04 Practice Set #4-1

```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 ππππ thick at a bending radius of 6 ππππ. Determine
the length of the neutral axis in the bending area in “ππππ”. The bend angle is given as 0.6 ππππππ.
Assume that the bending allowance constant is 0.5.
Solution #1
πΏπΏππ = πΌπΌ(ππ + ππππ)
πΏπΏππ = 0.6(6 ππππ + 0.5 ∗ 2 ππππ ) = 4.2 ππππ
Question #2
A 2 ππππ thick sheet of aluminum alloy is to be punched. The punch will perform the action of
material removal on a 100 ππππ2 equilateral triangular cross sectional area. The punch force is
estimated to be 12000 ππ. Determine the ultimate tensile strength of aluminum alloy to be punched.
Solution #2
πΉπΉππππππ = 0.7 πππ’π’π’π’ π‘π‘π‘π‘
πππ’π’π’π’ =
πΉπΉ
0.7π‘π‘π‘π‘
π΄π΄π΄π΄π΄π΄π΄π΄ ππππ ππππ ππππππππππππππππππππππ π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘π‘ =
π΄π΄ = 100 ππππ2 → ππ = οΏ½100 ∗
4
√3
√3 2
ππ
4
= 15.1967 ππππ, ππ = πΏπΏπΏπΏπΏπΏπΏπΏπΏπΏβ ππππ ππ π π π π π π π π π π π π  π π βππππππππππ ππππππππ
πΏπΏ = ππππππππππππππππππ = 3(ππ) → 3(15.1967 ππππ) = 45.59 ππππ
πππ’π’π’π’ = οΏ½
12000 ππ
οΏ½ → πππ’π’π’π’ = 188 ππππππ
0.7 ∗ 2 ππππ ∗ 45.59 ππππ
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&deg;, a feed of 0.010 ππππ/ππππππ and a depth of cut
= 0.100 ππππ. The shear strength of the work material is known to be 50,000 ππππ/ππππ2 , 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 (ππ) determines the unreformed
chip thickness (π‘π‘0 ) and depth of cut (ππ) determines the width of engagement between the tool and
workpiece (π€π€) which is also known as width of cut. Knowing this,
ππ = tan−1 οΏ½
ππβππππ π‘π‘βππππππππππππππ ππππππππππ: ππ = 0.4
ππ ππππππ πΌπΌ
0.4 cos 10
οΏ½ = tan−1 οΏ½
οΏ½ → ππ = 22.9&deg;
1 − ππ π π π π π π  πΌπΌ
1 − 0.4 sin 10
π΄π΄π π  =
π‘π‘0 π€π€
0.01 &times; 0.1
=
= 0.00257 ππππ2
sin 22.9
π π π π π π  ππ
Knowing the shear strength of the material, the force acing on the shear plane can be calculate as
follows:
It is also known that
ππ =
πππ π  =
πΉπΉπ π
→ πΉπΉπ π  = πππ π  &times; π΄π΄π π  = 50000(0.00257) = 128 πΌπΌπΌπΌ
π΄π΄ππ
ππ πΌπΌ π½π½
ππ
+ − → π½π½ = + πΌπΌ − 2ππ = 90 + 10 − 2(22.9) → π½π½ = 54.1&deg;
4 2 2
2
Instructor: Dr. Sayyed Ali Hosseini, P.Eng.
3
MANE 3190U Manufacturing and Production Processes
Practice Set #4
Knowing πΌπΌ, π½π½, ππ and using trigonometry:
πΉπΉππ =
πΉπΉπ π  cos(π½π½ − πΌπΌ)
→
cos(ππ + π½π½ − πΌπΌ)
πΉπΉππ =
128 cos(54.1 − 10)
= 236 πΌπΌπΌπΌ
cos(22.9 + 54.1 − 10)
πΉπΉπ‘π‘ =
πΉπΉπ π  sin(π½π½ − πΌπΌ)
→
cos(ππ + π½π½ − πΌπΌ)
πΉπΉπ‘π‘ =
128 sin(54.1 − 10)
= 229 πΌπΌπΌπΌ
cos(22.9 + 54.1 − 10)
Instructor: Dr. Sayyed Ali Hosseini, P.Eng.
4
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