (c) 11. (a) 12. (d) 13. (d) 14. (a) 15. (a) 16. (c) 17. (b) 18. (d) 19. (d) 20. (a) 21. (b) 22. (c) 23. (c) 24. (c) 25. (c) 26. (d) 27. (c) Explanations 1. (a) Reactions = Weach. 4. (c) C IW wt H.- a -1. L/2-0-1-4- LJ2 Fig. 4.37 tw 3a -I- a wic - SFD - Fig. 4.34 2. (c) x= 3. 2M x 2 wL2 ) (p 4. 2 )2 2x4 PL wL2 wL2 - Px +-+ 2 4 8 -Px + -w Fig. 4.35 -PI wL2 8 0 wL2 8P 5. (d) ABCD 4 2 3 5 6. (b) (d) BMD Fig. 4.36 SFD Fig. 4.38 Shear Force and Bending Moment Diagrams Strength of Materials 7. 41 67 12. (d) (d) M a .1. (L — 2a) M L SFD Fig. 4.39 8. Fig. 4.43 (c) 13. (d) dM =0 dx Fig. 4.40 dM 0 end moment M , dx 9. (c) At point of contraflexure, bending moment changes its algebraic sign and is zero at that point. 14. (a) 10. (c) 30kNm cvmeme_v_w, 1 50 kNm L L 2 Fig. 4.41 wL) L wL2 - P x 1.5L +(P+— - 2 2 8 - 1.5PL+ L + wL2 - wL2 2 4 8 PL = A 0 20 50 kNm 30 = 0 wL2 8 WL P= 2 Fig. 4.45 15. (a) Uniform 2 kN 8 4 kNm 11. (a) dM dx = a constant = SF is constant Ii M .1. 1m (-) 4 kNm BMD SFD Fig. 4.42 6 kNm Fig. 4.46 68 0. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions 16. (c) 2 only 2. Point of contraflexure is the point where bending moment changes sign. 20. (a) 4 kN DrIE 0.5 m A 17. (b) 11114111 0.5 m RA = 2 kNt RB = 2 kN 2m Fig. 4.50 (-) 6 kNm R8= 4 kNm 10 = 2.5 kA RA = 1.5 kN BMD z Fig. 4.47 18. (d) 3 and 4 are correct statements (+) ko-- 1.5 m SFD 5 kN 2 kN Fig. 4.51 21. (b) SF at fixed end. A 6 kN }-0— 2 m --144— 2 m —1±1- 1 m-0-1 -2 kNm BMD SFD Fig. 4.48 19. (d) Parabolic Fig. 4.55 22. (c) 25% I-0— 2 m 2m Fig. 4.56 MA — BMD Fig. 4.49 M, - w/ 2 2 wit 8 Strength of Materials 23. (c) Shear Force and Bending Moment Diagrams 4 69 25. (c) List-I A. Cantilever beam B. Overhanging beam C. Fixed beam D. Simply supported beam List-II 4. BM exists, deflection is zero at the supports 3. Shear force and BM exist at the supports 2. Deflection is zero, but BM exists at the supports 1. At the supports, shear force exists, BM is zero A. 3. B. 1. C. 4. w/ 2 2 w/ 2 8 w/ 2 6 26. (d) 1. Beam has supports at A and D. 4. Beam caries a UDL of 80 kN over portion AC. 27. (c) 3 kNm 1 kNm 24. (c) RA = 1 kN /4 1 kNm wit = 12 (-) , 4 RB = 1 kN 1 kNm SFD Fig. 4.57 wi i wi i wit wg c 4 X 2 4 x 6 8 24 2m 1 kNm (-) 3 kNm BMD Fig. 4.58 NM MI 05 Theory of Simple Bending CHAPTER Q.5.1 State the restrictions or assumptions made in deriving the formula for theory of bending M a I y E R [CSE-Mains, 1998, ME : 15 Marks] Solution: Following assumptions are taken while deriving the flexure formula: 1. The material of the beam is homogeneous and isotropic. 2. Young's modulus of the material in tension = Young's modulus in compression. 3. Material is considered as consisting of layers. 4. Each layer is independent to extend or contract irrespective of the layers above or below it. 5. Beam is initially straight. 6. Elastic limit of the material i.e., ae is not exceeded. 7. Beam is symmetrical about plane of bending. 8. Transverse sections which are plane before bending remain plane after bending. Q.5.2 A wire of diameter d is wound round a cylinder of diameter D. Determine the bending stress produced in the cross-section of the wire. Hence or other wise find the minimum radius to which a 1 cm diameter circular rod of high tensile steel can be bent without undergoing permanent deformation. Take yield stress = 17000 kg/cm2 and E. 2 x 106 kg/cm2. What is the magnitude of BM necessary for this? [CSE-Mains,1991, ME : 20 Marks] Solution: Wire diameter = d R= Drum radius, 2 Bending stress, ab _ E _ 2E d/2 R D b 20 b 2E d Gb E-5 Bending stress in wire 71 Theory of Simple Bending Strength of Materials E= 2 x 106 ab = 17000 kg/cm2 d= 1 cm Now, = Ed ab 2x106 x1 =117.647 cm 17000 It X13 1°3- =17000 x M = ab Z = b X 32 32 M = 1668.97 kg-cm Q.5.3 Compare the moments of resistance for a given maximum bending stress of a beam of square section placed (i) with two sides vertical and (ii) with a diagonal vertical. The bending in each case is parallel to vertical plane. [CSE-Mains, 2012, CE : 12 Marks] Solution: M = Zab Section-I (with two sides vertical) a3 -1 z1 = 6 Section-II (with a diagonal vertical) a 4 2 a3 ,___ Z2 12 X ,/2a 6J Say ab = Maximum bending stress a3 M1 = bX 6 a3 M2 = ab x 6,//2 a —h II M2 = =1.414 Fig. 5.1 Q.5.4 Compare the bending strengths of three beams of same material, same weight and same depth, if one of them has solid rectangular area 6 x 20 cm, second beam is a hollow rectangular section having a wall thickness of 2 cm. The third beam has I-section of equal flanges have web and flange thickness equal to 2 cm. [IFS 2011, ME : 10 Marks] Solution: For solid rectangular area 6 x 20 cm 6 x 202 = 400 cm3 6 Hollow rectangular section (assume width 8) 208 - (8- 4) x 16 = 6 x 20 = 120 cm2 (area same) 20B- 16B + 64 = 120 4B = 56 = 14 cm Z1 = 4- [14x203 10x163 ] 10 12 12 1 1 = — x — [112000 - 40960] = 592 cm3 10 12 [-4-- 6 mm - 20 mm Fig. 5.2 ...(ii) 7 2 a. MADE EASY 1AS & IFS (Objective & Conventional) Previous Solved Questions I B 2 cm I I 20 cm 16 cm 20 cm 2 cm B TT Fig. 5.4 Fig. 5.3 I-section 28 + 2B + 16 x 2 = 120 cm2 4B= 88 = 22 cm - = 63] 203 = [176000 - 81920] = 784 120 10 [2212 2012 Beam strength is proportional to section modulus Zi : Z2 : Z3 = 400: 592: 784 = 1 : 1.48 : 1.96 Q.5.5 Compute second moment of area of the plane lamina shown in the figure 5.5 about an axis parallel to the base and passing through the centroid. [CSE-Mains, 2012, ME : 20 Marks] Solution: Area, Al = 22 x 16 = 352 cm2 A2 — 22 x 9 = 99 cm2 2 A3 = TcR2 n x 82 = = 100.531 cm2 2 2 [4-16 cm --+-- 9 cm Fig. 5.5 4R 4 x 8 = = 3.395 cm (from diameter) 3n 3 x7c Location of centroid (from bottom) 7 for semi circle _ 99 x22 +100.531(22+3.395) 3 352+99+100.531 16 x 22 x11+ 111 3872 + 726 + 25529.85 551.531 7150.985 - 12.966 cm 551.531 Moment of inertia of semi circular section about its own centroidal axis. irR 4 1x/x = itR2 (412 8 2 37r Theory of Simple Bending Strength of Materials 73 nR 4 nR 2 16R2 ER 2 8R 4 x 8 2 9112 8 911 = R4 [0.3927 - 0.2829] = 0.1098 R4 = 0.1098 x 84 = 449.741 cm4 1182 2 - 100.531 cm2 22 2 16 x 223 x 223 2 +352(12.966 11) + 36 + 99112.966 - -3-/xx = 12 ) +449.741 + 25.1327 (22 + 3.395 - 12.966)2 = 14197.333 + 1360.535 + 2662 + 3141.338 + 449.741+ 15530.03 = 37340.98 cm4 = 3.7341 x 10 cm4 Q.5.6 The shear force diagram for rectangular cross-section beam is shown in the figure 5.6. Width of the beam in 100 mm and depth is 250 mm, determine the maximum bending stress in the beam. [CSE-Mains, 1995, ME : 30 Marks] Solution: SF diagram is only on one side. This is the case of a cantilever 3 m long AB, fixed at A and free at end B carrying UDL of 5 kN/m over BC= 1 m, a point load of 5 kN at Cand a point load of 5 kN at D. Loading diagram is shown in the figure as above. Maximum bending moment occurs at fixed end A MA = 5(2.5) + 5 x 2 + 5 x 1 = 12.5 + 10 + 5 = 27.5 kNm = 27.5 x 106 Nmm = ob x Z Z = Section modulus 5 kN 5 kN T 5 kN 5 kN LW-4/ 1-0-1 m -1+-1 m -44-1 m Fig. 5.6 2 100 x 2502 - bd = 1.04166 x106 mm3 6 6 27.5 x106 = 26.4 N/mm 2 (Max. bending strtess in beam) as 1.04166 x 10' Q.5.7 A cantilever beam of cross-section 50 mm x 100 mm in subjected to a compressive force parallel to the longitudinal axis of the beam at the free end. The point of application of the load is at 10 mm above the bottom surface of the vertical centroidal axis. Span of the beam is 4 m. Calculate the maximum stress in the beam. [CSE-Mains, 2012, CE : 15 Marks] Solution: b = 50 mm, d = 100 mm Eccentricity of loading, bd 3 = 50 x 1003 = 4.1667 x106 mm4 12 12 e= 50 - 10 = 40 mm 74 ► MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions Pin Newton, A —0-150 50 Bending moment M = P x 40 Nmm 8 P 10 mm 4° 4m Maximum bending stress, My / = + 50 x 40P Fig. 5.7 4.1667 x 106 = +480 x 106 P = ± 4.8 x 10-4 P NImm2 Direct shear, = 2 x 10-4 P N/mm 2 (Comp.) a = area 50 x 100 d amax at edge B = —(4.8 + 2) x 10-4 P. —6.8 x 10-4 P N/mm2 (Comp.) 6max at edge A = +4.8 x 10-4 — 2 x 10-4 = 2.8 x 10-4 P N/mm2 (Tensile) Q.5.8 Compare the flexure strength of the following three beams of equal weight per unit length. I-section 300 mm x 150 mm with flanges 20 mm thick and web 12 mm thick. 1. Rectangular section having depth twice the width 2. 3. Solid circular section [CSE-Mains, 2012, ME : 15 Marks] Solution: Having same material and equal weight per unit length means same area of cross-section for all three beams. 8 20 I-section Area of cross-sections = 2 x 150 x 20 + 260 x 12 = 6000 + 3120 = 9120 mm2 260 150 x 3003 138 x 2603 zr 12 20 12 337.5 x 106 — 202.124 x 106 = 135.376 x 106 mm4 135.376 106 150 —150 --xd Fig. 5.8 = 0.9025 x 106 mm3 Rectangular section Area = 282 = 9120 (same area of cross-section) B = 67.53 mm D = 135.055 m 4 BD2 67.53 x 135.0552 6 6 = = 0.20529 x 106 mm3 Circular section Fig. 5.9 Tcd2 4 Diameter, D = 28 — 9120 d = 107.758 mm nd3 Z3 x (107.7583 ) — — 32 0.122845 x 106 mm4 32 : Z2 : Z3 = 0.9025: 0.20529: 0.122845 = 7.346 : 1.671 : 1 Theory of Simple Bending Strength of Materials 4 75 Q.5.9 A beam is of square section with diagonals 60 mm long, vertical and horizontal, as shown in figure 5.10. Shear force at a particular section is 5 kN. What is the shear stress at layer AB. [CSE-Mains, 2009, ME : 15 Marks] Solution: Square section Diagonal, d = 60 mm 60 mm d4 60 4 = 27 x 104 mm4 = 48 48 '-J Fig. 5.10 Layer, AB, breadth = 30 mm = b = d/2 1 OAB = - x 30 x 15 = 225 mm2 2 Area, y 15 = 15+ — = 20 mm 3 ay = 225 x 20 = 4500 mm3 F= 5 kN = 5000 N TAB = Fay lb 5000 x 4500 27 x104 x 30 = 2.778 N/mm2 Q.5.10 Two long wooden planks form a T-section of a beam as shown in figure 5.11. If this beam transmits a constant vertical shear of 3000 N, find the necessary spacing of the nails between the two planks to make the beam act as a unit. Assume that allowable shear force per nail is 700 N. Solution: Location of G of section 200 50 r•-• # a a co 4G x 250 = 162.5 200 x 50 x100+200 x50 x 225 2 x200 x 50 = 50 + 112.5 = 162.5 mm y2 = 250 - 162.5 = 87.5 mm I Fig. 5.11 Moment of inertia, I 2 200 x 503 50 x 2003 + 50 x 200(162.5 100) + +200 x 50 (87.5 - 25)2 12 12 = 33.33x 106 + 78.125 x 106 + 2.083 x 106 = 113.538x 106 mm4 Shear stress along a-a, nail section = ti Fay aa = Ib b= 50 mm, I =113.538 x 106 mm4, F= 3000 N ay about neutral axis = 200 x 50 x (87.5 - 25) = 200 x 50 x 62.5 = 6.25 x 105 mm3 3000 x6.25 x10 5 x 50 - 0.33 N/mm2 as 113.538 x106 Allowable shear force = 700 N 76 ► MADE EASY 1AS & IFS (Objective & Conventional) Previous Solved Questions Spacing between nails = S mm (say) 700 = S x 0.33 x 50 S, spacing = 42.42 mm Nails are fixed at spacing of 40 mm Q.5.11 An I beam with the following dimensions is subjected to a shearing force of 20 kN. Flange: breadth = 50 mm, web thickness = 3.5 mm Thickness = 5.5 mm, web depth = 189 mm Area of cross-section = 9.4 x 102 mm2 Ad/ = = 220 x 104 mm4 Calculate the value of the transverse shear stress at the neutral axis x-x and at the top of the web. [CSE-Mains, 2012, ME : 12 Marks] Solution: F= 20 kN Shear force, = 220 x 104 mm6 b = breadth = 3.5 mm Shear stress at neutral axis b= 3.5 mm I 5.5 mm I I 189 mm x 5.5 mm 1-4— 50 I Fig. 5.12 a)7 = 50x 5.5 x(100 - 2.25)+ 3.5 x 189(189) = 26743.75 + 1562.9375 = 42371.69 mm3 20000 x 42371.69 = 110.056 N/mm2 220 x104 x3.5 Shear stress at top edge of web= 20000)26743.75 = 69.46 N/mm2 220x104 x3.5 Q.5.12 A horizontal beam of square cross-section is so placed that the loading in the transverse plane is along one of the diagonals of length d. If the shear A force at a section of the beam is S. Draw shear 2s 2 stress distribution diagram for the section and T a a d 2.25s indicate the position and magnitude of maximum T 2 d d -t shear stress on it. ir [CSE-Mains, 1998, ME : 30 Marks] I. j M8X Solution: Square section Shear stress Diagonal = d distribation d Fig. 5.13 Side = 4 (d ) = Shear force = S Take a section at a distance y. from NA, xx Area Aaa (distance of centroid) from xx x 1 d4 12 - 48 Theory of Simple Bending Strength of Materials Height of the triangle Aaa = 4 77 (d -y ) 2 Base of the triangle = 21d -y) 2 Area = Ay = Breadth, b= Transverse shear stress, T= id 2 )1 (d ) (d )2 y 2 x2 2 y 2 y 2 2y d 2 - Y ) (V d ( d 2( 2-y ) Say Tu b s(d _ )(2yd ) 48 x + y 3 6) d4 x2(2-y) = d At- = Y 4 ' At — = y Outer edge, 2 ' At y = 0, T= [ d 2 4_ dY 2 v 2 d 4 ' 2 - ' 8S[ d2 d d d2] — +— x— -2 x 16 d4 4 2 4 8S [d2 d2 d2] TA = d4 4 4 2 8S TNA = 8S [d2 d 2 d 2 d4 4 8 8 2S = d2 =o d 2 2S d 4 X 4 —d2 tmax For Tma,„ d'r _ 0 — d - 4y) = 0 dy = (2 Y = Tom _ Refer equation (i) d 8 8S [d2 d d d2] 9S s = 2.25x— , — +— x— -2x— = 64 4d2 4 2 8 d' d" Q.5.13 An I-beam has flanges 10 cm wide and 1 cm thick, and the web 12 cm high and 1 cm thick. At a section of this beam act a bending moment of 1000 kg-m and a shear force of 10,000 kg, find the normal and shear stresses at the following points on the vertical center line. (i) Top of the flange (ii) In the web at the junction with the flange (iii) At the neutral axis [CSE-Mains, 1999, ME : 20 Marks] 78 0. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions Solution: I-section shown in the figure 5.14 I 1cm I I = 10 x 143 9 x 123 12 12 About neutral axis, 12 cm = 2286.667 - 1296 = 990.667 cm4 M = 1000 kg-m = 1000,00 kg-cm (i) Bending normal stress 1 cm GA = • 1000,00 x 7 = 706.6 kg/cm2' 990.667 1000,00 x 6 = 605.65 kg/cm2 = 990.667 I-4— 10 —0-1 I Fig. 5.14 ac = 6NA=Q (ii) Shear stresses For Breadth, TA = 0 TB, ay = 10 x 1 x 6.5 = 65 cm3 b = 1 cm F = 10,000 kg Fay _ 10,000 x 65 Ib 990.667 x 1 tic 656.12 kg/cm2 ay = 10 x 1 x 6.5 + 1 x 6 x 3=65 + 18 = 8.3 cm3 b= 1 cm 10000 x 8.3 = 837.82 kg/cm2 T C 990.667 x 1 Q.5.14 Simply supported beam of 3 m span is subjected to loads as shown in figure 5.15(a) the beam is of I-section and all of the dimensions are in mm. Determine the principal stresses at point D in the web. The section is located at a 2m 1m distance of 1 m for right hand support. -4 5.667 kN 4.333 kN [CSE-Mains, 2013, ME : 25 Marks] (a) Solution: Fig. 5.15 (a) Reactions, 2 x 3x 1.5 +4 x 2= 3RB 17 = 3RB Reaction, RB = 5.667 kN Reaction, RA = 2 x 3 + 4 - 5.667 10 mm = 10 kN - 5.667 = 4.333 kN 40 mm Mc = Bending moment at C 50 mm 180 mm = 4.333 x 2-2 x 2 x 1 = 8.666 - 4 = 4.666 kN-m 15m = 4.666 x 106 Nmm (Sagging) 10 mm Shear force at C F°— Fc = 4.333 - 2 x 2 = 0.333 (just left to C) (b) Fc = 0.333 - 4 = -3.667 kN (just right to C) Fig. 5.15(b) Jump shear force at C = 4 kNI I I Theory of Simple Bending 4 Strength of Materials c x Y E) ab, bending stress - M l x„ Mc x 50 75 x 2003 60 x 1803 12 12 = 20.84 x 106 mm4 xr = (ado - 79 50 x106 -29.16 x106 4.666 x 106 x 50 20.84 x 106 = -11.20 N/mm2 (sagging moment) compressive Shear stress at D 'co = Fay lb when, b = 15 mm ay (about neutral axis) = 15 x 40 x 70 + 75 x 10 x 95 = 113250 mm2 F = 4000 N 113250 x 4000 = 1.45 N/mm2 - 20.84x106 x15 Principal stresses P1 , P2 = 11.20 44 11.20)2 l 2 +0.45)2 2 11.385 N/mm 2 1.45 N/mm = -5.6 ± V31.56 + 2.1025 = -5.6 ± 5.79 p1 = -11.385 N/mm2 p2 = +0.19 N/mm2 2 1.45 N/mm Objective Questions Q.1 Q.2 A cantilever beam of rectangular cross-section is 1 m deep and 0.6 m thick. If the beam were to be 0.6 m deep and 1 m thick then the beam would (a) be weakened 0.5 time (b) be weakened 0.6 time (c) be strengthed 0.6 time (d) have the same strength on the original beam because the cross-sectional area remains the same [CSE-Prelims, ME : 1999] If the T-beam cross-section shown in the given figure 5.16 has bending stress of 30 MPa in the top fibre, then the stress in the bottom fibre would be (G is centroid) (a) zero (c) -80 MPa Q.3 Fig. 5.16 (b) -30 MPa (d) 0.50 MPa The distribution of shear stress of a beam in shown in the figure 5.17. The cross-section of the beam is 80 0. Fig. 5.17 (b) T (d) Q.4 py 0 0 (a) 0 qy 0 By conjugate beam method, the slope at any section of actual beam is equal to (a) El times the SF of the conjugate beam (b) EI times the BM of the conjugate beam (c) SF of the conjugate beam (d) BM of the conjugate beam [CSE-Prelims, ME : 2002] Q.8 Select the correct shear stress distribution diagram for a square beam with a diagonal in a vertical position /\ (b) 0 (b) (c) (d) qy 0 ry py 0 0 0 0 (d) 0 qy 0 0 0 qy py, py (a) py 0 0) 0 0 0 0 0 0 0, py 0 0 [CSE-Prelims, ME : 2001] Q.5 Q.7 A uniform bar lying in x-direction is subjected to pure bending. Which one of the following tensors represents the strain variation when bending moment is about z-axis (p, q, and x constant) (c) Two identical planks of wood are connected by bolts at a pitch distance of 20 mm. The beam is subjected to a bending moment of 12 kNm and the shear force in bolts will be I [CSE-Prelims, ME : 2002] Q.9 A pipe of external diameter 3 cm and internal diameter 2 cm and of length 4 m supported at its ends. If carries a point load of 65 N at its center-section, modulus of pipe will be 657E (a) 10 10 I (a) zero (c) 0.2 kN Q.6 MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions Fig. 5.18 (b) 0.1 kN (d) 4 kN [CSE-Prelims, ME : 2001] A straight bimetallic strip of copper and steel is heated. It is free at ends. The strip will (a) expand and remain straight (b) not expand but bend (c) expand and bend also (d) first only [CSE Prelims, ME : 2002] (c) 3 64 cm 657t 96 cm 3 65/r 3 (b) 32 cm (d) 657t 124 cm 3 [CSE-Prelims, ME : 2002] Q.10 In a loaded beam under bending (a) both the maximum normal and maximum shear stresses occur at the skin fibres (b) both the maximum normal and maximum shear stresses occur at the neutral axis (c) the maximum normal stress occurs at the skin fibres and the maximum shear stress occurs at the neutral axis (d) the maximum normal stress occurs at the neutral axis while the maximum shear stress occurs at the stress fibres [CSE-Prelims, ME : 2003] Theory of Simple Bending Strength of Materials Q.11 Consider the following statements: Two beams of identical cross-section but of different materials carry same bending moment at a particular section, then 1. the maximum bending stress at that section in the two beams will be same 2. the maximum shearing stress at that section in the two beams will be same 3. maximum bending stress at that section will depend upon the elastic modulus of the beam material 4. curvature of the beam having greater value of E will be larger Which of the statements given above are correct? (a) 1 and 2 only (b) 1, 3 and 4 (c) 1, 2 and 3 (d) 2, 3 and 4 [CSE-Prelims, ME : 2002] Q.12 For the shear force to be uniform throughout the span of a simply supported beam, it should carry which one of the following loadings? (a) A concentrated load at mid span (b) UDL over the entire span (c) A couple anywhere within its span (d) Two concentrated loads equal in magnitude and placed at equal distance from each support [CSE-Prelims, ME :2007] Q.13 Which one of the following is the preferable cross-section of a beam for bending loads? (a) Circular (b) Annular circular (d) I-section (c) Rectangular [CSE-Prelims, ME : 2006] Q.14 Consider the following statements for simple bending of beams: 1. Neutral axis always passes through the centroid of the beam cross-section 2. Bending stress in a fibre is a longitudinal stress 3. Bending stress is zero at the neutral axis 4. Shearing stress is always zero on the neutral axis Which of the statements given above are correct? (a) 1, 2 and 3 only (b) 1, 2, 3 and 4 (d) 1 and 4 only (c) 2, 3 and 4 only [CSE-Prelims, ME : 2006] 81 Q.15 Three beams of 100 mm x 100 mm section are made from RCC, aluminium and timber respectively. These three beams are loaded identically with similar supports. Maximum bending stress will occur in which beam? (a) Timber beam (b) Aluminium beam (c) RCC beam (d) The three beams will have same maximum bending stress [CSE-Prelims, ME : 2008] Q.16 A simply supported beam of span Land flexural rigidity EI carries a unit point load at its centre. What is the strain energy in the bema due to bending? (a) L3 16E/ L3 (c) 96E1 (b) L3 48 E1 (d) L3 192 E1 [CSE-Prelims, ME : 2009] Q.17 For a triangular section with base 'b' and height ' h' , the ratio of the moment of inertia about an axis passing through its vertex and parallel to its base to the moment of inertia about an axis passing though its centre of gravity and also parallel to its base would be (a) twelve to one (b) nine to one (c) six to one (d) four to one [CSE-Prelims, CE : 2001] 0.18 A beam has rectangular section 100 mm x 200 mm. If it is subjected to a maximum BM of 4 x 107 Nmm, then the maximum bending stress developed would be (a) 30 N/mm2 (b) 60 N/mm2 (c) 90 N/mm2 (d) 120 N/mm2 [CSE-Prelims, CE : 2001] Q.19 Consider the following statements: For each component in a flitched beam under the action of a transverse load. 1. the radius of curvature will be different. 