16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #2 - Stress and Momentum balance Stress at a point We are going to consider the forces exerted on a material. These can be external or internal. External forces come in two ?avors: body forces (given per unit mass or volume) and surface forces (given per unit area). If we cut a body of material in equilibrium under a set of external forces along a plane, as shown in Fig.1, and consider one side of it, we draw two conclusions: 1) the equilibrium provided by the loads from the side taken out is provided by a set of forces that are distributed among the material particles adjacent to the cut plane and that should provide an equivalent set of forces to the ones loading the part taken out, 2) these forces can now be considered as external surface forces acting on the part of material under consideration. The stress vector at a point on ΔS is de?ned as: t = lim ΔS→0 f (1) ΔS If the cut had gone through the same point under consideration but along a plane with a di?erent normal, the stress vector t would have been di?erent. Let’s consider the three stress vectors t (i) acting on the planes normal to the coordinate axes. Let’s also decompose each t (i) in its three components in the coordinate system e i (this can be done for any vector) as (see Fig.2): t (i) = σ ij e j (2) 1 n body forces Ds f surface forces body forces n Figure 1: Surface force f on area ΔS of the cross section by plane whose normal is n σ ij is the component of the stress vector t (i) along the e j direction. σ11 σ12 σ13 σ 21 σ22 σ23σ31 σ32σ33 t t t (1) (2) (3) x1 x2 x3 e e e1 2 3 Figure 2: Stress components Stress tensor We could keep analyzing di?erent planes passing through the point with di?erent normals and, therefore, di?erent stress vectors t (n) and one might wonder if there is any relation among them or if they are all independent. The answer to this question is given by invoking equilibrium on the (shrinking) 2 ? ? tetrahedron of material of Fig.3. The area of the faces of the tetrahedron are ΔS 1 , ΔS 2 , ΔS 3 and ΔS. The stress vectors on planes with reversed normals t(?e i ) have been replaced with ?t (i) using Newton’s third law of action and reaction (which is in fact derivable from equilibrium): t (?n) = ?t (n) . Enforcing equilibrium we have: t (n) ΔS ? t (1) ΔS 1 ? t (2) ΔS 2 ? t (3) ΔS 3 = 0 (3) t (3)? t (2)? t (1)? ?e ?e ?e 1 2 3 n t(n) Figure 3: Cauchy’s tetrahedron representing the equilibrium of a tetrahedron shrinking to a point where ΔV is the volume of the tetrahedron and f is the body force per unit volume. The following relation: ΔSn i = ΔS i derived in the following mathematical aside: By virtue of Green’s Theorem: ?φdV = nφdS V S 3 ? ? ? ? ? applied to the function φ = 1, we get 0 = S ndS which applied to our tetrahedron gives: 0 = ΔSn ? ΔS 1 e 1 ? ΔS 2 e 2 ? ΔS 3 e 3 If we take the scalar product of this equation with e i , we obtain: ΔS(n · e i ) = ΔS i or ΔS i = ΔSn i can then be replaced in equation 3 to obtain: ΔS ? t (n) ? (n · e 1 )t (1) + (n · e 2 )t (2) + (n · e 3 )t (3) ? = 0 or ? ? t (n) = n · e 1 t (1) + e 2 t (2) + e 3 t (3) (4) The factor in parenthesis is the de?nition of the Cauchy stress tensor σ: σ = e 1 t (1) + e 2 t (2) + e 3 t (3) = e i t (i) (5) Note it is a tensorial expression (independent of the vector and tensor com- ponenents in a particular coordinate system). To obtain the tensorial com- ponenents in our rectangular system we replace the expressions of t (i) from Eqn.2 σ = e i σ ij e j (6) Replacing in Eqn.4: t (n) = n · σ (7) or: t (n) = n · σ ij e i e j = σ ij n · e i e j = σ ij n i e j (8) t i = σ ki n k (9) 4 ? ? ? ? ? ?? ? ? ? ? Transformation of stress components Consider a di?erent system of cartesian coordinates e ? i . We can express our tensor in either one: σ = σ kl e k e l = σ ? mn e m e n (10) We would like to relate the stress components in one the two systems. To this end, we take the scalar product of (10) with e ? i and e ? j : ? ?? ? ? ?? ? mn e i · e m e n · e j = σ ? ij e i ? · σ · e j ? = σ kl e i ? · e k e l · e j ? = σ ? ? ? ? ? mn δ im δ nj = σ ? or σ ? ij = σ kl e i · e k e l · e j (11) The factors in parenthesis are the cosine directors of the angles between the original and primed coordinate axes. Principal stresses and directions Given the components of the stress tensor in a given coordinate system, the determination of the maximum normal and shear stresses is critical for the design of structures. The normal and shear stress components on a plane with normal n are given by: t N = t (n) · n = σ ki n k n i t S = ?t (n) ? 2 ? t 2 N It is obvious from these equations that the normal component achieves its maximum t N = ?t (n) ? when the shear components are zero. In this case: t (n) = n · σ = λn = λIn or in components: σ ki n k = λn i σ ki n k = λδ ki n k (12) σ ki ? λδ ki n k = 0 5 ? ? ? ? ? ? ? which means that the principal stresses are obtained by solving the previous eigenvalue problem, the principal directions are the eigenvectors of the prob- lem. The eigenvalues λ are obtained by noticing that the last identity can be satis?ed for non-trivial n only if the factor is singular, i.e., if its determinant vanishes: ? ? ? σ 11 ? λ σ 12 σ 13 ? ? ? σ 21 σ 22 ? λ σ 23 ? ? = 0 ? σ 31 σ 32 σ 33 ? λ ? which leads to the characteristic equation: ?λ 3 + I 1 λ 2 ? I 2 λ + I 3 = 0 where: I 1 = σ ii = σ 11 + σ 22 + σ 33 (13) 1 ? ? ? ? I 2 = σ ii σ jj ? σ ij σ ji = σ 11 σ 22 + σ 22 σ 33 + σ 33 σ 11 ? σ 2 23 + σ 2 12 + σ 2 31 (14) 2 I 3 = det[σ] = ?σ ij ? (15) are called the stress invariants because they do not depend on the coordinate system of choice. Linear and angular momentum balance We are going to derive the equations of momentum balance in integral form, since this is the formulation that is more aligned with our “integral” approach in this course. We start from the de?nition of linear and angular momentum. For an element of material at position x of volume dV , density ρ, mass ρdV which remains constant, moving at a velocity v, the linear momentum is ρvdV and the angular momentum x × (ρvdV ). The total momenta of the body are obtained by integration over the volume as: ρvdV and x × ρvdV V V respectively. The principle of conservation of linear momentum states that the rate of change of linear momentum is equal to the sum of all the external forces acting on the body: D ρvdV = f dV + tdS (16) Dt V V S 6 ? ? ? ? ? ? ? ? ? ? ? ? where D is the total derivative. The lhs can be expanded as: Dt D D ?v ρvdV = Dt (ρdV )v + ρ dV Dt V V V ?t D but Dt (ρdV ) = 0 from conservation of mass, so the principle reads: ?v ρ dV = f dV + tdS (17) V ?t V S Now, using what we’ve learned about the tractions and their relation to the stress tensor: ? V ρ ?v ?t dV = ? V f dV + ? S n · σdS (18) This is the linear momentum balance equation in integral form. We can replace the surface integral with a volume integral with the aid of the diver- gence theorem: ? ? n · σdS = ?· σdV S V and then (18) becomes: ?v ρ ? f ??· σ dV = 0 V ?t Since this principle applies to an arbitrary volume of material, the integrand must vanish: ?v ρ ? f ??· σ = 0 (19) ?t This is the linear momentum balance equation in di?erential form. In com- ponents: ?v i σ ji,j + f i = ρ ?t Angular momentum balance and the symmetry of the stress tensor The principle of conservation of angular momentum states that the rate of change of angular momentum is equal to the sum of the moment of all the external forces acting on the body: D ρx × vdV = x × f dV + x × tdS (20) Dt V V S 7 ? ? ? ? ? ? ? It can be conveniently written as x i t j ? x j t i dS + x i f j ? x j f i ? dV = ? x i ?v j ? x j ?v i ? dV S V V ?t ?t Using t i = σ ki n k , the divergence theorem and (19), this expression leads to (see homework problem): (σ ij ? σ ji )dV = 0 V which applies to an arbitrary volume V , and therefore, can only be satis?ed if the integrand vanishes. This implies: σ ij = σ ji (21) 8