APPLIED MECHANICS DEPARTMENTS V NATIONAL INSTITUTE OF TECHNOLOGY SURAT- 395007 THEORY OF ELASTICITY & PLASTICITY - Prof. Johny 1 TEXT BOOKS: 1) Theory of Elasticity - Timoshenko S P and Goodier J N (Mc graw Hill) 2) Adv. Strength of material - E. Volterra & Gaines (Prentice Hall) 3) Theory of Elasticity - M. Filonenko , Borodich ( Dover Publ, N.Y ) REF. BOOKS: 1) Elasticity ( Tensor, dyadic and Engg. Approaches) - P C Chau & N J Pagano, C D Van Norstvand co.) 2) Structural mechanics with introduction to Elasticity and Plasticity - B. Venkatraman , Sharad A Patel (Mc Graw Hill) 3) Elasticity for Engineers - D. S. Dugdale, C. Ruiz (Mc Graw Hill ) 4) Matrix –Tensor method in continuum mechanics - S. F. Borg (C D Van Norstvand co. N.Y.) 5) Applied Elasticity -Chi-Teh-Wang (Mc Graw Hill N. Y. ) 2 INTRODUCTION Almost all engineering material possess to a certain extent the property of Elasticity. Assumptions - 1) The deformations are assumed to be infinitesimal. 2) The body is assumed to be perfectly elastic. 3) The body is assumed to be homogeneous. (i.e. the properties are same at each point.) 4) The body is assumed to be isotropic (i.e. the properties are same in all directions.) 3 4 . 5 . shows the tensile test diagram for a bar of constant c/s .γ Єx -------------2) The factor of proportionality γ represents a constant for each material and is called Poisson’s Ratio. γ = 1/m = Poisson ratio. Is accompanied by equal lateral contraction. Єy = Єz = . then the longitudinal elongation Єx. if the stress is bellow proportional limit. Єy and Єz. Fig. m = 1/ γ = Poisson No.Elastic properties of the body can specified by E = modulus of elasticity. i. 1) initial portion OA is straight . stress is directly proportional to strain σx = E Єx ------------------1) 2) portion A of the diagram correspond to the stress σp is called proportional limit.e. This diagram demonstrated the relation between tensile stress σx and unit elongation Єx. Inertia forces etc . X.g.are forces distributed over the surface of the body. g. Pressure. z Their components on x. Hyd. FORCES The Elastic body will be referred to a system of orthogonal Cartesian coordinate oxyz. y. Their components on x. Pressure Magnetic forces. . Y. y 2) External forces- o x Internal forces z External Forces e.g. axis will represented by- . z axis will represented by. e.. . Atm. Cohesion between Surface Forces Body forces particles. pressure. Gravitational forces. The forces acting on a body will be - 1) Internal forces. Exerted by one body to another. the volume of the body e. y. Z 6 .are forces distributed over . consider now element of area ∆A ( figure (b) ) with vertical intensity ∆R.τ Consider the body shown in figure (a) which is in equilibrium under the action of system of external forces P1 .ve in compression 7 .normal to ∆A ∆Rt -----------------------. At point P the center of ∆A normal component of stress at point P σ = lim ∆Rn --------------------------------------------------------------------(3) ∆A 0 ∆Rt Shear component of stress at point P τ = lim ∆Rt --------------------------------------------------------------------(4) ∆A 0 ∆A Normal stress σ + ve in tension .tangent in the plane. P2. ………… Pn. STRESS NORMAL STRESS – σ SHEAR STRESS . Now the body is cut in to two parts by a plane ( part 1 and part 2 ) In part (1) & (2) the resultant of internal force +R and –R resp. ∆R can be decomposed into two components- ∆Rn ----------------------. Figure-a Figure-b 8 . 