1 Differential Equations for Solid Mechanics

Transcription

1 1 Differential Eqations for Solid Mechanics Simple problems involving homogeneos stress states have been considered so far, wherein the stress is the same throghot the component nder std. An eception to this was the varing stress field in the loaded beam, bt there a simplified set of elasticit eqations was sed. Here the qestion of varing stress and strain fields in materials is considered. In order to solve sch problems, a differential formlation is reqired. In this Chapter, a nmber of differential eqations will be derived, relating the stresses and bod forces (eqations of motion), the strains and displacements (strain-displacement relations) and the strains with each other (compatibilit relations). These eqations are derived from phsical principles and so appl to an tpe of material, althogh the latter two are derived nder the assmption of small strain. 1

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3 Section The Eqations of Motion In Book I, balance of forces and moments acting on an component was enforced in order to ensre that the component was in eqilibrim. Here, allowance is made for stresses which var continosl throghot a material, and force eqilibrim of an portion of material is enforced. One-Dimensional Eqation Consider a one-dimensional differential element of length and cross sectional area A, Fig Let the average bod force per nit volme acting on the element be b and the average acceleration and densit of the element be a and. Stresses act on the element. () A b,a ( ) Figre 1.1.1: a differential element nder the action of srface and bod forces The net srface force acting is ( ) A ( ) A. If the element is small, then the bod force and velocit can be assmed to var linearl over the element and the average will act at the centre of the element. Then the bod force acting on the element is Ab and the inertial force is Aa. Appling Newton s second law leads to ( ) A ( ) A ba aa ( ) ( ) b a (1.1.1) so that, b the definition of the derivative, in the limit as 0, d b a 1-d Eqation of Motion (1.1.) d which is the one-dimensional eqation of motion. Note that this eqation was derived on the basis of a phsical law and mst therefore be satisfied for all materials, whatever the be composed of. The derivative d / d is the stress gradient phsicall, it is a measre of how rapidl the stresses are changing. Eample Consider a bar of length l which hangs from a ceiling, as shown in Fig Kell

4 Section 1.1 l Figre 1.1.: a hanging bar The gravitational force is F mg downward and the bod force per nit volme is ths b g. There are no accelerating material particles. Taking the ais positive down, an integration of the eqation of motion gives d g 0 g c (1.1.3) d where c is an arbitrar constant. The lower end of the bar is free and so the stress there is ero, and so Two-Dimensional Eqations g l (1.1.4) Consider now a two dimensional infinitesimal element of width and height and nit depth (into the page). and Looking at the normal stress components acting in the direction, and allowing for variations in stress over the element srfaces, the stresses are as shown in Fig (, ) (, ) (, ) (, ) Figre 1.1.3: varing stresses acting on a differential element Using a (two dimensional) Talor series and dropping higher order terms then leads to the linearl varing stresses illstrated in Fig (where, and the partial derivatives are evalated at, ), which is a reasonable approimation when the element is small. 4 Kell

5 Section 1.1 Figre 1.1.4: linearl varing stresses acting on a differential element The effect (resltant force) of this linear variation of stress on the plane can be replicated b a constant stress acting over the whole plane, the sie of which is the average stress. For the left and right sides, one has, respectivel, 1, 1 (1.1.5) One can take awa the stress ( 1/ ) / from both sides withot affecting the net force acting on the element so one finall has the representation shown in Fig (, ) Figre 1.1.5: net stresses acting on a differential element Carring ot the same procedre for the shear stresses contribting to a force in the direction leads to the stresses shown in Fig (, ) b, a (, ) Figre 1.1.6: normal and shear stresses acting on a differential element Take a, b to be the average acceleration and bod force, and to be the average densit. Newton s law then ields 5 Kell

6 Section 1.1 b a (1.1.6) which, dividing throgh b and taking the limit, gives b a (1.1.7) A similar analsis for force components in the direction ields another eqation and one then has the two-dimensional eqations of motion: b b a a -D Eqations of Motion (1.1.8) Three-Dimensional Eqations Similarl, one can consider a three-dimensional element, and one finds that b b b a a a 3-D Eqations of Motion (1.1.9) These three eqations epress force-balance in, respectivel, the,, directions. 6 Kell

