The following paper gives an insight into the proposal to construct tents put forward by Alpine Trekking Pvt. Ltd. The proposal is carefully examined and all considerations are taken into account. Based on the data given, a basic framework of the problem is constructed.
Further in the report, some iterations of the tent design are considered. The best one is selected and justified.
For the purpose of designing, analytical method is employed in which total space needed is calculated using all the constraints provided.
For maximum space utilization, the height of the tents is fixed at the maximum limit of 4 m.
So, accordingly the angle from base, width and length are varied for the iterations.
The paper is presented according to the sections in the client brief.
Section 1 includes details of the floor plans and designated areas.
Section 2 tells us about the internal structure.
Section 3 lets us know about the roof.
The following technical report gives a summarized technical analysis about design considerations of a hut structure. The demand for the designs was made by Alpine Trekkers Pvt. Ltd.
The basic requirements of the hut were that they should be able to accommodate 12 persons with 12 m2 of free space for movement and separate space for luggage and cooking area along with some other constraints.
The project also needed a heater which would be run by a micro - hydro generator. The cost considerations for the project were also to be taken into account.
The report also shows other designs which fulfill the criteria and discusses their pro’s and con’s.
SECTION 1 :
Section 1 deals with the floor plan and arrangement of designated areas. To maximize space and comfort, the default height for the structure is set at 4 meters which is the maximum allowable vertical height.
According to the given constraints, the floor clearances needed are:
Bunk beds of area (1.8 m * 0.8 m) = 14.4 m2. Since, 12 persons are to be accommodated considering 2 person bunk beds, we have a total of 6 beds.
Clearance to beds on one side = 0.8 m2.
Total luggage area of 3 m2.
Total cooking space of 2 m2.
Free space for movement of about 12 m2.
Door space for a 0.8 m wide door. Considering a 90o arc of door, we have the area at about 2.02 m2.
Considering the above requirements, the minimum requirement for floor space is about 33 m2. To facilitate calculations, values are rounded to next most convenient value.
The following table shows the iterations that can be achieved.
[ Table no. 1 : Table showing variations in design that can be achieved by varing angle with ground]
Width is the amount of space between side walls of 1.2 m height.
Base is the total distance between the anchor points of the tent.
The structure with angle 45o is selected as the best structure. The reasons are specified later. All further calculations are done with respect to this design consideration.
The designs are shown below.
[Image 1 : Image shows top view of selected design.]
[Image 2 : Image shows the front view of the selected design. All dimensions are in mm]
SECTION 2 :
Section two deals with amount of plywood required for the internal surfaces of the hut. In this section we also find the rate of heat loss through the walls, door and windows.
Then, we need to subtract the area of the windows and the door.
Once we do this, we can find the total amount of ply needed for the interior of the structure.
Area of ply wood = area of walls neglecting roof and base area - area of door - area of window.
= 44.24 m2 - 1.44 m2 - 1.44 m2.
= 41.36 m2 .
So, Area of the wall = AW = 41.36 m2.
Rate of heat loss is calculated using the formula,
Q = k*A*(T1 - T2)/d
Using data from the Client Brief, we have :
Ambient temperature = -10o C.
Inside temperature = 20o C
1. Ply wood:
Thickness = 18 mm.
k = 0.015 W/ m K.
So, rate of heat transfer is 1034 W from inside to out.
2. Windows :
Thickness = 8 mm
k = 1.4 W/ m K
Rate of heat transfer is 3780 W from inside to outside.
Since there are two windows, total losses are 7560 W.
3. Door :
Thickness = 36 mm
K = 0.015 W/m K
Rate of heat transfer through the door is 18 W from inside to outside.
So, total heat loss from inside to outside is 8612 W. (Without considering the roof)
QW = 8612 W.
SECTION 3 :
In this section, we were asked to find the quantity of ply needed for the internal ceiling and the quantity of steel sheets for the external roof of the structure.
Also, the amount of heat loss from ceiling was to be found out.
1. Plywood :
The total plywood needed for the ceiling apart from the walls is found out by trigonometry to be 49.77 m2.