9 . (5) . τxy dy dz . On face ABEF normal stress σx (σ for normal stress and x for normal stress is applied on a face of the cube perpendicular to the x. z now stresses – σx = .τxy .σx dy dz ---------------------------------------------------------.axis and y or z for shear stress component is directed in y or z dir.axis ) Shear stresses τxy and τxz ( τ for shear stress and x for the stress is applied on a face of the cube perpendicular to the x.dy dz τxz = .τxy dy dz --------------------------------------------------------. Consider a small cube OABCDEFG with sides dx . Hence faces are σx dy dz .(6) . τxz are +ve but area OCDG is –ve .(7) . dy and dz . Face OCDG the direction of stresses are shown these stresses σx .dy dz τxy = . Other three remaining faces are –ve . The outward normal to the face ABEF is considered as + ve ( because in the direction of x axis +ve ) BCDE and DEFG are also +ve. σx τxy τxz Stress tensor = τyx σy τyz ----------------------------------(8) τzx τzy σz 10 .dy dz this agrees that normal stresses are positive quantities and +ve in tension . y.τxz dy dz ---------------------------------------------------------. τxz dy dz directed in the –ve direction of x. 11 . shear stress τxy and τyx are acting .(9) τzx = τxz 12 .( τyx dx dz ) * dy = 0 therefore. τxy = τyx similarly. Taking moment about z axis .On the face perpendicular to x and y axis . τyz = τzy -----------------------------------.( force x distance)- ( τxy dy dz ) * dx . STRAIN AND STRAIN DISPLACEMENT RELATIONS (CAUCHY’S EQUATIONS ) 13 . y and z. Parallel to x. y and z axis . y+dy. z+dz ) The coordinate of all four points before and after stretching are given in table as below - 14 . z ) P2 ( x. y. y. v and w. z ) P3 ( x. points P1 ( x+dx. consider an unstrained elastic solid points P of the coordinates x. The small displacement of the points in a deformed body will be resolved into Components u. STRAIN AND STRAIN DISPLACEMENT RELATIONS (CAUCHY’S EQUATIONS) Before deformation After deformation Points coordinates Points Coordinates P x P’ x+u y y+v z z+w P1 x+dx P1’ x + u + dx + ∂u dx ∂x y y + v+ ∂v dx ∂x z z + w+ ∂w dx ∂x P2 x P2’ x + u+ ∂u dy ∂y y+dy y + v+ dy +∂v dy ∂y z z+ w+ ∂w dy ∂y P3 x P3’ x + u+ ∂u dz ∂z y u+ v+ ∂v dz ∂z z+dz z + w+dz+ ∂w dz ∂z 15 . ( 10) ∂y εz = ∂w ∂z 16 .The stretching of P P1 in the x direction is given by- ( x + u + dx + ∂u dx ) – ( x + u + dx ) ∂x = ∂u dx ∂x Therefore we get. εx = ∂u ∂x Similarly. εy = ∂v ------------------------------------. The change in angle due to elastic deformation 17 . considering the deformation of the angle P’2 P’ P3 and P’1 P’ P’3 γyz = ∂v + ∂w ∂z ∂y ------------------------------------(11) γxz = ∂u + ∂w ∂z ∂x εx γxy γxz Strain tensor = γyx εy γyz -------------------------------(12) γzx γzy εz 18 . Before deformation angle P1P P2 = 90° After deformation angle P’1 P’ P’2 = 90°. Consider an angle P1P P2 as shown in figure. (shear strain ) γxy = ∂u + ∂v --------------------------------------(11) ∂y ∂x similarly.γxy while . γxy = γyx γyz = γzy γzx = γxz The signs of strains εx . The signs of shear components of the strains γxy . γyz and γzx are assumed to be + ve if they correspond to the reduction of the angle during deformation and – ve if the corresponding angle increases during deformation. εy and εz are assumed to be + ve if they corresponds Extension and – ve if they corresponds contractions. 