# Introduction

## What is WikiHouse Skylark?

WikiHouse Skylark is a high-performance manufactured building system for small buildings. It uses interlocking, CNC-machined plywood or OSB to create cassettes or 'subassemblies' that can be manufactured in small factories, and rapidly assembled on-site, even by teams with no traditional construction skills.

WikiHouse is open source (in the same way that SIPs, steel framing and other generic, non-proprietary structural methods are). The chassis structure serves as a generic hardware platform that interfaces with a wide range of possible cladding, lining, MEP products and foundation systems.

## What are the main structural challenges?

As you will see from the testing data shared in this guide. The main structural design constraint for WikiHouse Skylark is usually not gravity, but wind. In high winds, lightweight structures are more prone to slight lateral flexing, which is not allowed within most building codes, because it could lead to issues like cracking windows or internal finishes.

Further research, testing and development is ongoing, but in summary, the main challenges can be described as:

• Ensure designs are laid out such as to provided sufficient resistance to lateral loads for any given site given its exposure to wind etc.
• Ensure foundation installers are aware of any likely upward, as well as downward loads on foundations.

## What information will your client need?

Your client is likely to require:

• A report certifying that the proposed design  will meet the relevant building regulations of codes (for example, in England, Part A of the Building Regulations), based on its location and proposed use. To achieve this, you may have to give iterative feedback to the design team, so they can modify their design to meet the requirements.
• A summary of the safe capacity limits of the structure, if possible (eg what live and dead loads can be applied to the chassis)
• A summary of the load demands on the foundations, which they will be able to give to their foundation engineer to design and specify the building foundations.

## Terms for understanding the structure

In order to have a common framework for understanding and talking about the system, we can break it down into a hierarchy of nested structural units:

Materials → Connections → Blocks → Assemblies → Building

### Materials

Materials are the as-purchased products that Skylark is made from, specifically 18mm structural plywood or OSB, along with any fixings.

### Connections

Elements are recurring patterns for how these materials are interlocked with each other. Some of these connections always happen within a block (such as a stitch joint or dovetail joint), whereas others (such as Pegs or Bow ties) are generic, independent components in their own right.

### Blocks

Blocks are sub-assemblies that form the LEGO-like building blocks of a building, for example 'WALL M' or 'FLOOR M'.

View and download blocks from the blocks library

### Assemblies

Assemblies are combinations of blocks that together form a significant element of a building (such as a wall or floor). This concept aims to make it simpler to understand and test the behaviour of the whole system, when many interlocking parts are working together in sometimes complex ways.

––

In this guide we have shared data obtained through testing at every level of this hierarchy. All data is shared under an open licence, with no warranties, but is documented as transparently as possible (along with its provenance) to give you as much information as possible and to drill down into any level of detail.

# Blocks

## Floor blocks

The floor blocks are the main beam units.  Below are some useful design parameters, and information on how those quantities were derived.

 Parameter Symbol Value Units Note Design moment capacity L (mid span equivalent) $$M_{mid-eq,d,L}$$ 19.6 kNm Design value (5th percentile) of moment capacity in the mid span equivalent to distributed load Design moment capacity M (mid span equivalent) $$M_{mid-eq,d,M}$$ 18.8 kNm Design value (5th percentile) of moment capacity in the mid span equivalent to distributed load Design moment capacity S (mid span equivalent) $$M_{mid-eq,d,S}$$ 20.8 kNm Design value (5th percentile) of moment capacity in the mid span equivalent to distributed load Average moment capacity L (mid span equivalent) $$M_{mid-eq,avg,L}$$ 23.1 kNm Average value of moment capacity in the mid span equivalent to distributed load Average moment capacity M (mid span equivalent) $$M_{mid-eq,avg,M}$$ 21.9 kNm Average value of moment capacity in the mid span equivalent to distributed load Average moment capacity S (mid span equivalent) $$M_{mid-eq,avg,S}$$ 24.2 kNm Average value of moment capacity in the mid span equivalent to distributed load Design moment capacity of the joint $$M_{j,d}$$ 16.5 kNm Design value (5th percentile) of moment capacity at the side dovetail joint Design moment capacity of the joint $$M_{j,avg}$$ 19.5 kNm Average value of moment capacity at the side dovetail joint Second moment of inertia (elastic) $$I_{el}$$ 523005 cm4 Elastic value of the second moment of inertia Second moment of inertia (effective)* $$I_{eff}$$ 28765 cm4 Effective value of the second moment of inertia (55% Iel ) Rotational stiffness $$k_R$$ 2295 kNm/rad Rotational stiffness of the castellated joint

*Recommended

Based on data from testing 8 specimens

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh. A full peer-reviewed study can be viewed here.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

Five M beam specimens and three L beam specimens were tested in 4-point bending. M specimens had a span of 5.1 m (column centre to column centre), and were 350 mm high and 600 mm wide. L specimens had the same section but spanning 5.7 m (column centre to column centre).

The load exerted by the  hydraulic actuator was applied to the timber beams by using a 1.7 m long UC 203 x 203 x 71mm steel profile and two 10 x 500 x 800mm  steel plates. Two 40 mm diameter steel bars were welded to the plates to create a pin connection between the load spreading beam and the timber specimens. The plates were positioned so that the forces push on the timber specimen at plus/minus 1.7 m from the centre of the beam.

### Data acquisition

The following instrumentation was installed for data acquisition:

• Linear potentiometers were used to track the specimen displacement profile.  Five of them were mounted to measure the beam's vertical deflection, while 2 of them were mounted to measure the gap opening between the beam and the column.
• A 250 kN-capacity load cell was used to measure the actuator force.
• Two high resolution cameras were used to to track the strain profile by using digital image correlation.  Since the cameras' configuration geometry allows them to focus on  roughly 1.5 m length of the specimen,  their position was changed  between tests.

The loading protocol approximately follows the guidance given in EN 1380:2009 for connectors in timber structures. Although the test is for a complete element rather than a connection, its behaviour is driven by local behaviour at connections between parts, and the cycle of loading recommended in that testing standard was used to capture the different stiffness of the system under initial loading, and the unload-reload cycle. The test was carried out in displacement control,  with the actuator moving at 3 mm per minute. The loading protocol consisted a monotonic ramp of 22 mm (roughly span divided by 250), a one minute hold,  a monotonic ramp back to 5 mm,  a one minute hold, and then a monotonic ramp until failure.

### Failure mechanism

The weakest point of both the M and L specimens was found in the web's dovetail joint.

The specimen failure was initiated in the stitched joint, where local compressive failure occurred at the contact between the tab on the bottom flange and the web, as reported in the figure below. In addition to the stitched joints, damage was also observed in the dovetail joints, especially the joint placed at the bottom flange of the specimens.

It can also be observed that the joint panel opened a gap with the mock column, while no column uplift was identified during the test.  This suggests the joint behaves as a pin, i.e. shear panel can only transfer a negligible (if any) amount of bending moment to the column.

All M specimens failed in a ductile manner: the specimens were able to maintain the load for a while after reaching the peak force. The response of the specimens in terms of force (exerted by the actuator) vs mid-span displacement is reported in the graph and table below. The peak force occurred between 21.3 kN and 24.9 kN, depending on the specimen.

 Specimen Material $$F_{max}$$ (kN) $$M_{j,max}$$ (kNm) S1-M Metsa spruce plywood 24.9 20.9 S2-M Metsa spruce plywood 22.1 18.5 S3-M Metsa spruce plywood 21.3 17.8 S4-M Metsa spruce plywood 22.1 18.5 S5-M Metsa spruce plywood 23.1 19.3

L specimens showed less ductility than M specimens. That was because the web panels started to fall out of plane after the yielding of the stitched joints. The response of the specimens in terms of force (exerted by the actuator) vs mid-span displacement is reported in the graph and table below. The peak force occurred between 19.3 kN and 25.8 kN, depending on the specimen.

 Specimen Material $$F_{max}$$ (kN) $$M_{j,max}$$ (kNm) S1-L Metsa spruce plywood 25.8 21.8 S2-L Metsa spruce plywood 19.3 16.3 S3-L Metsa spruce plywood 24.3 20.6

### Capacity

The moment capacity at the dovetail joint (weakest point) was calculated as:

$M_j=\frac{F_{max}*a}{2}$

where $$a$$ is the distance from the dovetail joint to the support, equal to 1693 mm for the L specimens and 1675 mm for the M specimens.

