Designing Shafts to Withstand Shear Stress

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Discussion Overview

The discussion revolves around the design of shafts to withstand shear stress, specifically focusing on the implications of a maximum allowable angle of twist of 1° over a length equal to 20 diameters. Participants explore the relevant equations and concepts related to shear stress, torque, and the modulus of rigidity.

Discussion Character

  • Homework-related
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants present the rule regarding the angle of twist and its implications for shear stress, referencing the modulus of rigidity.
  • There is confusion regarding the correct formula for calculating shear stress in circular shafts, with some participants questioning the validity of the equations provided in their lecture.
  • One participant clarifies that the limiting twist is indeed 1 degree over the specified length, addressing earlier ambiguity about the units.
  • Participants discuss the need to determine the applied torque and how it relates to the angle of twist and shear stress.
  • There are suggestions to use algebra to eliminate variables in the equations to find the shear stress corresponding to the maximum angle of twist.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the correct formulas for shear stress and the application of the angle of twist criterion. There is no consensus on the approach to solving the problem, as some participants challenge the adequacy of the provided equations.

Contextual Notes

Limitations include potential misunderstandings of the equations related to shear stress and torque, as well as the need for clarification on the criteria for maximum allowable twist.

Who May Find This Useful

Students and professionals involved in mechanical engineering, materials science, or any field requiring knowledge of shaft design and torsional mechanics may find this discussion relevant.

loststudent123
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Homework Statement


A rule often used in shaft design states that the angle of twist shall not exceed 1° on a length equal to 20 diameters. What stress in the material does this imply if the modulus of rigidity is equal to 80 × 103 N/mm2 ?

Homework Equations


T/J=Gθ/L=τ/r
J=π?32(D^4-d^4)[/B]
G=shear stress/shear strain
shear stress=F/A
F=Wx9.81

The Attempt at a Solution


so i found that the length of the beam is D(20) letting D equal its diameter
so i got the equation
(80x10^3)/(D(20)=Wx9.81Xπ(D/2)/(D/2)
but I am not sure if that's right or how to solve when there is two unknowns?
 
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loststudent123 said:

Homework Statement


A rule often used in shaft design states that the angle of twist shall not exceed 1 on a length equal to 20 diameters. What stress in the material does this imply if the modulus of rigidity is equal to 80 × 103 N/mm2 ?

Homework Equations


T/J=Gθ/L=τ/r
J=π?32(D^4-d^4)[/B]
G=shear stress/shear strain
shear stress=F/A
F=Wx9.81

The Attempt at a Solution


so i found that the length of the beam is D(20) letting D equal its diameter
so i got the equation
(80x10^3)/(D(20)=Wx9.81Xπ(D/2)/(D/2)
but I am not sure if that's right or how to solve when there is two unknowns?
It's not clear what criterion you are intended to use here. The text of your post doesn't correctly display the words, 'the angle of twist shall not exceed 1 [squiggle] on a length equal to 20 diameters.' What are the units of 1 [squiggle]?

If you know anything about calculating the shear stress of circular shafts, it should be that shear stress ≠ F / A.
What is the correct formula for calculating the shear stress for a circular shaft in torsion?
 
my lecture gave only those equations so I am not sure of anyothers and i fixed the angle of twist however.
 
loststudent123 said:
my lecture gave only those equations so I am not sure of anyothers and i fixed the angle of twist however.
Well, I must say that was a poor lecture then.

For a circular shaft, the angle of twist θ = TL / G J, where

T = applied torque
L = length of the shaft
G = modulus of rigidity of the shaft material
J = polar moment of inertia for the shaft, J = π D4 / 32 for solid shafts; J = π (Do4 - Di4) for hollow shafts,
Do = outer diameter of shaft
Di = inner diameter of shaft

The torsional shear stress, τ = T ⋅ r / J, where

T = applied torque
r = location of the point where shear stress is calculated, measured from the center of the shaft cross section
J = polar moment of inertia for the shaft cross section (see above)

You still haven't answered the question about the units of the limiting twist in the shaft (1 [squiggle] in 20 diameters).
 
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its degree so its one degree is the max angle of twist and its over the length of 20 diameters
 
loststudent123 said:
its degree so its one degree is the max angle of twist and its over the length of 20 diameters
You should be all set.
 
what should i put in for applied torque?
 
loststudent123 said:
what should i put in for applied torque?
You know the angle of twist produced by a certain value for T. The max. shear stress in the outer fiber of the shaft also depends on this value of T.

What you want is to find the shear stress in the shaft so that the angle of twist is no more than 1 degree in 20 diameters.

Use algebra to eliminate T from the two formulas.
 
thank you :)
 

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