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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.