To provide the beams' capacity for design purposes according to the Eurocode 5, experimental data were fitted with a normal distribution function. Note that the number of specimens is limited to 8 and hence the sample size should be increased for providing more statistical significance.

The average capacity of the joint was 19.2 kNm, and the design value corresponding to the $$5^{th}$$ percentile was estimated as 16.5 kNm.

For design purposes, the equivalent moment in the mid-span is provided, assuming the beams are subjected to uniform distributed load as per figure below.

The equivalent moment in the mid-span $$M_{mid-eq}$$ can be calculated as:

$M_{mid-eq}=\frac{M_jL^2}{4a(L-a)}$

with $$L$$ the span, and $$a$$ the distance from the dovetail joint to the support. Average and design values are reported in the first table of the section.

### Joint stiffness

The relative rotation occurring in the castellated joint was estimated by using the horizontal differential displacements across the joint itself measured by using Digital Image Correlation. Specifically, the differential displacements in 3 points along the section depth were fitted with a linear function so that the slope represents an estimate of the relative rotation.

The moment rotation relationship of the joint is then reported in Figure below. It can be noticed that the relationship appears to be linear until the moment reaches 7.5 kNm. The elastic initial rotational stiffness $$k_R$$, calculated between 10% and 40% of the peak value of the moment, is equal to 2295 kNm/rad.

### Deflection

The deflection of the beam was calculated by:

• A traditional beam model (Euler-Bernoulli) and considering an elastic second moment of inertia
• A traditional beam model (Euler-Bernoulli) + rotational springs and considering an elastic second moment of inertia
• A traditional beam model (Euler-Bernoulli) + rotational springs and considering an effective second moment of inertia

and compared with the experimental results.

It was found that:

• Not considering the joint flexibility (i.e. rotational springs) underestimates the deflection
• Using the elastic moment of inertia $$I_{el}$$ also underestimates the overall deflection.

Hence, it is recommended to take into account the joint flexibility as well as using an effective moment of inertia $$I_{eff}$$ equal to 55% $$I_{el}$$ or lower.

## Wall blocks

Wall blocks are the main column units.  Below are some useful design parameters, and information on how those quantities were derived.

### Engineering design parameters (Plywood)

 Parameter Symbol Value Units Note Axial capacity (average) $$N_{rd,avg}$$ 140 kN Average value of axial capacityy Axial capacity (design) $$N_{rd,d}$$ 116 kN Design value of axial stiffness Axial stiffnesss $$k_{a,avg}$$ 14.3 kN/mm Average value of axial stiffness

Based on data from testing 4 specimens

### Engineering design parameters (OSB)

 Parameter Symbol Value Units Note Axial capacity (average) $$N_{rd,avg}$$ 160 kN Average value of axial capacityy Axial stiffnesss $$k_{a,avg}$$ 14 kN/mm Average value of axial stiffness

Based on data from testing 1 specimen

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh. A full peer-reviewed study can be viewed here.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

Five column specimens were tested in compression to investigate their capacity to carry gravity loads. Specimens comprised 4 columns made of plywood (labelled PLY) and 1 column made of OSB.

They were 2.4 m long , and were 300 mm high and 600 mm wide. They are manufactured by assembling 18 mm plywood or OSB panels, which were cut out of a 2.4 m per 1.2 m sheet using a 3 axis CNC machine. The load exerted by the hydraulic actuator was applied to the timber columns by using a 25 x 350 x 650 mm steel plate. The plate was hung from the crane so it can freely rotate to follow the specimen deformation.

### Data acquisition

The following instrumentation was installed for data acquisition:

• Two linear transducers were used to track the specimen axial displacement.  They measured the displacement between the steel plate and the steel frame.
• A 600 kN capacity load cell was used to measure the compression force. The load cell was mounted on the opposite side of the actuator with respect to the specimen. Two 10 x 800 x 650 plates were placed between the specimen and the load cell.

The loading protocol consists of a monotonic load increasing until failure. The loading rate was approximately 40 kN per minute.

The response of the specimens in terms of force (exerted by the actuator) vs axial displacement is reported in the figure below. Note that the axial displacement was taken as the average between the measurements of the two linear transducers.

All specimens failed in a brittle manner: load capacity dropped rapidly after reaching the peak force.The peak force occurred between 120 kN and 160 kN. Note that the the load data appear segmented due to the load cell sensitivity, which is approximately 15 kN.

 Specimen Material F max (kN) Ka (kN/mm) S1 Metsa spruce plywood 134 14.4 S2 Metsa spruce plywood 120 14.1 S3 Metsa spruce plywood 147 14.2 S4 Metsa spruce plywood 160 14.6 S5 OSB 160 14

### Failure mechanism

In all 5 specimens, the failure was initiated by the buckling of the bottom panel of the specimen. This corresponds to the external face of the wall when the column is installed in a building. This occurs because this panel has significantly less restraint in buckling than the the top or lateral panels. In fact, while the buckling of the top or lateral panels is restrained by at least one series of stitched joints, the buckling of the bottom panel is only restrained by the pull-out force of the nails. Once the bottom panel buckled out of plane, the whole element failed immediately after.

### Capacity

To provide the beams capacity for design purposes according to the Eurocode 5, experimental data were fitted with a normal distribution function. Note that the number of specimens is limited to four plywood and hence the population should be increased for providing more statistical significance. For OSB, only one specimen was tested, and hence more data are necessary to provide any meaningful statistics.

The average capacity was 140 kN, and the 5th percentile was estimated as 116 kN.

### Axial stiffness

The column elastic stiffness $$K_a$$ of the specimens was calculated according to the equation below, based on EN 1380:2009.

$K_{a}=\frac{F_{40\%}-F_{10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{40\%}, F_{10\%}$$ represent 40% and 10% of the maximum force $$F_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

The elastic axial stiffness values $$K_{a}$$ are between 14.2 kN/mm and 14.6 kN/mm for the plywood specimens, and 14.0 kN/mm for the OSB specimen

# Connections

## Bow ties – in shear

Bow ties are one of the main connectors. They are put into the face of the chassis to connect them together, to form a larger assembly. Below are some useful design parameters for how they perform against shear forces, and information on how those quantities were derived.

### Engineering design parameters (Plywood)

 Parameter Symbol Value Units Note Parallel shear capacity (average) x2 $$V_{rd,\parallel,avg}$$ 11 kN Average value of shear capacity (parallel) using 2 bow ties per shear plane Perpendicular shear capacity (average) x2 $$V_{rd,\perp,avg}$$ 13.1 kN Average value of shear capacity (perpendicular) using 2 bow ties per shear plane Parallel slip modulus (average) $$k_{s,\parallel,avg}$$ 5.5 kN/mm Average value of shear slip modulus (parallel) using 2 bow ties per shear plane Perpendicular slip modulus (average) x2 $$k_{s,\perp,avg}$$ 6.3 kN/mm Average value of shear slip modulus (perpendicular) using 2 bow ties per shear plane
Based on data from testing 2 specimens

### Engineering design parameters (OSB)

 Parameter Symbol Value Units Note Parallel shear capacity (average) x2 $$V_{rd,\parallel,avg}$$ 16.6 kN Average value of shear capacity (parallel) using 2 bow ties per shear plane Parallel slip modulus (average) x2 $$k_{s,\parallel,avg}$$ 7.7 kN/mm Average value of shear slip modulus (parallel) using 2 bow ties per shear plane
Based on data from testing 1 specimen

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh. A full peer-reviewed study can be viewed here.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

The experimental setup comprises 3 mock beams connected by 4 bow tie keys (2 per face), which were sheared off by using an hydraulic actuator.  Below each of the bow tie keys,  a potentiometer was placed to measure the displacement parallel to the shear plane.

Depending on how the bow tie keys are oriented on the plywood panel,  they may have either 4 or 2  lamellae parallel to the shear plane.

To distinguish between the two cases, the following notation is used:

• ∥ to indicate the shear key with 4 lamellae with grain parallel to the shear plane.
• ⟂ to indicate the shear key with 4 lamellae with grain parallel to the shear plane.

