2. STRESS, STRAIN, AND CONSTITUTIVE RELATIONS

Transcription

1 . STRESS, STRAIN, AND CONSTITUTIVE RELATIONS Mechanics of materials is a branch of mechanics that develops relationships between the eternal loads applied to a deformable bod and the intensit of internal forces acting within the bod as well as the deformations of the bod. Eternal forces can be classified as two tpes: 1) surface forces produced b a) direct contact between two bodies such as concentrated forces or distributed forces and/or b) bod forces which occur when no phsical contact eists between two bodies (e.g., magnetic forces, gravitational forces, etc.). Support reactions are eternal surface forces that develop at the support or points of support between two bodies. Support reactions ma include normal forces and couple moments. Equations of equilibrium (i.e., statics) are mathematical epressions of vector relations showing that for a bod not to translate or move along a path then F = 0. For a bod not to rotate, M = 0. Alternativel, scalar equations in three-dimensional space (i.e.,,, z) are: F = 0 F = 0 F z = 0 M = 0 M = 0 M z = 0 (.1) Internal forces are non eternal forces acting in a bod to resist eternal loadings. The distribution of these internal forces acting over a sectioned area of the bod (i.e., force divided b area, that is, stress) is a major focus of mechanics of materials. The response of the bod to stress in the form of deformation or normalized deformation, that is, strain is also a focus of mechanics of materials. Equations that relate stress and strain are known as constitutive relations and are essential, for eample, for describing stress for a measured strain. Stress If an internal sectioned area is subdivided into smaller and smaller areas, A, two important assumptions must be made regarding the material: it is continuous and it is cohesive. Thus, as the subdivided area is reduced to infinitesimal size, the distribution of internal forces acting over the entire sectioned area will consist of an infinite number of forces each acting on an element, A, as a ver small force F. The ratio of incremental F force to incremental area on which the force acts such that: lim is the stress which A 0 A can be further defined as the intensit of the internal force on a specific plane (area) passing through a point..1

2 Stress has two components, one acting perpendicular to the plane of the area and the other acting parallel to the area. Mathematicall, the former component is epressed as a normal stress which is the intensit of the internal force acting normal to an incremental area such that: F = lim n (.) A 0 A where + = tensile stress = "pulling" stress and - = compressive stress = "pushing" stress. The latter component is epressed as a shear stress which is the intensit of the internal force acting tangent to an incremental area such that: τ = F lim t A 0 A (.3) The general state of stress is one which includes all the internal stresses acting on an incremental element as shown in Figure.1. In particular, the most general state of stress must include normal stresses in each of the three Cartesian aes, and si corresponding shear stresses. Note for the general state of stress that + acts normal to a positive face in the positive coordinate direction and a +τ acts tangent to a positive face in a positive coordinate direction. For eample, (or just ) acts normal to the positive face in the positive direction and τ acts tangent to the positive face in the positive direction. Although in the general stress state, there are three normal stress component and si shear stress components, b summing forces and summing moments it can be shown that τ = τ ;τ z = τ z ;τ z = τ z. z τ z τ z z τ τ τ τz z Figure.1 General and complete stress state shown on a three-dimensional incremental element..

3 Therefore the complete state of stress contains si independent stress components (three normal stresses, ; ; z and three shear stresses, τ ;τ z ;τ z ) which uniquel describe the stress state for each particular orientation. This complete state of stress can be written either in vector form or in matri form z τ τ z τ z τ τ z τ τ z τ z τ z z The units of stress are in general: Pa = N N or MPa = 106 m psi = lb f in or ksi = 103 lb f in Force Area = F L. (.4) (.5) In SI units, stress is m = N and in US Customar units, stress is mm = kip in. Often it is necessar to find the stresses in a particular direction rather than just calculating them from the geometr of simple parts. For the one-dimensional case shown in Fig.., the applied force, P, can be written in terms of its normal, P N, and tangential, P T, components which are functions of the angle,, such that: P N = P cos P T = P sin (.6) The area, A, on which P N and P T act can also be written in terms of the area, A, normal to the applied load, P, and the angle,, such that A = A /cos (.7) The normal and shear stress relation acting on an area oriented at angle,, relative to the original applied force, P are: = P N A = P cos A / cos = P A cos = cos τ = P T = P sin A A /cos = P cos sin = cos sin A where is the applied unidirectional normal stress. (.8).3

