GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

Cylindrical bar:
Diameter (D) = 200 mm
In ASA system:
Tool Geometry ( Tool 1)⇒0∘−9∘−7∘−8∘−15∘−30∘−0.05 inch Tool Geometry (Tool 2) ⇒0∘−9∘−7∘−8∘−15∘−0∘0.05 inch.
For Tool 1: Side cutting edge angle (Cs,1)=30∘
For tool 2: side cutting edge angle (Cs,2)=0∘
Since Cs+λ=90∘
Thus, λ1=60∘
λ2=90∘
Since specific energy consumption (Uc)=‌

Fc

1000fd

=U0(t1)−0.4
Fc=U0(t1)−0.4×1000fd
Now t1=fsinλ
Fc=U0(f‌sin‌λ)−0.4×1000fd
Fc∝(sin‌λ)−0.4
Thus,
={(‌

sin‌λ2

sin‌λ1

)−0.4−1}×100
={(‌

sin‌90∘

sin‌60∘

)−0.4−1}×100
=−5.59%