GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

Question : 62 of 65

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The thickness of a steel plate with material strength coefficient of 210 MPa, has to be reduced from 20 mm to 15 mm in a single pass in a two-high rolling mill with a roll radius of 450 mm and rolling velocity of 28 m\/min. If the plate has a width of 200 mm and its strain hardening exponent, n is 0.25, the rolling force required for the operation is _________ kN (round off to 2 decimal places).
Note: Average Flow Stress = Material Strength Coefficient

x \;{(\;\text " True Strain "\;)^{n}}/{(1+n)}

Your Answer:

Validate

Solution:

Given

{K}=210 {MPa}

{\Hi}=20 {mm}

{H}_{ {f}}=15 {mm}

{R}=450 {mm}

( {V})_{ {R}}=28 {m} {/} {min}

{B}=200 {mm}

{n}=0.25

(σ)_{ {o}}=\;{ {KE}_{T}^{ {n}}}/{ {n}+1}

{E}_{ {T}}=

True Strain

=\ln \;{ {A}_{ {i}}}/{ {A}_{ {f}}}=\ln \;{ {I}_{ {f}}}/{ {L}_{ {i}}}

{A}_{i}= {B} {H}_{ {i}}

{A}_{ℓ}= {B} {H}_{ {f}}

ε_{ {T}}=\ln \;{ {H}_{ {i}}}/{ {H}_{ {f}}}=\ln \;{20}/{15}

∊_{ {T}=0.2876

σ_{0}=

Average flow stress

=\;{210 ⋅(0.2876)^{0.25}}/{1.25}

σ_{0}=123.028 {mPa}

Rolling Force

=σ_{0} . I . B

I=

Contact length

=√{R ∆ h}

I=√{450 × 5}

I=47.43 {mm}

f=123.028 × 47.43 × 200

f=1167.04 {KN}

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