PROBLEM 6.50
Two alternatives for the design of a beam were presented. Determine which one will withstand a moment of M = 150 kN.m with the least amount of bending stress. What is this tension? At what percentage is it most effective?
- For each beam alternative, follow the procedures below:
Divide the cross section into three rectangles (soul, upper table and lower table) and identify the location of the centroid of each rectangle in relation to the base of the cross section.
Having located the centroids of each rectangle, now apply the formula for the moment of inertia of the cross section of the beam:
I = Σ(I' + A.d²), where:
I' = moment of inertia of each rectangle with respect to its centroid
A = area of ​​each rectangle
d = distance between the centroid of the cross section and the centroid of each rectangle
- Okay, now we have all the necessary data to calculate the maximum bending stress developed in the beam, which is calculated using the following formula:
σmax = M.c/I, where:
c = perpendicular distance from the neutral axis to a point farther from the neutral axis
M = internal bending moment applied around the neutral axis (it was given in the problem)
✅ Comparing both results of maximum bending stress, we observe that the beam alternative in (b) is the one that generates the smallest amount of bending stress, being, therefore, 53% more effective in relation to the beam alternative in (a).
📚 Source: Mechanics of Materials 7th Ed. (R.C. Hibbeler)
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