Super-Ken 2000

Dear friends of mine,

I am thankful for my mother because she has been taking care of my favor tortoises everyday, she moves my tortoises from their accommodation containers in the living room to the bathroom everyday at least twice per head per time. It is likely to cause damage or pain in her back if she is careless or uses incorrect posture to lift the tortoises. In order to keep his body from getting any damage, I got down to design a crane several days ago and hope it can help my mother.

The idea is shown below:

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Materials for the crane:
For the boom & mast(Steel Pipe- 50mm Diameter (2” water pipe)---weight 5.4kg/m
For the boom extension (Steel Pipe- 38mm Diameter (1-1/2” water pipe)---weight 4Kg/m
For the fixing brackets (Mild steel plate- 2-3mm thick)
Accessories:Manual Winch, Diverting Pulley, wire rope & lifting hook, trolley and base bracket for the mast.
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Calculation & Consideration

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Ww =(1Kg+5.4Kg/m x 0.15m)=1.81Kg(17.76N) (Weights mixed with the Hand Winch & the boom between A1 & D)

Wb =(5.4Kg/m x 0.5m)=2.7Kg (26.49N) (Weight of the boom between A1 and B)

We =(4Kg/m x 0.15m)=0.6Kg (5.89N) (Weight of the boom extension between B and C)

Wm =100Kg(981N) (The desired capacity is considered to be lifting 100Kg and for future purpose)
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A free-body diagram of boom DC Shown in Figure 2

Determining the pin reaction at E

Take Moment at Point A1
+ ↘(Σ M)A1 = 0:
-(Wb x 0.25m)-(We x 0.575m)-(Wm x 0.65m)+(ESin45 x 0.25m)+(Ww x 0.075m)+(Ecos45 x 0.05m)=0

-(2.7kg x 0.25m)-(0.6kg x 0.575m)-(100kg x 0.65m)+(Esin45 x 0.25m)+(1.81kg x 0.075m)+(Ecos45 x 0.05m)= 0
-0.675kgm-0.345kgm-65kgm+0.177m(E)+0.1358kgm+0.0354m(E)= 0
0.2131m(E)= 65.88kgm
E= 309.15kg
E= 3032.8N (309.15kg x 9.81)
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Determining the normal force (Fx) and the transverse shear forces (Vx):

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Next step I have to figure out the internal resultants such as F(x), V(x) and M(x) for the crane through the following calculations. As with all engineers and designers, I have been experienced engineering research and design practices that require aggressive thinking and problem-solving skills. And the most important to me is that this mission can enable me to develop a better understanding of science and engineering. Through this as well, it revealed that I am still weak in solving problems, although I had been exposed to the subjects such as Deformation of materials, Beam analysis and Basic Mechanics when I studied at the University, I have a few chance to feel truly and practice in the real sense.
I am not sure the approach whether is exactly right what I am working for, but I think it has kept close and not far from the real. To check whether the size and material of the pipe is able to saddle with a tortoise weighting 100Kg when the time I use the crane to lift it, a rigid conduit with 50mm (2” steel pipe) in diameter and 0.5m long whether is satisfied with these internal resultant forces, let’s working and finding what will be seen.
I assume that there are some internal resultant forces yielded in the boom when lifting, and assume that these now are put on an arbitrary cross section in the boom A1-C at a distance “X” from end A1. On the section at X (see Figure 3) you can see the unknown normal force (Fx), the unknown transverse shear force (Vx) and the unknown bending moment (Mx).
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Take Moment at Point A1 (for the unknown normal force, Fx)
+ →(Σ F- horizontal) = 0:
-Fx+E(Cos45)=0
-Fx + 309.15Kg(Cos45)=0
Fx = 218.6Kg (2144.47N)

(Note: Fx, normal force=Ax=2144.47N)

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To figure out the vertical shear forces, I begin to check by starting at the point A1 (A1 is the pivot point, is also the maximum shear force, I named it to be Ay) and then cut the beam a distance “X” from the A1, where x is a distance greater than zero and less the position where the loading of the beam changes in some ways. In this expression, the maximum shear force at A1, and other internal shear forces in the beam between A1 to B and in the beam between B and C, can be seen.

