Super-Ken 2000 is only for Tortoises

Today is 26th September,

Hello everyone
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It has been a long time since my last journal but this is certainly a heartsome and fascinating one, over the last few weeks I have been working behind the blogger on the crane design that you ALL should now able to see. This jouinal is in celebration of a few things. Firstly, my dream crane "Super-Ken 2000" has been finished, and it will be coming into use at my home a while later. Secondly, I learnt a lot as designing a crane that brought real-world design problems into my life, each problem coming up in the working stages provided good opportunity with me to review and apply knowledge essential for a job in science and engineering. The third thing which is the last one and is worth celebrating one, I gained rich experience throught the hands-on activities.. including market research, cost management, work planing and whatnot here.

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(1) Introduction

Super-Ken 2000 is a family crane having a kind and sincere heart, its body is made of mild steel, and is composed of steel pipes, manual winch, steel wire and a custome-made trolley. In all honesty, Super-ken 2000 is designed mainly for helping my mother to pick up the giant tortoises (2 Sulcata tortoises), my mother who needs to take them back and forth twice per day from the living room to the bathroom. Speaking of the handling posture, I have ever suffered from severe backache several months ago. Indeed, pain in my back gave me a unhappy experience, not only backache was physical painful, it badly affected my daily life, mood, working capacity and efficiency. The people said: back is an essential and vital bone of the body also known as the life bone, I don't want my worst experience in back that will probably recur to my mother.

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(2) Super-Ken 2000 can become an expectant helper

My dream crane has enough capacity to lift a tortoise weighting 110 lbs (at approximately 50Kg), its completion will enable to prevent my mother from getting any possible damage in back, hand or any parts of body. Wishfully, I hope Super-Ken 2000 can become an expectant helper.

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Unbelievable but true, when I put my hand on the crane's body, I can feel the crane's heart beating. Shh.... Don't tell other people, Super-Ken 2000 is a humanoid crane, it has feeling, heartbeating and has a kind heart!..... I made it live!
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Photo Sharing:

To view the large picture, please click on the thumbnail.

- Hand winch equipped with spur gear and locking device for load control.

1. - Super-Ken 2000 can be folded down for storage or transport
2. - Super-Ken 2000 can rotate different degrees and its boom can extend 150mm more for special case.
3. - Super-Ken 2000 is a family crane but only for tortoises.
4. - Super-Ken 2000 is a kind crane, it likes working with elderly or housekeeper.
5. - Super-Ken 2000 is made in Wong Tai Sin and designed by Ken Ngan.
   

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(3) What I learnt most and challenges I faced

The most important challenge I faced was the intense workload accompanied with limited time. I frequently faced the timed works and felt truly that they were pretty heavy. The so-called "One-Man-Band Project" was nothing more than one person to do one thing until it to the end. Encouragingly, a man behind the "One-Man-Band Project" is called "Frank Lai", he is one great friend of mine, Lai gave me more productive ideas and led me on finding out the right or possible right things. Needless to say anything, all of his ideas of course were coming from his views and hands-on experience. He merely was a talker, not to take part in any work, but his advice did work very well to my thought and did help pushing forward to my works. I realized a lot by doing this so-called One-Man-Band Project, from a drafting to a working drawing, from ambiguity in the beginning to understanding in the end, and from a dream to realization as well. It was worthwhile to experience all these!

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Apart from the above-mentioned, I learnt how to handle time on the manufacturing process to a lifting crane, yet I learnt most in the working stages should be sending understanding and accepatance to the manufactuer's works, not to discredit their workmanship, most of all did not follow my drawing to do... for example: (i) a 2" steel pipe, it was considered to be main mast, they turned the pipe into 1-1/2" diameter, and (ii) a 200mm x 200mm x 5mm thick steel plate for crane's footing they random minimized the size to be 150mm x 200mm (see picture). What a have!

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Sometimes, not only to me, all of yours living in the world have need to accept and understand to other parties, you must plunge your heart into their hearts, adjust or amend the standard a little bit in order to create a harmonic space. As a matter of fact, I could not address any unsatisfaction or objection to the manufacturer as the price they quoted before was very cheap.

