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 width of the orthogonal cutting operation w = 3.0 mm. Based on these data, determine the shear strength of the work material.

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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  [tex]\sigma = 245.09 N/mm^2[/tex]

Explanation:

From the question we are told that the

    The the cutting force is [tex]F_c = 1559N[/tex]

     The thrust force is [tex]F_t = 1271\ N[/tex]

    The width of the octagonal cutting operation is  [tex]w = 3.0\ mm[/tex]

     The thickness is [tex]t = 0.5mm[/tex]

     The shear plane angle is  [tex]\theta = 25.4^o[/tex]

The shear strength of the work materiel is mathematically represented as

                       [tex]\sigma = \frac{F_s}{A}[/tex]

Where  [tex]F_s[/tex] is the shear force which is mathematically evaluated as

                   [tex]F_s = F_c cos\ \theta -F_t sin\ \theta[/tex]

Substituting values we have

                  [tex]F_s = 1559 cos (25.4) - 1271sin(25.4)[/tex]

                        [tex]= 857.841N[/tex]

While A is the area which is mathematically evaluated as

                        [tex]A = \frac{t* w}{sin \theta }[/tex]

Substituting values

                      [tex]A = \frac{0.5 *3}{sin (25.4)}[/tex]

                         [tex]= 3.50[/tex]

Substituting this into the equation for shear strength

               [tex]\sigma = \frac{857.8}{3.5}[/tex]

                  [tex]\sigma = 245.09 N/mm^2[/tex]

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