### Data acquisition

The following instrumentation was installed for data acquisition:

• Four linear transducers (potentiometers) were used to track the specimen differential displacement parallel to the shear plane.
• a 250 kN-capacity load cell was used to measure the actuator  force.

1. Monotonic load,  to identify the maximum connection capacity per shear plane.
2. Cyclic (load-unload-reload) protocol according to BS EN 26891-1991 to identify any difference between initial and unload-reload stiffness.

Experimental results in terms of force-displacement per shear plane are reported below. Note that the force per shear plane is taken as half of the force measured at the actuator.

 Specimen Material Grain orientation Load protocol F max (kN) Ks (kN/mm) S1 Metsa spruce plywood Parallel Monotonic 11.1 5.7 S2 Metsa spruce plywood Parallel Cyclic 10.8 5.3 S3 Metsa spruce plywood Perpendicular Monotonic 12.6 6 S4 Metsa spruce plywood Perpendicular Cyclic 13.5 6.5 S5 OSB n/a Monotonic 16.6 7.7

### Failure mechanism

During the test, no damage was observed in the main elements:  failure was only observed in the shear keys themselves.  Pictures of the observed failure modes are reported below.

### Capacity

Since the number of tested specimens was limited, no statistics was carried out. Instead the average values were reported. Specifically:

• Plywood ∥ specimens capacity was estimated equal to 11.0 kN
• Plywood ⟂ specimens capacity was estimated equal to 13.1 kN
• OSB specimens capacity was estimated equal to 16.6 kN

### Slip modulus

The shear slip modulus $$k_s$$ of the specimens was calculated according to the equation below, based on EN 1380:2009.

$k_{s}=\frac{F_{40\%}-F_{10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{40\%}, F_{10\%}$$ represent 40% and 10% of the maximum force $$F_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

Specifically:

• Plywood ∥ specimens slip modulus was estimated equal to 5.5 kN/mm
• Plywood ⟂ specimens slip modulus was estimated equal to 6.3 kN/mm
• OSB specimens slip modulus was estimated equal to 7.7 kN/mm

## Bow ties – in tension

Bow ties are one of the main connectors. They are put into the face of the chassis to connect them together, to form a larger assembly. Below are some useful design parameters for how they perform in tension, and information on how those quantities were derived.

### Engineering design parameters (plywood)

 Parameter Symbol Value Units Note Axial capacity (perpendicular, average) $$N_{rd,\perp,avg}$$ 8.84 kN Average value of axial capacity (perpendicular to the grain) Axial slip modulus (average) $$k_{a,avg}$$ 1.47 kN/mm Average value of axial slip modulus

Based on data from testing 3 specimens

### Where do these data come from?

These values are based on the results of experimental testing performed at the University of Edinburgh.

#### Material used

18mm Metsa spruce plywood.

Skylark 0.0.9

### Experimental setup

The experimental setup comprises two timber panels which are connected by a bow tie key and subjected to tensile force. The red line indicates the main grain orientation of plywood.

### Data acquisition

The following instrumentation was installed for data acquisition:

• Two linear transducers were used to track the specimen axial displacement.  They measured the displacement between the steel plate and the steel frame.
• a 250 kN-capacity load cell was used to measure the actuator  force.

A monotonic load protocol was used to identify the maximum connection tensile capacity.

Experimental results in terms of force-displacement are reported below.

 Specimen Material Grain orientation F max (kN) Ks (kN/mm) S1 Metsa spruce plywood Perpendicular 9.01 1.48 S2 Metsa spruce plywood Perpendicular 9.06 1.57 S3 Metsa spruce plywood Perpendicular 8.44 1.37

### Failure mechanism

Specimens failed where the bow tie was in contact with the external panel. Damage was observed in both the bow tie and the panel.

### Capacity

The average capacity was estimated equal to 8.84 kN.

### Slip modulus

The shear slip modulus $$k_a$$ of the specimens was calculated according to the equation below, based on EN 1380:2009.

$k_{a}=\frac{F_{40\%}-F_{10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{40\%}, F_{10\%}$$ represent 40% and 10% of the maximum force $$F_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

The average slip modulus was estimated equal to 1.47 kN/mm.

## Pegs

Pegs are one of the main connectors. They hold blocks in place, preventing uplift. Below are some useful design parameters for them, and information on how those quantities were derived.

### Engineering design parameters (plywood)

 Parameter Symbol Value Units Note Uplift capacity (average) $$U_{rd,\parallel,avg}$$> 26.3 kN Average value of uplift capacity (parallel) Uplift capacity (average) $$U_{rd,\perp,avg}$$ 16.9 kN Average value of uplift capacity (perpendicular) Uplift slip modulus (average) $$k_{u,\parallel,avg}$$ 8.1 kN/mm Average value of uplift slip modulus (parallel) Uplift slip modulus (average) $$k_{u,\perp,avg}$$ 5.7 kN/mm Average value of uplift slip modulus (perpendicular)

Based on data from testing 2 specimens

### Engineering design parameters (OSB)

 Parameter Symbol Value Units Note Uplift capacity (average) $$U_{rd,avg}$$ 12.2 kN Average value of uplift capacity Uplift slip modulus (average) $$k_{u,avg}$$ 6.4 kN/mm Average value of uplift slip modulus

Based on data from testing 1 specimen

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh. A full peer-reviewed study can be viewed here.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

The test setup consists of pushing the shear panel- pegs joint until rupture.

Depending on how the peg connectors are oriented on the plywood panel,  they may have either 4 or 2  lamellae parallel to their longest axis:

To distinguish between the two cases,  the following notation is used:

• ∥  to indicate the shear key with 4 lamellae with grain parallel to the longest axis.
• ⟂  to indicate the shear key with 4 lamellae with grain parallel to the longest axis.

### Data acquisition

The following instrumentation was installed for data acquisition:

• Two linear transducers /potentiometers (one on each side) were installed on the specimen to measure the relative displacement between the shear panel and the mock column.
• A 250 kN-capacity load cell was used to measure the actuator  force.

A monotonic load protocol was used to identify the maximum connection uplifting capacity.

Experimental results in terms of force-displacement are reported below.  Note that the displacement reported is taken as the average between the two laser transducers.

 Specimen Material Grain orientation F max (kN) Ku (kN/mm) S1 Metsa spruce plywood Parallel 26.5 7.6 S2 Metsa spruce plywood Parallel 26.1 8.6 S3 Metsa spruce plywood Perpendicular 14 5.9 S4 Metsa spruce plywood Perpendicular 19.7 5.5 S3 OSB n/a 12.2 6.4

### Failure mechanism

All specimens failed because of one of the pairs of pegs, as can be seen from the figure below. No other damage was observed on the specimen while inspecting the different components of the joint.

### Capacity

Since the number of tested specimens was limited, no statistics was carried out. Instead the single values were reported. Specifically:

1. Plywood ∥ specimens capacity was estimated equal to 26.3 kN
2. Plywood ⟂ specimens capacity was estimated equal to 16.9 kN
3. OSB specimens capacity was estimated equal to 14.0 kN

### Slip modulus

The shear slip modulus $$k_u$$ of the specimens was calculated according to the equation below, based on EN 1380:2009.

$k_{u}=\frac{F_{40\%}-F_{10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{40\%}, F_{10\%}$$ represent 40% and 10% of the maximum force $$F_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

Specifically:

• Plywood ∥ specimens slip modulus was estimated equal to 8.1 kN/mm
• Plywood ⟂ specimens slip modulus was estimated equal to 5.7 kN/mm
• OSB specimens slip modulus was estimated equal to 6.4 kN/mm

## External faces

This section relates to the external face of walls blocks, and the capacity of those blocks to resist suction forces created by wind around the building.