19 . RELATIONSHIP BETWEEN STRESSES AND STRAINS:- ( GENERALIZED HOOK’S LAW ) 20 . ∊y = (σy – γ (σz + σx) --------------------------. subjected to the normal stress σx. e = ∊x + ∊y + ∊z And. ∊x´´. ∊x´´´ .due to σz) ∊x = –γ -γ Where.(13) ∊z = (σz – γ (σx + σy) Now. γ = = poison ratio ∊x = (σx – γ (σy + σz) Similarly. Normal strain ∊x = ∊x´ + ∊x´´ + ∊x´´´ (∊x´. Consider a small cube. σy and σz.due to σx.due to σy. ψ = σx + σy + σz e = ∊x + ∊y + ∊z = (σx + σy + σz) 21 . σz)} = . (15) in eqn.(15) Substituting eqn.e= ) ψ -------------------------------------------. + ∊x + = (1+ γ) σx = (∊x + ) 22 .(14) (σy + σz) = – σx ----------------------------------. (13) ∊x = {σx – γ (σy + σz)} = {σx – γ ( . σy = λe + 2G ∊y -------------------------------. λ= =Lame’s constant. G= = shear modulus or modulus of Rigidity. And. σx = λe + 2G ∊x Similarly.(16) σz = λe + 2G ∊z Where. 23 . p – p = . Now suppose hydrostatic pressure P acting on the cube- σx = σy = σz = p Eqn.(17) -p = = (λ + )e=Ke Where.3p e= --------------------. (14) is- ∊x + ∊y + ∊z = e = )ψ ψ = σx + σy + σz = . K = (λ + ) = compressibility of fluid / modulus of volume expansion = Bulk Modulus 24 .p . G G. γ E. γ E. G Λ-Lame’s λ K- constant G-Modulus of rigidity G G G G K- Compressibilit λ+ K y of fluid E-Modulus of elasticity 2G(1+γ) E E γ. RELATION BETWEEN ELASTIC CONSTANTS constant G. λ K.Poisson’s ratio γ γ -1 25 . EQUATIONS OF EQUILLIBRIUM :- (NAVIER’S EQUATIONS / LAME’S EQUATIONS) 26 Consider a small cube and express the condition of equilibrium of the forces (Body forces and surface forces) acting on it. In x- direction – X dx dy dz – σx dy dz + (σx + ) dy dz – τyx dz dx + (τyx + ) dx dz - τzx dy dx + (τzx + ) dy dx = 0 Dividing equation by dx dy and dz. + + +X = 0 + + +Y = 0 ------------------------ (18) + +Z = 0 Eqn. (18) are Naviers equations. of equilibrium for an elastic solid. 27 Applying D’Alemberts principle to the elastic body in motion. = mass density of the elastic body. Inertia forces in x, y, z direction Resp. F=ma = (density x volume x accln.) - ρ dx dy dz - ρ dx dy dz - ρ dx dy dz 28 + + +X= + + +Y= ----------------------.By adding body forces to equation (18) we get. (16) σz = λe + 2G ∊x 29 . σx = λe + 2G ∊x σy = λe + 2G ∊x -------------------------------.eqn.(19) + +Z= Equation (19) are Navier’s Equations of motion. Now. (21) G ∆w + (λ + G) +Z= Where. ∆ = Laplacian operator = + + 30 . G ∆u + (λ + G) +X=0 G ∆v + (λ + G) + Y = 0 --------------------------------.(20) G ∆w + (λ + G) +Z=0 Substituting the values in equation (19) G ∆u + (λ + G) +X= G ∆v + (λ + G) +Y= -----------------------. τxy = G γxy τyz = G γyz τzx = G γzx Substituting in eqn. ∊x= γxy = ∊y= γyz = ∊z= γzx = We get Lame’s equation of equilibrium in terms of components elastic displacements . 31 .And Hook’s law for shear stresses and strains. (18) and. ) They are first derived by Saint Venant and are valid if body is simply connected. From equation (10) & (11) differentiating twice - 32 .e. z but are related by relationship called compatibility Eqns. they can not be taken as arbitrary functions of the variables x. (10) and (11) ∊x= γxy = ∊y= γyz = ∊z= γzx = It follows that the components of strain are not independent. y. (I. COMPATIBILITY EQUATIONS :- (SAINT VENANT’S EQUATIONS) BELTRAMI-MICHELL’S EQUATIONS Refer eqns. = = And. Hence.(22) = 33 . = Similarly. = --------------------------. From equation (10) and (11) Hence. . + =2 =2 34 . (23) [ . + ]=2 [ . + ]=2 Substituting ∊x = (σx – γ (σy + σz) ∊y = (σy – γ (σz + σx) ∊z = (σz – γ (σx + σy) γxy = = τxy γyz = = τyz γzx = = τzx 35 . [ . + ]=2 -----------------------. + ) ------------------------------. Beltrami. And putting the values in Eqns. + + )-2 Δ σy + =. + ) ∆ τzx + =. (22) and (23). + ) 36 .