Page 272 - Introduction to Continuum Mechanics

256 Simple Extension

It acts on normal cross-sectional planes, and the maximum shearing stress is

and it acts on planes making 45° with the normal cross-sectional plane.

Let the undeformed length of the bar be / and let A/ be its elongation. Then EH = ~r and
from Eqs. (5.12.2a) and (5.12.4), we have

Also, iid is the undeformed length of a line in the transverse direction, its elongation M is
given by

The minus sign indicates the expected contraction of the lateral dimension for a bar under
tension.
In reality, when a bar is pulled, the exact nature of the distribution of surface traction is
often not known, only the resultant force is known. The question naturally arises under what
conditions can an elasticity solution such as the one we just obtained for simple extension be
applicable to real problems. The answer to the question is given by the so-called St. Variant's
Principle which can be stated as follows:
If some distribution of forces acting on a portion of the surface of body is replaced by a different
distribution of forces acting on the same portion of the body, then the effects of the two different
distributions on the parts of the body sufficiently far removed form the region of application of the
forces are essentially the same, provided that the two distribution of forces have the same resultant
force and the same resultant couple.
The validity of the principle can be demonstrated in specific instances and a number of
sufficient conditions have been established. We state only that in most cases the principle has
been proven to be in close agreement with experiments.
By invoking Saint-Venant's principle, we now regard the solution we just obtained for
"simple extension" to be valid at least in most part of a slender bar, provided the resultant
force on either end passes through the centroid of the cross-sectional area.

Example 5.12.1
A steel circular bar, 2 ft (0.61 m) long, 1 in. (2.54 cm) radius, is pulled by equal and opposite
axial forces P at its ends. Find the maximum normal and shear stresses if P = 10,000 Ibs
6
(44.5 kN). E Y = 30x 10 psi (207 GPa.) and v = 0.3.