Chapter 3 Torsion Introduction -- Analyzing the stresses and strains in machine parts which are subjected to torque T Circular -- Cross-section Non-circular Irregular shapes -- Material(1) Elastic (2) Elasto-plastic -- Shaft(1) Solid (2) Hollow

3.1 Introduction  T is a vector  Two ways of expression -- Applications: a. Transmission of torque in shafts, e.g. in automobiles

Assumptions in Torque Analysis: a. Every cross section remains plane and undistorted. b. Shearing strain varies linearly along the axis of the shaft.

3.2 Preliminary Discussion of the Stresses in a Shaft Free-body Diagram Where  = distance (torque arm) Since dF =  dA The stress distribution is Statically Indeterminate.

-- Must rely on “deformation” to solve the problem. Analyzing a small element:

3.3 Deformations in a Circular Shaft  =  (T, L) -- the angle of twist (deformation) Rectangular cross section warps under torsion

A circular plane remains circular plane

(in radians) Determination of Shear Strain  The shear strain   

 = c = radius of the shaft Since

3.4 Stresses in the Elastic Range Hooke’s Law  Therefore,(3.6)

(3.1) (3.9) But Therefore,Or, (3.6)

Substituting Eq. (3.9) into Eq. (3.6) (3.10) (3.9) These are elastic torsion formulas. For a solid cylinder: For a hollow cylinder:

Since   (3-13)

Mohr’s Circle (Sec. 7.4) -- Pure Shear Condition

Ductile materials fail in shear (90 o fracture) Brittle materials are weaker in tension (45 o fracture)

3.5 Angle of Twist in the Elastic Range (3.3) (3.15) Eq. (3.3) = Eq. (3.15)  Therefore, Hence,

For Multiple-Section Shafts:

Shafts with a Variable Circular Cross Section

3.6 Statically Indeterminate Shafts -- Must rely on both (1)Torque equations and (2) Deformation equation, i.e. Example 3.05

3.7 Design of Transmission Shafts -- Two Parameters in Transmission Shafts: a. Power P b. Speed of rotation where  = angular velocity (radians/s) = 2   = frequency (Hz) [N.m/s = watts (W)](3.21)

(3.9) For a Solid Circular Shaft: Therefore, 

3.8 Stress Concentrations in Circular Shafts

3.9 Plastic Deformation sin Circular Shafts (3.4) c = radius of the shaft

c  dd dA = 2  d  c  (3.1) Knowing dF =  dA Where  =  (  ) (3.26)

(3.9) If we can determine experimentally an Ultimate Torque, T U, then by means of Eq. (3.9), we have R T = Modulus of Rupture in Torsion

3.10 Circular Shafts Made of an Elasto-Plastic Material Case I:  <  Y Hooke’s Law applies,  <  max Case I Case II Case II:  <  Y Hooke’s Law applies,  =  max T Y = max elastic torque

Since Case III: Entering Plastic Region Case III 0     Y :  Y – region within the plastic range  Y    c: (3-29)

(3.26) By evoking Eq. (3.26) (3.31) 

Case IV Case IV -- Fully Plastic  Y  0: = Plastic Torque (3-33)

3.11 Residual Stresses in Circular Shafts

3.12 Torsion of Noncircular Members A rectangular shaft does not axisymmetry.

From Theory of Elasticity:

3.13 Thin-Walled Hollow Shafts