In a tensile test on the sample of cross-sectional area A as illustrated in Figure 6.1a, the normal to the slip plane lies at an angle ϕ $ \phi $ https://s3-euw1-ap-pe-df-pch-content-public-u.s3.eu-west-1.amazonaws.com/9781315114910/a1f6aa37-c8b0-4273-bc79-d03791cd487d/content/inline-math6_1.tif"/> to the force F, and the slip direction is at an angle λ $ \lambda $ https://s3-euw1-ap-pe-df-pch-content-public-u.s3.eu-west-1.amazonaws.com/9781315114910/a1f6aa37-c8b0-4273-bc79-d03791cd487d/content/inline-math6_2.tif"/> to the tensile axis. The area of the slip plane is A / cos ϕ $ A/\cos \phi $ https://s3-euw1-ap-pe-df-pch-content-public-u.s3.eu-west-1.amazonaws.com/9781315114910/a1f6aa37-c8b0-4273-bc79-d03791cd487d/content/inline-math6_3.tif"/> and the force in the slip direction is F cos λ $ F \cos \lambda $ https://s3-euw1-ap-pe-df-pch-content-public-u.s3.eu-west-1.amazonaws.com/9781315114910/a1f6aa37-c8b0-4273-bc79-d03791cd487d/content/inline-math6_4.tif"/> so that the shear stress on the slip plane in the slip direction is τ = F A cos ϕ cos λ ⏟ Schmid factor . $$ \begin{aligned} \tau = \frac{F}{A}\underbrace{\cos \phi \cos \lambda }_{\text{ Schmid} \text{ factor}}. \end{aligned} $$ https://s3-euw1-ap-pe-df-pch-content-public-u.s3.eu-west-1.amazonaws.com/9781315114910/a1f6aa37-c8b0-4273-bc79-d03791cd487d/content/math6_1.tif"/>