In orthogonal turning of a cylindrical tube of wall thickness 5 {mm}, the axial and the tangential cutting forces were measured as 1259 {N} and 1601 {N}, respectively. The measured chip thickness after machining was found to be 0.3 {mm}. The rake angle was 10^{\circ} and the axial feed was 100 {mm} \/ {\min}. the rotational speed of the spindle was 1000 {rpm}. Assuming the material to be perfectly and Merchant's first solution, the shear strength of the material is closest to

GATE Mechanical Engineering (ME) 2019 Shift 1 Solved Paper

Question : 47 of 65

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In orthogonal turning of a cylindrical tube of wall thickness

5 {mm}

, the axial and the tangential cutting forces were measured as

1259 {N}

and

1601 {N}

, respectively. The measured chip thickness after machining was found to be

0.3 {mm}

. The rake angle was

10^{∘}

and the axial feed was 100

{mm} \/ {min}

. the rotational speed of the spindle was

1000 {rpm}

. Assuming the material to be perfectly and Merchant's first solution, the shear strength of the material is closest to

Solution:

{N}=1000 {rpm}

, rake angle

(α)=10^{∘}

Feed (f)

=100 {mm} {/} {min}=100 {/} 1000=0.1 {mm}{/} {rev}= {t}

Chip thickness ratio (

{r})=\;{\;\text " wncut chip thickness "\;}/{\;\text " cut chip thickness "\;}=\;{0.1}/{0.3}=\;{1}/{3}

Shear strength is given by

τ_{s}=\;{F_{s} \sinϕ}/{b t}

Where,

F_{s}=

shear force

ϕ=

Shear angle

b=

width

h=5 {mm}

Using

\tan ϕ=\;{r \cos α}/{1-r \sin α}=\;{0.33 \cos 10}/{1-0.33 \sin 10}=0.345

Or,

φ=19.02

F_{s}=F_{c} \cos φ-F_{t} \sin φ=1606 \cos 19.02-1259 \sin 19.02=1103.29 {N}

So,

τ_{s}=\;{1103.29 \sin 19.02}/{5 × 0.1}=719.12 {MPa}

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