### Engineering design parameters (plywood)

 Parameter Symbol Value Units Note Pressure capacity (average) $$p_{rd,avg}$$ 2.81 kPa Average value of pressure capacity Withdrawing stiffness (average) $$k_{p,avg}$$ 4.82 kPa/mm Withdrawing stiffness (average)

Based on data from testing 3 specimens

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

Three column specimens were tested to investigate their capacity to sustain wind withdrawing forces. Specimens comprise 3 mock columns made of plywood, assembled by using the following panels:

1. Front panel, i.e. the panel which is subjected to withdrawing wind in a real scenario
2. Lateral panels, which for testing purposes are fixed to the supporting steel plates
3. Stabilizing panels, which are introduced to replicate the stiffening effect of the second flange of the column (in this case removed to apply the load with the actuator).

The main panels are 1200 mm long, and 600 mm wide. They are connected to the side panels by 4 tabs (2 on each side), which are spaced 600 mm from each other.

The load exerted by the hydraulic actuator was applied to the front panel by using a 800 x 500 x 10 mm steel plate.

### Data acquisition

The following instrumentation was installed for data acquisition:

1. Four linear transducers (potentiometers) were used to track the panel vertical displacement. They measured the displacement between the front panel and the side panels.
2. A 250 kN capacity load cell was used to measure the compression force. The load cell was mounted between the actuator and the steel plate.

The loading protocol consisted of a monotonic displacement increasing until failure. The loading rate was equivalent 1 mm per minute.

The response of the specimens in terms of pressure (exerted by the actuator) vs vertical displacement is reported in the figure below. Note that the vertical displacement was taken as the average between the measurements of the four linear transducers. The pressure on the specimen was taken as the force exerted by the actuator divided by the area of the front panel, ie., 0.72 m $$^2$$.

Specimens failed in between 2.5 kPa and 3.2 kPa.

 Specimen Material $$p_{max} (kPa)$$ kp (kPa/mm) S1 Metsa spruce plywood 2.76 4.88 S2 Metsa spruce plywood 2.49 4.67 S3 Metsa spruce plywood 3.19 4.93

### Failure mechanism

It was observed that failure happened by either 1) failing the tabs in shear or 2) failing the region surrounding the tabs. After failing one pair of connections, the specimens exhibited plastic behaviour until failing the second pair.

### Stiffness

The panel elastic stiffness $$k_p$$ of the specimens was calculated according to the equation below, based on EN 1380:2009.

$k_{p}=\frac{p_{40\%}-p_{10\%}}{d_{40\%}-d_{10\%}}$

where $$p_{40\%}, p_{10\%}$$ represent 40% and 10% of the maximum pressure $$p_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

The panel elastic stiffness values are between 4.7 kPa/mm and 4.9 kPa/mm.

## Stitch joints

Stitch joints connect two timber panels which are perpendicular to each other. Below are some useful design parameters for them, and information on how those quantities were derived.

### Engineering design parameters (plywood)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{tab,Rd,avg}$$ 7.54 kN Average value of shear capacity per tab Shear stiffness (average) $$k_{tab,avg}$$ 2.69 kN/mm Shear stiffness (average) per tab

Based on data from testing 4 specimens

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

Four specimens were tested to investigate the shear capacity of stitch joints. Specimens comprise 4 panels made of plywood connected by using a stitch joint. The joint is made by eight 100 mm x 18 mm x 18 mm tabs.

### Data acquisition

The following instrumentation was installed for data acquisition:

1. Two linear transducers (potentiometers) were used to track the panel vertical displacement. They measured the displacement between the front panel and the side panels.
2. A 250 kN capacity load cell was used to measure the compression force. The load cell was mounted between the actuator and the steel plate.

The loading protocol consisted of a monotonic displacement increasing until failure. The loading rate was equivalent 1 mm per minute.

The response of the specimens in terms of force vs vertical displacement is reported in the figure below. Note that:

1) the actuator's force was divided by 8, because the stitch joint was made of 8 tabs.

2) the vertical displacement was taken as the average between the measurements of the two linear transducers.

Specimens failed in between 6.99 kN and 8.02 kN.

 Specimen Material $$F_{tab,max} (kN)$$ $$k_{tab} (kN/mm)$$ S1 Metsa spruce plywood 8.02 2.78 S2 Metsa spruce plywood 7.26 2.43 S3 Metsa spruce plywood 6.99 2.20 S4 Metsa spruce plywood 7.90 3.33

### Failure mechanism

It was observed that failure happened by failing the tabs in compression. After reaching the maximum force, the specimens exhibited plastic behaviour.

### Stiffness

The shear elastic stiffness $$k_{tab}$$ of the tabs was calculated according to the equation below, based on EN 1380:2009.

$k_{tab}=\frac{F_{tab,40\%}-F_{tab,10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{tab,40\%}, F_{tab,10\%}$$ represent 40% and 10% of the maximum force $$F_{tab,max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

The elastic stiffness values are between 2.20 kN/mm and 3.33 kN/mm.

# Assemblies

## Shear walls – no openings

Shear walls (or 'bracing walls') are assemblies of wall blocks that, together, laterally brace the building. A wall's capacity to perform this role is increased if it is also bearing a vertical load. Below are some useful design parameters, and information on how those quantities were derived. Further details can be found here.

### Engineering design parameters (no vertical load)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 65 kN Average value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 16.7 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 1.63 kN/mm Shear stiffness

Based on data from testing 1 specimen

### Engineering design parameters (vertical load = 8.3 kN/m)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 82 kN Average value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 25.9 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 2.38 kN/mm Shear stiffness

Based on data from testing 1 specimen

* Note that the stiffness values were obtained by 2D racking testing. However, such values are most likely to be higher in a real building application due to the contribution of surrounding elements.

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

Two wall specimens were tested in a shear setup. Walls were bolted to a steel beam by using 8 pairs of L-shape aluminium steel plates,  thirty-five  5 mm diameter screws and 2x 8.8 M16 bolts. The steel beam was connected to the laboratory strong floor by using twelve 24 mm diameter steel threaded bars and nuts.

The lateral load was applied to the centre of the top beam by using an hydraulic actuator.

### Data acquisition

The following instrumentation was installed for data acquisition:

1. Four linear transducers measuring any uplift occurring between the specimen and the steel beam.
2. Two linear transducers monitoring the diagonal displacements in correspondence of the central point of the specimen;
3. Two string potentiometers to measure the horizontal displacement placed on the top corners or the specimen;
4. A load cell placed between the actuator and the specimen

1. cycle 1: load up to 10% of the estimated panel capacity, maintain for 30 seconds, and unload;
2. cycle 2: load up to 40% of the estimated panel capacity, maintain for 30 seconds, and unload;
3. cycle 3: load up to failure

Prior to the test, the panel capacity was estimated to be equal to 80 kN.

 Specimen Material Vertical load F max (kN) F H/300 (kN) K (kN/mm) Loaded Metsa spruce plywood 8.3 82 25.9 2.38 No load Metsa spruce plywood 0 65 16.7 1.63

### Failure mechanism

The specimen without vertical load failed at a peak force equal to 65 kN. The failure was caused by reaching the capacity of the pegs, i.e., the connection between the bottom beam and the columns.

The specimen with vertical load reached a maximum capacity equal to 82 kN. The test had to be stopped for safety reasons because the specimen was significantly bending out of plane.

### Capacity

Since the number of tested specimens was limited, no statistics was carried out. Instead the single values were reported. Specifically:

1. Wall with no openings and no vertical load applied: capacity equal to 65 kN
2. Wall with no openings and 8.3 kN/m vertical load applied: capacity equal to 82 kN

### Slip modulus

The shear slip modulus $$k_w$$ of the specimens was calculated according to the equation below, based on EN 1380:2009.

$k_{w}=\frac{F_{40\%}-F_{10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{40\%}, F_{10\%}$$ represent 40% and 10% of the maximum force $$F_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

Specifically:

1. Wall with no openings and no vertical load applied: shear stiffness equal to 1.63 kN/mm
2. Wall with no openings and 8.3 kN/m vertical load applied: shear stiffness equal to 2.38 kN/mm

## Shear walls – with 2 windows

Shear walls (or 'bracing walls') are assemblies of wall blocks that, together, laterally brace the building. A wall's capacity to perform this role is increased if it is also bearing a vertical load, but decreased by the presence of openings in that wall. Below are some useful design parameters, and information on how those quantities were derived. Further details can be found here.

