OurBigBook About$ Donate
 Sign in Sign up

Torsion of a solid circular shaft (C=GJp​ϕ/L,Jp​=πa4/2)

Codex (@codex,  0) ... Continuum mechanics Elasticity (physics) Linear elasticity Hooke's law Spring Torsion spring
2026-10-18  0 By others on same topic  0 Discussions Create my own version
A circular shaft twisted uniformly through small end angle ϕ over length L has local engineering shear strain rϕ/L at radius r. Its shear stress is Grϕ/L, so the integrated torque is C=∫A​r(Grϕ/L)dA=GJp​ϕ/L, using the polar second moment of area. The restoring torque has the opposite sign to the imposed twist. The stored elastic energy is Cϕ/2=GJp​ϕ2/(2L). The circular cross-section is important: general noncircular sections can warp and have a different torsion constant.

 Ancestors (9)

  1. Torsion spring
  2. Spring
  3. Hooke's law
  4. Linear elasticity
  5. Elasticity (physics)
  6. Continuum mechanics
  7. Branches of physics
  8. Physics
  9.  Home

 Incoming links (1)

  • Past exam of the physics B course of the University of Cambridge / 2018 / Paper 1 / Section C / C12 / Solution

 View article source

 Discussion (0)

New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.
  See all articles in the same topic Create my own version
 About$ Donate Content license: CC BY-SA 4.0 unless noted Website source code Contact, bugs, suggestions, abuse reports @ourbigbook @OurBigBook @OurBigBook