Download STRESS CIRCLEs Lecture Note

Introduction
 The transformation equations for plane stress can be
represented in graphical form by a plot known as Mohr’s
Circle.
 This graphical representation is extremely useful because it
enables you to visualize the relationships between the normal
and shear stresses acting on various inclined planes at a point
in a stressed body.
 Using Mohr’s circle you can also calculate principal stresses,
maximum shear stresses and stresses on inclined planes.
 Mohr’s circle is named after the famous German civil
engineer Otto Christian Mohr (1835-1918), who developed
the circle in 1882.
Principal stresses
 Planes that have no shear stress are called as principal
planes.
 Principal planes carry only normal stresses.
 In real life, stresses do not act in normal direction but rather
in inclined planes.
Normal Plane
Oblique Plane
Stresses in oblique plane cont’d
Notations:
𝜎 = Stress (Nmm-2) or (KNmm-2) or (MPa)
𝜎n = Normal stress (Nmm-2) or (KNmm-2) or (MPa)
𝜎t = Shear stress (Nmm-2) or (KNmm-2) or (MPa)
Stresses in oblique plane cont’d
 Member subjected to direct stress in one plane.
Member subjected to direct stress in two
mutually perpendicular plane.
 Member subjected to simple shear stress.
 Member subjected to direct stress in two
mutually perpendicular directions +
simple shear stress.
General stress system
When a body is subjected to axial bending and shearing
stresses, then the element in the body experiences a general
two-dimensional stress system. The resultant of these stresses
on any plane in the body can be resolved into a normal stress
and shearing stress. Consider a small element subjected to
two-dimensional stress system, as shown in Figure 1.0.
 Normal Stress in an inclined plane
In this diagram, we have three stresses acting on an element,
i.e. σx, σy and τ. To develop a relationship between the stresses
acting on an inclined plane AC and the stresses σx, σy and τ.
Consider the equilibrium of the element in Figure 1b. The
forces acting parallel and perpendicular to the plane AC
inclined at an angle θ with the horizontal are shown in Figure
1b.
Normal Stress in an inclined plane cont’d
Fig. 1.0: General stresses system
Normal Stress in an inclined plane cont’d
Considering the algebraic sum of forces perpendicular to the
plane, acting away from AC as positive, we get,
Substituting these values, we get
Normal Stress in an inclined plane cont’d
(1.1)
The expression (1.1) gives the normal stress acting on any inclined plane. Similarly
considering the algebraic sum of forces parallel to the plane acting downwards or along
CA as positive, we get
Normal Stress in an inclined plane cont’d
(1.2)
The above expression (1.2) gives the shear stress acting on any inclined
plane. Sign conventions: The tensile normal stresses are considered as
positive and shear stress developing clockwise rotation is treated as
positive.
Maximum Normal Stress on an Inclined Plane
Equation (1.1) can be written as
(1.1a)
Differentiating Eq. (1.1a), with respect to θ and equating to zero (maximaminima), we get
(1.3)
Maximum normal stress on an inclined plane cont’d
Equation (1.3) can be used to find the inclination of a plane for which the
maximum normal stress is acting on it. Substituting τθ= 0 in Eq. (1.2), we
get,
Hence, it is seen that on a plane where the normal stress is maximum,
the shear stress is zero or absent.
Principal planes and principal stress
In general stress system, the equation for normal and shear stress on an
inclined plane is given by
𝜎θ
The normal stress is maximum on a plane inclined at an angle θ, given by
Principal planes and principal stress cont’d
• The plane on which the normal stress is maximum and shear
stresses are absent is known as principal plane and the
corresponding normal stress is principal stress. In general, at
any point in a strained material, there are three principal
planes mutually perpendicular to each other. Out of the three
planes, the plane carrying maximum normal stress is called
major principal plane and corresponding stress as major
principal stress. The plane carrying minimum normal stress is
called minor principal plane and corresponding stress as
minor principal stress. In two-dimensional analysis, only two
principal planes exist.
• Consider Eq. (1.3), which represents inclination of principal
plane. This can be represented with a right-angled triangle
with an angle 2θ, as shown in Fig. 2.0.
Principal planes and principal stress cont’d
Fig. 2.0. Right-angled triangle with an angle 2θ
Fig. 2.1
From the triangle shown in Figure 2.0 and from trigonometry (Fig. 2.1), we know that
(1.4a)
Principal planes and principal stress cont’d
(1.4b)
Substituting Eqns. (1.4a & b) in Eq. (1.1a), the principal stress is given by,
Principal planes and principal stress cont’d
The principal stresses can be written as
Max. principal stresses
(1.5)
Min. principal stresses
(1.6)
Where σn 1 and σn 2 are known as principal stresses
Maximum shear stress
From Eq. (1.2), we have the shear stress on the inclined plane given by
From maxima-minima, the maximum shear stress is obtained by
differentiating Eq. (1.2) with respect to θ and equating it to zero.
Maximum shear stress cont’d
This can be represented with a right-angled triangle with an angle 2θ, as
shown in Figure 3.1. From the triangle shown (from trigonometry), we
know that (1.7b)
(1.7a)
(1.7b)
Fig. 3.1
Maximum shear stress cont’d
Substituting Eqns. (1.7a & b) in Eq. (1.2), we have
(1.8)
Comparing Eqns. (1.5), (1.6) and (1.8), we get
(1.9)
Maximum shear stress cont’d
From Eq. (1.9), it is clear that maximum shear stress is equal to
the half the algebraic sum of major and minor principal
stresses. The planes of maximum shear stresses are inclined at
45° to the principal plane as the product of tan 2θ is -1. Hence,
2θ differs by 90° or 0 differs by 450.
