Mechanics torsion analysis help

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SUMMARY

The discussion focuses on the torsion analysis of a 2-meter long shaft designed to transmit a torque of 1000 Nm, with a shear stress limit of 320 MPa. The shaft can be either a solid circular bar or a circular tube, with the tube's thickness specified as one-eighth of its outer diameter. The calculations involve determining the polar moment of inertia (J) using the formula J = π(D^4 - d^4) / 32, leading to the conclusion that the required outer diameter (D) of the tube is 28 mm after applying the torsion formula T/J = τ/r.

PREREQUISITES
  • Understanding of torsion mechanics and shear stress calculations
  • Familiarity with polar moment of inertia (J) for circular sections
  • Knowledge of torque transmission in mechanical design
  • Basic algebra and cube root calculations
NEXT STEPS
  • Study the derivation and application of the polar moment of inertia for various cross-sectional shapes
  • Learn about shear stress distribution in hollow versus solid shafts
  • Explore advanced torsion analysis techniques, including the use of finite element analysis (FEA)
  • Investigate material selection criteria based on allowable stress limits in mechanical design
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Mechanical engineers, students in engineering mechanics, and professionals involved in the design and analysis of rotating shafts and mechanical components.

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Homework Statement


the statement:
You are taked to design a 2m long shaft which is to be manufactured either as a solid circular bar or a circular tube. the shaft is required to transmit a torque of 100Nm without exceeding the alloqable strength of material.

Find the required outer diameter of the tube if the thickness of the tube is specified as one-eigth of the outer diameter?

shear stress = 320Mpa


Homework Equations


T/J = \tau /r

J= [pi(D^4 - d^4)] / 32


The Attempt at a Solution



thickness is 1/8
1/8 + 1/8 = 1/4

J= pi( D^4 - 0.75^4 D^4) / 32

work it out in sections
0.75^4 = 81/256 >>> 1 - 81/256 = 175/256 >>> 175/256 x pi = 2.14 >> 2.14 / 32 = 0.067D^4

then using the formula T/J = t/r

1000/0.067D^4 = 320x10^6 / D/2

get D on one side i get

1000 / 2 x 320 x 10^6 = 0.067D^4 / D

worked all that out then cube root it to get 0.028 which i then x1000 to get 28mm


I feel I'm on the right path ( i hope) but the whole 1/8 of the diameter has confused me and I'm wondering whether I've done that right. finding the "J" has to be my main concern i feel I've got the method right for finding D later on. If anyone can help my go through it and check everything is correct or if its wrong, advice on how to correct it would be fantastic.
thanks.
 
Physics news on Phys.org
Looks like you got it all correct, good step by step work. I assume the torque is 1000 Nm and not 100 Nm as originally stated.
 

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