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Given: Wx =(5.4Kg/m x (0.5-x)m)=(2.7-5.4x)kg (Weight of the boom between X and B

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Take Moment at Point A1 (for all unknown transverse shear forces, Vx)
+ ↑(Σ F- vertical) = 0:
Vx + E(Sin45)-(Wx+We+Wm)=0
Vx +309.15Kg(Sin45) – ((2.7-5.4x)Kg + 0.6Kg + 100Kg)=0
Vx +218.602kg-0.6kg-100kg-(2.7-5.4x)kg=0
Vx = -115.302kg-(2.7-5.4x)kg
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If x=0mm, Vx=-118.002kg (Note: Vx=Ay=-1157.6N)
(i.e: 1157.6N acting opposite to the direction shown on the free-body diagram)
If x=10mm, Vx=-117.95Kg (-1157.1N)
If x=100mm, Vx=-117.47Kg (-1152.4N)
If x=200mm, Vx=-116.92Kg (-1146.99N)

If x=500mm, Vx=-115.30Kg (-1131.11N)

If x=650mm, Vx=-114.49Kg (-1123.17N)

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Notice as X near zero, the shear force value in the beam goes to -1157.1N, it is the maximum shearing force, and as X approaches 500mm (Point B), the shear force value becomes -1131.1N. All of these values related to the vertical shearing force are a guide to help choosing a suitable pipe as the boom of my dream crane.

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To determine the bending moment expression by applying rotational equilibrium conditions, the values of bending moment (Mx) are listed below:

+↘(ΣMx)=0
-Mx–(Wx)(0.25m-x)-(We)(0.525m)-(Wm)(0.6m)+E(Sin45)(0.25m-x)=0
-Mx-((2.7-5.4x)Kgx(0.25-x)m)-(0.6Kgx0.525m)-(100Kgx0.6m)+(309.15KgxSin45x(0.25m-x))=0
-Mx–(0.25-x)(2.7-5.4x)kgm-(0.315Kgm)-(60kgm)+(218.6kg)(0.25-x)m=0
-Mx–(0.675-4.05x+5.4x^2)kgm-60.315kgm+54.65kgm-218.6xkgm=0
-Mx–(0.675-4.05x+5.4x^2)kgm-5.665kgm-218.6xkgm=0
-Mx-5.4x^2kgm+4.05xkgm-0.675kgm-5.665kgm-218.6xkgm=0
-Mx–〔5.4x^2 +214.55x-6.34〕kgm=0
Mx=-〔5.4x^2 +214.55x-6.34〕kgm

If x=0mm (x at A1), Mx=+6.34kgm (+62.2Nm)
If x=10mm, Mx= +4.19kgm (+41.10Nm)
If x=100mm, Mx= -15.17kgm (-148.82Nm)
If x=200mm, Mx= -36.79kgm (-360.91Nm)
If x=500mm (x at B), Mx= -102.29kgm (-1003.47Nm)
If x=650mm (x at C), Mx= -135.4kgm (-1328.27Nm, Max bending moment)
If x=800mm (x at the case using boom extension), Mx= -168.76Kgm (-1655.54Nm)

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Considering the bolted joint shown in Fig 4.

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As the force E has been found, it is 3.033KN, and the diameter of the locking bolt is considered as 10mm. Lacking any more precise information I can only do as per the 1st law of Super-Ken and assume that force E is equally divided between the sections a-a and b-b.

Consequently a force of 1/2(3.033 x 103)= E/2,1516.5N acts across either of these planes over a cross-sectional area:1/4 (π)d^2= 0.25 (3.1416) (10mm)^2 = 78.6mm^2. Thus the average shearing stress across either plane is Tau (Shear Stress).Tau= 1/2 (3032.8N/78.6mm^2) = 19.3Nmm^2. It is because the average shearing stress across either plane is Tau (Average Shear Stress), Tau can thus be found by 1/2 (E/A)

Tau (avg)= 19.293Mpa.