AC = 49.77 m2.
So, the rate of heat transfer from the ceiling is found, by the method shown earlier, to be 1244.25 W from interior to the exterior.
2. Steel sheets :
The total quantity of steel sheets needed are found in a similar manner to be 71.2656 m2
AR = 71.2656 m2.
So, the rate of heat transfer through the sheet is:
Thickness of sheet : 0.7 mm
k = 65.
So the rate of heat transfer is found to be 199 MW (199 * 106 W).
Total heat loss through the steel roof and ply ceiling is the summation of the two quantities, which is 199001244 W ≈ 199 MW.
But this loss is not from interior. This loss takes place from the space between the interior space and outer steel sheets.
Since, the heater is going to heat the interior of the hut, the losses from the wooden cavity only need to be considered.
So, the total losses add up to:
QW + losses through ply ceiling.
= 8612 W + 1244 W = 9856 W.
The heater coils need to compensate this loss.
SECTION 4 :
In this section, we had to explore options to reduce the heat loss.
Since major heat losses are taking place through the windows, we can replace the windows by double glazed windows.
Double glazed windows have a total thickness of 46 mm. (Two panes of glass and air gap)
Their co-efficient of thermal conductivity is k = 0.0263.
So, the amount of heat escaping the windows is
Q = (k * A * ∆T) / t
= 0.0263 * 0.72 m2 * (-30) / (46 * 10-3)
= 12.34 W.
For two windows, the quantity goes up to 25 W.
This is a very drastic reduction in heat losses.
So, heat losses without insulation : 9856 W
Heat losses with Double glazed windows = 2710 W.
QT = 2710 W.
Based on the above calculations of heat losses with insulating material, we need to select quantity of heaters.
Collectively, they should generate more heat than 2710 W.
So, we need 3 heaters of 1 kW each.
So, total number of heaters is 3.
NH = 3
MICRO - HYDRO GENERATOR :
The micro hydro generator needs to generate 2710 W of power. Since the generator operates at 100% efficiency, it can aid the calculation.
PG = 2710 W
Model of micro hydro generator :
MH - HG 2 with output of 2750 W.
SECTION 5 :
In this section, we need to check the cost of all the equipment and whether if fits the budget.
The following table summarizes the materials used and their respective quantities.
[Table No. 2 : Table summarizing key parameters of selected design]
The cost estimate:
Cost of Hydro generator : CHG 2 = $ 4000.
Cost of Electric heaters : CEH = 500 * 3 = $ 1500.
Cost of ply : CP = (92 * 50) = $ 4600.
Cost of roofing steel sheets : CR = (72 * 30) = $ 2160
Cost of windows : CW = 2 * 300 = $ 600
Grand total cost : C = $ 12860.
So, the proposed design fits the budget.
ALTERNATIVE DESIGNS :
The above design is suitable because the ratio of length to breadth of the structure is between 1 and 2.
Since, for maximum utility, the height is fixed at 4 m, the adjustments in the design can be made only by changing the angle θ with the horizontal.
Values of θ between 30 and 60 are needed.
Any value that tends to go near 30 makes the length to become shorter than the width. This hampers the aesthetic appeal as well as the structure’s stability in rough weather.
Stephen Ressler (2001) states, stability of triangles is a function of their angles. Make the angles to small and the structure may collapse. Conversely, make them too large and the members may crumble on loading.
Expert architect, Tod Williams (2013) believes the most stable structure in the world is the triangle. Structures of all shapes and sizes can be made with triangular shapes.
Similarly, any value that tends to go too near 60 leads to length becoming far bigger than the width. Due to this, the stability of the structure is affected and its appearance is distorted.
Hence the following structure was selected.
Key parameters of the selected design are:
1. Ressler, Stephen. 2001. Designing and Building Bridges. American Socity Of civil Engineers, U.S.A. . Available on: Google books [5th December 1997]
2. William, Tod. 2013. ‘Experts Discuss Structural Concepts’. Internet Journal. Available from <http://thetriangle.org/news/experts-discuss-structural-concepts/> [13th January 2013].