4 A = A cos A P T P N Figure. Unidirectional stress with force and area as functions of angle, For the two dimensional case (i.e., plane stress case such as the stress state at a surface where no force is supported on the surface), stresses eist onl in the plane of the surface (e.g., ; ; τ ). The plane stress state at a point is uniquel represented b three components acting on a element that has a specific orientation (e.g.,, ) at the point. The stress transformation relation for an other orientation (e.g., ', ') is found b appling equilibrium equations ( F = 0 and M = 0 ) keeping in mind that F n = A and F t = τa. The rotated aes and functions for incremental area are shown in Fig..3. The forces in the and directions due to F n = A and F t = τa and acting on the areas normal to the and directions are shown in Fig..4 B appling simple statics such that in the '-direction, F ' = 0 and ' = cos + sin + τ cos sin or P (.9) ' = + + cos + τ sin ' A= A cos ' A A= A sin Figure.3 Rotated aes and functions for incremental area. X.4

5 τ A ' A ' ' A τ'' A τ A X A Figure.4 Rotated coordinate aes and components of stresses/forces. Similarl, for the ''-direction, F ' = 0 and τ '' = ( )cos sin + τ (cos + sin ) or (.10) τ ' ' = Finall, for the ' direction, F ' = 0 and sin +τ cos ' = sin + cos τ cos sin or (.11) ' = + cos τ sin If the stress in a bod is a function of the angle of rotation relative to a given direction, it is natural to look for the angle of rotation in which the normal stress is either maimum or noneistent. A principal normal stress is a maimum or minimum normal stress acting in principal directions on principal planes on which no shear stresses act. Because there are three orthogonal directions in a three-dimensional stress state there are alwas three principal normal stresses which are ordered such that 1 > > 3. Mathematicall, the principal normal stresses are found b determining the angular direction,, in which the function, = f ( ), is a maimum or minimum b differentiating = f ( ) with respect to and setting the resulting equation equal to zero such that d d = 0 before solving for the at which the principal stresses occur. Appling this idea to Eq..9, gives.5

6 d d = 0 = ( )sin +τ cos sin cos = tan = τ (.1) There are two solutions for the principal stress angle (i.e., for maimum and minimum) so that. N1 = 1 τ tan-1 N = 1 τ tan-1 + π = N1 + π Using trigonometr on the geometr shown in Fig..5 results in τ tan = ( ) / sin = cos = τ + τ ( ) / + τ (.1) (.14) Substituting the trigonometric relations of Eq..14 back into Eq..9 gives for the plane stress case: = 1, = + τ tan p = ( ) / ± +τ (.15) ( ) + τ τ ( )/ Figure.5 Geometric representation of the principal direction relations.6

7 Note that for the plane stress case in the - plane, z = 0. Thus the 1 and subscripts in Eq..15 are onl for the - plane and are not necessaril 1 and for the threedimensional general state of stress. Therefore, ordering of 1 and of Eq..15 is onl preliminar, until the are compared to z and ordered according to convention 1 > > 3. For eample, for a particular plane stress state 1 and found from Eq..15 are 100 and 0 MPa, then the principal stresses are 1 = 100, = 0, 3 = 0 MPa. However, if 1 and found from Eq..15 are 15 and -5 MPa, then the principal stresses are 1 = 15, = 0, 3 = 5 MPa. Finall, if 1 and found from Eq..14 are -5 and -85 MPa, then the principal stresses are 1 = 0, = -5, 3 = 85 MPa. Performing a similar substitution of the trigonometric relations of Eq..14 back into Eq..10 gives for the plane stress case: τ ma = +τ ( ) (.16) ave = + and tan s = τ Note that the τ ma of Eq..15 is onl for the - plane. The maimum shear stress for the three-dimensional stress state can be found after the principal stresses are ordered 1 > > 3 such that: τ 1,3 = 1 3 Some general observations can be made about principal stresses. (.17) a) In a principal direction, when τ =0, then 's are maimum or minimum b) ma and min ( τ ma and τ min ) occur in directions 90 apart. c) τ ma occurs in a direction midwa between the directions of ma and min +, τ τ C= + φ = +,+τ R = ( - C) + τ ) tan φ = - τ ( - C) Figure.6 Mohr's circle representation of plane stress state.7

