For determining the normal and shear stresses on an inclined plane is by a graphical construction popularly known as Mohr’s stress circle or simply Mohr’s circle. For the sake of simplicity we take the co-ordinate axes along the principal stresses as shown in Fig. 3.5.
The equations of the previous section when referred to this co-ordinate system become as given below:
Two Dimensional Mohr’s Stress Circles:
Two dimensional cases arise, when the inclined plane is normal to X1 or X2 or X3. For instance, let the plane be parallel to X1 i.e. n1 = 0. The Eqn. (3.30) converts into a two dimensional Mohr’s stress circle giving relationship between σ2 and σ3. All the three two-dimensional cases that arise when the inclined plane becomes parallel to one of the axes are discussed below. When n1 = 0, Mohr’s stress circle equation (3.30) becomes
Figure 3.6(a) shows the position of the inclined plane on which stresses are to be found out. The corresponding Mohr’s circle is drawn in Fig. 3.6(b).
The procedure to find σn and σs from Mohr’s circle is as follows:
Similarly the other two-dimensional Mohr’s stress circles may be drawn corresponding to Eqns. (3.33) and (3.34). They correspond to the cases when the inclined planes are parallel to X2 and X3 respectively.
Three-Dimensional Mohr’s Stress Circles:
For three-dimensional Mohr’s circles we have to consider the Eqns. (3.30) to (3.32). These are the equations of circles. For instance the circle corresponding to Eqn. (3.30) has the center at [1/2(σ2 + σ3), 0] and radius equal to the square root of RHS of Eqn. (3.30). Similarly the other two circles have the centers at [1/2(σ1 + σ3), 0] and at [1/2(σ1 + σ2), 0]. The centers of these circles are same as those described by Eqns. (3.33) to (3.35), only the radii are different.
Figure 3.7 shows physical space in which OP is the direction of normal to the plane on which the normal and shear stresses are to be determined. In Fig. 3.8 the three circles corresponding to Eqns. (3.33) to (3.35) are drawn with centers at O1, O2 and O3 and, radii equal to 1/2(σ2 – σ3), 1/2(σ1 – σ3) and 1/2(σ1 – σ2) respectively. For all positions of P in the first quadrant of physical space (Fig. 3.7) the corresponding stress point P’ lies in the space between the three circles shown in Fig. 3.8.
ADVERTISEMENTS:
Now for determining the normal and shear stresses on any plane with normal OP Fig. 3.7 having direction cosines n1, n2, and n3, let us first consider the case when n3 remains constant and n1 and n2 can vary. Under this condition a conical surface OAPB is generated by OP (Fig. 3.7).
A generator of this cone in X2 – X3 plane is denoted by OB and that in X1 – X3 plane is denoted by O A. The stress state at the point A is represented by a point A’ on the circle with center O2. Because this is the Mohr’s stress circle which represents stress states for different positions of OA in X1 – X3 plane for different values of n1 and n3 and n2 = 0 for X1 – X3 plane.
Similarly the point B is represented by the point B’ on the circle with center O1. The two points A’ and B’ are connected by an arc of the circle with center at O3. This curve has been labeled as constant-n3 curve. Now if we keep n1 constant we get the points F and E in the physical plane and have corresponding points F’ and E’ on the Mohr’s stress circles with centers O2 and O3 respectively.
The two point E’ and F’ are joined by an arc of a circle with center at O1 and radius equal to RHS of Eqn. (3.30). The two curves F’ E’ and A’B’ intersect at the point P’ whose ordinate gives the shear stress and abscissa gives the normal stress. The procedure to arrive at point P’ is described below.
ADVERTISEMENTS:
Let n1 = cos θ, the angle θ is shown in X1-X2 plane in Fig. 3.7. When the inclined plane is perpendicular to OE the corresponding stress circle is that with center O3.
Therefore, from the center O3 draw a radius OE’ at an angle 2θ. Now with O1 as center draw the arc E’F’.
When OP is in X1 – X3 plane i.e., in position OA, the stress circle for planes perpendicular to OA is the circle with center O2. The point A lies on the generator of the cone obtained for constant value of n3. Hence let us take n3 = cos Ψ.
ADVERTISEMENTS:
The corresponding circle being one with center O2, we take an angle 2Ψ from σ3 side and draw a radius O2A’. Now with O3 as center draw an arc A’ B’, it intersects the first arc at the point P’. From P’ draw PN perpendicular to σn -axis. Then the normal and shear stresses on the plane perpendicular to OP are given by ON and NP’ respectively.