Design and fabrication of an inline pipe inspection robot

Abstract

A. Degrees of Freedom F = 3(n-1)-2f 1-f 2 n = no. of links of mechanism with fixed links. f 1 = no. of pin joints or revolute pairs or pairs that permits one degree of freedom. f 2 = no. of roll-slide pairs. F = 3*(8-1)-2(9)-0 F = 3 We are using a universal coupling so our robot has 6-degrees of freedom. IV. DESIGN Fig.1-Design B. Motor Total load on robot = 20N Power required to robot to carry weight of 20N with 0.1m/s speed is, P = W × v = 20 × 0.1 P = 2watts. In worst case if only one motor is working then it has to give total power. Power required to 6-DC motors to drive the robot is, P required = 6 × 2 P required = 12watts If a motor of 8v & 2amp current is selected then power provided by 6 motors is, P provided = 16watts Here, P provided > P required …….. Hence ok. So, select 6-DC motors of 6v, 1amp current & 35rpm each. Fig.2-Actuator C. Material Selection & Calculation of Spring Stiffness Material: stainless steel wire for normal corrosion resistance Type: Ground end Maximum elongation of spring is given as, δL = L2 L = 98.7368 75.4718 δL = 23.265 mm Spring force calculation: In vertical case, total load acting on robot is additional sum of weight of robot and frictional force i.e. 30N ∴ we have to design a spring which will hold the load of 30N ∴Design the spring for 30N force; Calculation of spring stiffness (K): Spring stiffness = spring force / maximum elongation of spring K = 30/23.265 K = 1.29N/mm Fig.3-Compression Spring D. Spring Proportions Calculation of spring wire diameter (d): average service Design stress is, τ = 48.5 kgf/mm2 τ = 485 N/mm2 but, shear stress in spring is given by formulae, τ = (8*F*C)/(π*d*d) 485 = (8*30*6)/ (π*d*d) d = 0.97mm