8 An interesting graphical relation occurs if the second equation in each of Eqs..9 and.10 are squared and added together: ' = = cos + τ sin τ ' ' =τ = sin +τ cos = + +τ = +τ ( h) + = r (.18) The result shown in Eq..18 is the equation for a circle (i.e., Mohr's circle) with radius, r= + τ + and displaced h= on the = ais as illustrated in Fig..6. Eamples of Mohr's circles are shown in Fig..7. A procedure for developing Mohr's circle for plane stress is shown in the following section. τ for - plane ma τ ma = 1 3 τ 1 Mohr's circle for stresses in - plane 3 τ 1 Mohr's circle for stresses in --z planes Figure.7 Eamples of Mohr's circle for plane and three dimensional stress states..8

9 Graphical Description of State of Stress -D Mohr's Circle Y τ In this eample all stresses acting in aial directions are positive as shown in Fig. M1. Fig. M1- Positive stresses acting on a phsical element. + X -face +τ + As shown in Figs. M and M3, plotting actual sign of the shear stress with normal stress requires plotting of the opposite sign of the shear stress with the normal stress on the Mohr's circle. Fig. M - Directionalit of shear acting on and faces. τ +, τ φ = +,+τ C= + R = ( - C) + τ ) tan φ = - τ ( - C) Fig. M3 - Plotting stress values on Mohr's circle. In this eample > and τ is positive. B the convention of Figs. M and M3, φ = on the Mohr's circle is negative from the + ais. (Mathematical convention is that positive angle is counterclockwise). Note that b the simple geometr of Fig. M3, φ = appears to be negative while b the formula, tan = τ /( - ), the phsical angle,, is actuall positive. In-plane principal stresses are: 1 = C+R = C - R Maimum in-plane shear stress is: τ ma =R=( 1 - )/.9

10 Y 1 Direction of + X The direction of phsical angle,, is from the - aes to the principal aes. Fig. M4 - Orientation of phsical element with onl principal stresses acting on it. Principal Ais Direction of Line of X-Y Stresses Fig. M5 - Direction of q from the line of - stresses to the principal stress ais. Note that the sense (direction) of the phsical angle,, is the same as on the Mohr's circle from the line of the - stresses to the aes of the principal stresses. Same relations appl for Mohr's circle for strain ecept interchange variables as and τ γ.10

11 Recall that all stress states are three-dimensional. Therefore, a more general method is required to solve for the principal stresses. One such method is to solve for the "eigenvalues" of the stress matri where, is the principal stress: τ τ z τ τ z τ z τ z z (.19) Finding the determinant for this matri and grouping terms gives: 3 I 1 + I I 3 = 0 (.0) where the stress invariants, ( I 1,I, I 3 ), do not var with stress direction: I 1 = + + z I = + z + z τ τz τ z I 3 = z + τ τ z τ z τ z τ z z τ (.1) Note that if principal stresses are used in Eq..1 for the terms then all terms containing τ will be zero since b definition, principal stresses act in principal directions on principal planes on which τ =0. Eq..0 can then be solved for the three roots which when ordered 1 > > 3 are the principal stresses. Eq..0 can be plotted as f the principal stresses are the values of which occur when f ( ) vs shown in Fig..9 where ( ) = 0. f( ) 3 1 Figure.9 Solving for cubic routes for principal stresses.11

