Chapter-1 Stress and Strain Page- 1 1. Stress and Strain 1.1 Stress (σ) Thery at a Glance (fr IES, GATE, PSU) When a material is subjected t an external frce, a resisting frce is set up within the cmpnent. The internal resistance frce per unit area acting n a material r intensity f the frces distributed ver a given sectin is called the stress at a pint. It uses riginal crss sectin area f the specimen and als knwn as engineering stress r cnventinal stress. P Therefre, σ = A P is expressed in Newtn (N) and A, riginal area, in square meters (m ), the stress σ will be expresses in N/ m. This unit is called Pascal (Pa). As Pascal is a small quantity, in practice, multiples f this unit is used. 1 kpa = 3 Pa = 3 N/ m (kpa = Kil Pascal) 1 MPa = 6 Pa = 6 N/ m = 1 N/mm (MPa = Mega Pascal) 1 GPa = 9 Pa = 9 N/ m (GPa = Giga Pascal) Let us take an example: A rd mm mm crss-sectin is carrying an axial tensile lad kn. In this rd the tensile stress develped is given by 3 P kn N ( σt ) = = = = 0N/mm = 0MPa A mm mm 0mm ( ) The resultant f the internal frces fr an axially laded member is nrmal t a sectin cut perpendicular t the member axis. The frce intensity n the shwn sectin is defined as the nrmal ΔF P stress. σ = lim and σavg = ΔA 0 Δ A A
Chapter-1 Stress and Strain Page- Tensile stress (σ t ) If σ > 0 the stress is tensile. i.e. The fibres f the cmpnent tend t elngate due t the external frce. A member subjected t an external frce tensile P and tensile stress distributin due t the frce is shwn in the given figure. Cmpressive stress (σ c ) If σ < 0 the stress is cmpressive. i.e. The fibres f the cmpnent tend t shrten due t the external frce. A member subjected t an external cmpressive frce P and cmpressive stress distributin due t the frce is shwn in the given figure. Shear stress (τ ) When frces are transmitted frm ne part f a bdy t ther, the stresses develped in a plane parallel t the applied frce are the shear stress. Shear stress acts parallel t plane f interest. Frces P is applied transversely t the member AB as shwn. The crrespnding internal frces act in the plane f sectin C and are called shearing P frces. The crrespnding average shear stress ( τ ) = Area 1. Strain (ε) The displacement per unit length (dimensinless) is knwn as strain. Tensile strain (ε t ) The elngatin per unit length as shwn in the figure is knwn as tensile strain. ε t = L/ L It is engineering strain r cnventinal strain. Here we divide the elngatin t riginal length nt actual length (L + Δ L)
Chapter-1 Stress and Strain Page- 3 Let us take an example: A rd 0 mm in riginal length. When we apply an axial tensile lad kn the final length f the rd after applicatin f the lad is 0.1 mm. S in this rd tensile strain is develped and is given by ΔL L L 0.1mm 0mm 0.1mm ( εt ) = = = = = 0.001 (Dimensinless) Tensile L L 0mm 0mm Cmpressive strain (ε c ) If the applied frce is cmpressive then the reductin f length per unit length is knwn as cmpressive strain. It is negative. Then ε c = (- L)/ L Let us take an example: A rd 0 mm in riginal length. When we apply an axial cmpressive lad kn the final length f the rd after applicatin f the lad is 99 mm. S in this rd a cmpressive strain is develped and is given by ΔL L L 99mm 0mm 1mm ( εc ) = = = = = 0.01 (Dimensinless) cmpressive L L 0mm 0mm Shear Strain ( γ ): When a frce