(24) Δ σz + =. + + )-2 ∆ τyz + =.In eqns.(25) ∆ τxy + =. + + )-2 -----------------------------.Michell compatibility Eqns:- Δ σx + =. (18) we get. e.(26) If there are no Body forces i.(27) Δ σz + =0 37 . (24) & (25).Now here. Δ σx + =0 Δ σy + =0 -------------------------------. X=Y=Z=0 substituting in Eqn. Δ σx = ( + + ) σx ∆= Laplacian operator = + + = σx + σy + σz Δ = + + ) -----------------------------. ∆ τyz + =0 ∆ τzx + =0 ----------------------------.(28) ∆ τxy + =0 38 . a concentrated load of intensity P is acting. In the third case. a variably distributed load of total value P is acting. these being uniformly distributed load of intensity po. But the longitudinal stress σx on the section at the center of the bar is the same and is expressed by- σx = P / A 39 . In 1855 Saint Venant enunciated the ‘Principle of the elastic equivalence of statically equipollent system of loads’. To explain the Saint Venant’s principle consider the different system of loading acting on bar with rectangular cross section of area A. Although the three system of external forces acting at two ends of the bar are different. According to this principle “ if system of forces acting on a portion of the boundary of an elastic body is replaced by a statically equivalent system of forces acting on the same portion of the boundary then the stresses. In the first case. strains and elastic displacements in parts of the body sufficiently far removed from this portion of boundary remain approximately same”. ( with total UDL = P) In the second case. SAINT VENANT’S PRINCIPLE ‘Principle of the elastic equivalence of statically equipollent system of loads’. SAINT VENANT’S PRINCIPLE :- SYSTEM-1 (U. L. D.) SYSTEM-2 (Concentrated load) SYSTEM-3 (Variable distributed load) 40 . G = and λ = [ +2 + + +X=0 Where. Refer Eqns. BOUNDARY VALUE PROBLEMS OF ELASTICITY:- Refer Eqns. e= + + 41 . γxy = . (18) of body forces. (13) and. (10) and (11) strain displacement relations. γyz = . Refer Eqns. γzx = And putting in equation (18) & (16) λ + G (2 + + +X=0 Now. Substituting- ∊x = . (18) 42 . ∊z = . ∊y = . γxy = γyz = γzx = [ + +X=0 Refer eqns. And solution of these equations. are referred as Boundary conditions.(29) + (1-2γ) ∆w + Z=0 Where. + (1-2γ) ∆u + X=0 + (1-2γ) ∆v + Y=0 -----------------------------------. ∆= Laplacian operator = + + Equation (29) gives Equilibrium eqns. 43 . and are known as Navier Equations. In terms of displacement. ρ = density of elastic body g = acceleration due to gravity. Y = Y0 = ρ g Hence Eqns. AIRY’S STRESS FUNCTION: It is assumed that in case of gravitational forces. (31) y + ve =0 =0 Equation of equilibrium In the absence of body forces X=Y=0 44 . Of Plane stress – Eqns. x X = 0. G. – y0 x 45 . y) This is related to components of stress by following equation σx = . ф = ф (x.(41) Equation of Compatibility.( + ) (σx + σy) = . Airy introduced a stress function. σy = τxy = . B.(1+γ) ( + ) ( + ) (σx + σy) = 0 -------------------------. 46 .