12 Strain Whenever a force is applied to a bod, it will tend to change the bod's shape and size. There changes are referred to as deformation. Size changes are known as dilatation (volumetric changes) and are due to normal stresses. Shape changes are known as distortion and are due to shear stresses. In order to describe the deformation in length of line segments and the changes in angles between them, the concept of strain is used. Therefore, strain is defined as normalized deformations within a bod eclusive of rigid bod displacements There are two tpe of strain, one producing size changes b elongation or contraction and the other producing shape changes b angular distortion. Normal strain is the elongation or contraction of a line segment per unit length resulting in a volume change such that = A' B' AB lim L f L o (.) B A along n AB L o where + = tensile strain = elongation and - = compressive strain = contraction Shear strain is the angle change between two line segments resulting in a shape change such that: γ = ( = π ) ' (for small angles ) (.3) h where +γ occurs if π > ' and -γ occurs if π < '. The general state is one which includes all the internal strains acting on an incremental element as shown in Fig..10. The complete state of strain has si independent strain components (three normal strains, ; ; z and three engineering shear strains, γ ;γ z ;γ z ) which uniquel describe the strain state for each particular orientation. This complete state of strain can be written in vector form z γ γ z γ z (.4).1

13 A γ = + Engineering shear strain, Figure.10 General and complete strain state shown on a three-dimensional incremental element. Alternativel, the complete state of strain can be written in matri form : γ γ z γ γ z γ z γ z z (.5) The units of strain In general: Length Length = L, L In SI units, strain is m m for and m or radian for γ m and in US Customar units, strain is in in for and or radian for γ in in Just as in the stress case it is often necessar to find the stresses in a particular direction rather than just calculating them from the geometr of simple parts. For the twodimensional, plane strain condition (e.g., strain at a surface where no deformation occurs normal to the surface), strains eist onl in the plane of the surface (e.g., ; ;γ ). The plane strain state at a point is uniquel represented b three components acting on a element that has a specific orientation (e.g.,, ) at the point. The strain transformation relation for an other orientation (e.g., ', ') is found b summing displacements in the appropriate directions keeping in mind that δ = L o and = γ h as shown in Fig..11 Simpl adding components of displacements in the ' direction, displacements in 'direction for Q to Q * (see Fig..1) gives ' = cos + sin + γ cos sin or (.5) ' = + + cos + γ sin.13

14 ' } d = γ d } ds Q ' Q* δ = d d Figure.11 Rotated coordinate aes and displacements for and directions } δ = Similarl, adding components of displacements in due to rotations of d' and d' rotation of d ' and d' (see Fig..1) gives d γ '' or = ( )cos sin + γ (cos + sin ) (.6) γ '' = sin + γ cos Finall, adding components of displacements in the ' direction, displacements in 'direction for Q to Q * (see Fig..1) gives ' = sin + cos γ cos sin or (.7) ' = + cos γ sin ' δ = ' Q { ' ds Q* = γ d δ = d cos = sin = d ds d ds δ = d Figure.1 Displacements in the ' direction for strains/displacements in the and directions.14

15 Just as for the stress in a bod which is a function of the angle of rotation relative to a given direction, it is natural to look for the angle of rotation in which the normal strain is either maimum or minimum. A principal normal strain is a maimum or minimum normal strain acting in principal directions on principal planes on which no shear strains act. Because there are three orthogonal directions in a three-dimensional strain state there are alwas three principal normal strains which are ordered such that 1 > > 3. Also as in the stress case, the principal strains for the plane strain case 1, = + ± tan p = γ + γ (.9) Note that for the plane strain case in the - plane, z = 0. Thus, the 1 and subscripts in Eq..9 are onl for the - plane and are not necessaril 1 and for the threedimensional general state of strain. Therefore, ordering of 1 and of Eq..9 is onl preliminar, until the are compared to z and ordered according to convention 1 > > 3. For the shear strain case γ ma = ave = + + γ τ, ( ) and tan s = γ (.30) γ / for - plane ma γ ma = 1 3 γ/ 1 γ/ Mohr's circle for strains in - plane Mohr's circle for strains in --z planes Figure.13 Eamples of Mohr's circle for strain

