The hitch in all of this is figuring out the two principal angles and which principal stress goes with which principal angle.

Transcription

1 Mohr s Circle The stress basic transformation equations that we developed allowed us to determine the stresses acting on an element regardless of its orientation as long as we know the basic stresses σx, σy, τxy, and the orientation of the element defined by an angle θ from the original x axis to the new x axis, designated by, say, x`. From these we developed the notion of principal stresses and maximum shear stress, all defined by relatively simple expressions. The hitch in all of this is figuring out the two principal angles and which principal stress goes with which principal angle. It turns out that the basic transformation equations define a circle, known as Mohr s circle, and any combination of stresses possible for an element face is determined as a point on the circumference of the circle. Once the circle is constructed, things like principal stresses and maximum shear stress become readily apparent from the geometry of the circle. In short, Mohr s circle provides a geometric approach to stress transformation. We begin with the basic stress transformation equations: (1) () x y x y cos ( ) xy sin ( ) x y sin ( ) xy cos ( ) In the first equation note that the first term to the right of the equal sign is just the numerical average of σx and σy so we define the average stress as (3) x y avg Now we move that term to the left of the equal sign in Eq. (1) and square both equations (1) and (): x y avg cos( ) xy sin ( ) x y sin ( ) xy cos( )

2 If we think in terms of adding these equations after expanding the right hand sides, and note that ( a b) a ab b we get terms like and x y cos( ) sin ( ) xy sin ( ) cos( ) xy x y The two terms corresponding to ab cancel because of the minus sign in the equation for τϕ. Therefore, on squaring and adding we get: avg But earlier we defined R x y xy x y Consequently the squared and added equations become avg R xy Compare this with the equation of a circle of radius r in x-y space: x x o y y o r where xo and yo are the x and y coordinates of the center of the circle and we conclude that avg R is the equation of a circle of radius R in σ-τ space with its center at σavg on the horizontal (σ) axis and 0 on the vertical (τ) axis. This circle is known as Mohr s circle.

3 As you know for a circle, any combination of x and y that satisfy the equation for the circle lies on the circumference of the circle. Therefore, any combination of normal (σ) stress and shear (τ) stress that satisfies avg R lies on the circumference of Mohr s circle. Stated another way, for a given stress state defined by stresses σx, σy, and τxy, any stresses σ and τ that satisfy the stress transformation equations (1) and () lie on the circumference of the Mohr s circle. Note from Calc III that Eqs (1) and () are the parametric equations for a circle where the parameter is the angle ϕ. The importance of this observation is that if we want to rotate an element through an angle ϕ, on the Mohr s circle the angle becomes ϕ. Thus, for example, the positive x and y faces on an element are 90 o apart. Therefore, on the Mohr s circle they will be 180 o apart. That is, those two points form the diameter of the circle. That makes constructing the Mohr s circle pretty easy. We begin by labelling the horizontal axis σ and the vertical axis τ then plot two points, one with coordinates (σx, τxy) and the other with coordinates (σy, -τxy). The minus sign for the second point arises because, from earlier work, we know that the shear stress on the y-face of the element rotates it in a direction opposite to that of the shear stress on the x-face. Having these two points plotted, connect them with a straight line. Since the line forms a diameter of the circle, its intersection with the σ axis locates the center of the circle (recall from above that the vertical coordinate of the center of the circle is 0 so it ll always be symmetric about the σ axis. Once the basic circle is constructed, used for many things, as the following examples will show.

4 Figure 1 shows a typical Mohr s circle. From the element shown off to the side of the figure note that on the x face, the normal stress, σx, is compressive while the shear stress on that face rotates the element CCW. The normal stress is therefore negative and the shear stress positive so the point representing the stresses on the x faces is located as shown on the graph. For the y face, the normal stress, σy, is tensile while the shear stress on that face rotates the element CW, resulting in a positive normal stress and negative shear stress. A point representing the stresses on the y face is therefore located as shown on the graph. Since these faces are 90 o apart on the element they ll be 180 o apart on the Mohr s circle so we connect them with a line and that defines the diameter of the circle. The intersection of this line with the σ axis defines the center of the circle so that point is used to construct the circle itself. Note that the circle is not centered on the coordinate system but lies to the right of the origin of the coordinate system. Notice that the circle could be located entirely to the left of the vertical axis, entirely to the right of it, or could span the vertical axis as is the case here. It just depends on the numerical values of the stresses used to construct the circle. Also note that the point representing the stresses on the x face is not located on the x or horizontal axis on the Mohr s circle. Again, its location is determined by the stresses acting on the x face. For similar reasons, the point representing stresses on the y face doesn t lie on the y or vertical axis. Regardless of how we rotate the element, the stresses on the resulting element faces (x` and y` faces, let s say) will be two points on the circumference of the circle. One for the x face and another for the y face. Since these faces are 90 o apart, the points on the circle will be diametrically opposite. Note that the points representing extreme values for normal stress lie where the circle intersects the σ axis. At these points the shear stress is zero and therefore these two points determine the principal stresses. The extreme vertical points on the circle give the maximum and minimum shear stress. Note that these points are not along the τ axis but along an axis to the right of it. You can see that the angle between the point of maximum shear and either of the principal stress point is 90 o. Since these points are 90 o on Mohr s circle they will lie on element faces that are 45 o apart, as we deduced in class. The Mohr s circle graph shows the shear axis as positive downward. By orienting the shear axis in that manner, a CCW rotation of the element translates to a CCW rotation (but twice as much) on the Mohr s circle. If you put the shear axis positive upward, then a CCW rotation on the element translates into a CW rotation (again, twice as much) on the graph and that can be very confusing. Although it hasn t always been that way, modern Mechanics of Materials textbooks seem consistent in orienting the positive shear axis downward. Once the basic Mohr s circle is drawn, principal stresses, max shear stresses, as well as the respective element orientations can be readily obtained from the geometry of the circle and a bit of trigonometry. And that s the beauty of Mohr s circle.