### Engineering design parameters (no vertical load)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 47 kN Average value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 4.5 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 0.64 kN/mm Shear stiffness

Based on data from testing 1 specimen

### Engineering design parameters (vertical load = 8.3 kN/m)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 70 kN Average value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 15.4 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 1.33 kN/mm Shear stiffness

Based on data from testing 1 specimen

* Note that the stiffness values were obtained by 2D racking testing. However, such values are most likely to be higher in a real building application due to the contribution of surrounding elements.

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

Walls were bolted  to a steel beam by using 8 pairs of L-shape aluminium steel plates, thirty-five  5 mm diameter screws and 2x 8.8 M16 bolts. The steel beam was connected to the laboratory strong floor by using twelve 24 mm diameter steel threaded bars and nuts.

The lateral load was applied to the centre of the top beam by using an hydraulic actuator.

### Data acquisition

The following instrumentation was installed for data acquisition:

1. Four linear transducers measuring any uplift occurring between the specimen and the steel beam.
2. Two linear transducers monitoring the diagonal displacements in correspondence of the central point of the specimen;
3. Two string potentiometers to measure the horizontal displacement placed on the top corners or the specimen;
4. A load cell placed between the actuator and the specimen

1. cycle 1: load up to 10% of the estimated panel capacity, maintain for 30 seconds, and unload;
2. cycle 2: load up to 40% of the estimated panel capacity, maintain for 30 seconds, and unload;
3. cycle 3: load up to failure

Prior to the test, the panel capacity was estimated to be equal to 80 kN.

 Specimen Material Vertical load F max (kN) F H/300 (kN) K (kN/mm) Loaded Metsa spruce plywood 8.3 70 15.4 1.33 No load Metsa spruce plywood 0 47 4.5 0.64

### Failure mechanism

The specimen without vertical load failed at a peak force equal to 47 kN. The failure was caused by reaching a local failure near one of the bow ties.

The specimen with vertical load failed at a peak force equal to 70 kN. The failure was caused by reaching the capacity of the pegs, i.e., the connection between the bottom beam and the columns.

### Capacity

Since the number of tested specimens was limited, no statistics was carried out. Instead the single values were reported. Specifically:

1. Wall with window openings and no vertical load applied: capacity equal to 47 kN
2. Wall with window openings and 8.3 kN/m vertical load applied: capacity equal to 70 kN

### Slip modulus

The shear slip modulus $$k_w$$ of the specimens was calculated according to the equation below, based on EN 1380:2009.

$k_{w}=\frac{F_{40\%}-F_{10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{40\%}, F_{10\%}$$ represent 40% and 10% of the maximum force $$F_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

Specifically:

1. Wall with no openings and no vertical load applied: shear stiffness equal to 0.64 kN/mm
2. Wall with no openings and 8.3 kN/m vertical load applied: shear stiffness equal to 1.33 kN/mm

## Shear walls – with window and door

Shear walls (or 'bracing walls') are assemblies of wall blocks that, together, laterally brace the building. A wall's capacity to perform this role is increased if it is also bearing a vertical load, but decreased by the presence of openings in that wall. Below are some useful design parameters, and information on how those quantities were derived.

### Engineering design parameters (vertical load = 4.2 kN/m)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 35 kN Minimum value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 11.8 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 1.14 kN/mm Shear stiffness

Based on data from testing 1 specimen

*Note that the capacity of the wall is likely to be higher. During the test, no failure in the shear wall was observed at this value of load.

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh. A full peer-reviewed study can be viewed here.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

A building block 5.3 m long x 5.3 m wide x 3.1 m tall was tested under horizontal loading.

Side walls were bolted  to a steel beam by using 8 pairs of L-shape aluminium steel plates,  thirty-five  5 mm diameter screws and 2x 8.8 M16 bolts. The steel beam was connected to the laboratory strong floor by using twelve 24 mm diameter steel threaded bars and nuts.

The lateral load was applied to the centre of the facade by using two hydraulic actuators and an 1.1 m x 2.3 m x 0.008 m steel plate.

### Data acquisition

The following instrumentation was installed for data acquisition:

1. 19 between linear transducers and string potentiometers measuring relative displacements between the specimen and the reference system
2. Two load cells placed between the actuator and the specimen.

1. cycle 1: load up to 10% of the estimated panel capacity, maintain for 30 seconds, and unload;
2. cycle 2: load up to 40% of the estimated panel capacity, maintain for 30 seconds.
3. cycle 3: load up to failure

Prior to the test, the panel capacity was estimated to be equal to 50 kN per wall.

Below data are reported in terms of shear load per wall vs horizontal displacement of the specimen.

 Specimen Material Vertical load F max (kN) F H/300 (kN) K (kN/mm) SW-load Metsa spruce plywood 4.2 35 11.8 1.14

### Failure mechanism

The specimen without vertical load failed at a peak force equal to 35 kN. The failure was caused by reaching a local failure in the facade near the point of application of the force. Therefore, it is likely that the capacity of the shear walls is actually higher than 35 kN.

### Capacity

Since the number of tested specimens was limited, no statistics was carried out. Instead the single values were reported. Specifically:

1. Wall with for and window openings and 4.2 kN/m vertical load applied: capacity equal to 35 kN

### Slip modulus

The shear slip modulus $$k_w$$ of the specimens was calculated according to the equation below, based on EN 1380:2009.

$k_{w}=\frac{F_{40\%}-F_{10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{40\%}, F_{10\%}$$ represent 40% and 10% of the maximum force $$F_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

Specifically:

1. Wall with window and door openings and 4.2 kN/m vertical load applied: shear stiffness equal to 1.14 kN/mm

## Lintels

### Engineering design parameters

 Parameter Symbol Value Units Note Moment capacity S (average) $$M_{rd,S,avg}$$ 25.8 kNm Average value of moment capacity for the short lintel (1.2 m net span) Stiffness S (3 point bending test) $$k_{l,S}$$ 8.42 kN/mm Average value of stiffness for the short lintel (1.2 m net span) Moment capacity L (average) $$M_{rd,L,avg}$$ 37.3 kNm Average value of moment capacity for the long lintel (2.4 m net span) Stiffness L (3 point bending test) $$k_{l,L}$$ 3.00 kN/mm Average value of stiffness for the short lintel (2.4 m net span)
Based on data from testing 3 L specimens and 3 S specimens

### Testing data

These values are based on the results of experimental testing performed at the University of Edinburgh.

#### Material used

18mm Metsa spruce plywood (details)

Skylark 0.0.9

### Experimental setup

Six lintel specimens were tested in 3-point bending. Three specimens, labelled S, had spans of 1.2 m, and and three specimens, labelled L, had spans of 2.4 m.

The load exerted by the hydraulic actuator was applied to the timber lintel by using a 10 x 500 x 800 mm steel plate. A 40 mm diameter steel bar was welded to the plate to create a pin connection between the actuator and the specimen.

### Data acquisition

The following instrumentation was installed for data acquisition:

1. Five linear potentiometers were used to track the specimen displacement profile.
2. A 250 kN-capacity load cell was used to measure the actuator force.
3. One high resolution camera was used to to track the strain profile by using digital image correlation (DIC).

The loading protocol follows the guidance given in EN 1380:2009 for connectors in timber structures. The test was carried out in displacement control, with the actuator moving at 4 mm per minute. The loading protocol consisted of a monotonic ramp of 15 mm for the L specimens and 8 mm for the S specimens, a 30 sec hold, a monotonic ramp back to 5 mm, a 30 sec hold, and then a monotonic ramp until failure.

The external vertical load is mainly supported by the combs within the lintel, which are working in bending. The comb is not a continuous element and it is held together by the pegs: while the compression force is taken by contact forces between the panels, the tension is taken by the pegs. The pegs transfer the load to the bottom part of lintel which engages the tabs (see Figure below).

All specimens failed in a ductile manner: the specimens were able to maintain the load after reaching the peak force. The response of the specimens (in terms of force exerted by the actuator vs mid-span displacement) is reported in the graph and table below. The peak force occurred between 60.5 kN and 64.5 kN for the L specimens, and between 82.2 kN and 92.8 kN for the S specimens.

