1. Transformation of Stress and Strain Ch. 7

3. And, 𝜏 = 𝑇𝑐𝐽 Where T is the torque, c is the radius of the cross section, and J = 𝜋2𝑐4 for circular cross section. If we took an element at pt. Q as shown and it is rotated with an angle 𝜃. The normal stress and shear stress will change. 𝜎𝑦 𝜀𝑥𝑦 𝜎𝑥 Σ 𝜎 𝜎𝑥 𝜃 𝜏𝑥𝑦 𝜎𝑦

4. Σ𝐹𝑦 = 𝜎 t ⅆ𝑙cos𝜃 - 𝜎𝑦t ⅆ𝑥 + Σ𝑥𝑦 tⅆ𝑦+𝜏tsin𝜃ⅆ𝑙 = 0 (dividing by tⅆ𝑙) 𝜎𝑑𝑥𝑑𝑙 + 𝜏𝑑𝑦𝑑𝑙 = 𝜎𝑦𝑑𝑥𝑑𝑙 - Σ𝑥𝑦𝑑𝑦𝑑𝑙 Σ𝐹𝑥 = 𝜎𝑥t ⅆ𝑦 - Σ𝑥𝑦 tⅆ𝑥 - 𝜎tsin𝜃ⅆ𝑙 + Σtcos𝜃ⅆ𝑙 = 0 𝜎𝑥ⅆ𝑦 - Σ𝑥𝑦ⅆ𝑥 - 𝜎sin𝜃ⅆ𝑙 + Σcos𝜃ⅆ𝑙 =0 𝜎sin𝜃 - Σcos𝜃 = 𝜎𝑥sin𝜃 - Σ𝑥𝑦cos𝜃 From 1&2 we can get that: 𝜎 = 𝜎𝑥+𝜎𝑦2 + 𝜎𝑦−𝜎𝑥2cos2𝜃 + Σ𝑥𝑦sin2𝜃 𝜏 = 𝜎𝑥−𝜎𝑦2sin2𝜃 + Σ𝑥𝑦cos𝜃 1 2 * *

5. Mohr circle used to facilitate calculations to find 𝜎𝑚𝑎𝑥 , 𝜎𝑚𝑖𝑛 , max. shear stress and (𝜎𝑥, 𝜎𝑦,Σ) at any angle of rotation as following: From Geometry we can find that: C = 𝜎𝑥+𝜎𝑦2 R = (𝜏𝑥𝑦)2+(𝜎𝑦−𝜎𝑥2)2 𝜏𝑚𝑎𝑥 𝜏𝑥𝑦 𝜎𝑚𝑎𝑥 𝜎𝑥 c 𝜎𝑚𝑖𝑛 𝜎𝑦

6. 𝜎𝑚𝑎𝑥 = C+R 𝜎𝑚𝑖𝑛 = C-R 𝜏𝑚𝑎𝑥 = R = (𝜏𝑥𝑦)2+(𝜎𝑦−𝜎𝑥2)2 tan2𝜃 = 2𝜏𝑥𝑦𝜎𝑦−𝜎𝑥

7. Thin walled pressure vessel 𝐹𝑝 = P*𝜋𝑟2 P𝜋𝑟2 = 𝜎𝑎*2𝜋𝑟𝑡 𝜎𝑎 = 𝑃𝑟2𝑡

8. 𝐹𝑝 = 𝑃∗2𝑟𝑙 𝐹h = 𝜎h∗2𝑡𝑙 2𝜎h𝑡𝑙 = 2𝑃𝑟𝑙 𝜎h = 𝑃𝑟𝑡 Note that: 𝜎h = 2𝜎𝑎 𝜏 = 0

9. Sketch Mohr circle 𝜏𝑚𝑎𝑥.1 = 𝜎h−𝜎𝑎2 𝜏𝑚𝑎𝑥.1 = 2𝜎𝑎−𝜎𝑎2 = 𝑃𝑟4𝑡 --------------------------------- Taking another plane 𝜏𝑚𝑎𝑥.2 = 𝑃𝑟2𝑡 𝜎𝑎 𝜎h 𝜏𝑚𝑎𝑥.2

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