STRESS-STRAIN DIAGRAM

Transcription

1 CHAPTER 7 STRESS-STRAIN DIAGRAM AND STRENGTH PARAMETERS 7.1 STRESS-STRAIN BEHAVIOUR OF MATERIA All engineering material do not how ame ort of behaviour when ubjected to tenion a well a compreion. There exit ome material like metal, alloy etc., which are more or le equally trong in both tenion and compreion. And thee material are generally teted in tenion again concrete, tone, brick etc., are uch type of material which are weaker in tenion and tronger in compreion. Hence, thee material are teted in compreion. Now the tre-train characteritic of mild teel are of pecific importance to the community dealing with baic engineering cience. 7. STRESS-STRAIN CHARACTERISTICS OF MID STEE (M.S) In order to obtain tre-train behaviour of M.S, a pecimen of uniform circular cro-ection i prepared following the pecification laid in IS 1608:005 identical to ISO 689:1998. A pecific length of maximum 4 inch or 100mm i generally elected in the well-middle part of the pecimen and thi length i deignated a gauge length, over which the amount of elongation i tudied. a Proportional imit b Elatic imit c Upper Yield Point d ower Yield Point e Ultimate Stre f Nominal Breaking Stre Fig. 7.1

2 186 Mechanical Science I Now the pecimen, uitably fitted in extenometer, mounted on the machine where loading i tarted gradually from zero till failure. Following i a tre-train curve of M.S pecimen having gauge length 100mm, teted in Amler Univeral Teting Machine of capacity 0T. Variou point on tre-train curve are marked in Figure PROPORTIONA IMIT It i the point on the tre-train curve, up to which the plot i a traight line and tre i proportional to train. Up to proportional limit, the material remain elatic and trictly follow Hooke aw. 7.4 EASTIC IMIT In the tre-train curve, it i the point jut beyond proportional limit. From proportional limit to elatic limit, the material remain elatic but doe not follow Hooke aw and o, tre and train are not proportional. 7.5 YIED POINT When the pecimen i loaded beyond elatic limit, it enter into elato-platic zone. In thi region, elongation of pecimen occur by coniderable amount without any perceivable amount of increae in load. Sometime thi yielding i accompanied by an abrupt reduction of load and thereby tre. In thi cae the upper and lower limit of tre are called upper yield point or tre and lower yield point or tre, repectively. ower yield tre i normally conidered a yield tre y of material, becaue upper yield tre i affected by peed of teting, form of pecimen and hape of cro-ection. 7.6 PROOF STRESS Some material like High Strength Deformed (HSD) teel, bra, duralumin etc., do not how any well defined yield point. For thee material, proof tre erve a analogou to yield tre. Bra Bronze Strain Strain Cat lron Timber O 0.00 Strain Fig. 7. Fig. 7. Proof tre i the tre that i jut ufficient to produce under load, a defined amount of permanent reidual train, which a material can have without appreciable tructural damage. Thi arbitrary value will be different for different material or different ue of ame material. It i determined from the tre-train curve by drawing a line parallel to initial traight part or tangent of the curve and at a ditance from the origin by an amount repreenting the defined reidual train (normally 0.1% or 0.%) thu determining the tre at which the line cut the curve. Strain

