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cricket20 [7]
3 years ago
12

A 75 mm diameter, 15 mm wide A80-M12V grinding wheel is used for traverse surface grinding of a 75 mm by 75 mm 1040 steel workpi

ece. The workpiece traverse speed is 1 mm/sec, the radial depth of grind is 1.5 mm, the (axial) intermittent cross-feed is 10 mm/stroke, and the wheel surface speed is 314 mm/sec. Estimate the required amount of grinding time (without spark-out) and the spindle power.

Engineering
1 answer:
Dmitrij [34]3 years ago
3 0

Answer:

Detailed solution is given and calculations are done

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The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.
zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity  G = 11 × 10³ Ksi =  11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft J\tau

shaft J\tau = \dfrac{\pi}{2}r^4

where ;

r = 1 in /2

r = 0.5 in

shaft J \tau = \dfrac{\pi}{2} \times 0.5^4

shaft J\tau = 0.098218

Now; the angle of twist at  B with respect to D  is calculated by using the expression

\phi_{B/D} = \sum \dfrac{TL}{JG}

\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}

where;

T_{CD} \ \  and \ \  L_{CD} are the torques at segments CD and length at segments CD

{T_{BC} \  \ and  \ \ L_{BC}} are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

L_{BC}} = 2.5  in

J\tau = 0.098218

G =  11 × 10⁶ lb/in²

T_{BC = -60 lb.ft

T_{CD = 0 lb.ft

L_{CD = 5.5 in

\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times  12 )}{ (0.9818)(11 \times 10^6)}

\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}

\phi_{B/D} = \dfrac{-21600}{1079980}

\phi_{B/D} = − 0.02 rad

To degree; we have

\phi_{B/D}  = -0.02 \times \dfrac{180}{\pi}

\mathbf{\phi_{B/D}  = -1.15^0}

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For  the angle of twist of C with respect to D; we have:

\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}

\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}

\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times  12 )}{ (0.9818)(11 \times 10^6)}

\phi_{C/D} = \dfrac{21600}{1079980}

\phi_{C/D} = 0.02 rad

To degree; we have

\phi_{C/D}  = 0.02 \times \dfrac{180}{\pi}

\mathbf{\phi_{C/D}  = 1.15^0}

3 0
3 years ago
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