4.4 Pricipal Stress Componets

4.4 Principal Stress Components finding max & min stresses by taking the derivative of the stress transformation eq's & setting them = 0 [σ_{max}/σ_{min}] [\frac{dσ_{x'}}{dθ}=-(σ_{x}-σ_{y})sin2θ+2τ_{xy}cos2θ=0] [\frac{sin2θ}{cos2θ}=tan2θ_{p}=\frac{2τ_{xy}}{(σ_{x}-σ_{y})/2}] direction of stress \[sin2θ=\frac{τ_{xy}}{\sqrt{(\frac{σ_{x}-σ_{y}}{2})^{2}+τ_{xy}^{2}}}\] cos 2θ \[{σ_{x}}'=\frac{σ_{x}+σ_{y}}{2}+\frac{(\frac{σ_{x}-σ_{y}}{2})^{2}+τ_{xy}^{2}}{\sqrt{(\frac{σ_{x}-σ_{y}}{2})^{2}+τ_{xy}^{2}}}\] \[σ_{1,2}=\frac{σ_{x}+σ_{y}}{2}\pm\sqrt{(\frac{σ_{x}-σ_{y}}{2})^{2}+τ_{xy}^{2}}\] \[tan2θ_{s}=-\frac{(σ_{x}-σ_{y})/2}{τ_{xy}}\] \[τ_{max}=\sqrt{(\frac{σ_{x}-σ_{y}}{2})^{2}+{σ_{xy}}^{2}}\] \[τ_{max}\] A could be but the magnitude for max & min Shear stress will be the same. \[tan2θ_{s}=\frac{τ_{xy}}{(σ_{x}-σ_{y})/2}\] \[σ_{1,2}=\frac{σ_{x}+σ_{y}}{2}\pm\sqrt{(\frac{σ_{x}-σ_{y}}{2})^{2}+{σ_{xy}}^{2}}\] \[tan2θ_{s}=-\frac{(σ_{x}-σ_{y})/2}{τ_{xy}}\] \[τ_{max}=\sqrt{(\frac{σ_{x}-σ_{y}}{2})^{2}+{τ_{xy}}^{2}}\] direction of max normal stress often corresponds to brittle failure due to fracture max Shear stress is critical for ductile materials which will fail due to plastic deformation Le note the direction to of the max shear stress is rotated by 45" from