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(4) Calculations and Decision Making
Free Body Diagram of the Lifting Crane:

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Weights:
Wt (Weight of the trolley)= 12Kg (Given by manufacturer)
Wb (Weight of the boom –1-1/2” pipe) = (4Kg/m x 0.275m) = 1.1Kg
We (Weight of the boom extension – 1-1/4” pipe) = (3.4Kg/m x 0.425m)= 1.445Kg
Wc (Weight of the Crane)= 20.162Kg
1. For 1-1/2” Main Boom (0.375m x 4kg/m)= 1.5Kg
2. For Winch System (6.5Kg)
3. For 2” Main Mast (0.865m x 5.41kg/m)= 4.671Kg
4. For Base Plate = 0.2m x 0.2m x 0.005m x 7874kg/m3 = 3.15kg
5. For 0.3m long Hollow Shaft for Base Plate (0.3m x 5.4kg/m)= 1.62Kg
6. For Metal Gill Covers (0.13m x 0.13m x 0.005m x 7874kg/m3 (Density) x 2) = 1.33Kg
7. For U channel (0.088m x 0.098m x 0.005 x 7874kg/m3 x 2)+(0.066m x 0.098m x 0.005m x 7874kg/m3 = 0.895Kg
8. For Fixing Plate (0.14m x 0.09m x 0.005m x 7874kg/m3)= 0.496Kg
Total Weight for Wc = 20.162Kg
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The weights of the load and pipes multiplied by their respective distances forward from the tipping point must be equalized by the weights of the counterweight and crane net capacity multiplied by their respective distances rearward from the tipping point (LHS = RHS, See Figure 1, 2 & 3). Any variation caused by changing the radius at which the load is lifted destroys this equilibrium.
Roughly speaking, a balanced condition of my lifting crane shall be achieved by equalizing forward and rearward moments, say LHS = RHS. Let’s see the conditions.
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Condition 1:
Weight of the tortoise = 100Kg (Maximum Working Load)
Distance between the centerline of the filler and the line of tipping point =0.375m (225mm + 150mm)
(Filler x 0.375m)+(20.162Kg x 0.225m)+(12Kg x 0.25m)=(1.1Kg x 0.1375m)+(1.5Kg x 0.2125m)+(100Kg x 0.425m)
(Filler x 0.375m)+4.537kgm+3kgm = 0.151kgm+0.319kgm+42.5kgm
(Filler x 0.375m)+7.537kgm = 42.97kgm
Filler = 94.49kg (not recommended)
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Condition 2:
Weight of the tortoise = 50Kg (Maximum Working Load)
Distance between the centerline of the filler and the line of tipping point =0.375m (225mm + 150mm)
(Filler x 0.375m)+(20.162Kg x 0.225m)+(12Kg x 0.25m)=(1.1Kg x 0.1375m)+(1.5Kg x 0.2125m)+(50Kg x 0.425m)
(Filler x 0.375m)+4.537kgm+3kgm = 0.151kgm+0.319kgm+21.25kgm
(Filler x 0.375m)+7.537kgm = 21.72kgm
Filler = 37.82kg
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By calculation, a 50Kg (see condition 2) load at a 425mm radius gives a forward load moment of 21.72kgm. This load moment would then be equalized or counterbalanced by a 37.82kg counterweight at a distance of 375mm, having rearward load moment of 21.72kgm.
According to the condition 1, if the weight of the load were to be doubled (say 100kg), the weight of the counterweight must also be tripled to keep the crane in balance. By so doing, adding a considerable amount of filler to keep the crane in balance is neither practical nor economic to my budget. Not to say money, a bulky weight will certainly make the crane unlovely, and it will be less likely to be accepted by any and every designer or engineer. As such, the maximum working load of my crane, in my final decision despite of how much i am unwilling, will be downgraded and only operate for a rated load of 50kg.
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Take Moment at Rf (Condition 2)
+ ↘(Σ M)Rf = 0:
(Wc x 0.225m)+(Wt x 0.25m)+(Filler x 0.375m)–(Rr x 0.45m)–(Wb x 0.1375m)–(We x 0.2125m)–(Weight of Tortoise x 0.425m)= 0
(20.162Kg x 0.225m)+(12Kg x 0.25m)+(Filler x 0.375m)–(Rr x 0.45m)–(1.1Kg x 0.1375m)–(1.5Kg x 0.2125m)–(50Kg x 0.425m) =0
4.536Kgm+3Kgm+(Filler x 0.375m)–(Rr x 0.45m)–(0.1513Kgm)–0.319Kgm–21.25kgm=0
-14.184kgm+(Filler x 0.375m)–(Rr x 0.45m)=0
Rr x 0.45m = (Filler x 0.375m) -14.184kgm
Rr = <(Filler x 0.375m) –14.184kgm> / 0.45m
Rr = 0.833Filler – 31.52kg
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Take Moment at Rr (Condition 2)
+ ↘(Σ M)Rr = 0:
(Rf x 0.45m)-(50kg x 0.875m)-(1.5kg x 0.6625m)-(1.1Kg x 0.5875m)–(20.162Kg x 0.225m)–(12Kg x 0.2m)–(Filler x 0.075m) =0
(Rf x 0.45m)- (43.75kgm)–(0.994kgm)–(0.6463kgm)–(4.536kgm)–(2.4kgm)–(Filler x 0.075m) =0
(Rf x0.45m)–52.33kgm–(Filler x 0.075m)=0
Rf x 0.45m = 52.33kgm + (Filler x 0.075m)
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As can be seen the calculation previously, a 50Kg load at a 425mm radius gives a forward load moment of 21.72kgm, this would then be equalized by a 37.82kg weight, (Filler = 37.82kg)
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When the weight at the filler's side is 37.82kg (Condition 2), Rf & Rr would be obtained:
Rr = 0.833Filler – 31.52kg
Rr = 0.833 (37.82kg) – 31.52kg
Rr = -0.016kg

