F. K. Kong MA, MSc, PhD, CEng, FICE, FIStructE, R. H. Evans CBE, DSc, D ès Sc, DTech, PhD, CEng, FICE, FIMechE, FIStructE (auth.)-Reinforced and Prestressed Concrete-Springer US (1987)

226 Shear. bond and torsion
(6.8-5)
where n is the membrane tension per unit length. Equation (6.8-5) is of
the same form as eqn (6.8-1). Therefore. the '1'f>1-hill' of eqns (6.8-4) must
be geometrically similar to the surface of the deftected membrane. Hence
we have the following conceptually useful membrane analogy for elastic
torsion:
If a thin membrane is mounted over the cross-section and inftated
by pressure, then St Venant's torsion constant K is proportional to
the volume under the membrane surface, and the torsional shear
stress VI in any specified direction at any point on the section is
proportional to the slope of the membrane at that point, the slope
being measured in a direction perpendicular to that of VI.
The membrane analogy makes it c1ear, for example, that the maximum
shear stress VI occurs in the direction tangential to the contour (see eqn
6.8-4(a» and that, normal to a contour line, the shear stress VI must be
zero, since that value of VI is numerically equal to the slope (i.e. = O) along
a contour.
It is interesting and useful to consider how the membrane analogy can be
extended to cover plastic torsion. If the torque T produces complete
plasticity, then VI is everywhere equal to the yield strength in shear
(assuming, for the time being, that an elastic-perfectly plastic material is
being considered);
r
eqn (6.8-4(a» is then interpreted as
tp = VI at yield = constant
That is, for a given section, the slope tp of the inflated membrane must
have a constant value everywhere equal to (KIT) times VI at yield.
Naturally, an inflated membrane cannot have such a constant slope;
however, if dry sand is poured over the cross-section, the definite angle of
repose of the sand will automatically lead to the formation of a heap having
a constant slope on aII faces. The membrane analogy is then modified as
the sand-heap analogy (or sand-hill analogy), which states that the torsion
constant K is twice the volume of the sand heap, provided that the yield
shear stress VI is interpreted at (TI K) times the slope tp of the faces. In
other words, eqns (6.8-4) for elastic torsion become applicable to plastic
torsion when the words '1'f>1-hill' are replaced by 'sand-heap'.
Consider a rectangular section in plastic condition. The sand heap (or
'1'f>1-hill') is then as shown in Fig. 6.8-2, in which I'f>lm denotes the height
at the ridge of the heap. The reader should verify that
volume of sand heap = !hminl'f>lm(hmilx - hmin) + ~h~inl'f>lm
= KI2 (from eqn 6.8-4(b» (6.8-6)
From Fig. 6.8-2, the slope of the faces of the sand heap is
tp = 21'f>Imlhmin (6.8-7)
Equation (6.8-4(a» states that VI = (TI K)tp. Substituting in the values
of K and tp from eqns (6.8-6) and (6.8-7) and simplifying,