Engineering Practice Questions from Competitive Exams, New Delhi (2026)

09/12/2021

IFS (Indian Forest Service) Year- 2019
Q-6.(a)During an orthogonal machining operation on mild steel, the results obtained are
uncut chip thickness, t1 = 0·25 mm ,
chip thickness, t2 = 0·75 mm
width, w = 2·5 mm
rake angle, α=00
cutting force, Fc = 950 N,
thrust force, FT= 475 N
(i) Determine the coefficient of friction between the tool and the chip.
(ii) Determine the ultimate shear stress τs of the work material. 15 Marks

Solution:-
Given that Chip thickness (tc) = 0·75 mm; depth of cut or uncut chip thickness (t) = 0·25 mm,
Width of cut (w) = 2.5 mm; Rake angle (α) = 0°
Tangential cutting force (Fc) = 950 N; Thrust force (Ft) = 475 N
Chip thickness ratio (r) = t_c/t = 0.25/0.75= 1/3=0.333
Coefficient of friction (μ)=?, Shear Stress (τs)=?

(i) the coefficient of friction between the tool and chip
The shear plane angle (Φ):-
We know that tanΦ =(r Cosα)/(1-r Sinα) = (r ×Cos 0)/(1-r ×Sin 0) = 1/3

tanφ= 0.333
Φ=18.43490≈ 18.4350
We know that
Coefficient of friction = (F_(c ) sinα+ F_t cos⁡α)/(F_c cosα-F_t sinα)

= (950 sin⁡0 +475 cos 0)/(950 cos0-475 sin 0)

= 475/950 = 0.5

μ= 0.5 (Ans.)
(ii) Ultimate shear stress of the work piece material
Shear Force (Fs) = F_c cosφ -F_t sinφ
= 950 cos 18.434 – 475 sin 18.434
= 901.254 – 150.2007 = 751.05 N
Shear stress = (Shear Force)/(Shear area) = (F_s SinΦ)/bt
= 751.05 x sin 18.434/2.5x10-3x0.25x10-3 = 379.986 MPa. (Ans.)