P is applied tangentially t the element shwn. Its edge displaced t dtted line. Where δ is the lateral displacement f the upper face f the element relative t the lwer face and L is the distance between these faces. Then the shear strain is ( γ ) = δ L Let us take an example: A blck 0 mm 0 mm base and mm height. When we apply a tangential frce kn t the upper edge it is displaced 1 mm relative t lwer face. Then the direct shear stress in the element 3 kn N (τ ) = = = 1N/mm = 1MPa 0mm 0mm 0mm 0mm 1mm And shear strain in the element ( γ ) = = = 0.1 Dimensinless mm 1.3 True stress and True Strain The true stress is defined as the rati f the lad t the crss sectin area at any instant. lad Instantaneus area ( σ ) = = σ( 1+ ε) T Where σ and ε is the engineering stress and engineering strain respectively. True strain
Chapter-1 Stress and Strain Page- 4 L dl L A d = ln ln 1 ln ln = l = + = = L A d ( ε ) ( ε) T L r engineering strain (ε ) = T e ε -1 The vlume f the specimen is assumed t be cnstant during plastic defrmatin. [ AL = AL] It is valid till the neck frmatin. Cmparisn f engineering and the true stress-strain curves shwn belw The true stress-strain curve is als knwn as the flw curve. True stress-strain curve gives a true indicatin f defrmatin characteristics because it is based n the instantaneus dimensin f the specimen. In engineering stress-strain curve, stress drps dwn after necking since it is based n the riginal area. In true stress-strain curve, the stress hwever increases after necking since the crss-sectinal area f the specimen decreases rapidly after necking. The flw curve f many metals in the regin f unifrm plastic defrmatin can be expressed by the simple pwer law. σ T = K(ε T ) n Where K is the strength cefficient n is the strain hardening expnent n = 0 perfectly plastic slid n = 1 elastic slid Fr mst metals, 0.1< n < 0.5 Relatin between the ultimate tensile strength and true stress at maximum lad max The ultimate tensile strength ( σ ) = u P A The true stress at maximum lad ( ) max P σ u = T A And true strain at maximum lad ( ) ln ε T A = A r A A = e εt max max εt Eliminating P max we get, ( σ ) = = = σ e u T P P A A A A Where P max = maximum frce and A = Original crss sectin area u
Chapter-1 Stress and Strain Page- 5 A = Instantaneus crss sectin area Let us take tw examples: (I.) Only elngatin n neck frmatin In the tensin test f a rd shwn initially it was A = 50 mm and L = 0 mm. After the applicatin f lad it s A = 40 mm and L = 15 mm. Determine the true strain using changes in bth length and area. Answer: First f all we have t check that des the member frms neck r nt? Fr that check AL = AL r nt? Here 50 x 0 = 40 x 15 s n neck frmatin is there. Therefre true strain L dl 15 ( εt ) = = ln = 0.3 l 0 ( ε ) T L A 50 = ln = ln = 0.3 A 40 (II.) Elngatin with neck frmatin A ductile material is tested such and necking ccurs then the final gauge length is L=140 mm and the final minimum crss sectinal area is A = 35 mm. Thugh the rd shwn initially it was A = 50 mm and L = 0 mm. Determine the true strain using changes in bth length and area. Answer: First f all we have t check that des the member frms neck r nt? Fr that check AL = AL r nt? Here A L = 50 x 0 = 5000 mm 3 and AL=35 x 140 = 400 mm 3. S neck frmatin is there. Nte here A L > AL. Therefre true strain A 50 ( εt ) = ln = ln = 0.357 A 35 But nt ( ε ) T L dl 140 = = ln = 0.336 (wrng) l 0 L (If n neck frmatin ccurs bth area and gauge length can be used fr a strain calculatin.) (After necking, gauge length gives errr but area and diameter can be used fr the calculatin f true strain at fracture and befre fracture als.) 1.4 Hk s law Accrding t Hk s law the stress is directly prprtinal t strain i.e. nrmal stress (σ) α nrmal strain (ε) and shearing stress ( τ ) α shearing strain ( γ ). σ = Eε and τ = Gγ The c-efficient E is called the mdulus f elasticity i. e i.e. its resistance t elastic strain. The c-efficient G is called the shear mdulus f elasticity r mdulus f rigidity. 1.5 Vlumetric strain( ε v )
Chapter-1 Stress and Strain Page- 6 A relatinship similar t that fr length changes hlds fr three-dimensinal (vlume) change. Fr vlumetric strain,( ) v ε, the relatinship is ( ε ) = (V-V 0 )/V 0 r ( ) v P ε v = V/V 0 = K Where V is the final vlume, V 0 is the riginal vlume, and V is the vlume change. Vlumetric strain is a rati f values with the same units, s it als is a dimensinless quantity. V/V= vlumetric strain = ε x +ε y + ε z = ε 1 +ε + ε 3 Dilatin: The hydrstatic cmpnent f the ttal stress cntributes t defrmatin by changing the area (r vlume, in three dimensins) f an bject. Area r vlume change is called dilatin and is psitive r negative, as the vlume increases r decreases, p respectively. e = Where p is pressure. K 1.6 Yung s mdulus r Mdulus f elasticity (E) = PL = σ Aδ 1.7 Mdulus f rigidity r Shear mdulus f elasticity (G) = τ γ = = PL Aδ Δp 1.8 Bulk Mdulus r Vlume mdulus f elasticity (K) = = Δv v Δp ΔR R 1. Relatinship between the elastic cnstants E, G, K, μ ( μ) ( μ) E = G 1+ = 3K 1 = 9KG 3K + G [VIMP] Where K = Bulk Mdulus, μ = Pissn s Rati, E= Yung s mdulus, G= Mdulus f rigidity Fr a linearly elastic, istrpic and hmgeneus material, the number f elastic cnstants required t relate stress and strain is tw. i.e. any tw f the fur must be knwn. If the material is nn-istrpic (i.e. anistrpic), then the elastic mdulii will vary with additinal stresses appearing since there is a cupling between shear stresses and nrmal stresses fr an anistrpic material. Let us take an example: The mdulus f elasticity and rigidity f a material are 00 GPa and 80 GPa, respectively. Find all ther elastic mdulus.
Chapter-1 Stress and Strain Page- 7 9KG = + = = 3K + G Answer: Using the relatin E G( 1 μ) 3K( 1 μ) we may find all ther elastic mdulus easily Pissn s Rati ( μ) : E E 00 1+ μ = μ = 1= 1= 0.5 G G 80 Bulk Mdulus (K) : = E μ E μ 00 3K K 1 3 1 3 1 0.5 = 133.33GPa 1.11 Pissn s Rati (μ ) = Transverse strain r lateral strain Lngitudinal strain = x y ( ) ( ) (Under unidirectinal stress in x-directin) The thery f istrpic elasticity allws Pissn's ratis in the range frm -1 t 1/. Pissn's rati in varius materials Material Pissn's rati Material Pissn's rati Steel 0.5 0.33 Rubber 0.48 0.5 C.I 0.3 0.7 Crk Nearly zer Cncrete 0. Nvel fam negative We use crk in a bttle as the crk easily inserted and remved, yet it als withstand the pressure frm within the bttle. Crk with a Pissn's rati f nearly zer, is ideal in this applicatin. 1.1 Fr bi-axial stretching f sheet 1.13 Elngatin Lf 1 = 1 ln L Original length L 1 L f = ln L f -Final length L Final thickness (t f ) = Initial thickness(t ) 1 e e A prismatic bar laded in tensin by an axial frce P Fr a prismatic bar laded in tensin by an axial frce P. The elngatin f the bar can be determined as