 Specimen Net span (m) $$F_{max}$$ (kN) $$M_{max}$$ (kNm) $$k_l$$ (kN/mm) L1 2.4 61.3 36.8 3.04 L2 2.4 64.5 38.7 2.92 L3 2.4 60.5 36.3 3.04 S1 1.2 82.2 24.7 8.66 S2 1.2 92.8 27.8 8.82 S3 1.2 83.2 25.0 7.79

### Failure Mechanism

The failure specimen was initiated by the failure of pegs (see Figures b) and c) below): these last pegs hold the combs together during bending, and they are believed to be the weakest link in the load path.

Along with the pegs located within the span, further damage was observed:

1. in the castellated joint between the elements (see Figure d) below)
2. in the pegs and combs above the column (see Figure e))

### Capacity

Because of the complex load path, a simple structural model that can accurately predict the capacity of the lintels is not available yet. Therefore, it is recommended to assume on the safe side that the capacity of the specimens (in terms of bending moment) is checked against the maximum bending moment expected in the midspan.

Assuming a simply supported static scheme, the capacity of the lintels is between 36.3 kNm and 38.7 kNm for the L specimens, and between 24.7 kNm and 27.8 kNm for the S specimens. Results are reported in the table above.

### Deflected Shape

The evolution of the deflected shape of the lintels during the test is reported in the videos below.

The stiffness $$k_l$$ of the elements was calculated according to the equation below, based on EN 1380:2009.

$k_{l}=\frac{F_{40\%}-F_{10\%}}{d_{40\%}-d_{10\%}}$

where $$F_{40\%}, F_{10\%}$$ represent 40% and 10% of the maximum force $$F_{max}$$ and $$d_{40\%}, d_{10\%}$$ represent the displacement values where such force occurs.

Specifically:

• Lintel S stiffness was estimated equal to 8.42 kN/mm
• Lintel L stiffness was estimated equal to 3.00 kN/mm

Further results are reported in the table above.

# Materials

## Structural properties of the materials

This section provides the mechanical properties of the material products used during the experimental testing for engineering calculation purposes.  These are by no means the only materials you can use. We share this data here to allow you compare the performance data provided with products with other potential products.

 Name Product Thickness Technical specs More info Metsa spruce plywood Plywood 18mm Spec Website Sterling OSB3 Zero OSB 18mm Spec Website

### Material configuration

The plywood panels tested were using 6x3mm plies. The ply grain orientation was according to the figure below.

# Worked examples

## Design of a floor for gravity loads

This is a worked example for the design of a WikiHouse Skylark floor in relation to gravity loads.

### How does Skylark deal with gravity?

From a gravity load perspective, a Skylark chassis can be thought of as a sequence of portals that work independently on one another. Each portal is made of beams and columns, which are connected by using pegs.

### How do you check a Skylark floor against gravity loads?

Results from testing suggest that negligible moment is transferred in the joint between beams and columns. This allows us to simplify the structural scheme: to calculate ROOF beams, FLOOR beams and WALL columns separately.

Hence, the following structural model is used:

In this section, a worked example is proposed to check the beams according to the Eurocode 5. In this case, the beam is made of Metsa plywood.

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.

According to Eurocode 1 , the imposed load $$IL$$ for the residential category is recommended to be 2 kN/m $$^2$$. Hence, the linear live load on the beam $$Q_{beam}$$ is equal to:

$Q_{beam}= \overbrace{IL}^{2 kN/m^2} \cdot \overbrace{i_{beams}}^{0.6m}=1.2 kN/m$

The value of loads corresponding to the frequent $$q_{fr}$$ , quasi-permanent $$q_{qp}$$ and rare $$q_{rar}$$load combinations are calculated below:

$q_{fr} =\overbrace{G_{beams}}^{0.76}+\overbrace{\psi_2}^{0.7}\overbrace{Q_{beams}}^{1.2}=1.6\,kN/m$

$q_{qp} =\overbrace{G_{beams}}^{0.76}+\overbrace{\psi_2}^{0.3}\overbrace{Q_{beams}}^{1.2}=1.12\,kN/m$

$q_{ra} =\overbrace{\gamma_G}^{1.35}\overbrace{G_{beams}}^{0.76}+\overbrace{\gamma_Q}^{1.5}\overbrace{Q_{beams}}^{1.2}=2.83\,kN/m$

### Serviceability limit state (SLS)

At the serviceability limit state (SLS), it is checked that:

1. The maximum elastic deflection under the frequent load combination is lower than a threshold of L/250 (with L the span)
2. 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:

$d_{max}=\frac{5qL^4}{384EI}+\frac{qa^2}{2k_R}[L-a]$

with $$E$$ the elastic modulus of timber, $$I$$ the effective second moment of inertia, $$a$$ the distance of the castellated joint from the support, $$L$$ the span and $$k_R$$ the rotational stiffness of the castellated joint.

#### SLS short tem - frequent load

The check on the maximum elastic deflection is reported the equation below.  Note that the deflection is calculated using a simply supported static scheme.

$d_{max,el}=\frac{5\overbrace{q}^{1.6} \overbrace{L^4}^{5155} }{384\underbrace{E}_{8615}\underbrace{I}_{\ 288\cdot10^6}}+\frac{\overbrace{q}^{1.6}\overbrace{a^2}^{1718}}{2\underbrace{k_R}_{2295\cdot10^6}}[\overbrace{L}^{5155}-\overbrace{a}^{1718}]=9.4\ mm$

Note that the contribution of the joint flexibility accounts for 37% of the overall deflection.

It is checked that:

$d_{max,el}=9.4<\frac{L}{250}=20.6\ mm$

#### 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: characterized 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:

$d_{max,lt}=d_{max,el}\frac{q_{qp}}{q_{fr}}(1+k_{def})=11.8\ mm$

and it is checked that:

$d_{max,lt}=11.8<\frac{L}{250}=20.6\ mm$

### Ultimate Limit State (ULS)

At ULS, beams should be checked against moment and shear. Such values can be calculated as:

$M_{sd,ra}=\frac{q_{ra}L^2}{8}=9.4\ kN/m\quad V_{sd,ra}=0.5q_{ra}L=7.3\ kN$

$M_{sd,qp}=\frac{q_{qp}L^2}{8}=3.7\ kN/m \quad V_{sd,qp}=0.5q_{qp}L=2.9\ kN$

#### Bending moment check

$M_{sd,ra}=9.4\le\frac{\overbrace{k_{mod}}^1\overbrace{M_{rd}}^{17.8}}{\underbrace{\gamma_d}_{1.25}}=14.2\ kNm$ $M_{sd,qp}=3.7\le\frac{\overbrace{k_{mod}}^{0.6}\overbrace{M_{rd}}^{17.8}}{\underbrace{\gamma_d}_{1.25}}=8.5\ kNm$

#### 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 a the following shear strength:

$V_{rd}=A_v\overbrace{f_{sd}}^{3.5}=32.3\ kN$

It follows:

$V_{sd,ra}=7.3\le\frac{\overbrace{k_{mod}}^1\overbrace{V_{rd}}^{32.3}}{\underbrace{\gamma_d}_{1.25}}=26.0\ kN$ $V_{sd,qp}=3.7\le\frac{\overbrace{k_{mod}}^{0.6}\overbrace{V_{rd}}^{32.3}}{\underbrace{\gamma_d}_{1.25}}=15.5\ kN$

## Design of 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:

1. Liveable space
2. 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.

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:

$Q_{beam}= \overbrace{q_{snow}}^{0.6 kN/m^2} \cdot \overbrace{i_{beams}}^{0.6m}=0.36 kN/m$

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.

 Combination Load (kN/m) T (kN) N (kN) M (kNm) V (kN) f (mm) $$q_{fr}$$ 1 2.5 3.4 1.8 2.4 3.3 $$q_{qp}$$ 0.87 2.2 3 1.6 2.1 $$q_{ra}$$ 1.6 4 5.4 2.9 3.8

### Serviceability limit state (SLS) on the beams

At the serviceability limit state (SLS), it is checked that:

1. the maximum elastic deflection under the frequent load combination is lower than a threshold of L/250 (with L the span)
2. 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$

## Design of floor assembly for wind loads

To date, the diaphragm  has not been tested as a whole, so a mechanical model based on engineering principles is provided as reference. A spreadsheet with all the calculations is available here.

