Naval Architecture and Marine Engineering PE Exam Preparation

[AK2021]

Hello there.

In OCT of 2022 I will be taking the Naval Architecture and Marine Engineering PE exam. I am a repeat test taker. In the past I have largely studied alone, but I thought while working through practice problems and prep course (through the society of naval architects and marine engineers) it would be nice to have a discussion board that could be used to share difficult questions, study techniques and other resources. I know this is one of the smaller test groups, but if anyone is interested in using this with me please do not hesitate to jump in. I will be using it as a record for my own study as well. Hope all is well!

Happy Thanksgiving

 

[AK2021]

I am having trouble with this question:
Applied Naval Architecture Ch. 3 #5

A shallow draft river vessel (L=310 ft, B=42 ft, T=8.0 ft, light ship conditions) Cb=0.75, Cm=0.92 Cw=0.697

The ship is drydocked to install a rubbing strip on the outer bottom. The wood strip is 220 feet long, 8 ft wide, and 1.5 feet thick, and it weighs 36 tons including all fasteners. Determine the vessel's draft from the bottom of the strip when it is refloated in freshwater after the strip has been installed.

I keep getting a draft=9.633 ft, but the book states the draft is 9.35 ft. The only thing I can think of is the wood plank must give the ship some sort of buoyancy. Otherwise, how can a ship that is drawing 8ft, have a final draft of 9.35 ft after a wooden plan of 1.5 thickness is attached to its keel?

 

[AK2021]

Having trouble with this question:
Applied Naval Architecture Ch. 3 #17
A Barge of a constant triangular cross-section is 180 ft long and has a beam of 24 ft at the waterline when floating in freshwater at 9-foot draft. A weight of 15 tons (already on board)is shifted 10 feet across the deck, causing the barge to heel 2.3 degrees. Determine the KG of the barge
a) As inclined
b) after removing the inclining weight, if it was 20 ft above the keel.

I think one of my main issues is how to calculate BM for a triangular vessel. The difference between the two KG values I am calculating is the same as the difference between the two the book provides, however mine are too shallow.

 

[AK2021]

Applied Naval Architecture Ch. 3 #23
A rectangular barge (L=67 m, B= 14 m, D=6 m, T=3 m, Seawater)
KG=3.7 m
List=6 degrees - STBD
If 250 tons of cargo are to be loaded on the main deck at 7 meters above the baseline, how far to port of centerline should it be loaded so that the barge will complete loading with no list?

This problem goes well until I try and calculate the distance. No matter what I do the value for "d" does not match the value in the book. I am trying to determine which calculated variable is incorrect.

 

[AK2021]

Applied Naval Architecture Ch. 3 #31
Rectangular Barge
L=120 ft, B=30 ft, D=13 ft, Seawater, T= 7 ft
Only weights on board are those for the inclining experiment.
It is then inclined and a transverse shift of 5 tons through 20 ft heels the barge 2 degrees. After the barge is returned to upright (the inclining equipment remains onboard), 480 tons of fuel oil having a specific gravity of 0.90 are loaded equally into three double bottom tanks, each 10 ft wide and 120 ft long, formed by two equally spaced longitudinal bulkheads. Free surface is created in each of the tanks. Calculate the virtual metacentric height (GvM) in this condition.

This one is a little less confusing, but I wanted to check. I think I am only subtracting the GGv only once when I should be subtracting it three times. This would give a GM corrected = 12.25-7.17-(0.21x3) which equals 4.45 m. This is only 0.03 m off from the answer the book gives.

 

[AK2021]

Applied Naval Architecture Ch. 3 #35
A large tanker displacing 145,000 tons in SW has a GM corrected for free surface of 15.0 ft with a center tank 100 ft long and 68 ft wide containing oil (38 ft^3/LT) to a depth of 70 ft. A cross-connecting valve is opened from this tank to the adjacent STBD wing tank (L=100 ft, B=42 ft) until the oil reaches equal levels in centerline and wing tanks, then the valve is closed. Calculate the final GM and angle of heel in this condition. (Tanks may assume to be rectangular-ignore bilge radius).

There are probably a few areas I am struggling with. I do not know if I am supposed to add the free surface correction (as is typical in loading problems), r reduce it since it is being spread out. Adding the correction gives a similar GvM to what the book says is the answer, however, when plugging this in to determine the angle of heel the value comes out much less than it should.

 

[RogerG]

Hello there.

In OCT of 2022 I will be taking the Naval Architecture and Marine Engineering PE exam. I am a repeat test taker. In the past I have largely studied alone, but I thought while working through practice problems and prep course (through the society of naval architects and marine engineers) it would be nice to have a discussion board that could be used to share difficult questions, study techniques and other resources. I know this is one of the smaller test groups, but if anyone is interested in using this with me please do not hesitate to jump in. I will be using it as a record for my own study as well. Hope all is well!