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For the sake of getting it easily, I want the bolt including its material and size can be popular and bought randomly in the local metal shop, Class 4.6 low-medium carbon steel M10 is considered.

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From the table, the bolt material (low –medium carbon steel, M5-M100, Class 4.6) has a shear strength of 225Mpa, the related factor of safety can be calculated by using the following simple equation:

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Factor of Safety= Shear Strength (M10 Bolt) / total shear force

Factor of Safety= (225Mpa)/(19.293Mpa x2)

Factor of Safety= 5.8(M10 bolt is acceptable, and a factor of safety of 5.8 is to apply to this case)
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Considering the size of the Diagonal Bar shown in Fig 4.

In like manner the size of the diagonal bar can be found, the bar (See Fig.4, the piece in yellow) is subject to an axial load of 3.03KN. The ultimate stress of bar material (Iron) is 210Mpa in accordance with the given information from the Internet.The factor of safety of this bar is designed in line with that of the hexagon bolt M10, say f=5.8:

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Factor of safety= Ultimate stress / Working Stress

5.8= 210Mpa/ Working Stress

Working Stress= 210Mpa / 5.8

Working Stress, σ=36.2Mpa

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To find the diameter of the bar, the equation σ=F/A is needed.

A=F/σ = 3033N/ 36.2 Nmm^2

A= 83.78mm^2

A= (pi x d^2)/4

4 x 83.78mm^2 =pi x d^2

d^2= 106.67 mm^2

d= 10.3mm
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Computing the torques acting on the mast.

The torques acting on the mast can be computed by the sum of vertical weights and strengths with the corresponding distance from the axis of the crane. In this design, the balanced weights must be calculated properly so that the base of the crane has to bear no bending moment any longer. (See Figure F-1, M1 + M2 must be zero at any time)

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The new idea is that it must be possible to make the crane free to bending moment when the crane operates at a side angle of 90 degrees or –90 degrees. In general, the rule in bending moment just looks at the need when the crane lifting the matter at 0 degree as the center of gravity is sit in the position that is little back from the center of the trolley. When solving the problem like the case in Figure F-2, making M1+M2≠0 to be M1+M2=0, is not a difficult problem but is a problem if the crane turns to 90 degrees despite of the direction in clockwise or not. This problem might be a major factor in causing the crane to overturn. A side force can have a drastic affect on the stability of the crane and cause my tortoise to drop from the crane.
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There are two possibilities to steady the crane’s body:

(1) First the counterweight of the crane (it is a fixed value in the crane) must be variable, however this is not realizable for lifting load every time.

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(2) Second, that is an alternative solution, is to preload the fillers into the counterweight frame, all fillers plus the weights of F2 and its counterweight frame need to saddle with a special portion of the rated load (at approximately half of the rated load, say 50Kg in total- for details, see Figure D). Speaking of the purpose, my dream crane “Super-Ken 2000” is designed mainly not for a short period of time, I hope it could be still working in the coming five to ten years because of a sulcata tortoise will about double in size every year for the first four or five years if it is well cared.

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I have two Sulcatas, they ALL are 4 years old, and the third largest species of tortoises in the world. I took them in my home in the year of 2003, on May 2nd, they were incredibly cute little sulcata babies in the beginning but reached at approximately 40cm in shell length, and 33 pounds (15Kg) in weight per head up to this now on 8th September in 2007. In fact, they will grow up and grow big increasingly in the years. That why the fillers with proper weights have to be preloaded!

(Note: Although sulcata babies are cute and tiny and will easily fit in the palm of your hand, they will grow up and grow big! My two sulcata babies were taken in for several years ago, in May the year of 2003. My nephew could put it up easily at that moment.)