PROVED
Construction of Mohr’s Circle
Consider a square element of a material subjected to plane stress (Fig.
2.0), and let σx , σy and τxy be the components of the stress exerted on the
element.
Fig. 2.0
Construction of Mohr's circle cont’d
Fig. 2.1
Construction of Mohr's circle cont’d
 Plot a point X of coordinates σx and –τxy and a point Y of coordinates σx
and + τxy (Fig. 2.1).
 If τxy is positive, point X is located below the σ axis and point Y above, as
shown in Fig. 2.1.
 If τxy is negative, X is located above the σ axis and Y below.
 Join X and Y by a straight line, define the point C of intersection of line
XY with the σ axis and draw the circle of center C and diameter XY.
Noting that the abscissa of C and the radius of the circle are respectively
equal to the quantities σave and R defined by Eqns.
Construction of Mohr's circle cont’d
 Thus, the abscissas of points A and B where the circle intersects the σ
axis represent respectively the principal stresses σmax and σmin at the
point considered.
 Also note that, since tan (XCA) = 2τxy/(σx – σy) the angle XCA is equal in
magnitude to one of the angles 2θp that satisfy Eqn.
Constructing Mohr’s Circle: Procedure
1. Draw a set of coordinate axes with σx1 as positive to the right
and τx1y1 as positive downward.
2. Locate point A, representing the stress conditions on the x
face of the element by plotting its coordinates σx1 = σx and
τx1y1 = τxy. Note that point A on the circle corresponds to θ =
0°.
3. Locate point B, representing the stress conditions on the y
face of the element by plotting its coordinates σx1 = σy and
τx1y1 = -τxy. Note that point B on the circle corresponds to θ =
90°.
Construction of Mohr’s circle: Procedure cont’d
Construction of Mohr’s circle: Procedure cont’d
4.Draw a line from point A to point B, a diameter of the circle
passing through point c (center of circle). Points A and B are at
opposite ends of the diameter (and therefore 180° apart on the
circle).
5.Using point c as the center, draw Mohr’s circle through points
A and B. This circle has radius R. The center of the circle c at the
point having coordinates σx1 = σavg and τx1y1 = 0.
Stress Transformation: Graphical Illustration
Explanation
 On Mohr’s circle, point A corresponds to θ = 0. Thus it’s the
reference point from which angles are measured.
 The angle 2θ locates the point D on the circle, which has
coordinates σx1 and τx1y1. D represents the stresses on the x1
face of the inclined element.
 Point E, which is diametrically opposite point D is located 180°
from cD. Thus point E gives the stress on the y1 face of the
inclined element.
 Thus, as we rotate the x1y1 axes counterclockwise by an angle
θ, the point on Mohr’s circle corresponding to the x1 face
moves clockwise by an angle of 2θ.
Explanation
 Principle stresses are stresses that act on a principle surface. This
surface has no shear force components (that means τx1y1=0)
 This can be easily done by rotating A and B to the σx1 axis.
 σ1= stress on x1 surface, σ2= stress on y2 surface.
 The object in reality has to be rotated at an angle θp to experience no
shear stress.
 The same method to calculate principle stresses is used to find
maximum shear stress.
 Points A and B are rotated to the point of maximum τx1y1value. This is
the maximum shear stress value τmax.
 Uniform planar stress (σs) and shear stress (τmax) will be experienced by
both x1 and y1 surfaces.
 The object in reality has to be rotated at an angle θs to experience
maximum shear stress.
Sign Conventions
 When the shearing stress exerted on a given face tends to rotate the
element clockwise, the point on Mohr's circle corresponding to that face
is located above the σ axis. When the shearing stress on a given face
tends to rotate the element counterclockwise, the point corresponding to
that face is located below the σ axis . As far as the normal stresses are
concerned, the usual convention holds, i.e., a tensile stress is considered
as positive and is plotted to the right, while a compressive stress is
considered as negative and is plotted to the left.
Worked Examples
1. Draw the Mohr’s Circle of the stress element shown below. Determine
the principle stresses and the maximum shear stresses.
Parameters:
σx = -80 MPa
σy = +50 MPa
τxy = 25 MPa
Coordinates of
Points
A: (-80,25)
B: (50,-25)
Example 1 cont’d
Example 1 cont’d
Principal Stresses
Example 1 cont’d
Maximum Shear Stresses
Assignment
For the state of stress shown, determine (a) the
principal planes and the principal stresses, (b)
the stress components exerted on the element
obtained by rotating the given element
counterclockwise through 30 degrees.
Class work
For the state of plane stress shown in Fig. below (a) construct Mohr's
circle, (b) determine the principal stresses, (c) determine the maximum
shearing stress and the corresponding normal stress.
Solution:
Class work cont’d
Class work cont’d
(a) Construction of Mohr's Circle
Class work cont’d
Class work cont’d
(c) Maximum Shearing Stress.
Since a further rotation of 90° counterclockwise brings CA into
CD in Fig. a further rotation of 45° counterclockwise will bring
the axis Oa into the axis Od corresponding to the maximum
shearing stress in Fig. We note from Fig. that τmax = R = 50 MPa
and that the corresponding normal stress is σ’ = σave = 20 MPa.
Since point D is located above the σ axis in Fig. , the shearing
stresses exerted on the faces perpendicular to Od must be
directed so that they will tend to rotate the element clockwise.
Class work cont’d
Thanks for
listening
REFERENCES:
1) A Textbook of Strength of Materials By R. K. Bansal
2) Strength of Materials S S Bhavikatti
3) Textbook of Mechanics of Materials by Prakash M. N. Shesha, suresh G. S.
4) Mechanics Of Materials by Ferdinand P. Beer.