Four Linear Continuum Theories - Continuum Mechanics of Anisotropic Materials - page 143

Fig. 6.3 A beam subjected to
a pure bending loading. The
bending moment M 1 is
applied about the 1-axis.
There are no other loads.
The origin of coordinates is at
the centroid of the cross-
section. The coordinate axes
coincide with the principal
axes of the area moment of
inertia

Solution : To obtain this solution, it is necessary to account only for the bending
moment about the x 1 axis, M 1 , and the fact that the lateral surfaces of this beam, that
is to say the surfaces other than the ends normal to the x 3 axis, are unloaded, that is
to say that there are no surface tractions applied. Because the lateral boundaries are
unloaded we are going to guess that stresses that could act on the lateral boundaries
are zero everywhere in the object, thus

T 11 ¼ T 22 ¼ T 12 ¼ T 13 ¼ T 23 ¼ 0 :

The only nonzero stress is then the axial stress T 33 . The moment M 1 must then be
balanced by a distribution of the axial stress T 33 in the beam. In general
T 33 ¼ T 33 ( x 1 , x 2 ) by observ-

ing that T 33 must be independent of x 3 . The argument for T 33 being independent of
x 3 is a physical one. Consider a free object diagram at any location along the length
of the beam in Fig. 6.3 . The reactive force system at that (any) location must be
equal to the moment M 1 applied at the end. Thus the stress distribution must be the
same along the entire length of the beam. The only nonzero strains due to T 33 are
then computed from ( 6.27 ),

T 33 ( x 1 ,x 2 , x 3 ); however this may be reduced to T 33 ¼

x 2 Þ¼ n 13
E 1

x 2 Þ¼ n 23
E 2

E 11 ð x 1 ; T 33 ð x 1 ; x 2 Þ; E 22 ð x 1 ; T 33 ð x 1 ; x 2 Þ; E 33 ð x 1 ; x 2 Þ

1
E 3 T 33 ð

¼ x 1 ; x 2 Þ:

When the compatibility equations (2.54) are applied to these three strains, the
following differential equations for T 33 are obtained:

2 T 33
@

2 T 33
@

2 T 33

@

x 1 ¼ @ x 2 ¼ @

x 2 ¼

0 :

@ x 1 @

c 2 x 2 ,
where c o , c 1, and c 2 are constants. The next step in the solution to this problem is to

The solution to this system of differential equations is T 33 ¼ c o þ c 1 x 1 þ