Mechanical Assignment Help

Notes:
Power transmission: 6+ 3X kW, where X is 9, thus 45kW

Q1:

For my student number S3849839, the required power is 45kW, and the smallest motor that can meet the required transmitted power requirement has a rated speed of 1475 RPM.

Q2:

Class 2, or Medium duty

Q3:

For my type of starts (Heavy) the service factor is 1.4, meaning my design power is 63kW

Q4:

SPB Belt

Q5:

Speed ratio is 1485/460, which is roughly 3.23. If we use a larger pulley of 637mm, we can use the 197mm pulley for the small one, we achieve a ratio of 3.228. this means the output shaft will spin at 459.752RPM

The small pulley pitch diameter is 190mm, the large pulley pitch diameter is 630mm, meaning the error in shaft speed is 0.148 RPM

Q6:     

For my belt system the small pulley diameter is 190mm, the small pulley shaft speed is 1485 RPM and the Power Rating (in kW) from AS2784 is 11.29 kW.

Q7:

The small pulley diameter is 190mm, the small pulley shaft speed is 1475 RPM and the speed ratio is 3.23, therefore the Power Increment per belt is 1.21 kW.

Q8:

Belt Length selected is 2500mm, since this gives us a centre length of 876.84mm, which is within the 1000mm limitation. I want to reduce the span slack as much as possible, whilst remaining a decent size.  If we take the next step smaller size, we won’t have enough belt to satisfy C >= D+d, therefore the smallest standardised belt size is 2500mm.

Q9:

The SP belt type is SPB, the belt length is 2500 mm, and therefore the Power Correction for Belt Pitch Length is 0.96.

Q10:

The small pulley diameter is 190 mm, the large pulley diameter is 630 mm, my selected centre distance is 876.84 mm, and therefore the Power Correction for Arc of Contact is 0.93.

Q11:

1. The Power Rating from AS2784 is 11.29 kW.

2. The Power Increment per belt is 1.21 kW.

3. Power Correction for Belt Pitch Length is 0.96.

4. The Power Correction for Arc of Contact is 0.93.

5. Therefore the total Power per Belt is 11.16 kW.

6. The design power for my system is 63 kW.

7. Therefore the total number of belts required is 5.645 (6) belts.

Q12:

System requires 6 belts, but there is 3.96 kW of excess power. This represents 6.3            % of the total required power.

We can reduce the excess power lost by using fewer stronger belts or use more weaker belts. However, changing either of these parameters will cause us to create a system that extends beyond given dimension limits or outside the standardised parts catalogue.

Q14:

Q15:

The belt system I have designed satisfies Condition 1.

The largest value of “Required deflection force”, P from table B1 of Appendix 1 AS2784 is 54 N.

The “Correction for centrifugal tension”, K (calculated by the equation give in Appendix 1 of AS2784 is 4.17.

For the belt system I have designed:

The “Static hub load”, Ws, is defined as the magnitude of the force that is exerted on the pulley by the belt system when the pulleys are NOT rotating. For the system I have designed, Ws is equal to 7841.996 Newtons.

The “Dynamic hub load”, Wr, is defined as the magnitude of the force that is exerted on the pulley by the belt system when the pulleys ARE rotating. For the system I have designed, Wr is equal to 7793.55 Newtons.

Q16:

Q17:

State the Static Load 8571.484 N and Dynamic 8518.53 N that you have calculated for the worst case loaded bearing.

If we require an L10 design life or 7,000,000 cycles using a rolling element bearing (k = 3.0), state the Static Load Rating 16396.66 N and Dynamic Load Rating 16295.36 N required for the worst case loaded bearing.

Q18:

Q19

Q20:

Q21:

For my design scenario, the maximum bending moment, Mq is 596.3 N and the maximum shaft torque, Tq is 1568.35 N. (TE is equal to 1705.86N). Therefore, from Appendix A of AS1403, the estimated shaft diameter using high strength steel is 35 millimetres.

Q22:

Given that the last digit of my student number is 9, the torque must be increased by 95%.

Q23:

Given that my shaft diameter is 35mm, according to AS2938, the smallest 1st choice normal module that can accommodate 21 teeth on the pinion with a pinion pitch diameter of no less than 2.5 times the shaft diameter (87.5mm) is 16 mm.

Q24:

For my student number a torque increase of 95% is required and I have selected a module of 12 mm for my design.

For 21 teeth on the pinion, the number of teeth on the mating gear that most closely achieves this required torque increase is 41 teeth.

For this number of teeth on the mating gear, and the module I have selected, the gear pitch dimeter will be 156.61 mm and according to the standard tooth profile of AS2938, the addendum diameter of the mating gear will be 158.61 mm and the dedendum diameter of the mating gear will be 154.11 mm.

Q25

Based on the number of teeth on my pinion 41, the Geometry Factor (found from the table below, based on the above assumptions) is 0.27.

For a pinion shaft speed of 1485 RPM, and pinion pitch circle diameter [pinion_pitch_diameter] mm, the pitch line velocity is 6.768 meters per second, and the Velocity factor, Kv, (found from the table below based on the above assumptions) is 1.4.

Q26:

Using the AGMA method (shown in equation below) and the assumptions above, the minimum required face width, b, for the pinion is: 63.1 mm.

Q27:

 

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