Draft:S Ghorai. 1 Body and surface forces. Stress Principle

Transcription

1 Stress Principle Body and surface forces Stress is a measure of force intensity, either within or on the bounding surface of a body subjected to loads. It should be noted that in continuum mechanics a body is considered stress free if the only forces present are those inter-atomic forces required to hold the body together. And so it follows that the stresses that concern us here are those which results from the application of forces by an external agent. Two general classes of force are considered. irst, those acting on the interior points of a body by the sorrounding medium. Gravitational and magnetic forces are of this type. Second, those forces which act upon and are distributed in some fashion over a surface element of the body, regardless of whether that element is part of the bounding surface, or an arbitrary element of surface within the body. These forces are called surface forces. 4 A 4 B A P A B 4 c igure : Illustration of stress.

2 onsider a body acted upon by a system of contact forces in equilibrium. If the body were cut along the shaded plane, the separate part will no longer be in equilibrium and would fly apart. However before cutting the portion of the body labelled A in figure is in equilibrium so an internal force must be exerted on the material in the plane π A by the material in the in the plane π B such that, and 4 form a system of forces in equilibrium. In the figure the plane π has a particular orientation but the same conclusion can be drawn whatever the orientation of the plane. Thus if any plane is drawn in a continuum body subjected to a system of forces, the material in this plane will be acted upon by an internal force exerted by adjacent material. The stress is used to quantify this internal force. Suppose now that instead of drawing the plane right through the body, we consider a small area π c which is part of a closed surface drawn in the interior of the body. The material in π c will be subjected to a force c by the adjacent layer of material outside the surface π c. This is the force which would have to be applied to π if the body were cut along this plane, to hold this portion of the body in equilibrium. We define the stress acting on this plane as as the magnitude of the force distributed over unit area. If we change the orientation of π, the magnitude and direction of the force will be changed. To define the stress completely, we must know it s value on a sufficient number of planes of different orientations. The stress vector is denoted by t (n) = t(x, t, n) where n implies that stress is meaningful only in conjunction with normal vector n. Principle of linear momentum states that time rate of change of the linear momentum on any part of a continuum body is equal to the resultant force acting upon that portion. or portion A and B of the body this principle may be expressed in integral form as D ρvdv = Dt V A D ρvdv = Dt V B ρbdv + t (n) ds V A S A ρbdv + t (n) ds V B S B where v is the velocity and b is the body force per unit mass acting on the body. Applying the linear momentum to the whole body we get D ρvdv = ρbdv + t (n) ds Dt V V Adding the previous two relations and using the relation for the whole body, we get S

3 [Note that if the normal to pi A is n then for π B is n. Let π A = π B = π] ( ) t(n) + t (n) ds = 0 π This relation must hold for arbitrary partitioning of the body i.e. for every imaginable cutting plane through point P. This implies that the integrand must vanish thus t (n) = t ( n) Stress tensor onsider the equilibrium of a small small portion of the body in the shape of a tetrahedron shown in the figure with the vertex P having a position vector x. Let n = (n, n, n ) denote the normal to the face AB and let ds, ds, ds and ds denote the area of the faces AB, PB, PA and PAB respectively. A x B P x n x igure : Tetrahedron element with base AB and one vertex at P. Now we have ds i = n i ds (i =,, ) Let t (n) denote the average stress vector on the area AB. Similarly t ( ê ), t ( ê ) and t ( ê ) are the average stress vector on PB, PA and PAB respectively. Also

4 4 let b is the average body force per unit mass acting on the tetrahedron. Since the tetrahedron is in equilibrium, the sum of all forces acting on the tetrahedron must be zero. Using the fact that t (n) = t ( n) we write this equilibrium condition as t (n) ds t (ei )ds i + ρ bdv = 0 or t (n) ds t (ei )n i ds + ρ bdv = 0 The volume of the tetrahedron is given by dv = h ds where h is the perpenducular distance of P from AB. Inserting this expression and canceling the common factor we arrived at t (n) = t (ei )n i ρh b = 0 Now let the tetrahedron shrinks to the point P by taking the limit h 0. Then the starred (averge) quantities take on the actual values of the quantities at the point P. And we have Let and thus we have Let t (n) = t (ej )n j t (ei ) = t (e i) j e j = ij e j where ij = t (e i) j t (n) = ji n j e i = ij e i e j Then t (n) = ji n j e i = T n This formula is known as auchy stress formula. Observation: The auchy stress formula expresses the stress vector assciated with the element of area having an outward normal n at a point P in terms of the stress tensor components ij at that point. Thus the state of stress (i.e. totality of stress vectors t (n) = t(x, t, n) at a fixed (x, t) B t (0, ) and for all possible direction n) at P can be determined if the stress vectors on the three coordinate planes or

