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MNF1_HW002 Homework 2

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שםעידו פנג בנטובניר קרל
ת”ז322869140322437203
דואר אלקטרוניido.fang@campus.technion.ac.ilnir.karl@campus.technion.ac.il

Question 1

Part a

We know that the maximum [[MNF1_004 Machining Processes#Turning#Key Parameters and Equations|cutting speed]] is given by:

Therefore the rotational speed is .
Substituting the given maximum allowed cutting speed, and its appropriate diameter of the workpiece, we get the following rotational speed:

Part b

For the minimal diameter, , and the same cutting speed , the rotational speed would need to be:

Therefore, the machine needs to allow rotational speeds in the following range:

By allowing the rotational speed to vary, we get shorter [[MNF1_004 Machining Processes#Turning#Key Parameters and Equations|cutting time]]:

Part c

From the drawings, we can see that the required surface finish is .
We also know that:

You can't use 'macro parameter character #' in math mode\begin{gathered} 3.95R_{a}=\dfrac{f^{2}}{8r} \\[1ex] f=\sqrt{ 3.95\cdot 8R_{a}r } \end{gathered} $$ Substituting $R_{a}$ and $r=\pu{0.8mm}$, we get: $$ \boxed { f=\pu{0.2mm} } $$ ### Part d The [[MNF1_004 Machining Processes#Turning#Cutting Force and Power|power required]] is given by:

\begin{aligned}
P_{c} & =F_{c}V_{c} \[1ex]
&= K_{s}fdV_{c}
\end{aligned}

\begin{gathered}
P=\pu{369N.mm^{-2}}\cdot \pu{0.2mm}\cdot \pu{5mm}\cdot \pu{250m.min^{-1}} \[1ex]
\boxed {
P=\pu{1.538kW}
}
\end{gathered}

You can't use 'macro parameter character #' in math mode If $N$ varies, the power stays constant in this value for the whole process. If $N$ stays constant, the cutting speed is at its maximum only in the beginning, meaning that the calculated power is true only for that time period. ### Part e The cutting is given by:

\begin{aligned}
t & =\dfrac{L}{fN} \[1ex]
& = \dfrac{\Delta D}{2fN}
\end{aligned}

t=\int_{D_{\text{min}}}^{D_{\text{max}}} \dfrac{1}{2fN} , \mathrm{d}D

\begin{gathered}
t =\dfrac{\pu{40mm}}{2\cdot \pu{0.2mm}\cdot \pu{1326.3rpm}} \[1ex]
\boxed {
t =\pu {4.524s }
}
\end{gathered}

\begin{aligned}
t & =\int_{D_{\min_{}}}^{D_{\max_{}}} \dfrac{1}{2fN} , \mathrm{d}D \[1ex]
& =\int_{D_{\min_{}}}^{D_{\max_{}}} \dfrac{\pi D}{2fV_{c}} , \mathrm{d}D \[1ex]
& =\dfrac{\pi}{2fV_{c}}\left[ \dfrac{1}{2}D^{2} \right]{D{\min_{}}}^{D_{\max_{}}} \[1ex]
& = \pu{16mm.min^{-1}.m^{-1}}
\end{aligned}

\boxed {
t=\pu {3.016s }
}

You can't use 'macro parameter character #' in math mode ## Question 2 ### Part a The angle between each tooth is:

\begin{gathered}
\beta=\dfrac{360^{\circ}}{z} \[1ex]
\boxed {
\beta=45^{\circ}
}
\end{gathered}

You can't use 'macro parameter character #' in math mode ### Part b ![[MNF1_HW002 Homework 2 2024-08-05 15.27.31.excalidraw.svg]] >Definition of $\alpha$ - the total arc angle that the tooth is in the material Some trigonometry: ![[MNF1_HW002 Homework 2 2024-08-05 15.30.35.excalidraw.svg]] >Geometry of the problem From the figure above, we can see that:

\begin{gathered}
\cos(180^{\circ} -\alpha)=\dfrac{w-R}{R} \[1ex]
180^{\circ} -\alpha=45.573^{\circ} \[1ex]
\boxed {
\alpha=134.427^{\circ}
}
\end{gathered}

You can't use 'macro parameter character #' in math mode ### Part c The number of teeth is given by:

\begin{gathered}
n=\dfrac{\alpha}{\beta} \[1ex]
\boxed {
n\approx 3
}
\end{gathered}

You can't use 'macro parameter character #' in math mode ### Part d The [[MNF1_004 Machining Processes#Milling#Cutting Force and Power|cutting force]] is *per tooth* is given by:

F_{c}=K_{s}f_{t}d\cos\theta

F_{c}=K_{s}f_{t}d(\cos\theta+\cos(\theta-\beta)+\cos(\theta-2\beta))

-\dfrac{\pi}{2}\leq \theta,\theta-\beta,\theta-2\beta\leq -\dfrac{\pi}{2}+\alpha

0\leq \theta\leq \dfrac{\pi}{4}

\boxed {
F_{c}=\pu {1892.74N }
}

You can't use 'macro parameter character #' in math mode ### Part e We can see that the minimum will be at $\theta=0$. Substituting the given parameters, we get:

\boxed {
F_{c}=1338.37N
}

You can't use 'macro parameter character #' in math mode ### Part f The average force is given by:

\begin{aligned}
F_{c}=K_{s}wd
\end{aligned}

\boxed{F_{c}=\pu {133.28kN }}

You can't use 'macro parameter character #' in math mode ### Part g The [[MNF1_004 Machining Processes#Milling#Cutting Force and Power|average power]] would be:

\begin{aligned}
P_{c} & =F_{c}v \[1ex]
& =F_{c}f_{t}zN \[1ex]
& =F_{c}f_{t}z \dfrac{V_{c}}{\pi D}
\end{aligned}

\boxed {
P_{c}=\pu{2.83kW}
}

You can't use 'macro parameter character #' in math mode ### Part h The [[MNF1_004 Machining Processes#Milling#Key Parameters and Equations|total milling time]] (if $L=\pu{200mm}$) is given by:

\begin{aligned}
t & =\dfrac{L+D}{v} \[1ex]
& =\dfrac{L+D}{f_{t}z (V_{c}/\pi D)}
\end{aligned}

\boxed{t=\pu {10.37s } }