2.810 Manufacturing Processes and Systems Quiz #1 Solutions

October 19, 2015

90 minutes

Open book, open notes, calculators, computers with internet off. Please present your work clearly and state all assumptions.

Problems: 1. How were these parts made? 2. Viscoelastic behavior of medieval glass. 3. Maximum extrusion rate for FDM. 4. Sheet metal springback. 5. Maximum material removal rate.

/ 20 points / 15 points / 20 points / 25 points / 20 points __________

/ 100 points

Quiz 1 Solutions

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Problem 1. How were these parts made? (20 points) Please examine the four parts you have been given (due to limited quantities, you will need to share the parts with other students, so please be considerate). For each part, please write down your observations on:

a) what material(s) the part is made of; b) the process(es) with which is was made (if multiple processes were used to make the part, try to identify all of them).

Please explain the reasoning behind all of your answers. Answers without an explanation will not receive full credit. 1. Pipe fitting

Material(s): Cast iron Reasons: Color and weight of the material, low cost. Process(es): Sand casting, machining to cut threads. Reasons: Sand casting is identifiable by the surface roughness, and a parting line is visible. Cast iron parts are often made by sand casting, since sand is able to withstand the high temperatures needed to melt cast iron (~1500℃). Machining was used to create the interior threads.

2. Window screen clip

Material(s): Aluminum Reasons: Color, extremely lightweight, easily formable. Process(es): Extrusion followed by shearing (punching and blanking) Reasons: Extrusion is an economical way to make a part with a constant cross section like this one. The extruded strip was then punched and blanked to form individual parts. Shearing marks are clearly visible on all the outside edges of the part and the hole.

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3. Bushing

Material(s): Bronze Reasons: Color, material often used in manufacturing of bearings. Process(es): Sintering (powder metallurgy) Reasons: The surface is smooth but has visible porosity. Sintering is commonly used for bushings in order to create pores that can be impregnated with lubricant. The end surfaces do not look cut off, as they would if the tube had been extruded.

4. Screw anchor (half)

Material(s): Zinc Reasons: Color, heavier than aluminum, material commonly used for die casting due to low melting point. Process(es): Die casting Reasons: Surface is smooth and has fine features. The shape lends itself well to a two-sided die. There is a visible parting line.

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Problem 2. Viscoelastic behavior of medieval glass. (15 points) It has been observed that the glass in the windows of ancient cathedrals in Europe is thicker on the bottom than on the top. Some have speculated that this difference in thickness is due to gravity and the viscous flow of glass over the approximately 800 years they have existed. Using the hypothetical data and trend lines from the figure below, please use your knowledge of viscoelastic phenomena to either support or reject this claim. Note: The trend in the figure is hypothetical, so you should not use any knowledge you may already have about this problem to influence your answer. There is, however, a calculation you have learned that could help resolve this issue.

The concept of viscoelasticity was described in the lecture on thermoforming. In polymers (and glass), temperature affects both the magnitude of the viscosity µ (strongly affected) and the elastic modulus E (only modestly affected). The viscoelastic time constant 𝜆, derived in the simple viscoelastic system model, is an indication of which of the two behaviors is dominant:

𝜆 =𝜇𝐸=

𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦𝑒𝑙𝑎𝑠𝑡𝑖𝑐  𝑚𝑜𝑑𝑢𝑙𝑢𝑠

The solution to the viscoelastic equation of motion (see Slide 12 from Thermoforming lecture) includes the term t/𝜆, i.e. time with respect to the viscoelastic time constant. For very fast times (t > 𝜆), it is viscous. To answer this question, we can compare the length of time this medieval glass has been installed in cathedrals (800 years) to the viscoelastic constant of medieval glass. Since the time constant depends on temperature, we need to assume an average temperature at which the glass was maintained. Assuming typical room temperature (25℃), we can read the viscosity of medieval glass (as a function of temperature) straight from the plot above:

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log 𝜇 25℃ ≈ 75 𝜇 25℃ = 10!"  𝑃𝑎 ∙ 𝑠 The elastic modulus for glass can be estimated from the table of mechanical properties in the textbook (Table 2.2) as 70-80 GPa. (The elastic modulus does not change much until the glass transition temperature, which for typical glass is in the range of 600℃.) Then the viscoelastic constant for medieval glass at room temperature is:

𝜆 =𝜇𝐸=10!"  𝑃𝑎 ∙ 𝑠80×10!  𝑃𝑎

=  10!"

