Page 241 - Marks Calculation for Machine Design
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                          January 4, 2005
                 Brown˙C05
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                                     PRINCIPAL STRESSES AND MOHR’S CIRCLE
                              U.S. Customary                      SI/Metric       223
                    Step3. FromFig.5.37,themaximumandmin-  Step3. FromFig.5.37,themaximumandmin-
                    imum shear stresses are shown 90 to the prin-  imum shear stresses are shown 90 to the prin-
                                          ◦
                                                                            ◦
                    cipal stresses, and equal to the average stress.  cipal stresses, and equal to the average stress.
                                 σ  12 kpsi                        σ  84 MPa
                      τ max = σ avg =  =  = 6 kpsi      τ max = σ avg =  =  = 42 MPa
                                 2    2                            2    2
                      τ min =−τ max =−6 kpsi             τ min =−τ max =−42 MPa
                    Step 4. As the uniaxial stress element is  Step 4. As the uniaxial stress element is
                    actually the principal stress element, the rota-  actually the principal stress element, the rota-
                    tion angle (2φ p ), and therefore the angle (φ p ),  tion angle (2φ p ), and therefore the angle (φ p ),
                    is zero.                           is zero.
                             2φ p = 0or φ p = 0                2φ p = 0or φ p = 0
                    Step 5. Using Fig. 5.38, the rotation angle  Step 5. Using Fig. 5.38, the rotation angle
                                             ◦
                    (2φ s )forthemaximumshearstressis90 clock-  (2φ s )forthemaximumshearstressis90 clock-
                                                                                ◦
                    wise, or negative, meaning         wise, or negative, meaning
                                         ◦
                                  ◦
                                                                           ◦
                                                                    ◦
                       2φ s = 2φ p − 90 = 0 − 90 =−90 ◦  2φ s = 2φ p − 90 = 0 − 90 =−90 ◦
                       φ s =−45 ◦                         φ s =−45 ◦
                      The important result from this example is that the given stress element is the principal
                    stress element, and that the maximum shear stress in (σ/2) acting at 45 .
                                                                       ◦
                    Biaxial Element. For a biaxial stress element, suppose σ xx = σ, σ yy = 2σ, and τ xy = 0,
                    where (σ) and (2σ) are the axial and hoop stresses in a thin-walled cylinder under an internal
                    pressure.
                      The first step in the process is to plot two points; one point having the coordinates
                    (σ xx ,τ xy ) and the other having the coordinates (σ yy ,−τ xy ), where for a biaxial stress element
                    these two points are (σ,0) and (2σ,0). This is shown in Fig. 5.39.





                                               (s,0)         (2s,0)
                                                                     s





                                  t (2q ccw)
                          FIGURE 5.39  Plot points (σ xx , t xy ) and (σ xx , −t xy ).

                      A line connecting these two points would cross the (σ) axis at the average stress (σ avg );
                    however, for a biaxial element both points are on the (σ) axis, so the average stress is
                    halfway between, that is, (3σ/2) as in Fig. 5.40.
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