(42) Airy’s Stress Function. (41) ( + ) ( + ) = =0 Δ Δ ф = 0-----------------------------.In the absence of body forces τxy = - Substituting in eqns. POLAR COORDINATES:- 47 . Consider also the two faces which are perpendicular to r and t axis . θ (x.axis. In polar coordinates r and θ will be used in showing two dimensional elasticity problems. Consider the system of Cartesian coordinates – o x y and Orthogonal system – o r t the r axis being inclined at the angle θ wrt x. Laplacian operator –Δ Δ + + PROOF- Laplacian operator Δ = + In polar coordinates r and θ. τrθ = τθr Airy’s stress function.(43) 48 .ф (x. y) = F [r (x. y). y) ΔΔ ф = 0 For polar coordinates. y)] --------------------------. consider functions F (x. on these faces the normal stresses σr and σθ and Shearing stresses τrθ = τθr will be acting. Now. r2 = x2+y2 θ = tan-1 = = cosθ = = sinθ = = = = Differentiating eqns. (43) = + = 49 . cosθ sinθ ( ( )+ ( cos2θ + sin2θ ( ) .(44) 50 . =( )( ) = cos2θ .2 sinθ cosθ ( Similarly- sin2θ + cos2θ ( ) + 2 sinθ cosθ ( Δf = + Δf + + Laplacian operator in polar coordinates Δ + + ------------------. ф (x. + + + + + + .(46) + . y) will be- ΔΔф=0 ( + + + + ) = 0 --------------------------. 51 .(45) . .Hence Airy’s stress function. are correlating derivatives in rectangular and polar coordinates. - Above eqns. -----. (47) = .From fig.it is clear that when θ = 0 σr = σx σθ = σy τrθ = τxy Hence eqns.( -------------------. (46) will be – σr = = + σθ = = τrθ = . + = 52 . (47) independent of θ . (45) and (46) will be Euler’s Eqns. =0 σr = σθ = -------------------. ( + )( + = + + =0 Putting in eqns.If ф and component of stress are independent of the variable θ then eqns.(48) τrθ = 0 53 . ) ∊r = + = (σθ. ) γrθ = - = =0 54 .And polar component of strain: ∊r = = (σr. Hence stress tensor = σx τxy τyx σy ----------------------------(30) As for example- The state of stresses in the thin plate which is subjected to force applied at the boundary. parallel to the plane of plate and uniformly distributed over the thickness In the case of plane stress the equilibrium equation (18) will becomes 55 . TWO DIMENSIONAL ELASTICITY PLANE STRESS One has a state of plane stress when stresses satisfy following condition σz = τxz = τyz = 0. (33) ∊z = . +x=0 = 0 ---------------------.(31) And the compatibility eqns.it should be noted that in the case of plane stress σz = 0 but εz # 0 Eqns. (σx + σy) Note.(34) 56 . of shear strains and shear stresses γxy = γyz = γzx = 0----------------. (22) will be = ---------------------.(32) Equation (13) will reduced to- ∊x = (σx – γ σy) ∊y = (σy – γ σx) --------------------------. (1+γ) ( + ) 57 .(35) In eqns. (35) ( + ) (σx + σy) = . with y and adding- =.( + )( + ) Substituting in eqns. (32) (σy . (31) Differentiating the eqns.γσx) + = 2(1+γ) ----------------. (33) and (34) in eqns.Substituting eqns. with x and the second eqns. BOUNDARY CONDITION- 58 . Ay and Az are the area of the faces OBC. Now = cos (nxˆ) = cos (nyˆ) = cos (nzˆ) If . z axis respectively. OAC. . Consider the small tetrahedron OABC if the n is the normal to the face ABC and nxˆ. OAB. nyˆ and nzˆ are the angle between n and the x. Following boundary condition can be derived- σx cos (nxˆ) + τxy cos (nyˆ) + τxz cos (nzˆ) = τyx cos (nxˆ) + σy cos (nyˆ) + τyz cos (nzˆ) = τzx cos (nxˆ) + τzy cos (nyˆ) + σz cos (nzˆ) = 59 . And if the A is the area of the face ABC and Ax. y. respectively. represents the components of the surface forces applied per unit area of face ABC. Hence in the case of plane stress the boundary condition will be- σx cos (nxˆ) + τxy cos (nyˆ) τyx cos (nxˆ) + σy cos (nyˆ) 60 . MOHR’S CIRCLE FOR STRESS:- Figure-a 61 . By expressing the condition of equilibrium of force acting on the small prisom of sides AC=dx. τxy= τyx σ= σx cos2 ф + σy sin2 ф -2 τxy sin ф cos ф------------------.