3 Stre-Strain Diagram and Strength Parameter 187 In pecifying proof tre, the amount of permanent train conidered, hould be mentioned, i.e.,0.1% proof tre, 0.% proof tre etc. 7.7 UTIMATE STRESS Yield point erve a the gateway to platic zone. Beyond yield point, due to udden decreae in load, material begin to train-harden and recover ome of the elatic property. And by virtue of that, gradual uprie of tre-train curve occur and terminate at a point, called ultimate tre. Thi i the maximum tre, the pecimen can withtand, without any appreciable damage or permanent deformation. 7.8 BREAKING STRESS While ultimate tre i the maximum tre with tanding capacity prior to failure, further increae of ultimate tre lead to failure of the pecimen and thi occur at breaking tre. Here the value of breaking tre lower than ultimate tre, a appearing in the tre-train diagram obtained during experiment, of ductile material, i omehow mileading. What happen in reality i that, beyond ultimate tre, there occur a reduction in area of cro-ection near at the middle of gauge length. Thi phenomenon i called formation of neck or formation of wait. A the grip of extenometer are attached at the end of gauge length, the effect of neck formation thereby the reduction in diameter of the pecimen cannot be taken into account. By reaon of which breaking tre exhibit value lower than ultimate tre. And thi breaking tre i called Nominal Breaking Stre. When the reduced cro-ectional area at neck i conidered to compute actual tre, it i found that breaking tre i pretty higher than ultimate tre. And thi i called True Breaking Stre. In cae of brittle material, ultimate tre i ame a breaking tre. 7.9 WORKING STRESS AND FACTOR OF SAFETY In practical deign of tructure, ome uncertaintie may be aociated in term of loading, material propertie etc. Not only that, in ome material, like concrete, non-ferrou alloy etc., Hooke aw doe not hold good. To encompa all thee apect, it i eential to limit actual tre generated to a value comparatively lower than yield tre of the material. And thi tre i conidered a a afe one. Thi afe tre i deignated at Working Stre ( w ). A pure number, higher than 1 (whole or fraction) that divide the yield tre to obtain working tre i called Factor of Safety. y w =, where n = factor of afety....(7.1) n Sometime working tre i computed deviding ultimate tre by factor of afety DUCTIITY It i the property of a material which allow of it being drawn out by tenion to a mall ection. Brittlene i the lack of ductility MAEABIITY It i the property of a material by virtue of which it can be turned to a very thin heet by the application of preure.

4 188 Mechanical Science I 7.1 TOUGHNESS It can be aid, in general, reitance to deformation. Thi deformation may be due to impact, abraive force, punch etc. 7.1 RESIIENCE et u conider a bar of length l and cro-ectional area A hanging vertically fixed at top, ubjected to a normal pull P. At zero-th intant of application of force, induced deflection i zero. Gradually load i increaed within proportional limit. At poition D, the amount of applied load i P which caue a diplacement δ. Y P D Force Diplacement X Fig. 7.4 (a) Fig. 7.4 (b) Here, external work done by the force, will be repreented by haded area in Figure 7.4 (b). 1 We = P δ... (7.) To reit the effect of external force, reactive work will be done by the force, generated internally due to deformation of body. Thi reactive work i defined a Internal Work or Strain Energy of the ytem, which i ymbolied by U. Amount of train energy i numerically equal to the external work done on the member. A loading i done within proportional limit, a oon a load i releaed, the ytem alo will loe energy and will come back to original poition. Thi property of an elatic material to aborb and releae energy with change in loading i called Reilience. Here total train energy of the ytem U = P δ = ( A)( ε l) = ( ε)( A l)...(7.) Therefore train energy per unit volume U 1 u = = ε=...(7.4) Al E Another very importent parameter in thi context i Proof reilience. It i the maximum train energy tored in a body before yielding occur. Strain energy will be maximum when the body will be treed up to elatic limit. And proof reilience per unit volume i known a Modulu of Reilience. u r p =...(7.5) E Sometime p tand for yield tre or proof tre, a the cae may be.