7 Section 1.1 Figre 1.1.7: from Cach s Eercices de Mathematiqes (189) The Eqations of Eqlibrim If the material is not moving (or is moving at constant velocit) and is in static eqilibrim, then the eqations of motion redce to the eqations of eqilibrim, b b b D Eqations of Eqilibrim (1.1.10) These eqations epress the force balance between srface forces and bod forces in a material. The eqations of eqilibrim ma also be sed as a good approimation in the analsis of materials which have relativel small accelerations Problems 1. What does the one-dimensional eqation of motion sa abot the stresses in a bar in the absence of an bod force or acceleration?. Does eqilibrim eist for the following two dimensional stress distribtion in the absence of bod forces? 3 48 / Kell

8 Section The elementar beam theor predicts that the stresses in a circlar beam de to bending are 4 M / I, V ( R ) / 3I ( I R / 4) and all the other stress components are ero. Do these eqations satisf the eqations of eqilibrim? 4. With respect to aes 0 the stress state is given in terms of the coordinates b the matri 0 ij 0 Determine the bod force acting on the material if it is at rest What is the acceleration of a material particle of densit 0.3kgm, sbjected to the stress 4 4 ij 4 and gravit (the ais is directed verticall pwards from the grond). 6. A flid at rest is sbjected to a hdrostatic pressre p and the force of gravit onl. (a) Write ot the eqations of motion for this case. (b) A ver basic formla of hdrostatics, to be fond in an elementar book on flid mechanics, is that giving the pressre variation in a static flid, p gh where is the densit of the flid, g is the acceleration de to gravit, and h is the vertical distance between the two points in the flid (the relative depth). Show that this formla is bt a special case of the eqations of motion. 8 Kell

9 Section The Strain-Displacement Relations The strain was introdced in Book I: 4. The concepts eamined there are now etended to the case of strains which var continosl throghot a material The Strain-Displacement Relations Normal Strain Consider a line element of length emanating from position (, ) and ling in the - direction, denoted b AB in Fig After deformation the line element occpies A B, having ndergone a translation, etension and rotation. (, ) (, ) B A B A * B Figre 1..1: deformation of a line element The particle that was originall at has ndergone a displacement (, ) and the other end of the line element has ndergone a displacement (, ). B the definition of (small) normal strain, In the limit 0 one has * AB AB (, ) (, ) (1..1) AB (1..) This partial derivative is a displacement gradient, a measre of how rapid the displacement changes throgh the material, and is the strain at (, ). Phsicall, it represents the (approimate) nit change in length of a line element, as indicated in Fig Kell

10 Section 1. B A * B A B Figre 1..: nit change in length of a line element Similarl, b considering a line element initiall ling in the direction, the strain in the direction can be epressed as Shear Strain (1..3) The particles A and B in Fig also ndergo displacements in the direction and this is shown in Fig In this case, one has * B B (1..4) (, ) A (, ) A B B * B Figre 1..3: deformation of a line element A similar relation can be derived b considering a line element initiall ling in the direction. A smmar is given in Fig From the figre, / tan 1 / provided that (i) is small and (ii) the displacement gradient / is small. A similar epression for the angle can be derived, and hence the shear strain can be written in terms of displacement gradients. 10 Kell

11 Section 1. Figre 1..4: strains in terms of displacement gradients The Small-Strain Stress-Strain Relations In smmar, one has 1 -D Strain-Displacement relations (1..5) 1.. Geometrical Interpretation of Small Strain A geometric interpretation of the strain was given in Book I: This interpretation is repeated here, onl now in terms of displacement gradients. Positive Normal Strain Fig. 1..5a, Negative Normal Strain Fig 1..5b, 1 0, 0, 0 (1..6) 1 0, 0, 0 (1..7) 11 Kell

12 Section 1. () ( ) () Figre 1..5: some simple deformations; (a) positive normal strain, (b) negative normal strain, (c) simple shear Simple Shear Fig. 1..5c, ( a) ( b) (c) 1 1 0, 0, (1..8) Pre Shear Fig 1..6a, 1 0, 0, (1..9) 1..3 The Rotation Consider an arbitrar deformation (omitting normal strains for ease of description), as shown in Fig As sal, the angles and are small, eqal to their tangents, and /, /. () () 1 Figre 1..6: arbitrar deformation (shear and rotation) 1 Kell

13 Section 1. Now this arbitrar deformation can be decomposed into a pre shear and a rigid rotation 1 as depicted in Fig In the pre shear,. In the rotation, the 1 angle of rotation is then. arbitrar shear strain pre shear rotation (no strain) Figre 1..7: decomposition of a strain into a pre shear and a rotation This leads one to define the rotation of a material particle,, the signifing the ais abot which the element is rotating: 1 (1..10) The rotation will in general var throghot a material. When the rotation is everwhere ero, the material is said to be irrotational. For a pre rotation, note that 1 0, (1..11) 13 Kell