2. the radius of curvature will be the same. 3. the maximum bending stress will be the same. 82 ► MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions 4. the maximum bending stress will be dependent upon the modulus of elasticity of the material of the component. Which of these statements are correct? (a) 1 and 3 (b) 1 and 4 (c) 2 and 3 (d) 2 and 4 [CSE-Prelims, CE : 2001] Q.20 A timber beam is simply supported at the ends and carries a concentrated load at mid span. The maximum longitudinal stress 'f is 12 N/mm2 and the maximum shear stress 'q' is 1.2 N/mm2. The ratio of span to depth would be (a) 10 (b) 6 (c) 5 (d) 4 [CSE-Prelims, CE : 2001] 0.21 The moment of inertia of a given rectangular area is minimum about (a) its longer centroidal axis (b) its polar axis (c) its axis along the diagonal (d) its shorter centroidal axis [CSE-Prelims, CE : 2003] Q.22 For a beam of rectangular section under bending the shear stress across the depth varies (a) Linearly (b) Exponentially (c) Hyperbolically (d) Parabolically [CSE-Prelims, CE : 2003] Q.23 A cantilever beam is loaded with a uniformly distributed load of intensity w along its entire length. The span of the beam is L. Which of the following Mohr's circle diagram correctly represent(s) the state of stress above the neutral axis of the beam? Shearing stress Shearing stress (i) Normal Tensile stress Normal Tensile stress (ii) Shearing stress Shearing stress Normal Tensile stress Fig. 5.20 Normal Tensile stress ( v) Select the correct answer from the codes given below: (a) Fig. (i) (b) Fig. (iii) (c) Figl. (ii) and (iv) (d) Fig. (iii) and (iv) [CSE-Prelims, CE : 2014] Q.24 In a thin uniform lamina having symmetrical central axis as shown below, the distance of centre of gravity from AD is A B oco 1.5 m K D T C —►] 2 cm I-4— 8 cm (a) 3 cm (c) 23/7 cm ►H Fig. 5.21 (b) 22/7 cm (d) 24/7 cm [CSE-Prelims, CE : 2004] Q.25 In the cross-section of a rectangular beam, what is the ratio of the average shear stress to the maximum shear stress? (a) 3/2 (b) 2/3 (c) 4/3 (d) 3/4 [CSE-Prelims, CE : 2006] Q.26 Which of the following are implied in the assumption of plane sections remaining plane (in simple bending)? 1. Stress is proportional to the distance from neutral axis. 2. Displacement is proportional to the distance from neutral axis. 3. Strain is zero across the cross-section. 4. Strain is directly proportional to the distance from neutral axis. Select the correct answer using the codes given below: (a) 1 and 4 only (b) 2 and 3 only (c) 3 and 4 only (d) 1, 2 and 4 Q.27 A beam of rectangular cross-section is to be cut from a circular beam of diameter D. What is Theory of Simple Bending Strength of Materials the ratio of the depth of the beam to its width for maximum moment of resistance? (b) 3 (d) [CSE-Prelims, CE : 2009] Q.28 A cantilever beam as shown in figure is subjected to end moment M0. What is the deflection at the free end? 1 21 4m -IA (a) (b) (c) (d) -41 83 top edge of the flange centre of the web junction of the flange and web bottom edge of the flange [CSE-Prelims, CE : 2010] Q.32 Two tubes of outer diameter 20 mm and inner diameter 16 mm each are combined with a solid bar of diameter 20 mm. Their centres make an equilateral triangle of side 20 mm. The section is of a cantilever, with solid tube on top, subjected to load at free end. If the maximum tensile stress developed in section is 100 MPa. What is the maximum compressive stress developed. 4 Fig. 5.22 10 M0 EI (b) 15 M0 El 20 M0 (c) El (d) 30 M0 El (a) [CSE-Prelims, CE : 2010] Q.29 A mild steel beam is simply supported. It has constant moment of inertia =106 mm4. The entire length of the beam is subjected to a constant BM of 107 Nmm, E = 2 x 105 N/mm2. What is the radius curvature of the bent beam in meters? (a) 200 (b) 2 (c) 20 (d) 0.2 [CSE-Prelims, CE : 2010] Q.30 In a beam of solid circular cross-section, what is the ratio of maximum shear stress to the average shear stress? 3 (a) 7 1 (c) 3 (b) 4 2 (d) 3 [CSE-Prelims, CE : 2010] Q.31 In case of a beam of I-section subjected to transverse shear force 'F, the maximum shear stress occurs a the (a) 136 MPa (c) 86 MPa Fig. 5.23 (b) 116 MPa (d) 73.2 MPa Q.33 A beam of uniform strength refers which one of the following? (a) A beam in which extreme fibre stresses are same at all cross-section along the length of the beam (b) A beam in which the moment of inertia about the axis of bending is constant at all crosssection of the beam (c) A beam in which the distribution of bending stress across the depth of cross-section is uniform at all cross-sections of the beam (d) A beam in which the bending stress is uniform at the maximum bending moment cross-section. Q.34 The cross-section of a beam in bending is as shown in the figure given below. It is subjected to a shear force acting in the plane of crosssection. Which among the following figures shows the correct shear stress distribution 84 0. IAS & IFS (Objective & Conventional) Previous Solved Questions (a) 3.2 (c) 3.4 across the depth of the cross-section of the beam? T 2 Fig. 5.24 MADE EASY (b) 3.3 (d) 3.5 [CSE-Prelims, CE : 2006] Q.37 The below diagrams show the details of two simply supported beams B1 and 82. EI is constant throughout the length and same for both the beams. The beams have the same area of cross-section and the same depth. What is the ratio of maximum bending stress in 82 to that in Bi? 4 kN kN/m X3 A1y-v-v-v111100.11 (b) m -+- 2 m-od B1 (a) 4 (c) 2 (d) [CSE-Prelims, CE : 2005] Q.35 A mild steel plate is subjected to a moment M each at its ends such that it bends into an arc of a circle of radius 10 m. The plate has width 60 mm and thickness 10 mm. E = 2 x 105 N/mm2. What is the maximum bending stress produced in the plate? (a) 100 MPa (b) 200 MPa (d) 400 MPa (c) 300 MPa [CSE-Prelims, CE : 2005] 1-4— 4 m B2 Fig. 5.26 (b) 1 (d) 1/2 [CSE-Prelims, CE : 2009] Q.38 Cross-sections of two beams A (600 mm x 200 mm) and 8(800 mm x 60 mm) are shown in the figure given below. Both the beams have the same material. By how many times is the beam A stronger than the beam B in resisting bending? 200 mm-1-1 i T 1 E E 0 co 800 mm A Q.36 What is the y-coordinate of the centroid of the area ABCDEshown in the figure given below? Fig. 5.27 (a) 80 (c) 50 Fig. 5.25 (b) 60 (d) 25 [CSE-Prelims, CE : 2007] Q.39 A mild steel structural beam has cross-section, which is an unsymmetrical I-section. The overall depth of the beam is 250 mm. The flange Theory of Simple Bending Strength of Materials stresses at the top and bottom are 200 N/mm2 and 50 N/mm2, respectively. What is the depth of the neutral axis from the top of the beam? (a) 50 mm (b) 100 mm (c) 150 mm (d) 200 mm [CSE-Prelims, CE : 2008] 1 85 Answers 1. (b) 6. (c) 11. (a) 16. (c) 21. (a) 26. (d) 31. (b) 36. (c) 2. 7. 12. 17. 22. 27. 32. 37. 3. (b) 8. (d) 13. (d) 18. (b) 23. (a) 28. (c) 33. (a) 38. (d) (c) (c) (c) (b) (d) (b) (b) (d) 4. (d) 9. (c) 14. (a) 19. (d) 24. (b) 29. (c) 34. (c) 39. (d) 5. 10. 15. 20. 25. 30. 35. Explanations 6. (c) 1. (b) Z 1- 0.6 x13 6 Z2 1 x 0.63 = 0.036 m3 6 Z2 Z 2. 0.1 m3 = 0.36 weakened by 0.64 times Fig. 5.29 Strip will bend and expand. (c) Y2 = 3° Y1 = 80 f2 = 30 7. (c) Slope = SF of conjugate beam. 8. (d) Square with diagonal vertical (d) is correct shear stress distribution = -80 MPa linearly proportional to distance y 3. 4. (b) Shear stress distributed is of T-section. d max (d) 9. at 8 from NA. (c) x z Fig. 5.28 G oc y Stress a Ey= 5. Fig. 5.30 = PY = E x = VG = _ vpy = qy (a) Zero, as BM is constant and shear force is zero. TC 65n / = 6 £31- 16) = 4 64 Z- 657C 1 657C 3 X = - cm 64 1.5 96 (a) (c) (d) (c) (b) (b) (a) 86 ► MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions 10. (c) = Ixx bh3 2bh3 bh3 36 9 4 =9 IGG - ;Mx I2h 3 1 Fig. 5.31 11. (a) 1 and 2 are correct statements. 5 b 12. (c) Fig. 5.33 18. (b) b = 100, h = 200 mm bh 2 100 x 2002 4 = x 106 = 6 6 6 M= 4 x 107 Nmm Z= RA a — Fig. 5.32 13. (d) I-section has maximum moment of inertia for same area of cross-section. w at = — Lx 12 4 bd` — tia = 1.5 L d W 2 L3 96EI L3 96E1 1 1 — =1.2 2 2 bd 10 = 1.5wL 2bd 12 x 1.5W bd 2 1.2 = or 16. (c) Strain energy, 48E1 6 = 60 MPa 106 19. (d) Flitched beam (Each component) 5. Radius of curvature will be the same 4. Maximum crb dependent on Eof component 15. (d) Stress depends on BM and section modulus independent of material properties. 3 1 wL =—WX x 20. (c) 14. (a) 1, 2, 3 are correct statements. U= 4 b SFD 2 W= 1 4 x 107 5 21. (a) MI minimum about the longer centroidal axis. D 17. (b) x bh 3 GG I = 36 bh3 bh 14h2 ) Ixx = 36 ± 2 9 2L d Fig. 5.34 BD 3 x 12 Theory of Simple Bending Strength of Materials 22. (a) Parabolically av "c max - 1 $7 2 3 26. (d) 1, 2 and 4 are correct statements. 27. (b) Fig 5.35 23. (a) Fig. 5.36 At a +ve normal stress shear stress Fig. 5.39 b2 +d2 = b = D sing, d =D cos() Z — dz = O ti Fig. 5.37 bd 2 6 1-1— 7 —""i = sin() = 8 cm —01 16x1+12x6 28 16 + 72 88 22 = = —cm 28 28 7 'ray — 1.5 3 0)) ( 0 sin — sin k 3 [cos() — 3 sin2 cos 0] = 0 6 1 core = 28. (c) Fig. 5.38 max 6 d_ b T 25. (b) Rectangular beam section 3 = dt 24. (b) D3 sinOcos2 0 6 D3 = — sinO(1—sin2 6 d_1 x — L L • G • G 21 Fig. 5.40 8— MnL L M L 3L x x —+ D. 2 2E1 2 M0L2 3M0L2 5 M0 42 M0 x 20 2E1 4E1 • 4 El El 88 P. IAS & IFS (Objective & Conventional) Previous Solved Questions 35. (a) R = 10 mm, b = 60 mm, t = 10 mm E = 2 x 105 N/mm2 29. (c) M I 107 106 MADE EASY E R 2 x 105 _20m R R a E — Y R 2 x 104 mm 30. (b) Beam of circular section 4 'c max 3 Tav 31. (b) I-section beam, Tmax occurs at centre of the web. y= _ 2x105 x5 10000 =100 MPa 36. (c) Y= 32. (a) Ai .7c x 100, solid A2 = n(100 — 82) = 367c (hollow) 2 =5mm 24 x 3+6 x 5 102 = 28 30 3.4 37. (d) M1 = 4 kNm IV12 — 1x16 8 = 2 kNm Same area, same depth as M 02 al = 0.5 38. (d) Fig. 5.41 1007cx17.32+36nx0 x2 10.07 172n y1 = 17.32 — 10.07 + 10 = 17.25 mm y2 = 10.07 + 10 = 20.07 Y- 6 20.07 x 100 =116 MPa "p 17.25 33. (a) Beam of uniform strength — in which extreme fibre stress are same at all cross-section along the length of the beam. 34. (c) For cross ZA - 200 x 6002 =12x106 mm3 6 ZB 800 x 602 = 48 x 104 mm3 6 ZA ZB = 25 39. (d) 200 MPa I 200 mm N A 50 mm 500 MPa T Fig. 5.44 Depth of neutral axis from top. •• • • Shear stress distribution Fig. 5.42 06 Deflection of Beams CHAPTER Q.6.1 A simply supported beam of length L carries a concentrated load Wat a distance 'a' from one end and 'b' from the other end (a > b). Find the position and magnitudes of the maximum deflection L and show that the position is always within — approx from the center of the beam. 13 [CSE-Mains, 2003, ME : 30 Marks] Solution: Wa = RB X L Wa L Reaction, RB = Reaction, RA = W x Wa Wb L L Take a section x-x at a distance x from A, in portion CB El 2y dx 2 Wb b RA = y R _ Wa B a>b Fig. 6.1 = RAx /- W(x - a) Wbx L W(x - a) ,r dy / W(x-a)2 2 Li — = Wbx ±ci dx 2L 2 Wbx 3 + Ci x +C2 / W (x-a)3 6 6L y = 0 at x = 0, therefore constant, C2 = 0 (second term not used) Ely = Moreover, , y = 0, x = L (second term -- v - a)3also used) 6 0- 0= 0= WbL3 W - (L - a)3 + C1 x L 6L 6 WL2 b 6 6 6 (L-a)3 + Ci L bL2 -W x b3 + Ci L 6 90 ► lAS & IFS (Objective & Conventional) Previous Solved Questions WbL Wb3 Wb r h2 6 ± 6L = 6L Constant, MADE EASY ,21 = 6b [b2 - a2 - b 2 -2ab]= Wb — [-a 2 - 2a b] 6L 6L Wab _ El dy dx = 6L [a + 2b] Wb x2 W 2L 2 (x a) 2 Wa b 6L (a + 2b) dy ymax will occur, where — = 0 dx n = Wbx2 W ( x2 2L = 2 2 k +a b 2 L ( 2 -zax) Wab — 6L (a + 2b) ab + a 2 -2ax)-- (a + 2b) 3L 0 = 3bx2 -3L(x2 + a 2 -2ax)-ab(a+2b) 0 = 3bx2 - 3Lx2 -3La 2 + 6axL-a2b-2ab2 0 = -3x2 (L -b)-3La2 +6axL-a 2b-2ab 2 or 0 = -3x2 -3La+6xL-ab-2b 2 3x 2 -6xL+3La+ ab+2b 2 = 0 3x2 -6xL+3a(a + b)+ab+2b 2 = 0 3x2 - 6xL+3a2 +3ab+ ab+2b 2 = 0 3x2 -6xL+3a 2 +4ab+2b 2 = 0 x= 6L + \1(602 - 4 x 3(3a2 +4ab+2b 2 ) 6 x= 6L + 6 1L2 -1(3a2 +4ab+2b2 ) 3 1, 6 L±/ L2 -(3a2 +4ab+2b 2 ) 3 X x X , = L±,\IL2 - - [a3 , L ( IL2 2L2 + a 2 3 i3L2 - 2L2 - a 2 = L+ \ 3 L± L2 - a 2 \1 3 2 + 2ab + b2 )] Deflection of Beams Strength of Materials x= , 91 2ab + b 2 3 L± Let us take a= 0.6L, b = 0.4L x= L± .12x.6x0.4L2 +(0.4) L 3 = L±LV0.16+0.0533 = L±LV0.21333 x = L±Lx 0.4619 x = -L x 0.4619 = 0.538 L= 0.5 L+ 0.038 L x = 0.5L + 0.038L -_-_ — from the center 26 ' L 26 or L -_) _ L , total 26 ) 13 II 2 _ a2 I- x= 3 Q.6.2 A cantilever beam of length L, is subjected to two concentrated loads of 2P and Pat its mid length L and free end respectively. If the deflection at its free end is limited to — . What should be the 500 value of P? Take flexural rigidity of the beam as El. [CSE-Mains, 2011, ME : 15 Marks] Solution: Flexural rigidity = El Deflection at free end, 3 2P ( L PL3 2P ( 1 x x 5 - —+- x - + 3E/ 3E/ 2 2E/ U) 2 Fig. 6.2 PL3 PL3 P L3 + + x3E112E1 El 8 = PL3 [1 1 + 1] + EI 3 12 8 PL3 r8+ 2 +3] 3E1 1_ 24 13 PL3 L x— = 24 E/ 500 P= El L2 X 24 1 24E1 X = 500 13 6500 L2 value of P Q.6.3 A cantilever 3 m long, and of symmetrical section 250 mm deep carries a uniformly distributed load of 30 kN/m run throughout together with a point of 80 kN at a section 1.2 m from fixed end. Find the deflection at the free end. E = 200 GPa, I = 54000 cm4 [CSE-Mains, 2012, ME : 15 Marks] 92 P IAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY Solution: Fig. 6.3 shows a cantilever, AB fixed at end B, Free at end A carrying a point load W at L1 from B end UDL of w throughout to length = flexural rigidity 8 = deflection at free end 80 kN = W w = 30 kN/m A L Fig. 6.3 WO1 wL4 WL3 + x L2 + 3E/ 2E1 8E1 where, L1 = 1.2 m, L2 = 1.8 m, L = 3 m, W= 80 kN and w = 30 kN/m EI = 200 x 106 kN/m2 x 54000 x 10-8 1114 = 108000 kNm2 80 x 1 22 30 x 34 80 x 1.23 + 1.8 + 2 x 108000 x 8 x 108000 3 x 108000 = 0.42666 x 10-3 + 0.96 x 10-3 + 2.8125 x 10-3 = 4.2 x 10-3 m = 4.2 mm 8- Q.6.4 A cantilever beam of length L and uniform flexural rigidity El is subjected to continuously distributed externally applied moment m kg-cm per cm of length of the beam. Using the area moment method show that deflection of the free end of the beam is 8 = mL3/3 El and explain why this is the same as that obtained for the case of a concentrated force P = is applied at the end of the beam. [CSE-Mains, 1991, ME : 20 Marks] Solution: kg-nn/m m Area of BM diagram = mL x Ditance of CG of BMD from A = L mL2 = 2 A 2L 2L mL3 = First moment of BM area from A = mL2 X 2 3 3 Deflection at the free end = BMD mL3 3E1 Fig. 6.4 (since deflection at fixed end is zero) Because in BM diagrams = mL = PL m= P Therefore same deflection. Q.6.5 Use Macaulay's method or area moment method to show that the deflection of a simply supported beam at point B with an off centre load at point A as shown in fig. 6.5 is given by pL2 x (L2 _ x2 _ L2 2 YB = 1. R x' *1. L, L2 R2 L Fig. 6.5 6 EL [CSE-Mains, 1989, ME : 30 Marks] 93 Deflection of Beams Strength of Materials1 Solution: Reactions, R1 = PL2 L R2 = PLi L Consider a section yy at a distance x' from point 1 Mx' = P(x - Li ) Bending moment, EI d y dx 2 = P(x' - L ) 1 PL2 L x' P(x' - Li ) dy PL2 'x 2 P 2 = x—-(x' - L1) + C1 dx L 2 2 PL2 x' 3 P 3 Ely = — x- --6(x'-Li) + L 6 C2 = O,x = 0, y = 0 x' = L, y = 0 El x0 = C1 = Ely = So, Deflection at B, (at distance x) PL2 2 L xL + C2 PL2 L2 P (1--1-1)3±C11- - P (1_32 )+4 6 L PL2L P 6xL 6 IL2 x x'3 px (x' 1-1)3 + 16 L 6 6 PL21-1,c, 6 x < x' ElyB = P al 6 x-- omitted term+ x — L 6 6 L 2 x3 PL PL2 x 3 PL32 x PL2Lx 6L Y 6L PL 2 x 112 s 6 Px1-2r 2 Lx +L22 -L2 ] 6L x 2 _ L21 6 EI L L Q.6.6 A beam ABCD, 5 m long is supported at A and C as shown in figure 6.6. It carries a point load of 2 kN at end D, and a moment of 2 kNm (CW) at B. What is the flexural rigidity of the beam (El), if the deflection at D is not to exceed 1 mm. [CSE-Mains, 2009, ME : 20 Marks] 2 2 kN-m :1410 1•-- 2 m RA = Solution: Reaction Taking moment about A x H x J, 2kN x B/ 2 m--4-1 m-R Rc = + 3kN 1kN Fig. 6.6 5 x 2 + 2 = 4Rc Rc = 3 kNT RA = 2- 3 = -1 kNL 94 0. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions Take a section x-x at a distance x from A, in portion CD (Fig. 6.6) d 2y El — = -1x x + 2(x - 2)° + 3(x - 4)) dx 2 Ely= Integrating two times, x3 / nN2 / + A y = 0, x = 0, C2 = 0, constant x = 4 m, support C, y = 0 64 0 = -- +4+0+4C1 6 0 = -10.6667 + 4 + 4C1 C1 = 1.666 3 Ely = 6 2 + (X — 2) -I- 3 2 (x — 4) +1.666x At end D, x = 5 m, Y = YD 125 1 E/YD = - — + 9 + - + 1.666 x 5 6 2 = -20.833 + 9.5 + 8.33 = -3.003 3.003 El m YD = 3.003 = 0.001 m EI 3.003 = 3003 kNm2 (Flexural rigidity) 0.001 El Q.6.7 A cantilever of length L, is loaded with uniformly increasing load, starting from zero at the free end and to a maximum of Wo at the fixed end. The free end is propped to the level of the fixed end. Determine the reaction at the prop and equation of elastic curve along with the slope at the propped end. EI is assumed constant. [CSE-Mains, 2006, ME : 30 Marks] Solution: Say prop reaction = R Rate of loading at a distance, Wx = woL x x L 2 Load uptox, = wo- X — wo x ( x2 x) BM at section x-x = WO — x 2L 3 X3 BM at section x-x = Rx - wo -E EI d2y (12 x3 = Rx -wo -6-E w Fig. 6.7 Deflection of Beams Strength: Of Materials El x 2 ,., vv ox4 - Rx +c dx 1 2 24L dy dx 0= woL3 24 95 41 0, x = L RL2 2 woL3 24 +1 RL2 Cl 2 Ely = Rx 3 6 wox 5 ( w oL3 120E + 24 RL21 r 2 )x + -2 x = 0, y = 0, C2 = 0 So, Ely = Rx 3 6 wox 5 ( w oL3 RL2 ) 120L + 24 2 )x x = L, y = 0 woo woo RL3 120 + 24 2 0= RL3 6 0 = _RL3 4.woL4 3 30 R= wol10 Ely = w°1-x3 w°x5 1w°1-3 w°L x L2 x 60 120L + 24 20 woi-x3 wox5 wo0 60 120L 120 Slope at free end x= 0 woL3 w01-3 w0L-3 20 120 EliA - Cl = 24 Slope at A, iA = woL3 120E1 0.3 m 10 kN 20 kN/m .._ Q.6.8 A simply supported beam is subjected to the loading as shown in the figure 6.8(a). Calculate the deflection at a section 2 m from end A. Assume E= 70 GN/m2, I= 830 cm4) [CSE-Mains, 1996, ME : 30 Marks] 11i1:3 101(N 141---- 1.5 m --144-- 0.9 m -414-0.6 m -44 20 kN/m 16 3 kNm Solution: Reaction 3xRB + 3 = 16 x 2.4 + 20 x 0.9 x(1.5 + 0.45) 3 + 3/38 = 38.4 + 18 x 1.95 =38.4 +35.1 RB = 23.5 kN T Total load = 20 x 0.9 + 16 = 34 kN 16 kN A D t 4-- 1.5 m '14 RA = 10.5 kN Fig. 6.8 x 0.9.m --4-}4- 0.6 m AB 9 6 0. lAS & IFS (Objective & Conventional) Previous Solved Questions Reaction, MADE EASY RA = 34 - 23.5 = 10.5 kN Figure shows the simplified loading diagram UDL extended to portion is portion DB upto x and negative UDL applied beyond D. 10.5x - 3(x -1.5)° - 2 (x - 1.5)2 +14-(x / - 2.4)2 -16(x - 2.4) 2 w = 20 kN/m El d2y dx 2 EI _ 10.5x - 3(x - 1.5)° -10(x - 1.5)2 + 10(x - 2.4)2 - 16(x - 2.4) dy 5.25x2 - 3(x -1.5)- dx 10 3 1 .5)3 + (x _ 3 (x - 2.4)3 - 8(x - 2.4)2 + 10 — (x -1.5)4 + 10 (x 2.4)4 8 (x 2.4.)3 12 12 3 Ely = 1.75x3 -1.5(x -1.5)2 CiX + C2 x = 0, y = 0, C2 = 0, Omitting II, Ill, IV and V terms x = 3 m, y = 0 0 = 1.75 x 27 - 1.5 x 2.25 - 10 2 4 (1.5) + \ 4 8 3 (0.6) - -(0.6) + 3C1 2 3 10 0 = 47.25 - 3.375 - 4.21875 + 0.108 - 0.576 + 3C1 = 39.188 + 3C1 C1 = -13.063 Ely = 1.75 -1.5(x -1.5)2 + 10 2 (x 1 1 4 5) ;(x - 2.4)4 - 3 (X - 2.4)3 - 1 3.063x x = 2m Sy2 . 1.75 x; - 1.5 x 0.25 + ;(0.5)4 - Omitted - Omitted + 13.063 x 2 = 2.333 - 0.375 - 0.05208 - 26.126 = -24.22 y2EI = 24.22 = 70 x 106 kN/m2 x 830 x 10-8 kNm 2 = 581 kNm2 24.22 Y2 = 581 = 0.04168 m = -41.08 mm Q.6.9 A simply supported beam carries a uniformly varying load with zero intensity at left support and an intensity of w at the right support. Calculate the maximum deflection and maximum slope and mention the position on the beam when these occur. [CSE-Mains, 1988, ME : 30 Marks] Solution: Figure 6.9 shows a beam AB of span length L, carrying uniformly varying load with maximum intensity w/m at right hand B. wL Total load on beam = 2 Deflection of Beams Strength of Materials Moment about A, 97 wL 2L -T x 3 = RB x L wL 3 Reaction, R8 = Reaction, wL wL wL RA = 2 — 3 6 RA = Consider a section xx at a distance x from A wL 6 wx = rate of loading = L x= Total load upto F2,-= Fig. 6.9 wx x = wx2 x L 2 2L CG of load ties at — from x-x 3 x 2 x x 2L 3 Wx = RAx Bending moment, wL Wx 3 -x x 6L 6 ci — = dx or Ely = wLx2 wx4 12 24L wLx3 36 + C1 (constant of integration) wx 5 120L + Ci x+ C2 (constant of integration) y= 0, x = 0, C2 = 0 y= 0 at x = L 0= wLx 3 wx 5 r , + + C2 36 120E 0 — 3) 4 wL A- CiL = 7 x wL4 +Ci L — (1 360 360 Content, 3 wL 360 x C1 o r.,dy wLx 2 wx 4 7 ", — = w dx 12 24L 360 dy Maximum deflection occurs, where — = 0 dx ,, 4 7 vv X 24L 360 w1-3 4 3) x y2 2L — 30 "' WL 2 0 = 12 W 0 = 4 or 41 LX2 X 2L 12 7 30 =0 wL 3 98 0. IAS & IFS (Objective & Conventional) Previous Solved Questions 4 7 I-2X2 - X 2 30 4 MADE =0 30L2x2 -15x4 - 7L4 = 0 15x4 - 30L2x2 + 7L4 = 0 2 + V900L4 - 420L4 x2 - 30L 30 2= x 300 ± NFINTIL71 30L2 ± L2 x 4.156 30 30 4 r301: 2 - , x- N/30 30 = L2 - 0.730L2 = 0.27L2 x = 0.5196L x2 ymax at x = 0.5196L EIy syma, = wLx3 36 7 ,3 WL x +1.1x 360 wL 3 -k(0.5196L) 5 7 3 x (0.51960 - — wL x 0.5196L -12 0L 360 = wL4 [0.003897 - 10003156 -0.0101] (Putting the value of C1) = 6.522 x 10-3 wL4 Ymax = 6.522 x10-3wL4 at x = 0.5196 from A EI 7wL3 1 360 xt aend A EI Slope at A Slope at B EBB = wL3 wL3 7 wL3 12 - 24 - 360 x (30 -15 - 7) wL3 = + 360 360 wL3 wL3 ja = 45E1 , (Hence, maximum slope at B) Q.6.10 Draw SF and BM diagram of the beam ACDB, where two couples are acting. Find the ratio of deflection at location C and D where couples are acting. [CSE-Mains, 2002, ME : 30 Marks] Solution: Reaction: Moment about A, 10 + 20 = 3RB RB = 10 kN RA = -10 kN Deflection of Beams Strength of Materials X = 10 kNm -10 kNm M2 = 20 kNm -10 kNm (c) BM diagram Fig. 6.10 SF diagram A vertically, SF is -10 kN constant BM diagram MA = Mc = -10 x 1 = -10 kNm (Hogging) (Just left of C) Mc = -10 + 10 = 0 kNm (Just right of C) MD = -10 x 2 + 10 = -10 kNm (Hogging) (Just left of D) Mc = -10 + 20 = +10 kNm (Sagging) (Just right of D) Me = Deflection Taking a section in the last portion DB of the beam Mx = -10x + 10(x -1)° + 20(x - 2)° El d2 Y = -10x + 10(x-1)0 + 20 (x - 2)° dx 2 dy EI— dx -5x2 +10(x-1)+20(x-2)+C1 Ely = - 5x 3 +5(x-1)2 +10(x-2)2 + Cpc + C2 3 y = 0, x = 0, constant, C2 = 0 y= 0, x = 3 = — — x3 3 3 +5x4+10x1+3Ci 0 = -45 + 20 +10 + 3C1 C1 = +5 4 99 100 10- lAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY Ely = -5c+ 5(x -1)2 +10(x-2)2 +5x 5 Elyc = 10 0 + Omitted+ 5 x 1= 3 = +3.33 - 3 x 8+ 5(1)+100)+5 x 2 = -13.33+5+0+ 10 = +1.667 Yc 3.33 =2 1.667 Yo Q.6.11 A beam of flexural rigidity 20 MN m2 is simply supported over a span of 6 m as shown in figure 6.11. It carries a concentrated load of 24 kN, 2m from the left hand support and a UDL of 6 kN/m on central 2 m part. The beam has vertical member welded onto it 2 m from right hand support which carry two horizontal loads of 24 kN as shown in the figure. Distance between these loads is 1 m. 24 kN 24 kN 6 kN/m WyTheThrY\ m C (a) 2m ► 1.1 2m D I18 kN 24 kN 2m '14 x A (b) Fig. 6.11 (i) (ii) Draw BM diagram of the beam. Calculate the vertical deflection at the central points of the beam. Solution: Reaction: Moment about A 24 x 2 + 24 + 6 x 2 x 3 = 48 + 24 + 36 = Reaction, 130 = Reaction, RA = BM diagram Just left to C, 6130 6R0 18 kN 24 + 12 - 18 = 18 kN MA = ° MB = 18 x 2 = 36 kNm M3 = 18 x 3 - 24 x 1- 6 x 1 x 0.5 = 54 - 24 - 3 = 27 kNm M4 = 18 x 4 - 24 x 2 - 6 x 2 x 1 = 72 - 48- 12 = 12 kNm Deflection of Beams Strength of Materials 4 101 M4' = 12 + 24 = 36 kNm (Just right to C) MD = ° 36 kNm A 27 36 kNm B BMD Fig. 6.12 Deflection at central point Extend UDL to portion CD and apply -w from C to Das shown. MC = 18x - 24(x - 2)- --t4 (x - 2)2 + 1.;1-(x - 4)2 + 24(x - 4)° 2 W = 6 kN/m El d y x2 = 18x - 24(x - 2)- 3(x- 2)2 + 3(x - 4)2 + 24 (x - 4)° d dy EI — = 9x2 - 12(x - 2)2 -(x- 2)2 + (x - 4)3 + 24(x - 4) + dx 3 Ely = 3x3 - 4(x -2) (x - 2) 4 4 + (x-44)4 +12(x -4)2 +Cix+ C2 x = 0, y = 0 Constant C2 = 0 y= Oatx= 6 m 24 +12x22 +6Ci 0 = 3 x 63 - 4(4)3 --•-• -+4 4 Constant, 0 = 648 - 256 - 64 + 4 + 48 + 6C1 0 = 380 + 6C1 C1 = -63.33 3 Ely = 3x3 - 4(x -2) 0y3 = 3 x 33 - 4 - 4 4 (x - 2) 4 4 + (x - 4) 4 +12(x -4)2 - 63.33x 4 + omitted + omitted term - 63.33x = 81-4-O.25-190=-113.25 81 -4El = 20000 kNm2 113.25 Y3 - 20000 = 5.66x10-3 m= - 5.66 mm Q.6.12 A cantilever beam AC of span '3a' and carrying a concentrated load 'P at distance '2a' from the fixed end A is strengthened by supporting its free end on the free end of another cantilever BC as shown in the figure 6.13. Determine percentage reduction in the moment at the support A. 10 2 IP. IAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY IC AI 2a —.4.-- 2a --id Fig. 6.13 [CSE-Mains, 2002, CE : 20 Marks] Solution: MA = -2Pa (without strengthening) Say reaction at C = RT Take a section at a distances of x from C d2y 2c1— = R xx- P(x- a) dx 2 dy Rx 2 P(x 2E1— = 2 dx 2 +Ci Fig, 6.14 dy = Oatx= 3a dx R9a2 P (4a2)+Ci 0= Ci 2 2 Constant, = 2Pa2 - 4.5Ra2 2 dy 2E1— = Rx dx 2 P (x -a)2 .1- 2Pa2 - 4.5a2R 2 Rx3 P 2Ely = — --(x - a)3 +2Pa 2 x - 4.5Ra2x + C2 6 6 y= Oatx= 3a 0 = 4.5Ra C2 = 9Ra 3 3 3 3 -- Pa +6Pa -13.5Re + C2 6 14 3 - 3Pa x3 2Ely = But or or P 3 14 2 --(x - a) 3 + 2Pa x — 4.5Ra2x +9Ra - — Pa 3 6 6 y= 0 at x = 0, from C 14 = 9Ra3 - -5-Pa- , Omitting Ilnd term R= 14P 27 Deflection of Beams Strength of Materials 1 03 14 14 4 - — MA = 27 x P x 3a - 2Pa = + 9 Pa - 2Pa = -- Pa Bending moment, Pa 4 -Reduction in BM = 12Pal H--Pa 9 149 14 7 1 % reduction = — Pax — x 100 = - x100 = 77.78% 9 9 2Pa 0.6.13 Compare the elastic deformations at the points of loading produced in a simply supported beam of length L., carrying a concentrated load Pet its mid-point to that of a cantilever of the length L, carrying a load P at its free end. The moment of inertia of the section of the simply supported beam is J in the left half and 0.5 J in the right half, For the cantilever, the moment of inertia of the section Is 2J upto a distance of 112 from the fixed end and J for the remaining portion. Assume that both the beams are of the same material. [CSE-Mains, 1990, ME : 20 Marks] Solution: BMD Fig, 0,15 For S,S, Beam Fig. 6.16 E d2y P = 2J dx2 dy c— = dx 2 4j P 0.5J p( L.) Px _2P( L = 2J J 2 - -j- x - +C 213 Cix + C2 3J k Y= 0, x = 0, C2 = 0 (Omitting lind term) Y= 0, X = L Ey = 0= • = Ey = 3 J P2 PL3 PL P L3 P L3 r, 3 r, X — - — X + = — - — + iL 12 J 3J 8 12J 24J PL3 1 _ - 24J x — 24J Px 3 P L)3 PL2 - 24J 104 10. IAS & IFS (Objective & Conventional) Previous Solved Questions Deflection at centre (x= L/2) PL3 PL3 PL3 EYc 96J 48J = 96J PL3 96EJ Y0 For Cantilever Beam d2 y E — xdx dx 2 Px 2dx px3 L/2 d - yi B E[xJ dx A E[(L.iB - y B L) - (0 x i A - y +Ey A = 3J o px3 L 6J PL3 PL3 24J 6J + L/2 PL3 48J 3 :J+ l a = , where iB= 0, YB = 0 PL3 2 -4 3 YA = i L Px2dx JI-I2 2J 16 x x J PL3 EJ Q.6.14 For the beam shown in the figure 6.17 compute the deflection of end B by force method. [CSE-Mains, 2013, CE : 15 Marks] Fixed end Solution: Consider a section at a distance x from A wx 2 - Rx M = bending moment = Rx2 where W= wL, total load, El d2y dx 2 w x2 L 2L Fig. 6.17 W 2 _ Rx-- x 2L dy = Rx 2 Wx 3 r , El — dx 2 6L dy = 0 at x = L dx 0= Constant, C1 dy El — = dx EIy = RL2 WL3 c 2 6L 1 WL2 RL2 6 2 Rx2 Wx3 RL2 2 6L 6 2 Rx3 Wx4 WL2 6 24L + 6 Y= 0, x = L RL2 ,-, x 2 x+u2 Deflection of Beams Strength of Materials 0= RL3 WL3 WL3 RL3 6 24 6 2 RL3 3 Constant, + WL3 8 +L.2 + C2 _ WL3 RL3 8 3 C2 Rx3 Wx 4 WL2 RL2 WL3 + x 6 24L 6 2 8 y = —8, x = 0 So, EIy At end A, 10 5 RL3 3 WL3 RL3 + 8 3 2(-5) R = k8 —81E/ + —E18 = WL3 k8L3 8 4- 3 k WL3 8 3 =— WL3 8= 81E/ + 1(12 3 Objective Questions Q.1 Q.2 The ratio of area under the bending moment diagram to the flexural rigidity between two points along a beam gives the change in (a) deflection (b) slope (c) shear force (d) bending moment [CSE-Prelims, ME : 1998] (a) (c) 3M/2 2E1 Q.3 For cantilever beam shown in the given figure 6.18, the deflection at C due to a couple M applied at B is equal to M A (b) EI (d) 2M/2 El [CSE-Prelims, CE : 2003] In a simply supported beam AB of span L, the mid point is C. In case-1, the beam is loaded by a concentrated load W. In case-2, the beam is subjected to a UDL of intensity w such the wL = W. The ratio of central deflection in case 1 to that in case-2 is 5 (a) — 3 (b) 3 5 (c) — 8 (d) 8 [CSE-Prelims, CE : 2003] El = Constant Q.4 Fig. 6.18 M/ 2 2E/ A simply supported beam AB of span L is subjected to a concentrated load W at the centre C of the span. According to Mohr's 10 6 0. IAS & IFS (Objective & Conventional) Previous Solved Questions moment area method, which of the following gives the deflection under the load? (a) Moment of the area of M/E1 diagram between A and C taken about C (b) Moment of the area of M1E1 diagram between A and B taken about B (c) Moment of the area of M/E1 diagram between A and B taken about A (d) Moment of the area of M1E1 diagram between A and C taken about A [CSE-Prelims, CE : 2005] Q.5 Q.6 (a) 12EIA 6EIA (b) /3 4E1A (C) 3 3E/A 3 / [CSE-Prelims, CE : 2006] (d) 3 A propped cantilever is acted upon by a moment M0 at the propped end. What is the prop reaction? Q.7 A cantilever has rectangular cross-section and supports concentrated load at its free end initially. If depth and width of the beam section are doubled, the deflection at free end of the cantilever will reduce to what percentage of the initial deflection? (a) 20% (b) 15.72% (c) 9.57% (d) 6.25% [CSE-Prelims, CE : 2005] Due to settlement of support at A of propped cantilever shown in the figure 6.19 given below, what is the vertical reaction at B? MADE EASY (a) 0.5M0 L (b) Mo (c) 1.5M0 (d) 2.0Mo L [CSE-Prelims, CE : 2007] A simply supported beam AB of span L carries a concentrated load Wat the centre of the beam with constant El throughout. What does the moment of the area of the M/Eldiagram between A and C, taken about A represent? [CSE-Prelims, CE : 2010] Q.8 B 1 El = Constant Answers Fig. 6.19 1. (b) 6. (d) 3. (d) 4. (d) 8. Deflection 2. (c) 7. (c) Explanations 1. (b) 3. El dx = xi Mdx (d) Case-1: Yc f" Mdx (L2' — L1') = 2. xl El = change of slope (c) + ML xL YOc 2E1 El 3 ML2 — 2 E/ Case-2: _ WL3 • 48E1 _ 5 wL4 Yc 3 • 84 E/ wL = W , 5 WL3 Yc —•384 x EI Yc 8 Yc' — 5 5. (d) Deflection of Beams Strength of Materials 107 7. (c) 4. (d) Moment of area of Mo diagram between A and C taken about A. --t) x Fig. 6.23 L 2 2 EI y 2 dx = Px + Mo dy El — dx Px 2 —+Mox+Ci 2 a y = 0 at x = L Fig. 6.21 dx r, PL2 A A 0 = — ivio...+L.,1 2 1 WL L(L) WL3 Moment about C= - — x- - = 2 4 2 6 96E/ 1 WL L( L) WL3 Moment about A= x— x- - = 48 2 4 2 3 5. PL2 C1 = ---MoL 2 Px 2 A A dy El — = + ivi o x dx 2 (d) 81 = WL3 3E1 P63 82 = 6. 2 AA I ivi oL p 2 L. T x-M°Lx+C2 ElY = I, = (2b)(2d)3 16bd3 = 16/ 12 12 = 2 PC y= 0,x=0, C2 =0 y = 0 at x = L 81 = 6.25% 0= (d) PL3 3 P L3 M L2 P L3 2 6 2 Moe - M01-2 2 Mo P = 1.5— Fig. 6.22 ARB = RBL3 8. (Deflection) Deflection at the centre C of the beam = moment 3E/ 3EIA of area of — between A and C taken about A. E/ M 07 Torsion CHAPTER Q.7.1 Design two solid circular shafts to transmit 200 HP each without exceeding a shear stress of 70 MPa at 20 rpm and other at 20,000 rpm. Give your inference about the final results from the view point of economy. Do you have any other suggestion to improve the economy further? [CSE-Mains, 2013, CE: 10 Marks] Solution: HP = 200 Shaft-1 RPM = 20 2.TE x 20 60 Torque, Ti _ = 2.0944 rad/sec 200 x746 2.094 = 71251.194 Nm = 71251.194 x 103 Nmm ic 3 it 71251194 = — xdi X T = — xdi3 x70 16 16 d13 = 5183990.724 mm3 d1 = 172.98 mm, shaft diameter Shaft-2 RPM = 20,000 27c x 20000 60 Torque, T2 - = 2094.4 rad/sec 200 x 746 2094.4 = 71.251 Nm = 71.25 x 103 Nmm 4 4 = 5183.0 X 70, 16 d2 = 17.3 mm, shaft diameter 71251 = 16 cll x Shaft 1 at 20 rpm Shaft diameter is too large. There is hardly any machine which runs at 20 rpm for power transmission. Shaft 2 at 20,000 rpm Shaft diameter is reduced to one tenth from 173 mm to 17.3 mm. There is saving in material by 99%. But at high speed of rotation, large centrifugal forces are developed if there is any eccentric mass on shaft, which will produce large deflection and slope in the shaft, rigidity is drastically reduced. Even to instal any gear, pulley for power transmission, keyways cannot be made on a shaft of 17.3 mm. Torsion Strength of Materials 1 109 For ordinary power transmission in simple machine speeds are 200-500 rpm, say at 400 rpm, shaft diameter will be approx. 47 mm. To install any gear, pulley on the shaft, keyways can be easily cut. Q.7.2 Find the diameter of a solid cylindrical shaft subjected to 100 rpm and transmitting 350 kW power, when is the shear stress not to exceed 90 N/mm2. What percent saving in weight would be obtained if this shaft is replaced by a hollow one, whose internal diameter equals to 0.65 of the external diameter, the length, the material and the maximum shear stress being the same. [IFS 2012, CE : 15 Marks] Solution: Power = 350 kW = 350 x 1000 Nm/s N, speed = 100 rpm Cylinder velocity, Torque transmitted, T 2rc x 100 = 10.472 rad/sec 60 350 x 1000 10.472 = 33422.54 Nm = 33422.54 x 103 Nm ti = 90 N/mm2, maximum shear stress T= 16 d 3 X d 3 x 90 16 d3 = 1.891298 x 106 mm3 d = 123.66 mm, shaft diameter (b) For same power transmission, same shear stress, polar modulus should be the same 33.422 x 106 = Zps for solid - rcd3 16 4 2 [ 4 Zph, for hollow = — D -(0.650 ]32 D —[0.82149]E)3 16 d3 =1,2173d3 0.82149 D = 1.06774 d Inter dia. = 0.694 d Saving in material depends on area of cross-section of solid and hollow shaft D3 - Area of solid shaft, As - Area of hollow shaft, Ah = itd 2 4 4 [(1.06774d)2 - (0.694d)2 ] = Ed2 [1.14 - 0.4816] 4 Ed2 x0.658 4 Ah AS = 0.658 Weight of hollow shaft is 0.658 x weight of solid shaft Saving in material = 34.2%. MADE EASY 110 10- IAS & IFS (Objective & Conventional) Previous Solved Questions Q.7.3 Compare the weights of equal lengths of hollow and solid shaft to transmit a given torque for the same maximum shear stress if the inside diameter is 2/3 of the outside diameter. [IFS 2012, ME : 10 Marks] Solution: Say, d = Diameter of solid shaft D = External diameter of hollow shaft 2D = Internal diameter of hollow shaft 3 T = torque = Maximum shear stress Solid shaft = 16T 3 70 ...(1) Hollow shaft 0.5D (D4 n = 16T TCD3 32T 81 32T = x 16 D.4) 704 65 81 ) 81 65 x ...