Complete Question
Suppose in the previous example (from Machining 1 lecture) that cutting force and thrust force are measured during an orthogonal cutting operation: Fc = 1559 N and Ft = 1271 N. The chip thickness before the cut =0.5mm and the width of the orthogonal cutting operation is 3.0mm . Rake angle is equal to 10 and shear plane angle is 25.4. Based on these data, determine the shear strength of the work material.
Answer:
The shear strength of the work material is

Step-by-step explanation:
From the question we are told that the
The the cutting force is

The thrust force is

The width of the octagonal cutting operation is

The thickness is

The shear plane angle is

The shear strength of the work materiel is mathematically represented as

Where
is the shear force which is mathematically evaluated as

Substituting values we have


While A is the area which is mathematically evaluated as

Substituting values


Substituting this into the equation for shear strength

