Multi-storey steel building

Footprint of building. The middle frame which includes main beam will be studied. For that purpose we work with assumption that all load within the red rectangle belongs to the frame

For multi-storey steel building, we have two main options:

• the stiffness is provided by moment resistant frame or
• the stiffness is provided by some kind of bracing system (and the structure uses simple connections).

There are also other approaches like for example partially restrained frames but these are complicated compared to the aboves.

This studied building has footprint as depicted above. The height is 9 floors, 4.5 m each. The columns are assumed to be simple supported at their footings. Only the middle frame will be taken and analysed. The main beam within each floor transmitts load from the secondary beams, which support slab (the secondary beam load will appear as a single force in the middle of the span of the main beam).

Wind load

While for small buildings wind load might be almost negligible and it is gravity load what determines sections, for tall buildings wind load is main component for analysis. Wind load increases with altitude and since it acts on a great arm (height of the building) it works with leverage.

For wind load the basic wind speed was taken as 30 m/s. Characteristic speed, which is related to a topology and height was determined as

height $\boldsymbol{z}$ [m]  $\boldsymbol{v_{n}(z)}$ [m/s]
10 22.6
20 27.0
30 29.7
40 31.6

The maximum characteristic wind pressure from the characteristic speed and with turbulences' impact

height $\boldsymbol{z}$ [m]  $\boldsymbol{q_{p}(z)}$ [Pa]
10 956
20 1215
30 1389
40 1517

Finally according to the shape/disposal of the building we can get load on the both front and back side. Wind load grows with height. If one side is taken as a front side (coefficient $c_{pe} = 0.8$), then wind pushes this wall while the opposite side is being sucked by the wind $(c_{pe} = -0.6)$.

height $\boldsymbol{z}$ [m]  $\boldsymbol{q_{w+}(z)}$ [kN/m]  $\boldsymbol{q_{w-}(z)}$ [kN/m]
10 765 -574
20 972 -729
30 1111-833
40 1214-910

Finally we want to convert Pascals to Newtons per meter. Since the wall—which is being assumed as a part for the main frame—has depth of 9 m, the wind load for the frame is stated as

height $\boldsymbol{z}$ [m]  $\boldsymbol{q_{w+}(z)}$ [kN/m]  $\boldsymbol{q_{w-}(z)}$ [kN/m]
10 6.9-5.2
20 8.8-6.6
30 10.0-7.5
40 10.9-8.2

Dead load

Dead load (or gravity load) is simplified to

kN/m2   $\boldsymbol{\gamma_f}$   $\boldsymbol{w_d}$ [kN/m2
Flooring system (slab)2.51.152.9
Equipment 0.71.250.8
Live load 3.01.253.8
$\sum$ 7.5

Since we want to process 2D frame, we have to convert the load to kN/m according to the depth of 9 m. Load is $q_d = w_d \times 9 = 67.5$ kN/m. Within one field (of 8 m) this load is transmitted in 3 places: in exterior column (2 m × 67.5 kN/m = 135 kN), interior column (270 kN) and in the middle of the beam where main beam supports the secondary beam (270 kN).

Model

A 2D model of the frame. For the start all members are considered with the same stiffness
Reactions for the above model

Sections

Based on the observed internal forces sections are suggested. This website was helpful to propose a design since for any profile a strength in tension, compression and bending is readily available.

• Exterior column: UPE 300×4. The built-up column 400 × 600 mm is assumed to form a new rigid section which is able to transmitt bending moments. Area of built-up section is 226 cm2, using tabled values and Steiner's theorem moment of inertia was determined as I = 69,100 cm4.
• Interior column: UPE 400×4. The built-up column 400 × 800 mm. A = 367 cm2, I = 110,600 cm4.
• The horizontal beam: IPE500×2: A = 232 cm2, I = 96,400 cm4
• For the upper part of the building there is no need to use such heavy members. Exterior column for upper part: UPE200×4, 300 × 400 mm: A = 116 cm2, I = 18,700 cm4
• Interior column for upper part: UPE270×4, 300 × 540 mm: A = 179 cm2, I = 27,840 cm4
• The horizontal beam for upper part was chosen as IPE400×2: A = 169 cm2, I = 46,200 cm4

These are not optimum sections. But we have quite meaningful sections to start with. We have to expect different internal forces once the model is calculated with non constant stiffnesses.

Model with sections assigned: results

The internal forces are now slightly different so we should consider new values, provide slightly different sections or more efficient sections, run the next iteration, make an assessment. But for the time being and our demonstrative purpose let us accept that the sections are designed appropriately since we want to step in to read deflections.

(moments)—within the beams the extremes are around 300 kNm except the ground floors where moments are around 1000 kNm due to the lateral load
(deflected shape)—the deflections (exaggerated) of the moment resistant frame on the top floor is almost half of meter from the assumed vertical load

The horizontal deflection on the top floor was found to be 0.38 m (see the above picture). The deflection from a vertical wind load is quite disturbing and it was taken only as a static load. For a case of resonance the deflections would be greater, perhaps much greater and with fatigue involved would threaten to bring a collapse. So, even if the frame could resist the load in terms of strength, the deflections would be unacceptable. The moment resistant frame is not feasible solution for tall buildings. The vertical deflection at the middle of the spans is almost 2 cm.

If moment resistant frame can not be used for tall buildings, what about third floor then? On the top of the 3rd floor the deflections observed are around 0.2 m. Even for three-storey building such displacement can not be allowed. The trick is we have to remove all floors above the 3rd floor firstly to show real displacement for three-storey building (the top's floors of the building act as a lever): vertical deflection is around 3.2 cm then.

(three-storey building only)
Leaving the same sections, the deflections of the moment resistant frame on the top floor is 3.2 cm from the assumed vertical load

Modified frame: braces

From the above results it is reasonable to conclude that the frame has to be equipped with braces. That improves stiffness significantly and likely gets the deflections under control.

For bracing, profiles L120x120x12 were considered. These are simply connected to the frame.

Once the frame is solved, it shows roughly 10 times smaller vertical displacement: at the top floor 4.4 cm (compare to 0.38 m on unbraced frame). According to the model, the most exposed brace has to resist 420 kN in compression. Profile L120x120x12 can transmitt 650 kN in pure compression ($\gamma_{M0} = 1$). It means that for most of the braces the angle is sufficient, only at the bottom floors section has to be adjusted to a stronger one (buckling). Besides that the frame is stiff as the whole, therefore the beams do not show excesive bending moments (as unstiffened frame does).

The deflections of the moment resistant frame with braces on the top floor is 4.4 cm from the assumed vertical load

The logical step is to remove the moment resistant connections (of beams to columns), which are expensive and replace them by simple connections. The maximum vertical deflection of the top floor is only slightly higher then (5.6 cm). But the structure is simpler and cheaper.

Internal moments on braced version of frame which uses simple connections between beams and columns

Summary

The example is highly simplified and its purpose is not to make design and/or assessment. The purpose is to demonstrate the difference between moment resistant frame, which is suitable for a few-storey buildings against braced frame which uses pinned connections. The latter is cost efficient for tall buildings.

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