DRAWING AND USING MOHR S CIRCLE TO SOLVE PLANE STRESS TRANSFORMATION PROBLEMS

Transcription

1 DRAWING AND USING MOHR S CIRCLE TO SOLVE PLANE STRESS TRANSFORMATION PROBLEMS DRAWING MOHR S CIRCLE Detailed Discussion Mohr s circle for plane stress is a graphical solution to the equations for stress transformation in the form of a circle. By convention, Mohr s circle is drawn to scale with normal stress (σ ) plotted on the abscissa (horizontal axis) and shearing stress (τ ) plotted on the ordinate (vertical axis). This coordinate system is also known as Mohr space. The scales for both the normal and shearing stress axes must be the same. The sign convention for normal stress is compression is positive and tension is negative. The sign convention for shearing stress is illustrated in the figure below. To draw Mohr s circle for plane stress, the state of stress (σ, τ ) acting on, and the orientation of, two different planes passing through a point in a material must be known. An additional requirement is that the center of Mohr s circle must lie on the abscissa (σ-axis). For the special case where the two planes on which the states of stress are known are perpendicular to each other, the σ-coordinate for the center of the circle can be obtained from the following equation: ave ( σ σ ) 1 σ = + 2 x y where σ x and σ y are the normal stresses acting on the two perpendicular planes. For this special case, the center of the circle can also be obtained graphically by plotting the two points representing the two known states of stress in Mohr space, and drawing a straight line between the two points. The intersection of this straight line and the σ-axis is the location of the center of the circle. For the general case where the two planes are not perpendicular to each other, the center of the circle can be obtained graphically as follows: Draw a straight line between the two stress points. Find the midpoint of the straight line Evert C. Lawton Page 1 of 7

2 Draw a perpendicular bisector from the midpoint of the straight line towards the σ-axis. The intersection of the perpendicular bisector and the σ-axis is the center of the circle. Once the center of the circle has been established, draw Mohr s circle by placing the point of a compass on the center of the circle and drawing a circle through the two stress points. Brief Summary The steps necessary to draw Mohr s circle by hand are briefly summarized as follows: 1. Draw σ and τ axes to the same scale on a piece. Select a scale such that your Mohr s circle will take up most of a sheet of paper. 2. Draw the two known (σ, τ ) points on the graph. Label these two points with their (σ, τ ) values. 3. Find the center of the circle, which must lie on the σ-axis. It can be found by trial and error by placing the metal point of your compass on a point on the σ-axis, placing the tip of the pencil on one of the (σ, τ ) points, and keeping the same radius see if the tip of the pencil goes through the other (σ, τ ) point. If not, keep trying until you find the correct center of circle such that the circle lies on the σ-axis and goes through both (σ, τ ) points. Once you have located the correct center of circle, simply draw the circle using your compass. The center of the circle can also be found by either of the two methods described in the detailed discussion above. USING MOHR S CIRCLE TO SOLVE STRESS TRANSFORMATION PROBLEMS Mohr s circle can be solved either graphically or analytically. In either case, there are two methods that can be used the double angle method and the pole method. No matter what combination of methods you use, the first step is to draw Mohr s circle as described previously. Graphical Solutions Since values of (σ, τ ) and angles between planes will be measured graphically, the Mohr s circle must be drawn to scale using the same scale for both axes. If drawn by hand, draw the circle to as large a scale as possible that will fit on the piece of paper you are using. Double Angle Method There are two rules that must be used to solve Mohr s circle problems using the double angle method: A radial line drawn from the center of Mohr s circle through a (σ, τ ) point on the circle represents the direction of the plane on which (σ, τ ) act. The direction of the radial line is not the same as the direction of the real plane, except by coincidence Evert C. Lawton Page 2 of 7

3 Angles between radial lines in Mohr s circle are twice the actual angles between the real planes represented by those radials lines, but are in the same relative direction (clockwise or counterclockwise). Several examples of these rules are illustrated in Example 7.02 shown in Figure 1 on the next page. Point X represents the state of stress acting on a vertical plane (+50, -40), so radial line CX represents the vertical direction. Point Y represents the state of stress acting on a horizontal plane (-10, +40), so radial line CY represents the horizontal direction. Radial line CA represents the direction of the principal plane on which σ 1 acts. By either measuring the angle between radial lines CX and CA or by determining the angle analytically as done in the example, it is known that the real angle between the planes represented by radial lines CX and CA is half the angle determined from the Mohr s circle. So this angle in Mohr s circle, designated 2θ p in the example, is found to be Therefore, the real angle is one-half of this value, and is designated θ p, and is calculated to be one-half of , which is equal to and is rounded to three significant figures in the example (26.6 ). This means that the plane on which σ 1 acts is oriented 26.6 counterclockwise from vertical, because radial line CA is oriented 53.1 counterclockwise from radial line CX, and radial line CX represents the vertical direction. The direction of σ 1 and the plane on which it acts are illustrated in Fig on p Note that in this figure, the direction of σ 1 is shown as θ p counterclockwise from horizontal, which means that the plane on which σ 1 acts (the major principal plane) is oriented θ p counterclockwise from vertical. It is known that the principal planes are always oriented perpendicular to each other, so the direction of the principal plane on which σ 3 acts (the minor principal plane) can be found as 90 from the plane on which σ 1 acts, as illustrated in Fig. 7.22, and is shown to be oriented θ p counterclockwise from horizontal. However, this direction can also be determined using the fundamental rules of the double angle method as follows: From geometry it can be readily determined that the angle between radial lines CY and CB is the same as the angle between radial lines CX and CA and is therefore equal to Radial line CB, which represents the direction of the plane on which σ 3 acts, is counterclockwise from radial line CY, which represents the horizontal direction. Therefore, the plane on which σ 3 acts is one-half of = counterclockwise from horizontal Evert C. Lawton Page 3 of 7