(ii) From eq. (i) and eq. (ii) 16T 81 x tD3 65 16T 7cd3 81 65 ...(iii) D = 1.0761 d Since length is same, material is the same (not given in problem), weight of shaft depends on area of cross-section. 3 u x Ah = .0761d)2 -(; x 1.0761d)2 ] = E[1.1580 - 0.51466]c/2 = -7-c10.643341d2 4 4 Ah W = 0.64334 = AS Ws Weight of hollow shaft is only 64.334% weight of solid shaft. 0.7.4 A stepped shaft ABC, is 0.8 m long. For a length AB = 0.4 m, shaft diameter is 40 mm and the length BC = 0.4 m, shaft diameter is 20 mm. Shaft is fixed at both the ends A and C. At the section B, a torque T is applied which causes a maximum shear stress of 100 MPa in stepped shaft. Determine the magnitude of torque T. Torsion Strength of Materials 41 111 Fixed A Fig. 7.1 [CSE-Mains, 2001, ME : 20 Marks] Solution: Torque Tapplied at section B is shared by shaft AB and BC, as shown ± T2 = T ...(i) (since eAc = 0) eAB = eBc T2 x0.4 T1 x0.4 GJ2 GJ1 J1 = 16 J2 as di = 40 mm, d2 = 20 mm T1 16J2 (d1 = 2d2 ) T2 J2 = 16T2 T1 = 0.94117 T T2 = 0.05883 T Assume that max. shear stress occurs in AB So, 3 — TC X 100 x 403 = 1256637 Nmm = 0.94117 T T = — T x 40 1 16 = 16 Torque, T = 1335186 Nmm = 1.335 kNm T2 = 78540.35 Nmm = 16 x r' x 20 3 Then, = 50 N/mm2 in portion BC. (Less than permissible hence, OK) T = 1.335 kNm Hence, Q.7.5 A shaft of 12 cm external diameter and 8 cm internal diameter is subjected to a bending moment of 300 kgf-m, twisting moment of 100 kgf-m and a direct thrust of 10,000 kgf. Determine the maximum principal stress and diameter in which it acts with reference to the axes of the shaft at the end points P and Q of diameter PQ as shown. [CSE-Mains, 1987, ME : 30 Marks] Solution: D = 12cm d = 8 cm Area of cross-section = 5- (D2 - d2 ) = 'It-(144 - 64) = 20 TG cm2 4 4 = 207c cm2 (Comp.) 10,000 = 159.155 kgf/cm2 2rc O Bending stress (sagging moment) Direct stress, ad = 64 (D4 -d 4 )= (12 4 -84 4 -d (D ) -1 ) 6 cm Q Fig. 7.2 112 MADE EASY ► IAS & IFS (Objective & Conventional) Previous Solved Questions -(20736 - 4096) = 260n cm4 = 1 64 ae = + My + 300 x 100 kgf-cmx 6 cm I - 2607c cm4 = ±220.37 kgf/cm2 Shear stress "C = x6 J = Polar moment of inertia = 21= 520n cm4 — 100x100 x6 520n = 36.73 kg/cm 2 Point P (normal stress = 22037 + 59.155 = 379.525 N/mm2 (comp.) and shear stress = 36.73 kgf/cm2) Principal stresses 379.525 + 2 /31, p2 379.125 2 2 ) (36.73)2 379.525 + ab 36.73 = -189.76 ± J36009.8 +1349.1 = -189.76 ± 193.285 ti = -383.04 kgf/cm2, + 3.525 kgf/cm2 36.73 Point Q (Normal stress = 220.37 - 159.155 = 61.215 kgf/cm2)(tensile) and sehar stress = 36.73 kgf/cm2) Principal stress Pi, P2 = 61.215 \I(61.215)2+ (36.73)2 2 ± = 30.605 ± V936.82 +1349.1 = 30.605 ± 47.81 = +78.416 kgf/cm2, - 17.205 kgf/cm2 36.73 kgf/cm2 Q.7.6 A brass tube fits closely over a steel shaft of 100 mm diameter. Find the thickness of the brass tube which would ensure that the torque applied to the assembly is shared equally by the two materials. Find the maximum shear stress in each material and the angle of twist in a length of 3 m. The torque applied is equal to 20 x 104 kgf-cm. Assume modulus of rigidity of steel = 8 x 105 kgf-cm2, modulus of rigidity of brass = 4 x 105 kgf/cm2. [CSE-Mains, 1990, ME : 20 Marks] Solution: GS = 2 GB Say outer radius of bras tube = R2 cm es e9 (Since its is parallel combination) But, TSLS TB LB GSJS GB JB Brass R Ts = TB as given Ls = TB Fig. 7.3 Torsion Strength of Materials GsJs = LIS GBJE3 GO -= 4 JE1 JD X Gs =2 or JI3 = 2Js Js' polar moment of inertia of steel shaft Js = EX 54 = 312.57c cm4 2 II (F? - 5 4 ) = 625 x TC JB = 2 x 312.5 p= 625x 7C CM 4 = 2 R24 - 625 = 1250 R24 = 1875 R2 = 6.58 cm Thickness of brass tube = 6.58 - 5 = 1.58 cm Steel shaft It x c x d3 Ts = 10x104 kg-cm= — Torque, 16 10x104 = EXTXr 3 =EXts X 53 2 Ts Shear stress, 2 - 10 x104 x 2 = 509.3 kgf/cm2 Icx125 Brass shaft TB = 10 X 104 kg-cm 4 _ 7c R -R/ x T B 2 R2 10 x 104 = n[6.58 4 -5 41 TC [1875 -625] 625ic x TB - 2 TB = XTB 2 6.58 6.58 6.58 Maximum shear stress in brass tube TB = 10 x104 x6.58 = 335.116 kg/cm2 625n 0 = TsL GsJs = 10x104 x300 - 0.0382 radian = 2.99° 8x105 x 312.5n Q.7.7 A shaft circular in section and of length L is subjected to a variable torque given by kx 2 , where x is the distance L measured from one end of the shaft and k is a constant. Find the angle of twist for the shaft by use of Castigliano's theorem. Torsional rigidity of the shaft is GJ, see fig. 7.4. [CSE-Mains, 1994, ME : 20 Marks] Solution: Take length dx along the length Strain energy due to twisting Fig. 7.4 113 114 ► MADE EASY, IAS & IFS (Objective & Conventional) Previous Solved Questions LT2dx u= IJO 2GJ au = 9 2Tdx 1 L kx2 1 GJ f0 L But T= k x2 L f L Tdx GJ 0 ar Jo 2a/ dx - kL2 kL3 3GJL 3GJ Q.7.8 Obtain the angle of twist of one end relative to the other end of a tapered shaft having radii ri and r2 at its ends and of length L. The shaft is subjected to equal and opposite torque Tat it ends. Modulus of rigidity of shaft is G. [CSE-Mains, 1995, ME : 30 Marks] Solution: r1 = radius at one end r2 = radius at other end I = radius at distance x (from r1 ) = + kx where k- ro I L = Polar moment of inertia, x angular fracture length, dx, T T L( dx kx) 4 Torque Fig. 7.5 [ 2Tdx dO = G1C(ri L 0 x `4 r4 = 7 2 x 2 T Here, dO = + kx)4 2Tdx A = Tdx] GJ 2T L S (ri + kx)-4 dx ° TtG(r1 + kx)- nG ° L = 2T [(ri + kx) 3 TcG = -3k = o 2T 1 37ckG[ ( ri + k x\3 1 1 0 1 11 2T [ 1 1 ] 2T [ =+ ankG (ri + kL)3 r13 3TckG r13 q 2T [r? - r1 3nkG ri3r? _ = L (r2 - ri )(ri2 + ri r2 + r22 ) 2T x anG (r2 -r1) r13r2 2TL [ rig + rir2 + anG 3 3 ri r2 d1 ] Q.7.9 A solid circular uniformly tapered shaft of length L with a small angle of taper is subjected to a torque T. The diameter at the two ends of the shaft are d and 1.2d. Determine the error introduced if the angular twist for a given length is determined on the basis of the uniform mean diameter of the shaft. [CSE-Mains, 1996, ME : 20 Marks] Torsion Strength of Materials 115 Solution: Tapered shaft 1.2d to d (di = d, d2 = 1.2d) +d1d2 3271 x 3nG 0- d2 +1.2d2 +1.44d2 ] d1 d2 d3 x(1.2d)3 32TL [ 3.64 d x 3itG 1.728 7.152 7± GJ 0', on the basis of mean diameter 1.1d 32TL TL 0' = rc(1.1 A G = 6.9571GJd 7.152 - 6.9531 x 100 = 2.78% 7.152 % Error _ Q.7.10 For the shaft loaded as shown in figure 7.6, calculate the maximum shear stress induced and the angle of twist for cross-section at A value of modulus of rigidity is G. 4M M C t 2t 3d t B O(A 2L 0-14 2L L Fig. 7.6 [CSE-Mains, 1994, ME : 20 Marks] Solution: Polar moment of inertia nd 4 J 1 it(2d)4 J,32 - 32 = 16,11, 7C (3d)4 J, = - 32 = 81J1 4M - T2 JZ A M2L Fig. 7.7 Let torque at 0 is Ti and at D is T2 as shown, Therefore in portion AB M- T1 = 4 M- T2 T2 - T1 = 3 M °Ao OBA + °DB = 0 x 2L (114-7-1)2L + T2 L L 1 J2 J3 Ji J2 =0 116 P. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions T1 x2L (M-Ti )x2L J1 16,11 Lx2Ti Lx2Ti - +T2 on- Ti)2L 16 M x2L 16 17 x 2LTi 16 + + IL 16J1 T2 x L 16 Tix2L 16 2ML 4. 16 17 8 + + T2L 16 L ) 81J1 T2 X L 81 + T2L 81 (81+16)T2L 16x81 M T1 + 8 + = 0 = 0 = 0 = 0 97T2 16 x 81 = M 97T2 17., /1 + 8 16 x 81 Also or 0 ...(ii) 8 M 8 97 1 8 — T1+- x- x-T2 = 8 x1 16 81 17 M Ti + 0.03522 T2 17 0.0588 M Ti + 0.03522 T2 T2 - Ti 3M T2 2.9547 M -0.0453 M 7-1 From eq. (ii) Ti x 2L = -0.0453M x 2L e AO GJ 0.0453M x 2Lx 32 = 0.9228ML Gxrcd 4 Gd 4 16T2 _ 16 x 2.9547M 1.881M tin portion BC n(2d)3 8icd3 d3 16T2 = 0.557 M ti in portion CD Hence, tmax is in portion on BC d3 TC(3d)3 1.881M - d3 Q.7.11 A cantilever ABC, 200 mm long and 20 mm diameter is fixed at end A as shown in the below figure 7.8. A horizontal axial load of 10 kN, a vertical load of 1 kN acts at end C. Torque T applied at section C, produces a maximum shear stress of intensity 50 MPa in cantilever. Determine principal stresses at point B of cantilever. 10 kN H 100 mm-01 100 mm Fig. 7.8 [CSE-Mains, 2001, ME : 20 Marks] Torsion Strength of Materials 41 117 Solution: Axial compressive force = 10,000 N 7C X 202 = 100 nmm2 A= — 4 as = compressive stress, uniform Area, 10,000 1007c 31.83 MPa (-ve sign indicate compressive nature) BM at B = 1 kN x 100 = 1000,00 Nmm Z = ird 3 IC 3 = x 20 = 0.7854 x 103 mm3 32 32 1000,00 = 127.324 MPa (at point B) CS, " - 0.7854 x 10° (Tensile at upper fibre) Net direct stress at B, (upper fibre) aB = 127.324-31.830 = 95.494 MPa Shear stress due to torque, ti = 50 MPa Bending stress, Principal stress at B, !Du p2 = \ (GB)2 2 - 2 = 47.747 ± V47.7472 + 502 = 47.747+ V2279.8 + 2500 = 47.747 ± 69.136 pi = 116.883 MPa 1)2 = -21.389 MPa Q.7.12 A solid circular shaft is encased in a hollow copper shaft so as to make a compound shaft. The diameter of steel shaft is 8 cm and outside diameter of the copper shaft in 11 cm. The compound shaft is 2 m long and is subjected to a torque of 8 kNm. Determine maximum shear stress in steel and copper. Gsteel = 2 Gcopper = 80 kN/rne [CSE-Mains, 2007, ME : 20 Marks] Solution: Solid steel shaft d = 8 cm rcd 4 Tcx8 4 J S — 32 - 32 128ncm4 Hollow copper shaft d= 8, D = 11 cm (114 - 84 ) = (14641- 4096) = 329.53n cm4 32 32 Gs = Gcu x 2 Angle of twist will be equal in both shafts JCL/ = °s = esu TsLs JsGs T„L„ J„G„ 118 0. lAS & IFS (Objective & Conventional) Previous Solved Questions Ls Ts Tcu = Lcu 2 128ir Js x Gs x = 0.778 Gcu 329.1341 1 Jcu Ts = 0.778 Tcu Ts +Tcu = 8 kNm 1.778 Tcu = 8 Tc„ = 4.50 kNm Ts = 3.50 kNm Also, Stress in solid steel shaft s 3.5 x106 7C x , = 43.51 N/mm2 - 16 128 x10- Tc Maximum shear stress in solid steel shaft Copper shaft Ta, = JCL, X 55 = 329.5371 x 104 tcu 55 X CU - 329.537c x104 Tcu 55 tai = 23.90 N/mm2, Maximum shear stress in hollow copper shaft. 4.5 x 106 Q.7.13 A mild steel shaft of 200 mm diameter is to be replaced by a hollow shaft of alloy steel for which the allowable shear stress is 25% greater. If the power to be transmitted is to be increased by 20% and speed of rotation increased by 5%, determine the maximum internal diameter of the hollow shaft taking its external diameter to be limited to 200 mm. [CSE-Mains, 1998, ME : 30 Marks] Solution: Tc Solid shaft 3 —(200) X = 16 P= coT =T PH = 1.2 P (4-/ = 1.05 w Hollow shaft T= 1.2P 1.05w it 1 .05 it [200 4 - d 4 ] x 1.25T P x = (200)3 ti [Since Ts = TH] — = w 16 1.2 16 200 or or ...(ii) 1.05 1.2 (2004 -d4 ) - 2003 200 1.09375[2004 - c/4] = 2004 (1.09375-1)2004 = 1.09375 d4 (0.09375)2004 = 1.09375 d4 x1.25 _ 4 0.09375 x 200 1.09375 d = 0.541 x 200 = 108.2 mm, internal diameter of hollow shaft. Torsion Strength of Materials 1 119 Q.7.14 The pulley A exerts a torque on shaft B as shown in figure 7.9. The total vertical tension on both sides of the belt on each pulley is 400 kg-f. Diameter of the shaft is 6 cm. If the tensile and shear stress intensities are not to exceed 3200 kgf/cm2 and 1600 kgf/cm2 respectively, what is the maximum power that can be transmitted by the shaft when running at 150 rpm. Shaft may be assumed to the simply supported at the bearings C and D. 400 kgf Rc 400 kgf RD = 66.667 kgf Fig. 7.9 [CSE-Mains, 1989, ME : 30 Marks] Solution: Reaction Moment about C RD x 0.9 + 400 x 0.3 = 400 x 0.45 60 RD = — 0 . 9 = 66.667 kgf Rc = 733.33 kgf BM diagram MA Mc = 400 x 0.3 = - 120 kgf-m (Hogging) MB = +66.67 x 0.45 = 30 kfg-m Mmax = 120 kgf-m = 120,00 kg-cm crmax = 3200 kgf/cm2 Amax = 1600 kgf/cm2 Me = Equivalent BM = 32 X d 3 X a ic 3 = --X 6 x 3200 = 67858.4 kgf/cm 32 Te = Equivalent TM = 16 x d 3 xTmax It 3 = — x 6 x 1600 = 67858.4 kgf-cm=VT2 + Mmax 2 16 (Putting the value of Mmax) = 46047626629 = 144000000 + T 2 T2 = 4460762629 T = 66789 kgf-cm = 667.89 kf-m Mi4T2 I- M 2 Me = 2 = 67858.4 120 N. MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions 2 x 67858.4 = 120,00 + VM2 +T2 123716.8 = Vm2 + T2 = V12000 2 + T2 T = 123133.4 kgf-cm = 1231.33 kgf-m or Thus value of Twill for greater than 667.89 kgf-m (hence permissible value for shear stress will reach first than permissib;e value of tensile stress) 2ic x 150 = 15.708 rad/sec 60 Power = 667.89 x 15.708 kgf-m/s = 10491.2 kfg-m/s = 140 HP, as 1 HP = 75 kfg-m = Objective Questions Q.1 Two steel shafts, one solid of diameter D and other hollow of outside diameter D and inside diameter DI2 are twisted to the same angle of twist per unit length. Ratio of maximum shear stress in solid shaft to that in the hollow shaft is 4 (a) 4 (c) 16 15 Q.3 A stepped solid circular shaft shown in the figure 7.11 is built in at its ends and is subjected to a torque To at the shoulder section. The ratio of reactive torque Ti and T2 at ends is (J1 and J2 are polarmoment of inertia) T, (b) 8 (d) 1 [CSE-Prelims, ME : 1998] Q.2 F A load perpendicular to the plane of the handle is applied at the free end as shown in the figure 7.10. The value of (shear force), bending moment (BM) and torque (T) have been determined respectively on 400 N, 340 Nm and 100 Nm by a student. Among the values these are (a) (b) (c) (d) Fig. 7.10 SF, BM and torque are correct SF and BM are correct BM and torque are correct SF and torque are correct [CSE-Prelims, ME : 1994] L2 -1.1 Fig. 7.11 J 2L2 (a) jiLi J1L2 (c) J2L1 Q.4 400 N 1" J2L1 (b) J1L2 J1L1 (d) J21-2 [CSE-Prelims, ME : 2001] Steel shaft and brass shaft of same length and diameter are connected by a flange coupling. The assembly is rigidity held at its ends and is twisted by a torque through the coupling. Modulus of rigidity of steel is twice that of brass. If torque of the steel shaft is 500 Nm, then the value of the torque in brass shaft will be (a) 250 Nm (b) 354 Nm (c) 500 Nm (d) 708 Nm [CSE-Prelims, ME : 2001] Q.5 A shaft is subjected to a bending moment of M = 400 Nm and a torque = 300 Nm. The equivalent bending moment is Torsion Strength of Materials A hollow shaft of length L is fixed at its both ends. It is subjected to torque Tat a distance of L/3 from one end. What is the reaction torque at the other end of the shaft? 2T (a) —3-- (b) (c) 3 (d) 4 2 [CSE-Prelims, ME : 2007] Q.7 121 (b) 700 Nm (d) 450 Nm [CSE-Prelims, ME : 2002] (a) 900 Nm (c) 500 Nm Q.6 4 Consider the following statements, in connection with a metallic rod of a circular section being subjected to equal and opposite torque Twithin elastic limit: 1. The transverse section of the rod does not experience warping. 2. The diameter of rod does not alter. 3. Angle of relative twist between two sections is proportional to the lengths between these sections. 4. A surface element of the rod is under pure shear state of stress. Which of these statements are correct? (a) 1, 2, 3 and 4 (b) 1, 2 and 3 only (d) 2, 3 and 4 only (c) 2 and 3 only [CSE-Prelims, ME : 2009] Q.8 Which one of the following is true for torsional shear stress at the axis of a circular shaft? (a) Minimum (b) Maximum (c) Negative (d) Zero Q.9 A circular shaft of length 'L' is held at two ends without rotation. A twisting moment 'T is applied at a distance l'/3 from left support as shown in the given figure. The twisting moment in the portion AB will be Fig. 7.12 (b) T/3 (d) 2T/3 (a) T (c) T/2 Q. 10 A circular shaft of length 'L' a uniform crosssectional area 'A' and modulus of rigidity '6' is subjected to a twisting moment that produces maximum shear stress in the shaft. Strain energy in the shaft is given by the expression T2AL/kG, where k is equal to (a) 2 (b) 4 (c) 8 (d) 16 [CSE-Prelims, CE : 2001] Q. 11 When subjected to a torque, a circular shaft undergoes a twist of 1° in a length of 1200 mm, and the maximum shear stress induced is. 80 N/mm2. The modulus of rigidity of the material of the shaft is 0.8 x 105 N/mm2. What is the radius of the shaft? (a) 90/7r mm (b) 108/Tc mm (c) 180/Tt mm (d) 216/7cmm [CSE-Prelims, CE : 2008] Q. 12 Power is transmitted through a shaft, rotating at 2.5 Hz (150 rpm). The mean torque on the shaft is 20 x 103 Nm. What magnitude of power in kW is transmitted by the shaft? (a) 501c (b) 120 TC (c) 100 Tc (d) 1507c [CSE-Prelims, CE : 2009] Answers 1. (d) 6 . (c) 11. (d) 2. (d) 7. (a) 12. (c) 3. (c) 8. (d) 4. (a) 5. (d) 9. (d) 10. (b) Explanations 1. (d) Ti J1 T22 G6 = _. T = L J2 Ti . Ji = TcD4 x T2 J2 32 32 16 04 04)= 15 It ( 16 12 2 0. lAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY 15 T, 16 ' T 2 Ti = T2 = 16 3 D T • ( -15 D4 )x_ 32 06 2 xi' ) D Fig. 7.14 • (15 D3 ), , t 16 x 16 —3 :7 16 16) 15 7C n Because, Ts Ls — TO LB LB TB GBJ B Ts _Gs TB GB = e, Ratio = 1 2. (d) SF = 400 N BM = 100 Nm 5. (d) m = 400 Nm T= 300 Nm Mc = 400+,4002 + 3002 2 = 450 Nm 400 N 6. (c) T= 100 Nm 80 Nm 400 N 400 N T= 100 Nm Fig. 7.15 Fig. 7.13 °BA = °BC 3. (c) 0 TA T1 L1 = T2 L 2 GJ1 GL/2 GJ x L 3 — TB X 2L GJ x 3 TA = 2TB T1 = J 1 L 2 T2 J2 L1 TA + TB = T TB = T/3 4. (a) T =G Ts s =2 TB GB Ts = 2TB TB = 0.5 Ts = 250 Nm 7. (a) All statements 1, 2, 3 and 4 are correct. 