Chapter-1 Stress and Strain Page- 8 δ = PL AE Let us take an example: A Mild Steel wire 5 mm in diameter and 1 m lng. If the wire is subjected t an axial tensile lad kn find its extensin f the rd. (E = 00 GPa) PL Answer: We knw that ( δ ) = AE Here given, Frce (P) = kn = 00N Length(L) = 1 m ( 0.005) π d π 5 Area(A) = = m = 1.963 m 4 4 9 Mdulus f Elasticity ( E) = 00 GPa = 00 N/m PL ( 00) 1 Therefre Elngatin( δ ) = = m 5 9 AE 1.963 00 Elngatin f cmpsite bdy ( ) ( ) 3 = =.55 m.55 mm Elngatin f a bar f varying crss sectin A 1, A,----------, A n f lengths l 1, l,-------- l n respectively. P l1 l l3 l n δ = + + + E A1 A A3 An Let us take an example: A cmpsite rd is 00 mm lng, its tw ends are 40 mm and 30 mm in area and length are 300 mm and 00 mm respectively. The middle prtin f the rd is 0 mm in area and 500 mm lng. If the rd is subjected t an axial tensile lad f 00 N, find its ttal elngatin. (E = 00 GPa). Answer: Cnsider the fllwing figure Given, Lad (P) =00 N Area; (A 1 ) = 40 mm, A = 0 mm, A 3 = 30 mm Length; (l 1 ) = 300 mm, l = 500 mm, l 3 = 00 mm E = 00 GPa = 00 9 N/m = 00 3 N/mm Therefre Ttal extensin f the rd P l 1 l l3 δ = + + E A1 A A3 00N 300mm 500mm 00mm = + + 3 00 N / mm 40mm 0mm 30mm = 0.196mm Elngatin f a tapered bdy Elngatin f a tapering rd f length L due t lad P at the end
Chapter-1 Stress and Strain Page- 9 4PL δ= (d π Ed d 1 and d are the diameters f smaller & larger ends) 1 Yu may remember this in this way, PL PL δ= ie.. π EA E dd 1 4 Let us take an example: A rund bar, f length L, tapers unifrmly frm small diameter d 1 at ne end t bigger diameter d at the ther end. Shw that the extensin prduced by a tensile axial lad P 4PL is ( δ ) =. πd1de If d = d 1, cmpare this extensin with that f a unifrm cylindrical bar having a diameter equal t the mean diameter f the tapered bar. Answer: Cnsider the figure belw d 1 be the radius at the smaller end. Then at a X crss sectin XX lcated at a distance x frm the smaller end, the value f diameter d x is equal t eq dx d1 x d d1 = + L x r dx = d1+ ( d d1) L d d1 1 = d1 ( 1+ kx) Where k = L d We nw taking a small strip f diameter 'd 'and length 'd 'at sectin XX. Elngatin f this sectin 'd ' length d ( δ ) x PL P. dx 4 P. dx = = = AE π d x π.{ 1 ( 1+ )} E d kx E 4 x x 1
Chapter-1 Stress and Strain Page- Therefre ttal elngatin f the taper bar ( δ) δ = d = x= L πed x = 0 1 4PL = πedd 4Pdx 1 ( 1+ kx ) Cmparisn: Case-I: Where d = d 1 4PL PL Elngatin ( δi ) = = πed1 d1 πed1 Case II: Where we use Mean diameter d1+ d d1+ d1 3 dm = = = d1 PL P. L Elngatin f such bar ( δii ) = = AE π 3 d1. E 4 16PL = 9π Ed1 Extensin f taper bar 9 = = Extensin f unifrm bar 16 8 9 Elngatin f a bdy due t its self weight (i) Elngatin f a unifrm rd f length L due t its wn weight W WL δ= AE The defrmatin f a bar under its wn weight as cmpared t that when subjected t a direct axial lad equal t its wn weight will be half. (ii) Ttal extensin prduced in rd f length L due t its wn weight ω per with ωl length. δ= EA (iii) Elngatin f a cnical bar due t its self weight ρ gl WL δ= = 6E A E max 1.14 Structural members r machines must be designed such that the wrking stresses are less than the ultimate strength f the material.
