Stress on a spaghetti rod balanced from its middle

[phantomvommand]

TL;DR Summary: A thin piece of spaghetti of diameter d is balanced horizontally from its middle.
It can have a length ℓ ≫ d before it snaps under its own weight. How does ℓ scale with d?

Let the spaghetti rod have density ρ, and consider ONLY its right half.

(1) The spaghetti rod experiences some horizontal stress, let it be $\sigma$. By taking moments about the pivot point, moment due to weight must be balanced by moments due to the horizontal stress. Therefore,
$\rho d^2 l (l) \sim \sigma d^2 (d)$, so $\sigma \sim \rho \frac {l^2} {d}$.
This agrees with the textbook solution.

However, the text makes an additional remark that:
(2) There must be a vertical normal force (at the pivot) $F \sim ρd^2ℓ$ to balance the weight. This vertical force is transmitted through the rod by a shear stress (i.e. an internal force per area, perpendicular to the rod) of order $σ_v$ ∼ F/A ∼ ρℓ.

It then compares the 2 stresses, and concludes that: Horizontal stress is much greater than $σ_v$, because of the miserably small lever arm, which is why thin rods usually break by snapping, not by shearing or pulling apart.

How did it reach this conclusion? I get that horizontal stress is much larger than vertical stress, but how does that lead to the claim that thin rods are more likely to break by snapping?

Doesn't it depends on the maximum horizontal and vertical stress for the material?

 

[Baluncore]

Doesn't it depends on the maximum horizontal and vertical stress for the material?

It depends on the ratio of tensile strength to shear strength for the material.

Is this homework?

Given tensile strength, how does an engineer find the shear strength for a material?

 

[phantomvommand]

It depends on the ratio of tensile strength to shear strength for the material.

Is this homework?

Given tensile strength, how does an engineer find the shear strength for a material?

Are both tensile and shear strength not fixed values for a particular material?

My view is that while horizontal stress is indeed larger in this case, it depends on the maximum horizontal and vertical stress that the spaghetti can withstand too.

 

[Lnewqban]

My view is that while horizontal stress is indeed larger in this case, it depends on the maximum horizontal and vertical stress that the spaghetti can withstand too.

Naturally, any structural material has less resistance to shear loads than to tensile-compressive loads (roughly around 1/2 ratio).

In this case,

The horizontal stress is "tensile stress (stress normal to the plane) - stress that tends to stretch or lengthen the material - acts normal to the stressed area".
"A normal force acts perpendicular to area and is developed whenever external loads tends to push or pull the two segments of a body."

The vertical stress is "shearing stress (stress perpendicular to the plane) - stress that tends to shear the material - acts in plane to the stressed area at right-angles to compressive or tensile stress".
"A shear force lies in the plane of an area and is developed when external loads tend to cause the two segments of a body to slide over one another."

Please, see:
https://www.physicsforums.com/threads/tensile-and-shear-capacity-of-metals.1045312/

https://www.engineeringtoolbox.com/stress-strain-d_950.html

https://www.engineersedge.com/materials/material_tensile_shear_and_yield_strength_15798.htm

https://www.eng-tips.com/viewthread.cfm?qid=69837

 

[phantomvommand]

Naturally, any structural material has less resistance to shear loads than to tensile-compressive loads (roughly around 1/2 ratio).

In this case,

The horizontal stress is "tensile stress (stress normal to the plane) - stress that tends to stretch or lengthen the material - acts normal to the stressed area".
"A normal force acts perpendicular to area and is developed whenever external loads tends to push or pull the two segments of a body."

The vertical stress is "shearing stress (stress perpendicular to the plane) - stress that tends to shear the material - acts in plane to the stressed area at right-angles to compressive or tensile stress".
"A shear force lies in the plane of an area and is developed when external loads tend to cause the two segments of a body to slide over one another."

Please, see:
https://www.physicsforums.com/threads/tensile-and-shear-capacity-of-metals.1045312/

https://www.engineeringtoolbox.com/stress-strain-d_950.html

https://www.engineersedge.com/materials/material_tensile_shear_and_yield_strength_15798.htm

https://www.eng-tips.com/viewthread.cfm?qid=69837

I see, then the ratio of tensile to shear stress in the spaghetti rod would far exceed 2:1 (generally), since $l >> d$. However, could you further explain why "thin rods usually break by snapping, not by shearing or pulling apart". Shouldn't a larger tensile stress cause it to pull apart instead of snap?

 

[Lnewqban]

However, could you further explain why "thin rods usually break by snapping, not by shearing or pulling apart". Shouldn't a larger tensile stress cause it to pull apart instead of snap?

Those are three different types of external loads, which are "felt" very differently by the same cross section.
Bending is always the king of destructive loads, mainly because it involves leverage-moments that other loads don't.
Once the area farther from the neutral axis starts separating, the cross-section area rapidly decreases, increasing the value of the internal tension stress (hence the snapping effect).

 

[Juanda]

TL;DR Summary: A thin piece of spaghetti of diameter d is balanced horizontally from its middle.
It can have a length ℓ ≫ d before it snaps under its own weight. How does ℓ scale with d?

View attachment 346203

I don't understand the bit where it asks "How does $l$ scale with $d$?".
Is it asking what's the greatest $d$ for a given $l$ or viceversa before it breaks or something like that?
I'd know how to do that but I don't understand the question in the way it's formulated.