5 5 x x igure : artesian stress components nine stress tensor components ij of any cartesian system at P are known. Note that ij (x, t) is the componenet in the direction e j of the stress vector which acts on a plane normal to e i at x and time t. The stress components are shown schematically acting on a block. But actually they act on a single point P. The three stress components shown by arrows acting normal to the coordinate plane and labelled,, are called normal stresses. The six arrows lying in the plane and pointing in the coordinate axes directions and labelled as,,,,, are called shear stresses. Note that the first components denote the coordinate plane and the second subscript denote the direction in which stress components act. x Mass conservation principle Let us consider P B be an arbitry region. Let ρ be the density. Then mass conservation principle implies that ρ 0 dv = ρdv Ω r Ω t rom this we get ρ 0 dv = JρdV Ω r Ω r

6 6 Since Ω r is an arbitrary region, we have ρ 0 = Jρ Now taking material derivative we get ρdiv v + ρ = 0 orollary: Let Φ be the Eulerian description of scaler, vector or tensor field. Then D ρφ dv = ρ DΦ Dt Ω t Ω t Dt dv Proof: onsider first Φ be a scaler. We have using the result of previous chapter D [ ] ρφ dv = ρφ + ρφ div v dv Dt Ω t Ω t [ ] = ( ρ + ρ div v)φ + ρ Φ dv = Ω t ρ DΦ Ω t Dt dv 4 Stress tensor symmetric Let P t B t be an arbitrary material region in the current configuration. The linear momentum M(P t ) of the material occupying the region P t in the current configuration is defined by M(P t ) = ρvdv P t If x is the position vector of a point in P t relative origin o, then the angular momentum of P t with respect to o is defined by G(P t ; o) = x (ρv)dv P t Now applying the conservation of linear momentum for the material in P t we get D ρvdv = ρbdv + t (n) da, Dt P t P t P t and applying the conservation of angular momentum we get D x (ρv)dv = x (ρb)dv + x t (n) da, Dt P t P t P t

7 7 Using the result of the previous section we get ρ(a b)dv = t (n) da P t P t and where a = v is the acceleration. ρx (a b)dv = x t (n) da, P t P t rom the first relation we get ρ(a b)dv = P t T nda = P t div dv P t rom this we derive div + ρb = ρa Using this relation in second relation we get x (div )dv = x ( T n)da P t P t Now from the right hand side, we get x ( T n)da = ɛ ijk x j pk n p e i da P t P t = ɛ ijk (x j pk ),p e i dv P t = ɛ ijk (δ jp pk + x j pk,p )e i dv P t = [ɛ ijk jk e i + ɛ ijk x j pk,p e i ] dv P t = [ɛ ijk jk e i + x (div )] dv P t Hence we get rom this we get P t ɛ ijk jk e i = 0 ɛ ijk jk = 0 Thus taking i =,, in turn we obtained = T

8 5 Principal stresses and principal directions 8 Let (x, t) be the stress at a particular point. Let there exists direction n such that stress vector t (n) = T n = n is parallel to n. Thus n = n The values is called principal stresses and the corresponding directions n are called the principal directions. Also the plane perpendicular to n is called principal stress plane. Since is symmetric, there exists three mutually perpendicular directions and three principle stresses. The stress vector at a point can be decomposed as a sum of normal force ( ) n t(n) n (n n)t(n) and a shearing force t (n) ( ) n t (n) n = (I n n)t(n) The magnitude of normal force is N = n t (n) = n (n) This is called normal stress and the normal stress is tensile if N is positive and compressive if N is negative. Similarly the magnitude of shearing force is S = t (n) N n = t (n) N Let us find the direction for which the normal stress froms an extremum. The normal stress is N = n (n) = ij n i n j Our goal is to find the unit vector n so that N is extremum. Using Lagrange multiplier method, we are interested in extremizing f = ij n i n j + λ(n i n i ) where λ is the Lagrange multiplier. The necessary condition is 0 = n s [ ij n i n j + λ(n i n i )] = ij n i δ js + ij n j δ is + λn i δ is = is n i + js n j + λn s = ( is + λδ is )n i

