# moment of inertia

Discussion in 'Boat Design' started by abhishek, Sep 28, 2009.

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### abhishekJunior Member

generally we find bottom shell parallel to base for big ships and even for small ships also but for some coastguard vessels like inshore patrol vessel we dont find the bottom shell parallel,(ie) bottom shell is with some small angle to the base, now to find out section modulus first when we check the moment of inertia, as the plate is a rectangle which is making angle with the base as said. now to find out moment of inertia for bottom shell if it is parallel to base i know moment of inertia is (width *(thickness cube))/12. but when is making some angle like 5 degree with base (ie) bottom shell plate , what is the formula for finding moment of inertia for that plate which is making angle with the base......

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It is a function of the total depth, in the vertical plane.

I = A.n^2/12

I = inetria
A = X.Sectional area
n = depth of section

So for a rectangluar shape, formula is
I = bd^3/12

Hence depth = n^2/12 = b^2/12

Therefore I = A.b^2/12

Where area = b.d

Ergo I = (b.d) . (d^2/12) = bd^3/12

QED

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### abhishekJunior Member

you r right Ad hoc sir, it is for vertical or horizontal plane, if it for an inclined plane what is that we do? i think we have to resolve that into vertical and horizontal components may be we get that in sin and cos angles. is that right?

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### jehardimanSenior Member

Do it by differential elemental area, the Parallel Axis theorem does not always make it easier.

If you want, you can think of a thin plate of span (s) and thickness (t) at some angle of inclination (theta) as a large number of thin layers that can be stacked up into a rectangle. In this case h(approximate) = s*sin(theta)+t*cos(theta) and b(approximate) = t*sin(theta)+s*cos(theta), but this is only approximate for t/s << 1.

Check here: http://en.wikipedia.org/wiki/Second_moment_of_area

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abhishek
The depth is the depth of section, in the vertical plane. (This is assuming you wish to find the interia about the vertical plane for ALL members). So, a plate that is horizontal, the depth is the normal section depth, or thickness. If the plate is at an angle, of 5 degrees, what is this depth, it is the sine of 5 degrees!

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### abhishekJunior Member

THANKS FOR YOUR REPLY AD hoc sir, (So, a plate that is horizontal, the depth is the normal section depth, or thickness. If the plate is at an angle, of 5 degrees, what is this depth, it is the sine of 5 degrees!) this is the answer u have given ,but i think when we resolve that into vertical and horizontal components, when we do for the horizontal component we get the answer in cos (theta), as cos (theta)=adj/hyp, so we her adj is the base and hyp is the plate, so for this M.I=bd^3/12*cos(5 degree0 and for vertical component bd^3/12*sin(5 degree). tell me whether i am correct or not, if correct how can i get the resultant moment of inertia which includes both horizontal and vertical components in the same formula.......

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### jehardimanSenior Member

Maybe we need to ask a question here...which moment of inertia are you trying to find?

Assuming standard x axis fwd, y axis Stbd, z axis down....

Iyy, the area moment of inertia of the midships section in vertical bending: the double integral of z^2*dArea

Izz, the area moment of inertia of the midships section in horizintal bending: the double integral of y^2*dArea

Iyz, the minimum area moment of inertia of the midships section in bending : the double integral of yz*dArea

Ixx, the mass moment of inertia of the ship in roll: the triple integral of yz*dVolume*rho

Last edited: Oct 3, 2009
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Your question is now getting into details...only way to be clear now, is with a diagram. See attached.

#### Attached Files:

• ###### Inertia inclined plate.jpg
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### abhishekJunior Member

ya this is what i want to explain you Ad hoc sir, here as the bottom shell is not a straight, which is inclined to the base as the plate is a rectangle as it makes an angle of 5 degree with the base , can i have an answer M.I=b*d^3/12cos(5degree) for x asix (ie) horizontal component and bd63/12sin(5degree) for y axis (ie) for vertical component. and i want what k refers in the diagram u have shown me?.............

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abhishek
I think you are getting confused.
please read my post #2 again and look at the diagram in post #8.

All you do is measure the vertical distance, what ever it is and however you calculate it.

The vertical distance, in correct terms, is the vertical distance that is perpendicular to the axis of the inertia that you are calculating.

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### abhishekJunior Member

thank you sir i got my answer...............

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