Rf x 0.45m = 52.33kgm + (Filler x 0.075m)
Rf x 0.45m = 52.33kgm + (37.82kg x 0.075m)
Rf = 122.59kg
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In the case of Condition 2, fillers weighting 37.82kg require being loaded on the counterweight frame; two front wheels Rf need to withstand a downward force 122.59kg. As per the calculation, each front wheel requires to saddle with the related downward force 61.3kg (122.59kg/2).
In terms of the value for Rr, it is a negative value and means that the rear side of the trolley may be easily raised because the entire loads are concentrated on the front wheels when the crane is operated. To make it steady, I will round off the filler to be 40Kg, then the values of Rr and Rf will be changed slightly to 1.8Kg and 120.77kg respectively.

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Alternatively, the counterweight can be moved further rearward from the centerline of tipping to compensate for the load, it can minimize the amount of the filler. Let's see the change.

(1st Trial) Moving back the whole crane about 50mm:

If one small adjustment is made in the design, the crane is shifted slightly back from the line of tipping point; a gain will be obtained at the rearward moment. Let’s see the condition 3 and 4.
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Condition 3:
Weight of the tortoise = 50Kg,
Distance between the centerline of the filler and the line of tipping point =0.425m (275mm + 150mm)
(Filler x 0.425m)+(20.162Kg x 0.275m)+(12Kg x 0.3m) = (1.1Kg x 0.0875m)+(1.5Kg x 0.1625m)+(50Kg x 0.375m)
(Filler x 0.425m)+5.54455kgm+3.6kgm = 0.09625kgm+0.24375kgm+18.75kgm
(Filler x 0.425m) = 9.94545kgm
Filler = 23.4kg
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Obviously, the gain at the rear moment has been obtained, moving the crane about 50mm rearward from the line of tipping point can free from 14.42kg for the filler (37.82Kg – 23.4Kg).
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Take Moment at Rf (Condition 3)
If the weight of the filler is 23.4kg (Condition 3), then Rf & Rr would be obtained:
(20.162Kg x 0.275m)+(12Kg x 0.3m)+(Filler x 0.425m)–(Rr x 0.45m)–(1.1Kg x 0.875m)–(1.5Kg x 0.1625m)–(50Kg x 0.375m) =0
Rr x 0.45m = (Filler x 0.425m) -10.812kgm
Rr = <(Filler x 0.425m)-10.812kgm> / 0.45m
Rr = <(23.4kg x 0.425m)-10.812kg> / 0.45m
Rr = -1.93kg

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Take Moment at Rr (Condition 3)
+ ↘(Σ M)Rr = 0:
(Rf x 0.45m)–(50kg x 0.825m)–(1.5kg x 0.6125m)–(1.1Kg x 0.5375m)–(20.162Kg x 0.175m)–(12Kg x 0.15m)–(Filler x 0.05m) =0
(Rf x 0.45m)–(41.25kgm)–(0.92kgm)–(0.59kgm)–(3.53kgm)–(1.8kgm)-(Filler x 0.05m) =0
(Rf x0.45m)–48.09kgm–(Filler x 0.05m)=0
Rf x 0.45m = 48.09kgm+(Filler x 0.05m)
Rf = <48.09> / 0.45m
Rf = 109.47kg
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Condition 4:
Weight of the tortoise = 15Kg
Distance between the centerline of the filler and the line of tipping point =0.425m (275mm + 150mm)
(Filler x 0.425m)+(20.162Kg x 0.275m)+(12Kg x 0.3m) = (1.1Kg x 0.0875m)+(1.5Kg x 0.1625m)+(15Kg x 0.375m)
(Filler x 0.425m)+5.54455kgm+3.6kgm = 0.09625kgm+0.24375kgm+5.625kgm
(Filler x 0.425m) = 9.94545kgm
Filler = -0.134kg
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Take Moment at Rf (Condition 4)
If the weight of the filler is –0.134kg (Condition 3), then Rf & Rr would be obtained:
(20.162Kg x 0.275m)+(12Kg x 0.3m)+(Filler x 0.425m)-(Rr x 0.45m)–(1.1Kg x 0.875m)–(1.5Kg x 0.1625m)–(15Kg x 0.375m) =0
5.545Kgm+3.6Kgm+(Filler x 0.425m)–(Rr x 0.45m)–(0.963Kgm)–0.244Kgm–5.625kgm=0
2.313kgm+(Filler x 0.425m)-(Rr x 0.45m)=0
Rr x 0.45m = (Filler x 0.425m)+2.313kgm
Rr = <(Filler x 0.425m)+2.313kgm> / 0.45m
Rr = <(-0.134kg x 0.425)+2.313kg> / 0.45m
Rr = 5.0134kg