8. (d) At the axis of the bar, torsional shear stress is zero. Torsion Strength of Materials 4 12 3 11. (d) 9. (d) TAB = L = 1200 mm 2T 0= 180 ti = 80 N/mm2 G = 0.8 x 105 N/mm2 T Bc = 3 Because TAB x L GJ TBc 2L 3 — GJ 3 TAB = 2 TBc T GO R L T1._ R= GO TAB 4- TBC = T TAB = = 80 x 1200 x180 0.8 x105 xic = 216/ir mm 2T 12. (c) 10. (b) Strain energy, U = 7C T2AL 4G co = 2.5 x 27c = 57r rad/sec T= 20 x 103 Nm Power = Tco Nm/s = 100 x 7C X 103 W = 100 7c kW 11•IM• 08 Springs CHAPTER Q.8.1 A close coiled helical spring is of 80 mm mean coil diameter. The spring extends by 37.55 mm when loaded axially by a weight of 500 N. There is angular rotation of 45° when the spring is subjected to an axial couple of 20.0 Nm. Determine the Poisson's ratio of the material of the spring. [CSE-Mains, 1996, ME : 20 Marks] Solution: Axial load, Stiffness, W = 500N Extended load = 37.75 mm k_ Axial couple, 500 = 13.265 N/mm 37 .75 Gd 4 8nD3 M = 20 x 103 Nmm 0 = Angular rotation = ML _ 7C El - 4 L = 27EnR =TcnD nd4 64 D = 80 mm / 13.245 - G= G x d4 8n x 803 13.245 x 8n x 803 n x 54251520 = d4 d4 ML El th 1. E= ML 20x103 xrcx nx80x4x64 = (t)./ it xlcd4 n = —, x130379724.14 cr Springs • Strength of Materials G= d" x 12 5 54251520 E 130379729.4 = 2.4032 = 2(1+v) G - 54251520 v = 0.2016, Poisson's ratio Q.8.2 A close coiled spring has coil diameter to wire diameter ratio of 6. Spring deflects 30 mm under a load of 500 N and maximum shear stress is not to exceed 350 MPa. Find diameter and length of wire required. Modulus of rigidity of wire material = 80 GPa. [CSE-Mains, 2012, ME : 10 Marks] Solution: G = 80,000 N/mm2 k = Stiffness = 300 = 16.6667 N/mm Gd 4 8nD3 D = coil diameter d = number of coil Where, 16.6667 = 80000 (d)3 xdx 8n D 1000 d d xi 1 )3 = 10000x — n 6 216 x n d — = 16.6667 x 216 = 0.36 n 10000 D 3 500x — - — 7c x d 2 16 Torque X "C 3 7E X d x 350 250 D = — 16 250 xr) = 6 d2 x350 116 d 16 1 x 350 — =21.82696 It d = 4.67 mm, wire diameter d2 = 250 x 6 x nLength of the wire required, d = 13 , number of turns 0.36 L = icDn = It x 6 x 4.67 x 13= 1144.36 mm Q.8.3 Design a suitable helical spring for a balance which is used to measure 0 - 100 kg over a scale of 80 mm. Spring is to be enclosed in a casing of 25 mm diameter. Approximate number of turns is 30. Also calculate the maximum shear stress induced, G = 0.85 x 105 N/mm2. [IFS 2011, ME : 10 Marks] 126 0. MADE, EA 1AS & IFS (Objective & Conventional) Previous Solved Questions Solution: Load = 0-100x9.8=0-980N Range = 80 mm 980 k = — = 12.25 N/mm 80 Stiffness of spring, k - Gd 4 8nD3 G = Shear modulus = 85000 N/mm2 Where, d = wire diameter n = number of coils = 30 Casing diameter = 25 mm D = 25 - d So, mean coil diameter Putting the values in k expression 12.25 - 85000 x d4 8 x30 xD3 D3 = 28.91156 d4 D = 3.069 C/4/3 25 - d = 3.069 d4/3 25 = d + 3.069 d4/3 Say, d= 4 mm RHS = 4 + 19.486 = 23.486 d = 4.1 mm RHS = 4.1 + 20.138 = 24.138 d = 4.2 mm RHS = 4.2 + 20.795 = 24.995 So, wire diameter, d= 4.2 mm Coil diameter, D= 25 - 4.2 = 20.8 mm 980 = 16 d3T 980 x 20.8 x 16 = 700 MPa 2xnx 4.23 For a closed coil helical spring under compression, illustrate the stress distribution across wire diameter. (ii) Differentiate between compound and composite helical spring. (iii) The spring load against which a valve is opened is provided by an inner helical spring arranged within and concentric with an outer helical spring. Free length of inner spring is 6 mm larger than that of outer helical spring. Outer spring has 12 coils of mean diameter 25 mm, wire diameter 3 mm and initial compression 5 mm when the valve is closed. Find the stiffness of the outer spring, if the greatest force required to open the valve of 10 mm is 150 N. If the radial clearance between the springs is 1.6 mm. Find the wire diameter of the inner spring, if it has 10 coils. For both springs, G = 82000 N/mm2. [CSE-Mains, 2002, ME : 30 Marks] Q.8.4 (i) Springs Strength of Materials 12 7 Solution: (i) On wire diameter, Ts = torsional shear stress = 16WR 703 R = mean coil radius 4W Td = Direct shear stress — icd2 At inner coil radius, resultant shear stress 'Cr = 'Cs + "Cd At outer coil radius, resultant shear stress, Tr = is — Td (ii) Compound spring Springs in parallel are called compound spring keq = k1 + k2 Composite spring Fig. 8.1 Shear stress distributions kik2 kequivalent = k1 + k2 Where k1 and k2 are stiffnesses of individual springs (iii) Initial compression in outer spring = 5 mm Initial compression in inner spring = 5 + 6 = 11 mm (As the free length of the inner spring is 6 mm more than the free length of the outer spring) Say k1 = Stiffness of inner spring in N/mm k2 = Stiffness of center spring in N/mm Initial load on valve = 11k1 + 5k2 Outer spring n= 12, R = 12.5, d = 3 mm k2 = Gd4 82000 x 34 64nR3 64 x12 x 12.53 = 4.428 N/mm The valve is opened by 10 mm, additional force required to open the value F0 = 10 ki + 10 k2 Ft = Total force = 10 + 10 k2 + 11 = 21 k1 + 15 k2 = 150 N k2 = 4.428 N/mm 15 k2 = 66.42 21 k1 = 150 — 66.41 = 83.58 N k1 — — 83.58 = 3.98 N/mm 21 Gd4 = 82000 x 16n1F3 16 x10 xl:q Radial clearance between the springs = 1.6 mm Outer spring, R2 = 12.5 mm Radial clearance = 1.6 mm Outer coil radius of inner spring = 12.5 — 1.6 = 10.9 mm mean coil radius of inner spring -= (10.9 — 0.5d1) mm + 5 k2 128 •• MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions 82000 x dj 3.98 — 160 x (10.9 — 0.5d)3 82000 d4 x 1 = 128.770 160 3.98 10.9 — 0.5d1 = 5.049 d11.33 5.049 d11.33 + 0.5 di = 10.9 (10.9 — 0.5d1)3 — say d1 = 2 mm LHS = 13.72 d1 = 1.8 mm LHS = 11.95 d1 = 1.7 mm LHS = 11.095 d1 = 1.68 mm LHS = 10.923 Wire diameter of inner coil spring is 1.68 mm Objective Questions Q.1 Two identical springs, each of stiffness k are assembled as shown in the figure 7.2. Combined stiffness of assembly is 1 2 1 2 (a) — kx2 - - kX1 2 2 (b) 2 k(xi — x2 )2 1 2, + xi)2 (c) — (d) k (x1 X2 2 [CSE-Prelims, ME : 1999] Q.4 A close coiled helical spring has wire diameter 10 mm and spring index 5. If the spring contains 10 turns, then the length of the spring wire would be (a) 100 mm (b) 157 mm (c) 500 m (d) 1570 mm Q.5 Match List-I (Type of spring) with List-II (Application) and select the correct answer List-I A. Leaf/helical spring B. Spiral springs C. Beleville springs List-II 1. Automobile/Railway coaching 2. Shearing machine 3. Watches Codes: A B C 2 3 (a) 1 2 (b) 1 3 1 2 (c) 3 3 1 (d) 2 [CSE-Prelims, ME : 2000] k2 Fig. 7.2 (a) k2 (c) k Q.2 (b) 2k (d) k/2 [CSE-Prelims, ME : 1998] Two close coiled springs are subjected to same axial force. If the second spring has 4 times the coil diameter, double the wire diameter and double the number of coils of the first spring, then the ratio of deflection of the second spring to that of the first will be (b) 2 (a) 8 (d) 1/16 (c) 1/2 [CSE-Prelims, ME :1998] 0.3 A spring of stiffness k is extended from a displacement x1 to a displacement x2 . Work done by the spring is Springs Strength of Materials Q.6 The equivalent spring stiffness for the system shown in figure 8.3. CS is the spring stiffness of each of the three springs. (a) S/2 (c) 2S/3 Q.7 Q.8 Fig. 8.3 (b) S/3 (d) S [CSE-Prelims, ME : 2001] If the number of turns in a spring are halved, its stiffness is (a) halved (b) doubled (c) increased from turns (d) not clayed [CSE-Prelims, ME : 2007] A closed coiled helical tension spring having 20 coils is cut in two parts such that part A has 8 coils and part B has 12 coils. What is the ratio of stiffness of the spring A to the stiffness of the spring B (a) 1 (b) 1/2 (c) 3/2 (d) 3/4 [CSE-Prelims, ME : 2006] Q.9 A close-coiled helical spring has 100 mm mean diameter and is made of 20 turns of 10 mm diameter steel wire. The spring carries an axial load of 100 N. Modulus of rigidity is 84 GPa. The shearing stress developed in the spring in N/mm2 is (b) 160//n (a) 120/it (c) 100/it (d) 80/7c [CSE-Prelims, CE : 2003] Q.10 For a laminated spring, 1= length, n = number of plates, width of each plate = b, thickness of each plate = t, central point load = W What is the expression for the maximum bending stress? (a) f = (c) f 3WI , 4nkt (b) f = 2W1 (d) f = (b) Same deflection 8 for both, k = k1 + k2 = 2k 3W1 Q.11 A close coiled helical spring is subjected to an axial moment M, producing an angle of rotation of 90° at free end with respect to the fixed end. Strain energy absorbed in spring is 1007t Nmm, what is M (a) 100 Nm (b) 200 Nmm (c) 300 Nmm (d) 400 Nmm Answers 1. (b) 6. (c) 11. (d) 2. (a) 7. (b) 3. (a) 8. (c) 2. (a) k, Gd 4 64nR3 W k1 = W = 88 k2 1 Fig. 8.4 3W1 2nb2t 2nbt 2 [CSE-Prelims, CE : 2007] 3nb2t Explanations 1. 129 4. (d) 5. (b) 9. (d) 10. (d) 130 0- k, = Gx16d4 7. 64 x 2n x 64R3 (b) Doubled Gd 4 Gd 4 16 64nR3 x 128 Gd 4 [11 64nR3 L8 3. MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions =x 64 Ran n is in denominator. 8. (c) (a) A = 8 coil B =12 coils kA oc 1 8 , r‘B c'c k A 12 , 3 = — = 1.0 = kB 8 -4'i [4 X2 Fig. 8.5 9. (d) D= 100 mm R 50 mm d= 10 mm n = 20 W= 100N G = 84 GPa — kx2 2 --kxi2 = Work done 2 2 4. (d) L= irDn.nx5dn = nx5x10x10 = 1570 mm 5. 6. (b) A. Leaf/helical spring B. Spiral Spring C. Belleville spring 1. Automobiles/Railway coach 3. Watches 2. Shearing machine 12 16WR rcd3 16 x100 x 50 Tcx103 L = Length n = number of beam W = Load t = thickness b = width WL n6 x 4 = 6 x' f_ 3 WL x , 2 nbt` 11. (d) 7r 1 2Mx2 = 10076 S8 `je 2S x S _ 2 s 2S+ S 3 80 10. (d) (c) Fig. 8.5 = M= 400 Nmm N/mm2 09 Struts and Columns CHAPTER Q.9.1 The critical buckling load of a cast iron hollow cylindrical column 3 m in length, when hinged at both the ends is equal to PkN. When the column is fixed at both the ends, its critical load increases to (P+ 300) kN. If the ratio of external diameter to internal diameter is 1.25, E= 100 GPa, determine external diameter of column. [CSE-Mains, 2012, ME : 12 Marks] Solution: Critical load, when both ends hinged = P Critical load when both ends fixed = 4P = P + 300 kN So P = 100 kN E = 100000 N/mm2 L = 3 m = 3000 mm P- 100000 - _ (Since effective length is half) n2EI 12 ic 2 X100000 XI 30002 30002 Tc 2 — 1 -t--(D4 64 d= 0.8D 911890.653 mm4 -d4 ) [ 4 - 0 4096D4 ] = 0.02898D4 = 64 D 0.02898 D4 = 911890.653 D4 = 3146.49 x 104 mm4 D = 74.89 mm, external diameter of CI column. I Q.9.2 A hollow column, 400 mm external diameter and 300 mm internal diameter, is hinged at both ends. If the length of the column is 5 m, E = 0.75 x 105 N /mm2, factor of safety 5. Rankine's constants 1/1600 and crushing stress 587 N/mm2, find the safe load, the column can carry without buckling. Use Euler and Rankine, formulae. [CSE-Mains, 2005, ME : 30 Marks] 13 2 P. IAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY Solution: E = 75 x 103 N/mm2 D = 40 cm d = 30 m 700 A = 4 (1600 — 900) = -iL = 175 5c cm2 / = 7 (402 — 302 ) = 8.59 x 104 cm4 = 8.59 x 108 mm4 64 1= 5m n2E1 5c2 x 75000 x 8.59 x108 P = Euler buckling load — 2 = 50002 = 25433970.54 N = 25433.97 kN FOS = 5 Pe safe = 5086.8 kN _ I _ 8.59 x108 = 1.5624 x 104 mm2 — A — 1755tx100 12 50002 k2 k2 PR— =1600 .A /2 = 1+ax--g— k2 587x1757cx100 1 x 1600 1+ 1600 = 161360.0 x 100 N = 16136 kN PR safe = 3227.2 kN Safe load without buckling = 3227.2 kN Q.9.3 A vertical column 6 m length is fixed at the base and a clockwise moment of 1.4 kNm is applied at the top of the column, a horizontal force of P is applied to the column at a height of 3 m above the base so as to give a CCW moment. Determine the value of force P so that horizontal deflection at the top of the column and at the point of application of P shall be equal (i) when the deflections are on the same side (ii) when the deflections are on the opposite sides of the vertical line through the foot of the column [CSE-Mains, 1993, ME : 20 Marks] Solution: Take a section yy at distance x from A Mx = 1.4 — P(x — 3) El d2y , = 1.4 — P(x — 3) Az' El dy = 1.4x --(x— 3)2 +ci 2 dx dy 0, at x = 6 dx = 0 = 1.4 x 6- 2 (3)2 +ci Constant, = 8.4 — 4.5 P+ c1 = 4.5 P— 8.4 Fig. 9.1 Struts and Columns Strength of Materials 4 133 dy El- = 1.4x - -(x - 3)2 + (4.5P - 8.4) dx 2 El y = 0.7x 2 - f(x - 3)2 + (4.5P - 8.4)x+ C2 y = 0, at x = 6, from top = 0.7 x 36-i x 27 +(4.5 P-8.4)x 6 +C2 0 = 25.2 - 4.5 P+ 27 P- 50.4 + C2 = 22.5 P- 25.2 + C2 C2 = 25.2 - 22.5 P Deflection at, x = 0, yA = 25.2 - 22.5P El Elyc = 0.7 x3 2 -0 + (4.5 P- 8.4) x 3 + 25.2 - 22.5 P = 6.3+ 13.5 P- 24.6 + 25.2 - 22.5 P At x = 3 m, = - 9P+ 6.9 6.9 - 9P Y, Deflections on same side 25.2 - 22.5P 6.9 - 9P El El 18.3P = 13.5P P = 1.335 kN Deflections on opposite sides 9P - 6.9 25.2 - 22.5P El El 31.5P = 32.1 = 1.019 kN (For deflection on opposite side) Q.9.4 A cast iron column of circular section 20.0 cm external diameter and 2.0 cm thickness and of height 4.0 meters is required to carry 15.0 kN at an eccentricity of 2.5 cm. If both the ends are fixed, find the extreme stresses In the column section. E. 100 GPa. [CSE-Mains, 2011, CE : 15 Marks] Solution: P= 15 kN, D= 20 cm, d= 16 cm A, area of cross-section, A = -71(202 -162 ) = = 113.1 cm2 4 15000 ad , direct shear - 132.626 N/cm2 = -1.32626 N/mm2 (Compressive) 113.1 eccentricity, e = 2.5 cm = 25 mm 4 .1p 2 E/- Pe sec = P-+MaX A Z le = equivalent length = I= 64 (D4- = 64 = 2 m (fixed ends) (204 -164 ) 134 ► MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions = it-[-I 60000 - 65536] = 4637 cm4 = 4637 x 104mm4 64 E I = 100000 x 4637 x 10+4 Nmm2 = 4637 x 109 N/mm2 1/EPI _ I 15000 = 1 I 15 _ 1 \ 4637 x 109 1000 4637 \ 1000 x 17.582 1 le X, = 1 \IP = 1000x 17.582 2 El 1000 x 17.582 sec 3.258° = 1.0016 Z= 100 0.05688 rad = 3.258° = 4637 x 104 mm4 = 4637 x 100 mm3 p 15000 x 25 x 1.0016 = +0.810 N/mm3 e sec3.258° = 463700 Amax = 1.32626 + 0.810 = 2.13626 N/mm2 (comp.) Amin = 1.32626-0.810 = 0.51626 N/mm2 (comp.) Q.9.5 A built up column consists of three ISWB 450 CD 0.794 kN per m connected effectively to act as one column (below figure). Determine the safe load carrying capacity of built up section, if the unsupported length of column is 4.25 m. Properties of ISWB 450 10 0.794 kN/m. Area = 101.15 cm2 I = 35057.6 cm4 /YY= 17067.7 cm4 tW = Thickness of web = 9.2 mm f = 250 MPa, yield stress. Slenderness Ratio -x 450 mm 10 20 30 40 50 60 70 550 148 145 139 132 122 112 Permissible stress in axial compression in MPa fr = 25 MPa. [CSE-Mains, 2013, CE : 10 Marks] Solution: Area of cross section = 3 x 101.15 cm2 = 303.45 cm2 I = 2 x 35057.6 + 1706.7 = 71821.9 cm4 /wPassing through G = 1 + 2Iyy + 2 x 101.15 (22.5 + 0.46)2 = 35057.6 + 2 x 176.7 + 202.3 x 527.1616 = 35057.6 + 3413.4 + 106644.8 = 145115.8 cm4 < I yy 71821.9 ,, - 236.684 cm2 K- 303.45 k = 15.38 cm L = 4.25 m = 425 cm L - = 27.63 k Struts and Columns Strength of Materials 4 13 5 From the table f = 145 + 30 - 27.63 10 Safe load = = = = x 3 = 145 + 0.711 = 145.71 N/mm2 145.71 x 303.45 x 100 N 4421570 N 4421.57 kN 4.42157 MN Objective Questions Q.1 Which one of the following statements is correct? (a) Eulers formula holds good only for short columns (b) A short column is one which has the ratio of its length to least radius of gyration greater than 100 A (c) column with both ends fixed has minimum equivalent length or effective length The equivalent length of column with one (d) end fixed and other end hinged is half of its actual length [CSE-Prelims, ME : 2000] Q.2 A strut's cross-sectional area A is subjected to a load Pat point S(h, k) as shown in the figure. Stress at the point Q(x, y) is Q.3 (a) S (c) R Which one of the pairs is not correctly matched (a) Slenderness ratio 1. Ratio of length of the column/least radius of gyration. (b) Buckling Factor 2. Ratio of maximum load to the permissible axial load on the column. (c) Short Column 3. A column for which slenderness ratio < 32. (d) Strut 4. A member of a structure in any position and carrying an axial compressive load. [CSE-Prelims, ME : 2003] Q.5 What is the cause of failure of a short MS strut under an axial load? x y P + Phy Pkx + (a) — A Ix I y (c) (d) P Phx Pky A Iy P _Phx Pky A Ix I y P + Phx Pky — A ly I x [CSE-Prelims, ME : 2000] (b) Q (d) P [CSE-Prelims, ME : 2002] Q.4 Q(x, y) S(h, k) (b) A short vertical column having a square cross section is subjected to an axial compressive force, centre of pressure of which passes through the point R, as shown in figure. Maximum compressive stress occurs at point. 13 6 0. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions (a) Fracture stress (c) Buckling (b) 1/2 (d) 1/4 [CSE-Prelims, CE : 2007] (a) 2 (c) 4 (b) Shear stress (d) Yielding [CSE-Prelims, ME : 2007] Q.6 Which one of the following columns has effective length twice the value of actual length? (a) Hinged-Hinged column (b) Fixed-Fixed column (c) Fixed-Hinged column (d) Fixed-Free column [CSE-Prelims, ME : 2009] Q.11 In an axially loaded compressive member with a circular cross-section of radius r, what is the radius of the core section which is proof against tensile stress? (a) r /2 (b) r /3 (d) r /6 (c) r /4 (CSE-Prelims, CE 2009] Q.7 Critical Euler buckling load for a long column of diameter D was evaluated as P. If the diameter of the section is reduced to D/2, what is the load carrying capacity of the modified column? (b) P/4 (a) P/2 (d) P/16 (c) P/8 [CSE-Prelims, ME : 2009] Q.12 A Short column of external diameter D and internal diameter 01 is subjected to an eccentric load P at an eccentricity of e thereby causing tensile stress at the extreme fiber. What is the magnitude of eccentricity? D2 +02 D2 + (a) (b) 8nD 8D Q.8 The axial load which just produces the condition of elastic instability in a column is (b) Euler load (a) Rankine load (c) Yield load (d) Crushing load [CSE-Prelims, CE : 2002] Q.9 If the stress on the cross-section of a circular short column of diameter D is to be wholly compressive, the load should be applied within a concentric circle of diameter (b) D/8 (a) D/2 (d) D/6 (c) D/4 [CSE-Prelims, CE : 2003] Q.10 Column C1 has both the ends hinged while the column C2 has one end hinged and other end fixed. What is the ratio of the critical load for C1 to that of C2 according to the Euler's formula? (c) D2 — D1 (d) 870 80 [CSE-Prelims, CE : 2010] Q.13 A column of length 4 m, an area of cross section 2000 mm2, moments of inertia, /xx = 720 cm4, l ye = 80 cm4, is subjected to a buckling load. Both the ends of the column are fixed. What is the slenderness ratio of the column (b) 120 (a) 200 (d) 80 (c) 100 Answers 1. (c) 6. (d) 11. (c) 2. (b) 7, (d) 12. (b) 3. (a) 8. (c) 13, (c) Explanations 1. (p) Only statement (c) is correct. In column if the both the ends fixed, has minimum equivalent length or effective length i.e. 2 . D2 —D-f 2. (D) All compressive in I-quadrant P Ph x_Pyk A 1), 1,, e =h 4. (b) 5. (d) 9. (0) 10. (b) Struts and Columns 4 Strength of Materials 13 7 10. (b) (x, y)Q (h, k)S --- C1,both ends hinged P1 = C2,one end fixed, other hinged P2 = = C2 e=k P Px ----xe -PY-xe A Iy x Y x 1 1 . (c) Column of circular section of radius r (Compressive) (a) p P(QR) x (QS) + °max = + ab =A I YY 4. (b) (b) is not correctly matched 5. External diameter D, 2 2 Internal diameter 01, A= 7C (D - D ) 4 = 1`— (D4 —D41 ) 64 (d) 4P a (d) n(D 2 - Of) c for a fixed, free column, effective length is twice the actual length. 7. 4 12. (b) ms strut short in length, fails by yielding 6. P = 0.5 P2 radius of cone = 3. n2E1 ab — (d) P.e x x 64 2 x n(D4 -D4) 32 Pe D _ m(D4 -D11) n(D2+EX)(02 -DD D is reduced to 2 I is reduced to — 32 Pe D 32 Pe D 4P n(D2+ DD(D2 - n(D 2 - Di ) 16 Pis reduced to 8. 