Chapter-1 Stress and Strain Page- 11 Wrking stress ( σ ) w σ y = n=1.5 t n factr f safety σ ult = n = 1 t 3 n 1 σ p = σ p = Prf stress n 1.15 Factr f Safety: (n) = σ r σ r σ y p ult σ w 1.16 Thermal r Temperature stress and strain When a material underges a change in temperature, it either elngates r cntracts depending upn whether temperature is increased r decreased f the material. If the elngatin r cntractin is nt restricted, i. e. free then the material des nt experience any stress despite the fact that it underges a strain. The strain due t temperature change is called thermal strain and is expressed as, ε = α( ΔT ) Where α is c-efficient f thermal expansin, a material prperty, and T is the change in temperature. The free expansin r cntractin f materials, when restrained induces stress in the material and it is referred t as thermal stress. ( ) σt = αe ΔT Where, E = Mdulus f elasticity Thermal stress prduces the same effect in the material similar t that f mechanical stress. A cmpressive stress will prduce in the material with increase in temperature and the stress develped is tensile stress with decrease in temperature. Let us take an example: A rd cnsists f tw parts that are made f steel and cpper as shwn in figure belw. The elastic mdulus and cefficient f thermal expansin fr steel are 00 GPa and 11.7 x -6 per 0 C respectively and fr cpper 70 GPa and 1.6 x -6 per 0 C respectively. If the temperature f the rd is raised by 50 0 C, determine the frces and stresses acting n the rd.
Chapter-1 Stress and Strain Page- 1 Answer: If we allw this rd t freely expand then free expansin δ = α ΔT L T ( ) 6 6 ( ) ( ) = 11.7 50 500 + 1.6 50 750 ( ) = 1.5 mm Cmpressive But accrding t diagram nly free expansin is 0.4 mm. Therefre restrained deflectin f rd =1.5 mm 0.4 mm = 0.705 mm Let us assume the frce required t make their elngatin vanish be P which is the reactin frce at the ends. PL PL δ = + AE Steel AE Cu P 500 P 750 r 0.705 = + π 9 π 9 ( 0.075) ( 00 ) ( 0.050) ( 70 ) 4 4 r P = 116.6 kn Therefre, cmpressive stress n steel rd 3 P 116.6 σ Steel = = N/m = 6.39 MPa A π Steel ( 0.075) 4 And cmpressive stress n cpper rd 3 P 116.6 σ Cu = = N/m = 59.38 MPa A π Cu ( 0.050) 4 1.17 Thermal stress n Mild steel cated with Brass A brass rd placed within a steel tube f exactly same length. The assembly is making in such a way that elngatin f the cmbinatin will be same. T calculate the stress induced in the brass rd, steel tube when the cmbinatin is raised by t C then the fllwing analgy have t d.
Chapter-1 Stress and Strain Page- 13 (a) Original bar befre heating. (b) Expanded psitin if the members are allwed t expand freely and independently after heating. (c) Expanded psitin f the cmpund bar i.e. final psitin after heating. Cmpatibility Equatin: δ = δst + δsf = δbt δbf Equilibrium Equatin: σ A = σ A s s B B Assumptin: 1. L = L s = L. α > α b s 3. Steel Tensin Brass Cmpressin B Where, δ = Expansin f the cmpund bar = AD in the abve figure. δ st = Free expansin f the steel tube due t temperature rise t C = α Lt s = AB in the abve figure. δ sf = Expansin f the steel tube due t internal frce develped by the unequal expansin. = BD in the abve figure. δ Bt = Free expansin f the brass rd due t temperature rise t C = α Lt b = AC in the abve figure. δ Bf = Cmpressin f the brass rd due t internal frce develped by the unequal expansin. = BD in the abve figure. And in the equilibrium equatin Tensile frce in the steel tube = Cmpressive frce in the brass rd Where, σ s = Tensile stress develped in the steel tube. σ B = Cmpressive stress develped in the brass rd. A s = Crss sectin area f the steel tube. A B = Crss sectin area f the brass rd. Let us take an example: See the Cnventinal Questin Answer sectin f this chapter and the questin is Cnventinal Questin IES-008 and it s answer.