5 Stre-Strain Diagram and Strength Parameter THERMA STRESS Any engineering material, when ubjected to change in temperature exhibit expanion in temperature rie and contraction in temperature fall. Thi change in temperature i often termed a Thermal oading. A tructural member of length l if ubjected to thermal loading of T, can expand or contract by an amount, δt = l α T, where α = thermal coefficient of material, when free expanion or contraction i allowed. In non-retrained ytem no tre will be developed, though there preent thermal loading. If free movement of the member i retricted partially or fully by omehow or other, ome amount of reactive force will be generated within the member, which will give birth to a reactive tre, termed a thermal tre. Multiple Choice Quetion Select the bet alternative (): 1. The impact trength of a material i an index of it (a) toughne (b) tenile trength (c) capability of being cold worked (d) hardne (e) fatigue trength. The property of a material which allow it to be drawn into a maller ection i called (a) platicity (b) elaticity (c) ductility (d) malleability. The lo of trength in compreion due to overloading i known a (a) hyterei (b) relaxation (c) creep (d) reilience (e) bauchinger effect 4. The maximum train energy that can be tored in a body i known a (a) impact energy (b) reilience (c) proof reilience (d) modulu of reilience (e) toughne 5. The total train energy tored in a body i termed a (a) reilience (b) proof reilience (c) modulu of reilience (d) toughne 6. Proof reilience per unit volume of a material i known a (a) reilience (b) proof reilience (c) modulu of reilience (d) toughne 7. The Figure 7.1 how the tre-train diagram for mild teel. The elatic limit, upper yield point, lower yield point and proportional limit are repreented by (a) A, B, C, D (b) A, C, D, B (c) B, C, D, A (d) C, B, D, A (e) B, C, A, D 8. Proof tre i (a) tre correponding to proportional limit (b) tre cauing material to break (c) tre cauing a pecific permanent deformation uually 0.1% or 0.%. (d) not related with engineering

6 190 Mechanical Science I 9. Thermal train caued in the material of a compoite body due to change in temperature will be (a) ame nature (b) oppoite nature (c) ame magnitude (d) none of above 10. True tre ( ) i related to imple tre () and train (ε) by (a) = ( 1 ε ) (b) ( 1 ) = +ε (c) = ε (d) = ε 11. Bulk modulu K in term of modulu of elaticity (E) and Poion ratio (µ) i given a equal to (a) E /(1 µ) (b) E(1 µ) (c) E (1 µ) (d) E (1+ µ) / (e) E(1 µ) / 1. The energy aborbed by a body, when it i trained within the elatic limit, i known a (a) train energy (b) reilience (c) proof reilience (d) modulu of reilience (e) toughne 1. Value of factor of afety i (a) greater than 1 (b) le than 1 (c) zero (d) none of above 14. Hooke aw i truly valid up to (a) elatic limit (b) proportional limit (c) platic limit (d) fatigue limit 15. In ductile material nominal breaking tre i (a) lower than true breaking tre (b) equal with true breaking tre (c) higher than true breaking tre (d) none of the above 16. In ductile material ultimate tre i (a) higher than true breaking tre (b) lower than nominal breaking tre (c) higher than nominal breaking tre but lower than true breaking tre (d) higher than true breaking tre but lower than nominal breaking tre 17. In a bra pecimen ubjected to tenion, which of the following can be obtained in tre-train diagram? (a) upper yield tre (b) lower yield tre (c) platic tre (d) proof tre 18. Thermal change of length of a metal i related to it thermal coefficient (a) inverely proportional (b) directly proportional (c) directly quare proportional (d) none of the above 19. Thermal train of a body doe not depend on (a) length (b) thermal coefficient (c) change in temperature (d) none of the above 0. In brittle material, normally, breaking tre i (a) higher than ultimate tre (b) lower than ultimate tre (c) equal with ultimate tre (d) none of the above

