casestudy-kuka6dofroboticarm-181013064718.pptx

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1. MACHINE DESIGN CASE STUDY Presented by- Khushal Nande TYB102 Sumedh Durge TYB114 Siddhant Kumawat TYB125 Guide- Bhavana Mariyappalavar Ma’am Venugopal Kulkarni Sir

2. KUKA 6 DOF ROBOTIC ARM MODEL NAME: KUKA 30-3

3. WHY ROBOTIC ARM? WHY KUKA KR 30-3?

4. ROLE OF 6 DOF IN ROBOTIC ARM

5. ROLE OF 6 DOF IN ROBOTIC ARM 6 degrees of freedom is the minimum needed to reach a volume of space from every angle. Because of this flexibility it able to reach most of the places and can perform a particular task. Therefore it plays an important role in the automobile assembly line, medical surgeries and many other fields.

6. ROLE OF 6 DOF IN ROBOTIC ARM

7. Working Envelope (Side View) Working Envelope (Top View) Working Envelope

8. PAYLOADS Rated Payload- 30kg Rated mass moment of Inertia- 9kgm 2 Rated total load- 65kg Nominal Distance to load center of Gravity Lxy = 180mm Lz = 150mm

9. PAYLOADS The load center of gravity refers to the distance from the face of the mounting flange on axis 6 Load Center of Gravity

10. PAYLOADS Load Center of Gravity

11. PAYLOADS Mounting Flange- ISO 9409-1-100-6-M8 Mounting Flange

12. PAYLOADS Mounting Flange

13. PAYLOADS Mounting Flange

14. PAYLOADS Foundation Load

15. TORQUE CALCULATIONS T=F*L Nm Where, F=force acting on the motor L=length of the shaft Force, F is given by, F=m*g N Where, m=mass to be lifted by the motor g= Acceleration due to gravity=9.8 m/s^2

16. TORQUE CALCULATIONS The values used for the torque calculations: W1=0.019 kg (weight of link 1) W2= W4=0.050 kg (weight of motor) W3= 0.020 kg (weight of link 2) W5= 0.027 kg (weight of gripper) W6= 250g (weight of load) L1= 0.07m (length of link 1) L2= 0.12m (length of link 2) L3= 0.145m (length of link 3) Taking the sum of forces in the Y-axis, ∑ Fy = (W6+W5+W4+W3+W2) g−CY=0

17. TORQUE CALCULATIONS CY= 11.24 N ∑Fy = (W6+W5+W4+W3+W2+W1) g−BY=0 BY= 11.43 N ∑Mc= - ( W3L2 / 2 ) − 𝑊5 (L2 + L3 / 2 )-W6 (L2+L3) − W4 (L2) + Mc=0 ∑Mb= -W6 (L1+L2+L3)−W5(L1+L2+ L3 2 ) −W4(L1+L2) −W3(L1+ L2 2 ) – W2(L1)-W1( L1 2 ) +Mb=0 Mc= 0.929 Nm Mb=1.792Nm

18. APPLICATIONS

19. APPLICATIONS

20. APPLICATIONS

21. REFERENCES KUKA- www.kuka.com http://nemertes.lis.upatras.gr/jspui/bitstream/10889/8840/1/MarkosKapeliotisMasterThesis.pdf

22. THANK YOU