Transformation of stress components Consider a different system of cartesian coordinates e. We can express our tensor in either one We would like to relate the stress components in one the two systems. To this end, we take the scalar product of (10) with e and e e;oe=ok(e; ek(el e=omn(e em)(e OI Ok(e;ek)(ere The factors in parenthesis are the cosine directors of the angles between the original and primed coordinate axes Principal stresses and directions Given the components of the stress tensor in a given coordinate system, the determination of the maximum normal and shear stresses is critical for the design of structures. The normal and shear stress components on a plane with normal n are given by n -Kinin It is obvious from these equations that the normal component achieves its maximum tN=lt( n)ll when the shear components are zero. In this case =An= al or in components Kink= asking (oki-Xoki nk=0 5� � � � � �� � � � � Transformation of stress components Consider a different system of cartesian coordinates e� i. We can express our tensor in either one: σ = σklekel = σ� mnemen (10) We would like to relate the stress components in one the two systems. To this end, we take the scalar product of (10) with e� i and e� j : � �� � � �� � mn ei · em en · ej = σ� ei ij � · σ · ej � = σkl ei � · ek el · ej � = σ� � � � � mnδimδnj = σ� or σ� ij = σkl ei · ek el · ej (11) The factors in parenthesis are the cosine directors of the angles between the original and primed coordinate axes. Principal stresses and directions Given the components of the stress tensor in a given coordinate system, the determination of the maximum normal and shear stresses is critical for the design of structures. The normal and shear stress components on a plane with normal n are given by: tN = t(n) · n = σkinkni tS = �t(n)�2 − t2 N It is obvious from these equations that the normal component achieves its maximum tN = �t(n) � when the shear components are zero. In this case: t(n) = n · σ = λn = λIn or in components: σkink = λni σkink = λδkink (12) σki − λδki nk = 0 5