7 Stre-Strain Diagram and Strength Parameter 191 Anwer 1. (a). (c). (c) 4. (c) 5. (a) 6. (c) 7. (c) 8. (c) 9. (c) 10. (b) 11. (a) 1. (c) 1. (a) 14. (b) 15. (a) 16. (c) 17. (d) 18. (b) 19. (a) 0. (c) Numerical Example EXAMPE 1 A teel bar of 5mm diameter wa teted in tenion and reult were recorded a, limit of proportionality = 196.kN, load at yield = 18.1kN, ultimate load = 78.0 kn. The elongation meaured over a gauge length of 100mm wa 0.189mm at proportionality limit, length of the bar between gauge mark after fracture wa 11.6mm and minimum diameter wa.64mm. Compute tre in the pecimen at variou tage, Young modulu, % elongation and % contraction. Determine permiible tre in the material for a afety factor of π SOUTION Initial c/ area of the bar = (5) mm 4 = Stree at proportionality limit Strain at proportionality limit Young modulu Stree at yield point Ultimate tree Final c/ area at fracture % elongation % contraction in area Permiible or working or allowable tree = = MPa = = = = MPa = = MPa = = MPa π = (.64) = mm = 100 = = 100 = = yield tre factor of afety = = 40. MPa 1.85

8 19 Mechanical Science I EXAMPE Steel railroad, 10m long, i laid with a clearance of mm at 15ºC. At what temperature will the rail jut touch? What tre will be induced in the rail at that temperature, if there were no initial clearance, while a = 117mm/mºC and E = 00 GPa. SOUTION et the deired temperature = T C Now, free thermal elongation from 15 C to T C = (T 15) = 0.117( T 15) A per condition provided, (T 15) = T = C 4 At no clearance condition, induced train will be = = and induced tre = ( ) MPa = 60MPa EXAMPE A teel rod ft long with c/ area of 0.5 inch i tretched between two fixed point. The tenile force i 100 lb at 40 F. Uing E = pi and a = inch/inch/ºf Calculate (a) the temperature at which the tre in the bar will be 10 ki, (b) the temperature at which the tre will be zero. SOUTION Initial tre in the rod = 100 = 4,800 pi 0.5 (a) Required tree i to be developed = 10,000 pi Additional tre to be developed = 10,000 4,800 = 5,00 pi Thi additional tre will be generated due to rie in temperature, ay at T º F 5,00 Strain correponding to additional tre = Elongation due to above train = ( )inch = inch Hence, (T 40) = = T = ºF (b) A tre i to be zero, temperature of the ytem will have to be reduced. et that temperature be T 1 F. 4,800 4 Now, train correponding to initial tre = = Thermal train due to reduction of temperature = (40 T ) 6 4 Here, (40 T ) = T 1 = F 1 1

9 Stre-Strain Diagram and Strength Parameter 19 EXAMPE 4 A bronze bar, m long with a c/ area of 0mm i placed between two rigid wall. At 0 C, the gap between bar and wall i.5mm. Find temperature at which compreive tre in the bar will be 5MPa. Take a = m/m/ºc and E= 80GPa. 5 4 SOUTION Strain correponding to 5MPa tre = = Elongation due to above train = = 1.15mm To generate above compreive tre, total elongation to be compenated by thermal rie will be = ( ) mm =.815 mm and let the final temperature be T ºC. 6 Now, ( T + 0) =.815 T = 50.6º C. EXAMPE 5 Calculate increae in tre for each egment of the compound bar, if temperature increae by 100ºF. Aume unyielding upport and bar i uitably braced againt buckling. If upport yield by 0.01inch, compute tree. SOUTION If free thermal elongation would be allowed, it will be for aluminium 6 = ( ) = inch 6 for teel = ( ) = inch = 15", Aa= 1.5in 6 E=9 10pi 6 = A S Aluminium Steel a= 10", A a=in 6 E a=10 10 pi -6 A =1,8 10 a and total = ( ) = inch...(1) A the upport are unyielding, to reit thi elongation ome mechanical tre will be developed and correponding contraction will be: a a for aluminium = a = 10 inch 6 E a for teel = = 15 inch 6 E 9 10 To atify compatibility for deformation, a = = 550 Another equation of compatibility, a A = A a a.0 = 1.5 a = (4) a...()... ()

