Moore's Circle Continued
Moore's Circle Continued MAE 3201
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This 15 page Class Notes was uploaded by Brandon Gutierrez on Friday April 1, 2016. The Class Notes belongs to MAE 3201 at 1 MDSS-SGSLM-Langley AFB Advanced Education in General Dentistry 12 Months taught by Dr. McBride in Spring 2016. Since its upload, it has received 21 views. For similar materials see Eng Mech:Strength of Materials in Mechanical Engineering at 1 MDSS-SGSLM-Langley AFB Advanced Education in General Dentistry 12 Months.
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Date Created: 04/01/16
More Mohr’s Circle Example Determine , , , and . 1 2 1 2 X 30 ksi (E.1) Y 0 (E.2) XY 12 (E.3) Draw Mohr’s circle, to scale. , 12 ) ( 0, 12 ) X Y 300 AVG 2 2 15.0 (E.4) MAE3201 17 1 R ( Y AVG ) XY 2 (015) (12) 19.2 2 (E.5) R 1519.2 4.2 ksi (E.6) 1 AVG R 15 19.2 34.2 ksi 2 AVG (E.7) 0.5 tan 112 19.3 2 15 (E.8) 90 19.3 90 109.3, 70.7 (E.9) 1 2 MAE3201 17 2 Example Determine 1 , 2 , 1 , and 2. 125 MPa X (E.1) 75 (E.2) Y XY 50 (E.3) Draw Mohr’s circle, to scale. (125 , 50 ) 75 , 50 ) MAE3201 17 3 X Y 12575 AVG 25.0 (E.4) 2 2 2 2 2 2 R ( X AVG ) XY (12525.0) ( 50) 111.8 (E.5) 1 AVG R 25.0111.8 137 MPa (E.6) 2 AVG R 25.0111.8 86.8 MPa (E.7) 50 2 0.5 18tan 1 76.7 (E.8) 100 1 20 76.7 90 166.7, 13.3 (E.9) MAE3201 17 4 ThreeDimensional State of Stress Reexamine the chunk of material used to determine the stresstransformation equations. Consider a general threedimensional state of stress, and include only the stresses that were not considered during the previous analysis. XZ ZY Z ZX YZ The shear stresses on the left vertical surface and on the bottom horizontal surface have no X or Y components. The three stresses on the front face of the material are balanced by three equalin magnitude, oppositeindirection stresses on the back face of the material that are not shown in the drawing. Therefore, if all these stresses had been included in the original analysis that required equilibrium of forces in the X and Y directions, the same results would have been obtained. However, there is an additional stress that is not shown, the component of shear stress on the inclined face of the material in the Z direction, perpendicular to the drawing. There was no such component of shear stress in the planestress problem, because there were no applied forces in the Z direction, and therefore that component of the shear stress was required to equal zero. In this problem, because of the other stresses shown in the drawing, it is necessary to include a component of shear stress on the inclined face of the material in the Z direction. This shear stress will also not affect the equilibrium equations in the X and Y directions. However, the maximum and minimum normal stresses computed from Mohr’s circle, will not be acting on planes with zero total shear stress. Therefore, these normal stresses are not principal stresses. In general, it is incorrect to use a twodimensional Mohr’s circle to analyze a threedimensional stress distribution. MAE3201 17 5 ThreeDimensional State of Stress, Special Case (Solve the homework using this technique) Consider a small element of material subjected to a planestress distribution plus a normal stress perpendicular to the plane of the drawing. Y XY Z X Z may be either tensile or compressive. It is possible to use a planestress Mohr’s circle to determine the two principal stresses acting on planes perpendicular to the drawing. Because equal in magnitude and opposite in direction normal stresses in the Z direction act on the front and back of the sectioned element of material, there will be no shear stresses in any direction on the inclined planes on which those two principal stresses act. The third principal stress is then equalZ, because that normal stress also acts on a plane on which no shear stress acts. This special case is an exception to the general rule that it is incorrect to apply Mohr’s circle to a threedimensional state of stress. After the three principal stresses have been determined, it is possible to determine the maximum shear stress in the material. This will be explained later in this lecture. MAE3201 17 6 Rework the planestress problem in a different manner. X uˆ XY Y Set the shear stress on the inclined plane equal to zero. We know that, if we start at the location on Mohr’s circle corresponding to X XY ) and move counterclockwise on the circle, we will arrive at a principal stress before we have moved on the circle. Therefore, in the material, we will reach a principal stress before we have rotated through. There is always a principal stress in the 0 90 range. The value o in the figure corresponds to that principal stress, although we do not yet know its value. Create a unit vector perpendicular to the inclined plane. u u i u j (17.1) X Y Note the following two equalities. uX = cos (17.2) uY= sin (17.3) A denotes the area of the inclined plane. The area of the bottom, horizontal surface is equal to Au Y and the area of the left, vertical surface is eqXa. to Sum forces in the X and Y directions. FX 0 AX X XYAu Y Au X (17.4) MAE3201 17 7 FY 0 Y AuY XYAuX Au Y (17.5) MAE3201 17 8 Rewrite equations (17.4) and (17.5), canceling a factor of A. 0( )X X XY Y (17.6) 0 u ( )u (17.7) XY X Y Y Equations (17.6) and (17.7) are simultaneous, linear equations in two unknown variandes, X u Y The equations are homogeneous. There are no terms that do not contain one of the two unknown variables. Therefore, a graph of the two equations results in two straight lines through the origin, as shown in the following sketch. Y uX The only simultaneous solution of two such homogeneous equations, a point on both lines, is the origin. This is referred to as the trivial solution. Not only is this an uninteresting solution, it is impossible, because the values of the two variables must be the components of a unit vector. The only solution to this dilemma is to choose a value for that makes the slopes of the two lines equal to each other. Then both lines will be the same line, and there will be an infinity of solutions. Every solution will have the same ratio ofY touX . The correct solution will be the valuYandfuX that create a unit vector. X XY (17.8) XY Y 2 0( )(X ) Y XY (17.9) This is the same equation that results from setting the determinant of the coefficient matrix equal to zero. 0 X XY (17.10) XY Y MAE3201 17 9 MAE3201 17 10 Demonstrate using the same values used with Mohr’s circle during the previous lesson. X 3 (17.11) Y 5 (17.12) 2 (17.13) XY 0 ( 3 )( 5 ) 4 (17.14) 0 2 2 19 (17.15) Use the quadratic formula. 2 ( 2) 2 4 (1) ( 19 1 4.47 5.47 , 3.47 (17.16) 2 (1) These are the same results previously obtained for the principal stresses. Substitute the value of 1 , equal to 5.47, back into either equation (17.6) or (17.7). It doesn’t matter which equation, because they will be the same equation with this particular choice of . Choose equation (17.6). 0 8.47u 2u (17.17) X Y u 4.24 u (17.18) Y X u sin Y tan 4.24 (17.19) u X cos = 103 .3 (17.20) This is the same result previously obtained for the plane on whi1, the algebraically larger principal stress, acts. Substitute the valu2, equal to 3.47, back into either equation (17.6) or (17.7), it does not matter which one, and obtain equal to , the same result previously obtained for the plane on which 2, the algebraically smaller principal stress acts. The two values of are referred to as eigenvalues. The values of the Xw uY pairs are referred to as eigenvectors. Problems of this type show up in many different areas of science. It is a theory of mathematics that real, symmetric matrices have real eigenvalues and mutually perpendicular eigenvectors. Therefore, the principal stresses, the eigenvalues, occur on planes that have mutually perpendicular normal vectors. MAE3201 17 11 Okay, that was fun, but Mohr’s circle is an excellent way to solve planestress problems, so why bother? The answer is that Mr. Hibbeler says that the analysis of threedimensional stress distributions is “beyond the scope of this text.” That may be true for him, but not for us. The previous analysis makes it possible to simply write the result. X XY XZ 0 (17.21) XY Y YZ XZ YZ Z To determine the principal stresses for a general, threedimensional state of stress, substitute values for the six stresses. Expand the determinant, yielding a cubic equation for . Solve that cubic equation, yielding three real values for the principal stresses. Because the matrix is real and symmetric, those principal stresses will act on three mutually perpendicular planes. Plot three principal stresses and draw three Mohr’s circles. 2 3 1 All possible pairs lie in the region outside the two smallest circles and inside the largest circle. The radius of the largest circle is the maximum shear stress in the material. MAX 1 3 (17.22) 2 MAE3201 17 12 It has been repeatedly stated that, in a planestress problem, R, the radius of Mohr’s circle, may or may not be the largest shear stress in the material. It is now possible to understand why. In a plane stress problem, the third principal stress is zero, acting on the plane of the drawing. Ther1ande, if 2re both positive, tension, the following situation results. 2 1 The maximum shear stress is clearly not the radius of the Mohr’s Circle representing the plane stress. The maximum shear stress in the material is the radius of the largest circle. The same situation exists if 1 and 2 are both negative, compression. 1 2 MAE3201 17 13 However, if is positive, tension, and is negative, compression, the following situation 1 2 arises. 2 1 The radius of the planestress Mohr’s circle is the maximum shear stress in the material. When analyzing planestress problems, always include the third principal stress, equal to zero, and always draw the other two Mohr’s circles. Relabel the three principal stresses, so that 1 is to the right,3is to the left, and2 is in between them. Then determine MAX using equation (17.22). Failure to follow these instructions is a major cause of deductions on tests. MAE3201 17 14 Homework Please note that these two problems do not require a general threedimensional treatment. They can be solved using the material on page 6 of this lesson. In Problem 990, the shear stress in the YZ plane is equal to 40 psi. For the test Going to have to solve a moores circle problem for 2D on the test Know that sigma 3 is 0 And now that the radius of the largest moores circle is (sigma 1 sigma 3)/2 MAE3201 17 15
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