Torsion of Beams - Structural and Stress Analysis - page 288

τ = τ Y = constant. Hence

τ Y π r e
2

2
3 πτ Y ( R 3

r e )

T = + −

which simplifies to

1

2 π R 3
3

r e
4 R 3

T = τ Y − (11.12)

Note that for a given value of torque, Eq. (11.12) fixes the radius of the elastic core
of the section. In stage three (Fig. 11.10(c)) the cross section of the bar is completely
plastic so that r e in Eq. (11.12) is zero and the ultimate torque or fully plastic torque,
T P , is given by

2 π R 3
3

T P = τ Y (11.13)

Comparing Eqs (11.11) and (11.13) we see that

T P
T Y =

4
3

(11.14)

so that only a one-third increase in torque is required after yielding to bring the bar to
its ultimate load-carrying capacity.

Since we have assumed that radii remain straight during plastic torsion, the angle of
twist of the bar must be equal to the angle of twist of the elastic core which may be
obtained directly from Eq. (11.3). Thus for a bar of length L and shear modulus G ,

TL
GJ e =

2 TL
π Gr e

θ = (11.15)

or, in terms of the shear stress, τ Y , at the outer surface of the elastic core

τ Y L
Gr e

θ = (11.16)

Either of Eq. (11.15) or (11.16) shows that θ is inversely proportional to the radius, r e ,
of the elastic core. Clearly, when the bar becomes fully plastic, r e →

0 and θ becomes,
theoretically, infinite. In practical terms this means that twisting continues with no
increase in torque in the fully plastic state.

11.4 T ORSION OF A T HIN-WALLED C LOSED S ECTION B EAM

Although the analysis of torsion problems is generally complex and in some instances
relies on empiricalmethods for a solution, the torsionof a thin-walled beamof arbitrary
closed section is relatively straightforward.

Figure 11.11(a) shows a thin-walled closed section beam subjected to a torque, T . The
thickness, t , is constant along the length of the beam but may vary round the cross