Second moment of inertia for a bent rectangle

In summary, the designer is trying to find the second moment of inertia for a beam with a cross-section of a small triangle and rectangle. The area of the profile is given by the equation, and should be approximately 1000 mm2. The designer is trying to find the deflection of the beam, and thereby the second moment of inertia.
  • #1
Ole Forsell
10
0
Hello.
I am currently working with a beam with the following cross-section:
upload_2017-3-20_8-47-17.png

It consist of three bended sections with the following parameters, alpha = 90 degrees, Thickness = 4 mm, Radius = 50.59 mm.
upload_2017-3-20_8-50-9.png

The top section consist of a small triangle and a rectangle. the triangle have a width = 4 mm and height = 2.81 mm. The rectangle have width = 4 mm and height = 19.60 mm.

The area of the profile is given by the following equation, and should be approximately 1000 mm2:
upload_2017-3-20_9-6-35.png


I need to find the deflection of this beam, and thereby the second moment of inertia. I have allready found this to be 4273323.41 mm4 in Section Properties in Solidworks, but this need to be substantiated by hand calculations. Does anybody know how to calculate the second moment of inertia for such a geometry?Ole
 
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  • #2
You want the second moment with respect to what axis?

Which ever axis it is, just apply the definition and carry out the integration.
 
  • #3
Dr.D said:
You want the second moment with respect to what axis?

Which ever axis it is, just apply the definition and carry out the integration.

Thank you for answering. I would like to find Iyy, which in this case mean in vertical direction.
Which definition do you mean?
 
  • #4
Your figure is unclear. Is the sinuous figure at the top truly a cross section, or is it a plan view? Where is the y-axis in that view? Which way is the loading applied?

You need a three dimensional presentation of your problem to make clear what the situation really is.
There are definitions online and elsewhere for the second moment of an area expressed in terms of integration.
 
  • #5
Dr.D said:
Your figure is unclear. Is the sinuous figure at the top truly a cross section, or is it a plan view? Where is the y-axis in that view? Which way is the loading applied?

You need a three dimensional presentation of your problem to make clear what the situation really is.
There are definitions online and elsewhere for the second moment of an area expressed in terms of integration.

Both figures are captured from the front plane, but shows the cross-section of the model. The x and y-axis is shown in both figures, and y is in vertical direction. The loading is applied in negative y-direction.

I have the 3D model, and can extract all information from SolidWorks, but these values need to be substantiated by calculations. But the second (area) moment of inertia only depends on the area, why would I need a 3D presentation?

Do you have any links, or formulas I can use to solve the problem?
 
  • #6
If the load is applied in the negative y direction as you say, this will tend to straighten the "crooked stick." This straightening action will involve bending (actually reducing the radius of curvature) in each of the arcs. Bending of this sort depends upon the second moment perpendicular to the x-y plane, Izz. You have not shown us what that section looks like.
 
  • #7
Dr.D said:
If the load is applied in the negative y direction as you say, this will tend to straighten the "crooked stick." This straightening action will involve bending (actually reducing the radius of curvature) in each of the arcs. Bending of this sort depends upon the second moment perpendicular to the x-y plane, Izz. You have not shown us what that section looks like.

The model shown in the figure is extruded 11 meters in the z-direction. In perspective, the height of the beam in y-direction is approximately 220 mm. So it is very long compared to the cross-section. I think I might have been a little unclear in the problem desciption. I want to analyse the deflection when the beam is fixed in both ends, and subjected to a uniform load which represents its own weight

I've been looking into a few other beam profiles, i.e, hollow tube, hollow triangle and T-profile, all with approximately the same cross-section area and the same length in z-direction. Assuming the same coordinate system as for this model, the deflection off the beam could be decided by only looking at Iyy for all these other beams. Why would it be any different for this model?
 
  • #8
With the z dimension at 11 m, it sounds like a plate bending problem, not a beam problem.

However, I seem to be unable to communicate with you clearly, so I'm going to drop out and let someone else try.
 

Related to Second moment of inertia for a bent rectangle

What is the second moment of inertia for a bent rectangle?

The second moment of inertia for a bent rectangle, also known as the moment of inertia or rotational inertia, is a measure of an object's resistance to changes in its rotational motion. It is a mathematical property that reflects the distribution of mass around an axis of rotation.

How is the second moment of inertia for a bent rectangle calculated?

The second moment of inertia for a bent rectangle can be calculated by multiplying the area of the rectangle by the square of its perpendicular distance from the axis of rotation. This can be expressed in the equation I = bh^3/12, where I is the second moment of inertia, b is the base of the rectangle, and h is the height from the axis of rotation.

Why is the second moment of inertia important for a bent rectangle?

The second moment of inertia is important for a bent rectangle because it determines how much torque or force is required to cause the object to rotate. A higher moment of inertia means the object is more difficult to rotate, while a lower moment of inertia means the object is easier to rotate.

How does the second moment of inertia change for a bent rectangle with different dimensions?

The second moment of inertia for a bent rectangle is directly proportional to the dimensions of the rectangle. This means that as the dimensions of the rectangle increase, the second moment of inertia also increases. Additionally, the distribution of mass within the rectangle can also affect the value of the second moment of inertia.

Can the second moment of inertia for a bent rectangle be negative?

No, the second moment of inertia for a bent rectangle cannot be negative. It is always a positive value, as it represents the object's resistance to rotation. A negative value would imply that the object has a net torque acting in the opposite direction of its rotation, which is not physically possible.

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