Production Engineering Test 5

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Production Engineering Test 5

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Weekly Quiz Competition

Question 1
1 / -0

A mild steel block of width 40 mm is being milled using a straight slab cutter 70 mm diameter with 30 teeth. If the cutter rotates at 40 rpm, and depth of cut is 2 mm, what is the value of maximum uncut chip thickness when the table feed is 20 mm / min?

A
0.00263 mm

B
0.00363 mm

C
0.00463 mm

D
0.00563 mm

Solution

Concept:
Maximum Uncut chip thickness in milling is given as, \(t_{max}=\frac{2f_m}{NZ}\sqrt{\frac{d}{D}}\)
where, fm = table feed, N = rpm, Z = number of teeth, d = depth of cut, D = diameter of cutter
Calculation:
Given:
D = 70 mm, d = 2 mm, fm = 20 mm/min, N = 40 rpm, Z = 30
Therefore, \(t_{max}=\frac{2~\times~20}{40~\times~30}\sqrt{\frac{2}{70}}\)
Maximum uncut chip thickness = 0.00563 mm
Question 2
1 / -0

In a orthogonal machining, the shear angle is 22° and rake angle is 11°, then the shear strain is _____

A
2.6-2.7

B
2.64-2.71

C
2.67-2.76

D
2.69-2.72

Solution

Concept:
γ = cot ϕ + tan(ϕ – α)
ϕ = shear angle, α = rake angle
Calculation:
Given:
ϕ = 22°, α = 11°
γ = cot 22° + tan (22° - 11°)
γ = cot 22° + tan 11°
∴ γ = 2.669 ∼ 2.67
Question 3
1 / -0

In a turning operation with orthogonal machining, the rake angle is 12° and the coefficient of friction at the tool chip interface is 0.60. If the spindle rotates at 300 rpm and tool is fed at 0.4 mm per revolution, then the value of chip thickness is _____ mm (Take depth of cut = 0.6 mm)

A
0.61-0.64

B
0.614-0.641

C
0.617-0.646

D
0.619-0.642

Solution

Concept:
In orthogonal machining, λ = 90°
Depth of cut (d) = b
From merchant’s theory
2ϕ + β - α = 90°
\(\tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }}\)
\(r = \frac{t}{{{t_c}}}\)
We need to calculate t_c.
Calculation:
Given:
λ = 90°, d = b = 0.6 mm, t = feed = 0.4 mm/rev, α = 12°
Now,
β = tan^-1 (μ) = 30.96°
2ϕ + β - α = 90°
∴ ϕ = 35.52°
Now,
\(\tan \phi = \frac{{r\cos \alpha }}{{1 - r\sin \alpha }}\)
\(0.7137 = \frac{{r\cos 12^\circ }}{{1 - r\sin 12^\circ }}\)
0.7137 = (0.7137 sin 12° + cos 12°) r
∴ r = 0.633
Now,
r = t/t_c
∴ t_c = 0.6319
Question 4
1 / -0

The surface of a metal slab of 278 mm length and 80 mm width is machined by a face milling cutter of 120 mm diameter, having 10 teeth, rotating at 50 rpm. The milling feed is 0.1 mm/tooth of the cutter. The combined time for job loading and unloading is 2 min. The approach and over travel of the tool are 1 mm each then

A
Compulsory approach is 2 mm

B
Total length to be machined is approximately 395 mm

C
Compulsory approach is 15.27 mm

D
The production rate per hour is 7.6 pieces

Solution

Explanation:
Given:
L = 278, w = 80 m, D = 120 mm, Z = 10 teeth, N = 50 rpm, Feed/tooth = 0.1 mm/tooth
T_{loading and unloading} = 2 min
AP₁ + OR₁ = 2 mm
Now,
\({\rm{Compulsory\;approach\;}}{\left( {{\rm{AP}}} \right)_{\rm{c}}} = \frac{1}{2}\left( {D - \sqrt {{D^2} - {w^2}} } \right)\)
∴ Compulsory approach (AP_c) = 15.27 mm
Now,
Total length = 278 + 15.27 + 2
∴ Total length = 295.27 mm
Now,
\({\rm{Time}}/{\rm{cut\;}} = \frac{{{L_{total}}}}{{fZN}} = \frac{{295.27}}{{0.1 \times 10 \times 50}}\)
∴ Time/cut = 5.9054 min
Now,
Time/part = 5.9054 + 2
∴ Time/part = 7.9 min
Now,
\({\rm{Production}}/{\rm{hr\;}} = \frac{{60\;}}{{7.9054}}\)
∴ Production/hr = 7.6 pieces
Question 5
1 / -0

In plunge surface grinding a steel specimen of 2.5 mm width and 200 mm length is ground with A-A-6-0-K-5-V-8 grinding wheel of 300 mm diameter and 25 mm wide. The wheel has undergone a radial wheel wear of 20 microns after removing material to a depth of 5 mm.

A
Volume of material removed = 471.2074 mm³

B
The grinding ratio is 5.3055

C
Volume of material removed = 2500 mm³

D
The grinding ratio is 6.5125

Solution

Explanation:
Given:
w = 2.5 m, L = 200 mm, d = 5 mm, Width of wheel constant = 25 mm
D_f = 300 – (2 × tool wear in radial direction)
D_f = 300 – 2(20 × 10^-3)
∴ D_f = 299.96
Now,
\({\rm{Grinding\;ratio\;}}\left( {{\rm{or}}} \right){\rm{\;Machining\;Ratio}} = \frac{{Volume\;of\;material\;removed\;on\;workpiece}}{{Volume\;\;of\;wheel\;wear}}\)
Now,
Volume of material removed = 200 × 2.5 × 5 mm³
∴ Volume of material removed = 2500 mm³
Now,
Volume of material removed on wheel \( = \frac{\pi }{4}\left( {d_i^2 - d_f^2} \right) \times w\)
Volume of material removed on wheel \(= \frac{\pi }{4}\left( {{{300}^2} - {{299.96}^2}} \right) \times 25\)
∴ Volume of material removed on wheel = 471.2074 mm³
Now,
\({\rm{Grinding\;ratio\;}}\left( {{\rm{or}}} \right){\rm{\;Machining\;Ratio}} = \frac{{Volume\;of\;material\;removed\;on\;workpiece}}{{Volume\;\;of\;wheel\;wear}}\)
\(\therefore {\rm{G}}.{\rm{R}}.{\rm{\;or\;M}}.{\rm{R\;}} = \frac{{2500}}{{471.2074}}\)
∴ Grinding ratio or Machining ratio = 5.3055

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Re-Attempt Weekly Quiz Competition

Production Engineering Test 5

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Production Engineering Test 5

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Rank

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SHARING IS CARING

Question 1

0.00263 mm

0.00363 mm

0.00463 mm

0.00563 mm

Solution

Question 2

2.6-2.7

2.64-2.71

2.67-2.76

2.69-2.72

Solution

Question 3

0.61-0.64

0.614-0.641

0.617-0.646

0.619-0.642

Solution

Question 4

Compulsory approach is 2 mm

Total length to be machined is approximately 395 mm

Compulsory approach is 15.27 mm

The production rate per hour is 7.6 pieces

Solution

Question 5

Volume of material removed = 471.2074 mm3

The grinding ratio is 5.3055

Volume of material removed = 2500 mm3

The grinding ratio is 6.5125

Solution

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Volume of material removed = 471.2074 mm³

Volume of material removed = 2500 mm³