Faculty of Engineering and Applied Science
MANE 3190U: Manufacturing and Production Processes
Instructors: Dr. Sayyed Ali Hosseini, P.Eng.
Practice Set # 4
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MANE 3190U Manufacturing and Production Processes
Practice Set #4
Question #1
In the process of bending a steel sheet of a 2 𝑚𝑚𝑚𝑚 thick at a bending radius of 6 𝑚𝑚𝑚𝑚. Determine
the length of the neutral axis in the bending area in “𝑚𝑚𝑚𝑚”. The bend angle is given as 0.6 𝑟𝑟𝑟𝑟𝑟𝑟.
Assume that the bending allowance constant is 0.5.
Solution #1
𝐿𝐿𝑏𝑏 = 𝛼𝛼(𝑅𝑅 + 𝑘𝑘𝑘𝑘)
𝐿𝐿𝑏𝑏 = 0.6(6 𝑚𝑚𝑚𝑚 + 0.5 ∗ 2 𝑚𝑚𝑚𝑚 ) = 4.2 𝑚𝑚𝑚𝑚
Question #2
A 2 𝑚𝑚𝑚𝑚 thick sheet of aluminum alloy is to be punched. The punch will perform the action of
material removal on a 100 𝑚𝑚𝑚𝑚2 equilateral triangular cross sectional area. The punch force is
estimated to be 12000 𝑁𝑁. Determine the ultimate tensile strength of aluminum alloy to be punched.
Solution #2
𝐹𝐹𝑚𝑚𝑚𝑚𝑚𝑚 = 0.7 𝑆𝑆𝑢𝑢𝑢𝑢 𝑡𝑡𝑡𝑡
𝑆𝑆𝑢𝑢𝑢𝑢 =
𝐹𝐹
0.7𝑡𝑡𝑡𝑡
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
𝐴𝐴 = 100 𝑚𝑚𝑚𝑚2 → 𝑎𝑎 = �100 ∗
4
√3
√3 2
𝑎𝑎
4
= 15.1967 𝑚𝑚𝑚𝑚, 𝑎𝑎 = 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿ℎ 𝑜𝑜𝑜𝑜 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
𝐿𝐿 = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 = 3(𝑎𝑎) → 3(15.1967 𝑚𝑚𝑚𝑚) = 45.59 𝑚𝑚𝑚𝑚
𝑆𝑆𝑢𝑢𝑢𝑢 = �
12000 𝑁𝑁
� → 𝑆𝑆𝑢𝑢𝑢𝑢 = 188 𝑀𝑀𝑀𝑀𝑀𝑀
0.7 ∗ 2 𝑚𝑚𝑚𝑚 ∗ 45.59 𝑚𝑚𝑚𝑚
Instructor: Dr. Sayyed Ali Hosseini, P.Eng.
2
MANE 3190U Manufacturing and Production Processes
Practice Set #4
Question #3
A turning operation is made with a rake angle of 10°, a feed of 0.010 𝑖𝑖𝑖𝑖/𝑟𝑟𝑟𝑟𝑟𝑟 and a depth of cut
= 0.100 𝑖𝑖𝑖𝑖. The shear strength of the work material is known to be 50,000 𝑙𝑙𝑙𝑙/𝑖𝑖𝑛𝑛2 , and the chip
thickness ratio is measured after the cut to be 0.40. Determine the cutting force and the feed force.
Use the orthogonal cutting model as an approximation of the turning process.
Solution #3
As can be seen in the following figure, in turning operation, feed (𝑓𝑓) determines the unreformed
chip thickness (𝑡𝑡0 ) and depth of cut (𝑑𝑑) determines the width of engagement between the tool and
workpiece (𝑤𝑤) which is also known as width of cut. Knowing this,
𝜙𝜙 = tan−1 �
𝑐𝑐ℎ𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟: 𝑟𝑟 = 0.4
𝑟𝑟 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼
0.4 cos 10
� = tan−1 �
� → 𝜙𝜙 = 22.9°
1 − 𝑟𝑟 𝑠𝑠𝑠𝑠𝑠𝑠 𝛼𝛼
1 − 0.4 sin 10
𝐴𝐴𝑠𝑠 =
𝑡𝑡0 𝑤𝑤
0.01 × 0.1
=
= 0.00257 𝑖𝑖𝑛𝑛2
sin 22.9
𝑠𝑠𝑠𝑠𝑠𝑠 𝜙𝜙
Knowing the shear strength of the material, the force acing on the shear plane can be calculate as
follows:
It is also known that
𝜙𝜙 =
𝜏𝜏𝑠𝑠 =
𝐹𝐹𝑠𝑠
→ 𝐹𝐹𝑠𝑠 = 𝜏𝜏𝑠𝑠 × 𝐴𝐴𝑠𝑠 = 50000(0.00257) = 128 𝐼𝐼𝐼𝐼
𝐴𝐴𝑆𝑆
𝜋𝜋 𝛼𝛼 𝛽𝛽
𝜋𝜋
+ − → 𝛽𝛽 = + 𝛼𝛼 − 2𝜙𝜙 = 90 + 10 − 2(22.9) → 𝛽𝛽 = 54.1°
4 2 2
2
Instructor: Dr. Sayyed Ali Hosseini, P.Eng.
3
MANE 3190U Manufacturing and Production Processes
Practice Set #4
Knowing 𝛼𝛼, 𝛽𝛽, 𝜙𝜙 and using trigonometry:
𝐹𝐹𝑐𝑐 =
𝐹𝐹𝑠𝑠 cos(𝛽𝛽 − 𝛼𝛼)
→
cos(𝜙𝜙 + 𝛽𝛽 − 𝛼𝛼)
𝐹𝐹𝑐𝑐 =
128 cos(54.1 − 10)
= 236 𝐼𝐼𝐼𝐼
cos(22.9 + 54.1 − 10)
𝐹𝐹𝑡𝑡 =
𝐹𝐹𝑠𝑠 sin(𝛽𝛽 − 𝛼𝛼)
→
cos(𝜙𝜙 + 𝛽𝛽 − 𝛼𝛼)
𝐹𝐹𝑡𝑡 =
128 sin(54.1 − 10)
= 229 𝐼𝐼𝐼𝐼
cos(22.9 + 54.1 − 10)
Instructor: Dr. Sayyed Ali Hosseini, P.Eng.
4