2 - DIMENSIONAL STRESS I. 2 DIMENSIONAL STRESS A. Determine the state of stress for an arbitrary plane --- the 2D case neglects the z direction B. q = Ð between the normal to the plane and the x - axis C. Develop an equation that will allow us to solve for sq and tq given any plane with an orientation of Ðq D. Neglect z, and only 4 components are used to describe the system s xx t xy t yx s yy remember: no rotation about z-axis, so t yx = t xy so there are only 3 stress components E. Remove the spatial triangle ABC F. This is a "free" body" diagram G. Want to determine s and t for plane AC H. To make calculations simple, choose a convenient length for AC (hypothesis) 1. Where: AC = 1, then: AB = cosq , BC = sinq note: there are 4 stresses (vector components) that make contributions to both the normal (s ) and shear (t ) stress these are: s xx, s yy, t xy and t yx I. To determine total stress, simply sum up all of these vector components in terms of q (s xx + t xy)cosq + (s yy + t yx)sinq = sq + tq II. SOLUTIONS A. Step 1 -- contribution of s xx to sq and tq 1. Normal component cos q = sq / s xx cos q ® sq = s xx cos2 q 2. Shear component sin q = tq / s xx cos q ® tq = s xx sinq cosq B. Step 2 -- contribution of s yy to sq and tq 1. Normal component sin q = sq 2 / s yy sinq ® sq 2 = s yy sin2 q 2. Shear component cosq = tq 2 / s yy sinq ® tq 2 = - s yy sinq cosq a. s yy negative due to direction C. Step 3 -- contribution of t yx 1. Normal stress sq 3 = t yx sinq cosq 2. Shear stress tq 3 = - t yx sin2q D. Step 4 -- contribution of t xy 1. Normal stress sq 4 = t xy sinq cosq 2. Shear component tq 4 = t xy cos2q E. Add all - sq total = s 1+2+3+4 q tq total = t 1+2+3+4q sq = s xx cos2q + s yy sin2q + 2t xy sinq cosq tq = (s xx - s yy)sinq cosq + t xy(cos2q - sin2q ) III. PRINCIPAL PLANES OF STRESS A. For any state of stress there are 3 orthogonal planes with no shear stress IV. PRINCIPAL STRESS AXES A. s 1 = max. principal stress B. s 2 = intermediate principal stress C. s 3 = minimum principal stress D. Substituting into the equations for sq and tq 1. s xx = s 1 2. s yy = s 3 3. t xy = 0 results in: s = s 1 cos2 q + s 3 sin2 q t = (s 1 - s 3 ) sin q cosq V. STRESS (graphic representations) A. Stress ellipse - 2D , ellipsoid - 3D B. Mohr circle plot VI. STRESS ELLIPSE - deals only with normal components A. s 1 and s 3 must be balanced by sq B. s 1 cos q = s xx C. s 3 sin q = s yy D. cos2q = s xx2 / s 1 2 and sin2 q = s yy2 / s 3 2 1. sin2 q + cos2q = 1 E. (s xx2 / s 1 2 ) + ( s yy2 / s 3 2 ) = 1 (defines an ellipse) VII. MOHR CIRCLE PLOT A. Remember principal stresses have no shear component associated with them 1. s xx = s 1 2. s yy = s 3 B. s q = s 1 cos2 q + s 3 sin2 q C. t q = ( s 1 - s 3 )sin q cosq D. sin2q = 1 - cos2q / 2 E. cos2q = 1 + cos2q / 2 F. sinq cosq = sin2q / 2 G. Substitute into: 1. s q = s 1 (1+cos2 q / 2) + s 3 (1- cos2 q / 2) = (s 1 + s 3 ) / 2 + (( s 1 - s 3 ) / 2)cos2 q x = center    radius      cosa 2. t q = s 1 - s 3 (sin2 q / 2) = ((s 1 - s 3 ) / 2)sin2 q y = radius        sina H. The angle 2q locates a point on circumference with coordinates (s q, t q ), which are the stress components on the inclined plane defined by q I. The coordinates at any point on the circumference represent the normal and shear stress for that state of stress J. Mean stress - average of greatest /least principal stresses = (s 1 + s 3 ) / 2 K. Deviatoric stress = range of stress within a body d = s 1 - s 3