10 194 Mechanical Science I Solving equation () and (4), = 1,46.5 pi, = 17, pi If the upport yield, equation of compatibility will be a = a = (5) Solving equation (5) and (4), a = 7,47.79 pi = 9,90.7 pi a EXAMPE 6 At 80ºC a teel tire, 1mm thick and 90mm wide i to be hrunk fit onto a locomative wheel, m in diameter, jut fit over the wheel which i at a temperature of 5ºC. Determine contact preure between tire and wheel at 5ºC. α = m/m/ºc and E = 00 GPa. SOUTION et diameter of tire at 5ºC = d and of wheel = D Conidering condition of compatibility, π D= π d[ 1+ α(80 5) ] 6 = π d[ ] D d =...(1) Circumferential train in the wheel πd πd D ε= = 1 πd d = Correponding tree in the wheel = ( ) MPa EXAMPE 7 = 19.6MPa In the adjoining figure, bar ABC i initially horizontal and vertical rod are tre-free. Determine tre in the aluminium rod, if temperature of the teel rod i decreaed by 40ºC. 9 E = N/m, E = N/m, a 9 α = 11.7µm/m/ºC, α µm/m/ºc, SOUTION From free-body diagram of the bar, taking moment at B, the condition of equilibrium will be, R 0.6 = R 1. A a = ( A ) = ( A ) t t al al C R A A B 0. m Aluminium c/ = 100 q.mm 0.9 m Steel c/ = 00 q.mm. A B C 0.6 m 1. m R C C

11 Stre-Strain Diagram and Strength Parameter = t al t = 8al...(1) Due to reduction of temperature, train in the teel rod = = Correponding tre teel rod ( t ) = ( ) = 9.6MPa From (1), 9.6 al = = 11.7MPa 8 EXAMPE 8 A teel tube of 4mm external and 14mm internal diameter encloe a copper rod of 1mm diameter. The aembly i held rigidly at both end at C. Compute (i) tree at 1 C, (ii) the maximum temperature the aembly can withtand. Aume E = 0GPa, 6 α = / ºC, 6 α = /º C, c ( ) max 0MPa, ( ) 115MPa, = c max = SOUTION A αc> α, copper rod will expand more than teel tube, if free expanion i allowed. A both end of aembly are fatened regidly, free expanion will not be permitted in either rod or tube. A thermal compromie will be happened. By virtue of it, free expanion of copper rod will be reduced and free expanion of teel tube will be increaed. At thi level, a mechanically induced compreive force will act at copper rod and a tenile force will act at teel tube. Both of thee force are equal in magnitude. Pc P Hence, αc T = α T + A E A E...(1) c c c (αc α ) T E c E...() 6 c ( 1 ) = c = 77...() From equation of compatibility, P = P A = A c c c C S

12 196 Mechanical Science I A c = A... (4) c 4 14 = =.69 1 Solving equation () and (4) c = 64.7 MPa 4.5 MPa} = To find maximum withtandable temperature T, we are to ubtitute maximum permiible tree in equation () 6 ( c) max ( ) max ( T ) = (5) 10 If ( c = 0MPa, according to equation (4), MPa ) max = If ( ) max = 115MPa, according to equation (4), c = 0.485MPa, which i higher than ( c ) max. So, maximum allowable tre in copper and teel will be 0MPa and MPa repectively. Subtituting in equation (5), ( T ) = T = 84.54ºC EXAMPE 9 A bar of uniform c/ A and length hang vertically, ubjected to it own weight. Prove that the train energy tored within the bar i Aρ U =. 6E SOUTION et u take a ection x ditance from bottom, of thickne dx. The elongation of length dx be dδ. Strain in length dx, ε x = dδ dx dx x W x W x W x x Stre in length dx, = Wx ρax x x A = A =ρ x ρxdx Now, E = = ε dδ Strain energy tored in dx x ρxdx dδ = E du = average weight elongation of dx 1 ρxdx = W x E 1 ρxdx Aρ = (ρ Ax) = x dx E E