16 c b a c b 60 a 45 Rectangular 60 Delta Figure..14 Rectangular and Delta rosettes. Note that the γ ma of Eq..30 is onl the - plane. The maimum shear strain for the three-dimensional strain state can be found after the principal strains are ordered 1 > > 3 such that: γ 1,3 = 1 3 (.31) As with stress, the complete strain state can be represented graphicall as a Mohr's circle. Eamples of Mohr's circles for strain are shown in Fig..13. Note that the same procedure for developing Mohr's circle for the plain strain case can be used as with the plane stress case with b making the following substitutions: and τ γ. An important application of the strain transformation relation of Eq..6 is to eperimentall determine the complete strain state since stress is an abstract engineering quantit and strain is measurable/observable engineering quantit. Equation.6 requires that three strain gages be applied at arbitrar orientations at the same point on the bod to determine the principal strains and orientations. However, to simplif the data reduction, the three strain gages are applied at fied angles usuall in the form of 45 rectangular rosette or 60 Delta rosette as shown in Fig..14. The resulting equations to determine the local (for the strain gage rosettes shown in Fig..14) - strains in preparation for determining the principal strains: 45 Rectangular 60 Delta Rosette Rosette = a = a = c = 1 3 ( b + c a ) (.3) γ = b ( a + c ) γ = 3 ( b c ) However, a limitation of measuring strains eperimentall is that stresses are often required to determine the engineering performance of the component. Thus, equations which relate stress and strain are required..16

17 Constitutive Relations If the deformation and strain are the response of the bod to an applied force or stress, then there must be some tpe of relations which allow the strain to be predicted from stress or vice versa. The uniaial stress-strain case is a useful eample to begin to understand this relation. During uniaial loading (see Fig..15) b load, P, of a rod with uniform cross sectional area, A, and length, L, the applied stress,,is calculated simpl as = P A (.33) The resulting normal strain, L, in the longitudinal direction can be calculated from the deformation response, L, along the L direction: L = L L (.34). Another normal strain, T, in the transverse direction can be calculated from the deformation response, D, along the D direction: T = D D (.35). A plot of vs. L (see Fig..16) shows a constant of proportionalit between stress and strain in the elastic region such that where E = L =m + b = E L (.36) is known as the elastic modulus or Young's modulus and Eq..36 is known as unidirectional Hooke's law. P D=Do Df Undeformed Deformed A P D=Df -D L=Lo Lf L=Lf - L Figure.15 Uniaiall-loaded rod undergoing longitudinal and transverse deformation.17

18 Ε ν Longitudinal strain, L Longitudinal strain, L Figure.16 Plots of applied stress vs. longitudinal strain and transverse strain versus longitudinal strain A plot of T vs. L (see Fig..16) shows a constant of proportionalit between transverse and longitudinal strains in the elastic region such that where ν = T L =m + b T = ν L (.37) is known as Poisson's ratio. For the results for three different uniaial stresses applied separatel (see Fig..17) in each of the orthogonal direction,, and z can give the general relations between normal stress and normal strain know as generalized Hooke's law if the strain components for each of the stress conditions are superposed: ( ( )) = 1 E ν + z = 1 ( E ν( + z )) z = 1 ( E z ν( + )) (.38) Since shear stresses are decoupled (i.e., unaffected) b stresses in other directions the relations for shear stress-shear strain are: γ = 1 G τ γ z = 1 G τ z (.39) where the shear modulus is G = τ γ = γ z = 1 G τ z E (1+ ν)..18

19 z Stress z E - ν = -νe - ν = -νe z -ν = -νe E -ν = -νe z z Y z - ν z = -νe z -ν z = -νe z E z X Z : Figure.17 Development of generalized Hooke's law. Generalized Hooke's law is usuall written in matri form such that: { } = [ S ] where { } is the strain vector, { }is the stress vector and S Epanded, Eq..40 is: { } (.40) [ ] is the compliance matri..19

