LECTURE 2 ANALYSIS OF STERSSES General State of stress at a point : Stress at a point in a material body has been defined as a force per unit area. But this definition is some what ambiguous since it depends upon what area we consider at that point. Let us, consider a point ‘q' in the interior of the body Let us pass a cutting plane through a pont 'q' perpendicular to the x - axis as shown below The corresponding force components can be shown like this dFx = sxx. dax dFy = txy. dax dFz = txz. dax where dax is the area surrounding the point 'q' when the cutting plane ^ r is to x - axis. In a similar way it can be assummed that the cutting plane is passed through the point 'q' perpendicular to the y - axis. The corresponding force components are shown below The corresponding force components may be written as dFx = tyx. day dFy = syy. day dFz = tyz. day where day is the area surrounding the point 'q' when the cutting plane ^ r is to y - axis. In the last it can be considered that the cutting plane is passed through the point 'q' perpendicular to the z - axis. The corresponding force components may be written as dFx = tzx. daz dFy = tzy. daz dFz = szz. daz where daz is the area surrounding the point 'q' when the cutting plane ^ r is to z - axis. Thus, from the foregoing discussion it is amply clear that there is nothing like stress at a point 'q' rather we have a situation where it is a combination of state of stress at a point q. Thus, it becomes imperative to understand the term state of stress at a point 'q'. Therefore, it becomes easy to express astate of stress by the scheme as discussed earlier, where the stresses on the three mutually perpendiclar planes are labelled in the manner as shown earlier. the state of stress as depicted earlier is called the general or a triaxial state of stress that can exist at any interior point of a loaded body. Before defining the general state of stress at a point. Let us make overselves conversant with the notations for the stresses.              We have already chosen to distinguish between normal and shear stress with the help of symbols s and t . Cartesian - co-ordinate system In the Cartesian co-ordinates system, we make use of the axes, X, Y and Z Let us consider the small element of the material and show the various normal stresses acting the faces Thus, in the Cartesian co-ordinates system the normal stresses have been represented by sx, syand sz. Cylindrical - co-ordinate system In the Cylindrical - co-ordinate system we make use of co-ordinates r, q and Z. Thus, in the Cylindrical co-ordinates system, the normal stresses i.e components acting over a element is being denoted by sr, sqand sz. Sign convention : The tensile forces are termed as ( +ve ) while the compressive forces are termed as negative ( -ve ). First sub – script : it indicates the direction of the normal to the surface. Second subscript : it indicates the direction of the stress. It may be noted that in the case of normal stresses the double script notation may be dispensed with as the direction of the normal stress and the direction of normal to the surface of the element on which it acts is the same. Therefore, a single subscript notation as used is sufficient to define the normal stresses. Shear Stresses : With shear stress components, the single subscript notation is not practical, because such stresses are in direction parallel to the surfaces on which they act. We therefore have two directions to specify, that of normal to the surface and the stress itself. To do this, we stress itself. To do this, we attach two subscripts to the symbol ' t' , for shear stresses. In cartesian and polar co-ordinates, we have the stress components as shown in the figures. txy , tyx , tyz , tzy , tzx , txz trq , tqr , tqz , tzq ,tzr , trz So as shown above, the normal stresses and shear stress components indicated on a small element of material seperately has been combined and depicted on a single element. Similarly for a cylindrical co-ordinate system let us shown the normal and shear stresses components separately. Now let us combine the normal and shear stress components as shown below : Now let us define the state of stress at a point formally. State of stress at a point : By state of stress at a point, we mean an information which is required at that point such that it remains under equilibrium. or simply a general state of stress at a point involves all the normal stress components, together with all the shear stress components as shown in earlier figures. Therefore, we need nine components, to define the state of stress at a point sx  txy txz sy tyx tyz sz tzx  tzy If we apply the conditions of equilibrium which are as follows: å Fx = 0 ; å M x = 0 å Fy = 0 ; å M y = 0 å Fz = 0 ; å M z = 0 Then we get txy = tyx tyz = tzy tzx = txy Then we will need only six components to specify the state of stress at a point i.e sx , sy, sz , txy , tyz , tzx Now let us define the concept of complementary shear stresses. Complementary shear stresses: The existence of shear stresses on any two sides of the element induces complementary shear stresses on the other two sides of the element to maintain equilibrium. on planes AB and CD, the shear stress t acts. To maintain the static equilibrium of this element, on planes AD and BC, t' should act, we shall see that t' which is known as the complementary shear stress would come out to equal and opposite to the t . Let us prove this thing for a general case as discussed below: The figure shows a small rectangular element with sides of length Dx, Dy parallel to x and y directions. Its thickness normal to the plane of paper is Dz in z – direction. All nine normal and shear stress components may act on the element, only those in x and y directions are shown. Sign convections for shear stresses: Direct stresses or normal stresses - tensile +ve - compressive –ve Shear stresses: - tending to turn the element C.W +ve. - tending to turn the element C.C.W – ve. The resulting forces applied to the element are in equilibrium in x and y direction. ( Although other normal and shear stress components are not shown, their presence does not affect the final conclusion ). Assumption : The weight of the element is neglected. Since the element is a static piece of solid body, the moments applied to it must also be in equilibrium. Let ‘O' be the centre of the element. Let us consider the axis through the point ‘O'. the resultant force associated with normal stresses sx and sy acting on the sides of the element each pass through this axis, and therefore, have no moment. Now forces on top and bottom surfaces produce a couple which must be balanced by the forces on left and right hand faces Thus, tyx . D x . D z . D y = txy . D x . D z . D y In other word, the complementary shear stresses are equal in magnitude. The same form of relationship can be obtained for the other two pair of shear stress components to arrive at the relations Goto Home