Beam bending of an angled flat plate

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SUMMARY

This discussion focuses on calculating stresses in an angled flat plate subjected to bending. Two methods are evaluated: one involving the calculation of new area moments of inertia (Ixx, Iyy) and the other resolving moments into parallel and perpendicular components. The first method yields lower stress values, while the second method, which incorporates the principal area moments of inertia, is believed to be more accurate. The importance of using the centroid as the origin and applying the parallel axis theorem for adjustments is also emphasized.

PREREQUISITES
  • Understanding of bending stress calculations in structural engineering
  • Familiarity with area moments of inertia (Ixx, Iyy)
  • Knowledge of the parallel axis theorem
  • Basic trigonometry for resolving angles in structural analysis
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  • Research the application of the parallel axis theorem in bending stress calculations
  • Learn about the derivation of area moments of inertia for various shapes
  • Explore methods for resolving bending moments into components
  • Study the impact of centroid location on stress calculations in angled plates
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Structural engineers, mechanical engineers, and students studying material mechanics who are involved in stress analysis of angled flat plates.

lordvon
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Hello everyone,
I am wondering how to properly go about calculating the stresses caused by bending on an angled flat plate. Two approaches seem valid but they give much different results:
-Calculate the new area moments of inertia for the angled flat plate (new Ixx, ymax).
-Resolve the moments into parallel and perpendicular components, then add the calculated stresses from the principal area moments of intertia (Ixx, Iyy).

I wrote a short script to compare the resulting stresses from these two methods and they give answers that are different by one or two orders of magnitude. The first method gives lower stresses (more stiff).
I cannot reason how which method is right. I do believe the second method is right, but I cannot explain why the first method is wrong. I know I am missing something with the first method.
 
Last edited:
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Here's how I did the first method, maybe they are similar as far as what you did.

t=thickness
theta=angle at which it is inclined
L=width of beam

I_{yy} = \int y^{2}\frac{t}{sin(\theta)}dy = y^{3}\frac{t}{3sin(\theta)}
y_{max} = Lsin(\theta)/2

I was very careful deriving this, so I hope this helps a little bit as far as checking to see if you got the same answer. I'm not sure how to do the second part. I'm not sure how to resolve bending moments into parallel and perpendicular components. As it turns out, the x direction moment area is also I_{xx} = \int x^{2}\frac{t}{sin(\theta)}dx = x^{3}\frac{t}{3sin(\theta)}, so at least that part is easy enough... However, note, you have to use x_{max} for the x direction component, and that will come out to be
x_{max} = Lcos(\theta)/2

Again, can't really help on separating the components, never had to do that before...

DISCLAIMER: Something doesn't quite seem right about this solution, though I've carefully checked it several times. If you see anything that looks wrong to you let me know and I'll look at it again.
 
Last edited:
Oh, hey, I have to add a correction. I did not use the centroid as the origin, but the corner (which is wrong). You can adjust for that by using the parallel axis theorem (subtracting A*∆x^2 or A*∆y^2 for Ixx or Iyy respectively, where the centroid is (∆x,∆y) ).
 

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