9 9 Thus we have λ assumes the role of principal stresses. Thus the principle directions are also the directions that extremize the normal stress. Hence the principle stresses include the maximum and minimum normal stress values. Since sigma is real and symmetric, the principal values and principal directions are real and so the principal values may be ordered as or convenience we take p, p, p as base vectors for cartesian coordinate system to be parallel to the principal directions of. Thus Now with direction n we get where n = n p i. Hence Hence = p p + p p + p p t (n) = n p + n p + n p N = n (n) = n + n + n S = t (n) t (n) N = n + n + n ( n + n + n ) Now we use the Lagrange multiplier method and form the function The stationary value is obtained from h = S λ(n j n j ) h n = n [ ( n + n + n ) λ] = 0 h n = n [ ( n + n + n ) λ] = 0 h n = n [ ( n + n + n ) λ] = 0 subject to the constraint n j n j =. One set of solution is n = ±p, N =, S = 0 n = ±p, N =, S = 0 n = ±p, N =, S = 0

10 0 Hence the stress assumes the absolute minimum value of zero on the planes whose normals are in the principal directions of stress. urthure we note that on these planes the normal stress N assumes its extremum values. A second set of solution is n = ± (p ± p ), S = n = ± (p ± p ), S = n = ± (p ± p ), S = Thus the maxumum value of shear stress is equal to one half of the difference of the maximum and minimum values of normal stress and it occurs in the plane whose normal bisects the angle between the normals to the plane of maximum and minimum normal stress. 6 Mohr s circles for stress onsider the state of stress with resepct to principal axes and let the principal stresses be ordered such that. We may express normal and shear stresses by and N = n + n + n N + S = n + n + n The unit normal vector satisfies n satisfies n + n + n = Supppose > >. Now the above three equations gives n = S + ( N )( N ) ( )( ) n = S + ( N )( N ) ( )( ) n = S + ( N )( N ) ( )( ) 0 0 0

11 Thus we get S + ( N )( N ) 0 S + ( N )( N ) 0 S + ( N )( N ) 0 These inequalities may be written as S + ( N ) R, where = + S + ( N ) R, where = + S + ( N ) R, where = + and R = and R = and R = The condition > > is assumed to avoid division by zero. But the same set of relations can be obtained even when there are repeated eigen values. All possible pairs of N and S at P must satisfy the above inequalities or within the shaded area shown in the figure below. In order to relate N and S in the Mohr s circle to O igure 4: Mohr s circle for the D showing three principal stresses normal direction n of the area element, we consider a small spherical body centrerd at P. As the unit normal n assumes all possible values, the point of intersection

12 of the line of actions will move over the entire sphere. Since N and S depends on the squares of the directions cosines, it value do not change upon reflection on the principal palnes. Thus we need to consider only the first octant of the spherical body. Let Q be the point of intersection and n = cos φê + cos βê + cos θê where the unit vector is expressed in terms of a pricipal base. A e* φ θ e* igure 5: Small spherical portion of the body with normal referred to a principal base. If n = ê then N =, S = 0. This corresponds to point point Q concides with e* A. Similarly N =, S = 0 and N =, S = 0 for point B and respectively. In the stress plane these points are a, b and c respectively. If we set θ = π/ and vary φ from 0 to π/ (β from π/ to 0), then it will move the quarter circle AB (90 o ) on the sphere and semicircle ab (80 o ) in the stress plane. Similarly for the quarter circle B and A we have semicircle bc and ca respectively. Let the angle φ be given some fixed value φ = φ f less than π/ and let β and θ take all possible values. This gives rise to the movement of Q along the circle arc KD. Thus the equations describing n = cos φ becomes Q β n B ( N ) + S = R + ( )( ) cos φ f = R This circle has centre concided with and radius R > R. Thus as Q moves along KD, the stress point q moves along the arc kd. If φ = π/ then R = R.

