University of California at Berkeley Engineering 36Department of Civil and Environmental Engineering J. LublinerStressUp to now, when we studied contact forces distributed over a surface, we considered only forces thatwere normal to the surface and compressive, and we referred to the magnitude of the force per unit area aspressure. More generally, a force distributed over a surface has an arbitrary orientation with respect to thethe surface, and must be represented by a vector, usually called the traction but sometimes also the stressvector.The value of the traction on a surface element around a given point of the body will in general, of course,depend on the location of the point but also on the orientation of the surface. Imagine yourself pullingaxially on a bar of cross-sectional area A with a force F . At any point in the bar, you would expect thetraction on an area element that perpendicular to the axis of the bar to have an axial direction and be, atleast approximately, of magnitude F/A. But if you consider an area element that is parallel to the axis,there is no a priori reason why the traction should not be zero.It follows that, at any point, the traction can be regarded as a function of the unit normal vector n of asurface ele ment about that point, and we can write it as T(n). Now, if the surface divides the body into twoparts, then the immediately adjoining surface has the orientation −n, and the traction there will be T(−n).The law of action and reaction demands thatT(−n) = −T(n) (1)Consider, now, a triangular area element ∆A whose sides are in the yz-, xz- and xy-planes, respectively,and whose orientation is given by the unit normal vector n = nxi + nyj + nzk. The components nx, nyandnzare of course the direction cosines of the normal direction to the area element. More over, the projectionsof the area element on the yz-, xz- and xy-planes are triangles with the respective areas∆Ax= nx∆A, ∆Ay= ny∆A, ∆Az= nz∆A. (2)If the four triangles can be thought of as constituting a tetrahedron, then the outward unit normal vectorsof the projected triangles are −i, −j and −k, as shown in Figure 1.zxypp ppbbbr6-@@@@@@*n−i?−j−kFigure 1zxypp ppbbbr6-@@@@@@BBBBM:T(n) ∆AT(−i) ∆AxT(−j) ∆AyT(−k) ∆AzFigure 2Figure 2 shows the surface forces acting on the tetrahedron. If the tetrahedron is sufficiently small, thenthe effect of body forces (such as gravity) on the equilibrium of the element can be neglected, because suchforces would be proportional to the volume ∆V , which is a higher-order differential than the area ∆A (thatis, in the limit of an infinitesimal element, ∆V/∆A → 0). Consequently the equilibrium of forces is given byT(n)∆A + T(−i)∆Ax+ T(−j)∆Ay+ T(−k)∆Az= 0In view of Equations (1) and (2), this may be rewritten as[T(n)∆A − nxT(i) − nyT(j) − nzT(k)]∆A = 0,and since the area ∆ A is positive, the quantity in brackets must be zero, that isT(n) = nxT(i) + nyT(j) + nzT(k).Finally, special symbols will be given for the components of T(i), T(j) and T(k:T(i) = τxxi + τxyj + τxzk,T(j) = τy xi + τy yj + τy zk,T(k) = τzxi + τzyj + τzzk,from which it follows thatT(n) = (nxτxx+ nyτy x+ nzτzx)i + (nxτxy+ nyτy y+ nzτzy)j + (nxτxz+ nyτy z+ nzτzz)k.The components τxx, τxyetc. are known as stress components or simply as stresses, and the arrayτxxτxyτxzτy xτy yτy zτzxτzyτzzis said to represent the stress tensor. The components τxx, τy yand τzz(which are often written as σx,σyand σz, respectively) are the normal stresses, while the remaining components (those with unequalsubscripts) are the shear stresses. Note that, with n being the outward normal, a normal stress is positiveif it represents tension and negative if compression. In a fluid at rest, therefore, if the pressure is p thenthe normal stresses are all equal to −p while the shear stresses are zero. Such a state is known as one ofhydrostatic stress.It is somewhat easier to visualize stresses in two dimensions than in three. Consider a very small cub oid(rectangular parallelepiped) with sides ∆x, ∆y, ∆z parallel to the axes. The stresses acting in the x- ory-direction on the planes that are parallel to the z-axis can be visualized in the xy-plane as in Figure 3:∆x -∆y6?τxxτxxτxyτxyτy xτy xτy yτy y-6??6-Figure 3The directions of the arrows representing the stresses are consistent with these stresses being positive.The eleme nt is assumed to be small enough so that the variation of the stress values through it can beneglected, so that the resultants of the tractions can be taken as acting at the midpoints of the sides, andforce equilibrium is obviously satisfied. As regards moment equilibrium, the forces due to the stresses τxy,acting over the area ∆y ∆z, form a counterclockwise couple with moment arm ∆x, and therefore the momentis τxy∆x ∆y ∆z. Similarly the stresses τy xcreate the clockwise moment τy x∆x ∆y ∆z, and therefore, formoment equilibriumτxy= τy x.Analogous analyses in the yz- and xz-planes lead to the relationsτy z= τzy, τxz= τzx.In words: the shear stresses on two mutually perpendicular planes and acting in mutually perpendiculardirections are equal.If there is an axis (say the z-axis) such that σz(or τzz, τxz(= τzx) and τy z(= τzy) are all zero, then thestate of stress is called plane stress.If there is an axis (say the x-axis) such that the only nonzero stress component is σx(or τxx, then thestate of stress is called uniaxial (a special case of plane stress).If there is a pair of axes (say x and y) such that the only nonzero stress components are τxyand τy x(andthey are, of course, equal), then the state of stress is called pure shear, another special case of plane stress.If the components in a state of plane stress are known with respect to a given set of axes (say x andy), then they can be determined for any other set of axes in the plane by the so-called wedge methodillustrated in Figure 4.JJJJJJJJJθ3σxτxyτy xσyσθτθ3JJJ]??Figure 4Problems1. From the force equilibrium of the wedge in Figure 4, show thatσθ= σxcos2θ + σysin2θ + 2τxysin θ cos θ,τθ= τxy(cos2θ − sin2θ) + (σy− σx) sin θ cos