LECTURE 4 Material subjected to combined direct and shear stresses: Now consider a complex stress system shown below, acting on an element of material. The stresses sx and sy may be compressive or tensile and may be the result of direct forces or as a result of bending.The shear stresses may be as shown or completely reversed and occur as a result of either shear force or torsion as shown in the figure below: As per the double subscript notation the shear stress on the face BC should be notified as tyx , however, we have already seen that for a pair of shear stresses there is a set of complementary shear stresses generated such that tyx = txy By looking at this state of stress, it may be observed that this state of stress is combination of two different cases: (i) Material subjected to pure stae of stress shear. In this case the various formulas deserved are as follows sq = tyx sin2 q tq = - tyx cos 2 q (ii) Material subjected to two mutually perpendicular direct stresses. In this case the various formula's derived are as follows. To get the required equations for the case under consideration,let us add the respective equations for the above two cases such that These are the equilibrium equations for stresses at a point. They do not depend on material proportions and are equally valid for elastic and inelastic behaviour This eqn gives two values of 2q that differ by 1800 .Hence the planes on which maximum and minimum normal stresses occurate 900 apart. From the triangle it may be determined                                                            Substituting the values of cos2 q and sin2 q in equation (5) we get This shows that the values oshear stress is zero on the principal planes. Hence the maximum and minimum values of normal stresses occur on planes of zero shearing stress. The maximum and minimum normal stresses are called the principal stresses, and the planes on which they act are called principal plane the solution of equation will yield two values of 2q separated by 1800 i.e. two values of q separated by 900 .Thus the two principal stresses occur on mutually perpendicular planes termed principal planes. Therefore the two – dimensional complex stress system can now be reduced to the equivalent system of principal stresses. Let us recall that for the case of a material subjected to direct stresses the value of maximum shear stresses Therefore,it can be concluded that the equation (2) is a negative reciprocal of equation (1) hence the roots for the double angle of equation (2) are 900 away from the corresponding angle of equation (1). This means that the angles that angles that locate the plane of maximum or minimum shearing stresses form angles of 450 with the planes of principal stresses. Futher, by making the triangle we get                                Because of root the difference in sign convention arises from the point of view of locating the planes on which shear stress act. From physical point of view these sign have no meaning. The largest stress regard less of sign is always know as maximum shear stress. Principal plane inclination in terms of associated principal stress: We know that the equation yields two values of q i.e. the inclination of the two principal planes on which the principal stresses s1 and s2 act. It is uncertain,however, which stress acts on which plane unless equation. is used and observing which one of the two principal stresses is obtained. Alternatively we can also find the answer to this problem in the following manner Consider once again the equilibrium of a triangular block of material of unit depth, Assuming AC to be a principal plane on which principal stresses sp acts, and the shear stress is zero. Resolving the forces horizontally we get: sx .BC . 1 + txy .AB . 1 = sp . cosq . AC   dividing the above equation through by BC we get Goto Home