Basic Equations: Differential Form - Mechanics of Structures: Variational and Computational Methods - page 51

EXAMPLE 1.5 Torsion of a Shaft of Circular Cross-Section
In the case of a shaft of circular cross-section, where radial symmetry is involved and
deformations of a cross section are the same when viewed from either end of the shaft
segment, plane cross-sections normal to the x axis will in fact remain normal to the x axis
and the radii will remain straight and unextended. The deformations (Eq. 1.139)

v =− φ xz, w = φ xy, u = 0 (1)

lead to an exact solution for a twisted bar of circular cross-section.
From Eqs. (1.18), the strain components are

x = ∂

u

y = ∂v
∂

z = ∂w
∂ x =

0,

y =

0,

z =

0

∂

(2)

= ∂v
∂ x + ∂

= ∂w
∂ x + ∂

= ∂w
∂ y + ∂v

u u

∂ y =− φ z, z = φ y,

γ γ γ

z =

0

xy xz yz

∂ ∂

The conditions of compatibility are satisfied as these strains are obtained from an admissible
displacement field. These strains reduce the constitutive relations of Eq. (1.34a) to

φ z, φ y, σ = σ = σ = τ = 0 , τ = G γ =− G τ = G γ = G (3)

x y z yz xy xy xz xz

The stress components of (3) satisfy the equations of equilibrium of Eq. (1.51), provided the
body forces are equal to zero.
The surface conditions of Eq. (1.57) must be satisfied on the circumferential surfaces
of the bar. The direction cosines of the unit normal to the circumferential surfaces are
( a x

= 0 ,a y , a z ) . For this torsion problem the surface tractions on this surface are zero, i.e.,

yz are zero. It follows that the sec-
ond and third of Eqs. (1.57a) are satisfied and the first equation becomes

p x = p y = p z = 0 . Also, the stress components σ

x ,

σ

y ,

σ

z ,

τ

a z τ + a y τ = 0 (4)

xz xy

As seen in Fig. 1.17c

y
a ,

z
a ,

a y = cos φ = a z = sin φ = (5)

Substitution of (3) and (5) into (4) shows that the surface conditions on the circumferential
surfaces are satisfied.
On the ends of the bar, the resultant shear forces can be shown to be zero although the
resultant moment is not zero. This moment is expressed as (Fig. 1.18)

(τ xz y M t = − τ xy z ) dy dz (6)

Substitute (3) in this expression to find

φ φ

y 2 z 2 r 2 dA

=

A (

+ ) =

M t G dA G (7)

A

This integral is the polar moment of inertia J of the circular cross section. Thus

d dx = M t

GJ

(8)