Finding second moment of inertia about x axis

In summary, the conversation is about finding the second moment of inertia about the x-axis for a cubic divided into four parts with a thickness of 1mm each. The attempt at a solution involves using the formula Ixx = Ix + A(d^2)/B, but the correct calculation is not clear. The person providing the summary suggests using the parallel axis theorem to simplify the calculation.
  • #1
werson tan
183
1

Homework Statement


The cubic is divided into 4 parts , A, B and C, D , each with thickness of 1mm , i am sked to find thesecond moment of inertia about x -axis

Homework Equations

The Attempt at a Solution


i' m using the formula Ixx = Ix +A(d^2)[/B]
for part CD, Ixx = 50(1^3) / 12 + 50 ((0.5-0.5)^2)
or Ixx = 50(1^3) / 12 + 50 ((25-0.5)^2)
is correct ?
 

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  • #2
Can anyone tried to ans? I need it urgently.
 
  • #3
werson tan said:

Homework Statement


The cubic is divided into 4 parts , A, B and C, D , each with thickness of 1mm , i am sked to find thesecond moment of inertia about x -axis

Homework Equations

The Attempt at a Solution


i' m using the formula Ixx = Ix +A(d^2)[/B]
for part CD, Ixx = 50(1^3) / 12 + 50 ((0.5-0.5)^2)
or Ixx = 50(1^3) / 12 + 50 ((25-0.5)^2)
is correct ?
If you're trying to calculate the second moment of area about the x-axis, no, your calculations are not correct.

Hint: Since everything is symmetrical about the x- and y-axes and no material is located inside the 1 mm thick plates, wouldn't it be easier to dispense with the parallel axis theorem and just subtract the MOI of the interior from the MOI of the exterior?
 
  • #4
SteamKing said:
If you're trying to calculate the second moment of area about the x-axis, no, your calculations are not correct.

Hint: Since everything is symmetrical about the x- and y-axes and no material is located inside the 1 mm thick plates, wouldn't it be easier to dispense with the parallel axis theorem and just subtract the MOI of the interior from the MOI of the exterior?
Ixx = 50(1^3) / 12 + 50 ((25-25)^2) this is the correct one??
 
  • #5
werson tan said:
Ixx = 50(1^3) / 12 + 50 ((25-25)^2) this is the correct one??
Like your original calculations, that expression doesn't make sense

What is (25 - 25) ?
 

1. What is the formula for finding the second moment of inertia about the x axis?

The formula for finding the second moment of inertia about the x axis is Ix = ∫(y^2 + z^2)dA, where y and z represent the perpendicular distances from the element of area to the x axis, and dA represents the differential area element.

2. How is the second moment of inertia about the x axis used in engineering?

The second moment of inertia about the x axis is used in engineering to determine the resistance to bending of a structural element, such as a beam or column. It is also used in calculating the deflection of a structure under load.

3. What is the significance of the second moment of inertia about the x axis?

The second moment of inertia about the x axis is a measure of an object's distribution of mass around the x axis. It indicates how resistant an object is to changes in its rotational motion about the x axis.

4. How do you calculate the second moment of inertia about the x axis for a complex shape?

The second moment of inertia about the x axis can be calculated for a complex shape by dividing the shape into smaller, simpler shapes and using the parallel axis theorem to sum up the individual moments of inertia. This can then be integrated to find the total second moment of inertia.

5. Can the second moment of inertia about the x axis be negative?

No, the second moment of inertia about the x axis cannot be negative. It is always a positive value since it represents the distribution of mass around the x axis.

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