13 Stre-Strain Diagram and Strength Parameter 197 Total train energy tored, U = du = 0 0 A ρ x dx E Aρ Aρ = = E 6E EXAMPE 10 In the adjacent figure, load i allowed to drop on the collar from a height h, find the expreion of tre induced in the rod due to impact. δ SOUTION Strain in the bar = E Vertical rod W (oad) δ = E...(1) Work done by the load = W (h + δ)...() h Strain energy tored by the rod Equating () and () following condition of equilibrium, W( h+ δ ) = ( A) E... (4) Subtituting the expreion of (1) into (4), we have, W WEh = 0 A A = ( A )...() E Collar Solving, AEh Sometime, W W W AEh = 1 1 A + + W term i deignated a equivalent tatic load, i.e., AEh We = W N...(5) (i) if δ h, from equation (4), Wh = E ( A) EhW =... (6) A

14 198 Mechanical Science I (ii) If h 0, from equation (4) EXAMPE 11 W =... (7) A Find the train energy of thee two member, loaded and hown in the figure. SOUTION Strain energy (S.E) of a bar ubjected to loading P For the firt member, S.E 1 1 Pl = P δ =. AE d d 4 10 d d l l l P 1 P P = + + π π π ( d) E ( d) E ( d) E P P 1 4 Pl 1 = + + d E π Pl 7 7 Pl = = πd E 16 8 πd E For the econd member, S.E 9l l 9l P 1 P P = + + π π π ( d) E ( d) E ( d) E Pl π d E = + + = Pl 1 Pl πd E 5 = 5 πd E. Exercie 1. A ma of 00 kg fall through a height of 500 mm on a concrete column of mm ection. Determine the maximum tre and deformation in the 4.0m long column, conidering modulu of elaticity of concrete 0.0GPa. An. [ 9.06MPa, 1.81mm]

15 Stre-Strain Diagram and Strength Parameter 199. A rod i m long at a temperature of 15 C. Find the expanion of the rod, when the temperature i raied to 95 C. If thi expanion i prevented, find the tre induced in the material of the rod. Take E = N / mm and α = per degree centrigrade. An. [ 0.88cm, 96N/mm ]. A teel rod 5cm diameter and 6m long i connected to two grip and the rod i maintained at a temperature of 100 C. Determine the tre and pull exerted when the temperature fall to 0 C if (i) the end do not yield, and (ii) the end yield by 0.15cm. Take E = 00GPa, α = / C. 4. Compute the maximum force a 00mm long compound bar compriing a copper rod of 18 mm diameter encloed in a mild teel tube of 00mm inner and mm outer diameter can utain. Aume Young modulii for copper and teel to be 10GPa and 195GPa, repectively. What will be the reduction in the maximum load, if the bar temperature rie by 40K? Find the reduction in the trength when the temperature fall by 40K. Allowable tree are 80MPa and 140MPa in copper and teel, repectively; α = K 1 for copper and K 1 for teel. 5. A.5m long teel column of cro-ectional area 5000mm i ubjected to a load of 1.6 MN. Determine the afety factor for the column, if the yield tre of teel i 550MPa. Determine the allowable load on the column, if the deformation of the column hould not exceed 5mm. 6. A compound bar comprie a 1.5mm diameter aluminum rod and a copper tube of 14.5mm inner and 5mm outer diameter. If the Young modulii of aluminium and copper are 80GP and 10GPa, repectively, determine the tre in the aembly when ubject to (i) a temperature rie of 95K, and (ii) a temperature fall of 5K ; α = K 1 for aluminium and K 1 for copper. 7. A teel rod of 0mm diameter pae centrally through a copper tube of 40mm external diameter and 0mm internal diameter. The tube i cloed at each end by rigid plate of negligible thickne. The nut are tightened lightly home on the projected part of the rod. If the temperature of the aembly i raied by 60 C, calculate the tree developed in copper and teel. Take E for teel and copper a 00GN/m and 100GN/m and α for teel and copper a per C and per C. An. [16., 8.4 N / mm ]