Designing Shafts to Withstand Shear Stress

AI Thread Summary
The discussion focuses on designing shafts to withstand shear stress, specifically addressing the maximum allowable angle of twist of 1° over a length equal to 20 diameters. The modulus of rigidity is given as 80 × 10^3 N/mm², and participants clarify the formulas needed for calculating shear stress and the polar moment of inertia. The correct approach involves using the angle of twist equation and the torsional shear stress formula to relate torque and shear stress. Participants emphasize the importance of correctly interpreting the parameters and units involved in the calculations. The goal is to derive the shear stress that ensures the angle of twist remains within specified limits.
loststudent123
Messages
6
Reaction score
0

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?
 
Last edited:
Physics news on Phys.org
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).
 
  • Like
Likes loststudent123
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 :)
 
Back
Top