4 Figure 1. Example Illustrating the Use of the Double Angle Method to Solve Mohr s Circle Problems. (Modified from Mechanics of Materials by F. P. Beer, E. R. Johnston, Jr., J. T. DeWolf, and D. F. Mazurek, 5 th edition, McGraw-Hill, New York, 2009, pp Evert C. Lawton Page 4 of 7

5 Pole Method There is only one fundamental rule in the pole method, which is described as follows: A straight line drawn in any direction from the pole will intersect the Mohr s circle at a point representing the state of stress on a plane with the same orientation in space as the line. This rule is more clearly explained by the following three sub-rules: 1. The pole, also known as the origin of planes, can be found by drawing a straight line from one of the points on Mohr s circle for which the state of stress (σ, τ ) is known, in the direction of the plane on which (σ, τ ) acts. The point where this line intersects the Mohr s circle is the pole. 2. The state of stress for a plane of any orientation can be found by drawing a line from the pole in the direction of orientation of the plane. The point where this line intersects the Mohr s circle represents the state of stress acting on that plane. 3. There is only one pole for any Mohr s circle. When answers are to be determined graphically, the pole method is generally easier to use than the double angle method because the direction of any line from the pole is the same as the orientation of the plane represented by that line. The solution to Example Problem 7.02 using the pole method is illustrated in Figure 2 on the next sheet. The letters used to designate points in Figure 2 are the same as those used in Figure 1, with the addition of point P corresponding to the pole. All values for the points are scaled off the figure. The solution to this problem using the pole method is described by the following steps: The pole is established first by drawing a line from point Y in the direction of the plane that the state of stress at point Y acts on (the vertical direction). This vertical line intersects the Mohr s circle at (50, 40), which are the coordinates for the pole. The location of the pole can be checked by drawing a line from point Y in the direction that the state of stress at point Y acts on, which is the horizontal direction. This horizontal line also intersects the Mohr s circle at (50, 40), confirming that this is the correct location for the pole. The values for σ 1 and σ 3 are scaled from the drawing at points A and B and are found to be +70 MPa and -30 MPa, respectively. The orientations of the planes on which σ 1 and σ 3 act are found by drawing straight lines from the pole to points A and B, respectively, and measuring the orientation of the lines referenced to either horizontal or vertical (by convention). The plane on which σ 1 acts is found to be oriented at counterclockwise from vertical, and the plane on which σ 3 acts is found to be oriented at counterclockwise from horizontal Evert C. Lawton Page 5 of 7

6 FIGURE 2. Solution of Example 7.02 Using the Pole Method Evert C. Lawton Page 6 of 7

7 The values for τ max and -τ max are scaled from the drawing at points A and B and are found to be +50 MPa and -50 MPa, respectively. The orientations of the planes on which τ max and -τ max act are found by drawing straight lines from the pole to points D and E, respectively, and measuring the orientation of the lines referenced to either horizontal or vertical (by convention). The plane on which τ max acts is found to be oriented at clockwise from horizontal, and the plane on which -τ max acts is found to be oriented at clockwise from vertical. Note that the angle between the two principal planes is 90, and the angle between the principal planes and the planes on which τ max and -τ max act are 45, as expected. In addition, all values determined using the pole method are the same as those found by the double angle method and by the analytical method used in the solution to Example Analytical Solutions Mohr s circle can be used, along with geometry and trigonometry, to derive equations to solve for angles and values of stresses at selected points. Either the double angle method or the pole method can be used for the analytical solutions. Use of the double angle method to obtain the appropriate equations needed to solve Example 7.02 is illustrated in the solution provided in Figure 1. It is important to note that general equations for solving stress transformation problems provided in reference books are based on knowing the state of stress on two planes that are perpendicular to each other. If the problem you are trying to solve does not meet this criterion, then none of the equations can be used. However, the necessary equations can still be derived using geometry, trigonometry, and either the double angle method or the pole method. Either the double angle method or the pole method can be used to solve any problem. The choice of which method to use should be based on which method can achieve the desired results with the least amount of effort for that particular problem. For problems involving known stresses on two perpendicular planes, the double angle method is usually easier to use than the pole method for analytical solutions. However, as noted previously, the pole method is usually easier to use than the double angle method for graphical solutions Evert C. Lawton Page 7 of 7