In an orthogonal cutting operation, shear angle = 11.31°, cutting force = 900 N and thrust force = 810 N. Then the shear force will be approximately (given sin 11.31° = 0.2) 

Concept:

Draw the merchant circle

Here

R = Resultant cutting load, Fc = Cutting force component or tangential cutting force

FT = Thrust force component or axial cutting force, Fs = shear force component, Fn = normal force component

F = Frictional force component on tool plane, N = Normal force component on tool plane, α0 = Orthogonal rake angle

β = Friction angle, Φ = Shear angle, t1 = Uncut chip thickness

The shear force in terms of other forces and shear angle is given by

F_s = F_c cos ϕ - F_T sin ϕ 

Calculation:

Given F_c = 900 N, F_T = 810 N, ϕ = 11.31°, sin ϕ = 0.2;

\(cos ϕ = \sqrt {1-{sin^2{ϕ}}}\)

⇒ cos ϕ = 0.98

F_s = 900 × 0.98 - 810 × 0.2 = 719.816 N ≈ 720 N