### SlideShare for iOS

by Linkedin Corporation

FREE - On the App Store

Torsion or twisting is a common concept in mechanical engineering systems. This section looks at the basic theory associated with torsion and examines some typical examples by calculating the main ...

Torsion or twisting is a common concept in mechanical engineering systems. This section looks at the basic theory associated with torsion and examines some typical examples by calculating the main parameters. Further examples include determination of the torque and power requirements of torsional systems.

- Total Views
- 10,941
- Views on SlideShare
- 10,941
- Embed Views

- Likes
- 0
- Downloads
- 187
- Comments
- 0

No embeds

- 1. Introduction to Torsion HNC in Engineering- Mechanical Science Edexcel Unit: Engineering Science (NQF L4) Author: Leicester College Date created: Date revised: 2009 Abstract: Torsion or twisting is a common concept in mechanical engineering systems. This section looks at the basic theory associated with torsion and examines some typical examples by calculating the main parameters. Further examples include determination of the torque and power requirements of torsional systems. Contents Introduction to Torsion..................................................................................................................................1 Torsion...........................................................................................................................................................2 Transmission of power...................................................................................................................................3 Credits............................................................................................................................................................4 These files support the Edexcel HN unit – Engineering Science (mechanical) Unit Key words outcome Stress 1.1 Stress, strain, statics, young’s modulus introduction BM, shear 1.1 Shear force, bending moment, stress force diagrams Selecting 1.2 Beams, columns, struts, slenderness ratio beams Torsion 1.3 Torsion, stiffness, twisting introduction Dynamics 2.1/2.2 Linear motion, angular motion, energy, kinetic, potential, rotation introduction For further information regarding unit outcomes go to Edexcel.org.uk/ HN/ Engineering / Specifications © Leicester College 2009 This work is licensed under a Creative Commons Attribution 2.0 License.
- 2. Introduction to Torsion Torsion Torsion is the term used for the twisting of a structural member when it is acted upon by TORQUE so that rotation is produced about the longitudinal axis at on end of the member with respect to the other. Torque - Fr - Twisting moment Fixed end The amount of twist ( torsion) that the shaft experiences will increase as we move away from the fixed end of the shaft. Assumptions; • The shaft has a uniform cross section • The shaft material is uniform throughout and the shear stress is proportional to the shear strain (Elastic region) • The shaft is straight and initially unstressed • The axis of twisting moment is the axis of the shaft • Plain transverse sections remain the same after twisting General equation for torsion of cross sectioned circular shafts; T = τ = Gθ Where T = torque (Nm) J r L J = polar 2nd moment of area τ = Max. shear stress (MPa) r = radius of shaft (m) G = Modulus of Rigidity (GPa) θ = Angle of twist (radians) L = length of shaft (m) Torsional stiffness - T/ θ (Applied torque per radian) Page 2 of 4
- 3. Introduction to Torsion Transmission of power P = 2πnT (n = revs/sec, T = applied torque Example 1 Calculate the torsional stiffness of a 0.5m long shaft, 15mm diameter. G = 90GPa. Using; T = Gθ so that T/θ = GJ / L J L And J = πd4 / 32 for a shaft J = (π x 0.0154)/ 32 = 4.97 x 10-9 T/θ = (90 x 109 x 4.97 x 10-9 ) / 0.5 = 894.62Nm / rad Example 2 A solid steel shaft is 2.5m long and 40mm in diameter. The maximum stress in the shaft must not exceed 60 MPa. Determine the maximum torque that can be applied and the angle of twist at this torque (in degrees) Assume G = 80 GPa J = (π x 0.044)/ 32 = 2.514 x 10-7 Using T = τ J r Therefore T = J τ = (2.514 x 10-7 x 60 x 106) / 0.02 r = 753.98 Nm Using; τ = Gθ (NOTE we can neglect one term) r L θ= τL = (60 x 106 x 2.5) / (0.02 x 80 x 109) rG =0.09375 rads Change radians to degrees ; 0.09375 x 57.3 = 5.372 degrees Note – Conversion Rads to degrees - multiply by 57.3 (360/2π ) = 57.3 Page 3 of 4
- 4. Introduction to Torsion Credits This resource was created Leicester College and released as an open educational resource through the Open Engineering Resources project of the Higher Education Academy Engineering Subject Centre. The Open Engineering Resources project was funded by HEFCE and part of the JISC/HE Academy UKOER programme. © 2009 Leicester College This work is licensed under a Creative Commons Attribution 2.0 License. The JISC logo is licensed under the terms of the Creative Commons Attribution-Non-Commercial-No Derivative Works 2.0 UK: England & Wales Licence. All reproductions must comply with the terms of that licence. The HEA logo is owned by the Higher Education Academy Limited may be freely distributed and copied for educational purposes only, provided that appropriate acknowledgement is given to the Higher Education Academy as the copyright holder and original publisher. The Leicester College name and logo is owned by the College and should not be produced without the express permission of the College. Page 4 of 4

Full NameComment goes here.