Page 343 - Introduction to Continuum Mechanics

Bending of a Incompressible Rectangular Bar. 327

2
a
On the top face in the Fig. 5.19, there is a normal stress (<p2& ) °d a shear stress,
(K(tpi — <P2))- On the bottom face, an equal and opposite surface traction to that on the top
face is acting. On the right face, which at equilibrium is no longer perpendicular to the x\ axis,
but has a unit normal given by

therefore, the surface traction on this deformed surface is given by

Thus, the normal stress on this surface is

and the shear stress on this same surface is, with 67* =

We see from the above equation that, in addition to shear stresses, normal stresses are needed
to maintain the simple shear state of deformation.
We also note that

This is a universal relation, independent of the coefficients #>/ of the material.

5.37 Bending of a Incompressible Rectangular Bar.
It is easy to see that the deformation of a rectangular bar into a curved bar shown in Fig. 5.20
can be described by the following equations

where (X,Y,Z) are Cartesian material coordinates and (r, 0, z) are cylindrical spatial coor-
dinates. Indeed, the boundary plane X = -a and X = a deform into cylindrical surfaces
rj = V-2aa + ft and r 2 = V2a a -I- ft and the boundary planes Y = ±b deform into the
planes 6 = ±cb ,