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By Victor E Saouma

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56) into Eq. 57-b) 43 Points (σxx , σxy ), (σxx , 0), (σyy , 0) and [(σxx + σyy )/2, 0] are plotted in the stress representation of Fig. 6. 59-b) then after substitution and simplifiation, Eq. 61) σ xx = σ xy We observe that the form of these equations, indicates that σ xx and σ xy are on a circle centered at 12 (σxx + σyy ) and of radius R. Furthermore, since σxx , σyy , R and β are definite numbers for a given state of stress, the previous equations provide a graphical solution for the evaluation of the rotated stress σ xx and σ xy for various angles α.

Solution: From Eq. 55-b) Plane Stress Plane stress conditions prevail when σ3i = 0, and thus we have a biaxial stress field. e when one of the dimensions is much smaller than the other two. 2 Mohr’s Circle for Plane Stress Conditions The Mohr circle will provide a graphical mean to contain the transformed state of stress (σ xx , σ yy , σxy ) at an arbitrary plane (inclined by α) in terms of the original one (σxx , σyy , σxy ). 56) into Eq. 57-b) 43 Points (σxx , σxy ), (σxx , 0), (σyy , 0) and [(σxx + σyy )/2, 0] are plotted in the stress representation of Fig.

14) 10 e 7 3 Symmetry of Stress Tensor From Fig. 16) We now consider moment equilibrium (M = F×d). The stress is homogeneous, and the normal force on the opposite side is equal opposite and colinear. The moment (∆X2 /2)σ31 ∆X1 ∆X2 is likewise balanced by the moment of an equal component in the opposite face. Finally similar argument holds for σ32 . 17) which must be zero, hence σ12 = σ21 . 18) σij = σji A more rigorous proof of the symmetry of the stress tensor will be given in Sect. 1. 4 Principal Stresses 2–7 We should note that this theorem is analogous to Newton’s famous third law of motion To every action there is an equal and opposite reaction.

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