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Mohr Circle Example

Mohr Circle Example. Point b. Point d. Find the principal stresses and their directions, then find the maximum shearing stresses and the directions of the planes on which they occur. In the above figure the shearing forces on the vertical

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Mohr Circle Example

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  1. Mohr Circle Example Point b Point d Find the principal stresses and their directions, then find the maximum shearing stresses and the directions of the planes on which they occur. In the above figure the shearing forces on the vertical surfaces are + and those on the horizontal surfaces are -. The normal stresses on all faces are +. Hence are points b and d on the Mohr circle would be given by Point b (12000,8000) and Point d (15000,-8000), remember on a Mohr circle the ordinate is the shearing stress and the Abscissa is the normal stress. Note that it is very important that you identify points b and d as well as l and m as shown in the figure above since this will allow you to determine which stresses act on which surface on the transformed shape.

  2. Principal Normal Stresses The prinicpal stresses are represented by points g and h. Point c or oc = .5 (15000+12000) = 13500 Hence ck = 15000 - 13500 = 1500 and then we can find cd = cg = sqrt(15002 +80002) = 8150 Therefore the minimum principal stress at point g is given by min = oc – cg = oc – cd = 13500 – 8150 = 5350 psi , note that point b becomes point g The maximum principal stress at point h is given by max = oc + cd = 13500 + 8150 = 21650 psi , note that point d becomes point h The angle 2p = arctan (kd / ck) = arctan (8000/1500) = 79.38o So the principal stress at point h acts on a plane that is oriented 39o 41’ from the original x axis

  3. Point b Point d 129.6o The principal stresses act as shown in the figure above. Remember that the shearing stresses in this case are equal to zero. Also note that on the more circle that point h represents the side originally identified as point d and that point g represents the side originally identified as point b

  4. You can also just work the equations instead of making a Mohr’s circle. For this example we have the following then: Eqn I Point d For 2p = 259.3o the value of  = 5350 psi Point b Note that point d is 39.3o from axis and point b is 129o as shown on the figure of the previous slide

  5. Maximum shearing stress The maximum shearing stress is given by cl in the Mohr’s circle shown above. This is the same as the radius of the circle found earlier as cd , so max= 8150 psi. Recall that 2s = 90 + 2p so that s = 84o 41’ Also note that point c for the maximum shearing stress is the corresponding value of the normal stress which is equal to oc which is  = 13500 psi

  6. Analytical Approach -10.65o II Point m For 2S = 349.3o the value of  = 8150 psi Point l Which is point c or the center of the Mohr’s circle

  7. Point l Point m Note that the normal stress is the same on all faces of the object.

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