14 Section Fiing Displacements The strains give information abot the deformation of material particles bt, since the do not encompass translations and rotations, the do not give information abot the precise location in space of particles. To determine this, one mst specif three displacement components (in two-dimensional problems). Mathematicall, this is eqivalent to saing that one cannot niqel determine the displacements from the strain-displacement relations Eample Consider the strain field 0.01, 0. The displacements can be obtained b integrating the strain-displacement relations: d 0.01 d g( ) f ( ) (1..1) where f and g are nknown fnctions of and respectivel. Sbstitting the displacement epressions into the shear strain relation gives f ( ) g( ). (1..13) An epression of the form F( ) G( ) which holds for all and implies that F and G are constant 1. Since f, g are constant, one can integrate to get f ( ) A D, g( ) B C. From 1..13, C D, and A C B C (1..14) There are three arbitrar constants of integration, which can be determined b specifing three displacement components. For eample, sppose that it is known that ( 0,0) 0, (0,0) 0, (0, a) b. (1..15) In that case, A 0, B 0, C b / a, and, finall, 0.01 ( b / a) ( b / a) (1..16) which corresponds to Fig. 1..8, with ( b / a) being the (tan of the small) angle b which the element has rotated. 1 since, if this was not so, a change in wold change the left hand side of this epression bt wold not change the right hand side and so the eqalit cannot hold 14 Kell

15 Section 1. b a Figre 1..8: an element ndergoing a normal strain and a rotation In general, the displacement field will be of the form A C B C (1..17) and indeed Eqn is of this form. Phsicall, A, B and C represent the possible rigid bod motions of the material as a whole, since the are the same for all material particles. A corresponds to a translation in the direction, B corresponds to a translation in the direction, and C corresponds to a positive (conterclockwise) rotation Three Dimensional Strain The three-dimensional stress-strain relations analogos to Eqns are, 1,, 1 1, 3-D Stress-Strain relations (1..18) The rotations are 1 1 1,, (1..19) 1..6 Problems 1. The displacement field in a material is given b A 3, A where A is a small constant. 15 Kell

16 Section 1. (a) Evalate the strains. What is the rotation? Sketch the deformation and an rigid bod motions of a differential element at the point ( 1, 1) (b) Sketch the deformation and rigid bod motions at the point ( 0, ), b sing a pre shear strain sperimposed on the rotation.. The strains in a material are given b, 0, Evalate the displacements in terms of three arbitrar constants of integration, in the form of Eqn , A C B C What is the rotation? 3. The strains in a material are given b A, A, A where A is a small constant. Evalate the displacements in terms of three arbitrar constants of integration. What is the rotation? 4. Show that, in a state of plane strain ( 0 ) with ero bod force, e where e is the volmetric strain (dilatation), the sm of the normal strains: e (see Book I, 4.3). 16 Kell

17 Section Compatibilit of Strain As seen in the previos section, the displacements can be determined from the strains throgh integration, to within a rigid bod motion. In the two-dimensional case, there are three strain-displacement relations bt onl two displacement components. This implies that the strains are not independent bt are related in some wa. The relations between the strains are called compatibilit conditions The Compatibilit Relations Differentiating the first of 1..5 twice with respect to, the second twice with respect to and the third once each with respect to and ields 3, 3, It follows that -D Compatibilit Eqation (1.3.1) This compatibilit condition is an eqation which mst be satisfied b the strains at all material particles. Phsical Meaning of the Compatibilit Condition When all material particles in a component deform, translate and rotate, the need to meet p again ver mch like the pieces of a jigsaw ple mst fit together. Fig illstrates possible deformations and rigid bod motions for three line elements in a material. Compatibilit ensres that the sta together after the deformation. deformed - compatibilit ensred ndeformed deformed - compatibilit not satisfied Figre 1.3.1: Deformation and Compatibilit 17 Kell

18 Section 1.3 Kell 18 The Three Dimensional Case There are si compatibilit relations to be satisfied in the three dimensional case:,,,. (1.3.) B inspection, it will be seen that these are satisfied b Eqns Problems 1. The displacement field in a material is given b, A A, where A is a small constant. Determine (a) the components of small strain (b) the rotation (c) the principal strains (d) whether the compatibilit condition is satisfied