9. e- P iF 6 D2 + 8D 13. (b) Yield load produces elastic instability in a column Imin = 80 cm4 A = 20 cm2 (c) kmin (c) D , core of section. Concentric circle of diameter _ 4 = 180 = 2 cm ,\ 20 L le = - (Fixed ends) = 200 cm 2 le = 200 =100 kmin 2 II • III • 10 Theories of Failure CHAPTER Q.10.1 Explain what do you understand by theories of failure. Compare any three failure theories graphically for an element subjected to two mutually perpendicular direct stresses. [CSE-Mains, 1989, ME : 20 Marks] Solution: A body may be subjected to various combinations of M (bending moment), T (twisting moment) and F(axial force) and may fail. At the critical section of the body principal stresses p1 > p2 > p3 can be worked out. Then three principal stresses individually or in combination are compared with the uniaxial tensile stress for a bar, under which the bar fails. For a brittle material it is no ultimate tensile strength but for ductile material it is aYP' yield point stress. Generally for any plane stress problem, two principal stresses p1 > p2 and p3 = 0 are determined as shown. There are five theories of failure but we let us take following 3 theories of failure 1. Maximum principal stress theory 2. Maximum shear stress theory 3. Strain energy theory P2 P2 YP YP 0J 0 P1 —= x aYP (c) (b) (a) Fig. 10.1 Maximum principal stress theory (Rankine's theory) P11 Gyp P2 5' Gyp p1 < ±1 YP P2 < YP e.A as shown in figure (a). Theories of Failure Strength of Materials 4 13 9 Maximum shear stress theory (Tresca theory) P1 P2 YP 2 - 2 1P1 - P21 5- cryp If p1 and p2 both are positive P2 Pi or — i t=— 2 2 Then If p1 > 0 and p2 < 0, then (P1 P2) < cryp Fig. (b) shows graphical representation of maximum shear stress theory. Strain energy theory [p12 p22 - 2vp1p2] 0. 2 yp 2 2 P1 + P2 2v P1P22 2 2 a YP a aYP YP or or < 1 (x2 + y2 - 2vxy) 5_ 1, ellipse where, x- P1 a YP = P2 a YP Fig. (c) shows graphical representation of strain energy theory. Q.10.2 What is failure theory? Discuss its importance. Enumerate various failure theories and mention their fields of application. [CSE-Mains, 1997, ME : 15 Marks] Solution: There are various theories of failure depending upon various types of stresses on a body and different types of strain energies absorbed by a body. These theories of failure are based on three dimensional response of any component due to external load, and at a particular section of interest, these principal stresses pi> p2 > p3 are determined. Analysis is based on the parameters of a simple tensile test on the material. In case of ductile material, yield point stress Gyp and in case of brittle material but are determined. Principal stresses or strain energies are compared with stresses in simple tensile test and strain energies in simple tensile test. There are 5 theories of failure 1. Maximum principal stress theory < used for brittle materials 2. Maximum shear stress theory G YP , used for ductile materials 2 P1 P3 2 3. Maximum principal strain theory (pi - vp2 - vp3 ) or [P1 - v(P2 P3)] < YP E ayp theory is not acceptable to designers 140 0. lAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY 4. Maximum strain energy theory 2E 2 3 [Pi + P2 + P3 - 2 v(PiP2 1 a via P2P3 + P3P1)] - '- 2 2E v, Poisson's ratio This theory is also not acceptable to designers 5. Shear strain energy theory 2ayp2 [(P1 - P2)2 + (P2 - P3)2 + (P3 For ductile materials, the theory is in good agreement with the experimental results and is used for the design of shafts. Q.10.3 Explain the salient features of maximum distortion energy theory of failure and discuss how it compares with maximum strain energy theory and maximum shear stress theory. Solution: Maximum distortion energy theory It principal stresses are p1 > p2 > p3 at a point of a body, then as per distortion energy theory aryl! [(P1 - P2)2 + (P2 - P3)2 + (P3 - P1)2] Where aYP = yield point stress of the material in simple tensile test. This theory is used for ductile materials and specially in the design of shafts. Maximum strain energy theory [P12 + p22 + p32 2V(P1 P2 + P2P3 P3P9] 5 ayp2 where, v = Poisson's ratio However this theory is not used by designers. Maximum shear stress theory of p1 > p2 > p3 Pi -P3 < a YP 2 - 2 This theory is used for the design of components of ductile materials. This theory provides more factor of safety and a conservative design of a component is obtained. Q.10.4 State and deduce the strain energy of distortion theory of elastic failure and compare it with maximum shear stress theory with respect to the field of application and suitability for optimization. [CSE-Mains, 1998, ME : 15 Marks] Solution: u = total strain energy per unit volume = —1 [Pi +14 + - 2 v(P 1P2 P2P3 P3P1)] 2E = uv + vs = Volumetric strain energy + Shear strain energy = due to volumetric stress, here pv = Pi + P2 + P3 = Pm 3 uv 2E 3,2 -2v(3p,i.1 )] = t'm (1 2v) 2E Theories of Failure Strength of Materials 3 = —X 2E 4I 141 (Pi + P2 +P3 )2 (1— 2v) u = u— uv, taking (1 + v) = G , taking v = 0.5, volumetric strain energy becomes zero. 2 us = (1+ v) 3E + p2 + /313 — (PiP2 P2P3 P3P1)] 1 [ 2 + P32 / kP1P2 + P2P3 + P3P1)] = — Pi+ P2 6G 1 = 12G [4 0 D,2 2p2 ' [(P1 — P2 )2 + (P2 + P3)2 + (P3 a_ 0 2 ,p3 — 0 — 2p2p3.— 2p3pi] a yp 2 6G 2ayp2 Maximum shear stress theory is used for ductile material and gives the maximum size of the component. Distortion or shear strain energy is also used for ductile material, but gives slightly lower value of dimensions of a component, as diameter of a shaft. Using this theory, shaft diameter can be optimized. Q.10.5 (a) What is the drawback of maximum principal stress theory? (b) A body is under action of two principal stresses of +40 N/mm2 and —70 N/mm2 and the third principal stress is zero. If the elastic limit in simple tension as well as in compression is 200 N/mm2, find the factor of safety based on the elastic limit according to (i) maximum shear stress theory (ii) maximum strain energy theory (iii) maximum shear strain energy theory Take v = 0.3 [CSE-Mains, 2008, ME : 20 Marks] Solution: (a) Maximum principal stress theory is used for brittle materials. It is not advisable to use this theory for ductile materials If principal stresses p1 > 0, P2 > Then this theory coincides with maximum shear stress theory p1 > 0, p2 < 0, then If Pi + p2 "Cmax GYP 2 — 2xFOS av„ , (Factor of safety is reduced) + P2) p1 = +40 N/mm2 (b) p2 = —70 N/mm2 cse = 200 N/mm2 (i) Maximum shear stress theory FOS — 40 — (-70) 2 FOS 200 2 xFOS 200 =1.82 = 110 14 2 0. (ii) IAS & IFS (Objective & Conventional) Previous Solved Questions MAINEEA Maximum strain energy theory 2 63q3 FOS [(p12 + p22-2v(p1p2)] = G 2 yp [702 + 402 - 2 x 0.3(-70)(40)] = 1— FOS 2 (4900 + 1600 + 1680) = GY FOS) a „ 90.44 - FOS FOS - 200 = 2.211 90.44 (iii) Maximum shear strain energy theory Factor of safety, )2 ( a [ p 1 2 p 2 2 p p 1 ] FOYPS 2 ( [(-70)2 + (40)2 + 70 x 40] )2 a YP FOS 2 YP (4900 + 1600 + 2800) < FOS 2 96.43 6 YP FOS 200 FOS - 96.43 = 2.074 Q.10.6 A round member is subjected to a direct tensile load of 20 kN and shear load of 12 kN. The yield stress in tension is 25 kN/cm2, and Poisson's ratio 0.3. Determine the diameter of the member, using a factor of safety of 2 according to (i) Maximum principal stress theory (ii) Maximum shear stress theory (iii) Maximum distortion energy theory Solution: Direct stress, 61 = Shear stress, T - Principal stress, p1 = 20 kN A 12 kN A 10 kN + (10 kN)2 (12k12 A A ) 15.62 kN -5.62 kN P2 = A A ' tmax - A 25. 62kN A Theories of Failure Strength of Materials -4I 143 25 kN/cm2 Gyp FOS = 2 Gyp all = 12.5 kN/m2 (i) Maximum principal stress theory = 25.62 kN - 12.5 kN/cm2 A A = 2.0496 cm2 = 4 d 2 d= 1.6 cm = 16 mm (ii) Maximum shear stress theory 15.62 A 12.5 kN 2 cm 2 A = 2.4992 cm2 F d= 1.78 cm = 17.8 mm (iii) Maximum distortion energy theory (pi 2 2_ p2 2 2 F25.621 +1-5.621 + [25.62 5.62] - (12.5)2 L A 1_ A L A2 p p i ) 2 156.25 A2 = = A= d= a 2 yp 656.3844 + 31.5844 + 143.9844 831.9532 2.3075 1.714 cm = 17.14 mm Q.10.7 A hollow shaft 30 mm inner diameter and 50 mm outer diameter is subjected to a twisting moment of 800 Nm and an axial compressive force of 40 kN. Determine the factor of safety according to theories of feature based on normal stress theory, shear stress theory and distortion energy theory. The tensile and compressive yield strength of material is 280 N/mm2 and Poisson's ratio is 0.3. [CSE-Mains, 2006, ME : 20 Marks] Solution: Shaft, D. 50 mm, II (502 A Area, d = 30 mm 3U -2) _ 400 7t mm2 4 Torque, 504 3-4) U = 53.40 x104 mm4 32 ( T= 800 Nm = 800 x 1000 Nmm T J Shear stress, 25 T 800 x 103x 25 ti = — x 25 = 53.4 x 104 = 37.45 N/mm2 a, axial compressive stress - P = A = 33.45 40,000 400/1 = -31.83 N/mm2 31.83 144 ► IAS & IFS (Objective & Conventional) Previous Solved Questions MAD EAU Principal stresses P1' P2 = (31.83)2 31.83 +37.452 2 ± 2 = -15.915 ± V253.28 +1402.5 = -15.915 ±,1655.7825 = -15.985 ± 40.691 p1 = +27.776 N/mm2 (tensile) p2 = -56.606 N/mm2 (comp.) Itmaxi = ±40.691 N/mm2 yp = 280 N/mm2 Maximum normal stress theory 4:3 YP P2 — FOS - FOS 280 56.606 = 4.946 Maximum shear stress theory — max 40.691 - FOS - CT YP 2 x FOS 280 2 x FOS 140 40.691 = 3.44 Distortion energy theory a )2 YP [(27.776)2 + (-56.606)2 - (27.776X-56.606)] = IFOS (771.506 + 3204.24 + 1572.29) = or p FOS yp 15548.036 - FOS FOS - 280 = 3.759 74.485 Q.10.8 A hollow circular steel shaft is subjected to a torque of 800 Nm and a bending moment of 1200 Nm. The internal diameter of the shaft is 60% of the external diameter. Determine the external diameter of the shaft according to (i) maximum principal stress theory (ii) maximum shear stress theory (iii) shear strain energy theory Take factor of safety as 2 and the yield strength of the material as 27 kN/cm2. Theories of Failure Strength of Materials Solution: aYP = 27000 N/cm2 = 270 N/mm2 T= 800 Nm M = 1200 Nm FOS = 2 ayp = allowable = 135 N/mm2 External diameter = D, internal diameter = 0.6 D 1 = 6 [D4 —(0.604 ]= 0.042726D4 4 J = 21= 0.08545 D4 ab, maximum bending stress = M D x T, maximum shear stress = 1200x103 xD 14043 x 103 2 — 0.042776 x 2D4 D3 T D 800 x 10 3 D 1 4681x 10 3 = x = x x 0.08545 J 2 2 D4 D3 Principal stresses ab 2 (5b2 2 'T 2 7021.5 8438.8 15460.3 D3 D3 D3 7021.5 8438.8 —1417.3 P2 (I) — D3 D' = , x 10 X 3 103 Maximum principal stress theory 103° x 15460.3 < 135 D3 D3 = 114.5153 x 103 D = 48.55 mm (ii) Maximum shear stress theory (iii) 8438.8 x 103 _ 135 —2 D3 D3 = 125.02 x 103 D = 50 mm Shear strain energy theory (p 1 115460.3 x10D3 2 + p 2 2 p p 1 ) 2 2 1-141,3 x101 (15460.3 x1417.3 x106 = 1352 ) D3 D3 x D3 ) 2 1352 x D6 = 106[239020.8761 + 2008739.3 + 21911883.2] 1352 x D6 = 106[262941498.6] 06 = 106 x 14427.517 D = 49.34 mm 4 145 146 0. lAS & IFS (Objective & Conventional) Previous Solved Questions MADE EASY Q.10.9 A hollow shaft of outside diameter 50 mm and inside diameter 20 mm is subjected to a torque of T Nm and a bending moment of 0.5 T Nm. If the tensile yield stress of the shaft material is 250 N/mm2, what is maximum permissible value of Tto avoid failure according to Tresca's failure theory? Take a factor of safety of 2.0 for a given application. [CSE-Mains, 2011, ME : 15 Marks] Solution: aYP= 250 N/mm2 FOS = 2 a = 125 N/mm2 Allowable, Allowable shear stress, ti = 2 = 62.5 N/mm2 Shaft, D= 50 mm d= 20 mm Torque, T- it 16 x (D4 -d 4 ) D it (50 4 - 20 4 ) x 62.5 - 16 x 62.5 x 50 71 609 x x 62.5 x 104 Nmm = 149.47 x 104 Nmm = 1494.7 Nm 16 50 T = T Nm M' = 0.5 T Nm _ Torque, BM, Tv equivalent torque = VT2 +(0.5T)3 = 1.118 T Nm = 1494.7 Nm T = 1336.9 Nm Q.10.10 The load on a rod consists of an axial pull of 10 kN along with a transverse shear force of 5 kN. Determine the diameter of the rod by using the following theories of failure Strain energy theory (i) Shear strain energy theory (ii) Elastic limit in tension is 270 N/mm2 and a factor of safety of 3 is to be used. Poisson's ratio = 0.3 [CSE-Mains, 2011, CE: 15 Marks] Solution: 6- 10,000 A 5000 A A = area of cross-section Principal stresses are 2 5000 + 2 a 11(1 — = — + 2 +i = A 2 P2 = 2 2J ± T2 = 5000 A (i) Strain energy theory ae = 270 N/mm2 x 5000 A .4 x 5000 A 2.414 x 5000 A 12070 A 0.414 x 5000 = A 2070 A Theories of Failure Strength of Materials 147 FOS = 3 270 = 90 Nimm2 3 Poisson's ratio = 0.3 [p12 + p22 - 2v pi p2 ] 5. al aAllowable [( = 12070)2 +( 2070)2 2 x 0.3 x 12070 x 2070] =902 ) + A A ) A2 [14.5684900 + 4284900 - 14990940] = 8100A2 A2 = 20365.523 —d A = 142.708 mm2 = 4 Rod diameter, (ii) Shear strain energy theory 2 d = 13.48 mm [P12- P1P2 + ,022] = aa2 12070)2 112070 X 2070) ( 2070)2 ] = 902 + + [1 A ) A2 A 1 --2 [145684900 + 24984900 + 4284900] = 902 A A2 = 18823.23 4 A = 137.2 mm2 = 1-t-d2 Rod diameter, d = 13.22 mm Objective Questions Q.1 0.2 A certain steel has proportionality limit of 300 N/mm2 in simple tension. Under three dimensional stress system, the principal stresses are 150 N/mm2 (Tensile), 75 N/mm2 (Tensile) and 30 N/mm2 (Compressive), p. = 0.3. The factor of safety according to maximum shear stress theory would be (a) 1.4 (b) 1.5 (c) 1.3 (d) 1.66 [CSE-Prelims, CE : 2001] Which one of the following theories of failure is widely used in the design of a machine element made of a ductile material? (a) Maximum normal stress theory (b) Maximum strain theory (c) Strain energy theory (d) Maximum shear stress theory [CSE-Prelims, CE : 2003] Q.3 Which of the following is the most appropriate theory of failure for mild steel? (a) Maximum principal stress theory (b) Maximum principal strain theory (c) Maximum shear stress theory (d) Maximum strain energy theory [CSE, Prelims-CE : 2009] Q.4 A shaft is subjected to a torque and an axial compressive force. Shear stress due to torque is 30 MPa and compressive stress due to axial force is 80 MPa. If yield strength of material is 255 MPa, what is factor of safety as per the maximum principal stress theory? (b) 3.01 (a) 3.19 (c) 2.83 (d) 3.64 Q.5 The principal stresses developed at a point are +80, -80 MPa, 0.0. Using shear strain energy 148 0 theory factor of safety is 2. What is yield strength of the material in MPa. Q.6 MADE EASY 1AS & IFS (Objective & Conventional) Previous Solved Questions (a) 150 MPa (b) 80fd MPa (c) 240 MPa (d) 277 MPa The principal stresses at a point are 70, 60 and -80 MPa. If is the yield point stress of the material, using maximum principal stress theory factor of safety is 3.6. What is FOS if the maximum principal strain theory is used. Poisson's ratio of material is 1/3 (a) 3.29 (b) 3.9 (c) 4.5 (d) 5.5 Q.7 A thick cylinder of an internal diameter 100 mm and external diameter 200 mm is subjected to internal pressure p. Yield strength of the material is 240 MPa. Using maximum shear stress theory FOS is 2. What is the maximum value of internal pressure p? (a) 120 N/mm2 (b) 90 N/mm2 (c) 72 N/mm2 (d) 60 N/mm2 Gyp Answers 1. (d) 6. (a) 2. (d) 7. (b) 3. (c) 4. (c) 5. (d) Explanations 5. 1. (d) 150+30) Gyp (P12 - P1P2 + P22) = FOS -Maximum shear stress = 90 2 ae 2 802 + 802 + 802= 3 x 802 = = i8 = 150 ayp 150 =1.66 FOS 90 2. 3. 4. (d) Shear strain energy theory p1 = 80, p2 = -80, p3 = 0 a = 300 N/mm2 Principal stress are +150, +75, -30 N/mm2 Poisson's ratio = 0.3 6. (d) For ductile material (d) Maximum shear stress theory is widely used. = 2 138.56 x 2 = 277 MPa (a) Maximum principal stress = 70 FOS = 3.6 ayp 70 x 3.6 = 252 MPa Maximum principal strain = 1 1 70 - 3(60)+ 3(80) (c) Mild steel — a ductile material (c) Maximum shear stress theory E FOS = (c) 7. 80 252 76.66 = 3.287 (b) R22 + Ri2 5 ac ax = P x 2 2 = R2 - Ri 3P 5 3 P+P tmax pm = -40 -,402 +302 = -90 N/mm2 a = 255 255 FOS = — = 2.83 90 = 2 4 -p = 120 3 p = 90 N/mm2 -1-A p 3 = 76.66 E 11 Strain Energy Methods CHAPTER Q.11.1 The L-shaped bar shown in figure 11.1 is of uniform cross-section 60 mm x 120 mm. Calculate the total strain energy. Take E = 2 x 105 MPa, G = 0.8 x 105 MPa. b= 60 mm d= 120 mm Fig. 11.1 [IFS 2011, CE : 10 Marks] Solution: Total strain energy U= UB + Us + UA = Strain energy due to bending of BC + Strain energy due to shear of BC + Strain energy due to axial load W w20 3w2L w2L _ 1+ 2 6EI 5Gbd 2AE Where, W = 10,000 N, L1 = 2000 mm, L2 = 1000 MM, E = 2 x 105 MPa, G = 0.8 x 105 MPa /- bd3 12 60 x1203 = 8.64 x106 mm4 12 b = 60 mm, d = 120 mm A=bxd= 60 x 120 = 7200 mm2 Putting the values Strain energy, (10000)2 x1000 + 6 x2 x105 x8.64 x106 0.8x105 x60 x120 2 x7200 x 2 x105 = 77160.5 + 208.33 + 357.143 Nmm = 77725.967 Nmm = 77.725 Nm U- (10000)2 x(2000)3 + (10000)2 x 0.6 x 2000 150 N. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions Q.11.2 The cross-sectional area of each member of the truss shown in figure 11.2 is A = 400 mm2 and E= 200 GPa. Determine the vertical displacement of joint C if a 4 kN force is applied. Solution: Reactions 4 kN RBV= RAV RAH 12 8 = 1.5 kN T 1.5 kN T 4 kN Fig. 11.2 Force in members 2 FAB - cosO 2 2 = — = -2.5 kN 0.8 FcA + FAB +2 kN (tensile) cos() = +2.5 kN(tensile) 4 kN Unit load diagram (Fig. 11.3) 1 kN 1.5 kN 2 kN 0.5 kN 0.666 kN 0.5 kN 0.5 kN Members n N nNL CB -0.833 -2.5 kN 10.4125 FCA FCB -0.833 CA -0.833 FAB = +0.666 AB +0.666 1nNL = 10.656 kNm A = 400 mm2 = 400 x 10-6 m2 = 4 x 10-4 m2 E = 200 kN/mm2 = 200 x 106 kN/m2 AE = 4 x 10-4 x 200 x 106 = 800 x 100 kN = 80000 kN +2.5 kN -10.4125 +2 kN 10.656 Fig. 11.3 Vertical deflection at C, Z nNL = 10.656 kNm EnNL 10.656 0.1332 x10-3 m 80000 = 0.1332 mm AE Q.11.3 A shaft is supported by two anti-friction bearings with loads of 140 Newton each acting at points B and F as shown in Fig.11.4. Portion of the shaft between B and C has diameter of 2D as Strain Energy Methods 4 Strength of Materials compared to a diameter D for the portion of the shaft between A and B and between C and F. Using Castigliano's theorem, determine the deflection of the shaft at points B and F. [CSE-Mains, 1992, ME : 20 Marks] 151 = —P Solution: Let us take length L, with AB = BC = CF= RA =P Moments about A PL 1.