In the composite filament winding process as shown in the figure, the mandrel diameter is 700 mm and is rotating at a speed, $ N = 6 $ rev/min.
If a $ 45^\circ $ helical winding angle is needed, the axial velocity, $ v_c $, of the slider should be ______ m/s (rounded off to two decimal places).
Assume $ \pi = 22/7 $. Ignore the thickness of the composite layers already wound on the mandrel.

Note: Figure is not to scale.

Question

A through hole of 10 mm diameter is to be drilled in a mild steel plate of 30 mm thickness. The selected spindle speed and feed for drilling hole are 600 revolutions per minute (RPM) and 0.3 mm/rev, respectively. Take initial approach and breakthrough distances as 3 mm each. The total time (in minute) for drilling one hole is ______. (Rounded off to two decimal places)

Answer 1

Step 1: Understanding the Question:
In a filament winding process, a continuous fibre is wound onto a rotating mandrel. The angle of the fibre (helical angle) results from the combined vector velocities of the rotating mandrel surface and the linearly translating slider feeding the fibre.

Step 2: Key Formula or Approach:
If the helical winding angle relative to the axis of rotation is $ \alpha $, the relationship is:
\[ \tan(\alpha) = \frac{\text{Surface velocity of mandrel}}{\text{Axial velocity of slider}} = \frac{V_m}{v_c} \]
Surface velocity $ V_m = \pi D N $.

Step 3: Detailed Explanation:
Given data:
Mandrel diameter $ D = 700 $ mm $ = 0.7 $ m.
Rotational speed $ N = 6 $ rev/min.
Helical angle $ \alpha = 45^\circ $.
$ \pi = \frac{22}{7} $.
First, calculate the circumferential surface velocity of the mandrel:
\[ V_m = \pi D N = \left( \frac{22}{7} \right) \times 0.7 \times 6 \]
\[ V_m = 22 \times 0.1 \times 6 = 13.2 \text{ m/min} \]
Now, apply the velocity vector relationship to find $ v_c $:
\[ \tan(45^\circ) = \frac{13.2}{v_c} \]
Since $ \tan(45^\circ) = 1 $:
\[ 1 = \frac{13.2}{v_c} \implies v_c = 13.2 \text{ m/min} \]
The question requires the velocity in m/s. Convert the units:
\[ v_c = \frac{13.2 \text{ m}}{60 \text{ s}} = 0.22 \text{ m/s} \]

Final Answer:
The required axial velocity is 0.22 m/s. Be careful with the definition of the winding angle in the diagram, sometimes it is defined relative to the transverse plane instead of the axial plane. Here at 45 degrees the tangent is 1 regardless, making it foolproof, but for other angles like 30 or 60 degrees, verify which axis the angle is drawn against. \[ \boxed{v_c = 0.22 \text{ m/s}} \]

Show Hint

Correct Answer: 0.22

Solution and Explanation

Questions Asked in GATE PI exam