Quiz
P2 L1
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Questions in this quiz (33)
1What's this?
- ALathe
2What's this?
- ACNC lathe
3What's this?
- AMilling machine
4Name the common machining processes
- AStraight turning
- BSlab milling
- CEnd milling
- DCutting off
5Name the model
- AMerchant model
6The Merchant model is known as [blank_start]orthogonal cutting[blank_end].
1. It is [blank_start]two[blank_end]-dimensional and the forces involved are perpendicular to each other.
2. The cutting tool has a [blank_start]rake[blank_end] angle of α and a [blank_start]relief[blank_end] or [blank_start]clearance[blank_end] angle.
3. Shearing takes place in a [blank_start]shear zone[blank_end] at [blank_start]shear[blank_end] angle Φ- Aorthogonal cutting
- Btwo
- Crake
- Drelief
- Eclearance
- Fshear zone
- Gshear
7Label the Merchant model: words are purple, variables are pink
- AChip thickness
- BShear angle
- CWorkpiece
- DClearance angle
- EDepth of cut
- FVelocity
- GChip
- HRake angle
- Ip
- Ja
- Kto
- Ltc
- MV
8State the difference between these two models of metal cutting
- AShear plane well-defined
- BShear plane not well-defined
- CShear plane well-defined
- DShear plane not well-defined
9Major [blank_start]independent[blank_end] variables in the cutting process include:
1. Tool [blank_start]material[blank_end] and coatings;
2. Tool [blank_start]shape[blank_end], surface finish, and [blank_start]sharpness[blank_end]
3. Workpiece material and [blank_start]condition[blank_end]
4. Cutting [blank_start]speed[blank_end], feed and [blank_start]depth[blank_end] of cut
5. Cutting [blank_start]fluids[blank_end]
6. [blank_start]Characteristics[blank_end] of machine tool
7. Work [blank_start]holding[blank_end] and [blank_start]fixturing[blank_end]- Aindependent
- Bmaterial
- Cshape
- Dsharpness
- Econdition
- Fspeed
- Gdepth
- Hfluids
- ICharacteristics
- Jholding
- Kfixturing
10[blank_start]Dependent[blank_end] variables in the cutting process include:
1. Type of [blank_start]chip[blank_end] produced
2. [blank_start]Force[blank_end] and [blank_start]energy[blank_end] dissipated during cutting
3. [blank_start]Temperature[blank_end] rise in the workpiece, the tool and the chip
4. [blank_start]Tool[blank_end] wear and failure
5. [blank_start]Surface finish[blank_end] and surface integrity of the workpiece- ADependent
- Bchip
- Cenergy
- DForce
- ETemperature
- FTool
- GSurface finish
11What does this diagram depict?
- AChip formation
- BVelocity diagram of the cutting zone
12Label the velocity diagram for the cutting zone where purple boxes are velocities and pink boxes are angles (note that p = φ, a = α)
- A90 - p + a
- BVs
- Cp
- DV
- E90 - a
- Fa
- Gp - a
- HVc
13What is this equation for?
- Acutting ratio
14The cutting ratio is related to the two angles by:
r = [blank_start]to / tc[blank_end] = sin p / cos (p - a)
- As the chip thickness is always [blank_start]greater[blank_end] than the depth of cut, the value of r is always [blank_start]less[blank_end] than unity
- The [blank_start]reciprocal[blank_end] of r is known as the [blank_start]chip compression ratio[blank_end]
- This is a measure of how [blank_start]thick[blank_end] the chip has become- Ato / tc
- Bgreater
- Cless
- Dreciprocal
- Echip compression ratio
- Fthick
15What does this expression represent?
- Ashear strain
16The [blank_start]shear strain[blank_end] that the material undergoes can be expressed as
γ = AB/OC = AO/OC + OB/OC
γ = [blank_start]cot[blank_end]Φ + [blank_start]tan[blank_end](Φ - α)
Large shear strains are associated with low [blank_start]shear[blank_end] angles or low or negative [blank_start]rake[blank_end] angles.
