Tubular Column Moment of Inertia

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SUMMARY

The moment of inertia for a tubular column is derived using the formula Ix = Iy = (π/4)(r24 - r14), where r2 and r1 represent the outer and inner radii, respectively. The discussion highlights a common misconception regarding the factor of 2 in the moment of inertia calculation, emphasizing that the material's uniform mass density leads to a proportional relationship between mass and area. The derivation process involves substituting the thickness of the ring into the formula and simplifying it by eliminating terms with exponents of thickness.

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Familiarity with tubular column geometry
  • Knowledge of polar area moment of inertia
  • Basic calculus for derivation processes
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  • Explore the differences between polar and axial moments of inertia
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Structural engineers, mechanical engineers, and students studying mechanics of materials who need to understand the calculations related to tubular columns and their moment of inertia.

SALMAN22
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Can anyone explain why the moment of inertia for a tubular column in that textbook is like so? (take a look at the attachments). It should be (I = MR^2), as far as I know.
 

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I think they are assuming the material has uniform mass density. So the mass is proportional to the area but then they are off by factor of 2.
 
hutchphd said:
I think they are assuming the material has uniform mass density. So the mass is proportional to the area but then they are off by factor of 2.
Can you derive it?
 
hutchphd said:
but then they are off by factor of 2.
For polar area moment of inertia, yes, but for Ix , or Iy, the approximation should be as given.
 
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SALMAN22 said:
Can you derive it?
For a ring,
Ix = Iy =( π/4 ) ( r24 - r14 )

For a thin ring of small thickness t, r Ξ r2 Ξ r1, but r2 = r1 +t.

Substitute into the formula for the ring, process, and eliminate all elements where t has an exponent.

Edit - corrected the formula for a ring
 
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