Stop at the Sign
A 22 inch by 34 inch speed limit sign is supported on a 3 inch-wide, 5 ft-long pole. Estimate the bending moment in the pole at ground level when a 30 mph wind blows against the sign. Assume aspect ratio for the sign is less than 0.1. Assume the pole is a square rod with sharp corners. Also assume the density of air is 0.00238 slug/ft^3. Note the drag force coefficient of a squared rod with sharp corners is 2.2.
Expand Hint
The drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$$D=\frac{1}{2}\rho U^{2}C_{D}A$$$
where
$$C_D$$
is the drag coefficient,
$$U$$
is the flowing fluid or moving object’s velocity,
$$\rho$$
is the fluid density, and
$$A$$
is the projected area of blunt objects with axes perpendicular to the flow.
Hint 2
Take the moment about the origin/base of the pole:
$$$\sum M_{o}=0=Force \times Distance$$$
For equilibrium:
$$$\sum M_{origin}=0=M_{B}-M_{P}-M_{s}$$$
where
$$M_B$$
is the bending moment,
$$M_P$$
is the moment on the pole, and
$$M_s$$
is the moment on the sign.
Remember that
$$M=Force \times Distance$$
, so the bending moment can be written as:
$$$M_{B}=2.5ft\cdot D_{P}+(5+\frac{17}{12})ft\cdot D_{s}$$$
where
$$D_P$$
is the drag force on the pole, and
$$D_s$$
is the drag force on the sign.
The drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$$D=\frac{1}{2}\rho U^{2}C_{D}A$$$
where
$$C_D$$
is the drag coefficient,
$$U$$
is the flowing fluid or moving object’s velocity,
$$\rho$$
is the fluid density, and
$$A$$
is the projected area of blunt objects with axes perpendicular to the flow.
For the sign,
$$\ell/D< 0.1$$
so,
$$C_{D_s}=1.9$$
. The post acts as a squared rod with sharp corners, so
$$C_{D_p}=2.2$$
.
Thus, with
$$U=30mph=44ft/s$$
$$$D_s=\frac{1}{2}\rho U^{2}C_{D_s}A_s=\frac{1}{2}(0.00238\frac{slug}{ft^{3}})(44\frac{ft}{s})^2(1.9)(\frac{22\cdot 34}{144}ft^2)=22.7\:lb$$$
$$$D_P=\frac{1}{2}\rho U^{2}C_{D_p}A_p=\frac{1}{2}(0.00238\frac{slug}{ft^{3}})(44\frac{ft}{s})^2(2.2)(\frac{3\cdot 5}{12}ft^2)=6.34\:lb$$$
Finally,
$$$M_{B}=2.5ft\cdot (6.37lb)+(5+\frac{17}{12})ft\cdot(22.7lb)=162\:ft\cdot lb $$$
$$$162\:ft\cdot lb $$$
Time Analysis
See how quickly you looked at the hint, solution, and answer. This is important for making sure you will finish the FE Exam in time.- Hint: Not clicked
- Solution: Not clicked
- Answer: Not clicked
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