Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
1. Greater Design Freedom : Any changes that are required in design can be incorporated at any design stage without worrying about any delays since there would hardly be any in ain integrated CAM environment.
2. Increased Productivity: In view of the fact that the total manufacturing activity is company organization through the computer it would be possible to increase the productivity of the plant .
3. Greater Operating Flexibility: CAM enhances the flexibility in manufacturing methods and changing of product lines.
4. Shorter Lead Time : Lead times in manufacturing would be greatly reduced.
5. Improved Reliability: In view of the better manufacturing methods and controls at the manufacturing Turing stage the products thus manufactured as well as of the manufacturing system would be highly reliable.
6. Reduced Maintenance: Since most of the components of CAM system would include integrated diagnostics and monitoring facilities they would require less maintenance compared to the conventional manufacturing methods.
7. Reduced Scrap and Rework: Because of the CNC machines used in production and the part programs being made by the stored geometry from the design stage the scrap level would be reduced to the minimum possible and almost no rework would be necessary.
8. Better Management control: Since all the information and controlling functions are attempted with help of computer a better management control on the manufacturing activity is possible.
difference between coplanar and noncoplanar force system
Q. Molding methods according to the method of making a mold? Following are the molding methods according to the method of mold making: • Open mold method: the method is su
Types in trusses
Q. Radiographic examination of welded joints? Radiographic examination (RT) of welded joints shall be performed as required by the Code. Acceptance criteria shall be in accord
Computers control system in flexible manufacturing system
Determine number of coils and length of the spring: A close coiled helical spring ought to extend by 120 mm under an axial force of 1200 N. If mean coil radius is equal to 40
Determine maximum stress and modulus of elasticity: A composite of glass fibres and epoxy has all of fibres laid along the length and is needed to carry a stress of 12 MPa. T
A flat belt, 8 mm thick and 100 mm wide transmits power between two pulleys, running at 1600 m/min. The mass of the belt is 0.9 kg/m length. The angle of lap in the smaller pulle
Fatigue crack propagation: If it is assumed that no plastic deformation occurs around crack tip and law of fatigue crack propagation for place of above Example is da/dN = 10
Q. What do you mean by Alloys? Metallic materials that contain alloying elements, including chromium (Cr), cobalt (Co), Columbium or niobium (Cb or Nb), copper (Cu), iron (Fe),
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +1-415-670-9521
Phone: +1-415-670-9521
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd