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- 1. AD572 Individual Project 32-33 Great James Street Osarieme Isokpan and Eleonore Champenois
- 2. The Extension is too noisy especially the upper ground floor. On our visit to the site, we took along a sound metre to record accurate sound measurements within the spaces, however before doing so, we started off by questioning those who occupied that particular space. The general response to this area was that it was too noisy as conversations travelled along the room and echoed around the space. After this, we then took our recording at the measurements were as follows Ground Floor 78dB with distinct conversation Vault 40dB with no conversation and 65dB with disctinct conversation 1st Floor 60dB with distinct conversation. From the measurements carried out above, it can be seen the the sound quality in the office spaces are quite high. The figures below show what the sound measurement of a office spaces should be Cellular offices = 35-40 dB Open plan offices = 40 - 50dB A cellular office space is an office whereby a large space is divided into various sections by the use of permanent walls An Open plan office is a space whereby office activities are spread across a single floor. Unlike cellular office spaces, there are no wall barriers to demarcate between the various operations within it. The extension in great James streets utilizes an open plan office space. The initial aim of the architects was to encourage more relations between the users of the building. However, the sound measurements in the building were up to 78dB. This accord- ing to the specified requirements sound level for an office space is too high. Aim The aim of our project is therefore to isolate the problems causing the excess amount of sound travelling across the building and provide a solution to that particular problem faced within it. As stated earlier, the sound level in 78dB which is too high for an open plan office space Problem Pictures showing the open plan layout of the Upper ground floor
- 3. Possible Cause of High Sound Level within the space In order to provide a solution to the problem of noise, we have first isolated the likely causes of noise within it. They are as follows 1) The High Inclined Ceiling 2) The choice of materials within the space 3) The Choice of Acoustic Material present within the building. 1) The High Inclined Ceiling - The Top floor of the extension has an inclined ceiling due to the triangulated shape of the roof, with the highest point of the inclination being 3.9m tall. The large volume of space caused by this high inclined ceiling caused echoing within it. the persistence of a sound in a space after it is produced is known as reverberation. In rooms, reverberation is created when sound produced in the enclosed space causes a large number of echoes to build up and then slowly decay as the sound is absorbed by the ceiling, floor, walls, other room surfaces and air. Inclined ceilings have both a sound spreading and a sound concentrating effect. In most cases, the sound is concentrated because the sound regulation of the area around the inclined ceiling has not been considered carefully. 2)The Choice of Materials within the Space - Different materials have different sound absorbing qualities. In the exten- sion, we will be looking at the materials utilised mainly on the ceiling. The ceiling is made out of sheets of plasterboard. Plasterboard which is an inner layer of gypsum sandwiched between two outer layers of lining paper including various additives in the gypsum layer and varying the weight and strength of the lining paper. Plasterboard however being quite a light material vibrates when it comes in contact with sound. This causes sound to spread more across the room, leading to a rise in the noise level within the spaces. Also due to the fact that the plaster board ceiling is not in a whole sheet across the ceiling but rather in separate panels, this also cause more vibration within the space. Plaster board at a frequency of 500Hz, 1000Hz and 4000Hz have an absorption co-efficiency of 0.05, 0.04 and 0.1 respectively. These are quite low as the closer the absorption coefficiency is to 1, the higher its acoustic properties. 3) The Choice of Acoustic Materials present within the Building - Within the ceiling panel and the roof, there are acous- tic panels presents however these panels are only a single layer. The architects did not take into consideration the shape of the roof before providing a basic acoustic control within the building. Studies show that the wall area opposite the in- clined ceiling should also be equipped with sound absorbing materials. As a principal rule, all surfaces above the normal ceiling height (2.60 m) including the end walls should be equipped with sound absorbers. However acoustic panelings in the extension are only present between the ceiling and roof. 3D Section of the space above showing the high inclined ceiling. and the coice of ceiling material. Section showing the material properties of the ceiling Bronze copper Bitumous layer Timber Joist Plasterboard
- 4. Solution In order to first provide a solution to the problem of noise within the building, we must first have an understanding of the following terms. As through the calculating and understanding the building in these aspects, it will enable us to recom- mend the best possible solution to reduce the noise level in the space. From the research carried out above however, it can be seen that the problem of the building is not predominantly as a result of the shape of the ceiling, but the architects not providing adequate acoustic solution to accommodate this shape 1) Reverberation Time 2)Sabines Law/Alpha Sabine 1)Reverberation Time - The reverberation time is perceived as the time for the sound to die away after the sound source ceases, but that of course depends upon the intensity of the sound. The optimum reverberation time for an auditorium or room of course depends upon its intended use. Around 2 seconds is desirable for a medium-sized, general purpose audito- rium that is to be used for both speech and music. A classroom should be much shorter, less than a second. Due to the fact of out building being an office space, research shows that the recommended reverberation time in an office is 0.50 seconds with that of an open plan office extended up to 1.2 The reverberation time is strongly influenced by the absorption coefficients of the surface and also the volume of the room as show in the Sabines formula below. 2) Sound Absorption Coefficient - When a sound wave in a room strikes a surface, a certain fraction of it is absorbed, and a certain amount is transmitted into the surface. Both of these amounts are lost from the room, and the fractional loss is characterized by an absorption coefficient a which can take values between 0 and 1, 1 being a perfect absorber. 3)Sabines law -The product of the reverberation time multiplied by the total absorptivity of the room is proportional to the volume of the room Diagrammatic representation of how sound bounces off surfaces within the space
- 5. Calculations 5.3m 1.7m 1.8m 13m V1 V2 V3 V4 V5 V6 H1 H2 H3 H4 H5 H6 H7 Total 268.71 4.47 44.02 68.95 30.61 8.78 234.26 3.40 33.45 60.11 23.26 7.65 220.48 3.20 29.63 56.57 20.61 7.2 192.92 2.30 22.69 49.50 15.78 6.3 165.36 1.69 16.67 42.43 11.59 5.4 158.47 1.55 15.31 40.66 10.65 5.18 144.69 1.30 12.76 37.13 8.87 4.73 197.84 2.56 24.93 50.76 17.34 6.46 299.89TOTAL V = LxWxH or V = ((bxBxH)/2) x H T = (0.163x299.89)/A In order to calculate the reverberation time, we first have to find the volume of the space. Because of the varying ceiling heights and also the shape being of an irregular geometric form with varying lengths of its side.We divided the shape into various sections that will enable us to find a volume. The various heights of the ceiling are as follows 3.9m,3.4m.3.2m,2.8m,2.4m,2.3m and 2.1m. To calulate the total volume we calculated the volume of each individual shape with the varying heights of the ceiling (3.9m,3.4m.3.2m,2.8m,2.4m,2.3m and 2.1m) and derived an average in order to have an estimated volume. After finding the volume, the absoption coefficient of each space is then requred to be known in order to calculathe reverberation time. Absorption surface is the whole surface in the room which can absorb sound waves. This includes walls, floors, windows and doors, ceilings and furniture. We decided to only take into account 1)the walls, 2)the floor, 3)the ceiling and 4)the doors and windows As we didnt have the precise measurements of the furniture and we considered that the architects had worked on the acoustic with- out the furniture to make it more efficient. Table showing how the volume of the space was calculated. Diagramtic representation of how the plan was divided into sections in order to calculate a volume
- 6. 1) WALLS : S = H x L with S the surface in m^3 , H the height in m and L the length in m The total surface is the addition of all the surfaces of the walls. H = 2.8m RESULTS : Knowing that the walls are made out of bricks we found their absorption coefficient at three different frequen- cies. When f = 500 Hz, a = 0.03 When f = 1000 Hz, a = 0.04 When f = 5000 Hz, a = 0.07 We can then calculate the absorption surfaces : When f = 500 Hz : A = S x A = 61.18 x 0.03 = 1.8354 When f = 1000 Hz : A = 61.18 x 0.04 = 2.4472 When f = 5000 Hz : A = 61.18 x 0.03 = 4.2826 2) FLOOR : To calculate the surface of the floor, we kept the division from the previous calculation. For rectangles : S = L x W For trapeziums : S = ((b+B)/2) x h Figure Length l (m) Width W (m) Height h (m) Big Base B (m) Small Base b (m) Surfaces (m^2) 1 2 3 4 5 6 2.81 11.67 3.45 1.5 0.94 2.8 2.1 1.5 2.66 1.5 1.25 1.5 0.47 1.1 2.64 32.70 2.29 7.25 1.95 2.25 TOTAL 49.08 We know the floor is made out of wood and this gives us the three absorption coefficient following : When f = 500 Hz, a = 0.10 When f = 1000 Hz, a = 0.07 When f = 5000 Hz, a = 0.06 Which can allow us to calculate the absorption surfaces : When f = 500 Hz : A = S x A = 49.08 x 0.10 = 4.908 When f = 1000 Hz : A = 49.08 x 0.07 = 3.4356 When f = 5000 Hz : A = 49.08 x 0.06 = 2.9448 3) CEILING : The ceiling is composed of triangles so S = (B x h) / 2 Triangle Base B (m) Height (m) Surfaces (m^2) 1 2 3 4 5 6 7 8 9 10 11 12 13 5.31 2.11 5.60 6.25 2.03 6.34 2.4 6.02 7.22 2.34 6.25 7.31 6.25 1.88 5.88 6.25 3.75 11.72 3.59 1.88 3.37 4.49 1.56 3.66 5.16 4.53 11.69 1.5 4.53 3.40 4.22 4.38 9.24 4.22 5.94 12.58 3.13 3.91 6.12 TOTAL 94.08 As the ceiling is made out of plaster board : When f = 500 Hz, a = 0.05 When f = 1000 Hz, a = 0.04 When f = 5000 Hz, a = 0.1 So when f = 500 Hz : A = S x A = 94.08 x 0.05 = 4.704 When f = 1000 Hz : A = 94.08 x 0.04 = 3.7632 When f = 5000 Hz : A = 94.08 x 0.1 = 9.408 4) DOORS : For this calculation we are considering that the doors height is 2.1m. S = (L x W) x2 with L the length in m and W the width in m S= (2.1 x 0.94) x 2 = 3.948 As the doors are made out of wood : When f = 500 Hz, a = 0.17 When f = 1000 Hz, a = 0.09 When f = 5000 Hz, a = 0.10 So when f = 500 Hz : A = S x A = 3.948 x 0.17 = 0.67116 When f = 1000 Hz : A = 3.948 x 0.09 = 0.35532 When f = 5000 Hz : A = 3.948 x 0.10 = 0.3948 Measurements in m Surfaces in m^2 11.80 4.80 0.3(x3) 0.5(x3) 0.7 0.65 1.1 0.4 33.04 13.44 2.52 4.2 1.96 1.82 3.08 1.12 TOTAL SURFACE 61.18m^2
- 7. 5) WINDOWS : The open office space is composed of four windows, two rectangular ones on the walls and two triangular-shaped on the ceiling. For the rectangles : S = L x W For triangles : S = (B x h) / 2 Figure Length l (m) Width W (m) Height h (m) Base (m) Surface 1 2.1 4.06 2 3 4 2.1 2.81 8.53 5.90 2.58 2.81 3.6249 3.28 3.44 5.6416 TOTAL 23.6965 As these are double-glazed windows : When f = 500 Hz, a = 0. 03 When f = 1000 Hz, a = 0.03 When f = 5000 Hz, a = 0.12 So when f = 500 Hz : A = S x A = 23.6965 x 0.03 = 0.7109 When f = 1000 Hz : A = 23.6965 x 0.03 = 0.7109 When f = 5000 Hz : A = 23.6965 x 0.02 = 0.4739 Finally we can then calculate the reverberation time for each frequency : When f = 500 Hz : A = 1.8354 + 4.908 + 4.704 + 0.67116 + 0.7109 = 8.1255 TR = (0.163 x 299.89) / 8.1255 = 6.02 s When f = 1000 Hz : A = 2.4472 + 3.4356 + 3.7632 + 0.35532 + 0.7109 = 10.7122 TR = (0.163 x 299.89) / 10.7122 = 4.56 s When f = 4000 Hz : A= 4.2826 + 2.9448 + 9.408 + 0.3948 + 0.4739 = 17.5041 TR = (0.163 x 299.89) / 17.5041 = 2.79 s In conclusion, the reverberation time is way too high for each of the frequencies thus showing a flaw in the acoustic insulation. The reverberation we derived were 6.02seconds , 4.56 seconds and 1.79 seconds.. These figures show that the time is takes for sound to travel round the space is too long, Thereby, making conversa- tions spread across the room and leading to a noisy atmosphere unsuitable for a working environment. These calculations will help us to suggest a suitable design adjustment that will aid in reducing the noise level within this area.
- 8. Design Iteration As it can be seen from the calculations above the Reverberation time for the space at the following Frequencies ; 500hz, 1000Hz and 4000Hz are 6.02seconds , 4.56seconds and 2.79seconds. From studies carried out, the ideal reverberation time for an open plan office space is between 0.50 seconds and 1.2 seconds. The results we derived from our calculations hoever, show that the time it takes for sound to travel back to its source is too long. In order to improve the sound quality of the building, we have to decided to look into the materi- als utilised in the construction of it. We habe chosen to replace the plasterboard on the ceiling with acoustic tile panels The panels are suitable for open plan offices or other premises where strict demands are made on good acoustics and speech intelligibility, and where demountability is vital. Ecophon Master A has an exposed grid system which each tile easily demountable. The system consists of Ecophon Master A tiles and Ecophon Connect grid systems, with an approx- imate weight of 5 kg/m族. The tiles are manufactured from high density, 3rd generation glass wool. The visible surface has an Akutex FT coating and the back of the tile is covered with glass tissue. The edges are primed and the grid is manufactured from galvanized steel. The tiles withstand a permanent ambient RH up to 95% at 30属C without sagging, warping or delami- nating. The glass wool core of the tiles is tested and classified as non-combustible The buildings on the right all utilise this system of tiling on their ceilings These are all office spaces / Private-public spaces that aimed to reduce the amaount of noise level within them. By incorporating the tiles on the ceiling, they were able to promote communication, collaboration and creativity. As stated earlier, the tiles are not ony demoutable, but they are also flexible and can be utilised on ceilings regardless of its shape and size. Top Left - Sumitomo Electric Bordnetze GmbH Office Top Right - Areco Sweden AB - SE Bottom Left - Ulster Hospital, Innovation and Medical Centre Bottom Right - Belgorod airport
- 9. 1. Ecophon Edge A panel 2 Main Runner 3 Cross Tee 4 Cross Tee 5 Connect Adjustable hanger 6 Connect Hanger clip 7 Connect Direct Fixing Bracket 8 Connect Angle Trim 9 Connect Shad- ow-line Trim The following Diagrams show how the acoustic tiles attach to the ceiling. As stated earlier, the panels can be cut into any desired shape. After cutting the panels into the shape of the ceil- ing, The Contact Angle Trim (8) and the connect shadow-line trim are placed directly on the ceiling. Afterwards the structure is braced with the main runner (1) which is then connected to the connect hanger clip (6). The Main runner and also Cross tees are fixed to the walls by the Connect Direct Fixing Bracket (7). Once the structure is secured with the use of screws, an adjustable hanger is hooked to the hanger clip. This hanger determines the distance between the acoustic tiles and also the ceiling itself. It enabled there to be a space between these two. However, our tiles will be placed as close to the ceiling as possible. Suspension with adjustable hanger and clip Suspension with direct bracket
- 10. Diagramatical representation of the Acoutic Panels in the shape of the ceiling Conclusion RESULTS AFTER IMPROVEMENT : The ceiling is not made with Ecophon Master A tiles which leads to a change in the absorption coef- ficient With a frequency of f = 500 Hz, a = 0.72 f = 1000 Hz, a = 0.92 f = 4000 Hz, a = 0.75 Thereby, when f = 500 Hz : A = S x a = 94.08 x 0.72 = 67.7376 When f = 1000 Hz : A = 94.08 x 0.92 = 86.5536 When f = 4000 Hz : A= 94.08 x 0.75 = 70.56 Thus the new reverberation time is : At f = 500 Hz : A = 1.8354 + 4.908 + 67.7376 + 0.67116 + 0.7109 = 75.86306 TR = (0.163 x 299.89) / 75.86306 = 0.64 s At f = 1000 Hz : A = 2.4472 + 3.4356 + 86.5536 + 0.35532 + 0.7109 = 90.50262 TR = (0.163 x 299.89) / 90.50262 = 0.54 s At f = 4000 Hz : A= 4.2826 + 2.9448 + 70.56 + 0.3948 + 0.4739 = 78.6561 TR = (0.163 x 299.89) / 78.6561 = 0.62 s From the results above, it can be seen that by changing the material property of the ceiling through the addition of acoustic panels, we were able to achieve a reduction in the reverberation time within the space. As stated earlier, the required reverberation time for an open plan office space is between 0.50 seconds and 1.2 seconds. In the results we derived, it can be seen the the reverberation at the following frequencies 500Hz,1000Hz and 4000Hz are 0.64 seconds, 0.54seconds and 0.62 seconds respec- tively. These reverberation times at the following frequencies are ideal for an office / open plan office space and by the inclusion of this design change there will be a signif- icant reduction in the level of noise experienced.