University of Saskatchewan
College of Engineering
Dept. of Mech. Engineering
ME 330.3 Final Exam Solution
April 16, 2014
Time: 2 Hours Instructors: Chris Zhang
Open Book Exam.
Name: ______
Student Number: ______
This exam question set consists of two parts. Part I consists of 10 questions. Part II consists of 3 questions. There is an extra question asked for a peer evaluation of your group members.
Attempt all questions.
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Peer Evaluation
Group member (name) / Score (1-5) (1: poor; 5: excellent)Part I (40 marks)
Each sub-question is given 4 marks.
Question I.1 [2 marks]
Define the process parameter and system parameter for any manufacturing process. List all the process parameters for the water-jet process.
Answer:
Standoff distance, nozzle opening diameter, water pressure, and cutting feed rate.
Question I.2 [2 marks]
Suppose a non-round hole of hard metal is to be fabricated. What manufacturing process may be chosen to make the hole? Why?
Answer: ECM (text: p 686). (1) material is hard, (2) non-round hole. EDM is also possible. Depending on the equipment and production requirement, the ECM is slower than EDM in terms of material removal.
Question I.3 [2 marks]
Take it a reference the diagram of classification of manufacturing processes (see slide 3 in the revision slide file of the course). Please explain which category of process the rapid prototyping technology should fall into or fall into none of them, and why.
Answer: none of them. There should be a category of manufacturing process called “deposit” (see the attachment A at the end).
Question I.4 [2 marks]
Suppose there is a mold made of a material that is very hard but the material is a conductive material. The mold has a very complex interior feature. What manufacturing process do you recommend and why?
Answer: EDM or ECM but EDM is better due to the need to cut into a complex interior feature. If ECM is used, the electrode is difficult to be made.
Question I.5 [2 marks]
List two roles of electrolyte in ECM and explain why we need to add some chemicals (e.g., salt) to reduce the resistance of electrolyte.
Answer: 1. Carry off the depleted material and 2. Remove heat and bubble. Reduce the resistivity of the electrolyte, as small resistance between the anode and cathode will be conducive to the depletion process.
Question I.6 [2 marks]
In EDM, why is the fluid flowing between the work and tool must be dielectric fluid?
Answer: there is a need to separate the work and tool in terms of electric path. The dielectric fluid becomes a larger between the work and tool, which essentially cuts off any electric path between the work and tool.
Question I.7 [4 marks]
List two principles that two parts can be welded. Discuss pros and cons with the principles (in particular list one advantage and one disadvantage for each principle).
Answer: (1) on the interface site, two parts are in the molten state, which is also called “fusion”, and (2) two parts on the interface are not melted, which is also called “solid state”. For fusion, the advantage is that the strength of weld on the interface is higher and closer to the parts welded, but the disadvantage is that there may be more change to have defects. For solid state, the strength of the weld is lower and more dissimilar to the parts welded, but there are few defects.
Question I.8 [4 marks]
List all the information you can extract from the notation of 50 F7/h8 on an engineering drawing.
Answer: (1) the nominal dimension (50). (2) basic shaft system (h). (3) manufacturing accuracy required (the number 7, 8). (4) type of fit (F, h). (5) tolerance for the hole and shaft (50 F7, 50h8).
Question I.9 [4 marks]
Explain the most important role of the riser in the casting process. However, is there any need the riser in the investment casting process? Why yes or no?
Answer: the most important role of the riser is to overcome gaps owing to shrinkage during the liquid state of material changes to the solid state in a room temperature. For investment casting, we still need a riser.
Question I.10 [4 marks]
Explain the difference between quality inspection and quality management.
Answer: Quality inspection is taken place on a particular product with a tool or manually. Quality management is taken place at the enterprise level. Quality management refers to the organizational behavior regarding the quality, involving their awareness, planning, coordinating and controlling.
Question I.11 [4 marks]
Explain the most unique feature with the powder metallurgy process, which can never be achieved by any other manufacturing process discussed in the class. Please list two applications of this unique feature in making products.
Answer: pores.
Question I.12 [8 marks]
Explain the failure criterion of the rolling process and failure criterion of the drawing process. Explain their difference in failure criterion and how the difference is related to the working principle of the two processes, respectively.
Answer: The failure criterion for the rolling includes: (1) the draft is too large, larger than the critical draft which is determined by the friction and radius of the roll and (2) over the power capacity limit. The failure criterion for the drawing includes: (1) the maximum compression stress in the work is over the yield and (2) over the power capacity limit.
It is shown that the second failure criteria for both the processes are the same. The difference in the first failure criteria for the two processes, respectively, is related to their work principle, respectively. The work principle of rolling process is the friction-driving, while the work principle of drawing process is the pulling force-driving.
Part II (60 marks)
Question II.1 (25 marks)
A casting product is a 2 in×4 in×6 in rectangular plate. Suppose that n=2 in the Chvorinov equation for TST. The riser takes the cylindrical form. TST of the casting is 3.5 minutes, and the solidification time of the riser is 25% longer than that of the casting. Please do the following:
(a) Suppose that the riser is put on the top of the casting and is open to the air (see Figure 1 b). Determine the diameter of the riser (assume: the height of the cape or upper box of the mold is 5 in). [10 marks]
(b) Suppose that the riser is put aside the casting and is blind in the mold (see Figure 1a). Determine the optimal geometry of the riser (its diameter and height). The optimal means that the volume of the riser is a minimum. [10 marks]
(c) Compare the volume of the riser for (a) and (b) and comment on pros and cons of the type of the riser. [5 marks]
Answer: see the attachment B.
