Friday, 30 April 2010
Wednesday, 28 April 2010
Calculations For Buckling
Monday, 26 April 2010
Material Decision
After a group meeting, the final decision between Aluminium 6061, Aluminium 6063, Cast Carbon Steel, and Carbon Steel 1023 was discussed. The group as a whole agreed that using any form of Steel would be too heavy of a material. This left the choice between the two grades of Aluminium. After some final research, we found that the cost of Aluminium 6063 is slightly more than 6061 but their properties are almost identical, however Aluminium 6063 is slightly heavier also. Both materials were had good joining properties, both could be welded through TIG (Tungsten Inert Gas) and MIG (Metal Inert Gas) welding.
This ultimately led to the decision that Aluminium 6061 should be used.
Aluminum 6061 can be found in a very wide range of products. A very common product would be a bicycle with a tubular frame. Here is an example of a bike using aluminium 6061.
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For the crane, a hollow tubular frame would be welded together to form the Truss' that will join together to create the crane.
Here is an example of a tubular frame used on a race car. The frame is designed to be strong but lightweight which makes it good for its application.
http://www.galmerinc.com/images/dsr_frame.png
Sunday, 25 April 2010
Material Analysis II
Aluminium
First to be analysed was the different types of Aluminium available. There is a wide variety of Aluminium available with varying grades of aluminium mixed with other materials. The beginning number, ranging from 1-8 of the 4 digit number represents the strength of the material. 1 being the weakest form, and 8 being the strongest.
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Steel
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The data will be presented to the group in the next meeting, we will discuss further the choice between Aluminium 6061, Aluminium 6063, Cast Carbon Steel, and Carbon Steel 1023.
Friday, 23 April 2010
Forces/Stress Calculation (i)
To find the counter balancing weight required for the crane beam to lift 1500kg and the total force going through to the boom, we use moments -
Equate the clockwise and anticlockwise moments:
(About C clockwise) - Xg x 1 = 55.2g x (1.5 - 1) + 1000g x 2
Xg = 27.6g + 2000g = 2027.6Kg mass required for balance
2027.6 x 9.8 = 19870.48N force required for balance
Equating the forces in a vertical direction:
R = Xg + 55.2g + 1000g = 2027.6g + 55.2g + 1000g = 3082.8Kg Mass
4582.8 x 9.8 = 30211.44N Force - Both of which is the total amount acting through to the boom.
Forces/Stress Calculation (ii)
Finding the area of the circular support –
Π x [(140-120)/2]2 = 314mm2 = 0.314m2
Therefore the stress acting on the surface area of the circular support is (Using s = F/A)
s = 30211.44/0.314 = 96214.78Nm-2
As there are four supporting strut ‘legs’, the total force would be equally distributed along them as such:
30211.44/4 = 7552.86N per ‘leg’
a = 7552.86 x cos30 = 6540.96N
b = 7552.86 x cos60 = 3776.43N
c = 7552.86 x cos60 = 3776.43NThe second boom/box support has the combined forces of the previous boom/box supports weight as well as the total weight of the boom with maximum load and counterweight.
Using previously done calculations of structure weight –
Therefore, the approximate mass of the boom support can be assumed to be ~ 27.6kg = 270.48N
So the total force acting down on the second boom support is 270.48 + 30211.44 = 30481.92N
Overall force acting downwards on the structure is – 52.5 + 27.6 + 27.6 + (13.4x4) = 161.3kg
Which is 1580.74N of force, as well as the force of the weight and counterweights which gives
9800 + 19870.48 + 1580.74N = 31251.22N