Strength of Materials- Asymmetric, simply supported beam problem

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SUMMARY

The discussion focuses on analyzing an asymmetric, simply supported beam problem, specifically calculating the second moments of area (Iy, Iz, Iyz) and determining maximum stress. The calculated values are Iz = 48.112 in4, Iy = 13.163 in4, and Iyz = -14.266 in4. The reaction force at the support is 6562.5 lbf in the negative Y direction. The analysis emphasizes the importance of calculating support reactions, constructing shear force and bending moment diagrams, and utilizing Mohr's circle for maximum stress determination.

PREREQUISITES
  • Understanding of second moments of area (Iy, Iz, Iyz)
  • Knowledge of shear force and bending moment diagrams
  • Familiarity with Mohr's circle for stress analysis
  • Basic principles of beam mechanics and support reactions
NEXT STEPS
  • Study the calculation of shear force and bending moment diagrams for asymmetric beams
  • Learn about Mohr's circle for determining maximum stress in non-symmetrical sections
  • Research the concept of shear center and its implications in beam analysis
  • Explore advanced topics in beam mechanics, including torsional effects in asymmetric sections
USEFUL FOR

Structural engineers, mechanical engineers, and students studying strength of materials who are working on beam analysis and stress calculations in asymmetric structures.

dorado29
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I'm working on a problem but I've hit a road block..

problem.jpg


Then the cross section of the piece is here:

crosssection11.jpg


The first part of the problem is to determine Iy, Iz, and Iyz of the member.. I found the y value of the centroid to be 2.8587" and the z value of the centroid to be 1.0734" using the bottom left corner as the origin. From there I did the following:

Iz = 1/12(5)(.5)3 + 2.5(.25 - 2.8587)2 + 1/12(.625)(7.5)3 + 4.6875(.3125 - 1.0734)2

and something similar for Iy and Iyz. All said and done I got

Iz = 48.112 in4
Iy = 13.163 in4
Iyz = -14.266 in4

Anywho, I then used that to get:

axes.png


Where I1 = 53.196 in4
and I2 = 8.0788 in4

Radius of gyration:
Rmax = 2.7205 in
Rmin = 1.0602 in

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The next part of the problem is to find the max stress (be it tensile or compressive) at the far left hand side. I calculated the reaction force of the support to be 6562.5 lbf in the negative Y direction (the opposite of Vy in picture 2) and no moments at all which makes sense considering that the far left side is essentially "exposed" to air.

Am I correct to treat this like Vy is simply negative? Another part of me wants to say that the reaction force from the support is applied to the bottom of the member and I don't know if that changes anything or not..

It seems that the max stress will be solely from the shear stress caused by the support force? I'm trying to figure that out but my book only covered shear stress for symmetrical beams. I remember my professor saying something about using just the webbing for the shear analysis(?) Like treat it as if only the vertical, 8 inch high, .625 inch thick webbing was there then do a simple shear analysis on that piece.. Can somebody point me in the right direction please?
 
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In order to analyze these beam problems properly, you should calculate the support reactions and construct the shear force and bending moment diagrams. The maximum bending moment will occur at a location where the shear force is zero. However, since you are looking to calculate the max. stress, there may be a location along the beam where the combined stress due to bending and shear may be a maximum. The magnitude of the max. stress can be determined using Mohr's circle. Analyzing an L-section is tricky because of the asymmetrical nature of the beam. The shear center for an L-section lies outside of the beam itself, and applying a shear force anywhere except thru the shear center leads to the creation of a torsional moment on the beam.
 

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