wiz-icon
MyQuestionIcon
MyQuestionIcon
17
You visited us 17 times! Enjoying our articles? Unlock Full Access!
Question

In orthogonal turning of a low carbon steel bar of diameter 150mm uncoated carbide tool the cutting velocity is 90m/min. The feed is 0.24mm/rev and the depth of cut is 2mm. The chip thickness obtained is 0.48mm. If the orthogonal rake angle is zero and the principal cutting edge angle is 90o, the shear angle in degree is

A
20.56
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
26.56
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
30.56
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
36.56
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 26.56
Given that,
Feed rate: f=0.24mm/rev
Chip thickness: t2=0.48mm
Principal cutting edge angle,Ψ=90o
Rake angle:α=0

We know that
Uncut chip thickness,
t1=fsinΨ
=0.24sin90o
t1=0.24mm

Cutting ratio: r=t1t2=0.240.48=0.5
tanϕ=rcosα1rsinα
=0.5cos0o10.5sin0o=0.5

Shear angle:ϕ=tan1(0.5)=26.56o

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Analysis of Orthogonal Machining
PRODUCTION
Watch in App
Join BYJU'S Learning Program
CrossIcon