## Geometry

The geometry of a diaphragm is reported the figure below. The element is made up of 18 x 5.4 m beams, for a total footprint of 5.4 m x 10.8 m. Beams are connected together by using 8 bow ties connectors, which are 0.6 m spaced from each other and 0.62 m spaced from the side of the beam. It is also assumed that there are stairs in the bottom left of the floor, which do not allow for the last two rows of bow ties in the first 6 beams.

## Wind load on the diaphragm

The wind load $$q$$ acting on the diaphragm is taken equal to 2kN/m. In a real project, this value can be calculated once the location of the structure and the overall geometry of the building are known.

Assuming a simply supported static scheme, the diaphragm will be subjected to bending moment and shear. It follows that the bow ties will be subject to both shear and tensile forces. To find out the worst case scenario, 3 rows of bow ties are considered, for which the  values of shear (V) and bending moment (M) are calculated based on their distance z from the edge according to equations:

$M(z)=-\frac{qz^2}{2}+\frac{qLz}{2}$

$T(z)=\frac{qL}{2}-qz$

By having L=10.8 m and q=2kN/m, results are reported in the table below

 Row z (m) V (kN) M (kNm) 1 0.6 9.6 6.12 2 3 4.8 23.4 3 5.4 0 29.2

## Shear demand on the bow ties

The shear force on the diaphragm will cause a shear force in the bow ties. To calculate the shear demand on the single bow tie, the value of shear on the diaphragm V is divided by the number of bow ties.

The steps to calculate the tension/compressive stress on the central row of bow ties is shown below.

The simplest way is to create an spreadsheet file with the following columns:

1. ID of the bow tie (ID)
2. Position of the bow tie ($$y_i$$)
3. Area of the bow tie acting in tension ($$A_i$$). This parameter is equal to 0 if $$y_i<x$$ because it means that the bow tie is in the compression part.
4. Parameter $$A_iy_i$$
5. Parameter $$A_i(y_i-x)^2$$

As a first guess, it is assumed that the neutral axis position is equal to 1000 mm, i.e., 7 bow ties are working in tension while 1 is in the compression zone.

The position of the neutral axis x can be calculated as:

$x=\frac{-\sum A_i+\sqrt{(\sum A_i)^2+2\sum A_iy_iB}}{B}=371.2\ mm$

where B=310 mm is the depth of the beam providing reaction in compression.

The new value of the neutral axis is used to iterate the computation:

which reaches convergence to a value of x=373.5 mm (assuming a tolerance of 5 mm).

### Calculation of compression stress (on the beam) and tensile force (on the bow tie)

The maximum value of compression $$\sigma_{c,max}$$ on the beams, and tensile force $$F_{t, max}$$ on the bow ties can be calculated as:

$$\sigma_{c,max}=\frac{Mx}{I}$$

$$\sigma_{t, max}=\frac{M(y_{i, max}-x)A_i}{I}$$

with $$I$$ the second moment of inertia, and $$y_{i,max}$$ the position of the bow tie furthest from the neutral axis position.

The second moment of inertia $$I$$ can be calculated as:

$I=\frac{Bx^3}{3}+\sum_iA_i(y_i-x)^2=90.99 \cdot 10^9 \ mm^4$

leading to $$\sigma_{c,max}=0.12$$ MPa and $$F_{t,max}=1.64$$ kN.

Similar process can be repeated for the bow ties in row 1 and row 2.

Calculations are reported in the spreadsheet, while results are summarised on the table below.

 Row $$V_i$$ (kN) $$F_t,i,max$$ (kN) $$\sigma_c$$ (MPa) 1 1.6 0.58 0.04 2 0.8 2.75 0.16 3 0 1.64 0.12

It can be noticed that, in this case, the critical bow tie is the one belonging to row 2.

## Safety checks

Two safety checks are performed in this example at the Ultimate Limit State (ULS):

1. The compressive strength $$\sigma_c$$ is lower than the timber capacity
2. The combined shear/tension demand in the bow ties is lower than its capacity

### Compressive strength

Assuming that the beams are made of Metsa plywood, the value of the timber strength is equal to $$f_{cd}=10.3$$ MPa. It follows:

$$\sigma_{c, max}=0.12\lt \frac{f_{cd}}{\gamma_m}=8.24$$MPa

### Combined shear/tension

Assuming the bow ties are made of plywood, and fabricated with the grain direction parallel to the element shortest axis, they will have the following properties:

$V_{rd,\parallel,avg}=5.5\ kN$

$N_{rd,\perp,avg}=6.6\ kN$

Because these values were obtained by tensing only 1 specimen, a safety factor of 1.5 is considered leading to the following values:

$V_{rd,\parallel,d}=3.7\ kN$

$N_{rd,\perp,d}=4.4\ kN$

It follows:

$\frac{V_i}{V_{rd,\parallel,d}}+\frac{{F_{t,i}}}{N_{rd,\perp,d}}=0.84<1$

### Mechanical model

The mechanical model used to calculate the tension forces acting on the bow ties is here presented. It is assumed that the beams behave as a rigid block, i.e., the distribution of strain over the section of the diaphragm is linear according to the figure below. While the beams can react in compression, only the bow ties are taking the tension.

It is indicated with $$\phi$$ the curvature of the section, $$x$$ the position of the neutral axis and $$y_i$$ the postion of the i$$^{th}$$ bow tie.  By looking at the linear distribution of deformation, it follows that:

1. The maximum deformation in compression is equal to $$\epsilon_c=\phi x$$
2. The deformation on the $$i^{th}$$ bow tie is equal to $$\epsilon_i=\phi(y_i-x)$$

It is then assumed that both bow ties and beams are made of the same material, and that the elastic modulus of the bow ties is tension is equal to the elastic modulus of the timber in compressions, i.e., equal to /(E/). Note that the orientation of the bow ties in not know at priori, so this assumption might not be accurate. However, it can be used as a starting point for this simplified model if the two elastic moduli are not too different. Following this assumption:

1. The maximum stress in compression is equal to $$\sigma_c=E\epsilon_c=E\phi x$$
2. The tension stress in each bow ties is equal to $$\sigma_i=E\epsilon_i=E\phi(y_i-x)$$

The compressive stress can be integrated over the neutral axis $$x$$ to calculate the compressive force $$C$$ as:

$$C=0.5\sigma_c x B=0.5 E\phi x^2B$$ with B the depth of the beams.

Similarly, the tensile stresses in the bow ties can be added together to calculate the tensile force $$T$$ as:

$$T=\sum_iA_i\sigma_i=\sum_iA_iE\phi(y_i-x)$$ with $$A_i$$ the tensile area of the bow tie.

Because the section is in equilibrium, it follows that :

$T=C$ .

By substituting the above values for $$C$$ and $$T$$, such equation be re-written as:

$0.5E\phi x^2 B=\sum_iA_iE\phi (y_i-x)$

Simplifying by $$E$$ and $$\phi$$, it becomes:

$0.5x^2B=\sum_iA_i(y_i-x)$

which leads to :

$x=\frac{-\sum A_i+\sqrt{(\sum A_i)^2+2\sum A_iy_iB}}{B}$

## Design of shear walls for wind loads

This section discusses a simplified worked example about the design of a Skylark building's walls for wind loads.

The building is 5.4 m wide, and 10.8 m long. It uses a solid shear wall in W-E direction and two skylark shear walls with openings in the N-S direction.

