King Saud University – College of Engineering – Industrial Engineering Dept.
IE-352
Section 1, CRN: 5022/5030/5041
Section 2, CRN: 32997/32999/32998
Second Semester 1433-34 H (Spring-2013) – 4(4,1,1)
MANUFACTURING PROCESSES – 2
Wednesday, Mar 13, 2013 (01/05/1434H)
MIDTERM 1 ANSWERS [10 POINTS ]
Name:
Ahmed M. El-Sherbeeny, PhD
Student Number:
4
Section:
8:00 / 10:00
Place the correct letter in the box at the right of each question [
1
2
Point Each]
E
1.
Plastic molding is classified as what type of manufacturing process?
A. material removal – shaping – processing operation
B. deformation – shaping – processing operation
C. mechanical fastening – assembly operation
D. coating – surface – processing operation
E. solidification – shaping – processing operation ( slides 1.33, 34 , 35 )
C
2.
Press fitting is classified as what type of manufacturing process?
A. material removal – shaping – processing operation
B. deformation – shaping – processing operation
C. mechanical fastening – assembly operation ( slide 1.32, 42 )
D. coating – surface – processing operation
E. solidification – shaping – processing operation
D
3.
Material whose molecules form a rigid structure that can’t be reheated is …
A. a ceramic
B. a thermoplastic polymer
C. an elastomer
D. a thermosetting polymer ( slide 1.26
)
E. a composite
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 1

King Saud University – College of Engineering – Industrial Engineering Dept.
4.
Material consisting of a semi-metallic – nonmetallic compound (e.g. 𝑺𝒊𝑶
𝟐
) is …
A. a ceramic ( slide 1.24
)
B. a thermoplastic polymer
C. an elastomer
D. a thermosetting polymer
E. a composite
A
5.
H, J, and K represent what systems (respectively)?
A. manufacturing; quality control; production
B. production; quality control; manufacturing
( slide 1.54
)
C. quality control; production; manufacturing
D. production; manufacturing; quality control
E. manufacturing; production; quality control
K
H
J
B
6.
Respectively, the fixed scales on the micrometer and Vernier caliper are called:
A. sleeve; Vernier scale
B. main scale; Vernier scale
C. main scale; sleeve
D
D. sleeve; main scale ( micrometer: slides 2.8, 12; Vernier caliper: slide 2.32
)
E. thimble; Vernier scale
7.
A snap gage (as the one shown below) is used to check the size of a shaft of nominal size and tolerance: 𝟒
𝟏
𝟒
𝒎𝒎 ± 𝟐% .
C
Respectively, determine the go, no-go, and an acceptable size [ 𝒎𝒎 ] from below:
A. 4.335
; 4.164
; 4.155
B.
4.164
; 4.335
; 4.155
C. 𝟒. 𝟑𝟑𝟓 ; 𝟒. 𝟏𝟔𝟒 ; 𝟒. 𝟏𝟕𝟓
( see slide 2.40;
- go gage should allow the following maximum size to go:
𝟒. 𝟐𝟓 + 𝟐%(𝟒. 𝟐𝟓) = 𝟒. 𝟑𝟑𝟓 𝒎𝒎 ;
- no-go gage should not allow the following minimum size to go:
𝟒. 𝟐𝟓 − 𝟐%(𝟒. 𝟐𝟓) = 𝟒. 𝟏𝟔𝟓 𝒎𝒎 ; i.e. no-go should be: 𝟒. 𝟏𝟔𝟒 𝒎𝒎
- in summary: 𝒏𝒐 − 𝒈𝒐 < 𝒂𝒄𝒄𝒆𝒑𝒕𝒂𝒃𝒍𝒆 𝒑𝒂𝒓𝒕 ≤ 𝒈𝒐
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 2

