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Question 8: Several forces are applied to the pipe assembly shown. If the inner and outer diameters of the pipe are equal to 1.50-in. and 1.75-in. respectively, determine: a-The state of the stress at point H on the top outside surface of the pipe. Show the results on a volume element. b-The principal stresses and maximum shear stress. c-The principal planes and plane of maximum shear. 6 in. a: 12 in -- 1,982-psi = 0.0 T= 670-psi 10 in. b: 0 = 205-psi =-2187-psi -- 1196-psi in-plane o min τ max 30 lb 5011 Sin. c c: = -17° CW or 73° CCW = +28° CCW Sin. E 50 lb 30 lb

Respuesta :

The difference between the primary stresses divided by one-half is the maximum shear stress. It should be noticed that two angles between 0° and 360° are produced by the primary planes equation, 2p.

Principal stress axes are perpendicular to zero shear stress planes. Orthogonal primary axes are present. The greatest shear stress is 45 degrees away from the main stress vector. The difference between the primary stresses divided by one is the maximum shear stress. Principal stresses are the highest and lowest normal stress values on a plane that is rotated through an angle and does not experience shear stress. Primary Plane On that plane, the main stresses are present, and there is no shear stress.

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