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Design of a pitched roof for gravity loads
The behaviour of the roof under gravity loads is currently conceptualised as a truss system.
How do you check a Skylark pitch roof against gravity loads?
Pitched ROOF beams have similar section properties to FLOOR beams, and they are connected at the top by using a hinged connection. Therefore, the roof requires a tie element (for example a cable) to help in transferring the vertical loads to the walls.
Where should the tie be place?
The positioning of the tie depends on two main factors:
- Liveable space
- Structural efficiency
From a liveable space point of view, the higher is the positioning of the tie, the better.
From a structural efficiency point of view, the lower is the positioning of the, the better. The governing factor is likely to be the connection between the tie and beams. Therefore, the higher is the positioning of the tie, the stronger such connection needs to be to transfer the load into the timber.
Below, a worked example is reported where the tie is placed at 600 mm height above the walls.
Structural model of the roof
The roof was modelled by using beam elements with rotational springs. The properties of the beams and the rotational stiffness of the joints can be found elsewhere in this guide.
The tie was modelled as a truss element, i. e., it is able to carry only carrying axial forces. The load was applied as distributed load to the beam elements, and it was calculated following the steps below.
Load definition
The permanent load \(G_{beam}\) acting on the beam can be calculated according to the following equation:
\[G_{beam}= \overbrace{SW_{beam}}^{0.16 kN/m}+\overbrace{G_{imposed}}^{1 kN/m^2} \cdot \overbrace{i_{beams}}^{0.6m}=0.76 kN/m\]
where \(SW_{beam}\) the self-weight of the beams, \(G_{imposed}\) super imposed load and \(i_{beams}\) the inter-axis between the beams.
The value of the snow load \(q_{snow}\) depends on the location. In this example, \(q_{snow}\) is assumed to be equal to 0.6kN/m \(^2\). Hence, the linear accidental load on the beam \(Q_{beam}\) is equal to:
\[Q_{beam}= \overbrace{q_{snow}}^{0.6 kN/m^2} \cdot \overbrace{i_{beams}}^{0.6m}=0.36 kN/m\]
Load combinations
The value of loads corresponding to the frequent \(q_{fr}\) , quasi-permanent \(q_{qp}\) and rare \(q_{rar}\) load combinations are calculated below:
\[q_{fr} =\overbrace{G_{beams}}^{0.76}+\overbrace{\psi_2}^{0.7}\overbrace{Q_{beams}}^{0.36}=1.0\,kN/m\]
\[q_{qp} =\overbrace{G_{beams}}^{0.76}+\overbrace{\psi_2}^{0.3}\overbrace{Q_{beams}}^{.36}=0.87\,kN/m\]
\[q_{ra} =\overbrace{\gamma_G}^{1.35}\overbrace{G_{beams}}^{0.76}+\overbrace{\gamma_Q}^{1.5}\overbrace{Q_{beams}}^{0.36}=1.6\,kN/m\]
Structural demand
The maximum values of:
- tensile force on the tie \(T_{sd}\)
- compression force on the beam \(N_{sd}\)
- bending moment on the beam \(M_{sd}\)
- shear force on the beam \(V_{sd}\)
- deflection on the beam \(f\)
are reported in the table below. Results were found by modelling the roof according to the structural scheme presented above into finite element software.
Serviceability limit state (SLS) on the beams
At the serviceability limit state (SLS), it is checked that:
- the maximum elastic deflection under the frequent load combination is lower than a threshold of L/250 (with L the span)
- the long term deflection under the quasi permanent load combination is lower than a threshold of L/250.
The elastic deflection of the beam can be calculated by using the following equation:
SLS short tem - frequent load
It is checked that:
\[d_{max,el}=3.3<\frac{L}{250}=12.9\ mm\]
SLS long tem - permanent load
A check on the long term deflection \(d_{max,lt}\) is reported below. It is assumed that the beams are working in Service Class I, defined by Eurocode 5 as: characterised by a moisture content in the material corresponding to a temperature of 20°C and the relative humidity of the surrounding air only exceeding 65% for a few weeks per year
which leads to the following value of creep factor \(k_{def}=0.8\).
It follows that:
\[d_{max,lt}=d_{max,el}\frac{q_{qp}}{q_{fr}}(1+k_{def})=5.1 \ mm\]
and it is checked that:
\[d_{max,lt}=5.1 <\frac{L}{250}=12.9 \ mm\]
Ultimate Limit State (ULS) on the beams
At ULS, beams should be checked against moment and shear.
Bending moment check
\[M_{sd,ra}=2.9\le\frac{\overbrace{k_{mod}}^1\overbrace{M_{rd}}^{17.8}}{\underbrace{\gamma_d}{1.25}}=14.2\ kNm\]
\[M{sd,qp}=1.6\le\frac{\overbrace{k_{mod}}^{0.6}\overbrace{M_{rd}}^{17.8}}{\underbrace{\gamma_d}_{1.25}}=8.5\ kNm\]
Shear check
Assuming the shear is carried by the beam’s web, the shear area of the beams can be calculated as:
\[A_v=\frac{5}{6}\overbrace{A_{web}}^{2\cdot310\cdot18}=9300\ mm^2\]
That leads to the following shear strength:
\[V_{rd}=A_v\overbrace{f_{sd}}^{3.5}=32.3\ kN\]
It follows:
\[V_{sd,ra}=3.8\le\frac{\overbrace{k_{mod}}^1\overbrace{V_{rd}}^{32.3}}{\underbrace{\gamma_d}_{1.25}}=26.0\ kN\]\[V_{sd,qp}=2.1\le\frac{\overbrace{k_{mod}}^{0.6}\overbrace{V_{rd}}^{32.3}}{\underbrace{\gamma_d}_{1.25}}=15.5\ kN\]
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