# Which method is better for calculating shear force and bending moment of ship by usi

Discussion in 'Boat Design' started by xichyu, May 18, 2016.

1. Joined: Sep 2011
Posts: 6,985
Likes: 554, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

You surprise me, I thought you more rigorous.

Last edited: Aug 31, 2017
2. Joined: Sep 2011
Posts: 6,985
Likes: 554, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

In post # 2 there is only a graphic representation of a system of forces. There is no indication of how BMs or SFs are calculated along the beam, nor does it indicate how the beam is supported. I think those graphs are a part of the explanation but totally insufficient to know how these calculations are performed. On the other hand, the graph corresponding to the forces of buoyancy, seems to corroborate what I say in post # 29 and 31, totally different to the "method" proposed by Gonzo to calculate the buoyancy only in certain points. If you agree with Gonzo's "method", please explain why and help me out of my mistake. Thank you.

3. Joined: Mar 2002
Posts: 1,607
Likes: 355, Points: 83, Legacy Rep: 158

### BarrySenior Member

I expect that it would be almost impossible to predict on an instantaneous basis the distribution of buoyant forces that are developing in this picture.
Ad Hoc, when large ships are designed, is there a method or set of rules that you use to establish a max distribution of downward forces and upward buoyancy forces, then work out only max bending loads, and then stresses. And then would you apply a factor of safety to deal with the unknown.

Ie half the weight of the boat downward at say the bow and stern of the boat with the upward equal force in the middle. Obviously unable to create in the natural sea state but would give a basis to which a factor of safety or GAEP, generally accepted engineering practices or to Lloyds etc?

Also, do you just allow the factor of safety, whether it be max stress in the skin, frames or wherever, take care of the forces imparted into the hull due the angular acceleration of the hull.

There are a lot of forces that a static X-Y plane force analysis , ie side profile of the hull, the beam shot so to speak, will not take care of. Torsion, angular acceleration forces, perhaps slamming

4. Joined: Oct 2008
Posts: 7,292
Likes: 1,180, Points: 113, Legacy Rep: 2488
Location: Japan

Not really, as like everything in design, it is very methodical. One just imagines different scenarios and then apply them. After apply them ascertain which has the greatest effect. Simple.

It is shown in the post #2. The image fig 6.3 shows the weight distribution of the vessel. Major lumps, such as engines, generators, or fuel tanks etc end up as peaks, whereas the general structure is a laborious method to establish but now generalised into a simple trapezoidal shape. All this weight is surmised and broken down into small sections along the length of the vessel, hence the step/jagged look. Then apply a large wave profile, also shown. The wave profile can be assumed to be anywhere along the vessel, i.e. crests at ends or midships or anywhere. It is the responsibility of the NA to check which gives the worst case.

Then fig 6.4, shows the summation of the downward forces, weight with the upward forces, buoyancy, along each section or step. Hence the appearance of the graph. This, as shown, is the loading curve.

Finally fig.,6.5 integrate along the length to obtain the SF and BM for each wave loading condition.

Classification societies have their own set of rules for establishing the SF and BM, but large vessels must perform this graphical type approach as it indicates major SF and BM during loading and unloading, especially in grain carriers ORE carriers etc.

The FoS is a combination of, the max 1 in a 100 years wave, along with a any FoS that the Classification societies apply. Added to this is whipping loads, which can be significant too.

Investigating loads from all direction linear and non-linear is the role of the NA…and on large vessels, is very time consuming.

5. Joined: Aug 2017
Posts: 23
Likes: 0, Points: 1
Location: Ukraine

### AlekseyJunior Member

I have following data:

Displacement of the ship (including cargo + tanks)
Lightship displacement
LBP
LOA
Weight of cargo in Bays
Weight of all tanks

First, I found weight of the vessel per 1 meter = Lightship displacement / LBP
Second, I found buoyancy forces per 1 meter = Displacement of the ship (including cargo + tanks) / LBP

Then, I divide vessel for 10 parts (from AP until FP).
I chose by drawing which tanks and bays are inside of these divided parts.
From AP to frame 20 I have Superstructure and some small tanks, and from frame 20 already I have cargo bays and other tanks

So, I calculate weight of cargo and weights of tanks inside every part and divide on the length of part - in this case I found deadweight of cargo per 1 meter in every divided part.