-150 cm-44-150 cm-44-150 cm L/3 L/3 L/3 = Re xL 3 2PL 3 Fig. 11.4 2L Re x — 3 Rc = RA = +PT E BMD (Fig. 10.5) MA = 0 PL 3 PL MB +3 MC - C B 2PL PL PL 3 3 - 3 Fig. 10.5 MF = 704 1AB = 18c = 64 CF 7c(204 1670 4 64 64 18c = 16 /AB f L/3 (Px)2 Strain energy, UAB — UCF = JO 2EI AB dx L/3 P2x2 2EIAB p 2x3 L/3 xdx _ P 2 L3 6EIAB 0 -162E/ AB 2 (11) UBc U— x3 = P2L3 ' 27 x 2 x E I Eic 2EIBc 2P2L3 162E/AB + P2L3 54 E/Bc = P2L3 E I AB 81 + P2L3 1 1 11,667P2L3 DAB 1_81 864 j /AB - (:a 704 64 DAB x 864 P2L3 54 x16 E BAB F 15 2 0. IAS & IFS (Objective & Conventional) Previous Solved Questions U— MADE EASY 11.667P2L3 64 746.667P2L3 237.67P2L3 0.275P2L3 x = E x 704 864 ED47c x 864 ED4 x864 ED4 Say deflection at B and F is 8, then 8— au 0.275 x 2PL3 = ED4 ap 8 — 0.550 PL3 ED4 P= 140N L= 4500 mm, D in mm 8— 0.550 x140 x 45003 7.02 x1012 ED4 ED4 mm Q.11.4 A steel tube having outside and inside diameters of 10 cm and 6 cm respectively is bent into the form of a quadrant of 2 m radius as shown in fig. 10.6 One end is rigidity attached to a horizontal base plate to which a tangent to that end is perpendicular, and the free end supports a load of 1000 N. Determine the vertical and horizontal deflection of the free end under this load using Castigliano's theorem. E. 2 x 1011 N/mm2. [CSE-Mains, 1993, ME : 20 Marks] 1000N = P Solution: Vertical deflection Take a small element Rd 0 at an angle 0 from vertical Bending moment at small element = P.R sin() rt/2(M9)2Rdo U, strain energy = So 2E/ — Ht--- 2 m Fig. 10.6 2 fir/2 p 2R 2 sin ORA 2EI au = sv , vertical deflection ap = = PR 3 fola EI sine Ode = n12 1 PR3 fo 2 So E1 X de n/2 ic/2 PR3 (1— cos 20) d0 E, PR3 cos 20 E/ x 2 7EPR3 PrcR 3 . 0= 4EI 4EI = 64 0 0 4 _ 6 4) = 427.256 cm4 E= 2 x 107 N/cm2 = 854.513 x 107 Ncm2 1000 x icx 2003 2 - 4x 854.513x10',= 0.1838 cm d0 Strain Energy Methods Strength of Materials Horizontal deflection Put frictitious horizontal load, 41 153 H = 0 at A M0 = P(R sine) + H(R - R dose) l 2 (M ) Rd() 171/2 J. 2E1e U cir/2 1 au OH = .10 aH 2E1[PRsine][R-Rcos0] x Rd() /r/2 2 §-1 = So 2EI [PR3][sin0-sin0cose]d0 3 r it/2 PR 3 f rE/2 sin20 = PR c/0 sin0d0 2 EI Jo El Jo _ PR 3 cos0 El =+ n/2 0 PR 3 cos20 + EI 4 nI2 0 PR3 PR [cosic cos01 + 4 j El El 4 PR3 PR3 [ 1 1] PR3 . +— - - = El 4 4 2E1 El 2 §4 = it- x8 =0.117 cm Q.11.5 Using the Castigliano's theorem, calculate the vertical deflection 8 at the middle of a simply supported beam which carries a uniformly distributed load of intensity w over the full span. The flexural rigidity El of the beam is constant and only strain energy of bending is to be considered. [CSE-Mains, 1991, ME : 20 Marks] Solution: Let us apply a fractious load P. 0 and the centre of the beam reactions, wL P RA = Rg = —+ — 2 BM at section x, MX= P= 2 0 l'x —1-I w wL P wx 2 Tx+ V 2 Beam is symmetrically loaded about center P Strain energy, U= 2 U2 (Mx) 2 dx .10 2E1 2 j am u2 2m x x—x dx 2E/ 0 DP = El 0 [ wL P wx 2 ]x .dx 2 x+ 2 x 2 2 But P = 0 8= 2 ulwLx 2 wx3 d i El o 4 4 )" Flg. 10.7 154 0. lAS & IFS (Objective & Conventional) Previous L/2 2 wLx3 wx4 EI 12 16 0 = MADE EASY Solved Questions 2 wL4 wL4 96 256 El 2 x 8-3 xwL4 = 5wL4 El 768 384E/ Q.11.6 A circular rod of diameter d is bent at right angle. It is fixed at one end and a load W is applied at the other end as shown in the fig. 10.8. Determine the deflection under the load W, if E and G are elastic and shear modulii of the material. [CSE-Mains, 1996, ME : 20 Marks] Solution: w 2b3 Strain energy due to bending in BC 6E/ 2a Strain energy due to twisting in member AB - (Wb) 2GJ W2a3 Strain energy due to bending in member AB - 6E1 IA/ 4,3 Total strain energy, au aw Li 2,3 a + vv m/a 2GJ 6EI 8 - Wb3 Wb2a Wa3 + 3E1 3E1 GJ J= 21 8- Q.11.7 (a) (b) vv U = " 6E1 Wb3 3E/ + Wa3 3E/ + Wb2a 2GI I - ird 4 64 8- W [ b3 a3 b 2al 64W + + = I 3E 3E 2G 704 r + a3 3E b 2a 2G ] State the theorem of Castigliano's. Using the above theorem, find the horizontal displacement along the load line of the frame shown in fig. 10.7. Considering the deflection due to bending only. The moment of inertia is the same for all sections. B Pa) B ' 10-P P44 P ' Pa c) P8 Pe 1-4-- b a D Fig.10.7 [CSE-Mains, 1996, ME : 30 Marks] Strain Energy Methods Strength of Materials 4 15 5 Solution: Castigliano's Theorem If a body is acted upon by forces F1 , F2, F3 .. F and U is the strain energy stored in the body due to these forces (causing, axial stresses, bending moments or twisting moments etc. then partial derivative of the strain energy with respect to force F., gives the displacement of the body in the direction of au aF; 8i, deflection along force F; Free body diagram of the figure is given P 2a3 Uco 6 E1 Member CD Member BC p2 a 2 x b UBC 2E1 Member AB UAB P2a3 6E/ P2a3 P 2 a 2 b 2E1 3E./ Total strain energy, au ap = 8 = deflection along P P 2Pa3 Pa 2 b Pa2 _ [2a + 3b] 3E1 EI 3E1 Q.11.8 Determine the vertical deflection of the joint C of the frame shown in fig. 10.10. Area of crosssection of each member is 850 mm2. E = 200 kN/mm2. 5 kN 10 kN RDH = 1 kN Unit load at C 3 RAV =15kN RDV = 10 kN Fig. 10.10 [IFS 2013, CE : 10 Marks] Solution: Reaction: Taking moments about D 10 x 3 + 5 x 3 = RAvx3 RAC = 15 kN 156 1. MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions RDH = 10 kN Roy = 10kN J. Forces in the members FED = -10 kN (comp.) FDD = 0 FAB = -15 kN (comp.) = +14.14 kN(Tensile) FBD = +1 Unit load at C FDD = -1 (comp.) FBD = o o =0 FBD FAB Members N n L NnL AB —15 0 3 0 BC —10 0 3 0 CD 0 1 3 0 BD +14.14 kN 0 3'I2 0 E nNL = 0 Vertical deflection at joint C is zero Vertical deflection at point C is zero. Q.11.9 The steel truss shown in fig. 11.11 is anchored at A and supported on roller at B. If the truss is so designed that under the given loading all tension members are stressed to 100 N/mm2, and all compression members to 80 N/mm2, find the vertical deflection of the point C. E= 2 x 105 N/mm2. Find also the lateral displacement of the end B. [CSE-Mains, 1998, CE : 40 Marks] 6 kN V E cq Fig .11.11 0.444 2 F 0.888 0-) 0 D RB = 5 6 kN [4— 2.4 m —44-- 2.4 m —44— 2.4 m Solution: There are 9 members in truss. Zero force member is FD Compression member AE, EF, FB, ED, FC Tension member AD, DC, CB Draw unit load diagram, unit load at Cas shown below: = tan-1 0.75 = 36.87° sine = 0.6 cos() = 0.8 For unit vertical load at C, Reactions at A and B RA = 6 kN 0.888 co. -1 2.4 m —.4.-- 2.4 m Fig. 10.12 2.4 m Strain Energy Methods Strength of Materials Joint A Joint B 0.444 0.888 Joint E FED = 0 FEF = 0.444 Joint F Joint C 0.444 (Net) 0.444 0.888 Members Type of force AE C EF n nN/A L(mm) -80 -0.444 +35.52 2400 85248 C -80 -0.444 +35.52 2400 85248 FB C -80 -0.888 +71.04 2400 170496 AD T +100 +0.555 +55.5 3000 166500 ED C -80 0 0 1800 0 DF 0 0 -0.555 0 3000 0 FC C -80 +0.333 -26.64 1800 -47952 CB T +100 +1.111 +111.1 3000 333300 DC T +100 +0.888 +88.8 2400 213120 En A N/A n(N/A) L L = 340992 + 499500-47952 + 213120 =1006160 1006160 NL = 5.03 mm Zn-- = AE 2 x 105 MUM 41 15 7 12 Miscellaneous Questions CHAPTER Q.12.1 Determine the bar forces in the members CB, BE and EF of the truss as shown. [IFS 2013, CE : 15 Marks] 10 kN 10 kN Solution: Force in various members can be determined by method of sections. (i) Let us break the frame by section x-x as shown Taking moments about B (force in BF passes through point B so no moment caused by it) [4-4 m -+- 4 m 10 x 4 + 20 x 4 = FEF X= FEF x distance B P) vL 4m Fig. 12.1 120 = FEF x 2.828 FEF= —42.443 (Comp.), Foo = +20 kN (Tensile), FOE = —10 kN (comp.) 10 kN (ii) Cut off by yy-section Taking moments about E FcB x 4 = 20 x 4 — 10 x 4 = 40 10 kN (tension) FCB Taking cut along section z-z as shown C y 10 kN 10 kN m —444-- 4 m Member BC, BE, BFare cut 10 kN (tensile) FCB Taking moments about point F 20 x 8 — 10 x 4 — 20 x 8 + FBE x 4 = 0 FBE = 10 kN (comp.) 20 Miscelaneous Questions Strength of Materials FBE 10 kN (comp.) FCB +10 kN (tension) FEF -42.443 kN (Comp.) 4 15 9 Objective Questions Q.1 Figure 12.2 shows the loading pattern on a truss. The force in the member AC is 2t 4t 4t 4t (a) 1 (c) 3 2t Q.4 8t (b) 2 (d) 4 [CSE-Prelims, CE : 2001] A pin-jointed lower truss is loaded as shown in the below figure. The force induced in the member DFis A 8t Fig. 12.2 (a) (b) (c) (d) Q.2 Zero 2t 8t Statically indeterminate [CSE-Prelims, ME : 2001] What is the vertical deflection of joint 'N' of the pin-jointed frame as shown in the figure below? 4m Fig. 12.5 (a) 1.5 kN (Tension) (b) 4.5 kN (Tension) (c) 1.5 kN (Comp.) (d) 4.5 kN (Comp.) [CSE-Prelims, CE : 2002] 0.5 (a) PL/6AE (c) PL/AE Q.3 FIg. 12.3 (b) PL/3AE (d) 2PL/AE [CSE-Prelims, CE : 2008] For the truss shown in the figure, which one of the following members has zero force induced in it? 20 kN The degree of static indeterminacy of the pinjointed plane frame shown in figure is w2 (a) BC (c) DE Fig. 12.4 Fig. 12.6 (b) AD (d) BD [CSE-Prelims, CE : 2003] 160 ► Q.6 MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions (b) 40 kN (d) 100 kN [CSE-Prelims, CE : 2006] (a) 0 (c) 80 kN The force induced in the vertical member CDof the symmetrical plane truss shown in the figure is 50 kN Q.9 In the plane truss shown above, how many members have zero force? V 50 kN A Fig. 12.7 (a) 50 kN (tension) (c) 50 kN (Comp.) Q.7 A (b) 100 kN (tension) (d) Zero [CSE-Prelims, CE : 2004] (a) 3 (c) 7 In a truss work as shown in the figure given below, what is the force induced in the member Fig. 12.10 (b) 5 (d) 9 [CSE-Prelims, CE : 2006] DE? Q.10 What is the force in the member EH for a pinjointed tower truss as shown in the figure? 50 kN 2 kN A T 3m 2 kN 3m Fig. 12.8 (a) 50 kN (tensile) (c) 50 kN (Comp.) Q.8 2 kN —110.E (b) Zero (d) 25 kN (Comp.) [CSE-Prelims, CE : 2005] 3m M A .4 Due to horizontal pull of 60 kN at C, what is the force induced in the member AB? C I ~-4 m H 60 kN Fig. 12.11 (a) 6.0 kN (Tension) (b) 6.0 kN (Comp.) (c) 7.5 kN (Comp.) (d) None of the above [CSE-Prelims, CE : 2008] Answers 3m Fig. 12.9 1. (a) 6. (a) 2. (c) 7. (b) 3. (d) 8. (d) 4. (c) 5. (a) 9. (d) 10. (c) Miscelaneous Questions Strength of Materials Explanations 1. FDF = 4.5 kN (T) FFE = 4 kN (T) (a) 2t At joint F 8t A 8t Zero force member at joint A. 2. (c) FDF = 5 kN 5 cos() = 4 kN Balance FEF 5 kN sine = 3 kN 3 + FDF = 4.5 kN FDF = 1.5 kN (Comp.) 5. (a) In members BC and CD, no force. Vertical displacement of N 1 [P 3. x FAB = FED = 0 Fa), = 0 L 1 L 1 P x + xx A E A AE 6. (d) Members = 21 = m Joints = 11 = j Ds = m + Re - 2j = 21 +(2 x 2 +1)- 2 x 11 = 4 Static indeterminacy 2. 4. (a) At joint D Fop = 50 kN (T) Tension 7. (b) 50 kN (c) RE- 12 + 6 4 = 4.5 kN 4 kN = 4.5 kN 4.5 kN At the joint E, FDE -= 0 161 16 2 0. lAS & IFS (Objective & Conventional) Previous Solved Questions 8. MADE EASY FEH cos° = 6 (d) At joint C 6 = 7.5 kN(Comp.) F = 84 0.8 Q.12.2 Analyse the continuous beam as shown in fig. 12.8 by moment distribution method. Draw the bending moment diagram. Assume El = constant. 20 kN 60 kN A' 4 sin° = - = 0.8 5 cos° = 0.6 2m 60 60 = = 100 kN F cos() 0.6 Bc FEic = FAB = 100 kN FBD = 0 9. 20 kN/m 4m 3m -40 IN/m +24 kN/m -20.488 (d) Fig.12.8 [CSE-Mains, 2012, CE : 15 Marks] Solution: Mp = 0 Mc = -40 kNm ME = None zero force members. a1.71 10. (c) 3 x 42 144_2 = + 8 = +6 kNm 8 24x 4 24x3x( x 2+1) 2 2 3 +2 = 32 + 108 = 140 kNm2 = a1.771' ( Origin at B) 24x3 2 x 2+ 24 x2 2 2 x3+3 11 72 + 24-) = 160 kNm2 3 a272 (origin at B or C) 9 kN 9 kN Considering span AA' (imaginary) of span zero. Clapeyron's theorem for spans A'A, AB sin° = 0.6 cos0 = 0.8 RH - 18+12+6 4 = 2 -x 6x 4x 2 = 32 kNm2 3 = 9 kN M A' X01-2MA(5-1-0)+5MB Miscelaneous Questions Strength of Materials = 6 x ai.71 0 5 -96 +40 +2.58 kNm MB = -5618MB = -56 - 18 x 2.58 5MA -56 - 46.44 = -102.44 -20.488 kNm MA +2.58 kNm MB -40 kNm MC -CO 10MA + 5MB = -192 Or 5MA + 2.5MB = -96 Spans AB and BC 5MA + 2MB(4 + 5)+ 4Mc 6 xaqi 6 xa 2x2 5 4 -6 x 140 6 5 4 163 5MA + 2.5MB 15.5% (origin at B) 10MA + 5MB= -5 x 160 —192 41 (origin at A) zE 0 x 32 = -168-48 =-216 5MA + 18MB - 4 x 40 = -216 5MA + 18MB = -56 A +2.58 , P2 points of contraflexure BM Diagram Fixed and Continuous Beams Q.1 A fixed beam as shown in the figure 12.15 has a span L and uniform flexural rigidity El. It is subjected to a concentrated clockwise moment at the centre. The deflection at the centre of the beam is (a) 0.5 (c) 1.5 (b) 1 (d) 2 [CSE-Prelims, CE : 2006] Q.3 A beam shown in fig. 12.16 of span 10 m and having El = 10000 kNm2 is subjected to a rotation of 0.001 radian at end B. What is the fixed end moment at A? 14e-- L/2 )-I4 L/2 —H Fig. 12.15 10m ML2 , upwards (a) 8 EI (a) 1.5 kNm (c) 3.0 kNm (b) ML2, downwards (c) Zero ML2 , downwards (d) 384E/ [CSE-Prelims, CE : 2002] Q.2 A fixed beam and a simply supported beam having same span and develop same maximum bending moment due to uniformly distributed load on entire span, what is the ratio of uniformly distributed load on fixed beam to that on simply supported beam Q.4 Fig. 12.16 (b) 2.0 kNm (d) 4.0 kNm A heavy weight W attached to a rod can slide in a grooved support as shown in fig. 12.17. What is the equilibrium sliding distance A? w L---4 Fig. 12.17 1 64 ► MADE EASY lAS & IFS (Objective & Conventional) Previous Solved Questions WL3 (a) 3E/ clockwise. What is the distance of the point of contraflexure from the left support? WL3 (b) 12 EI 3L (b) TC) WL3 (d) 48E1 WL3 (c) 24E1 (d) [CSE-Prelims, CE : 2010] Q.5 What is the bending moment at A for the beam shown in fig. 12.18 E= 200 GPa L 2L [CSE-Prelims, CE : 2006] Q.7 What is the fixed end moment for the beam shown in fig. 12.19 E= 100 GPa -I- L Fig. 12.18 PL (a) -3-(c) (b) PL (d) 3PL Fig. 12.19 2 (a) M0 2PL 3 [CSE-Prelims, CE : 2010] Q.6 A horizontal fixed beam AB of length L has uniform flexural rigidity El. During loading the right support A rotates through an angle 0 (c) (b) Mo 3 2M0 (d) 1. (c) 2. (c) 6. (a) 7. (b) 3. (b) (c) M L/2 — L/2 —al Zero deflection at centre. 2. 1-2 W2 L2 (c) w21-2 — = 1.5 W2 L/2 5. (b) Wi M -L2 24 wi MA = M1 = 12 = Mmax AA 8 ""max 12 - 8 Hi— L/2 60 Answers Explanations 1. 3 10m 4. (c) 5. (d) Miscelaneous Questions Strength of Materials Pi + P2 = P 2P P1= 3 Say equivalent load is W WL2 =0.001 0= ° 2E1 2P MA _ 3 Wx102 - 0.001 2 x 10000 W 0.2 kN MA = 0.2 x 10 = 2 kNm 4. 6. (a) (c) He— L/3 A fixed beam of length 2L, central load W, Ac 4E/0 P 2E10 L 5. 16 5 4 L I-1-4 2L/3 L Point of contraflexure lies at - from left support. 3 W (203 WL3 - 192E/ 24E/ 7. (b) (d) C L L -14 L-3 81 M P2 1-3 82 = 3E21 ['*-. 4 2/ ► ML 1 M0L x 21 - 4 21 = 82 P1 1 P2 21 E I= 2 E2 M= -Mo 3 P2 t C P2 FBD B 4 ML 1 ML 1 x x + 4 21 2 1 ML ML _ 3 MI_ + - 4/ 21 - 4 I P1 = 2P2 P 2 13 Rotational Stresses CHAPTER Q.13.1 A circular disc 50 cm outside diameter has a central hole and rotates at a uniform speed about an axis through its center. The diameter of the hole is such that the maximum stress due to rotation is 85% of that in the thin ring whose mean diameter is also 50 cm. If both are of the same material and rotate at the same speed, determine the diameter of the central hole and speed of the disc for the datas given Allowable stress = 900 kg/cm2 Specific weight = 7.8 gm/cm3 Poisson's ratio = 0.3 [CSE-Mains, 2006, ME : 30 Marks] Solution: In a thin disc, oc max occurs at inner radius, R1 If R2 is outer radius = 250 mm C max = k1 k2 - ,,„2 " [ki (2Fq + 3+v 8 = .3 8 k 2R1 = 0.4125 1+ 3v 1.9 = = 0.2375 8 8 2 PC° [0.4125(2 x 2502 = 0.85 x 2502 +Rn- 0.2375Rn xp [stress in ring = (02r2p1 g 0.85 x 2502 = 0.025 x 2502 = R1 = Allowable stress = 2 x 2502 7.8 x10-6x 0.85 x (0 9810 0.825 x 2502 + 0.175 R12 0.175 R12 94.5 mm 900 kg/cm2 = 900 x 9.81/cm2 = 88.29 N/mm2 88.29 N/mm2 (02 = 2090195.84 = 1445.175 rad/sec N = 13806 rpm (:4 Rotational Stresses Strength of Materials 1 16 7 Q.13.2 A steel rotor disc of uniform thickness 50 mm has an outside diameter 800 mm and a central hole of diameter 150 mm. There are 200 blades each of weight 2N at an effective radius of 420 mm pitched evenly around the outer periphery of the disc. Determine the maximum rotational speed such that maximum shearing stress in the disc does not exceed 375 MN/m2. Take (p) of steel as 7470 kg/m3. Following basic relations for radial stress (ar) and Hoop stress (a0) at radius of rotating disc at w rad/sec can be used with usual notations = A — —(3+v) pco 2 r 2 po) 2r 2 , Go = A+ 2 (i + 3V) 8 8 [CSE-Mains, 1995, ME : 30 Marks] Solution: t= 50 mm, R2 = 400 mm, R1 = 75 mm, timax= 375 N/mm2 and p = 7470 kg/m3 Number of blades = 200 Effective radius = 420 mm Weight of each blade = 2N Centrifugal force on the periphery due to blades 200 x 2 400 x co2 x 420 =17.1254t w2N 9810 Resulting area = it x 800 x 50 = 12.566 x 104 mm2 Radial stress at the periphery of the disc 17.1254(1)2 =1.363 x10-4 w2Nimm2 12.566x104 Radial stress at the inner radius 75 mm is zero pco 2r 2 = A——(3+v) 6r 8 x co2 xr Taking v = 0.3, Poisson's ratio for steel 2 2 or = 0 = A E3 (3 + 0.3)Pc° r 8 752 p = 7470 kg/m3 = 7470 x 10-9 kg/mm3 = 7.47 x 10-6 kg/mm3 3.3 8 = 0.4125 0 = A =A B 75“r, B 5625 0.4125x 7.47 x10-6 x(02 x752 1 1.7332 x10-2 co2 A B — 1.7332 x 10-20 Nimm2 5625 Radial stress at periphery of the disc = 1.363 x 10-4 (02 Nmm2 1.363 co2 x 10-4 = A — 1.363 co2 x 10-4 = A B 4202 176400 3+0.3 7.47 x10-6 w2 x 4202 1 8 0.543 w2 168 ► IAS & IFS (Objective & Conventional) Previous Solved Questions MADE 'EASY' B 176400 0 0.5431363 = A B 5625 1.7332 x 10-2 w2 = A ...from equations (i) B x 170775 0.525800)2 = 5625 176400 5625 x 176400 — 0.525800)2 x 5625 x 176400 — 3055 0)2 170775 B 176400 = 0.5431363 0 + 0.017318 0)2 = 0.560450 Maximum Hoop stress occurs at inner radius 75 mm. A 0.5431363a)2 + ao. = A+ 2 (1+ 3v) 8 x 30)2r 2 2 305542 1.9 6 x 032 x 752 0.560450) + 75, , 8 x 7.4 x 10= 0.56045 0)2 + 0.5431 0)2 — 0.998 x 10-2 (02 GA 'Cmax -max 2 = 375 mN/m2 Goma, = 750 N/mm2 = 1.09337 0 w = 685.827 rad/sec 27EN 60 N = 6550 rpm Q.13.3 (a) What do you understand by rotating disc of uniform strength? (b) A turbine disc is required to have a uniform stress of 150 MPa at a speed of 3200 rpm. The disc is to be 30 mm thick at the centre. What will be its thickness at a radius of 40 mm? Assume density of disc material = 7800 kg/m3. [IFS 2012, ME : 10 Marks] Solution: (a) Rotors of a steam or gas turbine, on the periphery of which turbine blades are attached are designed as disc of uniform strength, in which the circumferential and radial stresses, developed due to centrifugal forces on account of high angular speed ware equal and constant and independent of radius. Both radial and circumferential stresses are maximum at the centre of the disc, while radial stress becomes zero at outer radius but circumferential stress reduces to a minimum at the outer radius. So as, the stresses increase towards the centre of the disc, thickness of the disc also increases to maintain constant stress ar = Go = a as shown in figure. Relation between Fig. 13.1. Disc of uniform strength Rotational Stresses 4 Strength of Materials thickness t at any radius r and thickness