2112 A carbon steel bar 7.64 inch in diameter has a tensile strength of 65,000 l
ID: 2074244 • Letter: 2
Question
2112 A carbon steel bar 7.64 inch in diameter has a tensile strength of 65,000 lb/in' and a shear strength of 45,000 lb/in2. The diameter is reduced using a tuming operation at a cutting speed of 400 ft/imin. The feed is 0.011 in/rev and the depth ofcut is 0.120 in. The rake angle on the tool in the direction of chip flow is 13o. The cutting conditions result in a chip ratio of 0.52. Using the orthogonal model as an approximation of tuning, detennine (a) the shear plane angle, (b) shear force, (ccuting force and feed force, and (d) coefficient of friction between the tool and chip.Explanation / Answer
Let Fs= shear force for chip
Fn= normal force on chip
Fc = cutting force
Ft= feed force
N = normal force on tool
F= frictional force
A= rake angle= 13°
Q= shear angle
Feed rate = 0.011in/rev
Depth of uncut chip= 0.120in
Shearing area = ( feed rate ×depth of cut )/sin(Q)
Now,
a) Tan(Q) = rcosA/(1-rsinA)
Where r = chip thickness ratio= 0.52
Substituting the values
Tan(Q) = 0.573
Shear plane angle, Q=29.84°
Shearing area = (feed rate× depth of cut)/sin(29.84°)
= 0.011×0.120/sin(29.84)
= 2.652× 10^(-3) sq.in
b) Shear force, Fs= shear stress × shear area
= 450000×2.652×10^(-3) lb
= 119.377 lb
Also normal force on chip, Fn= tensile strength × area of chip
= 65000×2.652×10^(-3) lb
=172.434 lb
c) cutting force Fc = Fs cos(Q) + Fn sin(Q)
Substituting values of Fs, Fn and angle Q
Fc= 103.549 + 85.799
=189. 324 lb
Feed force,Ft = Fn cos(Q) - Fs sin(Q)
= 149.572 - 59.399
= 90.173 lb
d)
Coefficient of friction=(Fc sinA + Ft cosA)/( Fc cosA - Ft sinA)
= 130.441/164.186
=0.794
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