Robotics Fundamentals 1, Chapter 2: Challenge 3 solution (a servo arm and its screws)

Formulas: M = F d          play at the tip = L tan(angle)          A(t) = (pi / 4)(d - 0.9382 p)^2
          F(shear) = 0.6 x 800 x A(t)          F(pre) = 0.7 x 580 x A(t)          T = K F(pre) d, K = 0.2          bearing F = stress x d x t
(Strength figures for class 8.8 and the plastic are typical approximate values.)

(a) Arm: 0.08 kg x 9.81 = 0.785 N acting 0.125 m from the servo:      0.785 x 0.125 = 0.0981 N m
    Gripper: 0.05 kg x 9.81 = 0.4905 N acting 0.25 m from the servo:    0.4905 x 0.25 = 0.1226 N m
    Total to hold the arm and gripper horizontal: 0.0981 + 0.1226 = 0.2207 N m  (0.22 N m, about 15 % of the stall torque).

(b) Usable torque = 0.6 x 1.5 = 0.90 N m.   Left for the payload: 0.90 - 0.2207 = 0.6793 N m.
        Payload force = 0.6793 / 0.25 = 2.717 N     mass = 2.717 / 9.81 = 0.277 kg  (about 277 g)

(c) Play at the tip = 0.25 x tan(0.7 deg) = 0.25 x 0.01222 = 0.00305 m = 3.05 mm.

(d) M3: A(t) = (pi / 4)(3 - 0.9382 x 0.5)^2 = (pi / 4)(2.5309)^2 = 0.7854 x 6.4054 = 5.03 mm^2
        F(shear) = 0.6 x 800 x 5.03 = 2415 N
        F(pre)   = 0.7 x 580 x 5.03 = 2043 N
        T        = 0.2 x 2043 x 0.003 = 1.23 N m   (approximate; for a steel screw in a metal part)

(e) F(bearing) = 20 MPa x (3 mm x 3 mm) = 20 x 9 = 180 N per screw hole.
    The screw can carry 2415 N in shear, 2415 / 180 = 13.4 times more than the plastic hole. The PLA fails first (the hole elongates or cracks), long before the screw.
    Assembly: do not use the 1.2 N m figure. Tighten only until snug, put a washer under the head to spread the load, and use a heat-set insert (or a nut in a captured slot) for threads that will be undone. Better still, make the loaded region thicker or use a metal bracket where the load is high.

Check on method: the screw is not the weak link in a plastic part; the material around it is. A common slip is to size a joint from the screw's catalogue strength alone.