to at the centre is t = to.e pa 2 r 2 2ga Where, r= radius, at which thickness is t co = angular velocity of rotation of disc in rad/sec p = density of the material g= acceleration due to gravity a = constant stress r = 40 mm p = 7800 kg/m3 (b) 7800 x 9.8 109 = 0.07644 x10-3 N/mm3 27c x 3200 = 335.10 rad/sec 60 a = 150 N/mm2 g = 9810 min/sect — p(.02r2 0.07644 x 10-3 x 335.12 x 402 = 0.004666 2 x 9810+150 2ga pt 2r2 e 296 = e-0.004666 = 0.99534 t = to x 0.99534 = 30 x 0.99534 = 29.86 mm Note that radius given is too small Let us take 40 cm radius or 400 mm radius pco 2r 2 2ga — 43-0.4666 0.07644 x10-3 x 335.12 x 4002 = 0.4666 2 x9810 x150 = 0.6270 t = to x 0.6270 = 30 x 0.6270 = 18.8 mm 169 14 Unsymmetrical Bending CHAPTER Q.14.1 A cantilever beam of span 3 m is subjected to a vertical load of 1.0 kN at free end. The crosssection of the beam consists of equal angle 100 mm x 100 mm x 12 mm with one of its legs placed vertically. Find the magnitude and direction of the resultant deflection. Given /uu = 329.3 cm4. = 84.7 cm4, E= 2 x 105 N/mm2, centroidal distance = 29.2 mm. [IFS 2013, CE : 15 Marks] Solution: /„ = 329.3 cm4, /w = 84.7 cm4 Position of neutral axis / tam = tan() 100 0 =45° ‘, x I„ 0 = 45° and tang = 1 tang — 329.3 84.7 = 3.888 r a = 75.57° S = deflection at free end 29.2-1 00 Fig. 14.1 2 wL3 3E1„ sin2 0(iruu + cos2 0 W = 1 kN /„ = 329.3 x 10-8 m4 E = 200 kN/mm2 = 200 x 106 kN/m2 3EI„= 3 x 200 x 106 x 329.3 x 10-8 = 1975.8 kNm2 L = 3 m, = 45° 2 sin2 01W + cos2 0 = \10.5 x1329'312 +0.5 84.3 l'w ) Fig. 14.2 = V7.55765 + 0.5 = 2.8386 S— 3 1975 .8 x 2.8386 = 0.0388 m = 38.8 mm Resultant deflection in a direction perpendicular to neutral axis. Unsymmetrical Bending Strength of Materials 4 171 Q.14.2 Stress concentration factor is not considered harmful for ductile materials in static loading but for brittle materials, it has damaging effect in both static and dynamic loading. Justify the above statement giving illustrations. [CSE-Mains, 2011, ME : 15 Marks] Solution: In a brittle material, due to stress concentration, if ama, = SCF x aav is more than nut, ultimate stress, and the crack is developed in sample then it is carried through and specimen breaks. Figure shows a flat sample with a central hole, amax = 3 (Yaw a SCF = 3, crack may develop at inner edge of the hole and carried through axes aa. But in a ductile material, the plastic deformation at the tip of the fine crack blunts the edge of crack and further propagation of crack becomes difficult in a region where SCF is less. In fatigue load, brittle material should never be used. Sometimes in the presence of a notch in a machine member, ductile material behaves as a brittle material and machine member breaks. Fig. 14.3 Q.14.3 A cast iron sample when tested in compression fails along approximately 45° plane from its axis while when tested in torsion also fails along a 45° (approx) helical plane from its axis. Explain the reason for such failure and mention about the dominating stress causing failure. [CSE-Mains, 2011, ME : 10 Marks] Shear failure Solution: Cast iron is weak in tension and shear, but strong in compression. Its compressive strength is about three times its tensile strength. When CI sample is tested in compression, ac is the axial compressive stress, but shear stress "E is maximum on planes at ±45° to the axes of the sample. So sample breaks under shear on inclined plane at approx 45° to the axis as shown. Fig. 14.4 \ - ti \p 2 Failure under torsion Fig. 14.5 When a specimen is subjected to twisting moment, angular twist 0 and shear stress "C act on the sample. Shear stress is maximum at outer radius and gradually reduces to zero at the centre. Principal stress p1 = +T acts on a plane about 45° to the axis of the sample. Due to angular twist, fractured surface becomes helical. Q.14.4 How ductility of structural steel in evaluated from static tension? What is the effect of gauge length on the value of ductility? [CSE-Mains, 2011, ME : 10 Marks] MADE EASY 17 2 0- IAS & IFS (Objective & Conventional) Previous Solved Questions Solution: A standard sample of structural steel is tested under tension using Universal testing machine. Original diameter and original gauge length of the sample are noted down before applying tensile force. Tensile force on specimen is slowly and gradually increased till the specimen breaks into two pieces. Final gauge length is measured using an internal callipers. Find gauge length = L' Initial gauge length = L % elongation — L' —L L x 100 Ductility is measured by percentage elongation of the specimen d A 4 L Gage length Fig. 14.5 As per Barba's law Change in length SL = L'— L = bL+c,174 Where b and care Barba's constants L = gauge length, A= area of cross-section As the gauge length increases, bL increases but cJ remains constant. Therefore as the gauge length of the specimen increases, percentage elongation also increases. Q.14.5 (a) What is fatigue? Define fatigue limit and fatigue strength. (b) Explain how various factor of design effect the fatigue strength. (c) What is cumulative fatigue damage? Explain Miner's hypothesis and bring out the drawbacks of this hypothesis. [CSE-Mains, 2008, ME : 5 + 5 + 10 = 20 Marks] Solution: Behaviour of any material under fluctuating loads or stress cycle is termed as fatigue. For different stress levels, fatigue tests are performed on samples and cycles required for fracture of sample at a particular stress level S1, noted down say it is N1. Then S1 is known as fatigue strength of material at stress cycles N1. If a graph is plotted between S,. and then the stress Se at which the graph becomes asymptotic is known as stress limit as shown in the figure. Various factors affecting the fatigue life are (i) surface roughness (ii) size factor (iii) stress concentration factor (iv) factor of safety (v) order of reliability. s' S. I _ SB Fig. 14.5 Depending upon the surface roughness fatigue life of member is reduced. As the size increases life of the member decreases. As the stress concentration factor increases life of member decreases under fatigue loading. Finally depending upon the reliability factor more reliability factor less is the life. Unsymmetrical Bending 41 Strength of Materials 17 3 Cumulative fatigue Damage: There Is cumulative effect of stress levels and number of cycles put in at that level, that forms a fraction of total life upto failure k n. N 1.1 i=C ni = number of cycles accumulated at stress S, N= total number of cycles upto failure at stress level S, + n2 n, nk Nk = 1 N1 N3 ... or + C2 ... Ci + Ck = C is the fraction of life consumed by exposure to the cycles at different stress levels. This is known as Miner's rule or hypothesis. Drawbacks: Miner's rule becomes invalid if any stress level is more than than Se (endurance limit stress). 11.11111111 Cry p (yield point stress) or less 15 General Objective Type Questions CHAPTER Directions: The following 4 questions, items consist of two statements one labelled as 'Assertion A' and other labelled as 'Reason R'. You are to examine these two statements carefully and decide if Assertion A and Reason R are individually true and so weather the Reason R is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Q.1 Q.2 Assertion (A): I, T are channel sections are preferred for beams. Reason (R): A beam cross-section should be such that the greatest possible amount of area is as far away from the neutral axis as possible. A channel cross-section of the beam shown in the figure 15.1 carries a uniformly distributed load. w N/m w N/m • CG Fig. 15.1 Assertion (A): The line of action of the load passes through the centroid of the crosssection. The beam twists besides bending. Reason (R): Twisting occurs hence the line of action of the load does not pass through the web of the beam. Q.3 Assertion (A): A thin cylindrical shell is subjected to internal fluid pressure that induces 2-D stress state in the material along longitudinal and circumferential directions. Reason (R): The circumferential stress in the cylindrical shell is two times the magnitude of the longitudinal stress. Q.4 Assertion (A): Brittle material such as gray cast iron cannot be extruded by hydrostatic pressure. Reason (R): In hydrostatic extrusion billet is uniformly compressed from all sides by the liquid. [CSE-Prelims, ME : 2000] Q.5 Match List-I with List-II and select the correct answer. List-I A. Proof stress B. Endurance limit C. Leaf spring D. Modulus of rigidity List-I I 1. Torsion test 2. Tensile test 3. Fatigue test 4. Beam of uniform strength Codes: A (a) 2 3 4 1 3 1 4 (b) 2 2 4 1 (c) 3 2 1 4 (d) 3 Strength of Materials Directions: The following 4 (four) items consist of the statement, one labelled as Assertion 'A' and other labelled as Reason 'R'. You are to examine these two statements carefully and decide if the Assertion 'A' and Reason 'R' are individually true and of so whether the Reason is a correct explanation of the Assertion. Select your answers to these items using the codes given below and mark your answer sheet accordingly. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Q.6 Assertion (A): When an isotropic, linearly elastic material is loaded biaxially, the direction of principal stresses are different from those of principal strains. Reason (R): For an isotropic linearly elastic material, Hooke's law gives only two independent material properties. 0.7 Assertion (A): In theory of torsion, shearing strain increases radially away from the longitudinal axis of the bar. Reason (R): Plane transverse sections before loading remain plane after the torque is applied. Q.8 Assertion (A): For a thin cylinder under internal pressure, a minimum three strain gauges are needed to know the stress state completely at any point on the shell. Reason (R): If the principal stress directions are not known, the minimum number of strain gauges needed is three in a biaxial field. Q.9 Assertion (A): Buckling of long column causes plastic deformation. Reason (R): In a buckled column, the stresses do not exceed the yield stress. [CSE-Prelims, ME : 2001] Q.10 Match List-I (stress induced) with List-II (situation/location) and select the correct answer using the code given below the lists: List-I A. Membrane stress B. Torsional shear stress General Objective Type Questions 1 17 5 C. Double shear stress D. Maximum shear stress 1. Neutral axis of beam 2. Closed coil helical spring under axial load 3. Cylindrical shell subject to fluid pressure 4. Rivets of double strap butt joint Codes: A B C D (a) 3 1 4 2 (b) 4 2 3 1 (c) 3 2 4 1 (d) 4 1 3 2 [CSE-Prelims, ME : 2007] 0.11 Match List-I (Type of thread) with List-II (Use) and select the correct answer using the code given below the lists: List-I A. Square thread B. Acme thread C. Buttress thread D. Trapezoidal thread List-II 1. Used in vice 2. Used in lead screw 3. Used in screw jack 4. Used in power transmission devices in machine tool Codes: A BCD (a) 2 3 4 1 (b) 2 3 1 4 (c) 3 2 1 4 (d) 3 2 4 1 [CSE-Prelims, ME : 2007] 0.12 Match List-I (Natural of failure) with List-II (Nature of member) and select the correct answer using the code given below the lists: List-I A. Structural member fails in axial compression B. A member fails along a 45° helical plane subjected to torsion C. A structural member bends and collapse under axial compression load D. A member fails in double shear for a joint 176 ► lAS & IFS (Objective & Conventional) Previous Solved Questions List-II 1. Knucide joint 2. Long column 3. Strut 4. Cast iron round bar subjected to torsion Codes: A BCD (a) 3 1 2 4 (b) 2 4 3 1 (c) 3 4 2 1 (d) 2 1 3 4 [CSE-Prelims, ME : 2006] Q.13 What type of fracture occurs when a brittle material is under torsion? (a) Cup and cone (b) Granular transverse (c) Granular helicoidal (d) Smooth transverse [CSE-Prelims, ME : 2006] Q.14 Which one of the following is not correctly matched? (a) Torsional rigidity : Torque per unit angle of twist (b) Modulus of rigidity : Strain energy per unit volume (c) Creeping effect : Loss of mechanical energy [CSE-Prelims, ME : 2008] Directions: The following items consist of the statement, one labelled as the 'Assertion (A)' and the other as 'Reason (R)'. You are to examine these two statements carefully and select the correct answers to these items using the codes given below: (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Q.15 Assertion (A): Plane state of stress does not necessarily result in a plane state of strain. Reason (R): A uniaxial state of stress in general produces a 3-dimensional strain state. [CSE-Prelims, ME : 2008] MADE EASY Q.16 Which one of the following is true regarding the fatigue life of a set of identical ball bearings? (a) Directly proportional to load (b) Inversely proportional to load (c) Inversely proportional to the square of load (d) Inversely proportional to the cube of load [CSE-Prellms, ME : 2009] Q.17 Match List-I with List-II and select the correct answer: List-I A. Partial derivative of strain energy w.r.t. load B. Derivative of deflection C. Derivative of slope D. Derivative of moment List-II 1. Equation for shear force 2. Equation for slope 3. Equation for bending moment 4. Deflection under the load Codes: A BCD 3 2 (a) 4 1 (b) 3 2 4 1 (c) 4 2 3 1 1 4 2 (d) 3 [CSE-Prelims, ME : 2003] Directions: The following item consists of two statements, one labelled as the 'Assertion (A)' and the other as 'Reason (R)'. You are to examine these two statements carefully and select the correct answers to these items using the codes given below: (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true Q.18 Assertion (A): To a cantilever beam of circular cross-section, if a moment is applied with its axis perpendicular to the axis of beam, no shear stress will be induced in the beam. Reason (R): To the above beam, if moment is applied with its axis along the axis of the beam, no bending stress will be induced in the beam. [CSE-Prelims, CE : 2004] General Objective Type Questions Q.19 Match List-I with List-II and select the correct answer using the code given below the lists: List-I A. Beam B. Column C. Circular section of diameter d of a shaft D. Close-coiled helical spring List-II 1. Member subjected to twisting 2. Member used to store strain energy 3. Member subjected to buckling 4. Member subjected to bending Codes: A BCD (a) 4 3 1 2 (b) 1 2 4 3 (c) 4 2 1 3 (d) 1 3 4 2 [CSE-Prelims, CE : 2009] Q.20 Given that F +(3.12 ) where F denotes force and t time; how is r3 described dimensionally? (a) MLT-3 (b) MLT-2 (c) LT-4 (d) MLT-4 [CSE-Prelims, CE : 2009] List-II 1. T-1 2. M LT-2 3. mL.2-r2 4. ML-1T-2 Codes: A BCD (a) 1 3 2 4 3 2 1 (b) 4 2 3 4 (c) 1 1 (d) 4 2 3 [CSE-Prelims, CE : 2010] Q.23 Match List-I with List-I I and select the correct answer using the codes given below the lists: List-I A. Moment of inertia about diametral axis B. Moment of inertia about an axis tangent to the perimeter C. Moment of inertia about an axis through centroidal axis D. Polar moment of inertia List-I I Q.21 What is the unit vector of the resultant of the following two forces? = 21+31+4k; (a) 61 + 61+ 6k (c) -21+ 2k j + + (d) 21- 2k [CSE-Prelims, CE : 2009] Q.22 Match List-I with List-II and select the correct answer using the codes given below the lists: List-I A. Modulus of elasticity B. Work C. Force D. Frequency 1. ico4 32 2. 17704 64 3. 5704 64 4. 704 64 Codes: A (a) 4 (b) 1 (c) 4 (d) 1 F2 = 41+31+2k (b) 41 177 BCD 3 2 1 3 2 4 2 3 1 2 3 4 [CSE-Prelims, CE : 2010] Answers 1. (a) 2. (c) 3. (b) 4. (b) 5. (a) 6. (d) 7. (b) 8. (d) 9. (c) 10. (c) 11. (c) 12. (c) 13. (c) 14. (b) 15. (b) 16. (d) 17. (c) 18. (b) 19. (a) 20. (d) 21. (b) 22. (b) 23. (a) 178 N. MADE EASY IAS & IFS (Objective & Conventional) Previous Solved Questions Explanations 1. (a) Both A and R are true, R is the correct explanation of A. 2. (c) A is true but R is false, Line of action of the load must pass through the shear centre. 3. (b) Both, A and R are true but R is not the correct explanation of A. 4. (b) Both A and R are true but R is not the correct explanation of A. 5. (a) A. Proof stress B. Endurance limit C. Leaf spring D. Modulus of rigidity 2. Tensile test 3. Fatigue test 4. Beam of uniform strength 1. Torsion test 6. (d) A is false, R is true. 7. (b) Both A and R are true but R is not the correct explanation of A. 8. (d) A is false but R is true. 9. (c) A is true but R is false. 10. (c) A. Membrane stress 3. Cylindrical shell subjected to fluid pressure coiled B. Torsional shear stress 2. Close helical spring under axial load 4. Rivets of double C. Double shear stress strap butt joint D. Maximum shear stress 1. Neutral axis of beam 11. (c) A. Square thread B. Acme thread C. Buttress thread D. Trapezoidal thread 3. Used is screw jack 2. Used is lead screw 1. Used in vice 4. Used in power transmission devices in machine tool 12. (c) A. Structural member falls in axial compression B. A member falls along a 45° helical plane subjected to torsion C. A structural member bends and collapse under axial compression load D. A member falls in double shear for a joint 3. Strut 4. Cast iron round bar subjected to torsion 2. Long column 1. Knuckle joint 13. (c) Granular helicoid - Type of fracture in brittle material under torsion. 14. (b) Modulus of rigidity strain energy per unit volume. 15. (b) Both A and R are individually correct but R is not the correct explanation of A. 16. (d) Fatigue life of a ball bearing is inversely proportional to cube of load. 17. (c) dU A. dW dW B. C. D. dy dx d2y dx 2 dM dx 4. Deflection 2. Slope 3. Equation of BM 1. Shear force General Objective Type Questions Strength of Materials 18. (b) Both A and R are individually true but R is not the correct explanation of A. 19. (a) A. Beam B. Column C. Shaft D. Spring 4. 3. 1. 2. Bending Buckling Twisting Strain energy storage 22. (b) A. Modulus of elasticity B. Work C. Force D. Frequency 4. 3. 2. 1. 17 41 9 ML-1T-2 mL2T-2 MLT-2 T-1 23. (a) A. MI about diametral axis 4. B. AN about an axis 3. itD4 64 20. (d) MLT-2, = MLT-4 21. (b) 6i+61+6k C. Micentroidal axis 2. D. Polar moment of inertia 1. -11)E3 61+61+6k 6i I 64 tangent to perimeter + F2 = 61+61+6k Unit vector — 5704 k = - + - + - 17704 64 TcD4 32 MINIM