Circular Representation of State of Stress:
From a state of stress defined by the components sx, sy and txy, we express the stress components on an arbitrary plane as
t= - (sx-sy )/2)sin 2q + txy cos 2q
which we might rewrite in general as
s = a + b cos 2q + c sin 2q ---------(i)
t = - b sin 2q + c cos 2q -------------(ii)
Now Let us try to establish direct relationship between s and t by eliminating q between
Eqs. (i) & (ii).
To make simplify the effort, let us take the origin of coordinate at (a, 0), so that the new variable = (s - a) is let for building the relationship.
--------------(iii)
t = - b sin 2q + c cos 2q -------------- (iv)
By Squaring & adding, we obtain
+ b2 sin 2 2q + c2 cos2 2q - 2bc cos 2q sin 2q
= b2 (cos2 2q + sin 2 2q) + c2 (sin 2 2q+ cos2 2q)
i.e.
Since b and c are constants let b2 + c2 = r 2
∴
Eq. shows that if s and t, the normal and shear stress components on any arbitrary plane are plotted as coordinates, the locus of the point will be a circle whose centre will
(sx + s y)/2 , 0 and radius will be
In other terms, the state of stress at particular point might be represented by a circle & a point on the circle represents the normal & shear stress components, on some plane as horizontal & vertical coordinates.