Chapter-1 Stress and Strain Page- 14 1.18 Maximum stress and elngatin due t rtatin (i) max ρω L 8 σ = and ( L) ρω L δ = 1E 3 (ii) max ρω L σ = and ( L) ρω L δ = 3E Fr remember: Yu will get (ii) by multiplying by 4 f (i) 3 1.18 Creep When a member is subjected t a cnstant lad ver a lng perid f time it underges a slw permanent defrmatin and this is termed as creep. This is dependent n temperature. Usually at elevated temperatures creep is high. The materials have its wn different melting pint; each will creep when the hmlgus Testing temperature temperature > 0.5. Hmlgus temp = Melting temperature > 0.5 A typical creep curve shws three distinct stages with different creep rates. After an initial rapid elngatin ε, the creep rate decrease with time until reaching the steady state. 1) Primary creep is a perid f transient creep. The creep resistance f the material increases due t material defrmatin. ) Secndary creep prvides a nearly cnstant creep rate. The average value f the creep rate during this perid is called the minimum creep rate. A stage f balance between cmpeting strain hardening and recvery (sftening) f the material. 3) Tertiary creep shws a rapid increase in the creep rate due t effectively reduced crsssectinal area f the specimen leading t creep rupture r failure. In this stage intergranular cracking and/r frmatin f vids and cavities ccur.
Chapter-1 Stress and Strain Page- 15 c Creep rate =c 1 σ Creep strain at any time = zer time strain intercept + creep rate Time + c σ t c = 0 1 Where, c1, c are cnstants σ = stress 1.19 If a lad P is applied suddenly t a bar then the stress & strain induced will be duble than thse btained by an equal lad applied gradually. 1.0 Stress prduced by a lad P in falling frm height h σ d h = σ 1+ 1 + σ, being stress & strain prduced by static lad P & L=length f bar. L A AEh = 1+ 1+ P PL 1.1 Lads shared by the materials f a cmpund bar made f bars x & y due t lad W, AE x x Px = W. AE + AE x x y y AE y y Py = W. AE + AE x x y y 1. Elngatin f a cmpund bar, δ = PL AE + AE x x y y 1.3 Tensin Test
Chapter-1 Stress and Strain Page- 16 i) True elastic limit: based n micr-strain measurement at strains n rder f x -6. Very lw value and is related t the mtin f a few hundred dislcatins. ii) Prprtinal limit: the highest stress at which stress is directly prprtinal t strain. iii) Elastic limit: is the greatest stress the material can withstand withut any measurable permanent strain after unlading. Elastic limit > prprtinal limit. iv) Yield strength is the stress required t prduce a small specific amunt f defrmatin. The ffset yield strength can be determined by the stress crrespnding t the intersectin f the stress-strain curve and a line parallel t the elastic line ffset by a strain f 0. r 0.1%. (ε = 0.00 r 0.001) The ffset yield stress is referred t prf stress either at 0.1 r 0.5% strain used fr design and specificatin purpses t avid the practical difficulties f measuring the elastic limit r prprtinal limit. v) Tensile strength r ultimate tensile strength (UTS) σ u is the maximum lad P max divided by the riginal crss-sectinal area A f the specimen. Lf L vi) % Elngatin, =, is chiefly influenced by unifrm elngatin, which is dependent n L the strain-hardening capacity f the material.
