# How to analyze the general formula of hull beam calculation?

Discussion in 'Boat Design' started by sun, May 21, 2024.

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### sunSenior Member

This is the most basic expression of the calculation of the internal force of the hull beam, the normal basic theoretical derivation, how can we discuss it? Is there any need to discuss it? I totally don't understand. Is the basic skills of an engineering student, what can be discussed?
I really can't find the direction, very confused, is there anyone who can enlighten me, thank you very much!

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### TANSLSenior Member

A distinguished member of this forum, Pablo Sopelana, teaches courses on various disciplines of naval architecture/engineering, including one on structural calculation. I'm sure he can guide you in your search for answers.

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### sunSenior Member

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### Pablo SopelanaSenior Member

Hi @sun ,

Yes, it seems to me quite normal that you don't understand equations (3 - 12) based on the image you attached since the derivation of the term within the brackets (EI d2v/dx2) does not appear on it.

First of all, it is important to understand that the following:

M - bending moment
N - shear force
r - radius of curvature of the deflection
I - moment of inertia about the neutral axis (cross-sectional property)
E - Young's module (material property)
v- deflection (displacement) of the neutral axis from the equilibrium point (beam without any load)

Are interrelated in the following way: M/I = E/r

Or: M = EI/r

(You can find information about this by searching the internet for beams bending moment and shear force deduction)

If we express r in terms of the deflection, v, and substitute above, we obtain: M = EI d2v/dx2

(You can find information about this last expression by searching the internet for beam deflection derivation of equations)

Now, going to the upper equation of your figure, this equation is the one resolving the vertical forces shown in the figure considering equilibrium:

N + q dx = N + dN => dN = q dx

Then: dN/dx = q (which is equation 3 -11a)

(q is the force per unit length, in the case of the figure, q is distributed load)

Resolving now the moments through the middle section of dx considering equilibrium:

M + dM = M + N dx/2 + (N + dN) dx/2

Or: dM = Ndx + dN dx/2

Or considering equation dN above: dM = Ndx + q d2x/2

Now, since dx is infinitesimally small, we can write:

dM = Ndx => dM/dx = N (which is equation 3 -11b)

Now, you can read equations (3-12) as:

d (M) /dx = N

And:

d2(M)/d2x = d (N) / dx = q

On the other hand, we have several courses on Craft Design and particularly one on Structural Design and Scantling with ISO 12215, that you might find interesting:
- The courses: https://navalapp.com/courses/
- The Structural Design & Scantling with ISO 12215 course: https://navalapp.com/courses/structural-design-and-scantling-with-iso-12215/

Don't hesitate to check them out.

Ad Hoc and C. Dog like this.
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### sunSenior Member

Thank you, professor, for your patient answer, which gives me the feeling of going back to school to be guided by teachers again, and inspires me to think deeply and be interested in related issues. I will carefully study these materials, thank you again, and wish you a smooth work and good health!

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### Pablo SopelanaSenior Member

Thank you @sun for your best wishes. All the best also for you.