Continuum Formulations of Conservation Laws - Continuum Mechanics of Anisotropic Materials - page 61

Fig. 3.5 A uniform bar
subjected to a uniform
stress s

σ

e 3

n

θ

e 2

σ

t ( G ) Q T T ( L ) n ( G ) that since t ( G ) T ( G ) n ( G ) , it follows that T ( G ) Q T T ( L ) ¼ Q ¼ ¼ Q .

Since this is the transformation rule for a tensor (A83), T is a tensor.
Example 3.3.1
Determine the stress tensor representing the state of stress at a typical point in the
uniform bar subjected to a uniform tensile stress (Fig. 3.5 ). The applied tensile
stress is of magnitude s and it is assumed that the stress state is the same at all points
of the bar. Determine the stress vector t acting on the plane whose normal is n ,
where n is given by n e 3 . Determine the normal stress on the

plane whose normal is n and the shear stress in the direction m , m

¼ cos y e 2 þ sin y

n ¼ 0, m ¼

e 3 , on the plane whose normal is n .
Solution : The components of the stress tensor T at a typical point in the bar and
relative to the coordinate system shown in Fig. 3.5 , are given by

sin y e 2 þ cos y

2
4

3
5 ;

000
000
00

T ¼

s

thus the only nonzero component of the stress tensor is T 33 . The stress vector
t acting on the plane whose normal is n is then given by

2
4

3
5

2
4

3
5 ¼

2
4

3
5 ;

000
000
00

0
cos

0
0 t ¼ Tn ¼ y

s y s y

sin sin

thus this vector has only one nonzero component, namely, t 3 ¼ s sin y . The normal

sin 2

stress on the plane whose normal is n is given by t n ¼ s y . The shear stress on

the plane whose normal is n in the direction m , m n ¼ 0, m ¼ sin y e 2 þ cos y e 3

. (Note that one could choose m 0 ¼ is given by t m ¼ s cos y sin y sin y e 2

e 3 , where m 0 ¼ m 0 ¼s cos m and the direction of

the shear stress is reversed). From these results it is seen that when the normal to
the plane n coincides with the e 3 direction (

y m , then t cos y sin y ¼ t

y ¼ p /2), the stress component t 3 is