Happy Thanksgiving

Hi There,

I would be interested in studying with you. I'm a naval architect from the days before there was PE for our discipline outside of 2 States.. I was just looking at the SNAME review course.

Cheers

 

[DJC]

I am having trouble with this question:
Applied Naval Architecture Ch. 3 #5

A shallow draft river vessel (L=310 ft, B=42 ft, T=8.0 ft, light ship conditions) Cb=0.75, Cm=0.92 Cw=0.697

The ship is drydocked to install a rubbing strip on the outer bottom. The wood strip is 220 feet long, 8 ft wide, and 1.5 feet thick, and it weighs 36 tons including all fasteners. Determine the vessel's draft from the bottom of the strip when it is refloated in freshwater after the strip has been installed.

I keep getting a draft=9.633 ft, but the book states the draft is 9.35 ft. The only thing I can think of is the wood plank must give the ship some sort of buoyancy. Otherwise, how can a ship that is drawing 8ft, have a final draft of 9.35 ft after a wooden plan of 1.5 thickness is attached to its keel?

The strip will provide the buoyancy that it displaces. I can't see the givens, but look for whether in salt or fresh water and whether the strip's weight is ST or LT.

 

[DJC]

I am having trouble with this question:
Applied Naval Architecture Ch. 3 #5

A shallow draft river vessel (L=310 ft, B=42 ft, T=8.0 ft, light ship conditions) Cb=0.75, Cm=0.92 Cw=0.697

The ship is drydocked to install a rubbing strip on the outer bottom. The wood strip is 220 feet long, 8 ft wide, and 1.5 feet thick, and it weighs 36 tons including all fasteners. Determine the vessel's draft from the bottom of the strip when it is refloated in freshwater after the strip has been installed.

I keep getting a draft=9.633 ft, but the book states the draft is 9.35 ft. The only thing I can think of is the wood plank must give the ship some sort of buoyancy. Otherwise, how can a ship that is drawing 8ft, have a final draft of 9.35 ft after a wooden plan of 1.5 thickness is attached to its keel?

AK, I'm the same as Gerard, graduated ahead of PE availability but will finally pursue this year. I plan to pop in-out as time permits but thought about this earlier this morn and ultimately, its the buoyancy of the strip that is greater than the weight of it that allows for the rise. Use the Cw to back out the TPI (fw) .692*310*42/(36*12) = 20.86 LT/in. With a strip delta of 220*8*1.5/36 = 73.33 - 36 = 37.33 LT. Rise = 37.33/20.86 = 1.79" (.15'). So, T = 8-.15 = 7.85' (from original baseline) With strip, T = 7.85+1.5 = 9.35'

 

[DJC]

Having trouble with this question:
Applied Naval Architecture Ch. 3 #17
A Barge of a constant triangular cross-section is 180 ft long and has a beam of 24 ft at the waterline when floating in freshwater at 9-foot draft. A weight of 15 tons (already on board)is shifted 10 feet across the deck, causing the barge to heel 2.3 degrees. Determine the KG of the barge
a) As inclined
b) after removing the inclining weight, if it was 20 ft above the keel.

I think one of my main issues is how to calculate BM for a triangular vessel. The difference between the two KG values I am calculating is the same as the difference between the two the book provides, however mine are too shallow.

Its been a while since I’ve done these hand calcs, software’s fault, but I’m looking to solve for BM, as you, use 2/3 T to back out KB and ultimately get KM. then use the wt’s moment and heel angle to get GM and subtract GM from KM to get KG.V = 180*24*9*1/2 = 19440 cf. D = 540 LT (19440/36). The inertia is that of the waterplane about centerline, which for a constant cross section, is simply a rectangle – LB3/12 or 207,360 ft4. Thus BM = I/V = 207360/19440 = 10.67’ and KB = 2/3 * 9 = 6’; so, KM = 16.67’. With an overturning moment of 15LT @ 10’ = 150 ft-LT and GM = w*d/ (D * tan Q) = 150/(540*.0402) = 6.91’. KG = KM – GM = 16.67-6.91 = 9.76’. The second part is pretty straight forward, I think. Please let me know if I’ve erred

 

[dkcoeng]

Hello there.

In OCT of 2022 I will be taking the Naval Architecture and Marine Engineering PE exam. I am a repeat test taker. In the past I have largely studied alone, but I thought while working through practice problems and prep course (through the society of naval architects and marine engineers) it would be nice to have a discussion board that could be used to share difficult questions, study techniques and other resources. I know this is one of the smaller test groups, but if anyone is interested in using this with me please do not hesitate to jump in. I will be using it as a record for my own study as well. Hope all is well!

Happy Thanksgiving

Just joining how did you do? What do you recommend for a mach pe looking to take the NAME PE