# How Bridges Work 6, Exercise 1: a cantilevered footbridge arm
# A steel cantilever arm projects 6 m from its support. It carries a uniform
# load of 12 kN/m and a 20 kN point load at its tip. I = 1.5e8 mm4, E = 200 000 MPa.
# Find the moment and shear at the root, and the tip deflection (add the two
# loads: uniform load w L^4 / (8 E I), tip load P L^3 / (3 E I)). Compare the
# deflection with an assumed limit of L/180, and find the I that would pass.
E = 200e3
w, P, L_m, I = 12, 20, 6, 1.5e8
L = L_m * 1000

m_root = w * L_m**2 / 2 + P * L_m
shear = w * L_m + P
d_udl = w * L**4 / (8 * E * I)
d_tip = P * 1000 * L**3 / (3 * E * I)
total = d_udl + d_tip
limit = L / 180
print(m_root, shear)
print(round(d_udl, 1), round(d_tip, 1), round(total, 1))
print(round(limit, 1), total <= limit)
print(round(I * total / limit))

# Output when run:
# 336.0 92
# 64.8 48.0 112.8
# 33.3 False
# 507600000
#
# Printed lines: root moment (kN m) and shear (kN); deflection from the
# uniform load, from the tip load and in total (mm); the limit L/180 and
# whether it passes; the I needed (mm4). The root carries 336 kN m. The
# uniform load alone gives 216 kN m there (wL^2/2), four times the 54 kN m
# (wL^2/8) of a simply supported beam of the same length and load. The
# tip deflects 112.8 mm against a limit of 33.3 mm, so it fails by a factor
# of 3.4; reaching the limit would need I of about 5.1e8 mm4 (507,600,000), more
# than three times the section. Cantilevers are stiffness-critical. The limit
# of L/180 is an assumed illustrative value.