Based on the assumption that the shear angle adjust itself to [blank_start]minimise[blank_end] the cutting force,
Φ = 45 + [blank_start]α[blank_end] - [blank_start]β[blank_end] (when [blank_start]μ[blank_end] = 0.5 - 2)
Note that
- β = friction [blank_start]angle[blank_end]
- μ = tanβ = [blank_start]coefficient[blank_end] of friction- Ashear strain
- Bcot
- Ctan
- Dshear
- Erake
- Fminimise
- Gα
- Hβ
- Iμ
- Jangle
- Kcoefficient
17Name these here chips
- AContinuous chip
- BSecondary shear zone
- Cbuilt up edge chip
- DSegmented chip
- EDiscontinuous chip
18Continuous chips
- Formed with [blank_start]ductile[blank_end] materials machined at [blank_start]high[blank_end] cutting speeds and/or [blank_start]high[blank_end] [blank_start]rake[blank_end] angles
- Deformation takes place along a narrow shear zone called the [blank_start]primary[blank_end] shear zone
- High friction at the [blank_start]tool chip interface[blank_end] may cause a [blank_start]secondary[blank_end] shear zone to develop.- Aductile
- Bhigh
- Chigh
- Drake
- Eprimary
- Ftool chip interface
- Gsecondary
19Built up edge chips
- Consist of layers of material from the [blank_start]workpiece[blank_end] that are deposited on the tool tip
- As it grows larger, the BUE becomes [blank_start]unstable[blank_end] and eventually [blank_start]breaks apart[blank_end]
- BUE can be reduced by:
- Increasing the [blank_start]cutting[blank_end] speeds
- Decreasing the [blank_start]depth of cut[blank_end]
- Increasing the [blank_start]rake[blank_end] angle
- Using a [blank_start]sharper[blank_end] tool
- Using an effective cutting [blank_start]fluid[blank_end]
- Using a cutting tool that has a lower [blank_start]chemical affinity[blank_end] for the workpiece material- Aunstable
- Bworkpiece
- Cbreaks apart
- Dcutting
- Edepth of cut
- Frake
- Gsharper
- Hfluid
- Ichemical affinity
20Serrated chips
- Also called [blank_start]segmented[blank_end] or nonhomogenous chips
- They are [blank_start]semicontinuous[blank_end] chips with large zones of low shear strain and small zones of high shear strain (shear [blank_start]localization[blank_end])
- Chips have a [blank_start]sawtooth[blank_end] appearance- Asegmented
- Bsemicontinuous
- Clocalization
- Dsawtooth
21Discontinuous chips
Consist of [blank_start]segments[blank_end] that are attached firmly or loosely to each other
Form under the following conditions:
- [blank_start]Brittle[blank_end] workpiece materials
- Materials with hard [blank_start]inclusions[blank_end] and impurities
- Very low or very high [blank_start]cutting speeds[blank_end]
- Large [blank_start]depths of cut[blank_end]
- Low [blank_start]rake[blank_end] angles
- Lack of an effective [blank_start]cutting fluid[blank_end]
- Low [blank_start]stiffness[blank_end] of the machine tool- Asegments
- BBrittle
- Cinclusions
- Dcutting speeds
- Edepths of cut
- Frake
- Gcutting fluid
- Hstiffness
22In the images below, depicting the cutting of mild steel,
a) depicts a discontinuous chip;
b) depicts a chip with BUE;
c) depicts a continuous chip with a well-developed secondary shear zone.
Are the tool velocities in each case low, medium, or high?
- Alow
- Bmedium
- Chigh
23What is the tool depicted below?
- Achip breaker
24Knowledge of the cutting forces and power involve:
- Data on [blank_start]cutting forces[blank_end]
- [blank_start]Power[blank_end] requirements
The cutting force acts in the direction of the [blank_start]cutting speed[blank_end], and supplies the [blank_start]energy[blank_end] required for cutting.- APower
- Bcutting forces
- Ccutting speed
- Denergy
25Label this diagram of cutting forces (pink boxes are for words, purple for variables)
- Acutting force
- BFc
- CThrust force
- DFt
- ENormal force
- FN
- GFriction force
- HF
26What do these equations represent?
- AResultant force
- BForce normal to resultant
- CShear force
- DNormal force
- ECoefficient of friction
27The tool holder, work holding devices, and machine tool must be [blank_start]stiff[blank_end] to support thrust force with minimal [blank_start]deflections[blank_end].
The effect of [blank_start]rake[blank_end] angle and [blank_start]friction[blank_end] angle on the direction of thrust force is [blank_start]Ft[blank_end] = R sin(β-α).
The magnitude of the [blank_start]cutting force[blank_end] is always positive as the force that supplies the work is required in cutting.- Astiff
- Bdeflections
- Crake
- Dfriction
- EFt
- Fcutting force
28The power input in cutting is:
-Power = [blank_start]FcV[blank_end].
The power dissipated in the shear plane is:
-Power for shearing = [blank_start]Fs[blank_end]Vs
Denoting the width of cut as w, the specific energy for shearing is
-μs = [blank_start]FsVs[blank_end] / wtoV
The power dissipated in friction is:
-Power for friction = [blank_start]FVc[blank_end]
The specific energy for friction is:
-μf = [blank_start]FVc[blank_end]/wtoV = [blank_start]Fr[blank_end]/wt0
The total specific energy is:
- μt = μs + μf
The sharpness of the tool tip also influences forces and power.- AFs
- BFsVs
- CFVc
- DFcV
- EFVc
- FFr
29In cutting, nearly all of the energy dissipated in [blank_start]plastic deformation[blank_end] is converted into [blank_start]heat[blank_end] that in turn raises the temperature in the cutting zone. Since the heat generation is closely related to the plastic deformation and [blank_start]friction[blank_end], we can specify three main sources of heat when cutting.
1. Plastic deformation by [blank_start]shearing[blank_end] in the [blank_start]primary shear zone[blank_end]
2. Plastic deformation by shearing and [blank_start]friction[blank_end] on the [blank_start]cutting face[blank_end]
3. [blank_start]Friction[blank_end] between chip and tool on the [blank_start]tool flank[blank_end]
Heat is mostly dissipated by:
1) The [blank_start]discarded chip[blank_end] carries away about 60-80% of the total heat
2) The [blank_start]workpiece[blank_end] acts as a heat sink, drawing away 10-20% heat
3) The [blank_start]cutting tool[blank_end] will also draw away ~10% heat- Aplastic deformation
- Bheat
- Cfriction
- Dshearing
- Eprimary shear zone
- Ffriction
- Gcutting face
- HFriction
- Itool flank
- Jdiscarded chip
- Kworkpiece
- Lcutting tool
30In cutting, nearly all of the energy dissipated in plastic deformation is converted into heat that in turn raises the temperature in the cutting zone.
Since the heat generation is closely related to the plastic deformation and friction, we can specify three main sources of heat when cutting. Match these to Q1, Q2 and Q3.
a) Plastic deformation by shearing and friction on the cutting face
b) Plastic deformation by shearing in the primary shear zone
c) Friction between chip and tool on the tool flank
Heat is mostly dissipated through three routes. Match these to q1, q2 and q3.
x) The cutting tool will draw away ~10% of heat
y) The workpiece acts as a heat sink, drawing away 10-20% heat
z) The discarded chip carries away about 60-80% of the total heat.
- Ax
- By
- Cz
- Dx
- Ey
- Fz
- Gx
- Hy
- Iz
- Ja
- Kb
- Lc
- Ma
- Nb
- Oc
- Pa
- Qb
- Rc
31High cutting temperatures have several negative effects:
- Affect the [blank_start]wear[blank_end] of the cutting tool. Cutting temperature is the [blank_start]primary factor[blank_end] affecting tool [blank_start]wear[blank_end].
- Can induce [blank_start]thermal damage[blank_end] to the machined surface.
- Causes [blank_start]dimensional errors[blank_end] in the machined surface.
Any reduction in the cutting temperature will require substantial reduction in either the cutting [blank_start]speed[blank_end] or the cutting [blank_start]feed[blank_end], or both. As a negative result, [blank_start]cutting time[blank_end] and therefore [blank_start]production rate[blank_end] will also decrease.
To avoid these problems, the best solution to the problem of cutting temperature is using [blank_start]coolants[blank_end].- Awear
- Bprimary factor
- Cwear
- Dthermal damage
- Edimensional errors
- Fspeed
- Gfeed
- Hcutting time
- Iproduction rate
- Jcoolants
32Name the operation
- Amilling
33In milling,
[blank_start]V = πDN[blank_end]
- V = [blank_start]peripheral[blank_end] speed of the cutter
- D = cutter outer [blank_start]diameter[blank_end]
- N = [blank_start]rotational[blank_end] speed of cutter
Three types of feed in milling can be identified:
Feed per tooth [blank_start]fz[blank_end]: the basic parameter in milling equivalent to the feed in turning:
- Selected with regard to the surface finish and dimensional accuracy required.
- In the range of 0.05-0.5mm/tooth
Feed per revolution [blank_start]fr[blank_end]: Determines the amount of material cut per one full revolution of the milling cutter
- Calculated as fr~fz.z
- z is the number of the cutter's teeth
Feed per minute [blank_start]fm[blank_end]:
- Calculated taking into account the rotational speed N and number of cutter's teeth
- fm = fz.zN = frN- AV = πDN
- Bperipheral
- Cdiameter
- Drotational
- Efz
- Ffr
- Gfm
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- Quiz on P2 L1, created by Lizzie O on 22/03/2017.
- Author
- Lizzie O
- Language
- English
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