King Saud University – College of Engineering – Industrial Engineering Dept.
D. 4.164
; 4.335
; 4.175
E. 4.335
; 4.164
; 4.345
8.
The correct reading in the … shown below is …
A. Vernier caliper; 1.42 𝑚𝑚
B. Vernier micrometer;
C. Vernier micrometer;
1.42 𝑚𝑚
10.42 𝑚𝑚
= 1 ∗ 1 𝑚𝑚
+ 0 ∗ 0.5 𝑚𝑚
+ 42 ∗ 0.01 𝑚𝑚
= 𝟏. 𝟒𝟐 𝒎𝒎
D. micrometer; 10.42 𝑚𝑚
E. micrometer; 𝟏. 𝟒𝟐 𝒎𝒎
E
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 3

King Saud University – College of Engineering – Industrial Engineering Dept.
9.
The correct reading in the … shown below is …
A. Vernier caliper; 2
63
128
𝑖𝑛
B. Dial caliper; 2
119
128 𝑖𝑛
C. Dial caliper; 𝟐
𝟔𝟑
𝟏𝟐𝟖
𝒊𝒏
63
D. Universal bevel protractor; 2
128
𝑖𝑛
E. Vernier micrometer; 2
119
128
𝑖𝑛
= 2 ∗ 1 𝑖𝑛
+
+
8
7
16
7
∗
𝑖𝑛
1
16
𝟔𝟑
= 𝟐 𝑖𝑛
𝒊𝒏
𝟏𝟐𝟖
C
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 4

King Saud University – College of Engineering – Industrial Engineering Dept.
10.
The correct reading in the … shown below is …
A. Vernier caliper; 𝟒𝟒. 𝟑𝟓 𝒎𝒎
B. Vernier micometer; 44.35 𝑚𝑚
C. Vernier caliper; 44.35 𝑖𝑛
D. Vernier caliper; 44.70 𝑚𝑚
= 4 ∗ 10 𝑚𝑚
+ 4 ∗ 1 𝑚𝑚
+ 35 ∗ 0.01 𝑚𝑚
= 𝟒𝟒. 𝟑𝟓 𝒎𝒎
E. Vernier caliper; 4.435 𝑚𝑚
A
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 5

King Saud University – College of Engineering – Industrial Engineering Dept.
11.
The correct reading in the … shown below is …
A. Vernier caliper; 0.854 𝑖𝑛
B. Vernier caliper;
C. Vernier caliper;
𝟎. 𝟖𝟕𝟐 𝒊𝒏
8.72 𝑚𝑚
D. Vernier micometer; 0.854 𝑖𝑛
= 0 ∗ 1 𝑖𝑛
+ 8 ∗ 0.1 𝑖𝑛
+ 2 ∗ 0.025 𝑖𝑛
+ 22 ∗ 0.001 𝑖𝑛
= 𝟎. 𝟖𝟕𝟐 𝒊𝒏
E. Vernier caliper; 8.72 𝑖𝑛
B
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 6

King Saud University – College of Engineering – Industrial Engineering Dept.
12.
The reading in the gage shown below is …
A. 18° 03′
B. 𝟏° 𝟓𝟕′
C. 1° 03′
D. 2° 57′
E. 18° 57′
B
1°
+ 57′
= 𝟏° 𝟓𝟕′
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 7

King Saud University – College of Engineering – Industrial Engineering Dept.
13.
What is the reading indicated by the red line below in 𝒊𝒏.
and 𝒎𝒎 ?
A. 𝟎. 𝟒 𝒊𝒏 ≈ 𝟓𝟏
𝟏𝟐𝟖
𝒊𝒏 ≈ 𝟏𝟎. 𝟐 𝒎𝒎
B. 4 𝑖𝑛 ≈ 3 63
64
𝑖𝑛 ≈ 10.2 𝑐𝑚
C. 0.4 𝑖𝑛 ≈ 13
32
𝑖𝑛 ≈ 10.2 𝑚𝑚
D. 0.4 𝑖𝑛 ≈ 51
128
𝑖𝑛 ≈ 10.2 𝑐𝑚
E. 4 𝑖𝑛 ≈ 3 63
64
𝑖𝑛 ≈ 10.2 𝑚𝑚
A
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 8