8  𝑠 = 1.25×10!"  𝑠 = 1.45×10!"  𝑑𝑎𝑦𝑠 =  3.96×10!"  𝑦𝑒𝑎𝑟𝑠

Compared to this time constant, the time period of 800 years is negligible (800 years

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Problem 3. Maximum extrusion rate for Fused Deposition Modeling (FDM). (20 points) Slide 29 from John Hart's presentation, reproduced below, shows the nozzle for the so-called Fused Deposition Modeling (FDM) Additive Manufacturing process for thermoplastics. An ABS filament approximately 1 mm in diameter is pushed through the heating zone (Zone I) where it melts, and is then reduced in area for printing. We are concerned with the maximum rate at which plastic can be pushed through this zone and still be guaranteed that the output is fully melted (at the point where it enters Zone II). Referring to the figure, assume the diameter of the heating zone is dl = 1.0 mm and its length is ll = 1.25 cm. Its walls are maintained at 260°C with an electric resistance heater. Please make an estimate for the maximum rate at which the ABS filament can be fed into the heating zone (in meters/second). Note: The chart from the textbook by Lienhard for conductive heat transfer in cylinders is given for reference.

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Transient temperature distributions in a long cylinder of radius ro at three positions. r/ro = 0 is the centerline. (Lienhard, Fig 5.8)

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This problem can be approached by considering the ABS filament as a cylinder that’s being heated by the nozzle walls that are held at constant temperature. Initially, the filament is at room temperature (Ti =  25℃). The walls remain at Tw =  𝑇! = 260℃.  In order to ensure that the ABS melts before it enters the transition zone, we can look at the time needed for the center of the filament to reach melting temperature. Because ABS is an amorphous polymer, it does not have a well-defined melting point where it changes phase and thus no latent heat of fusion; instead, it changes from solid to rubbery after it passes the glass transition temperature and then to a fluid with decreasing viscosity as it rises beyond the melting point. From John Hart’s other slides on Fused Deposition Modeling and from Boothroyd’s Ch. 8, Table 8.5 on injection molding parameters, the recommended forming temperature for ABS is in the range of 240-260℃. (This solution assumes T = 250℃; other 𝜃’s in the range 0-0.1 are OK.) The transient temperature distribution chart provided above provides a graphical way to estimate the time needed to reach a given temperature at a given radial position within the cylinder. We need to calculate the dimensionless temperature (θ) and Biot number (Bi), and use the top chart (for r0/r = 0, since we are looking at the center of the cylinder) to find the Fourier number (Fo). Note that the charts are provided for a body that’s being cooled; in order to adapt them to a body being heated, we can just change the signs in the temperature calculation so that θ remains between 0 and 1:

𝜃 =𝑇! − 𝑇𝑇! − 𝑇!

=260 − 𝑇260 − 25

=260 − 250

235= 0.043

For the inverse Biot number, Bi-1 = k/hr0, knowing that thermal conductivity k is very low (~0.2 W/mK for polymers), we can assume that Bi-1 ~ 0. At the intersection of this θ and Bi-1, the Fourier number on the x-axis is approximately 𝐹𝑜 = 0.45, which we can use to calculate time (using typical thermal diffusivity 𝛼 for polymers):

𝑡 =𝑟!!𝐹𝑜𝛼

=0.05  𝑐𝑚 !(0.45)

10!! !!!

!

