What is the total moment of inertia using the negative area method?

[Benjamin_harsh]

Homework Statement

What is the area moment of inertia for this problem?

Relevant Equations

$I_{2} = I_{3} = \large\frac{1}{12}\normalsize bh^{3} = \large \frac{1}{12}\normalsize *35*(180 mm)^{4} = 1.701*10^{7} mm^{4}$

$I_{total} = (I_{1} + A_{1}d_{1}^{2} ) + (I_{2} + A_{2}d_{2}^{2}) + (I_{3} + A_{3}d_{3}^{2})$(d is the distance between centroid of the individual box and centroid of whole shape.)

By negative area method by joining the gaps in I section.

$I_{total} = (I_{1} + A_{1}d_{1}^{2} ) - (I_{2} + A_{2}d_{2}^{2}) -(I_{3} + A_{3}d_{3}^{2})$

Centroid of two new sections matches with centroid of final shape. So $d_{1}$, $d_{2}$ and $d_{3}$ are zero.

$I_{total} = (I_{1} - I_{2} - I_{3})$

Moment of Inertia of a Rectangle 1,$I_{1} = \large \frac {1}{12}\normalsize bh^{3} = \large \frac {1}{12}\normalsize 110*260^{3} = 1.611 * 10^{8}mm^{4}$

Moment of Inertia of a Rectangle 2 and 3 are equal because they have similar dimensions $I_{2} = I_{3} = \large\frac{1}{12}\normalsize bh^{3} = \large \frac{1}{12}\normalsize *35*(180 mm)^{4} = 1.701*10^{7} mm^{4}$

Area moment of Interia, $I_{total} = I_{1} - 2I_{2} = 1.272*10^{8} mm^{4}$

 

[PhanthomJay]

Problem assumes I guess that the web width is 40 mm (same as flange width) but the dimension should have been shown on the drawing.

 

[Benjamin_harsh]

Problem assumes I guess that the web width is 40 mm (same as flange width) but the dimension should have been shown on the drawing.

Are you saying that taking 35 as base is wrong?

 

[PhanthomJay]

No, I am saying it is correct if the width of the middle vertical rectangle is assumed to be 40mm. The drawing does not show that width, which it should have, because the problem otherwise could not be solved with incomplete information given.