Robotics Fundamentals 1, Chapter 2: Challenge 1 solution (deflection, stress and shape)

Formulas: I = b h^3 / 12          delta = F L^3 / (3 E I)          sigma = 6 F L / (b h^2)          SF = yield / sigma
Data: E = 69 GPa = 69e9 Pa, yield = 275 MPa, F = 8 N, b x h = 25 mm x 4 mm, L = 0.20 m

(a) Flat (h = 4 mm in the direction of the load, b = 25 mm):
        I     = 0.025 x 0.004^3 / 12 = 0.025 x 6.4e-8 / 12 = 1.333e-10 m^4
        delta = 8 x 0.20^3 / (3 x 69e9 x 1.333e-10) = 0.064 / 27.6 = 0.00232 m = 2.32 mm
        sigma = 6 x 8 x 0.20 / (0.025 x 0.004^2) = 9.6 / 4.0e-7 = 24 MPa
        SF    = 275 / 24 = 11.5

(b) On edge (h = 25 mm, b = 4 mm):
        I     = 0.004 x 0.025^3 / 12 = 0.004 x 1.5625e-5 / 12 = 5.208e-9 m^4
        delta = 0.064 / (3 x 69e9 x 5.208e-9) = 0.064 / 1078.1 = 5.94e-5 m = 0.059 mm
        sigma = 6 x 8 x 0.20 / (0.004 x 0.025^2) = 9.6 / 2.5e-6 = 3.84 MPa
        SF    = 275 / 3.84 = 71.6
    I ratio = 5.208e-9 / 1.333e-10 = 39.06 = (25 / 4)^2, so the deflection is 39 times smaller (2.32 / 0.0594 = 39.1).

(c) Flat bar, L = 0.30 m (1.5 times longer):
        delta = 2.32 x 1.5^3 = 2.32 x 3.375 = 7.83 mm
        sigma = 24 x 1.5 = 36 MPa
        SF    = 275 / 36 = 7.6

(d) Same section in ABS (E = 2.2 GPa): deflection scales with 1 / E, so the factor is 69 / 2.2 = 31.4.
        delta = 2.32 x 31.4 = 72.7 mm (the stress is unchanged at 24 MPa; ABS yields at roughly 40 MPa, so SF is only about 1.7, another reason not to copy a metal design in plastic).

(e) Largest tip load for a deflection of 1 mm (flat bar, L = 0.20 m):
        F = 3 E I delta / L^3 = 3 x 69e9 x 1.333e-10 x 0.001 / 0.008 = 0.0276 / 0.008 = 3.45 N
    (about 350 g; the load is limited by stiffness, not strength: the stress at 3.45 N is only 10.4 MPa.)

Check on method: h in the formula is the depth measured in the direction of the load, not "the longer side". A bar is stiffer standing on edge only if the load acts along its long side. A common slip is to use b and h the wrong way round.