= 1.1  𝑠

Thus the filament needs to spend a minimum of 1.1 seconds inside the heating element. Then the maximum feed rate can be calculated from (length of heating zone/time):

𝑣 =𝑙𝑡=1.25  𝑐𝑚1.1𝑠

= 1.13𝑐𝑚𝑠= 0.01

𝑚𝑠

Alternatively: The simplest way to solve this problem is by analogy with our one-dimensional analysis for injection molding. The same equation applies for heating and for cooling. Remember

we got that !!~ !

!!

! and that 𝑡 =!!

!

! was the exact solution when 𝜃 = 0.1  for constant wall

temperature and Bi-1 = 0 (Injection Molding lecture, Slide 14). We have a very similar situation here, except that the geometry is now a cylinder.

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Problem 4. Sheet metal springback. (25 points) An engineer is building a bicycle frame from titanium (elastic modulus E = 103 GPa, yield strength Y = 550 MPa) and needs to bend a 1.5 mm thick strip of titanium so that the resulting angle is θ = 60 degrees, as shown in the figure. The length of the arc formed by the bend must be s = 6 cm. He knows he needs to overbend the strip in order to compensate for springback. Please approximate the angle to which he should bend it initially.

First we need to identify the angle subtended by arc length s. Since the two radii forming that angle are perpendicular to the sides, and the third angle is 60°, the arc angle (call it α) must be equal to: α = 360 – 180 – 60 = 120° = 2.094 radians

From the geometric relationship between arc length, angle, and radius, 𝑠 = 𝛼!𝑅!. Therefore,

𝑅! =𝑠𝛼!

=6  𝑐𝑚2.094

= 2.87  𝑐𝑚  

This is the desired bend radius (𝑅!). We know that some amount of springback will occur after bending, so we need to overbend the sheet in order to end up with this radius. From the springback equation (in the form seen in the sheet forming lecture slides): 1𝑅!−1𝑅!

= 3𝑌ℎ𝐸

−   4𝑅!!𝑌ℎ𝐸

!

1𝑅!

= −3𝑌ℎ𝐸

+   4𝑅!!𝑌ℎ𝐸

!+1𝑅!

θ = 60°  

s  h = 1.5 mm  

120°  

60°  

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Plugging in known values: 1

0.0287= −3

550×10!  𝑃𝑎(1.5×10!!𝑚)(103×10!𝑃𝑎)

+   4𝑅!!550×10!  𝑃𝑎

(1.5×10!!𝑚)(103×10!𝑃𝑎)

!

+1𝑅!

34.84 = −3 3.56 +   4𝑅!! 3.56 ! +1𝑅!

34.84 = −10.68  +  180.45𝑅!!   +

1𝑅!

180.45𝑅!!   +

1𝑅!− 45.52 = 0

180.45𝑅!!   − 45.52𝑅! + 1 = 0

This is a cubic equation that can be solved via graphing, MATLAB, etc. The only (feasible) solution to the resulting equation is: 𝑅! = 0.0220  𝑚 = 2.2  𝑐𝑚 Alternatively, you could notice that the cubic term is negligible and ignore it in order to solve the equation more easily. (You know that 𝑅! should be pretty close to 𝑅!  so it is under 0.03 meters, which, when raised to the 3rd power, is on the order of 2.7×10!!.) Neglecting that term, the answer is similar:

𝑅! =1

45.52= 0.02197 ≈ 0.022  𝑚 = 2.2  𝑐𝑚  

Assuming the length of the arc stays the same, we can calculate the subtended angle using this initial radius:

𝛼! =𝑠𝑅!=

6  𝑐𝑚2.2  𝑐𝑚

= 2.727  𝑟𝑎𝑑𝑖𝑎𝑛𝑠 = 156°

Therefore, the angle between the two straight sides when bending should initially be: 𝜃! = 360 − 180 − 156 = 24°

