Torsion of Beams - Structural and Stress Analysis - page 281

T

B

g s

C

τ

A
D

A δ s

g

r

r

R

D

O

O

u R

T

r

δ

L

F IGURE 11.4
Torsion of a solid
circular section bar

(a) (b)

Figure 11.4(a) shows a circular section bar of length L subjected to equal and opposite
torques, T , at each end. The torque at any section of the bar is therefore equal to T and
is constant along its length. We shall assume that cross sections remain plane during
twisting, that radii remain straight during twisting and that all normal cross sections
equal distances apart suffer the same relative rotation.

Consider the generator AB on the surface of the bar and parallel to its longitudinal
axis. Due to twisting, the end A is displaced to A so that the radius OA rotates through
a small angle, θ ,toOA . The shear strain, γ s , on the surface of the bar is then equal to
the angle ABA in radians so that

AA
L =

R θ
L

γ s =

Similarly the shear strain, γ , at any radius r is given by the angle DCD so that

DD
L =

r θ
L

γ =

The shear stress, τ , at the radius r is related to the shear strain γ by Eq. (7.9). Then

r θ
L

τ
G =

=

γ

or, rearranging

τ
r

G θ
L

= (11.1)

Consider now any cross section of the bar as shown in Fig. 11.4(b). The shear stress,
τ , on an element r is tangential to the annulus,

is in the plane of the cross section and is constant round the annulus since the cross
section of the bar is perfectly symmetrical (see also Fig. 11.3). The shear force on
the element

δ s of an annulus of radius r and width δ

δ s of the annulus is then τ δ s δ r and its moment about the centre, O, of

the section is τ δ s δ rr . Summing the moments on all such elements of the annulus we