5 As an example, consider the element shown in Fig.. This is the one we ve been working with using the equations derived in class. On the x face the normal stress is compressive and the shear stress rotates the element counterclockwise and therefore σx = -9 MPa and τxy = 31 MPa. On the y face, the normal stress is compressive and the shear stress rotates the element clockwise so that σy = -47 MPa and τxy = -31 MPa. Axes for normal and shear stress are established on a sheet of graph paper so we can make a drawing that s reasonably to scale. Note that I ve emphasized the sign of the shear stress by indicating that values below the normal stress axis rotate the element CCW whereas those above the axis rotate the element CW. The stresses on the x and y faces, (-9, 31) and (-47, -31), respectively, are plotted on the graph and a line is drawn that connects the points. The point where this line intersects the normal stress axis is the center of the circle. With that established, the circle can be drawn on the graph. One of the major problems students have with the Mohr s circle is understanding how rotations of the element translate to rotations on the Mohr s circle, and vice-versa. An easy way to get a handle on this is to clearly identify the x axis on the circle. To do this, just extend a ray from the center of the circle to the point representing the stresses on the element s x face. Label that radial line x and put in a few hash marks (///) to indicate that that s your home base. Now, suppose the element is rotated 30 o CCW from its x axis. That rotation is represented on the Mohr s circle as 60 o CCW from the x axis shown on the circle. Conversely, if it s 0 o CCW from the x axis on the circle to the lowest point on the circle (max shear) then the element face on which the max shear stress occurs is a face whose outer normal is 10 o CCW from the element s x axis.

6 Note that with the Mohr s circle drawn with reasonable accuracy it s easy to use the geometry to figure out the radius and center location no need to remember equations. We can get the principal stresses and the element s configuration, as well as the maximum shear stress and the element s configuration, using Fig. 3a. The angle from the x axis on the Mohr s circle to the point (σp, 0) is θp CW. Therefore the normal for the element face for σp is located CW from the element s x axis an amount θp. From the geometry, sin(θp) = 31 / 38.3 and therefore θp = 7.0 o. The value of σp is, from the geometry, σavg R = MPa. Note that the stresses are measured from the origin of the coordinate system, not the center of the circle! A similar calculation gives σp1 = -31. MPa. The point representing (σp1, 0) is diametrically opposite the point (σp, 0) and is therefore the element faces for the two principal stresses are 90 o apart so at this point we can draw the element in its principal configuration, as shown in Fig. 3b. From Fig. 3a, note that θp + θs = 90 o and therefore θs = 45 o - θp. Since θp has already been determine, it s easy to get θs as o. From the Mohr s circle, this rotation is CCW. Note that this is the angle to the maximum shear face so the orientation of the shear stress arrow on that

7 face will be such that it rotates the element CCW. Note also that the coordinates of the point at the bottom of the circle are (σavg, R) so there is a normal stress also acting on the max shear face. With this information, the element in its max shear configuration can be drawn, as shown in Fig. 3c. Now, suppose we wish to know the stresses on the faces of an element rotated 40 o CW from its original x-y configuration. On the Mohr s circle that will be an angle of 80 o CW, as shown in Fig. 4a. The points where a line 80 o CW from the line representing the x axis intersects the circle at two diametrically opposite points. Therefore, the stresses represented by coordinates of these two points will be on element faces that are 90 o apart and one of these faces (represented by the left-most point) will be face rotated 40 o CW. Since we ve already calculated the CW angle from the x axis on the circle to the σ axis we can use it to get the angle from the σ axis to our 80 o line. That angle comes out to be 5.96 o. With that angle the vertical coordinate of our x point, the shear stress on the x face, is R sin(5.96 o ) = Since it s above the normal stress axis, it s negative. Therefore, on the x face the

8 shear stress is τxy = MPa. Because of the negative sign, the shear stress arrow will be drawn to that it rotates the element CW. The normal stress on the x` face is given by σx` = σavg R cos(5.96 o ) = MPa. Similarly, σy` = σavg + R cos(5.96 o ) = MPa. As a quick check, note that σx` + σy` = MPa as does σx + σy. The element in this configuration is shown in Fig. 4b. It s interesting to note the progression of the various stresses from the original configuration to the principal configuration (rotation of 7.0 o CW) to the 40 o CW configuration to the max shear configuration (rotation of 90 o o = 7.0 o CW). Note that nothing we did with Mohr s circle required use of the stress transformation equations and everything could be worked out strictly from the geometry of the circle. Note also that, unlike the approach using the transformation equations, there was no ambiguity regarding which principal stress went on which principal face. It was clear from the geometry of the circle. To summarize, Mohr s circle is a valuable graphical tool for doing stress transformation, either general transformations or specific transformations to find principal stresses and max shear stress as well as the respective element configurations. It involves only drawing a reasonably accurate graph and then using the geometry of the graph to obtain whatever results you re after. Points to remember are: Make a good drawing Orient the positive shear axis downward so that rotations of the element and corresponding rotations on Mohr s circle have the same orientation Angles double in going from the element to the circle Identify and label the x axis on Mohr s circle, and measure angles on the circle from that reference. Angles measured from the x axis on the element are measured from the x axis identified on the circle and are in the same direction (i.e. CW or CCW). However don t forget to double the element angle when working with the circle. Use your knowledge of geometry and trigonometry to set up and work out calculations using the Mohr s circle.