(36) 62 . τxy .e.axis . σx and σy are + ve (tension). = cosф .is the +ve (clockwise) τxy .e. BC=dy. The stresses are σx. AB=ds and thickness unity. vary an a function of the angle ф. perpendicular to AB) ∑Fn =0 σ ds = σx dy cos ф+ σy dx sin ф-τxy dy sin ф. Suppose a cube of side dx and dy and unit thickness in z. σy And τxy . (Along the direction of n i.is the -ve (anti-clockwise ) The prob.direction.i.τyx dxcos ф but = sinф . Perpendicular to xy plane. Is to find the intensity of the normal stress σ and shear stress τ acting on the plane whose normal makes an angle ф with the x. trigonometry cos2ф = (1+ cos2ф) sin2ф = (1.(39) 63 .τyx sin2ф--------------(38) τ= sin2ф + τxy cos2ф -----------------------.(37) Similarly ∑Ft =0 τ ds = σx dy sin ф.σy sin ф cos ф + τxy cos2ф .τyx dx sinф τ = σx cos ф sin ф .cos2ф) as(cos2ф.from.σy dx cos ф + τxy dy cos ф .sin2ф=cos2ф) sin2 ф= 2sin ф cos ф hence equation (36) can be return as σ= + cos2ф – τxy sin2ф-----------------------. y= τ 64 . )2 = cos2ф – τxy sin2ф) Now squaring both the sides of equation (39) τ2 = ( sin2ф+ τxy cos 2ф )2 Adding both the equation (σ . )2 τ2 = )2 + τxy2 Compare with the equation of circle – (x-a)2 + y2 = r2 Hence x = σ. = cos2ф – τxy sin2ф Squaring both the sides- (σ .Rewriting the equation (37) σ. Figure-a figure-b 65 . CN σmax = + σmin = - Extreme value of shear stress τmax/min = ± r = ± 66 . stress will occur on planes where shear stress is zero. then point x represents the state of stress on the vertical plane passing through point o . And min.a= and r= each points on the circle the state of the stress on a plane passing through point o. from the figure we know that max. point n represents the state of stress on the plane passing through point o. σmax = OM = OC + CM σmin = ON = OC . y) w=o Components of strain.PLANE STRAIN:- u = u(x. (10) and (11) ∊x = .refer eqns. ∊z = =0 γxy = γyz = =0 γxz = =0 Strain tensor = ∊x γxy γyx ∊y -------------.(40) 67 . ∊y = . y) v = v(x. TYPICAL EXAMPLES- Constrained at both ends subjected to internal pressure 68 . (13) ∊x = (σx – γ (σy + σz) = (σx (1– γ) .Refer eqns. σy) ∊y = (σy – γ (σz + σx) = (σy (1– γ) .will be + +X = 0 + +Y=0 Refer eqns. (18) . σx) σz = γ (σx + σy) 69 . Refer eqns.(1+γ) ( + ) Now here. ∆= Laplacian operator Δ= + = σx + σy + σz = (1+ ) (σx+σy) Boundary condition- σx cos (nxˆ) + τxy cos (nyˆ) τyx cos (nxˆ) + σy cos (nyˆ) 70 . (26)- Δ = + + ) ( + ) (σx + σy) = . MOHR’S CIRCLE FOR STRAINS:- εmax = + εmin = - ( )max/min = ± 71 . Strain Energy 2. Strength theory or theory of failure 3. Bette’s and Maxwell’s reciprocal theorem BETTE’S RECIPROCAL THEOREM 72 . ENERGY PRINCIPLE AND GENERAL THEOREMS 1. Principle of virtual work 4. BETTE’S RECIPROCAL THEROM The indirect or mutual work done by a system of external forces A elastic solid during the application of a new system of external forces B. LAB = LBA MAXWELL’S RECIPROCAL THEROM:- The displacement of a point A in α direction produces by unit force acting at point B in β direction produces by unit forces acting at point A in the α direction. while forces A produce the mutual work LAB Total work L = LA + LB+LAB If the order is reversed L = LB+ LA +LBA Therefore. 73 . then force system B is applied. it will produced work LB . if they were already acting during the application of a new system of a external forces A. Assume the system A is applied first. it will produce work LA .