# Why is the maximum stress of a uniformly tapered round bar lesser than that of a inverse uniformly tapered round bar?

Given two round bars with uniformly tapering radii, one from a smaller to a larger radius and one from a larger to a smaller radius, the former bar experiences greater maximum stress.

Since stress depends solely on the weight and area of a bar, I don't understand how this occurs. The force while taking cross-sections of the bar remains constant, leading me to believe that the maximum stress would be equal, just at opposite ends of the bar.

• Sounds about right. Can you tell us where this statement came from that the two bars behave differently and what assumptions were made for that statement? Commented Apr 27, 2023 at 2:20
• I have added a diagram. The statement was given to me in class, with the assumptions that both beams have the same weight, shape and young's modulus. Commented Apr 27, 2023 at 3:32
• I don't understand what you mean by "the force while taking cross-sections of the bar remains constant". What can you say about the stress at the upper end of each bar? Think about that first, then try to work out the stress as a function of distance from the bottom.
– Puk
Commented Apr 27, 2023 at 8:28
• Are the bars hanging from the support under their own weight? Commented Apr 27, 2023 at 15:05
• Yes, the only force applied is the weight force. Commented Apr 28, 2023 at 7:16

## 1 Answer

The force while taking cross-sections of the bar remains constant

The error in reasoning lies here. The (axial tension) force isn’t constant but increases from zero at the bottom of a suspended object to the object’s weight at the top of the object. The stress at the top then depends on the weight and top cross-section and is larger for a smaller top cross-section (illustration 2).

• If this is the case, then why is it that while analyzing a two-section, horizontal rod, we take cross-sections and establish equilibrium for each of them in order to solve for stress and strain? If we attempt a method similar to the one depicted in this video, the force would remain constant throughout the beam. Is this the difference between weight and applied force? Commented Apr 28, 2023 at 18:03
• Yes. Draw a free-body diagram of any section. In the vertical case, the weight acts in the axial direction and thus affects the axial tension and stress. In a horizontal rod, the weight acts transversely, producing a shear stress that's ignored in the linked video. Make sense? Commented Apr 28, 2023 at 18:10
• I don't think I'm giving a good explanation. Why would the method depicted in this diagram not work for a loading of this type? If we utilize a method like this, then the maximum stress would be equal for both beams, given that the force utilized for stress at any point along the beam is simply the total weight. Commented Apr 28, 2023 at 18:24
• As I wrote above, the force at any point is not the total weight. I again suggest that you draw a free-body diagram of a section and calculate the forces acting on it. The tension corresponds to the weight below the section, not the total weight. Commented Apr 28, 2023 at 18:32
• Ah, that makes sense. Thank you! Commented Apr 30, 2023 at 12:31