# How Bridges Work 7, Exercise 2: how heavy must the anchorage be?
# Each cable of the bridge in Exercise 1 ends in an anchorage that must resist the
# horizontal pull H by friction alone (ignore the vertical component for this
# exercise). The friction coefficient is 0.6, the required factor of safety is 1.5,
# and concrete is 2400 kg/m3. For the three sag ratios of Exercise 1, find the
# anchorage's weight in kN, mass in tonnes and volume in m3. Then repeat for
# friction coefficients of 0.4 and 0.8 at the 1/10 sag.
def h_pull(w, span, sag):
    return w * span**2 / (8 * sag)

def anchorage(h_kn, mu=0.6, factor=1.5, rho=2400):
    weight = factor * h_kn / mu
    tonnes = weight / 9.81
    return round(weight), round(tonnes), round(tonnes * 1000 / rho)

for ratio in (8, 10, 12):
    print(f"1/{ratio}", anchorage(h_pull(60, 300, 300 / ratio)))

for mu in (0.4, 0.8):
    print("mu", mu, anchorage(h_pull(60, 300, 30), mu=mu))

# Output when run:
# 1/8 (45000, 4587, 1911)
# 1/10 (56250, 5734, 2389)
# 1/12 (67500, 6881, 2867)
# mu 0.4 (84375, 8601, 3584)
# mu 0.8 (42188, 4300, 1792)
#
# Output shows (weight kN, mass tonnes, volume m3) for each case. The size of an
# anchorage is directly proportional to H: the flat 1/12 cable needs 2,867 m3
# of concrete against 1,911 m3 for 1/8, 50% more. The friction coefficient
# matters as much: at 0.4 the 1/10 anchorage needs 3,584 m3, at 0.8 only
# 1,792 m3 (the value of 2,389 m3 at 0.6, scaled by 0.6/0.4 and 0.6/0.8).
# This is only a friction model: real anchorages also have to resist the cable's
# vertical pull, and many rely on being set into rock rather than on weight alone.