Torsion

1
Torsion

• Cylindrical shafts
• Compatibility relation between shear strain g
  and angle of twist b.
• Lg rb ? g (b/L)r (1)
• Elasticity relation between shear stress t and
  shear strain g.
• t Gg (Gb/L)r (2)
• Equilibrium relation between applied torque and
  internal shear stress distribution

2
Torsion

• Internal shear stress distribution is equivalent
  to the torque

where J is the torsion constant (equal to the
polar moment of area in this special case)

• Eliminating b between Eq. (2) and Eq. (3)

• Strength constant

3
Torsion

• Bars with non-circular cross section
• z axis of twist
• q angle of twist per unit length
• b qz
• Displacement components
• u qyz, v qxz, w qy(x,y)
• y(x,y) warping function

4
Torsion
Shear strain
Solution for stress Compatibility equation
Equilibrium equation
Prandtls stress function f
Equilibrium equation identically satisfied while
compatibility equation becomes
5
Torsion

• Boundary condition no traction on lateral
  surface S
• sPz szx cosg szy sing 0 on S
• This leads to
• f 0 on S
• Torque

With the stresses found as functions of x, y and
q, the above integral is reduced to the form
Thus the integration of r.h.s. of (4) produces
the torsional constant J.
6
TorsionSolutions for the stress function

• Elliptical section
• Assumed solution

• This satisfies the boundary condition. It also
  satisfies the differential equation if B is given
  by

• Stresses

7
TorsionSolutions for the stress function

• Elliptical section

• Maximum shear stress

• Torsional constant

• Strength constant (defined through T Ctmax)

8
TorsionSolutions for the stress function

• Equilateral triangle section
• Assumed solution

• This satisfies the boundary condition. It also
  satisfies the differential equation if B is given
  by

• Maximum shear stress

• Torsion constant

• Strength constant

9
Torsion Rectangular section

• Series solution for the stresses valid for b ? h
  leads to
• tmax szy( h,0) k3(2h)Gq (5)
• With only the first term in the series retained

• Substituting the stresses into the integral
  giving the torque

10
Torsion Rectangular section

• Again, sufficient accuracy is achieved by
  retaining the first term in the series (n 0)
• T ? k1 (2b)(2h)3Gq (6)

where

• From (5) and (6)
• T ? k2 (2b)(2h)2 tmax, k2 k1/k3
• ? Torsional constant J k1 (2b)(2h)3
• Strength constant C k2 (2b)(2h)2

11
Torsion composite section

• The section is divided into N sub-sections of
  known torsional properties.
• Total torque

where Ti is the torque acting on sub-section i.
Since twist of all sub-sections is the same,

• Section made up of rectangular sections
• Ti Ci (tmax)I
• Hence

12
Torsion composite section examples

• Open channel section (all dimensions in mm)
• Find tmax due to T 600 Nm
• 2h1 9, 2b1 100
• 2h2 5, 2b2 182
• 2h3 9, 2b3 100
• Since bi/hi gt 10 for all i
• k1 ? k2 ? 0.333, k3 ? 1

where ? tmax 96.11 MPa
13
Torsion composite section examples

• Open I-section (all dimensions in mm)
• Determine T for an allowable shear stress of 240
  MPa and a safety factor of 3
• 2h1 2h2 2h3 2h 6,
• 2b1 2b3 100, 2b2 138
• bi/hi gt 10 for all i
• k1 ? k2 ? 0.333, k3 ? 1

where
? T 324.5 Nm
14
Torsion closed thin-wall section

• Relative to a curvilinear frame of reference s-n
• szn ? 0, szs t(s)
• Equilibrium in z direction gives
• Shear flow q(s) t(s) t(s) constant.
• Torque

• where A is the area enclosed by the centreline
  l.
• Strain energy must be equal to the work done by
  the torque

Substituting from (Eq. 7) to Eq. (8)
15
Torsion closed section with internal walls

• Section considered as an assembly of N tubular
  sub-sections (compartments), each subjected to
  torque Ti.
• Total torque

Compartment i
Common twist for all compartments
where q? is the shear flow due to torque in
adjacent compartments.
16
Torsion closed section with internal walls

• Example 6.6
• Determine maximum torque for an allowable shear
  stress of 40 MPa.
• G 26 GPa

T T1 T2 2A1q1 2A2q2 A1 (60?103)2, A2
0.5p(30?103)2

• ? q1 / q2 1.22 BUT t1 / t2 1.5
• Therefore maximum shear develops in l2
• q2 t2tmax 120?103, q1 1.22 q2 ? max T
• For the wall with t3 1.5, check that (q1
  q2)/1.5 lt tmax