Calculating Second Moment of Aria for Hollow Square Beams

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To calculate the second moment of area for a hollow square beam, first determine the second moment of area for the outer solid square and then subtract the second moment of area for the inner square. The formula involves calculating the area moment for both the outer square (100mm x 100mm) and the inner square (70mm x 70mm). The correct approach is to compute the moment of area for the outer square and then subtract the moment of area for the inner square. This method provides the accurate second moment of area for the hollow square beam.
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Question:

How do you work out the second moment of aria for a hollow square beam?

Attempt:

I have looked over my notes and it says how to get the second moment of aria for all beams but a hollow square one. It has I beam, Circulare beams, Rectangular...

How do you go about working it out for a hollow square beam?
 

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Basically you just take the area moment of a solid square 100mm a side and subtract off the inside.
 
thanx.

So does that mean I would work out (100mm x 100mm) - (70mm x 70mm) and us that as the aria in the formula or work out the Moment of aria for the 100mm x 100mm and take off the moment of aria for a 70mmx70mm square?
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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