20 1 ν ν E E E ν 1 ν E E E z ν ν γ = E E E z G γ 0 0 τ z G 0 τ z γ z τ G z (.41) Although Eq..41 is a ver convenient form, it is not that useful since stress is rarel known with strain being the unknown. Instead, strain is usuall measured eperimentall and the associated stress needs to be determined. Therefore, the inverse of the compliance matri needs to be found such that { } = C [ ] { } (.4) where [ C]is the stiffness matri such that [ C] = [ S] 1. Epanding Eq..4 gives E (1+ν) 1+ ν νe νe (1 ν ) (1+ν )(1 ν) (1+ν)(1 ν) νe E z (1+ν)(1 ν) (1+ν ) 1+ ν νe (1 ν ) (1+ν )(1 ν) z τ = νe νe E (1+ν)(1 ν) (1+ν)(1 ν) (1+ν) 1+ ν (1 ν) γ G 0 0 τ z G 0 γ z τ z G γ z (.43) Generalized Hooke's can be simplified somewhat for the special case of plane stress in the - plane since z =0. Being orthogonal to the - plane, Since z is a also a principal stress b definition, all the shear stresses associated with the z-direction are also zero. Thus, the stress-strain relations for plane stress in the - plane become E νe (1 ν ) (1 ν 0 ) νe E = (1 ν τ ) (1 ν 0 ) 0 0 G γ (.44) For the special case of plane stress, although z =0, the strain in the z-direction is not zero but instead can be determined such that Plane stress : z = 0, z 0 = ν 1 ν ( + ) (.45).0

21 Similarl for the special case of plane strain, although, z =0, the stress in the z- direction is not zero but instead can be determined such that Plane strain : z = 0, z 0 = ν( + ) (.46) For the special case of hdrostatic pressure, no distortion takes place, onl size changes. In this case the hdrostatic stress is ( ) H = and the dilatation (i.e., volumetric change) is (.47) V = V V = ( + + ) (.48) The ratio between the hdrostatic stress and the dilation is a special combination of elastic constants called the bulk modulus. ( ) ( ) = E k = H = + + V (1 ν ) (.49) It is useful to know that elastic constants are related to the atomic structure of the material and thus are not affected b processing or component fabrication. For eample, E is related to the repulsion/attraction between two atoms. The force-displacement curve for this interaction is shown in Fig..18. Since this is the uniaial case, recall that = P L and = A L =. Therefore, since E is defined as E = e = d with d d = dp A d and = e then E = d d = e A dp d. Note from Fig..18 that dp d is the slope of the force-displacement curve, thus making the elastic modulus E fied b atomic interaction. From a materials standpoint, covalent and ionic bonds such as those in ceramics are stiff leading to high elastic moduli in those materials. Metallic bonds are intermediate in stiffness leading to intermediate elastic moduli in metals. Secondar bonds such as those found in polmers are least stiff leading to low elastic moduli in those materials. For Poisson's ratio, it is useful to consider the sphere model of atomic structure as shown in Fig..19. Before deformation the triangle between centers has a transverse side length of 3R, a longitudinal side length of R and a hpotenuse of R. After longitudinal deformation, the transverse side length is 3R -d, the longitudinal side.1

22 Figure.18 Repulsion-attraction force-displacement curve between two atoms. length is R+d and the hpotenuse is still R. Appling Pthagorean's theorem after deformation gives ( R ) = ( 3R d ) + ( R + d) (.50) Epanding Eq..50 and eliminating higher order terms leaves Rd = 3Rd d d = 3. (.51) Recalling that the longitudinal and transverse strains can be written in terms of the deformations ( = R + d ) R R ( 3R d ) 3R = 3R = d R d = R = d 3R d = 3R (.5) Combining the two terms for d and d in Eq..5 and equating them to Eq..51 gives d d = R 3R = 3 1 ν = 3 (.53) which shows that from a simple sphere model and epected deformations, Poisson's ratio is fundamentall linked to atomic interactions giving a value of 1/3 which is the range of man dense materials..

23 Figure.19 Undeformed and deformed sphere model for atomic structure and determination of Poisson's ratio.3