King Saud University – College of Engineering – Industrial Engineering Dept.
14.
The limits of interference in ANY shaft-hole 𝑭𝑵 fit: 𝒎𝒂𝒙 𝒊𝒏𝒕
= ⋯ ; 𝒎𝒊𝒏 𝒊𝒏𝒕
= ⋯
A. 𝑚𝑎𝑥. ℎ𝑜𝑙𝑒 𝑠𝑖𝑧𝑒 – 𝑚𝑖𝑛. 𝑠ℎ𝑎𝑓𝑡 𝑠𝑖𝑧𝑒 ; 𝑚𝑖𝑛. ℎ𝑜𝑙𝑒 𝑠𝑖𝑧𝑒 – 𝑚𝑎𝑥. 𝑠ℎ𝑎𝑓𝑡 𝑠𝑖𝑧𝑒
B. 𝑚𝑖𝑛. 𝑠ℎ𝑎𝑓𝑡 𝑠𝑖𝑧𝑒 – 𝑚𝑎𝑥. ℎ𝑜𝑙𝑒 𝑠𝑖𝑧𝑒 ; 𝑚𝑎𝑥. 𝑠ℎ𝑎𝑓𝑡 𝑠𝑖𝑧𝑒 – 𝑚𝑖𝑛. ℎ𝑜𝑙𝑒 𝑠𝑖𝑧𝑒
D
C. 𝑚𝑖𝑛. ℎ𝑜𝑙𝑒 𝑠𝑖𝑧𝑒 – 𝑚𝑎𝑥. 𝑠ℎ𝑎𝑓𝑡 𝑠𝑖𝑧𝑒 ; 𝑚𝑎𝑥. ℎ𝑜𝑙𝑒 𝑠𝑖𝑧𝑒 – 𝑚𝑖𝑛 𝑠ℎ𝑎𝑓𝑡 𝑠𝑖𝑧𝑒
D. 𝒎𝒂𝒙. 𝒔𝒉𝒂𝒇𝒕 𝒔𝒊𝒛𝒆 – 𝒎𝒊𝒏. 𝒉𝒐𝒍𝒆 𝒔𝒊𝒛𝒆 ; 𝒎𝒊𝒏. 𝒔𝒉𝒂𝒇𝒕 𝒔𝒊𝒛𝒆 – 𝒎𝒂𝒙. 𝒉𝒐𝒍𝒆 𝒔𝒊𝒛𝒆 ( see slide
3.6
)
E. 𝑚𝑎𝑥. ℎ𝑜𝑙𝑒 𝑠𝑖𝑧𝑒 – 𝑏𝑎𝑠𝑖𝑐 𝑠𝑖𝑧𝑒 ; 𝑚𝑖𝑛. ℎ𝑜𝑙𝑒 𝑠𝑖𝑧𝑒 – 𝑏𝑎𝑠𝑖𝑐 𝑠𝑖𝑧𝑒
Questions 15-19 . Consider a
3–
5
16
" nominal diameter, 𝐿𝐶 7 fit between a shaft and a hole.
VALUES SHOWN BELOW ARE IN THOUSANDTHS OF AN INCHES
Nominal Size
Range (Inches) Clearance
Class LC6
Standard
Tolerance Limits
Over To
Limits Hole Shaft
H9 f8
Clearance
Class LC7
Standard
Tolerance Limits
Limits Hole Shaft
H10 e9
Clearance
Limits
Class LC8
Standard Tolerance
Limits
Hole Shaft
H10 d9
0 0.12
0.12 0.24
0.24 0.4
0
1.9
0.4
2.3
0.5
2.8
0.6
0.4 0.71
1
0
1.2
0
1.4
0
1.6
-0.3
-0.9
-0.4
-1.1
-0.5
-1.4
-0.6
0.6
3.2
0.8
3.8
1
4.6
1.2
1.6 -0.6
0 -1.6
1.8 -0.8
0 -2
2.2 -1
0 -2.4
2.8 -1.2
1
2
1.2
4.2
1.6
5.2
2
1.6
0
1.8
0
2.2
0
2.8
-1
-2
-1.2
-2.4
-1.6
-3
-2
0.71 1.19
1.19 1.97
1.97 3.15
3.15 4.73
4.73 7.09
7.09 9.85
9.85 12.41
7.1
1.6
8.1
2
9.3
2.2
10.2
5.1
1.2
6
1.4
4
1
3.2
0.8
0
5
0
0
4
0
4.5
0
3
0
3.5
0
2
0
2.5
-3.6
-1.6
-4.1
-2
-4.8
-2.2
-5.2
-1.6
-0.8
-2
-1
-2.6
-1
-3
-1.4
5.6
1.6
7.1
2
8.5
2.5
10
3
11.5
3.5
13.5
4
15.5
4.5
17.5
0
5
0
6
0
8
0
7
0
0
4
0 -2.8
3.5 -1.6
-3.6
-2
0 -4.5
4.5 -2.5
-5.5
-3
-6.5
-3
-7.5
-4
-8.5
-4.5
-9.5
13.5
6
16
7
18.5
7
20
6.4
2.5
8
3.6
9.5
4
11.5
5
0
8
0
0
7
0
6
0
5
0
4.5
0
4
0
3.5
-8.5
-6
-10
-7
-11.5
-7
-12
-3.6
-2.5
-4.5
-3
-5.5
-4
-7
-5
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 9