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Calculate the value of Rf (Condition 4)
Rf + Rr = 15kg + 1.1kg + 1.5kg + 20.162kg + 12kg + (-0.134kg)
Rf = -Rr + 49.628
Rf = -5.0134 + 49.628
Rf = 44.615kg

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(2nd Trial) Turning the lifting crane at different degrees

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Condition 3a:
Weight of the tortoise = 15Kg
Distance between the centerline of the filler and the line of tipping point=0.31m.
(Filler x 0.31m)+(20.162Kg x 0.16m)+(12Kg x 0.21m) = (1.1Kg x 0.2025m)+(1.5Kg x 0.2775m)+(15Kg x 0.49m)
(Filler x 0.375m)+3.226kgm+2.52kgm = 0.223kgm+0.416kgm+7.35kgm
(Filler x 0.375m)+5.746kgm = 7.989kgm
Filler = 5.98kg
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Take Moment at Rf
+ ↘(Σ M)Rf = 0:
(Wc x 0.16m)+(Wt x 0.21m)+(Filler x 0.31m)-(Rr x 0.32m)–(Wb x 0.202m)-(We x 0.2775m)–(Weight of Tortoise x 0.49m)= 0
3.23Kgm+2.52kgm+1.854kgm–(Rr x 0.32m)-0.222kgm–0.4163kgm–7.35kgm =0
Rr x 0.32m = -0.384kgm
Rr = -0.384kgm> / 0.32m
Rr = -1.2kg
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Total upward forces = Total downward forces
20.162kg + 12Kg + 5.98kg + 1.1kg +1.5kg + 15Kg = -1.2kg + Rf
Rf = 56.942Kg
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In the case of Condition 3a, fillers weighting 5.98kg are needed to load on the counterweight frame; the front wheels Rf need to withstand a downward force 56.942kg. As per the calculation, each front wheel requires to saddle with the related downward force 56.942kg/2.

To get rid of the problem on counterbalance, the makeshift to make the crane steady was that I put a chain weighting 8Kg into the metal case (Counterweight), the values of Rr and Rf will be changed slightly to 0.76kg and 57Kg respectively. As it has been mentioned early, the negative in the value of Rr represents a little force that can easily overturn the crane.

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All in all, I must give tribute to all those who helped me to make my crane work, to make my dream coming true, as well to understand more on the principle of Mechanical Engineering Design.

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Best Regards

Ken Ngan

P.S.: Super-Ken 2000 is a humanoid crane, I can feel its heart beating!

EVERYTHING IS GOING WELL.

Dear friends

Today is Tuesday, 11 Sept 07.

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I think everything is going well and the working drawing for fabricating my cool crane finally got amazing smooth as well.. Tommorrow I will send my drawing to some factories and hope they could offer me a best price.

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I do agree that doing this mission can be challenging and attractive to me,.. because attractiveness is the key to keeping me busy. Likewise, a proper difficulty level (of course, not too high) is also the key to making me think in higher orders and better understand the subject matter. Some said: work while working, play while playing , I think this working principle does not work very well on me, because this mission is a funny game, not a boring job... Work and Play while playing here!

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The working principle of the Lattice Boom Crane is the source of my crane's idea, a load adding on the mast provides a balanced function while lifing up the load. It is a best evidence to tell my friend that my idea will be possible and feasible.

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Best Regards

Ken Ngan

Like a house on fire!

Dear friends
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Today is sunday & is a sunny day, 10th in Sep 07.

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My engine (say.. my heart) has started up now, and it is like a house on fire! I got parts of the stuff from the store for my cool crane yesteday, including 1 set of hand winch, 1 pulley, 1 eye hook and 1 steel wire. All of these were surprisingly expensive and cost me four hundred Hong Kong dollars.

The worker of the store told me that the hand winch can lift a boat weighting 1200lbs, and is good enough for general purposes. Bascially I desired to buy a winch merely for 100Kg - 200Kg capacity (220-440lbs) and wanted a slim size, but okay so long as I got these stuff.

As a matter of fact, the manufacturing cost and the feasibility of my idea are bounded up in the budget, the cost of this crane (so far) has reached about $400 including above-mentioned stuff. As well as I have taken about 40 to 50 hours to do, to think, to write, to draw and to plan the fabrication of my crane. As with all people or manufacturer, I don't want to do a crane which is too expensive.

Next step, I will start working for computer drawing, see you!Best Regards

Ken Ngan

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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