13 S c b a igure 6: Mohr s stress semicircle for the first octant of the sphere A φ e* K E e* e* φ D Q β n G β B N igure 7: Reference angle φ and β for the point Q (see Mase & Mase for better figure). Next let β = β f < π/. In this case θ (and φ take all possible values. Thus Q moves along the circle arc EG. In this case the equations for n = cos β f becomes ( N ) + S = R + ( )( ) cos β f = R which defines a circle of radius R < R. When β f = π/ then R = R. In this

14 4 case q moves along the arc eg. S k g d c b a igure 8: Mohr s stress semicircle for the last figure In summary for a specific n at a point P in the body, point Q, where the line of action of n intersects the spherical octant of the body is located at the common pointof the arcs KD and EG. At the same time the corresponding stress point q (having coordinates N and S ) is located at the intersection of the of the circle arcs kd and eg. 7 Plane Stress When one and only principal stress is zero, we have a state of plane stress in which the plane of the two non-zero principal stresses is the designated plane. Taking x axis as the direction of the zero principal stress the matrix reprersentation of the stress tensor is [] = q Let the unit normal n in the x x plane make an angle θ with the x axis. Let N and S be the components along the unit normal and tangential to it. Then taking the normal, tangential and x as a coordinate system we get using the transfomration e e N

15 5 law N = + S = Squaring and adding we get + ( N ) + + S = sin θ + cos θ cos θ + sin θ ( ) + To draw the Mohr s circle we use the following convention. irst draw the state S B (, ) A (, ) B N A igure 9: Mohr s stress circle for plane stress of stress at point A and then the state of stress at the point B. The shear stress is positive if it rotates the element in the clockwise directions and negative if it rotates the element in the counter clockwise direction. Draw the circle with AB as the diameter. The most highest and lowest normal stresses are ± R where is the centre of the circle and R is radius. These are also the nonzero principal stresses and the values are + ± ( ) +

16 6 Now we can also determine the orientation of the principal stresses. Since A is on the x axis, we must rotate anticlockwise by the angle A N = θ p to get to the point of maximum stress. Since the angles in the Mohr s circle are doubled, the orientation is θ p with respect to the x-axis in the anti-clockwise directions. 8 Piola-Kirchoff Stress We have seen that the equations of motion for the deformed configuration is given by ρa = div + ρb In elasticity, our final goal is to find the deformation field and the auchy stress tensor that arise in a body subjected to a given system of forces. In this respect the equations of motion in deformed state spatial variable are not much of help since x is itself unknown. Hence it is desirable to rewrite the equations of motion and boundary conditions with respect the reference configurations. Hence we introduce a stress tensor, first Piola-Kirchoff stress tensor, defined by T = J where J is Jacobian of the motion/deformation. Note the S is a two point tensor T = T Ai E A e i, T Ai = J Aj ji osider the corresponding volumes P t B t and P 0 B 0 in the current and reference configuration respectively. Now total surface force at time t can be calculated as t (n) da = P t nda = P t J T NdA = P 0 T T NdA P 0 Using localisation theorem we get nda = T T NdA = T T N = t (n)da da Thus Piola-Kirchoff stress tensor specifies the contact (surface) force in B t per referential area with respect to the orientation in B r. Also T Ai = E A Se i = T T E A e i = e i t(x, t, E χ A ) da da whrere E χ A is the image of E A under the motion x = χ(x, t). Hence T Ai is the represents the component in the i-direction of the force acting on an element of

17 7 surface in B t which corresponds to an element of surface in B r with unit normal along E A, the force being expressed per unit area of the element in B r. Sometimes T T is referred as first Piola-Kirchoff stress tensor. The first Piola-Kirchoff stress tensor is not symmetric i.e. T T T auchy s equations of motions in current variables can now be recast in terms of Lagrange variables. Thus ρa = div + ρb becomes ρ 0 A = Div T + ρ 0 b and = T becomes T = T T T Here A A(X, t) and b b(χ(x, t)) It is desirable to define a symmetric stress tensor because the constitutive equations then takes a simpler form. Hence we define the second Piola-Kirchoff stress tensor by S = T T = J T Now the equations of motion becomes ρ 0 A = Div S T + ρ 0 b and S T = S