King Saud University – College of Engineering – Industrial Engineering Dept.
15.
The basic size (BS) is … 𝟑–
𝟓
𝟏𝟔
" = 𝟑 +
𝟓
𝟏𝟔
= 𝟑 + 𝟎. 𝟑𝟏𝟐𝟓 = 𝟑. 𝟑𝟏𝟐𝟓
A. 0.2688 𝑖𝑛
B. 0.3313 𝑖𝑛
C. 0.9375 𝑖𝑛
D. 2.6875 𝑖𝑛
E. 𝟑. 𝟑𝟏𝟐𝟓 𝒊𝒏
E
16.
Respectively, 𝑠ℎ𝑎𝑓𝑡
𝑀𝑀𝐶
= ; 𝑠ℎ𝑎𝑓𝑡
𝐿𝑀𝐶
= …
A. 3.3125 𝑖𝑛; 3.3175 𝑖𝑛
B. 3.3175 𝑖𝑛; 3.3125𝑖𝑛
C. 𝟑. 𝟑𝟎𝟗𝟓 𝒊𝒏; 𝟑. 𝟑𝟎𝟔𝟎 𝒊𝒏
D. 3.3122 𝑖𝑛; 3.3119 𝑖𝑛
E. 3.3130 𝑖𝑛; 3.3175 𝑖𝑛 𝑠ℎ𝑎𝑓𝑡
𝑀𝑀𝐶
= 3.3125 − 0.003 = 𝟑. 𝟑𝟎𝟗𝟓; 𝑠ℎ𝑎𝑓𝑡
𝐿𝑀𝐶
= 3.3125 − 0.0065 = 𝟑. 𝟑𝟎𝟔𝟎
C
A
17.
Respectively, ℎ𝑜𝑙𝑒
𝑀𝑀𝐶
= ; ℎ𝑜𝑙𝑒
𝐿𝑀𝐶
= …
A. 𝟑. 𝟑𝟏𝟐𝟓 𝒊𝒏; 𝟑. 𝟑𝟏𝟕𝟓 𝒊𝒏
B. 3.3175 𝑖𝑛; 3.3125𝑖𝑛
C. 3.3095 𝑖𝑛; 3.3060 𝑖𝑛
D. 3.3122 𝑖𝑛; 3.3119 𝑖𝑛
E. 3.3130 𝑖𝑛; 3.3175 𝑖𝑛 ℎ𝑜𝑙𝑒
𝑀𝑀𝐶
= 3.3125 + 0 = 𝑏𝑎𝑠𝑖𝑐 𝑠𝑖𝑧𝑒 = 𝟑. 𝟑𝟏𝟐𝟓; ℎ𝑜𝑙𝑒
𝐿𝑀𝐶
= 3.3125 + 0.005 = 𝟑. 𝟑𝟏𝟕𝟓
E
18.
Respectively, 𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝒄𝒍𝒆𝒂𝒓𝒂𝒏𝒄𝒆 = …; 𝑴𝒊𝒏𝒊𝒎𝒖𝒎 𝒄𝒍𝒆𝒂𝒓𝒂𝒏𝒄𝒆 = ⋯
A. 0; 0.005 𝑖𝑛
B. 0.005 𝑖𝑛; 0.0035 𝑖𝑛
C. 0.010 𝑖𝑛; 0.0025 𝑖𝑛
D. 0.0035 𝑖𝑛; 0.005 𝑖𝑛
E. 𝟎. 𝟎𝟏𝟏𝟓 𝐢𝐧; 𝟎. 𝟎𝟎𝟑 𝐢𝐧
𝑀𝑎𝑥. 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = ℎ𝑜𝑙𝑒
𝐿𝑀𝐶
− 𝑠ℎ𝑎𝑓𝑡
𝐿𝑀𝐶
= 3.3175 − 3.3060 = 𝟎. 𝟎𝟏𝟏𝟓
𝑀𝑖𝑛. 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 = ℎ𝑜𝑙𝑒
𝑀𝑀𝐶
− 𝑠ℎ𝑎𝑓𝑡
𝑀𝑀𝐶
= 3.3125 − 3.3095 = 𝟎. 𝟎𝟎𝟑
19.
Respectively, 𝒔𝒉𝒂𝒇𝒕 𝒕𝒐𝒍𝒆𝒓𝒂𝒏𝒄𝒆 = …; 𝒉𝒐𝒍𝒆 𝒕𝒐𝒍𝒆𝒓𝒂𝒏𝒄𝒆 = ⋯
A. 0; 0.005 𝑖𝑛
D
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 10

