# How Bridges Work 6, Exercise 2: how the suspended span changes a cantilever
# Use the average dead load of the Forth Bridge from the chapter (51,324 tonnes over
# 2,467 m, in kN/m), a cantilever arm of 207.3 m on each side of a tower, and
# suspended spans of 0, 53, 106.7 and 160 m resting on the arm tips (each arm
# takes half the suspended span's weight as a tip load). For each case print the
# moment at the tower, the tower's reaction, and the chord force for a depth of 100 m.
G = 9.81
w = 51324 * G / 2467
a = 207.3
depth = 100

for s in (0, 53, 106.7, 160):
    p = w * s / 2
    m = w * a**2 / 2 + p * a
    reaction = 2 * (w * a + p)
    print(s, round(m), round(reaction), round(m / depth))

# Output when run:
# 0 4385196 84615 43852
# 53 5506350 95432 55064
# 106.7 6642313 106392 66423
# 160 7769813 117270 77698
#
# Columns: suspended span (m), moment at the tower (kN m), tower reaction (kN),
# chord force at 100 m depth (kN). With no suspended span (0) the arms still
# produce 4.39 million kN m at the tower; adding the 106.7 m suspended span of
# the real bridge lifts that to 6.64 million, and each extra 50 m of suspended
# span adds roughly a million more. The suspended span also loads the arm tips,
# so both the moment and the tower reaction grow with it. The arithmetic is a
# simple uniform-load model; the real bridge's weight is not uniform.