Chapter-1 Stress and Strain Page- 17 A Af vii) Reductin f Area: q = A Reductin f area is mre a measure f the defrmatin required t prduce failure and its chief cntributin results frm the necking prcess. Because f the cmplicated state f stress state in the neck, values f reductin f area are dependent n specimen gemetry, and defrmatin behaviur, and they shuld nt be taken as true material prperties. RA is the mst structure-sensitive ductility parameter and is useful in detecting quality changes in the materials. viii) Stress-strain respnse 1.4 Elastic strain and Plastic strain The strain present in the material after unlading is called the residual strain r plastic strain and the strain disappears during unlading is termed as recverable r elastic strain. Equatin f the straight line CB is given by σ = ttal E Plastic E = Elastic E Carefully bserve the fllwing figures and understand which ne is Elastic strain and which ne is Plastic strain
Chapter-1 Stress and Strain Page- 18 Let us take an example: A mm diameter tensile specimen has a 50 mm gauge length. The lad crrespnding t the 0.% ffset is 55 kn and the maximum lad is 70 kn. Fracture ccurs at 60 kn. The diameter after fracture is 8 mm and the gauge length at fracture is 65 mm. Calculate the fllwing prperties f the material frm the tensin test. (i) % Elngatin (ii) Reductin f Area (RA) % (iii) Tensile strength r ultimate tensile strength (UTS) (iv) Yield strength (v) Fracture strength (vi) If E = 00 GPa, the elastic recverable strain at maximum lad (vii) If the elngatin at maximum lad (the unifrm elngatin) is 0%, what is the plastic strain at maximum lad? π Answer: Given, Original area ( A ) = ( ) = 5 0 0.0 m 7.854 m 4 π Area at fracture ( A ) = ( 0.008) m = 5.07 5 m f 4 Original gauge length (L 0 ) = 50 mm Gauge length at fracture (L) = 65 mm Therefre L L0 65 50 (i) % Elngatin = 0% = 0= 30% L0 50 A0 A (ii) Reductin f area (RA) = q = f 7.854 5.07 0% = 0% = 36% A0 7.854 3 Pmax 70 (iii) Tensile strength r Ultimate tensile strength (UTS), σ u = = N/m = 891 MPa 5 A 7.854 3 Py 55 (iv) Yield strength ( σ y ) = = N/m = 700 MPa 5 A 7.854 3 PFracture 60 (v) Fracture strength ( σ F ) = = N/m = 764MPa 5 A 7.854 6 Pmax / A 891 (vi) Elastic recverable strain at maximum lad ( ε E ) = = = 0.0045 9 E 00 ε = ε ε = 0.000 0.0045 = 0.1955 (vii) Plastic strain ( ) P ttal E
Chapter-1 Stress and Strain Page- 19 1.5 Elasticity This is the prperty f a material t regain its riginal shape after defrmatin when the external frces are remved. When the material is in elastic regin the strain disappears cmpletely after remval f the lad, The stress-strain relatinship in elastic regin need nt be linear and can be nn-linear (example rubber). The maximum stress value belw which the strain is fully recverable is called the elastic limit. It is represented by pint A in figure. All materials are elastic t sme extent but the degree varies, fr example, bth mild steel and rubber are elastic materials but steel is mre elastic than rubber. 1.6 Plasticity When the stress in the material exceeds the elastic limit, the material enters int plastic phase where the strain can n lnger be cmpletely remved. Under plastic cnditins materials ideally defrm withut any increase in stress. A typical stress strain diagram fr an elastic-perfectly plastic material is shwn in the figure. Mises-Henky criterin gives a gd starting pint fr plasticity analysis. 1.7 Strain hardening If the material is reladed frm pint C, it will fllw the previus unlading path and line CB becmes its new elastic regin with elastic limit defined by pint B. Thugh the new elastic regin CB resembles that f the initial elastic regin OA, the internal structure f the material in the new state has changed. The change in the micrstructure f the material is clear frm the fact that the ductility f the material has cme dwn due t strain hardening. When the material is reladed, it fllws the same path as that f a virgin material and fails n reaching the ultimate strength which remains unaltered due t the
Chapter-1 Stress and Strain Page- 0 intermediate lading and unlading prcess. 1.8 Stress reversal and stress-strain hysteresis lp We knw that fatigue failure begins at a lcal discntinuity and when the stress at the discntinuity exceeds elastic limit there is plastic strain. The cyclic plastic strain results crack prpagatin and fracture. When we plt the experimental data with reversed lading and the true stress strain hysteresis lps is fund as shwn belw. True stress-strain plt with a number f stress reversals Due t cyclic strain the elastic limit increases fr annealed steel and decreases fr cld drawn steel. Here the stress range is σ. ε p and ε e are the plastic and elastic strain ranges, the ttal strain range being ε. Cnsidering that the ttal strain amplitude can be given as ε = ε p + ε e
Chapter-16 Riveted and Welded Jint Page- 1