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Problem 5. Maximum material removal rate. (20 points) Consider a very large Invar tool being built for curing a graphite epoxy composite (Invar is a high temperature nickel alloy). To make this tool, approximately 3 x 106 cm3 of this material needs to be removed by milling using a single 25 kW spindle. (a) With the given spindle power limitations, estimate the shortest cutting time possible to remove this material. Referring to Table 21.2 in the textbook, the specific energy required to machine nickel alloys is in the range of 4.8 - 6.7 Ws/mm3. Since we are estimating the shortest cutting time, we can take a value at the low end of that range, such as us = 5 Ws/mm3. (Note: Using other values in that range is OK.) The material removal rate (MRR), limited by the available spindle power, can then be calculated as:

𝑀𝑅𝑅 =𝑃𝑢!=25,000  𝑊

5 !"!!!

= 5000𝑚𝑚!

𝑠= 5

𝑐𝑚!

𝑠

The time required to remove 3×106 cm3 of material is:

𝑡 =𝑉𝑜𝑙𝑀𝑅𝑅

=3×10!  𝑐𝑚!

5 !!!

!

= 6×10!  𝑠  1  ℎ𝑟3600  𝑠

= 166.7  ℎ𝑟 = 6.9  𝑑𝑎𝑦𝑠

Note that this is just the machining time. Additional time would be spent on moving the spindle, changing the cutting tool, reorienting the part, etc. (b) Assume you are using a 4-tooth, 10 cm diameter milling cutter and the maximum general purpose cutting conditions for high temperature nickel alloys (from your textbook). Assuming that the depth of cut cannot exceed 6 mm, recalculate the shortest cutting time. The recommended milling settings for high-temperature nickel alloys are listed in Table 24.2 of Kalpakjian. As suggested, we take the maximum values from the general-purpose starting conditions:

Feed per tooth f = 0.18 mm/tooth Surface speed V = 370 m/min

(Alternatively, if you took the maximum values from the full range of conditions, you’d be using the higher values of f = 0.38 mm/tooth and v = 550 m/min.) We can first use these to calculate the rotational speed Ω in revolutions per minute (RPM), which is equivalent to the surface speed divided by the circumference of the tool:

Ω =𝑉𝜋𝐷

=370  𝑚/𝑚𝑖𝑛0.10𝜋  𝑚

= 1178  𝑅𝑃𝑀 Knowing the rotational speed, the number of teeth, and feed per tooth gives us the linear speed of the workpiece:

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𝑣 = 𝑓Ω𝑛 = 0.18𝑚𝑚𝑡𝑜𝑜𝑡ℎ

1178𝑟𝑒𝑣𝑚𝑖𝑛

4  𝑡𝑒𝑒𝑡ℎ = 848𝑚𝑚𝑚𝑖𝑛

= 84.8𝑐𝑚𝑚𝑖𝑛

The equation for MRR for this milling tool is then the product of linear speed, width of cut, and depth of cut. The width of cut wasn’t given but we can assume it’s equal to half of the cutter diameter:

𝑀𝑅𝑅 = 𝑣𝑤𝑑 = 84.8𝑐𝑚𝑚𝑖𝑛

5  𝑐𝑚 0.6  𝑐𝑚 = 254.4𝑐𝑚!

𝑚𝑖𝑛= 4.2

𝑐𝑚!

𝑠

This is lower than the MRR presumed in part (a) due to practical limitations on cutting conditions, so the machining time would increase to:

𝑡 =𝑉𝑜𝑙𝑀𝑅𝑅!

=3×10!  𝑐𝑚!

4.2 !!!

!

= 7.14×10!  𝑠  1  ℎ𝑟3600  𝑠

= 198  ℎ𝑟 = 8.3  𝑑𝑎𝑦𝑠

Note: If you were using the maximum values from the full range of conditions in Table 24.2, you would get Ω = 1751  𝑅𝑃𝑀, 𝑣 = 266 !"

!"#, 𝑀𝑅𝑅 = 13 !!

!

!. This is greater than the MRR from part

(a). In that case, your limiting factor would not be the cutting conditions but the spindle power, as before.