After, I fount total WF for every divided part = weight of the vessel per 1 meter + deadweight of cargo per 1 meter

Then, I found Load (t/m) for every part = buoyancy forces per 1 meter - total WF for every divided part

If Load (t/m) for every part < buoyancy forces per 1 meter,

After , I tried to find SF and BM and have totally different from Cargo Program.

In Cargo program BM and SF calculated for some frames, and I found info only now to calculate DB and SF between the frames.

And one more, in AP and FP, BM and SF is ZERO, but I have always more.

6. Joined: Aug 2017
Posts: 23
Likes: 0, Points: 1
Location: Ukraine

### AlekseyJunior Member

in which book?

7. Joined: Sep 2011
Posts: 6,985
Likes: 554, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

With this procedure you get a linear, uniform distribution of weights and thrust, which is very far from the actual distribution. The BM and SF values calculated in this way will be totally unrealistic. The distribution of weights and buoyancy should be somewhat similar to what is shown in post # 2.

8. Joined: Aug 2017
Posts: 23
Likes: 0, Points: 1
Location: Ukraine

### AlekseyJunior Member

In post #2 only pictures. I am really can't find procedures and explanations how to calculate like in post #2

Last edited: Sep 3, 2017
9. Joined: Aug 2017
Posts: 23
Likes: 0, Points: 1
Location: Ukraine

### AlekseyJunior Member

Here below is my calculation.....

Last page - cargo program results....

10. Joined: Sep 2011
Posts: 6,985
Likes: 554, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

Divide the boat into a series of cross sections, eg 1 m in length (or whatever you want) and calculate the weight of the structure of each section. By joining the points that represent these values throughout the length, you will get the structure weight distribution curve.
Add the weights of the rest of the items in the boat (engine, tanks, containers, ...). Each weight can be considered as a continuous load distributed throughout the length of the item.
The curve of areas of frames will be used to calculate buoyances at each point.
From there :
- consider a point at a distance "x" from the AP. Calculate the moment of the forces to the right of that point with respect to the point in question.
- Repeat for various values of "x"
- SF at each point is the difference between the weight and the buoyancy at that point.
This procedure, except for error or omission, is the one that I would use. It is very lavish. Therefore, if there is a simplified procedure, as some people seem to know, it would be interesting to know.
Sorry, my level of English does not allow me to be clearer or more didactic.
Edit : I had written this text before seeing post # 39

11. Joined: Aug 2017
Posts: 23
Likes: 0, Points: 1
Location: Ukraine

### AlekseyJunior Member

If I have between frames 20 and 41 (between 12.6 m and 27.3 m from AP) Bay 29, 27 and 14 tanks. Total weight of all - 1072 mt (for present loading condition). As I need to divide this section oа 14.7 m length in smaller sections? Like 14.7 m / 10 = 1.47 m. And then divide total weight of load of this section also by 10? 1072 t / 10 = 107.2 t/ per 1.47 m ???? Is it correct?

12. Joined: Sep 2011
Posts: 6,985
Likes: 554, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

I do not know if I understand you correctly. If a container measures 11.5 m, divide the weight of the container between 11.5 and place a uniform load of this value (weight / 11.5) distributed along the container. The same with the weight of the tanks and the same with the rest of items.

13. Joined: Aug 2017
Posts: 23
Likes: 0, Points: 1
Location: Ukraine

### AlekseyJunior Member

Yes, all this already done in post #39. The reason of my question - total different SF and BM and also Load curve if You compare with Cargo Program

14. Joined: Sep 2011
Posts: 6,985
Likes: 554, Points: 123, Legacy Rep: 300
Location: Spain

### TANSLSenior Member

If the load curves are totally different, do not continue with the calculations. Try, before anything else, to find out why the load curves are different.

15. Joined: Aug 2002
Posts: 15,818
Likes: 1,222, Points: 123, Legacy Rep: 2031
Location: Milwaukee, WI

### gonzoSenior Member

Basic Ship Theory by Rawson and Tupper

Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.