2. Reason for the Design. Introduction. Design analysis of power screw.

3. Available jacks that's uses for left the car are difficult to use and need to some strength while lifting the car, so it’s difficult to use especially for the elderly and women. Furthermore, available jacks are typically large, heavy and also difficult to store, transport, carry or move into the proper position under an automobile. There also reports on car jacks which lead to a serious number of accidents.

4. Therefore, our project will solve this problem radically; where we are going in our project to design a built in automatic car jack will lift the entire car or sides of it by pushing a button next to the driving seat. Our jack is self-locking, which make it safer than other jack technologies like hydraulic jacks which require continual pressure to remain in a locked position. So in use of our jack we will save effort, time and don’t need to storage area like traditional jacks.

5. Our jack will be designed as four power screw jacks. And will be connected to four positions in the car beside every wheel. These four jacks connected to one power source which internal car power source ( the car drive shaft ) with another shafts and controllers as bellow in the figure below. ( the car drive shaft ) with another shafts and controllers as bellow in the figure below.

7. Each jack will have a single button to control it from the driving seat, so there is four buttons for the four wheels. Our jack will be designed to lift a car with an average weight of 1.5 tons with factor of safety to be 3. our design is safe, reliable and able to raise and lower the height level by the control system. our design is safe, reliable and able to raise and lower the height level by the control system.

9. The total average weight of the car is to be as following ( 1500 kg * 9.81 = 14715 N = 3306.742 Ibf ) ; The total average weight of the car is to be as following ( 1500 kg * 9.81 = 14715 N = 3306.742 Ibf ) ; then this weight will be distributing to the four jacks and we will take the maximum load to be applied (1000 Ibf ) for each. then this weight will be distributing to the four jacks and we will take the maximum load to be applied (1000 Ibf ) for each. Our design steps will be according to satisfying the design factor of safety that’s we chosen to be 3 then we will make a trials with changing the major diameter for the power screw from table ( 8 – 3 ) and when we achieve or exceed our design factor of safety we will stop the trials and deciding to chose that’s parameter in the last trial. Our design steps will be according to satisfying the design factor of safety that’s we chosen to be 3 then we will make a trials with changing the major diameter for the power screw from table ( 8 – 3 ) and when we achieve or exceed our design factor of safety we will stop the trials and deciding to chose that’s parameter in the last trial.

10.  The power screw material is chosen to be steel, machine oil (to minimizing friction ) with nut material “ the gear “ of steel then the coefficient of friction ( ƒ ) for thread pairs from table ( 8 - 5 ) is to be ( 0.17 ) “ starting friction “.  The power screw is to be a single thread then the lead is the same as the pitch.  The power screw is an Acme thread with thread angle equal to 29 ْ = 2α.

11. Firstly we find the mean and root diameters then the lead … Then substitute it in the raising and lowering torque equations …

12. Check power screw for self locking condition by make sure that’s the following relation is satisfying ( where tan λ = l / π dm ) then for self locking … To find the efficiency during lifting the load we use … Then get the stresses by using these equations …

13. Then the three-dimensional stresses according to the following figure … And sub it in the von Mises stress equation … Finally check out for the factor of safety to be equal or more than 3 ( we chose from table ( A – 18 ) a steel G10060 (HR) with yield strength = 24 kpsi ) then ( we chose from table ( A – 18 ) a steel G10060 (HR) with yield strength = 24 kpsi ) then And get the factor of safety ( n ).

14. Trial F (Ibf) d (in) P (in) dr (in) dm (in) L (in)π fsec αλself locking 110000.250.06250.18750.218750.06253.1415930.171.03290.0909457yes 210000.31250.0714290.2410710.2767860.0714293.1415930.171.03290.0821445yes 310000.3750.0833330.2916670.3333330.0833333.1415930.171.03290.0795775yes 410000.50.10.40.450.13.1415930.171.03290.0707355yes 510000.6250.1666670.4583330.5416670.1666673.1415930.171.03290.0979415yes 610000.750.1666670.5833330.6666670.1666673.1415930.171.03290.0795775yes 710000.8750.20.6750.7750.23.1415930.171.03290.0821445yes Trial TL (Ibf. In) TR (Ibf. In)e Τ (psi) σ (psi) σB (psi) σb (psi) σ ` (psi) Sy (psi)n 19.11277429.625780.33576122889.49-36216.6-17694.461930.3794666.28240000.253522 212.7487236.191050.31411713156.31-21908.8-12236.242147.0660819.63240000.394609 315.7820643.131090.3075028853.198-14967.1-8708.9629859.2942398.5240000.566058 423.3034956.120980.2835934465.966-7957.75-5375.918143.6624428.17240000.982472 520.6750675.37860.3519013987.252-6061.04-2679.689500.68515240.66240001.574735 631.5641286.262190.3075022213.3-3741.77-2177.247464.82210599.63240002.264231 735.69641101.33490.3141171678.101-2794.49-1560.755375.97757.606240003.093738

15. Questions