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Viper Rooms Sign (Intermediate)
Suggested viewing: Viper Rooms sign (Beginner)
Imposed Loads (Variable Actions):
The critical imposed load incident on the sign is mainly the wind load. For analysis we will have to consider the column from another dimension than the one in the previous section. Except in really high structures, wind load is often uniform; that’s how we will model it here. Now to deal with loads spread over an area we can consolidate them, exactly like dealing with UDL acting in a line. A conservative assumption for the wind action is 1.2 kN/m2 and assuming the sign is 1.5X1.0 meters. We can consolidate the wind load into one point load acting on the centre of the sign: Force = Pressure * Area
Force = 1.2 * (1.5 * 1.0) = 1.8 kN
So now we have a force of 1.8 kN acting on the centre of the sign. Now it is important to notice that the sign is attached to the column using four connections. The connections are symmetrical so we can assume that the force 1.8kN is divided over the four connections, each carrying 0.45kN. The diagram we have now will help us further analyse this problem. We can see that although the two slender columns that the forces act on are connected at their top, the top isn't restrained and the two slender columns experience the same loads. Effectively, the presence of the top bar is redundant. Now, to construct the bending moment diagram we need to find the value of the moment at two points, at the lower load and at the axis. Because, we already know that the moment at the upper load is zero; we have two points of interest left, the lower load and the axis.
To find the moment at the axis we add the moments generated by both loads at that point.
Now we can construct the BMD.
A very important note to take away from this example is that structures have different BMDs in different dimensions. A structure needs to be able to support all critical applied loads in any direction. The total moment in the axis is the sum of both moments so it is 2250 Nm.
The horizontal arm transfers the load to the main column through horizontal shear, horizontal bending and torsion.
This moment needs to be resisted by an opposite moment. To figure out where this opposite moment acts, let’s imagine how the sign would deflect in extreme wind.
We can see that the single column also deflects, meaning it experiences moment.
The only point that doesn't move is the bottom of the column, which is where we expect the reaction moment to exist naturally.
We know that the moment at the top of the column is 2250 Nm, and that it needs to be transferred to the ground. There will also be the horizontal force (1800 N) transferred to the column so the moment will increase linearly as we move from the top of the column to the ground. Assuming the column is 2 meters high then the increase in the moment will be 3600 Nm; making the moment at the bottom of the column 5580 Nm.
OTHER THINGS TO THINK ABOUT
Think about what the BMD would look like parallel to the sign and not perpendicular to it.
Furthermore you might have come across a case where a concentrated moment just appears in a simply supported beam, like the one below.
Think about how the BMD would look like, and what loading case would produce such moments.
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