King Saud University – College of Engineering – Industrial Engineering Dept.
B. 0.005 𝑖𝑛; 0.0035 𝑖𝑛
C. 0.010 𝑖𝑛; 0.0025 𝑖𝑛
D. 𝟎. 𝟎𝟎𝟑𝟓 𝒊𝒏; 𝟎. 𝟎𝟎𝟓 𝒊𝒏
E. 0.0115 𝑖𝑛; 0.003 𝑖𝑛 𝑠ℎ𝑎𝑓𝑡 𝑡𝑜𝑙𝑒𝑟𝑎𝑛𝑐𝑒 = |𝑢𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 − 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡| = |−0.0030 − (−0.0065)| = 𝟎. 𝟎𝟎𝟑𝟓 ℎ𝑜𝑙𝑒 𝑡𝑜𝑙𝑒𝑟𝑎𝑛𝑐𝑒 = |𝑢𝑝𝑝𝑒𝑟 𝑙𝑖𝑚𝑖𝑡 − 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡| = |0.005 − 0| = 𝟎. 𝟎𝟎𝟓𝟎
B
20.
The figure below represents a fit of category: …; type: …; tightest fit is: …
A. location; transition; 𝐻
B. location; interference; 𝑲
J
( see slide 3.26;
H also: handout “Fits US tables…”, pg. 633 )
C. location; clearance; 𝐻
D. location; transition; 𝐾
E. force; interference; 𝐽
K
El-Sherbeeny, PhD Mar 13, 2013 IE 352 (01,02) - Spring 2013 Midterm 1 Answers Page - 11