Question

A circular sign has a diameter of 40 cm and is subjected to normal winds up to 150 km/h at 10°C and 100 kPa. Determine the drag force acting on the sign. Also determine the bending moment at the bottom of its pole whose height from the ground to the bottom of the sign is 1.5 m. Disregard the drag on the pole. The drag coefficient for a thin circular disk (sign) is CD = 1.1..

Answer 1

Answer:

147.7 N

221.55 Nm

Explanation:

P = Pressure = 100000 Pa

$R_s$ = Mass-specific gas constant = 287.015 J/kg k

T = Temperature = 10+273 = 283 K

C = Drag coefficient = 1.1

A = Area

r = Radius = 0.2 m

v = Speed of wind = $\frac{150}{3.6}\ m/s$

L = Length of pole

Density

$\rho=\frac{P}{R_sT}\\\Rightarrow \rho=\frac{100000}{287.058\times 283}\\\Rightarrow \rho=1.2309\ kg/m^3$

Drag force

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2309\times 1.1\times \left(\pi \times 0.2^2\right)\times \left(\frac{150}{3.6}\right)^2\\\Rightarrow F=147.7\ N$

Force on the circular sign is 147.7 N

$M=F\times L\\\Rightarrow M=147.7\times 1.5\\\Rightarrow M=221.55\ Nm$

Bending moment at the bottom of the pole is 221.55 Nm