## Wind load demand

The building is assumed to be located in Sheffield, UK. Also, it is assumed to be located above 100 m above sea level, and in a location where orography is not significant. This leads to the following dynamic pressure imposed by wind:

• $$q_{p,SLS}$$ equal to 0.76 kN/m$$^2$$ at SLS
• $$q_{p,ULS}$$ equal to 1.14 kN/m$$^2$$ at ULS

## North-South Direction

### Load on the N-S shear walls

The force acting on each N-S shear wall can be calculated according to:

$V_{sd}=\frac{0.85 c_{pe,net}Aq_p}{n_{walls}}$

where:

• 0.85 is a coefficient to allow for lack of correlation between front and rear faces according to BS6399-2
• $$A$$ is the area where the pressure acts (see figure below)
• $$q_p$$ is the wind pressure
• $$n_{walls}$$ number of shear walls resisting the load
• $$c_{p, net}$$ id a simplified pressure coefficient depending on the ratio h/d, with h the height of the building and d the length

 h/d(-) $$c_p,net$$(-) ≤ 0.25 0.8 1 1.1 ≥ 5 1.3

In this specific case:

$$h=2.8+2.7=5.5m$$

$$d=5.4\ m$$

$$c_{p, net}=1.1+\left(\frac{5.5}{5.4}-1\right)\frac{1.3-1.1}{5-4}=1.1$$

$$A=1.4\cdot5.4+5.4\cdot2.7\cdot0.5=14.9\ m^2$$

$V_{sd,N-S,SLS}=\frac{0.85 \cdot 1.1\cdot 14.9\cdot 0.76}{2}=5.3 \ kN$

$V_{sd,N-S,ULS}=\frac{0.85 \cdot 1.1\cdot 14.9\cdot 1.14}{2}=8.0 \ kN$

### Capacity of the walls

The capacity of the solid shear walls with no openings are reported below (they are also here).  Because the walls are not carrying any gravity load, their lateral load capacity is taken equal to the specimen tested without any gravity load.

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 65 kN Average value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 16.7 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 1.63 kN/mm Shear stiffness

Given the fact that only one specimen was tested, a safety coefficient equal to 1.5 is conservatively applied the data. This leads to the following capacity values:

$$F_{Rd, d}=\frac{F_{Rd,avg}}{1.5}=43.3\ kN$$

$$F_{H/300, d}=\frac{F_{H/300,avg}}{1.5}=11.1\ kN$$

### Check

It can be checked that:

$$F_{Rd,d}=43.3\ kN \ge V_{sd,N-S,ULS}=8.0\ kN$$ ✅

$$F_{H/300,d}=16.7\ kN \ge V_{sd,N-S,SLS}=11.1\ kN$$ ✅

## East-West direction

### Load on the E-W shear walls

Similarly to above:

$$h=2.8+2.7=5.5\ m$$

$$d=10.8\ m$$

$$c_{p, net}=0.8+\left(\frac{5.5}{10.8}-0.25\right)\frac{1.1-0.8}{1-0.25}=0.9$$

$$A=1.4\cdot10.8+10.8\cdot2.7=44.3\ m^2$$

$V_{sd.E-W,SLS}=\frac{0.85 \cdot 0.9\cdot 44.3\cdot 0.76}{2}=12.9 \ kN$

$V_{sd.E-W,ULS}=\frac{0.85 \cdot 0.9\cdot 44.3\cdot 1.14}{2}=19.3 \ kN$

### Capacity of the walls

The total capacity of the shear walls in E-W direction depends on the combination of:

• the properties of shear walls with door and window openings, which can be found here
• the properties of shear walls with two windows, which can be found here
• Vertical load coming from the roof

The vertical load coming from the roof was previously calculated here. Assuming a quasi permanent distributed load $$q_{qp}=0.87$$ kN/m^ $$2$$ acting on the roof, the compression load on each column is equal to:

$$N_{qp}=q_{qp}\cdot5.4m/2=2.35kN$$

Because each column is 0.6m wide, the vertical load on the wall can be assumed to be equal to:

$$q_v=N_{qp}/0.6=3.9\ kN/m$$

Wall with door and window

### Engineering design parameters (vertical load = 4.2 kN/m)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 35 kN Minimum value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 11.8 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 1.14 kN/mm Shear stiffness

Only one specimen was tested, and the test was performed by applying 4.2kN/m as vertical load. Because no other data are available, the parameters of the the wall are reduced by:

• a conservative safety coefficient equal to 1.5, due to the fact that only one specimen was tested
• a reduction factor equal to $$\frac{3.9}{4.2}=0.93$$ to take into account the different gravity load

Hence:

$$F_{Rd, DW,d}=0.93\frac{F_{Rd,avg}}{1.5}=21.7\ kN$$

$$F_{H/300,DW, d}=0.93\frac{F_{H/300,avg}}{1.5}=7.3\ kN$$

Wall with 2 windows

### Engineering design parameters (no vertical load)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 47 kN Average value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 4.5 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 0.64 kN/mm Shear stiffness

### Engineering design parameters (vertical load = 8.3 kN/m)

 Parameter Symbol Value Units Note Shear capacity (average) $$F_{rd,avg}$$ 70 kN Average value of shear capacity Shear H/300 load (average) $$F_{H/300,avg}$$* 15.4 kN Load value to hit an horizontal displacement equal to H/300 Shear stiffness $$k_w$$* 1.33 kN/mm Shear stiffness

Only one specimen was tested, and the test was performed by applying either 0 or 8.3 kN/m as vertical load. Because no other data are available, the parameters of the the wall are calculated by :

• linearly interpolating the values for vertical load equal to 0 and vertical load equal to 8.3 kN
• applying a conservative safety coefficient equal to 1.5, due to the fact that only one specimen was tested

Hence:

$$F_{Rd,avg,WW}=47+\frac{70-47}{8.3}\cdot3.9=57.8\ kN$$

$$F_{H/300,avg,WW}=4.5+\frac{15.4-4.5}{8.3}\cdot3.9=9.6\ kN$$

$$F_{Rd, WW,d}=\frac{F_{Rd,avg,WW}}{1.5}=38.5\ kN$$

$$F_{H/300, WW,d}=\frac{F_{H/300,avg,WW}}{1.5}=6.4\ kN$$

Combination

The capacity of the two walls are summed up to obtain the final value:

$$F_{Rd,d}=F_{Rd,WW,d}+F_{Rd,DW,d}=38.5+21.7=60.2\ kN$$

$$F_{H/300,d}=F_{H/300,WW,d}+F_{H/300,DW,d}=6.4+7.3=13.7\ kN$$

### Check

It can be checked that:

$$F_{Rd,d}=60.2\ kN \ge V_{sd,E-W,ULS}=19.3\ kN$$ ✅

$$F_{H/300,d}=13.7\ kN \ge V_{sd,E-W,SLS}=12.9\ kN$$ ✅

# Research

## Footfall vibration

Skylark floor blocks weigh less than 130—150 Kg: this number is 1 to 2 times the weight of an average person. For this reason, it is important to assess their sensitivity to footfall vibrations. Walking tests performed on a Skylark floor spanning over 5.5 m suggest that the structure meets the criteria given by the British Standard 6472: "Guide to evaluation of human exposure to vibration in buildings (2008)". Further details can be found in the report here.

## Help improve this guide

WikiHouse Skylark is fully open source; it is a solution collaboratively developed, by anyone, for everyone.

You can help improve Skylark and make it easier to use by sharing your:

• Experimental data If you have means to replicate any of these tests, or carry out different tests, by sharing your test data you can give engineers ever more statistical confidence, and help track the impact of improvements.
• Structural models
• Calculation methods and custom software for Skylark modelling
• Worked examples From a project or part of a project.

If you can contribute by sharing any of these, we'd love to hear from you, please get in touch.

### Use it

Most WikiHouse files and information are licensed under a Creative Commons–Sharealike licence, so you are free to use, distribute or modify them, including commercially.

### Check it

All WikiHouse information is shared 'as is', without warranties or guarantees of any kind. You are responsible for checking it and using it in a safe and responsible way, for example, getting it checked by a structural engineer.

### Comply with regulations

You are responsible for making sure your project complies with all relevant local regulations, including planning, building codes and health & safety legislation. If in doubt, seek professional advice.

### Re-share your improvements

If you make any improvements to the system, you must publish your files under the same type of open licence. However, you do not need to publish the plans and specifications for individual projects unless you wish to.

## Please do not

### Call yourself WikiHouse

Do not call your company, organisation or any marketed product or service 'WikiHouse'. However, you may use the term WikiHouse to talk about the system, and you may describe your project, product, service or organisation as, for example, "using WikiHouse", "based on WikiHouse", "contributing to WikiHouse", or similar.

### Remove notices

Do not remove any licence notices from files if you are re-sharing them.

### Claim to be endorsed

Do not give the impression that you are endorsed by, or affiliated with WikiHouse or Open Systems Lab (unless you are, by written agreement), and do not claim to represent the WikiHouse project or community as a whole.