# Mass required to prevent sign falling over with a set wind load - activity stations for disabled children

I'm currently working on my thesis and I'm stuck on a question. I'm designing activity stations for disabled children to be used for equine therapy.

The stand is 9ft tall and I've calculated the the wind load at 0.94lbs, now I need to calculate the mass required to stop the station falling over (the concrete base is not a foundation sunk in the ground - it needs to remain portable)

The frontal area of the stand is evenly spread. Any help would be appreciated.

• It depends on how the mass is distributed at the base. For example, if the mass is concentrated approximately at a single point directly below the sign on the ground, then you need a nearly infinite amount of mass. If the mass is on a leg that is far from directly below the sign, then you need very little mass. The distribution of the base mass affects the lever arm of the base mass, which affects the torque it can provide when the wind attempts to tip the sign. – Brionius Mar 22 '16 at 12:44
• Thanks for the reply, the mass is evenly distributed, would you know any formula for such an equation? – John Dooley Mar 22 '16 at 13:35
• Are you sure the wind load is only 0.94 lbs? That seems very low. Imagine a 1 lbs weight...that's not a lot of force. – Brionius Mar 22 '16 at 14:20

Assuming a square base of width $w$ with mass $M$, and a horizontal wind load $F_w$ at a height $h$, then the condition for static equilibrium is

$$\sum\tau = 0$$ $$\tau_{wind} + \tau_{base} = 0$$

Since if the sign tips, it would rotate around the edge of the base, that's a convenient axis about which to compute the torques: $$- F_w \frac{h}{2} + M g \frac{w}{2} = 0$$

Solving for M...

$$M = \frac{F_w h}{g w}$$

Plugging in your numbers (0.94 lbs-force = 4.18 N, 9 ft = 2.7 meters),

$$M = \frac{11.29 ~\rm Nm}{9.81 ~\rm{m/s^2} ~w}$$ $$M = \frac{1.15 ~\rm kg~m}{w}$$

If you prefer lbs and feet,

$$M = \frac{8.32 ~\rm lbs~ft}{w}$$

For example, if your square base has a width of 3 feet, then you need a minimum of $M = 2.77 ~\rm lbs$

Note 1: This analysis assumes the mass of the sign itself is negligible compared to the base. This is a conservative assumption, since extra mass on the sign will make it more stable for initial tipping.

Note 2: I'm very skeptical that a sign of any significant size would only experience a wind load of 0.94 lbs in any significant wind. I would double-check that figure.

EDIT: I revised my answer now that the OP made it clear that the sign is a rectangle that extends from the ground up to 9 feet.

• Brilliant thanks again for the reply, the front of the sign is a half sphere shape with low drag of .42, i recalculated the wind load for the rear which is flat, it was 3.18lbs, the wind speed was also low - 56mph, does this still seem wrong? – John Dooley Mar 22 '16 at 14:44
• 56 mph is very nearly hurricane-force winds. Unless your sign is very small, 3.18 lbs still seems quite low. For a 3 ft x 3 ft square in 56 mph winds, the empirical formula I found indicates that the wind pressure would be approximately 85 lbs. – Brionius Mar 22 '16 at 14:49
• the actual sign is only 9" wide, does this still seem out? – John Dooley Mar 22 '16 at 14:58
• @JohnDooley: how did you estimate or measure the 0.94 lb load figure? – Gert Mar 22 '16 at 15:03
• @JohnDooley: I also disagree with Brionius using the full height of the panel to calculate $\tau_{wind}$. The load doesn't act on the top of the panel, it acts on the CoG of the panel, at $\frac{H}{2}$. – Gert Mar 22 '16 at 15:08 The weight of the base needed to prevent toppling of the stand also depends on it's width $x$.

In order to prevent toppling about the point $P$, there must not be a net moment about that point.

This means mathematically that:

$$F\frac{H}{2}=mg\frac{x}{2}$$

Where $H=9\:\mathrm{ft}$ and $F=0.04\:\mathrm{lbs}$. I'm assuming the load you assigned acts on the centre of gravity of the vertical panel and that the mass of the panel is negligible.

So the mass $m$ required is:

$$m=\frac{FH}{gx}$$

Note that $F$ and $H$ need to be converted to S.I. units, if you want to use $g=9.81\:\mathrm{ms^{-2}}$ as the Earth's acceleration.

• Could you explain how the factor of $\frac{1}{2}$ arises in the left side of your first equation? – Brionius Mar 22 '16 at 14:51
• @Brionius: I assume the wind load to be homogeneous, so it can be replaced by a force acting on the centre of gravity of the panel. I'm surprised you didn't do the same. – Gert Mar 22 '16 at 14:53
• I assumed the center of gravity of the panel was 9 feet above the ground, since the OP says the "stand" is 9 feet tall, not the panel itself. Since the OP has said that the sign is 9" on a side, that seems reasonable. That is, unless the wind loading on the post holding up the sign is significant. – Brionius Mar 22 '16 at 15:08
• @Brionius: The OP now stated that the sign is a half-sphere. Strictly speaking we would have have to consider the wind moment to act on the CoG of that half-sphere. In reality it won't matter so much, because the OP will have to factor in a safety margin, e.g. make the actual mass about twice the calculated one, to cover unexpected gusts of wind and such like. – Gert Mar 22 '16 at 15:26
• Sure, but if the wind force on the stand is negligible, and the wind force on the sign is centered in the middle of the sign, then the wind force is applied 9 feet above the ground, not 4.5 feet above the ground. Which is why I don't think you should use that factor of $\frac 1 2$. – Brionius Mar 22 '16 at 15:29