(Note: Here are the same tortoises and same child (my nephew at right first), they weighted about 15KG (33 pounds) when this photo was taken, August in 2007. My nephew who tried using a force of “Nine Cows + Two Tigers“to lift it but eventually not successful.)

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Would you like me to tell you why I think a half amount of the rate load (say 50Kg included the weight of F2) is enough for this crane to overcome this problem? Practically Speaking, I don’t sure what I would view to this decision but my idea should be satisfied and in line with the requirement as stated in the 2nd law of Super-Ken.

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Figure F-3 gives a good look of the mechanical solution of the controlled counter-moment. On the back of the mast there is the metal case with its fillers. The displacement of the counterweight A3 times a reasonable weight F3 would be compensating for the deficiency on the counter-moment M2 .

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Choosing a correct size of steel rope:
Last Saturday, I got down to find my desired components for my crane at new reclamation street in Mongkok, most of stores in this street are well-known and they are specialized in selling industrial equipment and tools, like hand tools, bearings, steel wires, electrical and pneumatic parts… and whatnot here. I passed a store and saw some lifting ropes and winches there, basically, I desired to buy a steel wire but I was not sure exactly how big that is, then… I asked the salesman of the store, could you pass me a steel wire which can lift up a tortoise weighting 100Kg? It was funny that he got back to me no tools for lifting up tortoises especially here, all from here only for hoisting heavy substances but not for tortoises. That’s the funniest joke I have ever heard! The guy was rude but I only managed to smile in front of him, maybe it was my fault, I should say to him “Could you introduce a steel wire to me, the wire needs to saddle with a thing weighting a load of 100Kg?”
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Not to mention the guy, I need to choose a correct size of steel wire for my “Super-Ken 2000”, I don’t want to use the rope which is oversized and has big size in diameter. For this reason, I have opted for a hands-on approach, i.e. actually calculating for the wire by using the following formula and information:

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Maximum Rated Load, Wm=100Kg (assume that the tortoise will be increasingly growing til to 100)

Number of ropes, n= 1

Safety factor, f= 5

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The factor of safety f is given by the equation, from which the minimum breaking load N of the rope can be expressed:

N x n = F x f = Wm x gn x f

N x n = Wm x gn x f

N x n = 100kg x 9.81 x 5

N x n = 4905N

n = 1 rope

N x 1=4905N

N = 4905N (minimum breaking load, 4.9KN or 0.5T)

For the rope selection, I copied the relevant data out of the supplier catalogue, details for 6x19 and 7x19 construction group are listed below. (Extracted from MAINCO WIRE ROPES LTD)

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6x19 construction group
This rope is widely used for general-purpose engineering; it has good strength and flexibility with reasonably good resistance to abrasion and crushing.

Nominal diameter d= 2.5, total breaking load is 0.34t

Nominal diameter d= 3.0, total breaking load is 0.50t

7x19 construction group

A similar rope to 6 x 19 but the fibre core is replaced by a wire strand resulting in less flexibility, but a slightly higher breaking strain and better resistance to crushing.

Nominal diameter d= 2.5, total breaking load is 0.37t

Nominal diameter d= 3.0, total breaking load is 0.54t

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As seen in the information given above, the best choice seems to be a crane rope of 3mm in diameter of either 6 x 19 or 7 x 19 construction group (minimum breaking load 0.5T, or 0.54T respectively).

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In order to decrease the specific pressure in the winch drum and extend life of the rope. A crane rope of 3mm in diameter of 7x19 construction group will be given preference in this design.

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Summing up the above, I have laid holding of the desired things for my cool crane through the rules in science and engineering, and all works completed above are in accordance with the requirements as stipulated in Super-Ken Laws. After today, I will start up my engine to think, to draw and to work for my cool crane.
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See you next!
Ken Ngan
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The 1st law of Super-Ken: Wit and will strive for the victory

The 2nd law of Super-Ken: You must reap what you have sown.

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