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Bus stop Bench (Intermediate)
Suggested viewing: Bus stop bench (Beginner)
This is a really well designed and detailed stainless steel bench, located outside the fire station on Eyre Street in the centre of Sheffield (i.e. around the back of the new Market at the
DESCRIPTION AND LOCATION
bottom of the Moor, and not far from Decathlon), the postcode for which is S1 3FG.APPLIED LOADS
The first thing to do with any structure is to think about the loads which are likely to be applied to it. These are usually fairly obvious, but not always. Indeed, structures are sometimes subject to abnormal loads, which the designer needs to try and anticipate, and design for as appropriate.
In this case, obvious loads include the self-weight of the bench itself, and the weight of people sitting (or even standing) on the bench, as well as any bags and packages they may have. There will be some wind load on the bench (although the holes in the seat will reduced vertical wind loads), although this is likely to be nominal, and can therefore be ignored.
Here, there is only one support (i.e. at the base of the vertical post) so all loads applied to the bench (including its self-weight) must be transferred through the horizontal seat, to the vertical post, and down to this support.
Imposed Loads (Variable Actions)
Unlike dead loads, we can never guarantee that the imposed loads on one side of the seat will exactly balance those on the other side. This is where the ‘see-saw’ analogy falls down (quite literally). Hence, the seat must cantilever about either side of the central post.
Indeed, any structural member which is only supported at one end must have a fixed support, unless you can guarantee that the loads will always be balanced by an identical member on the other side of the support, which is highly unlikely in reality.
Not so obvious loads might include somebody trying to kick the bench over (in other words, a horizontal load applied to the seat or at the top of the post, which is likely to be far more onerous than any wind horizontal loading). Not that this would be an accidental load case, so EC1 allows use to apply a reduced load factor of 1.0 to it, as opposed to the standard load factor of 1.05 for an imposed load.
CONCEPTUAL/QUALITATIVE BEHAVIOUR
Supports:
The trick with imposed loads is twofold: Firstly, you need to identify what the appropriate loads are (e.g. 1.5 kN/m2 (unfactored) might be adequate for a house, but for an office a value of 2.5 kN/m2 (unfactored) would be more appropriate).
Secondly, you have to consider the potential load combinations/scenarios. This is not such an issue for members in pin ended, braced, frames (such as simply supported steel beams) when maximum loads on all members will normally be critical. In frames where there is continuity however (e.g. rigid/unbraced steel frames and in-situ reinforced concrete structures) careful consideration needs to be given to the potential imposed load combinations/scenarios in order to determine the most onerous member forces, frame reactions and deflections etc.
Balanced Imposed Load Scenario
In the bench example, probably the most obvious load scenario is the seat fully loaded on each side. In reality, we could probably expect no more than 3 ‘British Standard’ adults weighing 90kg (or 0.9kN) each (unfactored) sat on the bench, as shown in the loading case below. This will give the maximum moment, shear and deflection in the seat itself, as well as the maximum (unfactored) axial compression of 3 x 90 = 270kg (or 2.7 kN) in the vertical post. However, there will be no bending moment or shear force in the post for this scenario, we can see from the respective diagrams.
Note the bending moment (‘M’) at any position along the seat is equal to the imposed load multiplied by the distance from the end of the seat to the position in question (‘x’) = 0.9x (kNm). Note also that this is a linear relationship, and explains why bending moment diagrams due to point loads are an inclined straight line in shape.
Note the shear force (‘S’) at any position (‘x’) along the seat is equal to the sum of the imposed loads applied between the tip of the cantilever and the position in question. In the case of a 90kg (or 0.9kN) single point load applied at the tip, S = 0.9 kN. This explains why shear force diagrams due to point loads are a series of horizontal straight lines in shape (with vertical lines, equivalent to the values of the applied point loads or support reactions, connecting the horizontal lines at applied load positions and supports respectively).
Finally, let us consider the deflected form diagram. This is similar in shape to the deflected form diagram for dead loads. Again, the deflection at the central post will be zero, whilst maximum deflection will intuitively occur at either end of the seat. It can be shown that the deflection in a cantilever subject to a point load of ‘P’ kN at the tip being equal to PL3/3EI. Note that this cubed relationship between the applied load and the span is the reason for the deflected form diagram due to a point load applied at the tip of a cantilever being a curved shape, as we can see.
Unbalanced Imposed Load Scenario 1
However, we cannot guarantee that the seat will always be fully loaded. Indeed, it is quite likely that just one person will sit on the seat. But where will they sit - centrally or on the tip, or in between? Clearly someone sat on the tip will be the more onerous condition, as below (obviously the case with someone sat on the right hand tip instead should also be designed for).
The bending moment diagram for the loaded half of the seat will essentially be like the previous case, with a moment at the root of the loaded half of the seat being equal to 0.9L (kNm). But where does this moment go? This is essentially a moment distribution problem, with two potential members into which the moment could be shared, in proportion to their relative stiffnesses.
However, you can’t distribute loads or moments into a cantilever beam. In moment distribution terms, the unloaded half of seat has zero stiffness, and its moment distribution factor is zero. Just remember that we are trying to get all the loads down to the supports or foundations, but there is no support at the right hand end of the unloaded half of the seat. Hence, all the load, and the full moment of 0.9L, must go into the vertical post.
The shear force diagram for the loaded half of the seat will essentially the same again. This shear goes into the post, which resists it as an axial compression, as shown.
Once again, we will finish by looking at the deflected form, which, for the loaded half of the seat, will be similar to the above. However, because the loading about the post is unsymmetrical, the post must also rotate, with the head of the post remaining at 90 degrees to the supported end of the left hand cantilever/seat.
Hence, the end of the left hand seat will deflect more here than before. Similarly, the supported end of the right hand cantilever/seat must be at 90 degrees to the post. Since there is no imposed load applied to the right hand cantilever/seat, its deflected form considering imposed loads only will take the form a straight line (the angle of which is determined by the tangent of the deflected form curve to the left hand seat at its junction with the post) as illustrated below.
If it helps, you could consider the post as an isolated member subject to a downward load of 90kg/0.9 kN plus an anti-clockwise moment of 0.9L (kNm) applied at its head:
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