Respuesta :
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]