Question II.2 (35 marks)
A hot rolling mill has rolls with its diameter being 24 in. It can exert a maximum force equal to 400,000 lb. The mill has a maximum horsepower of 100 hp (39´106 in lb/min). It is desired to reduce a 1.5 in.-thick plate by the maximum draft in one pass. The width of the plate is initially 10 in. In the heated condition, the work material has a strength coefficient of 20,000 lb/in2 and a strain hardening exponent of 0. Furthermore, the friction coefficient between the work and the roll is 0.15. Please do the following:
(a) Determine the maximum draft and the final thickness of the plate associated with the maximum draft. [10 marks]
(b) Determine the maximum speed of the rolls (ft/min). [5 marks]
(c) Suppose it is possible to increase the friction coefficient between the roll and work to 0.20. Please analyze and explain whether the maximum draft of (a) may or may not be changed. [5 marks]
(d) Suppose that the required reduction of the thickness of the work is beyond the capacity of this one-pass rolling system (e.g., the required final thickness is 1.0 in). One solution is with a two-pass rolling with each pass having the same capacity as the one described in the above. Please outline the design procedure to determine the optimal allocation of the draft in the first pass and in the second pass (respectively). The optimal criterion is: the same exit speed for the first pass and second pass. (No number calculation is required). [15 marks]
Answer: attachment C
- THE END –
Attachment A:
Attachment B
Solution to Part II – Question 1
Question 1:
1(a):
Casting:
V=2(4)(6)=48
A=2(2x4+2x6+6x4)=88
Note: c: cast; r=riser; n=2.
TSTc = 3.5 min, so
Note: Let H=height of riser = 5 . D=diameter of the riser.
(1)
Note: A: surface area of the riser. The surface area of the riser includes three parts: the side surface area (A1), the top surface area (AT) and the bottom surface area (AB).
(2)
AB=0 (as it is connected with the casting).
AT is assumed to be 2 times the round area (). So AT= (3)
Note: TSTr=1.25(TSTc)=4.375 min.
According to the riser, we have
. This leads to
(4)
From Equation (4), we get D=3.244.
It is noted that in the above calculation, the top surface area of the riser may be assumed differently, which leaves some open-ended situations. Whatever assumption is, the top surface area cannot be assumed zero, as the zero surface area means that this surface does not play any role in the dissipation of heat through the surface (which is the case of the bottom surface, indeed).
1(b):
In this case, the riser is blind to the air and aside the casting. Therefore, the surface area of the riser is calculated by
(5)
By applying the equation of , we get the following equation:
(6)
Further, we have for the riser
(7)
Substituting (6) into (7), we get the equation for the volume of the riser in terms of the diameter of the riser. We will then differentiate the volume of the riser with respect to the diameter of the riser and let the derivative be zero. We then find the diameter to make the volume of the riser a minimum, that is,
.
We can subsequently find Hr from Equation (6), which is the same as .
In fact, we can generally prove that Dr=Hr in this case regardless of what casting is. This can be done by starting with the equation below.
By putting the expression of Ar and Vr to the above equation, leading to the equation that expresses the relation between Hr and Dr. From that equation, express Hr in terms of Dr. Then, making the derivative of Vr with respect to Dr from Equation (7) be zero, we can obtain:
Then from the aforementioned relation between Hr and Dr, we can show Hr=Dr=.
1(c):
For the case (a), the volume of the riser is 41.305 , and for case (b), the volume of the riser is 38.455 . Therefore, the volume of case (a) is larger than the volume of case (b). This result is reasonable, as in case (a), the heat dissipation rate of the fluid through the riser is higher than that in case (b).
The attributes describing the riser herein are: (1) the location of the riser (on the top of the casting or aside the casting) and (2) blind or open.
The open riser has an advantage of easy manipulation and control, but the heat dissipation rate is very high, which thus increases the volume of the riser.
Putting the riser on the top of the casting has an advantage of slow heat dissipation rate of the riser, so the volume of the riser tends to be small. The shortcoming is that in the area of the casting, which interacts with the riser, the surface quality is poor, and there is a need of secondary process on this area.
Attachment C
Solution to Part II - Question 2:
2(a):
From the friction point of view, we get
From the point of view of power, we get
The maximum draft must the smaller one about the two. Therefore, the maximum draft is 0.27.
To find , the following procedure is followed.
First, from the flow stress equation, we get average flow stress 2000.
Second, from the equation between F and L, i.e., , we get L=2.
Third, from the equation of , we get .
2(b):
The key is to determine F. Use of the following equation
.
=1.8.
P= , from this equation, we get N=9.5788 rev per min =60.2 ft /min2.
2(c):
According to the friction governing equation, we get
. However, with this draft, the power is not enough (with the force of 400000).
Therefore, the maximum draft is 0.33.
2(d):
According to the optimal criterion, N1 = N2 (1: pass 1; 2: pass 2). In the following, we assume that the widths for pass 1 and pass 2 are the same and we further assume that the draft for each pass is less than 0.27 (which means the friction plays a dominant role in this case).
For pass 1,
From the above equation, we get
Notice
(because n=0)