CONTINUUM MECHANICS for ENGINEERS

and θ range through all admissible values, point Q moves along the circle arc
EG of Figure 3.15A. For this case Eq 3.8-2b may be restructured into the form
⎡ 1 ⎛
⎞⎤
2 2 ⎡1
σ − ⎝σ
+ σ ⎠ + σ = ( σ −σ
) ( σ − σ ) cos β + ⎛
1 σ −σ
⎣⎢
⎦⎥
⎣
⎢ ⎝
2
2
(3.8-10)
which defines a circle whose center is coincident with that of circle C2, and
π
1
having a radius R2. Here, when β1 = , the radius R2 reduces to ( σI−σIII) ,
2
2
which is the radius of circle C2. As Q moves on the circle arc EG of
Figure 3.15A, the stress point q traces out the circle arc eg, in Figure 3.15B.
In summary, for a specific ˆn at point P in the body, point Q, where the
line of action of ˆn intersects the spherical octant of the body (Figure 3.15A),
is located at the common point of circle arcs KD and EG, and at the same
time, the corresponding stress point q (having coordinates σN and σS) is
located at the intersection of circle arcs kd and eg, in the stress plane of
Figure 3.15B. The following example provides details of the procedure.
Example 3.8-1
The state of stress at point P is given in MPa with respect to axes Px 1x 2x 3 by
the matrix
(a) Determine the stress vector on the plane whose unit normal is
1
ˆ ⎛ ˆ ˆ ˆ ⎞
n= e e e .
3 ⎝2
1+ 2 + 2 3⎠
(b) Determine the normal stress component σN and shear component σS on
the same plane.
(c) Verify the results of part (b) by the Mohr’s circle construction of
Figure 3.15B.
Solution
(a) Using Eq 3.4-8 in matrix form gives the stress vector
or
2
2
⎞⎤
2
N I III S II III II I I III R ⎠ 2
⎡25
0 0⎤
⎢
⎥
[ σ ij]=
⎢
0 −30 −60
⎥
⎣
⎢ 0 −60
5⎦
⎥
ˆ ⎡t
( n)
⎤
1 ⎡25
0 0⎤
⎡2
⎤
⎢ nˆ
t
( ) ⎥ ⎢
⎥ ⎢ 3 ⎡ 50⎤
⎥ ⎢ ⎥
⎢ 2 ⎥ =
⎢
0 −30 −60
⎥ ⎢1
1
3⎥
=
⎢
−150
⎥
⎢ nˆ
t
( ) ⎥
⎣ 3 −
⎦ ⎣
⎢
⎦
⎥ ⎢ ⎥ 3
0 60 5 2
⎣⎢
⎦⎥
⎣
⎢ −50⎦
⎥
3
t
( nˆ ) 1 ⎛ = ⎝50eˆ
−150eˆ −50eˆ
3
1 2
⎞
3⎠
t ( ˆn )
i
⎦
⎥ =