Project ForgeA straight-teeth horizontal slab milling cutter has 4 teeth and diameter (D) of 200 mm. The rotational speed of the cutter is 100 rpm and the linear feed given to the workpiece is 1000 mm/minute. The width of the workpiece (w) is 100 mm, and the entire width is milled in a single pass of the cutter. The cutting force/tooth is given by F=Ktcw, where specific cutting force K=10 N/mm², w is the width of cut, and tc is the uncut chip thickness. The depth of cut (d) is D/2, and hence the assumption of d/D≪1 is invalid. The maximum cutting force required is ________ kN (round off to one decimal place).
Explanation
Feed per tooth ft=1000/(4×100)=2.5 mm/tooth. Since d/D≪1 is invalid, the exact engagement (contact) angle must be used: cos(θmax)=1−d/R=1−(D/2)/(D/2)=0 (since d=D/2=R), giving θmax=90°. Uncut chip thickness varies as tc(φ)=ft·sinφ, maximized at φ=θmax: tc,max=ft·sin90°=2.5 mm. Maximum cutting force per tooth = K·tc,max·w = 10×2.5×100 = 2500 N = 2.5 kN. Since the 90° engagement span equals the 90° tooth pitch of this 4-tooth cutter, only one tooth is ever in the high-force region at a time, so this is also the maximum total cutting force.
Source: GATE ME 2022 Session 2 (ME-2) (2022)
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